A NNALES DE L ’I. H. P., SECTION A
D. M. O’B RIEN A. C ANT
A. L. C AREY
On characteristic identities for Lie algebras
Annales de l’I. H. P., section A, tome 26, n
o4 (1977), p. 405-429
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405
On characteristic identities for Lie algebras
D. M. O’BRIEN, A. CANT A. L. CAREY Department of Mathematical Physics, University of Adelaide Ann. Inst. Henri Poincaré,
Vol. XXVI, n° 4, 1977,
Section A :
Physique théorique.
ABSTRACT. - Let
U(L)
be the universalenveloping algebra
of acomplex semi-simple
Liealgebra L,
and let V be an irreducible finite dimensional L-module. Denoteby ~
the map fromU(L)
into(End V)
0U(L)
definedfor xeL
by
= 1 (8) x + x (8) 1. In a recent paper, Kostant[6]
hasshown
that
for each element z of the centre ofU(L)
there is aunique
monicpolynomial
p~ with coefficients from the centreof U(L)
such that = 0.For the case where z is the universal Casimir
element,
these identities have beenexploited
in thephysical
literature for many years, and weresystematically analysed
as « characteristic identities for Liealgebras » by
Bracken and Green[3]
and Green[4].
In this paper weprovide
a commonviewpoint
for these differentapproaches.
The main aim of this paper isto establish the existence and
properties
of the characteristicidentity
forthe Lie
algebra
of thegeneral
linear group, and to prove some results which indicate the relevance of these identities for infinite dimensional repre- sentationtheory.
RESUME. - Soient
U(L) l’algèbre enveloppante
d’unealgebre
de LieL
semi-simple complexe,
et V un L-moduleirreductible,
de dimension finie.On note par 9
l’application
deU(L)
dans(End V)
8>U(L),
definie pour tout x E L parè(x) Recemment,
Kostant[6]
a montreque, pour
chaque
element z dans le centre deU(L),
il existe unpolynôme pz
avec coefficients appartenant au centre de
U(L),
tel quepz(az)
= 0. Dansle cas
particulier,
ou z est 1’element deCasimir,
lesphysiciens
ont utilisedepuis longtemps
telles « identitescaracteristiques
». On peut voir BrackenAnnales de l’Institut Henri Poincaré - Section A - Vol. XXVI, n° 4 - 1977. 27
406 D. M. O’BRIEN, A. CANT AND A. L. CAREY
et Green
[3]
et Green[4]
pour uneanalyse systematique
des « identitescaracteristiques
pour lesalgebres
de Lie ». On donne ici un traitementunifié pour ces
approches
divers. Le butprincipal
de cet article est 1’etablis-sement de l’existence et des
proprietes
de « l’identitécaractéristique » .
pour
l’algèbre
de Liegl(n).
On demontrera aussi certains resultatsqui
indi- quent l’utilité de ces identites pour la theorie desrepresentations
de dimen-sion infinie.
1. INTRODUCTION
Suppose
thatQ
denotes the rational field and thatgl(n, Q)
denotes theLie
algebra
overQ
with basis and Lieproducts
Let U denote the universal
enveloping algebra
ofgl(n, Q)
and Z the centreof U.
Finally,
let A denote the n x n matrix over U with entriesThe aim of this paper is to demonstrate the
following
facts.1)
A satisfies an nthdegree
monicpolynomial identity,
whose coefficients cl, c2, ..., c~ are elements of Z.
2)
A does notsatisfy
apolynomial identity
of lowerdegree,
soequation (2)
is the
only
nthdegree
monicpolynomial identity
satisfiedby
A.p(A)
iscalled the characteristic
polynomial
of A andequation (2)
the charac- teristicidentity
of A.3)
U can be embeddedinjectively
in analgebra
U overQ
in whichp(A)
can be factorized : n
The roots of
p(A) satisfy
for all u E U.
Annales de l’Institut Henri Poincaré - Section A
ON CHARACTERISTIC IDENTITIES FOR LIE ALGEBRAS 407
4)
al, a2, ..., an are distinct elements of 0.5)
A has aspectral representation,
where the
projection
matricesE 1, E2,
...,En
over 0satisfy
6) Suppose
that F is any field which satisfiesIf À =
(~,1, ~2?
...,Àn) E Fn,
there exists(up
toisomorphism)
one andonly
one irreducible
UF-module
withhighest weight £,
denotedby Vp~),
andal, a2, ..., an can be reordered so that
7)
The values of the roots of the characteristicpolynomial
in aparti-
cular irreducible
representation determine,
and are determinedby,
theinfinitesimal character of the
representation.
8)
Let G be any field such thatQ ~
G ~ C. Then all the results1)
to7)
hold for the Lie
algebra gl(n, G)
if wereplace Q by
Gthroughout.
Notethe
following points :
a)
We shallonly
prove1)-7)
forgl(n, Q),
as the extension togl(n, G)
is immediate.
b)
Similar results hold for all the classicalsubalgebras
ofgl(n, G).
(Compare
Green[4]
and Kostant[6].)
c)
In aparticular representation
ofgl(n, G),
the matrix A may beregarded
as a « matrix of operators », in which case it may
satisfy
apolynomial identity
ofdegree
less than n. We do not pursue thispoint here,
but it isoften
important
inapplications [4].
Identities of the form
(2)
have adistinguished history.
The first person toexploit
them was Dirac[1],
who wrote down what amounts to the cha- racteristicidentity
for the Liealgebra so(3,1).
Thisparticular example
is
intimately
connected with theproblem
ofdescribing
the structure ofrelativistically
invariant waveequations. Later,
it was shownby
Lehrer-Ilamed
[2]
thatn2
elements chosen from the universalenveloping algebra
of any Lie
algebra satisfy n2
identities. He also noted that in certain cir- cumstancesthe n2
identities can be written as asingle polynomial identity
of
degree n
for an n x nmatrix, analogous
to theCayley-Hamilton identity
Vol. XXVI, n° 4 - 1977
408 D. M. O’BRIEN, A. CANT AND A. L. CAREY
satisfied
by
a matrix over a commutativealgebra.
The firstmajor
steps in theanalysis
of thegeneral problem
were takenby
Bracken and Green[3]
and Green
[4], [5],
who gave the identities for the classical Liealgebras,
and
employed
themtogether
with some of the results listed above to syn- thesize irreduciblerepresentations (over
thecomplex field)
for thesealge-
bras. The
techniques
devisedby
Bracken and Green arepowerful
and arequite straightforward
toapply. However,
the correctalgebraic
frameworkfor their results needs
clarification,
and this is carried out in this paper.Some recent
unpublished
workby
Hannabuss has shown how the matrix A overgl(n, Q),
or thecorresponding
matrix over any of the classicalsubalgebras
ofgl(n, Q),
can beinterpreted
as an operator on the tensorproduct
moduleV(À)
@V,
whereV(À)
and V are both finite dimensional.By
this means we show how to connect the above results with animportant
recent paper
by
Kostant[6].
In Section
2,
wegive
ageneral
discussion of Kostant’sresults,
andderive,
as a
corollary,
the mostgeneral
theorem on the existence and factorization of the characteristicidentity
of acomplex semisimple
Liealgebra.
At thesame time we discuss an extension of Kostant’s results to real
forms,
andconjecture,
on the basis of ouranalysis
ofgl(n, Q),
thatthey
extend toreductive Lie
algebras
over any field F which satisfiesQ~F~C.
In the
subsequent
sections the results listed at thebeginning
of the intro-duction are
proved.
A number of conventions areemployed
there and welist them below.
1)
F will denote a subfield of thecomplex
field C which includes the rational fieldQ.
2)
The statement « K is analgebra »
willimply
that K is associative and has anidentity
element.3)
If K is analgebra
overQ,
the statement « V is aKF-module »
willimply
that V is considered as a vector space over F.4)
Unlessexplicitly
statedotherwise,
Roman indices will take values1, 2,
..., n and the summation convention willapply
torepeated
Romanindices.
2. THE APPLICATION OF KOSTANT’S WORK TO POLYNOMIAL IDENTITIES
Our notation follows that of
Humphreys [7].
Let L be acomplex
semi-simple
Liealgebra
ofrank I,
letU(L)
be the universalenveloping algebra
of
L,
and letZ(L)
be the centre ofU(L).
Select a Cartansubalgebra
H of L, with dual spaceH*,
and let C denote the set of roots of L relative to H.Let A
= ~ a 1,
Ll2, ..., be a base of C. We denote the set ofpositive (negative)
roots relative to A and take 5 to be half the sum ofAnnales de l’Institut Henri Poincaré - Section A
409
ON CHARACTERISTIC IDENTITIES FOR LIE ALGEBRAS
the
positive
roots. Let A~ c H* be the set of dominantintegral
linearfunctions on H.
Finally,
let W denote theWeyl
group.We are interested in the structure of the L-module
V(£)
(8)V,
whereV(l)
is the finite dimensional irreducible L-module with
highest weight
and V is any L-module
(possibly
infinitedimensional)
which admits aninfinitesimal character x. This means
that,
for everyzeZ(L),
where
is an
algebra homomorphism.
This condition on V is not very restrictive because Dixmier[8]
has shown that every irreducible L-module admitsa character. A well-known theorem
proved by
Harish-Chandra[7]
assertsthat any character x is
equal
to thecharacter xu
of an irreducible L-moduleV(p,)
withhighest weight p
EH*, However, p
isonly
determined up to link- age, where ,u,/1’
E H* arelinked, written p - /1’,
if there is some U E Wsuch
that /1
+ ð = +~).
We denote the
representations
of L affordedby V(À)
and Vby
03C003BB and 03C0respectively,
and we let{ v l’ V 2’ ....~}
be the set of distinct
weights
ofV(~).
Kostant considers the
algebra
(End V(~))
(8)U(L)
and defines
by
He extends 0 to a
homomorphism
ofalgebras
and proves thefollowing
theorems.
THEOREM
4 . 9[6].
- Let z EZ(L).
Then there exists aunique
monicpoly-
nomial E
Z(L) [x]
such thatTHEOREM 5.1
[6].
- Let z EZ(L)
andput z
= (8)~c)(z).
Theoperator z
in
V(À)
(8) V satisfies theidentity
In order to relate these results to
previous
work on characteristic iden-Vol. XXVI, n° 4 - 1977
410 D. M. O’BRIEN, A. CANT AND A. L. CAREY
tities,
we take thespecial
case z = eL, the universal Casimir element of L.We know that
[7]
where we have used the bilinear form
( , )
induced on H*by
theKilling
form of L. The
identity (4)
then becomesFollowing
Hannabuss(unpublished),
we now define the operatorAv
on
V(Å)
Q Vby
1
It is
straightforward
toverify
thatAv
commutes with the L-action onV(À) Q9 V. Furthermore, by (6),
which reduces to
This is the
polynomial identity
which appears in the work of Bracken and Green[3],
Green[4]
andHannabuss,
under theassumption
that Vis a finite dimensional irreducible L-module with
highest weight
11. However Kostant’s results show that theidentity (8)
is valid for any V which admitsa character.
Similarly,
we can consider the operatoras an element of
(End V(~))
QU(L). Then, using (3),
we obtainWe call
(10)
the characteristicidentity.
When is the naturalrepresentation
of a classical Lie
algebra, AL
isessentially
the matrix over the Liealgebra
considered
by
Bracken and Green.We see that Kostant’s results link the various accounts of
polynomial
identities and
show,
inparticular,
that the characteristicidentity
doesAnnales de l’Institut Henri Poincaré - Section A
411 ON CHARACTERISTIC IDENTITIES FOR LIE ALGEBRAS
not
depend
on the choice of basis of L. This is so because CL isindependent
of the choice of basis.
Nevertheless,
it is of interest to writeAL
in termsof a
basis {
x2, ...,(k
= dimL),
of Land itsdual { x 1, X2, ..., xk ~
with respect to
Killing
form of L. Withwe obtain
For the
particular
case where L =gl(n, Q),
we choose thebasis {
and take for 7~ the n x n matrix
representation
in whichwhere is the n x n matrix whose
only
non-zero element isequal
toone and lies at the intersection of the ith row and the
jth
column. We findThis matrix agrees with the matrix A introduced
previously. (Note
thatin this case the number m of distinct
weights
of coincides with the rank n ofQ).)
Now suppose that
LR
is a real form of L. The universalenveloping alge-
bra of
LR
and its centreZ(LR)
arenaturally
embedded inU(L)
andZ(L) respectively.
It is clear from theproof
of theorem 4.9 in[6] that,
when the coefficients in the
polynomial pz
are also elements ofZ(LR). Thus,
Kostant’s theorem can be used to derive characteristic identities for the real forms ofcomplex semi-simple
Liealgebras.
On the basis of the results derived for
gl(n, Q)
in this paper, weconjecture
that Kostant’s results
apply
to reductive Liealgebras
over any subfield of thecomplex
field which includes the rational field.Henceforth we shall consider
only
the case ofgl(n, Q)
and adhere to the notational conventions established in the introduction.3. EXISTENCE OF THE CHARACTERISTIC IDENTITY Lehrer-Ilamed
[2]
has shown that the elements of any 3 x 3 matrixover a non-commutative
algebra satisfy
nine identities. Hisproof
can begeneralized
so that itapplies
to any n x n matrix. Thestarting point
is theclassical
Cayley-Hamilton
theorem.Vol. XXVI, n° 4 - 1977
-
412 D. M. O’BRIEN, A. CANT AND A. L. CAREY
THEOREM. -
Suppose
that K is a commutativealgebra
overQ.
Ifis any n x n matrix over
K,
then B satisfies whereand
This result is well known
(Greub [9] ),
but there are severalpoints
whichshould be noted.
1) bk
is apolynomial
with rational coefficients in traceB,
traceB2,
...,trace Bk. More
precisely,
+ trace Bk E
Q [trace B,
traceB2,
..., trace(12)
where
Q[Xl,
x2, ...,xk_ 1]
denotes thering
ofpolynomials
overQ
inindeterminates xl, x2, ...,
2)
Each termbkBn-k
is ahomogeneous polynomial
ofdegree
n in theelements of B.
3)
Thesingle
matrixidentity provides n2
identities between the elements of B. The(i, j) identity
can be writtenIn this
identity,
where
Annales de l’Institut Henri Poincaré - Section A
413
ON CHARACTERISTIC IDENTITIES FOR LIE ALGEBRAS
and
The
precise
form ofis not essential for the arguments which
follow ;
it is sufficient to know that for each(i, j)
it is an array of rational numbers.Equation (13)
possesses asimple generalization
which holds when theassumption
that K should be commutative is relaxed.Suppose
thatSi
1 is a finite dimensional vector space overQ
with basisSet
and construct
the tensor
algebra
based onS1.
Definewhere
and the summation is over all
permutations
of theset { 1, 2,
...,p }.
Ifvl, v2, ..., vp are
arbitrary
vectors inS1
withdefine
It is obvious
that
V1 V2 ... isindependent
of the order in which ... , vp areprescribed.
Vol. XXVI, n° 4 - 1977
414 . D. M. O’BRIEN, A. CANT AND A. L. CAREY
THEOREM 1. - If B =
(biJ
is an n x n matrix whose elements are chosenfrom
Si, possibly
withrepetitions,
thenProof
- SetS° = Q
and S 1 =S 1.
Thepth symmetric
power ofS~,
denoted
SP,
is thesubspace
of T with basisConstruct the
graded
vector spaceEach element of
S,
say may beregarded
as an infinite sequencewith cvp E
Sp(p
=0, 1, 2, ...)
andonly finitely
many 03C9p non-zero. Addition and scalarmultiplication
in S areperformed componentwise.
Define abilinear map x : SP x Sq --~
by
and then extend x to a bilinear map x : S x S - S with the further definition
/ ’rB B
Equipped
with thisproduct,
S is a commutativealgebra
overQ.
The matrix B is a matrix overS, provided bi j _ ~ b‘~ ~
is identified withso the
Cayley-Hamilton
theorem asserts thatSince
equation (14)
followsdirectly
from(15)
and theproof
of the theorem iscomplete.
As a
corollary
of thistheorem,
the elements of any matrix B over the universalenveloping algebra
of any Liealgebra
overQ satisfy
the n2identifies
(14).
To provethis, simply
letS1
be thesubspace spanned by
the elements of B. For the matrix A over
U,
this result can besharpened.
Annales de l’Institut Henri Poincaré - Section A
415
ON CHARACTERISTIC IDENTITIES FOR LIE ALGEBRAS
THEOREM 2.
- Set tk
= trace Ak. The matrix A over U satisfies thepoly-
nomial
identity
in which
p(A)
is called the characteristicpolynomial
of A.Proof
- SetThus, bl, b2,
...,bn
are the coefficients which appear in theCayley-
Hamilton theorem and
certainly satisfy (17). According
to theorem1,
the elements of Asatisfy
where x denotes a
permutation
of{ 1, 2,
...,n ~. Alternatively,
The commutation relations
(1) permit (19)
to be cast in the formwhere
and qij
is the sum of all those termsproduced by reordering
the factors in(18). hij
is ahomogeneous polynomial
in the elements of A withdegree
n, but
qij
is nothomogeneous
and itsdegree
is(n - 1)
at most. Writewhere
(qm)ij
denotes the contribution toq~~
fromVol. XXVI, n° 4 - 1977
416 D. M. O’BRIEN, A. CANT AND A. L. CAREY
is a sum of monomials of the form
The coefficient of each monomial ensures that one
superscript
isequal
to i, onesubscript
isequal to j,
and each of theremaining superscripts
is contracted with one of the
remaining subscripts.
Thus(qm)ij
is the(i, j)
element of a
polynomial
in A with coefficients inQ[tl, t2,
...,tn] ;
It follows from
(20)
thatqm(A)
hasdegree (n - m)
at most, soThe
only
term in(19)
which involves An has been absorbed intoh(A),
soFurthermore,
because has
degree (n - 1)
at most whenregarded
as apolynomial
in the elements of A. If
then A satisfies
(16). Finally,
it is clear that ci, c2, ..., cnsatisfy (17).
A similar
proof
can be devised for the case in which the matrix A isreplaced by
the matrixAL
E(End V(~))
(8) Uappropriate
to a tensor repre- sentation ofgl(n, Q)
inV(~). Naturally
theproof
is morecomplicated,
andso will not be
given
here.However,
Green[4]
does treat this extension.4. FACTORIZATION
OF THE CHARACTERISTIC POLYNOMIAL
The characteristic
polynomial
of A cannot besplit
into linear factors in thealgebra
of n x n matrices over U.Nevertheless,
it is in factorized form thatp(A)
is most useful for the construction of irreduciblegl(n, Q)-
modules. In this section we shall
injectively
embed U in analgebra
LJover
Q
so thatp(A)
factorizes in thealgebra
of n x n matrices over U.LEMMA 1. - If
q(A)
is anypolynomial
in A with rationalcoefficients,
then .
Annales de l’lnstitut Henri Poincaré - Section A
417
ON CHARACTERISTIC IDENTITIES FOR LIE ALGEBRAS
Proof
- The result is obvious when thedegree
ofq(A)
isequal
to one.Induction on the
degree
ofq(A)
establishes the resultgenerally.
LEMMA 2.
Proof - According
to lemma1,
Thus, tm E Z
and soQ [t]
£; Z.If z E
Z,
then z cannot have any free indices.Every fully
contracted ele-ment of U can be reduced to a
polynomial
in t1, t2, ...by repeated
use ofthe commutation relations. Because A satisfies an nth order
polynomial identity
with coefficients inQ [t],
the elements tm + 2, ... must also lie inQ [t]. Hence,
Z ~Q [t].
Now,
Z is anintegral domain,
becauseQ
is a field. The field ofquotients
of Z can be identified with
Q(t),
the field of rational functions of t with rational coefficients. Let Z denote the minimal extension field ofQ(t)
inwhich the
polynomial p(x) splits
into linear factors :Finally,
set a =(al, a2,
...,an).
LEMMA 3. - Z =
Q(a).
Proof
- Jacobson[10]
shows that Z =Q(t, a).
It is obvious thatQ(a) ~ Q(t, a) ,
so it is
only
necessary to prove theopposite
inclusion. Each of the coeffi- cients ci, c2, ..., c~ in(21)
can beexpressed
as asymmetric
function ofal, a2, ..., an, denoted
by
ck =
ck(a)
EQ(a) .
Theorem 1 asserts that
ci + t1 E
Q ~ Q(a).
Thus,
t 1 E Q( a) . Similarly,
2cz
so
.
Vol. XXVI, n° 4 - 1977
418 D. M. O’BRIEN, A. CANT AND A. L. CAREY
The argument can be
repeated
to proveHence,
and so
Although
thepolynomial p(x)
can be factorized inZ,
it is not correctto say that
p(A)
can also befactorized,
because ameaning
has not yet beenassigned
toproducts
uai and aiu, u E U.However,
this iseasily
remedied.T pt
where the tensor
product
is that of Z-modules. The mapsare
injections and,
once LJ is endowed with the obviousmultiplication, they
becomeinjective algebra homomorphisms.
The factorization ofp(A)
now follows
simply.
THEOREM 3. - The matrix
A,
considered as a matrix overU,
satisfiesal, a2, ..., an are called the
eigenvalues
of A.5. RELATIONSHIP BETWEEN EIGENVALUES OF A AND HIGHEST WEIGHTS
Let F be any field
satisfying Q ~
F ~C,
and let H be the Cartan subal-gebra
ofgl(n, Q)
with basisDEFINITION.
- Suppose
that V is an irreducibleUF-module
which con-tains a vector vo such that
vo is called a maximal vector of
highest weight ~,,
where A is the linear func- tion from H to Fgiven by aii
-~,i.
We shall writeVF(~,)
for such a V.The main result of this section is the
following theorem,
whoseproof
will be
given
in a sequence of lemmas.Annales de l’Institut Henri Poincaré - Section A
419
ON CHARACTERISTIC IDENTITIES FOR LIE ALGEBRAS
THEOREM 4. -
Suppose Å
=~,2, - - -, Up
toisomorphism
there is
only
one irreducibleUF-module VF(~,)
withhighest weight
/L Theeigenvalues
of A can be reordered so thatwhere
Note that the
expression (25)
iscompatible
with theexpression
whichwould be
expected
on the basis of Kostant’sresults, namely,
Here,
vi, v2, ... , vn are theweights
of thecontragredient
of the naturalrepresentation
ofgl(n, Q).
Let us
begin
with the case F = C. Harish Chandra has proven the exis- tence of an irreducibleUc-module
withhighest weight A [11].
LEMMA 4.
- Suppose
that vo is the maximal vector of and thats(x)
is a
polynomial
inZ [x].
1) [s(A)ij, akl] = 03B4kjs(A)il - 03B4ils(A)kj.
2) s(A)ijv0
= 0 ifK 7.
3) s(A)i ivo
= (JiVO ,(Ji E C (no summation).
Proof
- All three results are obvious if thedegree
ofs(x)
is one. Induc-tion on the
degree
ofs(x)
then establishes the resultsgenerally.
We now set
Lemma 4 shows that vo is an
eigenvector
of with someeigenvalue
E C. Because C is
algebraically
closed and isirreducible,
theeigenvalues
of A arerepresented by scalars ;
akv = for all v E
where ak
E C.LEMMA 5. - The
eigenvalues
ofsj(k) satisfy
the differenceequation,
and the
following boundary conditions,
Vol. XXVI, n° 4 - 1977
420 D. M. O’BRIEN, A. CANT AND A. L. CAREY
~’roof.
- We haveSince
alkvO
= 0 for 1k,
the summation over I can be restricted to the range 1 ~ k.Thus,
where
part (1)
of Lemma 4 has been used to commute the factors. Sinceit follows that
Hence,
It is clear that
Also,
satisfies the differenceequation
which has the solution
consistent with
(26).
We note now that the
ordering
of al, a2, ..., an isarbitrary,
becausethese
quantities
appear as the roots of apolynomial equation.
LEMMA 6. - We may reorder a1, a2, ..., an so that
Furthermore,
if1)
is non-zero for kj
n, thenProof
- Becauses 1 (A)
=0, ~ 1 (k)
= 0 for all k.Annales de l’Institut Henri Poincaré - Section A
421
ON CHARACTERISTIC IDENTITIES FOR LIE ALGEBRAS
Now,
so Àn
must beequal
to one of the scalars a2, ..., an. Relabel a 1, a2, ..., anso that
Àn
= an.Then,
Now use induction.
Suppose
that al, a2, ... , an have been reordered sothat = 0 for 1 Since
we
have,
for 1 k -1, Thus,
and so either
1)
=0,
oror both. In the first case, we can say
nothing.
In the second case, relabel the roots ..., ak - 1 so thatSince
we have in either case that
Note that the
relabelling
of ..., ak-1 does not alter thelabelling
of ak, ak+ 1, ..., an.
Hence,
the lemma is established..LEMMA 7.
- Suppose
that is such that butVol. XXVI, n° 4 - 1977 28
422 D. M. O’BRIEN, A. CANT AND A. L. CAREY
for
any j
> k. Then there is apolynomial
xk, ...,xn],
wherexk -1, xk, ..., xn are
indeterminates,
for whichIf C" is
equipped
with the Zariskitopology,
the set of Å E Cn such that-
1)
# 0 forany j
is dense in Cn.1)
is apolynomial
in ak, ak + 1, ~ ... , an and~,k _ 1, ~,k, . - . , ~,n
with coefficients in C. Because
1)
# 0 for kj,
lemma 6 asserts thatHence, 1)
can be rewritten as apolynomial
in~_i, Àk,
...,Àn.
Denote this
polynomial by
qk. LetIk
denote the ideal in x2, ...,xJ generated by qk
and letMk
denote the closed subset of Cn determinedby Ik.
n
Then Mk
is also closed. Itscomplement Mo.
which is open, is neces- k= 1sarily
non-empty, for otherwise we would cover Cnby
closed sets, none of which isequal
toCn, contradicting
a well known property of the Zariskitopology [7]. Mo consists
of those elements of Cn for which(Jij - 1) #
0for any
j. Any
non-empty open subset of Cn is dense in the Zariskitopology.
Define an
equivalence
relation on Fn as follows. If a,{3
EFn,
write a -{3
if and
only
ifwhere x is a
permutation
of theset {1, 2,
...,~}.
Let S denote theequi-
valence class
which
contains a, and letFn
denote the set of allequivalence
classes.
Lastly, if g
is asymmetric polynomial
function onFn,
defineg : Fn --~
Fby g(a)
=~(S).
Note that the symmetryof g
is neededif g
is to bewell defined.
We have constructed a map
fc :
Cn -~Cn
where =&, by taking
A E Cn to the
uniquely
determined moduleVe(À)
on whichwhere and a = a2, ...,
an) uniquely
determines &. We have shown that a 1, a2, ..., an can be reordered so thaton a
(Zariski)
dense subset of Cn. The next step will be toequip C"
witha suitable
topology
and show thatfc
is continuous.Let
C[x]
denote thering
ofsymmetric polynomials
in indeterminatesx =
(xi,
x2, ...,xn).
IfI
is any ideal inC [x],
letWe
impose
atopology
onCn by declaring
the setsM (I)
to be closed.Annales de l’lnstitut Henri Poincaré - Section A
423
ON CHARACTERISTIC IDENTITIES FOR LIE ALGEBRAS
LEMMA 8. -
fc
is continuous.Proof
- Consider thefollowing polynomial
inC[x, y],
where
It is well known that every
symmetric polynomial
in xi, x2, ..., xn canrearranged
as apolynomial
ins2(x),
...,sn(x).
Thusê[x]
=C [s(x)]
where s =
(S1,
s2, ..., Theorem 2 shows thatso
C [s(a)]
=CM.
HenceCM
=C[t].
If g E letg’
denote the samepolynomial regarded
as an element ofCM.
Suppose
thatM( I)
is a closed set inC"
and that M ~ Cn is itspreimage
under
fc. This,
if there exists such that If g EI,
thenBy
direct calculation we see thatand in
general
where E
C [x]. Hence, (L
1, z 2, ...,Let I denote the ideal in
C[x] generated by
the setWe have established
that,
if ~, EM,
thenso M is closed in the Zariski
topology
of Cn.Thus., fc
is continuous.Vol. XXVI, n° 4 - 1977
424 D. M. O’BRIEN, A. CANT AND A. L. CAREY
COROLLARY. - The roots al, a2, ..., ~ of the characteristic
identity
may be reordered so that
Proof.
DefineIt is clear that gc is continuous in the Zariski
topology.
gc coincides withfc
on an open
(and
thereforedense)
subsetof Cn,
denotedby
M. Lethc =
gc -fc
and suppose 0. The set
is open in
Cn,
because{ 0}
is closed inC"
andhc
is continuous.Thus,
we have written Cn as the union of twodisjoint,
non-empty, opensubsets,
namely,
.Since
M and N are also closed.
However,
Cn cannot be coveredby
two closedsubsets,
neither of which is Cn itself[7]. Consequently,
N is empty andfc
coincides with gc on Cn.
We have therefore established theorem 4 for the case F = C. We shall
now show that it remains true for any field F which satisfies
Q ~
F ~ C .LEMMA 9. - For any Å E
Fn,
there exists up toisomorphism only
oneirreducible
UF-module
withhighest weight A,
denotedVp(~).
The maximalvector vo with
weight A
isunique.
For anyzeZ, where (
e F.Proof
- Let denote the irreducibleUc-module
withhighest weight
A. isspanned by
the setBecause A E
Fn,
allweights
of lie in Fn. LetVF(~,)
denote the vectorspace over F
spaned by
B. It is clear thatVF(~,)
is aUF-module
and is irre-ducible. The
uniqueness
ofVF(~,)
follows from theuniqueness
ofThe maximal vector in is
unique. Suppose
vo and~o
are bothmaximal vectors in Extend the field of scalars from F to C. Both vo and
vi
remain maximal vectors and so vo =~o’
Annales de l’Institut Henri Poincaré - Section A
ON CHARACTERISTIC IDENTITIES FOR LIE ALGEBRAS 425
Suppose
zeZ and 0. Sinceis a maximal vector with
highest weight
/LHence,
where the
case ~
= 0 may now be included. BecauseVF(~,)
is an irreducibleUF-module,
thecommuting ring
of is a divisionring,
so(z
-~)
is either invertible or zero. The first
possibility
isobviously
not true, soThe
proof
of theorem 4 can now becompleted.
Define amap fF
from Fnto
C" by
the sequencewhich is to be understood as follows. A
uniquely
determinesVp~).
As inlemma
9,
is obtained from OnFinally,
a =(ai,
a2, ...,an) uniquely
determines 5.Consequently, Hence,
al, a2, ..., an can be reordered so thatIt follows from a result of Dixmier
[8] that,
if V is any irreducibleUc- module,
then Z isrepresented by
scalars in V. Because C isalgebraically closed,
Z will also berepresented by
scalars in V. Thisproperty
is so useful that it isadvantageous
to isolate thoseUp-modules
in which it continues to hold even when F # C.DEFINITION.
- Suppose
that V is an irreducibleUF-module
and that xis the
representation
of U affordedby
V. V will be called a scalarUF-module
’
(where lv
is theidentity
operator onV) .
This definition will prove useful in the next section.
6. CHARACTERS
AND THE SPECTRAL REPRESENTATION OF A If V is a scalar
UF-module,
thenakv = for all v E
V ,
where a~ E F. Theorem 4 shows
that,
if V has a maximal vector ofhighest weight ~,,
then the roots al, a2, ... , an may be reordered so thatVol. XXVI, n° 4 - 1977
426 D. M. O’BRIEN, A. CANT AND A. L. CAREY
This suggests
perhaps
that the numbers al, a2, ..., an,suitably ordered, might generalize
thelabelling
of irreducibleUp-modules by highest weights,
because ai, a2, ..., an arealways
definedand,
as will be shownbelow,
are distinct elements of Z.
However,
lemma 10 below shows that al, a2, ..., anprovide essentially
the same information as the character on V.LEMMA 10.
- Every
scalarUp-module
V admits a character x. If akv = for all v EV ,
then &
uniquely determines x and, conversely,
isuniquely
determinedby
x.Proof.
- Because V is a scalarUF-module,
z. v =x(z)v
for all v E Vwhere
x(z)
EF, implying
that V admits the character x : Z -~ Fgiven by
z -
x(z). As tk
is asymmetric polynomial
function of a, &uniquely
deter-mines
x(~) _ x(t2)~
...,which in turn
uniquely
determines x because Z =Q [t]. Conversely,
thecoefficients of the characteristic
polynomial
of A are elements ofZ,
so xuniquely
determines &.We now
proceed
to a discussion of some results of Green[5]
for whichthe
algebraic
frameworkdeveloped
above seemsappropriate.
LEMMA 11. - The
eigenvalues
al, a2, ..., an are distinct elements of Z.Proo, f. Suppose
thecontrary,
that ai = a~ for some i ~j.
There existsan irreducible
Uc-module Vc(~,)
whosehighest weight
~, satisfies~i_i~~~_.7. (27)
Theorem 4 asserts that
which contradicts
(27).
This result allows a
spectral representation
of A to be written downimmediately.
Define thefollowing n
x n matrices over LT :Note that these matrices are well defined
because ai
# a~ if i #j.
THEOREM 5. - The matrices
Ei, E2,
...,En satisfy :
Annales de l’Institut Henri Poincaré - Section A
427
ON CHARACTERISTIC IDENTITIES FOR LIE ALGEBRAS
The matrix A has the
spectral representation
Proof
-1)
It followstrivially
from theidentity (24)
thatIf
q(A)
is apolynomial
in A with coefficients inZ,
it iselementary
to esta-blish
by
induction on thedegree of q
thatIn
particular,
2)
TheLagrange interpolation polynomial
ofdegree (n
=1)
to the cons-tant
polynomial
1 inZ [x]
isThe
interpolation
is exact because thedegree
of the constantpolynomial
is zero and therefore
certainly
less than n.Thus,
The steps in the
proof
of the ancient result still hold inZ [A], provided
1is
reinterpreted
as the unit matrix over Z.Thus,
3)
Thespectral representation
of A followssimply
from(28) by
multi-plication by
A :Vol. XXVI, n° 4 - 1977