The Determination of the Admissible by 0.

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The Determination of the Admissible
Nilpotent Orbits in Real Classical Groups.
by
James 0. Schwartz
SUBMITTED TO THE DEPARTMENT OF
MATHEMATICS IN PARTIAL
FULFILLMENT OF THE
REQUIREMENTS FOR THE
DEGREE OF
DOCTOR OF PHILOSOPHY
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
June 1987
b
Massachusetts Institute of Technology 1987
Signature of Author
,4
Department ofYMathematics
May 1,
Certified by
Accepted
"
1987
David A. Vogan
rpýsis Supervisor
by
Sigurdur Helgason
by~
Chairman, Departmental Graduate Committee
AUG 22 1988
ARCHIVES
The Determination of the Admissible
Nilpotent Orbits in Real Classical Groups.
by
James O. Schwartz
Submitted to the Department of Mathematics
on May 1, 1987 in partial fulfillment of the
requirements for the Degree of Doctor of Philosophy
ABSTRACT
This thesis classifies the admissible nilpotent orbits,
in the sense of Duflo, in the following groups: SL(n,R),
Sp(2n.R). O(pq), U(p,q), SO(p,q) and SU(p,q). The
philosophy of coadjoint orbits suggests that the
admissible orbits should be the ones to which one can
attach representations.
Thesis Supervisor:
Title:
David A. Vogan
Professor of Mathematics, M.I.T.
Dedication
This thesis is dedicated to my parents
with love and gratitude.
Acknowledgments
Interacting with my advisor David Vogan over the
past four years has been a remarkable experience. He has
It
been tremendously generous with his time and ideas.
of
such
to
a
person
exposed
has been a privilege to be
deep commitment and excellence in his discipline. He is
an inspiring example of clear thinking and intellectual
honesty.
I am very grateful to Jerry Orloff for the many
hours he spent with me in the dungeon teaching me the
basics of Lie theory, and later, puzzling over the
meaning of D.V.'s remarks.
Nesmith Ankeny got me started at M.I.T. and his
presence and unique theories about the world have
enriched this experience.
I've spent many fruitful and not so fruitful hours
discussing math and other quandaries with Jeff Adams,
Jesper Bang-Jensen, Sam Evens, Bill Graham, Monty
McGovern and Roger Zierau.
Jamie Kiggen and Bill Schmitt have been a great
source of support, humor and affection over the past five
years.
My friendships with my brother Tony, and sister
Cassie, Jerry, Kathi, Steve, Peter B., Peter H., Annie
I have
and Jeff have meant an enormous amount to me.
shared many good times with Martin whose obsession with
movies and other forms of fun were a welcome relief from
I thank John Rhodes,
the colder beauty of mathematics.
the playful Spartan, for his comradeship during the years
we were tutors. Thanks to Phyllis Ruby for her
helpfulness and Maggie Beucler for typing my thesis.
Most of all I wish to express my love for Fluke,
Flip, Wig, Bertha, Fatface, and the Hund; many names for
the most important person, my wife, Ann Hochschild.
Table of Contents
Abs tract
0.
Introduction and Preliminaries
1.
Reduction to LnK
2.
Complex Structure on
3.
Classification of Nilpotent Orbits
4.
Explicit Forms for
5.
Admissibility in Sp(2n,R), O(p,q), U(p,q),
SO(p,q) and SU(p,q) .
44
SL(n,R)
76
6.
Admissibility in
References
g~/g E
0 0
L
and
LnK
85
-1INTRODUCTION
This thesis is a study of
the connection between
representations of a semisimple Lie group and the
partition of the dual of
its Lie algebra
under the coadjoint representation.
g
into orbits
Kirillov first
introduced this method in the study of simply connected
nilpotent groups and it was generalized by
Auslander-Kostant and Duflo to simply connected type I
solvable groups.
In these cases the orbit method has
been successful in setting up a correspondence between
equivalence classes of irreducible unitary
representations and coadjoint orbits of
G
on
go.
The
connection between the algebraic picture of the unitary
dual and the more geometric orbit picture has provided
useful insight and added coherence to the representation
theory.
To implement the correspondence discussed above,
Duflo introduced a method for picking out a subset of
orbits he calls admissible orbits in
go
.
He describes
a bijection between admissible orbits and unitary
representations for nilpotent and Type I solvable groups.
In the case of a real semisimple Lie group it is not at
present understood which nilpotent coadjoint orbits
should correspond to representations of the group.
-2Moreover, there isn't a general technique for attaching
representations to nilpotent coadjoint orbits -geometric quantization has not been applied successfully
to nilpotent orbits except in special cases.
The
admissible nilpotent orbits are good candidates on which
to attempt quantization.
We have classified the nilpotent orbits and
determined which of them are admissible in the following
real classical groups:
Sp(2n,IR) , O(p,q)
and
U(p,q)
-- the first two are the groups preserving a symplectic
and a symmetric form, respectively, on a real vector
space, and the latter is the group preserving an
Hermitian form on a complex vector space.
In addition,
we have determined the admissibility of the nilpotent
orbits in the following semisimple groups:
SU(p,q)
and
SO(p,q)
SL(n,R) .
Chapters 1, 2, and 4 describe the method used to
determine whether an orbit is admissible.
In Chapter 3
we recall how the real nilpotent orbits under each of the
real classical groups above are parametrized by
equivalence classes of representations of
SL(2,IR)
on a
vector space preserving the appropriate non-degenerate
sesquilinear symplectic or Hermitian form (Springer,
Steinberg).
In Chapter 6 we do the same in the case of
SL(n,R) , except we look at representations of
SL(2,IR)
-3preserving a multilinear n-form.
A nilpotent orbit under
a complex group may break up into a disjoint union of
several nilpotent orbits under a real form of the complex
group, but it turns out,
in the cases we have studied,
that the issue of admissibility of a nilpotent orbit
depends on its orbit under the complexification of
G
The final two Theorems in Chapter 5 show which complex
nilpotent orbits are admissible in terms of the
parametrization discussed above.
In Chapter 6, we show
all nilpotent orbits in
are admissible.
SL(n,R)
PRELIMINARIES
When
G
is semisimple, we can make use of the
(or any nondegenerate G-invariant
Killing form, B
bilinear form on
go), to identify
this identifies adjoint orbits on
orbits
on
We denote
go
.
{Adg(X)jg E G}
G
{g E GAdg(X) = X}
be called
go
go
with
go , and
with coadjoint
the orbit of
X E g
by
0
= (F E qo
X, G X
which fixes
The Lie algebra of
[X,F] = 0}
.
GX
will
The adjoint
representation gives us a smooth, transitive action of
on
0X
; thus
the coset space
OX
=
We will call the isotropy subgroup of
X , the subgroup of
GX
go
is a homogeneous space isomorphic to
G/G
G
-4We now introduce an example which shows why
integrality of an orbit is not a discriminating enough
criterion to use to pick out orbits which should
correspond to representations.
integral
An orbit
OX
is called
if there exists a finite dimensional,
irreducible unitary representation
v
GE
of
satisfying
the following condition:
iB(E,X)*I = dv e(X)
If
E
is nilpotent, then
Lemma 1).
VX E gq
E
B(E, q)
= 0
It follows immediately that
for all nilpotent orbits -- take
representation.
v
(Chapter 1,
E
is admissible
to be the trivial
But an orbit and the representation
attached to it should match in dimension in the following
sense.
The representation attached to a nilpotent orbit
should have Gelfand-Kirillov dimension equal
the dimension of the orbit.
shown that
to one half
Howe and Vogan [9] have
there are no such representations associated
to the minimal nilpotent orbit in
Sp(2n,IR)
.
This
example points out the need for an extended notion of
integrality.
Duflo's definition of admissibility is
closely related to integrality, but it makes use of the
symplectic structure of the orbits.
We now describe this
structure.
OX
is a symplectic manifold, i.e. there is a closed
2-form which is non-degenerate on each tangent space.
-5E E 0X , we construct a skew-symmetric
For a fixed
bilinear form on
*
smooth mapping
g
E
.
given by
-)) TE(OX)
g
--
Adg(E)
is onto and has
We define our bilinear form on
go/go
X )
OX
Thus we have a canonical isomorphism:
E
TE(X)
E(O
G -
:
consider the
as follows:
d~le :
Then the map
kernel
TE(OX)
and carry it over to
0X
by means of this
E
o
q0
0
isomorphism.
To begin, define the skew-symmetric mapping
WE :
R
90Xgo -
The radical of
as follows:
wE
E
is
go
wE(V,W) = B(E,[V,W])
, and therefore
wE
makes
sense when it is considered as a form on the quotient
E
space
g /g
go /go
E
Clearly,
We will omit
.
GE
The group
E
is non-degenerate on
the verification that
OX .
a closed 2-form on
wE
wE
gives us
(See Guillemin-Sternberg).
acts on the symplectic vector space
via the adjoint representation and preserves the
g /g
form
wE
.
For
g E G
E
E
E
g.(Y+g o ) = Adg(Y) + g o
and
It
Y + g
E
E
Eo/go
E
is easy to check
, we define
that
this
E
into
,
Sp(go/ao)
vector space
the group which preserves
g /g
o
wE
on the
E
o
The symplectic group of any vector space
V, Sp(V)
has a well-known two-fold cover called the metaplectic
cover, Mp(V) (Shale [6]).
let
T
Let
7 : Mp(V) --+ Sp(V)
and
-6denote our mapping
mp
(GE
,
G
E
-- SP(go
/o)0
E
the "pullback" cover of
.
G
(GE mp = {(x,y) 6 GE x Mp(go/o
E)(x
mapping
p
(GE) m p -G
:
E
We define
as follows:
=
)
r(y)}
.
The
defined by projection on the
first factor is a two-fold cover of
GE
MP(g o /E
(GE)mp
GEE
E
Sp(go /g)
T
We say a representation
vmp
of
(GE )mP
is genuine if
it is not trivial on the kernel of the map
pl
:
(GE)mp -, G E
representation
i-B(E,X)*I
Finally. we say a genuine
vmp
is admissible if dumPle(X) =
E
E
X Eg
. The group G
and the
for all
orbit are said to be admissible if
admissible representation.
that
i*B(E,X) = 0
nilpotent.
(GE )
z t
Thus
, and
(GE)mp
,
G
where
kernel of the map
For
E E g
admits an
It is not difficult to see
for every
vmp
(GE) m p
X E g
E
if
E
is
must be trivial on
will be admissible if and only if
z
pl
is the nontrivial element of the
:
(GE)mp --
GE
nilpotent, the algebraic group
be decomposed as follows:
GE = L k U , where
L
GE
is
can
-7reductive and
U
simply connected.
to
It follows that
L
and its two-fold cover
Possibly after replacing
we can embed E
may
(GE)mp
6
' where
take
is isomorphic
9t(2)
the Cartan involution of
0 (9I(2))
,
or
polar decomposition enables us
Because
as
L
LmP
by another element of
in a 6-stable copy of
is
=
X
E
to write
0E
inside
g
L = ZG(SL(2))
reductive subgroup in our decomposition of
L
, and
Lmp x U , which reduces the question of admissibility
to the subgroup
go
GE
is unipotent, normal in
GE
.
, as
.
Then we
the
The
G = Kexp(p )
is 9-stable and algebraic, we can decompose
L = (LnK)exp(lop o) , and, as above, the question
of admissibility reduces to a question about the two-fold
cover of
by mapping
LnK .
LnK
We recall that this cover is constructed
into
Sp(go/g)
.
Hence the question of
admissibility of an orbit finally reduces to analyzing a
representation of a compact group on a finite dimensional
vector space.
-8Chapter 1
We begin by showing how the question of
admissibility for the nilpotent orbit
0E
is equivalent
to a topological question about the two-fold metaplectic
cover
(GE )mp
GE C G .
of the subgroup
We begin with
an easy Lemma.
Lemma 1.
Let
Proof:
.
[go
[E,X] = 0
this
qf(2)
B([goli,
[o]j) = 0
s9(2)-theory tells us that
E
E C
i<
[Io
0
i
Proposition 1.
, where
i .
; therefore
A nilpotent orbit
We decompose
[go•
i
is
i = j .
unless
2
E
o)
0E
if and only if the kernel of the map
is not in the identity component
to embed
the
The go-invariance of
E E [Co
B(E,
go
implies that
{E,F,H} .
s~(2)-triple
under
i
implies
B(E,X) = 0
is the Killing form on
H-eigenspace with eigenvalue
B
Then
Use the Jacobson-Morosov Theorem
in a standard
=
B
We must show that
B(E,X) = 0
0
goq be nilpotent.
E
X E g0 , where
for all
E
E
The
and
= 0
.
C go
r
(GE mp
:
is admissible
(GE)mp of
(GE )mP
GE
-9Proof.
Let
(e,z}
M =
denote
order two subgroup of
GE
clear that
= 0
mp
vmp
Hence
Vmp(z) = I
Now assume
z C (GE) mp
an
it is
EGE
(GE )mp
then, by
= I
vmpI
vm p
and
of
.
In
is not genuine.
(G E mp , and we construct a genuine
z
representation
vmp
of
= I ; that
mp
r --
is not admissible, because if there were
dvmPI
particular,
If
(GE mP
an admissible representation
the Lemma,
the kernel of
(GE mP
is
vmp
such that
is admissible.
(GE mp
Let
M = {z,e}
component group
denote the image of
(GE)mP
=
(GE)mP/(GE)mp
M
.
in the
Then
SC (GE)mp , an order 2 normal subgroup, is contained in
the center of the group, hence
K C Z((GE)mp)
.
Let
a
be the non-trivial one dimensional representation of
M
Let
p
denote the right regular representation on
Ind(GEmp(a) .
M
p(z).f
:
Let
For
f
Ind
(E)mpa)
we calculate
M
g E
(GE
mp
,
then
p(z)f(g) = f(g z) = f(z g) = z.f(g) = -f(g)
Thus
(GE) m p
p(z) X I .
by setting
We define a representation
vmp(g) = p(g)
where
v
of
g E (GE) mp
andI
-10g
is the image of
vmp
g
(GE) m p
in
It is clear that
is an admissible representation of
GE
We now show
reductive subgroup
admissible.
E E go
is admissible if and only if a
L C GE
Proposition 2.
2
.
which we define below, is
Use the Jacobson-Morosov Theorem to embed
in a standard
L = ZG(sX(
))
GE
,
U
9s(2)-triple
{EF,H} C g 0
is isomorphic to
is
L K U
where
the closed subgroup of
corresponding to the Lie subalgebra
U
(GE )mp
u
0
= g
E
GE
n [E,
0
'0
o]
is unipotent, connected and simply connected.
Proof:
Assume
9X(2)-triple
u E U
.
is the nilpositive element in the
(E,F,H} C go .
g 6 GE
element
and
E
Fix
First we show that every
can be decomposed
g E G
{E.Adg-1 (F),Adg-1 (H)}
.
Then
g =
e E L
-*u with
{E,F,H}
and
are both 9s(2)-triples with the
same nilpositive element.
By Kostant (1959),
all triples
with the same nilpositive are conjugate by an element of
U .
Furthermore,
the set of neutral elements for all
triples with nilpositive
where
= [E, o] n g
u
00
U --+ H+u o
which sends
E
0
E
is the linear coset
H+u 0
and there is a 1-1 onto mapping:
u ---+ Adu(H)
.
Hence
there
is a
-11unique
u E U
such that
, Adu(Adg -1(H))
{E, Adu(Adg-1 (F))
Thus
u*g
-1
= L
E ZG(9t(2))
g =
Now we
, and we have decomposed
-1
(g.u-1 )u
show that
connected, simply connected, and
u
The exponential map:
u
--
the exponential map,
exp
nilpotent group
dimension of
dim n = 1
n .
is unipotent
is normal in
GE
U
:
is 1-1 and onto.
n --
N
To see
, for a connected
This statement is true for
because the exponential map is onto for a
n E N
n
has a nontrivial center
with
j , such that
exp(X+Zl)-exp(-Z
exp
there exists
exp(X+Z 1 ) = n-z
E
U
U
is a nilpotent Lie algebra.
therefore we may assume
Z2
,
N , is onto we use induction on the
connected abelian group.
Thus for
.
E LU
L n U = {e}
We will show below that
= {E,F,H}
2
exp Z 2
) = n
exp(X+Z -Z2) = n .
z E Z
n/g
---+ N/Z
X+g E n/j
, Z1 E
= z
j .
is onto.
such that
Also we
can find
; therefore
and this gives
To see
basis for the matrices
:
n
exp
is one-to-one, we find a
in which they are strictly
upper triangular (Engel's Theorem).
It is easy to see by
-12direct computation that
exp
matrices.
is homeomorphic to Euclidean
Therefore
space, i.e.
Let
U
U
is one-to-one on those
is connected and simply connected.
[go] k
be the H-eigenspace with eigenvalue
under the adjoint representation of
sI(2)
on
go
k
As
an easy consequence of the 9X(2)-theory, we get
u
[ o
C
Let
.
u C U
and
W E u
such that
k>O
exp W = u
If
X
E
$
[go
,
k
then
Adu(X) = X+Y
with
kŽO
Y E
[go
; thus
k
U
is unipotent.
By the same
k>O
argument, if
X C [go] 0
commutes with the
and
L n U = {e}
thus
GE
= L
enough to show
X E u
.
Adg(X) E
[E,W] = X
,
.
XI(2)
.
U
= u
0
go
.
0
On the
, because
.
U
.
is connected it
Since
[E,Adg(x)]
U
is normal
for all
g C GE
= Adg[E,X] = 0
We know there is a
W E go
AdR
This shows
We now show
in
GE
is
and
, hence
such that
so:
Adg(X) = Adg([E,W])
hence
action on
Adg(X) C u
We have
E
Adu(X) f [go]
e E L , AdR(X) E [go
other hand, for
that
then
Adg(X) E [E,go]
as desired.
= [Adg(e),Adg(W)] = [E,Adg(W)]
and we get
We note that
L
Adg(X) C gE n [E,go
is reductive, because
-13the centralizer of a reductive algebraic subgroup of an
algebraic group is reductive.
Recall
0
the covering map
GE
have shown that
(GE)m p --
:
GE
L x U .
is isomorphic to
We
.
Define
_-1
Lmp =
-1(L)
and we prove
Proposition 3.
Let
U
be the analytic subgroup of
Lie algebra
u
.
Since
an isomorphism of
by
L
, U
is
Lmp x U
isomorphic to
Proof:
(GE) m p
The metaplectic cover
U
7
is simply connected,
U
onto
U
is normalized by
Since
u
with
is
is normalized
o
Lmp
We have an injective map
must show this map is onto.
covering maps as
(GE)mp
i :
Let
(GE )m p
L mPU -
denote the
72
and
V1
and
in the diagram:
LmP U
(GE)mP
-
GE
Choose
x
1
E
(G
Emp
mp
show that either
i(y 1 ) ; i(y 2 )
and
{y, Y2
i(y 1 ) = x
I
or
=
-1
2
or
1
(X
i(y 2 ) = x
but they both maps to
xl
1
)
We will
1
under
7T
1
-14Since
r 1 (x 1 ) = x1
conclude either
there exists a
or
equals
0
= -F'
and
subalgebra spanned by
.
OH'
t'
9'
on the
V'E = -F
as follows:
e
to a Cartan
p'
' = [R(E-F)
with
0
0
and
p'
RH e IR(E+F)
=
involution
g
=
p"
" $
9") and
9"
V' C
o
Helgason).
.
One can extend
9'
on the whole Lie algebra
to a Cartan
go
so that
(the Cartan decomposition with respect to
f"
o
and
p' C p"
o
o
(Mostow, p.
277
Since all Cartan decompositions are
conjugate, we can find
and
{E' =
= -H'
{E,F,H}
sT(2) =
Then
satisfies the
This corresponds
decomposition of
.
such that the sf(2)-triple
We define a Cartan involution
8'H = -H
0
go
a Cartan involution on
g C G
OE'
xl1
be a standard
F' = Adg(F), H' = Adg(H)]
relations
and
i(Y 2 )
{E,F,H} C go
Let
sf(2)-triple and
Proof:
,1 is a two-to-one map, we
i(y 1 )
Proposition 4.
Adg(E),
and
Adg(p")
o
= p
decomposition on
where
g E G
such that
go=
p0
0
O
go with respect to
Adg(t") = f
o
is
8.
the Cartan
We have
-1
Adgog"oAdg
and the triple
{Adg(E),Adg(F),Adg(H)}
satisfies the desired relations.
O
o
=
-15We recall the polar decompositions for a reductive
G .
group
G = K exp(p o)
We can write
the Lie algebra decomposition of
the 1-eigenspace and
involution
8
for
G
C
=
t
p
is
the(-1)-eigenspace for a Cartan
go
defined on
compact subgroup of
.
po
go
which comes from
, and
K
the maximal
corresponding to the subalgebra
The following two Lemmas give this decomposition
L .
Lemma 2.
The centralizer of a O-stable subgroup
H C G
is O-stable.
Proof:
c
g
:
It suffices
G ---
G
be defined by
We observe that
c
c(eg) = Goc g0
g
Ad(Og) = OoAdgoO
(x)
= g*x*g
because
-1
.
Let
, x EG
.
c g(x) =
= O(g)-x-O(g1 ) = O(g(Ox)g- ) = O(cg(Ox))
(Og).x*(Og)for
to show that
x C G .
Differentiating this equation gives us what
we want.
0
Lemma 3.
Suppose
IH/HOI = n
Then
H
is a O-stable subgroup of
H = (H n K)exp(po n
§
)
G
and
where
§o = Lie(H)
Proof:
It is clear that
and we want
to
show
= (H n K)exp(po f
H
H11 = H
.
Assume
) CH
for a moment we
-16know
n
HO = (H
on)
K) 0 exp(po
can write
h = k exp X
-l
calculate
(Oh)-1*h = exp2X E H
get
(exp
2 X)n
Fix
h E k
, with
E exp(po0n
exp X
o)
ýo)
X E po
and
k E H
H' = (H n K) 0 exp(po n
We
IH/HnI = n . we
and, therefore,
o
K
So it remains to prove that
o0) .
.
X E po
Then
= exp(2nX) E H 0
, and hence
and
Since
exp(2nX) E H0 N exp po = exp(po
2nX E po
h E H , then we
So
.
O = (H f K) 0 exp(po
§o)
is an open set in
H0
because it contains a neighborhood of the identity,
therefore, assuming it's a group, it is both open and
closed which implies
group.
H' = H
.
Now we show
H'
is a
The following argument shows that it is
sufficient
to show for
expX expY E exp(po n
p,o
X,Y E ~
o) .
Let
kl'expX 1 ,k2 *expX
2
E H'
then
(k 1 exp X 1 )-(k 2 exp X 2 ) = kl(k
-1
2 k2
)*expX 1 *k 2 (expX 2 )
= kl'k2 expX 3 *expX 2
which is an element of
H'
expX 3 *expX 2 E exp(po n
So we set
o)
O(t) = exptX-exptY = k(t)exp(Z(t))
,
-17
k(t) E K , Z(t) C po
t
.
We want
to show
and
that
-
k(t),
Z(
a re analytic in
Z(t)
fn
t)
but it is enough to show for small
P0
t
for all
t
It is enough to
show that the coefficients in the power series expansion
of
Z(t)
belong
(t)
(00(t))-1
(e@(t))
p(t)
to
40
.
We
have
= exp(Z(t)) , hence
= exp(2Z(t)),
1
Z(t)
-
1
2
-1
log((eO(t))
0(t))
log(exptY*exptX-exptX.exptY)
By Campbell-Baker Hansdorf this last expression can be
written as a power series in
bracket in
ý
X
and
Y .
t
whose coefficients are
Since
X
and
, these coefficients do as well.
Proposition 5.
We have
n
L = (L
Y
belong to
0
K)exp(po n IO)
where
10
0 = Lie L
Proof:
Follows directly from Lemmas 2 and 3.
Proposition 6.
Proof.
The group
LmP
=
O
(L n K)mP exp(pO
nlO)
Apply the same argument used in Proposition 3. 0
Recall
that an orbit is admissible if and only if
-18the non-trivial element in the kernel of
7:
z
GE
(GE)mp -
the map
is not in the identity component.
be the non-trivial element in the kernel of
Proposition 3,
Therefore,
(GE)mp = LmP U
z E (GE) mp
Proposition 6,
LmP
and
U
By
is connected.
if and only if
= (LnK)mP.exp(ponl
v .
Let
z E (LmP)0
)
and
.
By
exp(p ol
is connected.
Thus,
z E (LK)p
We conclude that the admissibility of an
.
z E (LmP)O
)
if and only if
orbit can be decided by considering the compact subgroup
LAK c G
E
GE
We have seen that
representation into
Sp(2n)
maps under the adjoint
, where
LK C G E
The image of the compact subgroup
compact subgroup of
are conjugate in
Sp(2n) .
Sp(2n)
2n = dim (to/
E
E
must be a
Since all maximal compacts
, we can map
LnK
into
Sp(2n) n O(2n)
, a maximal compact subgroup of
isomorphic to
U(n) .
Sp(2n)
The following Proposition, which
we will not prove, enables us to compute the metaplectic
cover
U(n) m p
Proposition.
of
U(n)
The group
{(k,e i)
explicitly.
U(n)mp
is isomorphic to
E U(n) x S 1 I det(k) = e2i
Define a character
X
on
LnK
.
by composing the
-19following maps:
LnK
Sp(2n)nO(2n) ---S
We construct the pullback cover
metaplectic cover
Introduction.
(LAK)mP
Si
from the
as described in the
The next Corollary follows directly from
the description of
Corollary.
U(n)mp
U(n) det
U(n)mp
given above.
(LAK) m p
The pullback cover
is equal
to
1 1 x(g) = z2 }
{(g,z) E (LAK)xS
The next Theorem shows that an orbit
if and only if the character
there is another character
K
is admissible
\
is a square,
of
LAK
that is,
such that
S2
Theorem.
Let
kernel
the covering map
z
t
of
(LAK)0o
Proof:
<=>
be the non-trivial element in the
t (LnK) mop
0
7
the character
First, assume
be a path from
z
z
) = \2
the identity to
Let
{(g(O),e i)
:
(LAK)mp -
y
of
LAK
.
LAK
Then
is a square.
and we show there cannot
z
in
(LAK)o
E (LnK)xS11 0 <
,
that
e
is
< 7}
-20be a path from
(e,1)
, g(r) = e
g(O) = e
to
(e,-1)
, and "
P(g(O)) = e iO
Therefore
\(g(O))
=
\i(g(r))
= e
P(e)
i7n
= 1 ,
= -1
(g(O)) = \
(g(O))
\(g(o))
=
-e
or
p(g(O))
hence
which
(LnK) m0 p
in
= e
ie
2
W e have
= e2i 0
ie
But
Thus
is impossibl e because
g(7r) = e
Conversely, assume
a square.
Define a character
g E LnK , P(g) = {wj
check this
P(g)
and we sh ow
(g,w) E (LNK)o }
we have
as desired.
k(g) = w
It is easy to
LnK
; therefore
0
In the next Chapter we develop the machinery
necessary to compute
X .
x
P by setting, f or
is a well defined character of
(g,w) E (LK)m
k(g) =
z e (LnK)mp
For
-21Chapter 2
In order to construct a map from
LnK
into
U(n)
go/go
we will find an (LnK)-invariant hermitian form on
considered as a complex vector space with respect to a
complex structure
J .
The vector space
go/g
is
endowed with the natural GE-invariant symplectic form
WE .
We seek an (LnK)-invariant inner product
complex structure
J
Hermitian form on
g /g
such that
E
and
is a
, that is, the following
relation must be satisfied by
iT(x,Y) =
T = b + iwE
b
T(Jx,y)
J
all
and
T :
x,y E goa
0 0
.
Equivalently, the compatibility condition can be
expressed in terms of
J, b , and
b(x,y) = oE(Jxy)
If
b
and
T(x,y)
wE
all
wE
as:
E
x,y E go/o0
are (LnK)-invariant then so is
= b(x,y) +
iwE(x,y)
= b(e.x,e.y) +
= T(e.x,e.y)
iwE(e.x,..y)
T
a
-22Assume we have found an endomorphism
J
go/ E
on
following
four
properties:
satisfying the following four properties:
1)
J2 = -1
2)
WE(Jxy) = -WE(xJy)
3)
WE(Jx,x)
4)
J
x,y E go
all
E
with equality if and only if x = 0
_0
commutes with the action of
Then we define a symmetric bilinear form
LfK .
b
go/
on
0
def
b(x,y)
= oE(Jx,y)
by setting
0
and we make the
following
Claim.
on
= b + iwE
T
!o9 o
b
J
The necessary compatibility among
is built into the definitions.
property 2.)
as follows:
wE(Jy,x) = b(y,x)
.
b
b(x,x) = wE(Jx,x) > 0
(LOK)-invariant because
e
(LOK)-invariant hermitian form
viewed as a complex vector space with complex
structure given by
Proof:
is a
E LnK
we have:
b
J , wE , and
is symmetric by
b(x,y) = wE(Jx,y) = -WE(x,Jy) =
is positive definite because
if
x X 0 .
J
Finally, T
commutes with
LQK
is
.
For
-23-
T(x,y) = oE(Jx3y) + iWE(x'y)
= WE(e-JxCy) + iWE(e.x.~.y)
= WE(J(e.x).e.y) + iCE(e'x'e.y)
= T(e.x,e.y)
0
.
g /gq
We now exhibit a complex structure on
satisfies properties (l)-(4).
isotypic subspaces under
Decompose
sT(2)
o
go
into
W1
; go=
which
o
where
i=l
W1
m.V
Let
[W1],
Define
an irreducible 9X(2)-module, dim Vi = i
V
,
1
be the H-eigenspace with eigenvalue
J
on
Definition.
[Wile
for
x E [W ]e
For
•
-i-1
•
e
, define
i-3
e
as follows;
J(x) =
(Je.i) -1/ 2 .(adE(x))
where
depending on
e , and defined as follows.
i
and
the weight spaces of
constants
j
V
j2
are one dimensional,
such that for
.i
is a constant
.
Since
there are
x E [Vi]%
[F.[E,x]] = je.i-x .
We compute
and
F
j9,i
in the
by looking at the action of
el(2)
representation on homogeneous
polynomials of degree i-1.
0
1
operator
r((O 0 ) )
operator
x
8
ay
E
In this representation, the
corresponds to the differential
10
00 )
r( 0
to
, 8x ,0
y
8
,
and
1
7r(o
-1
0
)
to
-24-
a
8
So we have:
Y --
ax
i-l+e
i-1-
2
defn.i-1+
2
=
i
Ri
,i = J-e-2,i
2
y
)
2
i-1+e
i-1-e
2
2
..
conclude that
F.E.(x
i-3-P
i+1+e
ri-i-el
i-] .i..
2
=
y)
e,i(x
Proposition.
i-1-
2
=
i+1+
i-l-e
1 2
2
at
and note that
J , defined above, satisfies properties
(1)-(4).
Proof:
1.)
We
check
J2 = -1
.
It is enough
to check
this on a basis, and we choose a special basis made up of
eigenvectors spanning each isotypic component
x E [Wi
Pick
J
2
x = J((
(J
=
-1/2
,i )*O(adE(x))
-1/2
= (J,,i )J(O[E,x])
-1/2
,i )-J(O[E,x])
) 8[E,[ O E, O x]]
-1
= (-J~ i)-[F.[E,x]]
X
C q
;
-1/2
-1/2
S(J,i )-(J--2,i
--
W
since
,i
since O[E,x]E[W
and
1 --
E = -F
2
-25-
2.)
Check
w E(Jx,y) = -WE(xJy)
We
check this
equality on a basis of eigenvectors chosen as in 1.).
note that
W E(E o]i,[go]j] = B([E,[go
O
equals
unless
j = -i-2
we conclu de that for
x,y
-W E(xJy) = 0
WE(JXY)
.
i]l[¾o]j)
0
Since
We
which
J[[go]i] = [o]-i-2
in different eigenspaces
.
Now take
x,y E
[Wi]e
Then
.
WE(JXy) = B([E,Jx],y) = -B(Jx,[E,y]) =
-1/2
-1/2
B(O[E,x],[E,y]) = -(Ji
.~
-(Je,i
).B([E,x],6[E,y]) =
-WE(X Jy)
3.)
-WE(X,
Check
wE(Jx,x) > 0
0
x
-1/2
which is alway s positive and
non-zeroi 0([E,x],[E,x])
non-zero on
4.)
to
fixed by
0 0
go/g
Finally we must see that
LnK C GE
Certainly
se e why
8
conclude f or
WE(JX,x)
=
Jx) = -B([E,x],(jei )*-[E,x])
-1/2
want
if
and
8
adE
J
coimmutes with
commutes with
commutes with
LfK .
0oAd(g) = Ad(Og)oO
for
LnK , so we
Since
g E G
k E LfK , BoAd(k) = Ad(k)oe
K
is
we
as desired.
n
Now we decompose
g
=
(
[go]i
into a direct sum
i=n
of eigenspaces under
representation.
the H action of
Because
the 9s(2)
L = ZG(s~( 2 )) , L
preserves
o
-26[go]
the eigenspaces
i
for all
i
Thus for
e E L
we
have:
detR(Ade)
=
II de
t AdR[go]i]
i=l
For
g E GE
we let
Adg
Adg
on the
uotient
space
Ad(LAK)
as
the complex
onto
pace
E
g /g0
Ad(LOK) .
Proposition.
Proof:
.
Let
Then
V =
[go
, we have
go
01o-1
We will
h
detC(hlV)
transformations of
J
J
maps
i , hence
).
rrr
_detc[A-"•l~,]
)*detd
l[•oCg]i
m Eol)
i0-1
= ±1
o]_-i-2 =
i[
consider
The complex structure
(+1
go
preserves
= detR h
0o#
if
h E
J([oi
[go ] i
igo]
o
h E LAK ,
V = [g ].i[goi
]
detC(hlV)
([goli)e • Since
J
.
This is possible because
for all
[go]-i-2
det C(Ad- )=det C(Ad--
i X -1
E
go/
group of complex linear
commutes wi
[go ]
denote the action induced by
)
(LnK)o )
=
and commutes with
E R .
But
LAK
is compact, therefore its image under the determinant map
must be
±l
(and +1 on the identity component).
0
-27By the discussion leading up to the Proposition and
the Proposition we can conclude:
detC(Adh)
= ± detC[Adh
•
[o]_]
Our goal is to calculate this determinant.
We begin this
task by finding a parametrization for the nilpotent in
g0
(under certain groups).
this problem.
The next chapter addresses
-28Chapter 3
Classification of Nilpotents
Let
G
algebra
go
be a real classical Lie group with real Lie
We describe a correspondence between Lie
algebra maps from
st(2)
into
go
and nilpotent orbits
for which we need the following
Definition.
A Lie algebra map
equivalent to another such map
s~(2,R) --
:
*'
:
go
sl(2,R) --
is
go
if
-1
there exists
g E G
such that
O(X) = go'(X)g-
Vx E
s9(2,[R)
Proposition 1 (Kostant).
There is a bijection between
equivalence classes of maps from
nilpotent orbits under
Proof:
into
go
and
G .
By the Jacobson-Morozov Theorem we can embed any
nilpotent
1(2,IR)
sT(2)
.
E E go
in a subalgebra of
go
isomorphic to
Thus there is at least one equivalence class
of maps corresponding to each orbit.
To see there is
only one, we use the fact that any two copies of
91(2,R) C go
with the same nilpositive element are
conjugate by an element of the subgroup
U C GE C G
-29(Kostant, 1959), where
subgroup of
= [E,g ]
U
is the closed, connected
corresponding to the Lie subalgebra
GE
E
this establishes the bijection.
0to
u0
0
Thus, we conclude that the classification of
nilpotent orbits under (for the moment) an arbitrary rea 1
group
G
is equivalent to the classification of
equivalence classes of maps from
Now we assume
sT(2,R)
into
is a real linear Lie group which
G
preserves a non-degenerate sesquilinear form
vector space
G = G(wo)
in
M n(F)
defined over a field
V0
Then
.
go
is a real
(for some
n)
sT(2,R) - ~
from
,
--- G
orbits under
maps from
G
on a
We denote
wO
lift
.
Because Lie
to Lie group map s
the classification of nilpotent
is equivalent to the classification of
SL(2,R) -*
element of
F.
wO
matrix algebra containe d
preserving
algebra maps from
SL(2,IR)
g
G(wo)
up to conjugacy by an
G(wO)
We introduce a notion of equivalence among pairs
(o,V)
w
where
a vector space
is a non-degenerate sequilinear form on
V/F
.
Definition.
We define
an F-linear
isomorphism
W 1 (VW) = w2(Tv,Tw)
(wlV1)
1
T
:
for all
V1 -
2,V2)
V2
v,w E V 1
if there exists
such that
-30-
This isomorphism is unique up to left multiplication
by an element of
G(w 2 )
element of
.
T :
G(wl)
V 1 -- * V 2
isomorphism
OT
:
T
or right multiplication by an
Let
(wlV
1 )~(w2,V 2
)
and
implement the equivalence.
Then the
induces an isomorphism of groups
for
defined as follows:
G(w 2 ) --- G(wo)
-1
g
G(w 2 )
let
OT(g) = T
ogoT E G(w)
Fix an equivalence class
all maps
SL(2,R) --
G(wi)
, and let
(wo
0 ,V)
(WOVO) .
be a representative of
.
(w0 ,VO)
We consider the set of
such that
(wi,Vi) E (w0 ,Vo)
and define an equivalence relation on this set as
follows.
Definition.
0 :
Let
T
SL(2IR) ---- G(wl)
:
SL(2,R) --- G(w 2 ) .
there is an isomorphism
forms
w1
and
w2
and
We say
T
T
:
-1
V
1
7 ~
S
---- V 2
and
if and only if
preserving
oj(x)oT = T(x)
the
for all
x 6 SL(2)
We observe that the set of equivalence classes of
mappings of
SL(2) --
G(i) ,
i E
I ,
is in bijection
with the set of equivalence classes of mappings from
SL(2) ---
G(w 0 )
for
the fixed group
G(wO)
equivalence is defined by conjugacy within
where
G(O) .
the
-31Now consider the set of all t riples
form
w
and
on
SL(2)
a representation of
{(7r,o,V)
7r
is
leaving invariant the
V
Define an equivalence
(w,V) E (o0 ,VO)}
relation on this set as follows:
Definition.
(lW
isomorphism
T
1, V1))
:
V
V2
-
1
if there is an
2,( 2,V2)
which
action and preserves the forms
intertwines the
and
w1
SL(2)
in the sense
w2
already explained.
Clearly, the following two sets are just
descriptions of the same thing:
{(r,w,V)Ir
is a
representation of
preserving
o
and
(W,V) E (o 0 ,VO)}
(wo,Vo)
SL(2)
V
SL( 2)
{f :
---
equivalence classes of
and
G(wi)I(J i,Vi
As we've noted, the
.
, i E I}
on
set of
the same as
the latter set is
set of equivalence classes (under conjugation in
({
:
SL(2) --
G(wO)}
.
of nilpotent orbits under
Therefore,
(w,V) E (WO,VO)}
SL(2)
G(O)),
classifi cation
the
reduces to understa:nding
G
the partition of the set of triples
representation of
the
on
V
{(7r,w,V) •
preserving
w
is a
a nd
into equivalence classes.
First we choose a standard model for an irre ducible
es(2)
representation over
R
,
(ri,Vi)
,
where
Vi
the vector space of homogeneous polynomials of degree
is
-32i-i
/R
in two variables
(irreducible
fe(2)-representations
are determined by their dimens ion).
of
SL(2)
(aX+cY)
(1bXV+
(0 -1)
E = (O 0) and
A
A
to the operators
respectively.
form on
Vi
nil
is given by substitution
the L i
al
a
Xe
Also, we fix a n
(
sf(2)
a
3y
-y . and
correspond
a
ax
y --3
i-1
,Y determine)
= 1 .triples
i)
(up to equivalence)
representation of
(7i ,wii1,V
nn
V•
SL(2)
(wi,Vi)
H =
2)-invariant bilinear
It
i-
c d
'oVO)an d we wil 1 determine all
0
Now we use
wi(X
of
n)
y'*
(a b)X i j
gebra the elements
0
(
F =
ax
The action
a b
tcd ) E SL(2) ,
or
a
x
by setting
We fix
form and
X'YJ
on a monomial
of variables; tnat is I,
j
i
where
VV
,
triples
is a
an SL(2)-invariant
E (00'
the semisimplicity of
a representation
(r0 ,V0 )
91(2)
to break up
into its irreducible
components.
Lemma.
V 0 = @ Vi H i
i
IR
and
=
H
over
Hom(
2
as 9l(2)-modules (V i is as above)
, where
)(V',V)
H
i
is a vector
IF
Proof:
Define a map
r :
$ Vi®H
i
--
V
by
i
defn.
T(e
i
(v.Ti))
It is
clear
T(vi)
=
that
i
T
is
where
v.®T
1-1 and since
i
e
V O®H
dim V =
space
-33dim(G Vi H i ) , T
i
SVi OHi
i
is an isomorphism.
We consider
as an 9X(2)-module by making
9s(2)
first factor alone, and then it is clear that
intertwining map.
act on the
T
is an
0
We want to find all s1(2)-invariant forms on
V0
ViOH
i
We fix our canonical sI(2)-invariant form
.
i
nondegenerate sesquilinear form
Definition.
Let
S wieh
h
i
on
H
i
be a nondegenerate
i
*
sesquilinear form on
Vi H i
given by
i
i
1
where
V0
i
v.®T
,w.@S
1
The form
$
1
1
Vi H
iO.hi
on
$ Vi H i
o
as follows:
i
w(T(9 v.iT
T(e W.Si))
)
i
=
(70'9
gives a form
0VO)
and
*i.
i
(
*h
i
T
i
, $ iw0Si )
hi , $ ViH
i)
(r,
i
.
are equivalent
i
triples in the sense defined previously.
A choice of
on
-34nondegenerate sequilinear form
i ,
h
fixes an e1(2)-invariant form on
Proposition shows this gives all
on
e ViOH
i
Vi
H
for each
0 Vi H i .
i
The next
sf(2)-invariant forms
i
Proposition.
on
on
Fix the canonical sr(2)-invariant form
*
Then the SL(2)-invariant forms on
Vi H i
i
all of the form
@ o ®h
i
sequilinear form on
Proof:
hi
(2)Q ViH
is an arbitrary
H
SL(2) invariant forms on
correspondence
Hom
where
0 Vi H i
i
are in 1-1
with intertwining operators,
i
s((2)
i
Hom
9
(2)(
(G Vj@HJ)
)
.
j
ViHiV), $((Vj
We have
) *()))
ii
,J
(
Homs(2)(V i
,(V
)
(V i (Vi)
)
)
Hom(HiH)
ij
0 Hom o(
HomF(H. (Hi) )
)
i
W3
are
-35-
But we have fixed an
Hom
identifies
i
sT(2)
form
(V ,(V2 )) )with
sI(2)
i
o
on
V
i
which
IF , and we conclude
-36Chapter 4
We have seen
that nilpotent
orbits of G --
the group
which preserves a nondegenerate sesquilinear form on a
vector space V -- on its Lie algebra are in one to one
correspondence with equivalence classes of
representations of
sf(2)
on a vector space with an
9s(2)-invariant sesquilinear form.
representative
We fix a
(v, $ (Vi@Hi), $ Wi@hi)
i
(same notation
as in previous section), where we choose
hi
to be
Hermitian or skew-Hermitian so that the form
the vector space
$ Vi H i
$ w ihi
i
on
is Hermitian or symplectic.
i
This gives us the nilpotent orbits in
o(p,q)
or
u(p,q)
if
V
sp(2n,[)
and
is real or complex,
respectively.
Definition.
We say that a basis
{(Ti
is a
l<jidim H
standard basis for
form
Hi
3.)
hi
on
Hi
complex,
for
2.)
h
Hi
if
for
with respect to the sesquilinear
1.)
for
hi
hi
skew-Hermitian and
symplectic and
Hermitian, the matrix
(H)j
k
equals
1.)
P -iI
q
2.)
0O
-I
or
3.)
or
3.)
Hi
real, or
(hi(Ti ,Tk)
h ())
[
-Iq
-37-
In order
find a representative of the orbit
corresponding
(v,@Vi H i
the equivalence class
i@h
i
@ Vi H i
i
we choose a standard basis on
the linear transformation
respect to this basis.
of
0E
Then
.
Let
r(E)
g = (g)ij E G
Adg((E)ij)
and
and we realize
{(e
=
n
1
j=1
(E)i j
with
is a representative
g
-1
=(g
is realized by writing
respect to the basis
i
i
as a matrix
(E)i j E go
0E
-G
)ij
'r(E)
EG
with
g..e.}
31 3 l in
We now show how to construct a Cartan involution
0 , and a O-stable copy of
sT(2,R) C go
whose
nilpositive element is a representative
of
We define an inner product
0
b 1lb2
Q = @
1
i
E
@ Vi Hi
on
2
i
which satisfies the following two properties.
1.)
respect
There is a standard basis of
to
the G-invariant form
orthonormal with respect to
@
@ Vi®Hi
i
o ®hi
1
with
which is
Q = @ b l®b
i
2.)
Q(v(E)v,w) = Q(v,v(F)w)
Assume these
map
L
.
v,w E V
two conditions are satisfied.
be the matrix corresponding to
(E)ij E go
for all
v(E)
Let
in this basis;
The adjoint with respect to
Q
is given by conjugate transpose when
written in an orthonormal basis;
hence
(E)i j
of a linear
L
is
(E)ij
ij == (E)
Ei j
-38But condition 2.)
=
r(F)
.
e((E)ij) = -(F)i
{(E)ij,(F)ij,(H)ij} C go
Q .
We now define
Vi®Hi
T
b1
( i-l-sy
)X
s
i
and an inner product
b2
Let
Then the basis
i @hi
( i-
in
spanned
Hi
on
V
(i-1t
on
We define an inner product
Claim I.)
91(2)
SO<s<i-ll<k<dim H
H
, b
I
i
by setting
) xi-l-ty
t )
=
st
st)
i
H
b (T 1 T 1 ) =
2( m
n
v@T,w@S E Vi@Hi
or,
is e-stable.
i-1) 1/2 Xi-l-sys
bl( (i-1s
and the
j
r(E)
Choose the following basis for
Define an inner product
for
(E)j i = (F)i j
Hence we conclude that
other words,
by
gives us the following equation,
b Ob
by:
6
mn
on
b (vOT,w@S)
ViHi
as
follows:
b (v,w)b (T,S)
be a symplectic form on
Vi@Hi
1/2 xi--sOT
O<s<i-1 l<k<dim H
up to the sign and order of the elements is a Darboux
and orthonormal with
basis with respect to
w @hi
respect to
bl@b 2
on
II.) Let
hi
Vi@H i
be an Hermitian form on
ViHi
w1i h1
be an Hermitian form on
V OH1
II.)
Let
-39a)
If
i
is odd then (up to order)
1i
)
/2)(X
s
O<s<i-l,
1(-syS
l<k<dim Hi
i
with regard to
respect to
b)
If
i
i
Tk
w Oh
@
i-l-s
i
Tk)}
X sY
is a standard basis
i
and orthonormal with
b i@b
1 2
Hi
is even and
real,
then (up to
order)
i-i
s/2
O<s<i-1,
i-1-s sT i
k
l<k<m
where
1i
Xi-1-s@ T i
k+m
dim Hi = 2m
a standard basis with respect to
c)
If
i
is even and
Hi
i Ohi
b
and orthonormal with respect to
is
1
1
b2
2
complex then
(up to order)
{(i s)/2 (X i -
i
k-
i
-XSyi-l-ss
1
Y @ i *Tk ± Xs
i
1
is a standard basis with respect to
and orthonormal with respect to
Proof:
Obvious.
b 1 Ob 2
0
Hence we have shown that
(1),
O@hi
Q
satisfies condition
and we now show that it satisfies property (2).
suffices to check this on a basis for
Vi@H
i
.
For
It
-40k X e
and
m A n
both
Q(E.(k+)
i- 1 1/2(xi-l(k+l)yk+l
i-
1
1/2
Q((k+
i-l 1/2(Xi-l-eiyTi ))
i
m
n
(xi-l-(k+l)yk+l Ti)F.i-i
(X
Tm
e
equal zero.
Thus, it suffices
Q((i-
(X
1/2(Xi-k+
k+
+1'lT),F.(
(i-i
IkT
, k)
i
i-
F
lOwT)
i-1-eR
(X
@
and
Ti ) )
n))
Y
to check:
Q(E.(i-1)1/2(Xi-(k+1)k+T
Q(E(k+1 )
1/2 (
i-l-kyk Ti
1/2
{X
1/2
))=
i-1-kykTi
Y
(X))
i-1 1/2
k+l
(k+l)
The left hand side equals
1-1
hand side is
i-
and the right
i-i1) 1/2
1/2
1/2
It is easy to check
k+1
1/2
1/2
(k+1) /2(i--k)1 / 2
these are both equal to
We now describe an explicit realization of
matrices.
Let
V = $ Vi H i
L
as
be the decomposition of
V
i
under a e-stable copy of
nilpositive element.
linear
isomorphism
sT(2)
Any
g
of
V .
Hence,
L
.
sr(2)
g
as its
gives a
Furthermore, since
commutes with
intertwines the action of
lemma this implies that
E
g E L = ZG(zl(2))
V --* V
linear transformation
which has
9l(2)
V .
on
the
, g
By Schur's
preserves isotypic components
can be characterized abstractly as
the group of intertwining operators
T
:
V -- V
where
-41T
V .
preserves the bilinear form on
We realize these
intertwining operators as matrices with respect to the
basis chosen on
V
9s(2)
This gives
in
go
to determine the 8-stable copy of
L
as a 9-stable matrix
group.
Proposition.
L a Sp(2ml)xO(p 2,q
a.)
2
)xSp(2m 3 )x***
if
G
preserves a symplectic form on a real vector space
where
2m
i
= dim Hi
and
p.+q. = dim H
L a O(plql)xSp(2m 2 )xO(p ,q
3
b.)
3
j
)x.*.
if
G
preserves a Hermitian form on a real vector space
2m. = dim H i
where
L r
c)
and
V
V
p.+q. = dim Hi.
U(pl.,q)xU(p 2,q 2 )x***
if
G
Hermitian form on a complex vector space
preserves a
V , where
= dim H
p.+q
Proof:
We know
form
ciOhih
0S Vi H i
V
on
* Vi®Hi
.
with G-invariant bilinear
As we have observed
g E L
preserves
isotypic components, therefore we can write
g =
where
g
g
:
ViH
i
--
.
VH
Recall that
Hi
i
can be thought of as
linear transformation
of each map.
But
(V i V) •
Hom
i
gi
g E L
of
H
i
g
defines a
by acting on the range
preserves the sesquilinear form
-42on
V
and therefore
form on
H
i
.
E Sp(2ml)x**..
g = 2 (Id)@g
i
g
On the other hand, an element
specifies a linear
0 Vi H i
i
of
= (g
g
v.OT i E V iHi
i
.
where
N
let
with
i(vi®g
(Ti))
i
1
(g ,***,g
transformation
[as follows:
=
,gN)(@ v.®Ti)
(g ,'
must preserve the sesquilinear
)k
dim H
i
g 1 (T)
and
gkj T k
k=l
d
intertwine
g
Clearly,
preserves the form on
V .
invariant form on
g E ZG(sT(
2
))
Hi
,
therefore
g
preserves the
These two observations show that
O
We ve seen that the question of admissibility comes
down to the subgroup
LflK .
We now prove
the following
Lemma.
a.)
LAK E (Sp(2ml)NO(2ml))x(O(P2 ,q 2 )nO(p 2 +q 2 )x * * *
(V symplectic)
b.)
LNK E (O(Plql)NO(Pl+ql))x(
)(Sp(2m2)O(2m2))x
(V real hermitian)
c.)
LflK
(U(Pl'ql)AU(pl+q1 )) x(U(p 2 ,
(V complex Hermitian)
2
)U(p
2 +q 2
**
)) x *..
***
)
-43Proof:
GnK
is the linear group which preserves the
sesquilinear form
i
w =
i
w$ @h
on
V
as well as the
i
inner product
Q =
b
b
.
The proofs of all three
statements are the same, therefore we only do a.).
must show that
(gl'--.,gN) E (Sp(2ml)xO(2ml))x***
determines an element of
product
Q .
So we
L
which preserves the inner
But this is clear because
{T }
ljidim Hi
ii
Hi
for
is an orthonormal basis of
b.
2
On the other hand an element
element
(gl,'
e E LnK
',gN)
E Sp(2ml)xO(plql)x...
preserve the form
i
b2
on
H
i
for each
i .
determines an
which must
Hence
(gl,°,gN)E(Sp(2m)nO(2ml))x(O(P 2 q 2 )nO(p2 +q 2 ))x
**
.
-44Chapter 5
We now consider the case where
or Hermitian (with signature (p,q))
will show
is a matrix
which preserves the standard symplectic
C Mn ()
algebra
go
form on
Fn
We
.
is isomorphic in these two cases to the
go
space of symmetric or antisymmetric homogeneous
polynomials in
of degree two.
V
We begin with the
following:
Definition.
Let
be a vector space over
V
define
V
to be equal to
we let
C
act on
c E C
,
set
HomF(V,V)
, where
c-v
for
is
v C V
V .
V@V *
V
IF
V@V
IF
as G modules and go
is a vector space over
F
V
nondegenerate
G-invariant sesquilinear form
from
Proof:
V
and
the
modules where
inherits the
Then we
F = C
as sets, but when
as follows:
c.v = c-v
multiplication in
Lemma.
V
V
F .
with a
w , and
G-module structure and invariant form
V
w
V .
First, observe that the G-invariant form
w
V -
V
gives a G-equivariant. F-linear, mapping
on
-45, v
v,w E V
for
defined as follows;
---
v
where
def.
v (w) =
.(v,w)
b
V@V
V0V ---
Now define a map
v,w E V
.
. O(v@w) = v@w
by setting, for
The adjoint actions on
V.
give rise to the contragredient actions on
V
If we
take the representation on the tensor products arising
V
from these actions on
that
*
is a
(or go) equivariant isomorphism.
G
Finally, define a map
follows for
[T(v@f)](w)
it is easy to check
Vn
and
VOV
T
.
It
Hom(VV)
as
let
v,w E V , f E V
= f(w)-v
---
is easy to check that this map
defines an isomorphism which intertwines the adjoint
action of the group
G
as well as the adjoint action of
End V
the whole algebra
o
.
We now want to consider the symmetric or
antisymmetric elements in
V@V , by which we mean the
real span of elements of the form
,
<v®w-w®v>
because
V
respectively, in
and
V
VOV .
<v~w+w@v>
or
This makes sense
are identical as sets.
We note that
such elements are not closed under scalar multiplication
by elements of
the
following notation.
field
F = C
.
We introduce
the
-46-
Definition.
2
A2 (V)
Let
and
2
S2(V)
be the set of
antisymmetric and symmetric elements in
Proposition 1.
Let
preserving
go C Hom(V,V)
V®V
.
be a Lie algebra
Under the identification
Hom(V,V) 2 V@V
described above we have:
a.)
if
w
is skew-Hermitian, then
go
corresponds
to S2 (V) C VV.
b.)
if
is Hermitian, then
w
go
corresponds to
Ta
to be the unique
A2(V) C V®V .
Proof:
T E VOV
Let
and we define
element of
VO.
Vx,y E V .
We want to find
such that
T =
Write
I
Ow i
w(Tx,y) = -w(x,Tay)
(T E V@VIT = Ta}
and we compute
Ta
i
y,) = w(I W(wix)viy)
i
W(i v.iwi(x)
=
w(wix)w*(viyx)
1
W(=(wisvi(Y),X)
i
= e6.(x. w.
v. (y))
i
where
e -=
Ta = (-e) I
i
(
1
if
-1
if
w.Ov.
1
1
w
a
is Hermitian
is
skew-Hermitian
.
Thus
and the Proposition follows.
D
-47-
We need the following additional notation:
Definition.
Let
S(V,W)
be the symmetric elements in
Let
A(V,W)
be the antisymmetric elements
V®W $ WOV .
in
VOW $ WOV
We note that
isomorphic to
inside
V®W
on
V
submodules.
Suppose (r,V)
V = V 1V
and
are both
is a representation of
the direct sum of
2
sT(2)
Then
S2(V) = S2(V
b)
A2(V) = A 2(V)
Obvious.
Corollary.
A(V,W)
that is of interest to us.
a)
Proof:
and
, but it is the natural way each sits
(V$W) 0 (V$W)
Proposiition 2.
es(2)
S(V,W)
Let
1)
@ S(V 1 ,V 2 ) $ S2(V 2 )
$ A(V 1 ,V2 ) C A2(V 2 )
0
V = @ Vi®H
i
, Vi
the standard
i
irreducible 9s(2)-module over I with canonical
st(2)-invariant form
o
i
, H
i
a vector space with the
standard Hermitian or skew-Hermitian form
with respect to a
hi
defined
-48basis
{T}
Sl(idim
S2
a)
H1
Then
.
dim H'
=2(a
(Vi@Hi)
i
=
i
(v'@F*T 1 ))
qj
j=l
@ S 2 (Vi@F*Tk
i,k
, VJ OF*Te) $
S(V'iF-T k
i>j
k,e
i
i
* (S(V @F.Tk
i1
S2
A 2(9ViHi) = *** replace
b)
V
i
i
FT))
by
A2
and
A
by
S
,
Now we use the Clebsch-Gordon formula to decompose
Vi
S2
, S 2 (V i ) , and
2
A (V
i)
.
Proposition 3.
where
a)
Vi@V j = M(i+j-1)@M(i+j-3)@**-OM(i-j+1)
b)
S2(V
c)
A2(V ) = N(2i-3)$N(2i-7)$--**N(p)
where
q = 1
q = 3
i
) = P(21-1)$P(2i-5).***@P(q)
and
and
p = 3
p = 1
if
if
i
i
odd
even
and
M(r)
is an
irreducible sf(2)-module C Vi Vj with dimension r , P(r)
an irreducible 9X(2)-module C S2(V
and
N(r)
i)
of dimension r
an irreducible ol(2)-module C A2(Vi) of
dimension r.
-49-
We
let
{T1}
be the basis for
Hi
described in
Chapter 4 and we introduce the following additional
notation:
2
A (V
OWT )
2
ii
ii
ii
P= (2i-1)Pkk(2i-5)@,**@Pkk(q )
S 2 (V i®RT)
k
A(ViI-Tk
kk
kk
, VJ OR-T
(for
kk
)=Nkl(i+j-l)eNkf(i+j-3)E--..Nk(i-j+l)
,
k.T
V OR*T)
S(Vi
ii
ii
ii
= Nkk(2i-3)@Nkk(2i-7)$...@Nkk(P
=Pkl(i+j-l)@Pkf(i+j-3)@**-@Pk(i-j+l)
ifj or kXe)
k
(Vi O.Tk)@(VJOR-T)
R
First, assume
ke (1+j-1)@*..@Mk(i-j+1)
kes··M,
= Mk
i
V = OVi H
is a vector space over
Then, using the Corollary to Proposition 2 above, we
IR .
get:
S2(v) =
S (V)
=
ii
ii
D [Pkk(2i-1)@-***Pkk(q
i,k
a
[Pk
(i+j-1)@**-@P
)]
(i-j+l)] a
i>j
k,.e
(
k>e
and
[P"k(2i-1)@D* .@Pk
(
(1)]
-50A 2 (V) =
[Nkk(2i-3)@*..*DN(p)] $
i,k
( [N
i>j
(i+j-1)..***N
(i-j+l)] a
k,e
C [Nke(2i-1)C*** Nk (1)]
1
with
q
and
p
N
V = $ V @H
i=l
Now let
i.e.
Hi
u(p,q)
of
V
as in Proposition 3.
is complex.
be a complex vector space,
as an SL(2)-module.
We define a real form
as the real span of a subset of
Definition.
V=
VR
A2 (V)
We want to decompose
<Xi-1-s
s
V
V :
Ti
k 1<i<N
i
l<k<dim H
Proposition 4.
Proof:
and
vOw -
A2(V) = A2(VI)
v®w-w@v E A 2 (V)
Let
w = w 1 +iw 2
wOv =
with
(v 1 +iv
= V W
+
1
2
i[v2 w
.
We can write
v 1 ,V2 ,W 1
)1(wl+iw
-
+ iA 2(V )
2)
-
-
V 1 OW
2
-
E A2(V ) + i(S2(VR))
.
Then
(wl+iw 2 )1(vl+iv
Wl@V1 + v 2Ow 2
1
E V
2
v = v 1 +iv
-
w 2 ®v
.
w 2v
1
o
2
2
+ W1 V
2
]
)
2
-51Thus for
complex, we have
V
A 2 (V) =
[Nkk(2i-3)@...Nkk(p)]
i,k
i>j
k,e
[Nk(2i-1)@***@Nk (1)]
0
a
1
k>i
e
i'[Pkk(2i-1)@** @PkkP
(q)]
i,k
0
i*[P
(
(i +j-l)@**.- @P
(i-j+l)
i>j
k,e
0
i-[Pke(2i-1)@..-@Pkf(1)]
k>R
with
p
and
q
as in Proposition 3.
Proposition 5.
sI(2)
-
(r,V,w)
V = @W i
and
Proof:
where
w(Wi,Wj) = 0
Let
[V]k
be
w .
w(I[V]k,[V]_)
v
[W
= 0
j_1
W
is isomorphic to
for
i
w
copies of
.
the eigenspace with eigenvalue
unless
and
w E [W]
i > j
Assume
k = e .
c-w(F.E.v,w) = -c.-(E.v,F.w) = 0
because
m.
By the st(2)-invariance,
under the H-action.
let
be a representation of
be with an st(2)-invariant sesquilinear form
Then
V
Let
1
.
Then
for some
is a lowest weight vector in
w(v,w) =
c E IR
W
and
k
-52-
Proposition 6.
B :
The Killing form
S2(V) x S2(V) -- R
or
B
considered as a map
A2 (V) x A2(V) -
B :
IR
is given in terms of the canonical G-invariant
sesquilinear form
w
on
V
as follows:
for
v@w,v'@w'
E VOV
B(v@w,v'@w')
Proof:
= co(v,v')*-(w,w')
(c E IR)
Up to a mul tiple, there is a unique nondegenerate
G-invariant sesquil inear form on a simple Lie algebra.
We recall that we want to fineI the determinant of
the linear map induced by the adjojint representation of
(LnK)O
on the quotient space
results, for
E
go / E
By our earlier
e E (LnK)o , it suffiices to restrict our
attention to the (-1)-eigenspace ol
det,(Adjg
/go
a
) = detC(Ade I[o_ 1)
We now describe an explicit basis for
identification of
above.
go
with
S 2(V)
or
[go]A2(V)
using the
explained
a
-53The only irreducible submodules in
which contribute to the (-1)
and
S(Vi@Hi
i 4 j
1.)
If
V
is real,
= (Vi@Hi)@(VJ@HJ)
(t)
those
, VJOHj) or A(Vi®Hi,VJ HJ)
k,P
l<s<j
MiJ
M1
namely
with i > j
(mod 2)
Definitions.
Mij
A2(V)
eigenspace (under the
H-act ion) are those of even dimension;
compo sing
or
S2(V)
=
$
then we set:
ij
Mk
( i +j+l-2s)
Mkf(t)
k,e
Mij =
kR
Mke(i+j+1-2s)
1l<s<j
2.)
If
V
is complex, then we set:
Mij = (Vi@Hi)@(Vjo HJ)
[M
-=
(i+j+l-2s)@i-*M
(i+j+1-2s)]
k,e
Mi(t) = k(t)][Mke(t
k,e
MiJ
ke
l<slj
)
@ i-Mk
ij( i + lj +
[M ( i+j+l-2s)Di.M'kJ
i~j~l-s)$
kR
For the remaining calculations in
will make use of
:
VOW --
S(V, W)
A(V,
]
this Chapter we
the isomorphisms
{a : VOW --
-2 s )
W)
-54defined by:
a(v@w) = v~w + w®v
P(vOw) = vOw - w®v
We get:
j OH j )
(V'@H')@ (V
VJ@H
i
-(V-i@H
[A(Vi®H
i
Vj@H
j
)
j
)
and
(r)
P
M
M(r
Nke(r
We will often suppr ess the distinction be tween
)
(Vi@Hi)@(VOHj
j
S(Vi@Hi, VH
A(Vi®H i , VJOHj )
anLd
or
)
Claim 1.
WE(M i
a.)
,Mi
) = 0 unless i=i' and j=j'
b.) WE(Mij(t),Mij(t')) = 0 unless
Proof:
We have
V = C Vi®H
i
t=t'
with invariant sesquilinear
i
form
w
unless
.
Proposition 5 implies
i = j
B([E,MiJ],M
WE(Mij,M
i 'j
i'j
.
We observe
) .
) = 0
Since
that
that
[E,M
ij]
if and only if
w(Vi®Hi,VJ®HJ)
WE(M i,M
C M ij
i 'j
)
we conclude
B(MiJ,Mij)
= 0
But,
B(MiJ,M
j) = B((ViOHi) (V®HJ),(Vi
=cc(V'iH
i ,
V'i
Hi
®HJ))
Hi')®(V
).w(V O®H.V
Hj
)
0
-55and this last expression is
j = j'
Fix
i
V
Vi@H
V =
unless
i = i'
and
O
by Proposition 6.
Claim 2.
0
and
i > j,
and let
i
lk,k'<m i
If
,
and le,e'<m.
C S 2 (V)
Mk
,
then
J
for i even, j odd, m. = dim Hi and 2m. = dim H
a)
1
kM ej
B(Mk'~
X) 0 <=> k'=k
e'=e
and
for i odd, j even, 2m. = dim H i and m. = dim H J
b)
1
B(Me
If
j
,PM
M
e)
C A 2 (V) and
M
V
J
0 <=> k'=k
real,
e'=R
and
then
for i odd, j even, m. = dim Hi and 2m. = dim H J
c)
i
1
k''
M
C A 2 (V)
ke
dim H
e'=e
and
for i even, j odd, 2m. = dim Hi and m. = dim H J
d)
If
0 <=> k'=k
Mk,e+m.)
B(Mk'
J
= m.
e)
Proof:
B(M
k+m.,e
1
V complex, dim H
,
then
k
e,M k)
X 0 <=> k'
0
looks
on
i
m.
1
= k
and
and
e'
= e
O
Use Proposition 6.
In order to compute
know how
j
J
S2
S (V)
on
and
and
[•o]_1
A2 (V)
(V)
we will need to
-56Proposition 7.
X E go
S :
The Cartan involution
2
is given on
S (V)
A2
and
X --4 -Vt
for
by:
A (V)
for
(Xsyi-1-sOTk)O(X
i (Xj-1-ty t
t T ) E Mij
0((Xsy
i -l - s
tT
Ti )(Xj-1-t
=
V
j-1-t
(-1) s+t+l
hi
s 0k
i -l - s
V.
.
where
) @ (X
V
hJ(TJ,T
m
P
) = 5e
Pm
hi (T1 ,T ) =km
are the appropriate forms on
H
and
and
H
and
respectively.
Proof:
The Cartan involution
conjugate
transpose on
.
su(p,q)
We use
the
A 2 (V) .
and
X E S2(V)
basis on
(or
V
hais
bais{(
go = sp(2n,R)
V = $ViH H i
(under Q)
X.
and set:
or
Q = @ bi b
2
1
i
to compute
then
X
=
0
on
transformation
t , hence
OX = -X*
We take the usual orthonormal
k'
,, ,,.
O<s<i-1
Q
so(p,q)
with respect to an orthonormal
(i-1 ) 1/2.i-l-s
Y~ sT
OT i }
X
s
L
,
If we write a linear
A2(V))
So we now compute
is given by negative
inner product
(defined earlier) on
S2(V)
0
er
WII i
,
mn,
ULIU~I
-57s
and L =
v4 v3
i
is,
.
j
i X 1,2
and
= <v 3 v
Lv. = 0
.
,
Q(v
if
i
,L v)
j A 3,4
= 0
= 0
,L v 2 )
and
Q(Lv 4 ,v
1 )
if
Therefore,
.
for
i x 3,4
.
if and only if
Q(Lv 3,v 2 ) =
,L v 1 ) .
We have
4
and
Q(v 4 ,L*vl) =
Q(Lv 4 ,vl) = Q(v
Also,
j X 2
= Q(v
4
,L v
)
.
Finally,
Q(Lv 3 ,v 2 ) = Q(v 3 ,L*v 2 )
two cases:
go
preserves a symplectic form:
Q(Lv3', 2 ) = (-l)S(-l)i+lhi(Tk' T
3
= 0
if and only if
we break the verification that
Q(v
1
Thus it remains to show that
(-)tQ(v4v4) , hence
I.)
for all
i X 3,4 , and since the image of
Q(Lv 4 ,v 1 ) = (-l)tQ(vl,V1)
into
L v.
Q(Lv 4 vk) = Q(v 4 ,L Vk) = 0
Likewise,
3
for
j) = Q(v 3 ,L v)
Q(Lv 3
Q(v
Q(Lvi,vj) = Q(vi.,Lvj)
To begin, observe that
Q(Lvi,v.) = 0
k A 1
we want
1-
and
L
= (-1
t
that
,
k
j--tdemonstrate that
We
2
Ti
1/2 xi-1-s
i-1
1
,L v 2 ) = (-)-1j1
)Q(v2 v2 )
and
(TT)Q(v3v 3
)
-58V.
Since
hi(Tk
conclude
Q(Lv 3 ,v
Q(v
3
2
) = -(-1)s(-
,L v 2 )
k
1
,Tk)Q(v2 ,v 2 )
) i+lhi(T
. V .
1.
,Tk) = (-1)
3
and
,L v 2 )
We know that
0
h(T
k
k
Wk
A (V)
tells us,
Mij(k) C W k
.
J = cOoadE :
J :
Our
.
L
thus
o
and we're done.
preserves the isotypic parts of
g
9s(2) C qo ; i.e.
0
formula for
in addition, that
[-1
[go] -1
--
[Mij(k)]_l
on
0
S
2
(V)
and
preserves
, and thus we get
(of course
k
must be
[MiJ(k)]_l = {0}).
Our description of
2
A2 (V)
= (1)+land
,T)
We have already observed that
[Mij(k)]_
even or
Now
.
under the action of a G-stable
:
and
V
.
= -(-l)t+s(-l)t(-l)J+lh(TJ,T 3 ) .
Q(Lv3',v2 ) = Q(
8
, we
preserves an Hermitian form:
.
hi(T
h (Tk'Tk) = (-)
and
Q(Lv 3 ,v 2 ) = Q(v3,L v2 )
go
II.)
= (-1)
,T)
shows that
commutes with
L
the action of
91(2) , L
the (-l)-eigenspace
i
Mi.
preserves
preserves
[M i(k)]_l
.
L
on
and
Moreover, since
M i(k)
and also
Putting all of this
together we get the following conclusion:
e E (LnK)O
S2(V)
for
-59-
detC(Ade)
detC(Ad•
[MiJ]
i>j
i j(2)
detC (Ad- I
i>j
[Mij(t)] _-1
itj(2)
t
We want an explicit b asis for
[Mij(t)]_l
We
begin by finding a basis f or the irreducible submodule
MkU(i+j+l-2s) C M i j
We define a basis on
Mk(i+j+l-2s)
isomorphic to the standard basis for an
which is
st(2)
representation on homogene ous polynomials of degree
i+j-2s (= Si+j-2s(X,Y))
operator,
:
i+j-2s(X, Y) --
Xk (i+j-2s) = O(Xi+
basis on
Let
.
j- 2 s )
*
be an intertwining
MiJ(i+j+l-2s)
.
We
and this fixes a canonical
Mk(i+j+l-2s)
kP6
(i+j-2s-2m) = ,(Xi+j-
X
Definition.
2
s-mm)
0 < m < i+j-2s .
Lemma 1.
The highest weight vector
Mk(i+j+1-2s)
S
p=l1
()p-l
s-1
Xk(i+j-2s)
is a multiple of
i-p p-I1
p- l)(X 'P
k
)O(Xj-l-s+PyS -POT -J9
in
set
-60Xk
We know
Proof:
must be a linear
(i+j-2s)
combination of eigenvectors under the H-action with
eigenvalue
i+j-2s
; that is,
a sum of
the following
vectors in M (i+j+1-2s)
{(Xi-pYp-@T' )
(X j -
1- s + p y
s-p
T
)l<p<s
s
Xl
k (i+j-2s) =
Set
where
2
p=
c (Xi-p-1Ti)(Xj-1s+PYs-P®T)
P
c
are some constants.
Use the fact that
p
(i+j-2s)) = 0 to solve for
c
in terms of
P
adE(X
cl
0 = adE(XUk(i+j-2s))
s-1
Ss- [(s-p)c-
i
j-l-s+p+l s-p@Te)@(x
Y
lTe)
[(s-p)c pi(X
p= 1
s-p-1 Oj
Yp-10T 1 )i(Xj-l-s+p+l
+ pcp+(X
Therefore
(s-p)*c
implies that
c
p
+ p-Cp+
(-1)P-1
p=l
l)(XjPOT
p=l
s- 1
= 0
= (- 1 )-1(s-)
p-1
We normalize so that
s
1
i-pp-1 ~
)(X'-T
X
, which
1 < p < s-1
as desired.
a
(i+j-2s) =
(j--s+p
s-
j
-61X
Lemma 2.
(-i-j+2s)
canonical basis on
s
(-i)
( P-1
s. I
p=l
Proof:
Xiyj
to
0 1 .(Xi-p
0
Mk
the lowest weight vector in our
(i+j-1-2s)
p-
p)[(X
S-1
p-lk
1Y
i-
).(X
Y
p
(-1)i.jYi.
-1T
T
equals
i) @
T J)
-e
in
.
SL(2)
It
So
(X-l-s+
)yS
T
(X s - p Y j - l - s + p
(10 1 )
Consider the element
sends
-I
,
TJ)
goes to
goes to
(-1)i+j-1-s*(Xp-lyi-POT )@(XS-p j-1-s+P@T ) , and
(-1)i+j-1-s = (-1)
s
because
i
t j
We will also need a basis for
submodule
Set
X
.
(mod 2)
0
the irreducible
iM k(i+j+l-2s) C i-S(Vi®IRTk , V)*IR.T )
(i+j-2s)
=
S(-1)P-lP-)[(Xi-p pp=l
i.T k)@(Xj- -s+ps-p@T )]
The same reasoning used in Lemmas 1 and 2 shows that
XJ(i+j-2s)
is a highest weight vector in
kS
i'M k (i+j+l-2s)
s
(-1)s2
(-1)Pl
and that
- s
X(-i-j+2s) =
l
-1)[(XPlyi-P@iT
)(i XS-pyj-l-s+p Oj)]
p=l
is a lowest weight, where we define a canonical basis
s
{Xke i+j-2s-2m)}0m(i+j2
Lemma 3.
Fix
V = $ VP®H
as in the previous case.
,
i > j
,
$i@h
p
form on
V , and let
lk(<m
1
and
l<e<m.
j
i
canonical
-62-
If Mij C S2 (V)
a)
, then:
for i even, j odd, m. = dim Hi
,I
1
O(Xk(i+j-2s)) = -h (T
9mi
= dlim T 3
=
•
il
|1
e+m (-1-j+2s))
J
= dim Hi
, m. = dim H J
kTk)(X k
for
O(X
If
i odd, j even, 2m.
k
M i j C A 2 (V)
c)
is real,
then:
c~
,
((i+j-2s)) = -hi (TkTk)(X
,
k
k
k
((i+j-2s))
= dim H
J
is
~
complex,
zm
= aim n-
+m.(-i-j+2s))
k kR~m
hJ(TJ,T J)(Xlm
v
M i j C A 2 (V) , V
(e)
V
for i even, j odd, 2m.1 = dim Hi
O(X
If
and
SV
for i odd, j even, m. = dim Hi
1
O(X
d)
hJ (TJ,TJ)(Xlj
(i+j-2s))
Lv
i
, mj = dim Hi
•r 11
= dim Hi
and
then:
e(X i(i+j-2s)) =
-i.hJ
O(X
(T ,T )h'(Tk,Tk) (X
k'k)(ke(-i-j+2s))
ee
(i+j-2s))
=
i-hj (T ,T )h(T',Tk)(Xke(-i-j+2s))
Proof:
By Lemma 2 we have
Xk(-i-j+2s) =
(-)s I
(-1)p-1( s -
) [ (X p -
Y i - p OT
p=l
Compare this expression with:
)
(X s - p
J-
1- s + p
T
)
-63O(X k
(-1)
(i+j-2s)) =
j+s+1
) [ (XS-PYj-+P@TJ)®(Xp - 1y i- p®
-
(-1)
i)
p=1
Lemma 4.
a)
b)
c)
(Same setup as Lemma 3)
J(Xk (-1))
J(XkU(-1))
J(Xi(-1))
(-1))
i+j+1-2s
2
=(-1)
i+j-1-2s
2
=(-1)
hi(Tk,Tk)(Xk'
h (T ,T )(Xjm
i+j+1-2s
2
T k'Tk)
iT
ij +m
(Xk
=(-1)
i+j-1-2s
2
d)
J(Xk
e)
J(X k(-)) =
=(-1)
+m.J (-1))
th
I-
(-1))
to,10 k '
(TJ,TJ)(Xlj
i+j+1-2s
(-1)
2
2
i-h (TR,TR)hi (T ki ,T ki ) (X i
(-1))
J(X ke(-1)) =
i+j-1-2s
2
S
(-1)
i-hJ(TJ ,T)h i
kT
.Ljj
i
ijk
,Tk',T)(Xke(-1))
-64Proof:
ij
(for Case a) only)
J(Xk(-l))
= O(X
ij
-1/2
-1/2.eoadE(X k
= (-1)
i
(1))
(-l))
,
because
[E,Xk(-l)] = jl/2Xk (1)
i+j-1-2s
= 6(co(adF)
2
where cO
(Xke(i+j-2s))
i+j+l-2s
2
=
(i+j-2s)!
i+j-l-2s
i+j-l-2s
(-1)
2
2
c(adE)
O(Xk(i+j-2s))
Apply Lemma 3-a.):
i+j-l-2s
i+j-l-2s
-h
(Tk,
i+j+l-2s
2
(X
In each of
+.(-i-j+2s))
i
0
k,) Xk, e+m.
(-1)
Lemma 5.
2
co(adE)
2
= (-1)
in Lemma 3,
the cases
is a complex orthogonal basis for
{XkeC-1)}k,e
[Mij(i+j+1- 2 s)]_l
and
all the vectors in this basis have the same length with
respect to the inner product
Proof:
b
on
g /g
0 0
The orthogonality follows from Claim 2.
The
basis elements all have the same length because we
constructed them the same way on each submodule
M
kP(i+j+l-2s)
.
O
-65We recall
the following identification described in
a previous section:
If
then
LK
If
then
(Sp(2ml)nO(2m 1 ))x((O(p 2 )xO(q
L C O(p,q)
LNK
If
then
L C Sp(V)
2
))x...
,
(O(pl)xO(ql))x(Sp((2m 2 )nO(2m 2 ))x..L C U(p,q)
LNK
,
(U(pl)xU(ql))x(U(p 2)xU(q
The product of
2
))x--.
two elements on the left side of the
isomorphisms above is the usual product on a Cartesian
product of groups.
e E LAK
Therefore, if
corresponds
to
(gl,---,gN) E
(Sp(2ml)nO(2m 1 ))x(O(p 2 )x(O(q
2
))x
we conclude
det,(AdP) = 7 detC(Adg )
j
and we can comp'ute the determinant of any element in
if we know
Adg
det C(Adgj)
for all
invariant we must compute
j
Mij
Since
detC(Adg
Mij
M
i Z j mod (2) .
)
LnK
is
for all
-66Proposition 8.
I.
Assume
i > j
Symplectic Group
a)
BA (where
be given by the matrix
B
are
dimensional space
HJ
.
M
1
det(A+iB)even power if m. even
1
g
=
[
BA
E
Sp(2m i ) n
with the same conventions as above, then
detC(Adg
-,
c)
if m. odd
) =
b) For i odd, j even,
O(2m.)
2m.
Then
[det(A+iB)Odd power
i
and
A
m.xm. matrices) with respect to the
standard symplectic basis on the
detC(Adgj
n O(2mj)
gj E Sp(2mj)
For i even, j odd, let
ij)
For i even, j odd, let
= 1
gi C [O(pi )xO(q i)] O
then
det(Ad-gi
d)
ij)
For i odd, j even, let
then
= 1
gj
.
E [O(pj)xO(qi)]
0
'
-67-
detC(Adgi I
II.
=
Group preserving a symmetric
1
form on a real vector
space.
a)
For i even, j odd,
gi
i- A
C Sp(2m i )
dim H J = m.
3
,
Mij)
M la
b)
=
det(A+iB)
odd power
i ).
dim Hi = 2m.
"
For i odd, j even,
de t(Adgi
For i even, j odd,
if m. is odd
even power if m. is even
For i odd, j even, g
detC(Adg•j
d)
O(2m
then
det(A+iB)
det,(Adg.
n
= [_B
Mij)
E Sp(2m )nO(2m )
= 1
gi E [O(Pi)xO(qi)] 0
ij)
g
detC(Ad-gj I ij
= 1
E [O(p )xO(q j)]0
= 1.
'
-68III.
Group preserving an Hermitian form on a complex
vector space.
a)
For i even, j odd
gi = A E U(pi)(or U(q.))
,
dim H i
P+q.
dim H J = m.
I
det
[(detA)Odd power
(detA)even power
i)
(Adg
i
b)
dim H
I mij
)
(or U(qj)),
dim H
= p.
= m.
1
(detA)odd power
odd
S(detA)even
even
For i odd, j even,
detc(Ad-g
d)
is even
For i even, j odd
gj = A E U(pj)
det,(Adg
j
is odd
For i odd, j even,
detC(Ad-gi
power
gj E [U(p )xU(q )10
Mi)
= 1
.
gi E [U(pi)xU(qi)]
Mij)
= 1
0
-69We will only prove the statements for
Proof:
the
Symplectic group because the proofs for the other groups
are very similar.
!g . hence
gj
L
preserves isotypic submodules of
ij
M (i+j+l-2s)
preserves
detC(Adgj [Mi(i+j+1-2s)]_)
detC(Adg
Mij(i+j+l-2s)
)
.
which equals
"
m.
Case (a):
We have
g
(T
) =
2
m.
a
m=1
1 < e < m.
T
j
me m
+ m=I (-bm )Tm+mM.
m=1
i
Therefore
m.
A
We compute
a
2
ke(-1))
m.
Xim(-1) +
-1 (-b)Xi
m=1
me km
m=1
m.
3
ii
m.
3
mD)Xk,m+m.3 (-1)
)
k.J(Xj(-1
2 a Xij(-1) + 7,(-bme)
m8Fk~s km(-'))
m=1
where
m=1
i+j+l-2s
2
ak,s = (-1)
m.
=
m=1
i
i
hi(TkTk )
(-
(ame-ik sbme)Xj
1)
Thus
j
det C Adg j2m
=Mk(i+j+1-2s)]-1
R=1
and
=
I det(A-ia
s=1
B)
)
-70m.
k=1
j
s=1 det(A-iek,
k=l
s=l
1
det
Since
j
Adgjj I
C
C
[Mij]_l
is odd and
ek ,s
B)
5
s
we get:
k,s+l
m.
detC(Ad-gj
Case (b) :
T det(A-iek
k=1
Mij)
Same reasoning as part
"I
B)
(a) leads
to the
equation,
m.
det (Adg.
I
) =
case
J
I
TI det(A-ie OsB)
e=1 s=l
[Mij]_
In this
J
is even
(es
= -e,s+)
,therefore
m
IT
e=1
det(Adgi I
.
Case (c):
Set
U=
Lemma 2-c. ) we get
Adg i (U) C U
and
Sg
= 1
m. m.
0 $
(D [MUk(i+j+1-2s)]_
s=1 k=l e=1
[Mij ]-1
Ado
(JU
TT
''J-J
= U@J(U)
C JU
Since
we conclude
By
1
-71-
detC(Adgi [Mij
In fact
det[(Adgi U)
at the image in
R
) = detR(Adg-i U ) E R
equals
because we are looking
of a connected, compact set
containing the identity and
Case (d):
1
det(Ad(e))
= 1
Use the same reasoning as in part (c).
O
Theorem.
(I)
Let
G
be the symplectic group with same notation
as above, then
if j is even
1
odd power
det((Adg ) =
if j is odd and there
are an odd number of
even dimensional
irreducibles of
dimension > j in
( Vi H i
det(A+iB)even power otherwise
-72(II)
Let
be a group preserving an Hermitian form on
G
a real vector space, then
1
if j is odd
let(A+iB)odd power
det
(Adgj)
=
if j is even and there
are an odd number of
odd dimensional
irreducibles
of dimension
< j
let(A+iB)even power
III.)
Let
G
in
@ ViOH i
i
otherwise
be a group preserving an Hermitian form on
a complex vector space, then
odd power
power
(det A)
if j is odd and there
are an odd number of even
dimensional irreducibles
of dimension >j in
@ Vi H i
i
detC(Adgj)
=
d d power
(det A)o
if j is even and there
are an odd number of odd
dimensional irreducibles
of dimension
< j in @ Vi@H i
i
ev e
(det A)
n
power otherwise
-73-
Corollary.
G
.
Let
E E go
Fix
0
and
g = (gl'--gN) E LnK
N
U
det((Adg) =
the orbit of
E
under
. Then
n.
(detCgi)
and
1
is admissible <=> n.
E
i=l
is even for all
Proof:
i
The proof follows directly from the Theorem above
and the last Theorem in Chapter 1.
Theorem.
of
E
E
Fix a nilpotent
under
under
SU(p,q)
U(p,q)
with H J 0 {0}
D
E E su(p,q)
.
Then the orbit
is admissible <=> 1.)the orbit of
is admissible, or 2.)
If for all odd j
the number of even dimensional irreducibles
of dim ) j in 9 Vi H i is odd and for all even j with H
i
$
{0} the number of odd dimensional irreducibles of
dimension < j in
*
Vi®H i is even.
i
Proof:
of this element under
under
EE su(p,q) .
Fix a nilpotent element
SU(p,q)
U(p,q)
The orbit
is the same as the orbit
, and, as we've seen, this orbit
N
corresponds to a decomposition of
C
'
*
i
ViH
i=1
under the action of
N
H
i=1
(U(Pi)xU(qi))}
9s(2)
.
Let the set
be the subgroup
{(gl-'-.gN)
N
LnK C U(p,q)
-74determined by
E
(see Chapter 4).
LnK C SU(p,q)
determined by
E
Then the subgroup
is
N
{(gl' '
. ' gN)
.
i=l
I
(U(P.)xU(q.))
N
(det(gi))i = 1}
.
i=l
the preceding Corol lary we know that ther e is
By
sequence of integers (n 1 ,'- S,nN) which determine the
determinant character:
n.
detC(Adg) =
U
i=l
(det gi)
where g = (gl, '°°gN
g = (gl,'.-'gN) E LnK C SU( p,q)
For
, 9
(det gi
= 1
i
therefore on this subgroup the c haracter has
the
following form:
n.+ki
detC(Adg) = I (de t gi)
i
where k is an arbitrary integer
Thus
E C su(p,q)
is admissible
if and only if this
character is a square, which is true if and only if
is even for all
j
or even for all
j .
Theorem.
of
E
under
Proof:
Fix
under
O(p,q)
The
lemma because
and
n2j+1
n2j
is either odd for all
j
o
a nilpotent
SO(p,q)
.
E E so(p,q)
.
Then the orbit
is admissible <=> the orbit
of
E
is admissible.
SO(p,q)
case follows directly from the next
SO(p,q) 0 = O(p,q) 0o .
-75Lemma.
The orbit of a nilpotent
for
if and only if it is admissible
for
G
Proof of Lemma:
z i (GE mp
E
is admissible for
-0
GE
Lie(GE)
and only if
G O0
if and only if
(GE
= Lie(G)
0
, it
p= (GE) 0
z f (GE)p
.
mp
)E
E
G
above that
G
is admissible
We have the following coverings of groups:
.
(GE )mp D (
Since
E E go
.
follows from the inclusions
Therefore
z t (GE)mp
00
if
-76Chapter 6
SL(n,R)
In Chapter III we analyzed nilpotent orbits under a
group which preserves a nondegenerate symplectic or
orthogonal bilinear form on a vector space
V .
We now
consider nilpotent orbits under a group which preserves a
nondegenerate, multilinear, alternating top-dimensional
form on a vector space
V0
; that is, we are considered
nilpotent orbits in
sl(n,R)
is the dimension of
V0
under
SL(n,IR)
where
n
'
As before, Kostant's 1959 results give us a
bijection between nilpotent orbits in
se(n,IR)
under
SL(n,[R)
and equivalence classes of mappings from
SL(2,J)
into
SL(n,IR)
(where two maps are equivalent if
they are conjugate by an element of
SL(n,IR)).
We have a notion of equivalence among vector spaces
with non-zero top dimensional multi-linear alternating
forms.
(,wV)
Let the pair
be a vector space with such
a form.
Definition.
((1 .V1 )
isomorphism
T :
W2 (TV1'
n)
"
V1 where
(w2 .V2 )
V2
if there exists a linear
such that
{v 1 '.0*Tv n }
1
(vl.'
is a basis
.. V n) =
for
V1
-77The isomorphism
T
is unique up to right or left
multiplication by an element of
SL(n,R) , and the map
induces an isomorphism
2)
G(wi)
T
:
G(w
G(wl)
is the group preserving the form
(We note that
be a representative of
to
SL(2,IR)
n .
--
where
w. , i = 1,2
1
G(wl) = G(w 2 ) = SL(n,R)).
Fix an equivalence class
equal
---
T
(wO,VO)
(wo,VO)
and let
(o0 ,VO)
with dimension of
V0
We consider the set of all maps
G(wi)
where
(wi,Vi) E (wo,VO)
.
Define an
equivalence relation on this set in the same way we did
for the orthogonal and symplectic groups
III).
(see Chapter
The same reasoning used in the earlier cases shows
that the classification of nilpotent orbits in
under
SL(n,[R)
st(n,R)
reduces to understanding the partition of
the set of triples
{(7,w,V)
7r
is a representation of
SL(2,R)
on
V
which preserves a (dimension of V)-nondegenerate,
multilinear alternating form, and
(G,V) E (wo,VO)}
into equivalence classes (where the notion of equivalence
is analogous to the equivalence in Chapter III).
Let
(r,V)
We recall that
notation).
be a representation of
V
@ Vi®H i
i
et(2,R)
(see Chapter III for
on
V
-78-
Proposition.
a.)
If
Hi
4 {0}
for some odd i, then there is one
(up to equivalence) non-zero
SL(2,R)-invariant
multilinear alternating form on
b.)
Hi = {0}
If
( Vi@Hi
i
for all odd i, then there are two
equivalence classes of s uch forms.
V = $ Vi@Hi
i
Let
Proof:
intertwining operator
T
:
V ---
) C
iso typic
i
T(W
)
C Wi
and
[Wip
HJ s
First, assume
{v =X j-1 OT
Any
preserves
V
Therefore
components and eigenspaces.
T([W i]
i
W i = Vi@H
and
1
.
(0}
w(Tv 1 ' .,TVn)
V .
take
- 1OT v. *
v.=Yj
j
1' j+l
v2 =X j-2 YTTJ 1'
to be a basis for
We
If
is an n-form on
L
= (det T)w(v 1 '9
,v }
V
then
Also by the
nv) .
observations above
det T =
i det(TI
W
i
IT
=
i
Let
WO(v
1
,
' .Vn)
)
_i-1
)
be two n-forms on
O,0 1
= 1
and
arbitrary constant
(v,V,O0)~(-,V,Vl) .
c .
o(v
1
''..,v)
det(Tj
i-i1
i
V
[Wi
defined by putting
= c
,
for an
We want to show that
To do this we define a linear
-79V --4 V
:
T
isomorphism
as follows:
1
T(ve) = elc njIv
for
1 • e • j
T(vk) = Vk
for
j+1
It is clear that
calculate:
~0 (V
1
,
Thus we've shown
.
.,V)
for all odd i.
wO(v 1 .'
0,v
1
and
= c > O
n)
0(v'I
(v,V, 10 )~(vV,Wl) , o
0 (v 1
So we assume
,
n) = 1
e.
c < 0 , and we derive a contradiction.
n=
W l(V1---,Vn
Assume that
) =
An argument
(W,V,•0 )~(V,V,Wl)
similar to the one above shows that
where
.
= (det T)l 1 (V1'900v n)
Hi = {0}
Now assume
< n
is an intertwining operator, and we
T
wl(Tv 1l,**,Tv)
ec-11c = 1 =
< k
a = sign of c
and
T :
V 1 ---- V 2
intertwines the
s9(2)-representations and preserves the forms.
T
is an intertwining operator,
depend on the eigenvalue
det(T
2e)
W
p .
det(T
)
Because
does not
Thus,
= I det(T
2e
)
p
[W
]
)2e
= (det TI
> 0 , for all
[W2e]
Since
(r,V.w 0 )~(r,V,Gl)
we must have
1 = W00(v1.•-v ) = W 1 (Tv1,---Tv)
(det T)l( 1(vl*
.,vn)
=
= (det T)-c = [U (det TI
R
e)]*c
2
W
-80But this is impossible because
and
c
is negative.
T(det TI
e
is positive
2
We conclude there are two
equivalence classes of n-forms on
all odd i.
WC)
V
if
H i = {0}
for
o
N.
Fix
(v,V= @ VigH
,w)
(notation as above) and, as
i=l
we've shown, this triple corresponds to a nilpotent orbit
0E
in
sT(V)
We construct a nondegenerate
bilinear
form
T
= q
i
®i@h
i
on
V
9s(2)-invariant
, where
W
i
is our
i
canonical 9s(2)-invariant form on
inner product on
{T
i}
,T
Hi
Vi
and
defined as follows:
be a basis for
Hi
hi
is an
let
and set
dim H
6
hi(Tj ,Tk ) =
The form
V -- +
.
isomorphism
jk
T
gives us an SL(2)-equivariant mapping
This map gives us an SL(2)-equivariant
V@V --
Now we decompose
SL(2,IR)
have:
.
V-@V
gq(V) • V®V
under the action of
Using the notation introduced in Chapter 4, we
-81-
(V'O'@Vj~
i,j
ij
(V0
®VM~
i,j
=
8
i,j
We want to compute
order
Proposition.
g((V)
We write
denotes
the center of
sl(V)
eigenspace under
[we (v),go(v)]
0 , we conclude
sits inside the
for
p X 0 .
S[g(V),(V),(V)]
qg(V)
for
.
j
Clearly both summands are
as fo Slows:
13]p
=
(p(v)
(V)
[0(v)]
(v)
and
= 0
= [q
for
So we may now restrict our attention to
[(V)]_l
, where
Theref ore we may decompose the p th
adH
19(V)]p
In
given above.
qg(V) = g
ad(gX(V))-invariant.
(V) ]_1 ) .
detC((LnK)O)[1
[g~(V)1p = [sI(V)]p
Proof:
p
the
to do this we mus t see how
decomposition of
Since
M (i+j+l-2s)
e,k s=1
In order to compute
when
p
0.
[qI(V)]1
detC(Adh
l[s(V)]_l
h E LnK , we construct a basis for
([sI(V)]_I)c
0
-82To do
this we want
to know how
for
[
The calculation of
6
our canonical basis
o
-1/2
J = J-I
*oadE
-1
looks on
]-1
made in Chapter 4 goes
through here, and we have:
e((Xsy i-l-sOT i
®Wk)@(XJ
-1-tY t TJ
twJ)
=
(-1) s+t+l(Xtyj-1-t@T )@(Xi-l-sS
sTk
i)
We will need the following:
Lemma.
O(XUk(i+j-2s) = (-1)
Proof:
O(Xlk(i+j-2s)
Xk(-i-j+2s)
ij
S
=(
2
- 1
(-1)p-
p=l
(1) i + s + l
=
(-
)(X
p-l
p=1
(-1)s-l1
(-1)
(-1)
i+s+l
i-Pyp-1T i)@(Xj-I-s+Pys-P®Ti))
s-i
(-1)p-1 (Xp-Ij-p®T )O(XS-Pyi-s+POT i
p=l
(-l)i X
1(-i-j+2s)
Wk
Proposition.
T )(XP-lyi-PT i
p-1 )(XS-Py -1-s+P
J(Xlj(-1)>
13
e
3i+j-1-2s
2
o (-1)
k
Sji
k( -1)
-83Proof:
= (
J(Xk(-l))
=
-1
/2 .oadE(Xk
(-1))
(-1))i
O(xk•(-1))
[E,X
because
(-1)]
S(i+j-2s)1/2
1
X(1)
i+j-1-2s
= 0 cO(adF)
where
c
0
i+j-1-2s
2
Sc
=(-1)
i+j-1-2s
2
Sc
=(-1)
X'(
2
i+j-2s)
(i+j+1-2s
2
=
ij-2s
i+j-2s
O(adE)
O(adE)
i+j-1-2s
2
(-X) (i+ji+j-1-2s
2
s))
2
(-1) Xek(-i-j+2s)
2i+j-1-2s
2
=(-1)}
*ekX(-1)
Corollary.
Proof:
j([MJl-
Clear.
1)
= [Mji
D
An argument analogous t o the one given in
Proposition 4.1 shows
GL(1)x-**xGL(N)
subgroup of
Theorem.
L
and that
is isomorphic to a subgroup of
LnK
is isomorphic
to a
O(1)xo**xO(N)
The determinant character is trivial on
(LnK)O , therefore all
nilpotent orbits are admissible.
-84Proof:
We know
([Mij 1_l)
Fix
[Mi]_-1e[Mi]_-1
preserves
J
[Mij ]-1
[Mij ]-1eJ([MiJl
-1
E 0 (j) .
[Mji
1
_l
Because
and
Adg
commutes with
LnK
we conclude:
det(Adgj
If
and
g
=
ij
([M'
gj E (LnK)o , then
) = det (Adgjl
I ) CM
)CIR.
1-
= 1 .
det,(Adg I[([MiJ_
1)
-85References
1.
M. Duflo, "Construction de representations
unitairies d'un groupe de Lie," in Harmonic
Analysis and Group Representations, C.I.M.E.,
1982.
2.
V. Guillemin and S. Sternberg, Geometric
American Mathematical
Asymptotics, Math Surveys 14.
Society, Providence, Rhode Island (1978).
3.
S. Helgason, Differential Geometry, Lie Groups, and
Symmetric Spaces, Academic Press, New York, 1978.
4.
B. Kostant, "The Principal Three-Dimensional
Subgroup and the Betti Numbers of a Complex Simple
Lie Group," The American Journal of Mathematics 81
(1959), 973-1032.
5.
A.A. Kirillov, "Unitary Representations of Nilpotent
Lie Groups," Uspehi Mat. Nauk 17 (1962), 57-110.
6.
D. Shale, "Linear Symmetrics of Free Bozon Fields",
American Mathematical Society Transactions 103
(1962), 149-167.
7.
T.A. Springer and R. Steinberg, "Conjugacy Classes",
in Seminar on Algebraic Groups and Related Finite
Groups, Lecture Notes in Mathematics 131, SpringerVerlag, Berlin/Heidelberg/New York, 1981.
S.
D. Vogan, "Representations of Reductive Lie Groups",
to appear in Proceedings of International Conference
of Mathematicians (1986).
9.
D. Vogan, "Singular Unitary Representations",
Non-Commutative Harmonic Analysis and Lie Groups,
Lecture Notes in Mathematics 880, 506-536.
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