ON SUSANA A. SALAMANCA (1980) INFINITESIMAL CHARACTER

advertisement
ON UNITARY REPRESENTATIONS WITH REGULAR
INFINITESIMAL CHARACTER
by
SUSANA A. SALAMANCA RIBA
Sc.B. Universidad Autonoma Metropolitana
Mexico D.F. Mexico
(1980)
Submitted
of
in Partial Fulfillment
the Requirements
the
for
Degree of Doctor of Philosophy
the
at
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
MAY, 1986
@ Susana A. Salamanca Riba
The author hereby grants to M.I.T. permission to reproduce
and to distribute copies of this thesis document in whole
or in part.
Signature
of Author
..
,.
.....
Department
Certified by
....................
of Mathematics
May 9, 1986
.................
gDavid A. Vogan
Supervisor
Accepted by
............
MASSACHUSETTS INSTITUTE
OFTECHNOLOGY
SEP 16 1986
L!BRAP S
ARCHIVES
Chairman,
-
-
----...............
.............
Nesmith C. Ankeny
Departmental
Graduate
Committee
2
Dedication
Dedico esta tesis a mis padres y hermanos por el amor,
apoyo y
fe que me han prodigado durante
especialmente a mi
toda mi vida,
hermana Lourdes con quien he compartido
la misma experiencia de estudiantes en M.I.T. y que durante
estos afios fue mi ejemplo y mi apoyo y que generosamente me
dio su tiempo, comprension y numerosos consejos para el
desarrollo de mi
investigacion.
3
Acknowledgements
It
is a pleasure for me
to
thank my advisor David
Vogan for introducing me to representation theory,
for
teaching me how to do research, for being always willing to
share his knowledge, and
for being a
source of support and
enthusiasm.
I also want
to thank all my friends
constant encouragement and support,
for providing
and for
teaching me how
to do research, especially Jeff Adams, Jeremy Orloff, Roger
Zierau, my sister Lourdes Salamanca Young, Devra Garfinkle,
and Ranee Gupta.
The
latter
three,
together with Michele
Vergne have been outstanding female models
out
the time
I was a student at M.I.T.
Barbasch for helpful
for me
I also
through-
thank Dan
discussions and suggestions, and
generous advice and interest in my work.
Thank you Luis Casian for patiently discussing with me
numerous parts of my thesis as well as
problems
other interesting
in mathematics, for continuous advice and under-
standing, and constant
interest
in my progress.
Thank you again Roger Zierau and David Vogan for
making research in mathematics so much fun.
Finally I want to thank Phyllis Ruby for her constant
interest and help during my stay at M.I.T.,
for her excellent work in typing this
and Viola Wiley
thesis.
4
ON UNITARY REPRESENTATIONS WITH REGULAR
INFINITESIMAL CHARACTER
by
SUSANA A. SALAMANCA RIBA
Submitted to the Department of Mathematics on May 2, 1986,
in partial fulfillment of the requirements for the degree
of Doctor of Philosophy.
ABSTRACT
G = SL(nR),
For
SU(p,q),
and
SP(n,R)
we prove
that every irreducible unitary representation of
the same infinitesimal
character as
that
involve case-by-case arguments
features of
these groups.
So
with
of a finite dimen-
sional, arises as cohomological parabolic
one-dimensional unitary character.
G
induction from a
The techniques used
that do not use any
it seems reasonable
that these arguments could be extended in order
special
to hope
to solve
the problem for other groups.
G
For
of
G,
90 = k
Lie(G),
G
WK
as above let
+ P%,
(7il)
on which
K
be a maximal compact subgroup
the Cartan decomposition of
an irreducible Hermitian representation of
Z(g)
acts as
on a finite dimensional module,
the Harish-Chandra module of
K-type of
WK.
We prove
a Zuckerman module
A
that
(N),
q = t + u C g.
subalgebra
character
X
restricted
to
o=
of
V
Thesis Supervisor:
Title:
(p@V )
either
for
some
and
IK
(ji,V )
is
a
lowest
isomorphic to
8-stable parabolic
and one-dimensional unitary
L = NG(q),
G
7r
or else the Hermitian form
is indefinite.
Dr. David A.
Vogan, Jr.
Professor of Mathematics
5
Contents
Abstract
4
1.
Introduction
6
2.
Notation and Main Results
12
2.1
Structure Theory
12
2.2
Harish-Chandra Modules
14
2.3
Parabolic Subalgebras
16
2.4
Derived Functor Modules
17
2.5
The Modules A (X)
36
2.6
Reduction Step for
2.7
3.
the Proof of
Theorem 1.3
45
Methods
59
to Detect Nonunitarity
G = SL(n,R)
66
3.1
Preliminary Notation
66
3.2
Construction of
LY(X) for a Harish-
Chandra Module X
4.
5.
3.3
Lowest K-types
3.4
Proof
of
69
the Modules A
(X)
of Theorem 2.6.7 for G = SL(n,R)
72
75
G = SU(p,q)
82
4.1
Preliminary Notation
82
4.2
Computation of
Lv(X) and XV(X) for a
Harish-Chandra Module X
85
4.3
The Lowest K-types of
95
4.4
Parameters XV (AQ (X)) and LV(A ())
100
4.5
Proof of Theorem 2.6.7 for G = SU(pq)
104
the Modules A q(X)
G = SP(n,R)
130
5.1
Preliminary Notation
130
5.2
Consruction of
5.3
Lowest K-types of
5.4
Proof of Theorem 2.6.7 for G = SP(n,R)
Bibliography
LV(X)
for any Module X
the Modules A
(N)
133
138
142
154
6
Chapter
Introduction
1.
those irreducible unitary
The problem of classifying
representations of a Lie group whose
ter
is
the same as
al,
has
infinitesimal charac-
that of an irreducible
interested many people
for many years.
representations are
important because
esting applications
like
the
Here
is an example:
is a semisimple Lie group,
subgroup such that
F\G
finite dimensional
G-module,
on
L2 (F\G)
F
an irreducible
m(7r,F),
*
=
m(r,F)7r,
u
Matsushima's formula
(see Borel-Wallach
[1980], p. 223) is
H (F,F)
m (r,F)H*(g,K;WKOF
=
rEG
Here
g
G
irreducible, unitary representations
the set of
of G.
(irW)
a discrete
then the right action of
rEG
u
F
is compact, and
L2 (\G)
G
ch. VII.5-VII.6).
gives a Hilbert space decomposition, with
finite multiplicities
with
in inter-
theory of automorphic forms
for example, Borel-Wallach [1980],
G
These
they appear
(see
If
finite dimension-
is Lie(G) ® C,
u
and
WK
is
the Harish-Chandra
7
module of
The relative Lie algebra cohomology
(irt).
the
groups on the right are non-zero only when
r
mal character of
is
that of
the same as
XK
the modules
F.
algebraic construction
In Vogan-Zuckerman [1984], an
of
infinitesi-
with non-vanishing cohomology groups in
terms of cohomological parabolic
induction is given.
this way are conjec-
derived functor modules constructed
tured to exhaust a larger
The
family of unitary representations
described below.
to chapter 2 for precise definitions of
We refer
the
terms used below.
Let
pact
G
be a reductive Lie group,
subgroup,
90 = k
0
D
bra (cf.
L + u C g
2.3),
OL = L,
L =
and
that
be a
go.
e-stable parabolic
subalge-
is
OU = U,
L,
with
denoting complex conjugation
= L.
q, n o
Let
a maximal com-
the corresponding Cartan involution, and
the Cartan decomposition of
q =
Let
0
K
L
dimensional
be
the normalizer of
unitary character of
q
L.
in
A
G,
X
(g,K)-module
is a Harish-Chandra module constructed as
Chapter 6 by cohomological parabolic
and
a oneA
(N)
in Vogan [1981]
induction from the
8
one-dimensional unitary character
X.
(See Definitions
2.4.14, 2.5.2).
The conditions on the infinitesimal
character men-
tioned above can be weakened to include a
larger family of
representations.
If
X
is a Harish-Chandra module with infinitesimal
x,
character
and
g
h C
a Cartan subalgebra, then up
Weyl-group orbit considerations
-y E
I'
.
h
corresponds to a weight
Choose a positive root system
A+(g,h)
such
that
is dominant.
Conjecture
1.1.
Suppose
X
is an irreducible unitary
Harish-Chandra module such that
A +(g,h).
q
x
to
Then, there are a
r -
p
is dominant
for
e-stable parabolic subalgebra
and a unitary one-dimensional character
X
of
L
such
that
X
A
(M).
Some progress has been made when we assume that
regular and
T.
integral.
J. Enright
[1979]
then there exists a
X.
Also,
Namely, for
proved that
(g,K)
in Speh [1981],
G
if
module
is regular
(N)
following result
the same result for
is proved in this
integral
isomorphic to
G = SL(n,IR)
is proved.
The
is
a complex group,
i
A
-r
thesis.
9
Theorem 1.2.
and
-r
If
G
is
SL(n,R),
regular and integral,
SU(p,q),
then, for
or
some
q
SP(n,R)
and
X
X = A (N).
The proof
from Speh's
for
is new and quite different
original one.
The proof
involves
SL(n,R)
is by
induction on
the dimension of
choosing an appropriate proper
and embedding
9K
as
subgroup
that
unitarity of
G
and our
The
set up
It
L C G
the Langlands submodule of a derived
functor module induced from a representation of
such a way
G.
L,
in
the information about unitarity or non-
the representation of
representation
L
can be carried up to
IK'
thesis is organized as
follows.
In Chapter 2 we
the notation and results needed to restate and prove
the result
in the following form:
Theorem 1.3.
Suppose
X
is an irreducible Harish-Chandra
module with regular integral
infinitesimal character,
equipped with a non-zero Hermitian form
<
>.
Then,
K-type
V
,
either
a)
X
b)
There are a
V
® p,
= A
(X),
for some
q,
lowest-K-type
such that
A
as above, or
V6
and a
2
10
i = 1,2,
HomK(V .,X) A 0
1
<
and the restriction of
,
>
to
the sum
V5
G V
is
indefinite.
Sections 2.1
the results
through 2.4 are devoted to notation and
that will be needed for
main issues are
functor modules
the definition of
9
(Y)
the proof.
the Zuckerman derived
and Vogan's embedding of any
irreducible Harish-Chandra module
derived
functor module.
ties of
these modules.
into some Zuckerman
We also give some useful proper-
In Section 2.5 we define the modules
some nice
The two
A
(X)
and prove
features that we will use in later chapters.
Sections 2.6 and 2.7, and Chapters 3 and 5 are the
actual
proof of Theorem 1.3.
The main results
are Theorems
2.6.7 and 2.6.8, which say that we can exhibit
submodule of a derived functor module
way
that we can reduce the problem
XL
of
the group
L.
Chapters
I
X
(XL)
to the
as a
in such a
representation
3, 4 and 5 are
the proof of
Theorem 2.6.7.
We argue by contradiction:
embedding result we
find another
With the help of Vogan's
0-stable parabolic
algebra and another Zuckerman module containing
have
to check several
conditions that will
ensure
X.
the
subWe
11
reduction, but mainly 2.6.7 b)
and c).
the representation
isomorphic to a module
XL
is not
Then assuming that
OL
A O(N
)
we prove non-unitarity on
XL.
For
this we use
q
the properties of
the
A
(N)
modules discussed in 2.5 and
some techniques discussed in 2.7, primarily Lemma 2.7.1.
12
Chapter 2
In this chapter we set up notation, state
the basic
results we will need and our main result, and provide a
scheme for
the proof.
For undefined terms in this section see,
Vogan
[1981] Chapter 0.
2.1.
Structure Theory
for example,
We will denote Lie groups by upper case roman letters
such as
G,
H, L
letters such as
and complex Lie algebras by script
g,
h,
L.
We will make
the distinction
between the real Lie algebra of a Lie group and its
complexification as follows:
90 = Lie
Let
9 =
0
C,
etc.
U(g) = universal enveloping algebra of
= center of
Z(g)
(G)
and
U(g.).
Although we will eventually study connected
simple
g
real
linear Lie groups, we will consider connected
reductive linear Lie groups.
These are Lie groups
real
satisfy-
ing:
a)
G
b)
go
c)
G
is connected
is a real
has a
reductive Lie algebra
faithful
finite dimensional representation
13
Let
00
8
be a Cartan involution of
the Cartan decomposition of
eigenspaces of
go
and
+1
into the
go
go = k0
and
-1
8.
Fix once and for all a nondegenerate, invariant
symmetric bilinear form on
go.
We will denote
this and
its various complexifications, restrictions and
(
dualizations by
,
).
We may choose
Cartan decomposition of
go
it so that
the
is orthogonal and
> 0
(
,
)
1k0
( ,
Let
H
be a Cart an subgroup of
the roots of
ht
In general
subalgebra of
9
then
o
in
if
g9
A(V,o)
multiplicities).
< 0.
G.
A = A(g,h)
Denote by
g.
o, is an abelian reductive Lie
and
V
is an
ad(a)-stable subspace of
is the set of weights of
For any
B C A(V,o3)
o,
let
in
V
(with
p(B) =
a.
aEB
When there
no
confusion we will use
A(V)
H
is a
8-stable Cartan subgroup,
H = TA;
with
T = H
If
and
is
A (g,k)
is
8-stable.
n
K,
A = H
n
for
A(V,o).
then
(exp po) = exp(t 0fp 0 )
14
W = W(g,h)
Let
be
the Weyl group of
W(G,H) = NG(H)/H
Let
A+
& = h
p
=
+ n,
(resp.
c
the corresponding Borel
g
and
n K.
be a set of positive roots of
H
t
f
subalgebra and
Hc)
c
is
C k
0 -0
be a Cartan subalgebra.
to be
the centralizer in
in
p =
of
Tc = Hc
n
K
G.
the other extreme,
0
G)
K.
abelian subalgebra and
ft
(resp.
is called the fundamental or maximally compact
Cartan subgroup of
On
with
TcAc
a Cartan subgroup of
Hc
g
fc
Define
8-stable, so we can write
Hc
then
in
p(n).
Let
t0.
A +(g,)
=
NK(H)/H
f
h=
if
0
p
as
ts + a
0
0
is a maximal
is maximal abelian
is also a Cartan subalgebra of
centralizer
Hs
in
G
go.
Its
is a Cartan subgroup of
G,
the
maximally split one.
2.2.
Harish-Chandra Modules
Let
(r,l)
be a continuous complex Hilbert space
representation and
*K
the
subset of
W
of
K-finite
g,
15
vectors.
If
(irl)
is admissible,
K-isotypic components
the
are
is,
if all
the
finite dimensional,
then
limit
=
7r(x)v
exists
for all
is
XK
a
representation
complex).
in
tx)v-v) ,
v E AK
and defines a
*K'
module since we can complexify the
(g,K)
Also
(7r(exp
and
go
IK
1
lim
t-+O
x0 E
representation of
WK
WK
of
that
to a representation of
AK
is a
g
representation of
the Harish-Chandra module of
(r,l)
(W
K.
being
We call
(cf. Harish-
Chandra [1953].
Any
irreducible unitary representation of
G
is
admissible.
Denote by
by
d(g,,K)
A(g,K)
the category of
the category of admissible
An irreducible
(g,K)
module
(g,K)
(g,K)
modules and
modules.
is automatically
admissible.
A
g
scalars on
module
X.
X
is
quasisimple if
Z(g)
Then we have a homomorphism
x
: Z(g)
-- 4 C
acts by
16
x(z)x = V(z)x
called the
infinitesimal character of
Any
irreducible
If
(r,X),
are equivalent
element of
(g,K)
module is quasisimple.
(r'.X') E A(g,K)
L
Write
G
(g,K)
modules.
we say
(7r',*')
2.3
X
and
X'
the
defined by
= {L : X --
I
X
is complex linear and
i'L = Lr}.
set of equivalence classes of
If
(r,g),
(r',*')
irreducible
are representations of
they are infinitesimally equivalent
if
(rr
K)'
are equivalent.
Parabolic subalgebras
Let
tc
0
k
-
'
Fix
x E i(tc
We
say that
(9,K)-module maps
Hom9,K('r') = Homrn,K(X,X')
G
we
if there is an invertible map which is an
the set of
for
X.
define a
Let
*
e-stable parabolic subalgebra
q
as follows.
17
A(ttC) = {a E A(g,tc)
<a,x> = 0}
A(u) = A(.,tc) = (a E A(g,tc)
<a,x> > 0}
A(L)
=
t =
D
CX + t,
a
aEA(L)
then
q = L + u
2.4.
Derived Functor Modules
is
CX
@-stable.
In this section we consider
cation of Harish-Chandra modules
a certain set of parameters
Chandra module.
a
aEA(u)
that part of
the classifi-
that consists of
attaching
to an irreducible Harish-
We are going to exhibit each irreducible
module as a submodule of a derived functor module.
(gK)
set
We will first consider a particular
modules when
(g,K)
G
is quasisplit.
of irreducible
To define
these
groups we need some notation.
Let
a
be a maximal abelian subalgebra and
C p0
the corresponding connected subgroup of
Let
M = K
Define
in
As
=
centralizer of
s,a)=
A(g/(i+as
As
G.
in
K.
the nonzero roots
g.
Fix a positive system
A+
C As
As
and let
of
as
18
=
0
0
and
N
+
aEA~
the corresponding connected subgroup of
Definition 2.4.1.
S
^s
A
For a fixed representation
, define the Hilbert
{f : G
=
A6,V
m E M;
The action of
a
G
--
V
E A;
on
n
V
I
E N
f
G.
Definition 2.4.2.
G
abelian.
M
measurable;
f(g);
f K E L2 (K)}.
and
I(6@v)
= f(g
=
Ind
1g 0 )
G
(60v),
the
+ a
is
p
representation of
of
given by
(T 6,V(g))f(gO)
representation
(6,V)
space
= a -(v+p) 6(m)
f(gman)
defines a
G.
Ps = MAsN.
Define
and
RX
(D
is quasisplit
if
a
induced
19
Hence, if
subalgebra of
of
G
go
is quasisplit
= i0 + a0
0 0
k
HS = MAs = TsAs
and
is a Cartan
a Cartan subgroup
G.
Therefore
^S
H
=homomorphisms
Definition 2.4.3.
a)
d6
a0 n
: H -- + C
is
trivial on
b)
Consider
A
a E A
is
^s
^
M
xA
representation
^
M x
6 E M
s
(a)
is
fine if
[g.
the set
A = {a E AS
A root
x
real
if
! a f AS
a =
s
for some
0 E As
real.
Let
Ma
n
K.
| a(X) = 0}
aa = {X E a
and
Choose an injection
a
a
0
:
such that
a 0
-1
Oa 0G =
E a0
00o
GA
= MaAa;
Ka =
20
0
1
Oa .0
E a-root
[0
Put
c)
A
a
For
included in
ii)
1
ka
a -1
representatioi
i)
space.
0]
E K
real
a E
0
0
is
fine
G)
(for
has eigenvalues
p( iZa)
{0,+1}
For each complex root
a E i
is
1(k a n
K
6 E M
If
d)
|I
is
is
f ine,
a
Suppose
G
set
A(b)
= {i
E
p Im
See Vogan [1981] 4.4.8.
p
to
M
is a
p E K
sum of
is
fine.
Then
fine representa-
each occurring with multiplicity one.
p E A(6)
Say that
irreducible
occurs in
is quasisplit and
the restriction of
M,
6
trivial.
909 )
fine representation
and
Proposition 2.4.5.
tions of
if
(g.,K)
Then there
for some
6
module containing
is a character
fine.
the
V E As
Homg,K(XI(6 @v)) A 0.
Let
K-type
such that
X
be an
ji.
21
For an arbitrary linear reductive Lie group we will
define
the notion of minimal
(or
lowest)
K-type of a
representation and attach to it certain parameters.
next
section we will
then construct a
(g,K)
In the
module with
these parameters using a reduction to a quasisplit group.
The
irreducible representation with
a subquotient of
this module.
that
lowest
For proofs of
K-type is
these results
see Vogan [1981] Chapters 5 and 6.
Fix a Cartan subalgebra
t
of
kc ,
a positive root
system
A+(k) = A+(k,tc)
A+(k)-dominant
and a
for
its differential.
weight
hc
positive root
dominant.
as
in 2.1.
system
write
k))
E
)*
(t
E (Cc)*
Then there exists a
A+ (9,c)
which makes
See for example, Vogan [1981], p.
Fix a noncompact imaginary root
X
i
Define
2pc = 2p(A+
Let
Cc;T
for a root vector for
the root
x-
=-U
e-stable
p + 2p c
239.
E A(g9,hc).
f.
Put
Write
22
Z P= XP+
=
X_
centralizer of
|
(hc)= {x E h
@
S= (t )
0t3
Then
9
(ftc)'
is reductive,
Lie algebra
9P
E
is real
Cartan subalgebra of
g
0g
Z
in
go
(x) = O}
(2.4.6)
Z.
$ <Z13 >.
the subgroup
Gp
reductive linear and
and of
go.
of
G
with
h
is
a
See Vogan [1981], p.
235.
Proposition 2.4.7 (Vogan [1981], 5.3.3).
A+(k,tc)-dominant weight
ment
Xv(p)
ties:
G
-
fix a
Then
XV(p)
./..
a)
,3r} .it
E (tc M
there
having the following proper-
dominant; and write
is dominant
for
A+(g,hc),
If we put
-2<p
c>
, p+2p c>
A +(g,h
)
p = p(A + ( ,.c))
and
of imaginary roots,
A(g,hc)
each
is a unique ele-
e-stable positive root system
p + 2p
making
i C Tc,
For
there is a set
satisfying
23
c
V =
p
,
then
o < c. < 1,
and
XV(p) = XV(p) = p + 2pc
a E A (g,hC)
If
b)
<a ,P1> 4 0
then
c)
simple,
The root
or
there
P1
such tha t
=
Either al 1
e)
Write
A + (9.hc
satisfy
"
.1
t
n+9
a
of
is
A + (g,hc)
a + Oa.
d)
in equations
and either it
i s a complex simple root
= 0,
C
g~
as
i.
is noncompact;
Pi
(aXv(p)> = 0,
imaginary, and
is
for some
+ 2+v
v.
-p
1
ft
=~
(2.4.6).
1,h 1 )
or
Then
and
these same condi t ions
c1
A 0.
= hP
the positive system
its
for
subset
g1
and
{P2''13d
the weight
24
Definition 2.4. 8.
let
For a
A +(k)-dominant
G
qV(p)
be
the parabolic defined by
cribed in 2.3.
If
we also define
G
v
is a
V(7r)
-
(a)
is quasisplit
L
t
V( i-2p(unp)) =
(b)
(c)
If
r E K
G
'v(11)=
q
as desp
qV
qV(p)
L0
G
XV(p)
^c
p E
K-type of highest weight
Proposition 2.4.9 (Vogan [1981]
A+(k)-dominant and
weight
§5.3).
= L + a.
has highest weight
Suppose
p E T
then
is fine
is
Then
p,
r
.
We will now define a preordering on
K.
Def inition 2.4.10.
a)
as
If
7r
E K
has highest weight
in proposition 2.4.6.
the
If
X
is a nonzero
lambda=
E K
|
X(r) X 0
N =
C(
X,>.
(gK)-module define
r-isotypic component of
{r
put
Define
1111lambda
b)
p E T
and
X.
X(ir)
Then the set
vI"l lambda
is minimal}
to be
25
We define it
is nonempty.
types.
c)
LKT
We will refer to
XV(X)
Define
of
X
to be
such a
XV(p)
=
of
lowest
K-
as an LKT.
7
p
for
G
qV(X)
and let
the set
a highest weight of a
the parabolic associated
be
to
11.
will
When there is no confusion we
parameters as
Let
tL
qV
be
normalizer of
T
Li
tion of
VL
c
qV
,
so
in
these
XV.
the Levi
factor o
qV,
f
and
LV
the
G.
let
be
7r
the
irreducible representa-
generated by the
fl K
L
and
refer to
p-weight space,
inside
flK.
Let
Lv
V
=
V0
[A(R (Lfl4*
E
(LVfK)
(2.4.11)
R = dim u
Notice
p-2p(uflp)
that
L
and choose
= TA
H
6
is a highest weight of
is fine for
a maximally split
E T,
r V
By
and c)
proposition 2.4.9 b)
Fix
n o.
Lv.
Cartan subgroup of
fine occurring in
L
L
26
This
discrete
(qV,HV'
triple
VL)
is,
by definition a
0-stable data attached to
We have attached all
representation
X
X.
these parameters to a
with LKT
7r.
Now with these parameters we will exhibit a
module
that contains
X
set of
as a subquotient.
(9,K)
We need more
definitions.
Definitior
2.4.12.
quadruple
(q,H,b,v)
a)
q = L + a
b)
L
is a
c)
6 E T
d)
If
is
(G,a>
i)
G
ii )
that
<X
H = TA C L
L
and
is a
v E A.
L + p(A(a,tc))
> 0
, > = 0
E
t
6
and
C h
for all
a E A(a,h)
for all
13 E A(L,h).
0-stable data attached to
together with any character
of
0-stable
L.
the differential of
the discrete
)
i s a
the conditions:
Lv
(o,,VHV'b
G
0-stable parabolic subalgebra of
is fine for
XL E t
X E A(g,K)
0-stable data for
split Cartan subgroup of
XG =
Notice
of
with
is quasisplit, and
maximally
then
A set
0-stable data for
G.
v E A
some
are a set
27
In order to give a generalization of Proposition 2.4.5
for non-quasisplit groups we need to define the objects
in
which we are going to realize the Harish-Chandra modules of
G.
Definition 2.4.13 Zuckerman Functors.
q =
Let
L + a C g
be a
0-stable parabolic
subalgebra as defined in 2.3 and
subgroup corresponding to
t0'
is connected, then so are
G
Since
the reductive Lie
L C G
K, T,
L
and
L n K.
Let
Z
be any
(q,LK)
module.
Define
pro (Z) = pro9,LnK (Z) = Hom (U(g),Z)Kfinite
t
Q U9)ZLnfl
q LfK
Od
This
is a
(g,LK)
Now, if
W
module.
is any
(g,LfK)
I
FW = {v E W
FW
is a
is a
left exact
derived functors of
(g,LfK)
}.
F : A(9,LfK) --
A(9,K)
functor.
The Zuckerman functors
a
dim U(k)-v <
module and
(g,K)
module define
module
F
then
{(jflK
(written
W
r
).
admits an
1i
That
are the right
is,
if
W
is
injective resolution
28
0
W
1
----
10
then we have a cochain complex
0
W
Define
FW=ker
m i
Im
Definition 2.4.14.
Then
L
n
L
K
00
FW
-
_
.
Let
We will define
induction functor
(oq)i
.
0
q = t + u
L C G
and
i-th
as
in 2.3.
(connected) Lie group and
compact subgroup of
the
1
F W = FW.
So
is a reductive linear
is a maximal
-
F"l
1I 0
G.
cohomological parabolic
: A(L,LfK)
--
A(g,K),
as
follows.
Let
U
act
V E A(L,LfK).
We make
V E A(q,LfK)
trivially.
Let
prog.,LK
q, LK
VO Adim
C
U)
then
()(V)
Q(V) =
=
i(
F i(W).
= 9
(V)
by letting
29
We are now in a position to state the generalization of
Proposition 2.4.5.
moment
it
is convenient at
to mention some properties of
this
these derived functor
that we will need later.
modules
Suppose
such
However,
=
q
+ u
are two parabolic
2 -
1
L1
t2
n1
L
subalgebras
that
t 2'
t
set
U=
then
=
1
+ U C
C2
K C L2
K;
is a parabolic subalgebra of
Proposition 2.4.15 (Zuckerman [1977];
With notation as above if
that for
n
is an
W
L 2.
Vogan [1981] 6.3.10).
(L ,L fK)
module
such
q X q
[2 ]qW
=0
then
g
[ q21
[ g 2 ]qo0
pqq
1pq()
)
q
Proposition 2.4.16 (Zuckerman [1977]; Vogan [1981] 6.3.11).
Let
L,
q
= L + u
be a
9-stable parabolic
a Cartan subalgebra.
Let
Y
be an
subalgebra;
(L,LK)
h C
module
30
with infinitesimal character
1 (W)
has
infinitesimal
Definition 2.4.17.
Let
be a set
(definition 2.4.12) .
P = TAN
and for all
a
Then
X + p(u)
character
Standard Representations.
(q,H,6,v)
such that
X E h.
of
0-stable data for
H = TA C L
Let
and choose
is a minimal parabolic
G
N C L
subgroup of
L
in the corresponding positive system
A(n,a)
(Re v,a> < 0.
IL(60v)
Let
Ind
L
p
(60vol).
0-stable data
be
the principal
We define the standard
(q,H,6,v)
in 2. 4.14 where
We
A +(k)
module with
= !q(IL(60v))
s = dim a
n k.
will now state some properties of these standard
modules.
(0= h0
(g,,K)
by
G(qv)
as
series representation
Fix
n
k0
tC
C0
kf
a Cartan subalgebra containing
and a positive root system
= A + (tk,tc)
u A(unk).
A+ (Lfk,tc).
Set
31
Proposition 2.4.18 (Vogan [1981] 6.5.9).
be a
set
of
0-stable data for
G,
Let
(q,H,6,v)
d6 + p(u)
XG =
and
then
a)
b)
1i(IL(6®v))
If
7r
highest
has infinitesimal
is a
K-type occurring in
weight
C 7rL
(N
G
,v).
D
6
with
q (IL( 0v))
then there exists an
L n K-type rL
77
highest weight
6
chara cter
7L
of
such that
IT and
p = pL +
2p(unp) +
n aa.
aEA (A)
n a EIN
c)
Moreover if
term is
is
7r
zero and
7L
a
is
LKT,
then the
fine for
Suppose
L
(oV,H,6V)
and
to
(b)).
X.
s (I
X
Then there is a character
(6 Lv))
parameters
is
the standard
(qV,H,6
,VV),
(g)
is an irreducible
a set of discrete
and the proof of
(9,K)
module
0-stable data attached
V
E A
(g,K)
then
summation
L.
Proposition 2.4.19 (Vogan [1981] 6.5.9
6.5.12
last
such that if
module with
Homn,K(X,
V(
V)))
is one dimensional.
Lemma 2.4.20.
subalgebra and
Let
Y
q = C + a C g
an
(L,LK)
be a
module.
0-stable parabolic
Write
32
s
n
= dim a
k,
(Y)
XL =
<N +p(u),a> > 0;
X = 9
and
a E A(u).
Choose
(Y).
A +(k)
Assume
= A+(Lfk) U
A(unk).
L
Suppose
L
n
K)
of
Y
A+(tnk,tc)
with respect to
and
is dominant.
is the highest weight of a LKT
b)
p + 2pc
compatible with
A +(k)
2pc = U
+
for
A+(g),
where
a E A+
=
If
<a,-r>
0;
pLfk
A+(g)
+
=
A+(L) U A(U).
2p(ufnp) + 2p(unk) +
2
pLfk
is
simple,
then
)
<4+2p c,a> = <p +2ptnk'a> + <2p(u),a>
ii)
2
A+(k).
and
a E A +(g)
If
jL
+ 2p nk + 2p(u).
=
i)
so that
is dominant for
is dominant
A+(g)
Suppose
A+(L)
Then
p = p L + 2p(fnp)
ji +
the positive system
that we choose
a)
Proof.
(for
<pL+2p t, a>
a E A(u)
hence
<a,P>
then,
0
+ 0
for any simple root
for
3 E A+(L).
r,
is
33
{fp,} C A(Lfp)
Choose
as
in Proposition 2.4.7, such
that
xV
xL
L + 22p~fk
tn1
-p
c 43
1
+
Osc
(1
then
p + 2p
(2.4.21)
=
1
+
c
3
+ 2p(a)
then
V
<a , p + 2p c> =
V
V .x L + p(u.)> + (a,p
(a,
-
p
> 0 + 1.
This proves b) of
For
simple,
(a),
it is enough
lemma.
to prove
that
then
(
i)
the
If
V
,4
-Y E A+(tfk)
+ 2pc >
2.
if
r E A+(k)
is
34
<
(,p + 2pc > =
<( , L
V
<=
+ 2p(Lfk)>
L
,1
Vr,2
+
u)
+ 2p(Lfk)> + 0
> 2,
since
L
is dominant for
1,
-Y E A(ulk)
If
ii)
A+(Lflk).
<
then, as for b):
and it is an integer s ince
W + 2pc
,p + 2pc
exponentiates.
q.e.d.
Lemma 2.4.22.
A+(g),
dominant for
Proof.
In the
By 2.4.21,
setting of Lemma 2.4.20,
then
Xv(pG)
= x
P(U)
-
XV(p)
the condi tions
for
in Proposition 2.4.7:
Condition (a)
Since
A(u),
satisfies
XL + p(U)
is
c io .
C
We claim that
p
+ p(U).
p =XL +
pi + 2pc -
if
A(u)
holds by the definition of
= -A(u)
L + p(a).
(V XL+
<aX + p(u)> > 0
and
2.4.7 (b) holds because
V L
(a,X + P(u)) = 0
for
a E
implies
a E A(L).
But
then,
<a,p(u)> = 0;
Since simple roots
A(g.,hc),
for
2.4.7 (c) (d) (e)
hence
A(,h c)
V
<aP3.>
A 0
are
simple
for
some
for
hold.
q.e.d.
35
K-type occurring in
X
with highest weight
T1
lambda-norm of
77.
We want to estima te the
L
qV=
Let
L
06V(Y)=
=
Hv L VV)
,
and
Y
to
V
If
b)
for
in Definition 2.4. 12.
V
L
is a set
,VV)
is
a
K-type
8-stable data for
of
in
X,
then
G.
(<X(7 7 ),xV(n)>
G
( X>.
in b)
c)
If equality holds
77
a highest weight of a
a LKT of
then
77 = TL + 2p(ufnp)
L(LfK)T
of
Y
and
V
V
is
is
X.
d)
Conversely if
LKT of
X
Proof.
For
all
L
With no tation as in Lemma 2.4.20
L
a)
8-stable data for
as
V+
Lemma 2. 4.23.
a C Q
+
(Definition 2.4.12).
xG = d 6 L + p(V)
Let
G
+ U1
a set of
L H V, 6
attached
L1
be a
L.
1 C
a1 +
QV
(7r,Z)
let
In the setting of Lemma 2.4.20,
roots
-q =
77L +
2p(unp),
then
a
and equality holds.
in
(a) we only need
to see
that
A(u).
By hypothesis on
L
L
L
(qryH'V,
6vy),
xL = dL + p(
)
V
<XG ,a>
> 0
for
36
L V
, a> > 0
(
H = TA
for
is maximally split
a E A(ul),
Cartan of
6
and
L
is
fine.
Now
G
X
=
If
dL
v + P(aV) = d6 L +
+=
a E A(ul) 9 A(C)
+ p(U)
L
V
a E A(u),
L
+ p(U).
then
(XV + p(U),a> =
If
-
L V
Xv,
a> > 0.
by hypothesis in Lemma 2.4.20,
G V
(X ,a> > 0.
The proof
for b)
and c)
is exactly the proof of Lemma
6.5.6 in Vogan [1981].
q.e.d.
2.5.
The Modules
Let
group,
L
q =
be a connected real
C + u C g
the normalizer of
Let
of
G
A (N).
L.
X
:
Assume
--
4
that
a
q
C
reductive linear Lie
0-stable parabolic subalgebra and
in
G.
Then
L0 = Lie(L).
be a one-dimensional
representation
37
a)
X
(2.5.1)
L
We say
b)
is
the differential
character of
L
(XIc a> > 0
for all
that
X
Definition 2.5.2.
(call
of a unitary
it
X
also).
a E A(utc
is an admissible representation of
With notation as above, we define
Harish-Chandra module
A
(X)
with
the
by
= O
()
s = dim u
n
A (N)
L.
(Definition 2.4.14)
k.
Fix positive root systems
A+(Lfk)
A +()
Then
and
are positive
=
A+(L,L),
A +(k)
A +()
t-root
=
and
compatible with
A+(Lfk).
= A+(Lfk) U A(unk)
A+(q) = A+(L) U A(U)
systems
for
k
g,
and
Choose a fundamental Cartan subalgebra
hc
respectively.
c + ac
and a
38
positive root system
A+(g,hc)
A+(gh
i c
so
=
A +(g-).-
1
P =
Then
that
aEA+ (g,hc)
Proposition 2.5.3
I3EA'(g9)
(Vogan-Zuckerman [1984].
Speh-Vogan
[1980] and Vogan [1981]).
weight
(tc)*.
in
The
See also
Regard
c
X
as a
Let
p = N ic
a)
P1.
(g,K)
+ 2p(unp) E
module
A (N)
(tc )*
is the unique
irreducible module satisfying:
i)
As a
K-representation,
K-type with highest weight
ii)
+p:
is
Z(g)
Z(g)
where
contains
X
A (N)
(z)
b y the character
=
( X+p)(f(z))
and
the Harish-Chandra homomorphism.
iii)
Any
K-type occurring in
A (X)
highest weight of the form
T7 =
X
c
+
the
p.
acts on
-- + C;
A (N)
2p(up) +
n 0 P.
PEA(up)
n EIN
has a
f
39
b)
Proof.
is the unique LKT of
The infinitesimal
--
S:L
i
Moreover
C
character of
X + pt.
is
has infinitesimal
A (M).
the representation
Then by Proposition 2.4.16
X + P
character
+ p(c)
A (X)
So ii)
p.
holds.
If
(lowest)
L = Nl
L
Choose
n
K-type
L
is the highest weight of
the
C .
of
4.
A
(L)
L
xV
then
tc
=
making
L (4L
V(.
)
pL +
2
= i c + 2ptk
-
Then
dominant.
p tLk
Pt +
c
#i
|c
with
Q
But
a sum of roots
iL
in
= 0
so
a E A(u)
is
IA(L)
L.
A(L)
can be chosen so
that
Q
is dominant.
Then if
simple,
(L
V
+ p(),a)
> 0.
In
fact
(L
+ p(u),a> = (
<N
c
+ p(a),a> >
<p(u),a>
> 0.
By Lemmas 2.4.20, 2.4.22, and 2.4.23,
i)
and b)
hold.
40
The irreducibility and uniqueness of
A
work, and since we won't be using these facts we
Speh-Vogan
[1980].
take more
(A)
refer
See also Vogan [1981].
1.3 in Vogan [1984], w e have
By 2.5.4 and Theorem
following.
In the above
Proposition 2.5.5.
A
(A)
1,2,
A(q.)
Fix
be
A
(k).
Q A+(k)
A. E t
and
+ U . C g;
Li
admissible one-dimensional
(Definition 2.5.1).
A1(A )
A
=
We need a
Lemma 2.5.7.
= L.
1 1
Let
6-stable parabolic subalgebras such that
representations of
Proof.
setting, the modules
are unitarizable.
Proposition 2.5.6.
i =
to
X2
and
A(2
U1
2
Then,
)
nP = u2 n A.
few lemmas:
Suppose
Q =
L + U, q = t + U C g,
6-stable parabolic subalgebras, and
admissible representations
such that
A
:
-
C
are
the
41
(2.5.8)
Then
Proof.
1)
q 2 q,
2)
X .
3)
a
nafl
.
=
~to A (y).
By induction by
S
but
a D a.
and
A(L).
n
A()
L
t D
that is,
stages (Proposition 2.4.15),
Lf(unk) (y)
(dim
)
q f L = L + a n I
and,
u n L c k
by 3),
so
dim alt
.
(Cx)
qnfl
Hence
Ms(ex)
s(C)
=
A'~'(X)
this proves
the
lemma.
q
q.e.d.
By
this lemma, we may assume
proposition are maximal
Lemma 2.5.9.
that both
with respect
i
in
's
to conditions 1)
In the above setting
A(ct ink) = (a E A(g)
I <a,AX
+ 2p(u
op) >
=
01}.
the
-
3).
42
Proof.
Suppose
a E A+ (k,tc)
a)
a f A(L.fk).
b)
<a,pi.> = 0,
p
1
is a simple root so that
1
n
A(z) = Span(A(L.),a)
Let
A(u)
.fln).
1
= X. + 2p(u
=
A(g.)
A(u.)\A(T)
11
We want
to contr adict
Breaking up
A(u
nilb)
1*
the maximality of
in maximal
{ 0; 0+a;
(i.e.
-y0
a
qi
strings
+ra},
. ..
- a, -0 + (r+i)a f A(ulp))
and using representation theory of
oL(2)
we can conclude
that
<a,2p(u . Ap> > 0
and we have equality if and onl y if
under
the
the
a
three dimensional sub ialgebra
a-root vector
X
is
1
a
g.
a
But, by definition of
X.,
1
<a, A.>1
> 0.
invariant
that contains
43
So,
a
imply that
(a) and (b)
a. f P
is invariant under
and
<a,X.> = 0 = <a,2p(u .lop)>.
1
w
that
Now we want to prove
(2.5.10)
1 E A+(g,hc)
If
P
n
= A.
n p.
1
P3
and
=
a
then
tc
s
If
is not
)3
in
Hence
A(u. flp)
so it
for
A(k,t c),
a i A(Lfk)
and
a =
Y + 6
7
and
again.
-r -
In fact,
A+(g,hc).
and
a =
root of
-it
a
contradicts invariance under
is an imaginary root of
a
simple for
say,
c'
is complex, then the non-compact
also
E A+(g.,,c)
[
A+(g,he).
since
a
we can assume
is
that
a
is
simple
if
-,
then
E A(u inP),
-e 6 A(u .lo);
1
contradicting
invariance
5
44
Consider a simple factor
L.
L0
Then
is not orthogonal
be a set of simple roots
a = Pi
Say
n.P
=
for
n P A 0.
with some
So
6
the string
Hence
satisfying
t0
a.
not contained in
Let
ni 0+1
containing
6
a.
is adjacent to
> 0
a.
and such that
n.<a,p .> < 0.
is a non-compact root, and
through
{p1,02'...'13
Then there is a non-compact root
<a,P> =
a + P =
to
L0
1+1
P
and
t0
Suppose
1
t0 g T,
6 E A(u i po).
is not complete.
is compact and
(C q)
q
is not maximal
(2.5.8).
This proves Lemma 2.5.9.
q.e.d.
We are now able
to prove Proposition 2.5.6.
By Lemma 2.5.9,
ti
1
hence
X\
+ 2p(a 1)
<2p(u.),P> = 0
(X. ,a> > 0,
1
-
X2
for all
a E A(u 1.).
n
k
- 62
n
k
n
k
=
2
n
k
2
).
+ 2p(u
P E A(L.)
So
But
and
(X.P>
=
<2p(u.),a>
> 0,
45
A(Li)
= {1
A(ui)
=
cX. + 2p(u.),P> = O}
<
E A(g,,tc)
{a E A(g,tc)
(X
n
fl
Hence
and
X
+ 2p(u.),a> > 0}.
=
2
- X2'
This proves Proposition 2.5.6.
q.e.d.
2.6.
Reduction step for
the proof
of Theorem 1.3.
We are now in a position to prove the main result
stated in Chapter
1.
We will argue by contradiction and
reduction to a proper subgroup
Suppose
X E
d(g.,K)
Hermitian form
<
,
realized as an
A
(N)
>.
L C G.
is irreducible and has a
We will assume
module, but will
X
cannot be
exhibit
X
as a
Langlands submodule of some derived functor module induced
from an
(L,LK)
module
XL,
making sure
information can be carried over
We need to keep
track of
to
G
and
that
this
X.
the existence of Hermitian
forms at different
steps of
signatures on some
finite sets of
induction as well as of
K-types.
their
46
Recall
from Voga n [1984]
Hermitian dual of
a
Yh =
module
(9,K)
: Y -- ), C
If
(Definition 2.10)
I
is a
(g,K)
on a
(g.,K)
An invariant,
module
Y
<
such
is a pairing
,
>
Y x Y-
<x,ay+bz> = a(x,y> + b<x,z)
<ax+bw, y>
a,
b)
b E C,
X E C
x E Y}.
symmetric Hermitian form
that
a)
for
Ef(x),
< 00 ;
module.
2.6.1.
Def inition
Y
dim U(k)-f
f(Nx) =
the
x,
y,
z,
= a(x,y> + b<w,y>
w E Y.
<(U+iV)x,y> = -<x,(U-iV)y>
U,
V E go
c)
<k-x,y> = <x,k
d)
<x,y> = <y,x>.
-y>
y, x E Y.
k E K.
47
<
The radical of
>
Rad(,
It
on
Y
is clear
are given by
f = fh
.
Yh -- + Y.
Proposition 2.6.2.
Then
X
{x E Y
=
that
is
(x,Y> = 0}.
<
invariant symmetric Hermitian forms
(gK)
maps
:
f
Y
Yh
-
such that
Moreover we have
Suppose
X E sd(g,K)
is
irreducible.
admits a non-zero invariant Hermitian form
if and
only if
h
In this
any
case
the Hermitian form is non-degenerate and
two such forms differ by multiplication by a real
constant.
Proposition 2.6.3.
(4V,HV,5,vV)
a
Let
set of
X E d(g,K)
be
irreducible and
@-stable data attached to
X,
so
Xh
if
that
dim Hom ,K(X,
(see Proposition 2.4.19).
and only if
there
(I V(
Let
is an element
6
@V))
Hv = TA.
=
1
Then
X
48
w E W(L,A)
=
In
such that
WD = -v.
and
65
this case we get a Hermitian form on
X
from a
form on
L
g (I V(6V(v
)).
vV).
QV
This
result
is essentially due to Knapp and Zuckerman
[1976].
A
formulation close
to this
one is
in Vogan
[1984],
Corollary 2.15.
Corollary 2.6.4.
Let
X E 4(g,K),
with a non-zero Hermitian form
Let
q = t + u
that
L 3
be a
LV,
data attached
and
X.
X
,
>.
Write
0-stable parabolic
U C U
to
<
irreducible, endowed
qV =V(X).
subalgebra such
(qV,HV,6V,vV),
a
0-stable
Write
L = 9 t iq(IL
-
(Vvv)).
Then
xh
L
Proof.
This
has a Hermitian form
is a formal
<
,
>
consequence of Proposition 2.6.3.
q.e.d.
49
Proposition 2.6.5.
parabolic subalgebra.
with a
q =
Fix
L + u C g,
Suppose
(possibly degenerate)
Y
E
a
A(L,LnK)
8-stable
is equipped
invariant Hermitian
form
L
Then
G
< >
is a natural
invariant Hermitian form
s h h
[As(Y)]
on
Proof.
6.1.5
there
Recall from Vogan [1981] Chapter 6,
Definition
the functors
: A(L,LfK)
ind
--
+* A(g,LfK).
Q
ind9Y = U(g)
0_ Y.
1%
Q
Write
V
:
eiY = VY = p
where
yj
functors
ind9(Y@A t op a)
A (g ,LfK) -- + A (g,K)
(cf.
A(,K)
A(L,LfK) --
are the Zuckerman's
Definition 2.4.14).
By hypothesis,
Set
we have a map
L
0:Y-*y
h
Y = Y @ Atop
.
50
This induces a map
ind(Y) -
pro
h
and
s
G
h
By Theorem 5.3 (Enright-Wallach) in Vogan [1984]
L2s-e
h
h
Let
< , > : VY
+ (VsY) h
be the natural pairing given by Definition 2.10 in Vogan
[1984].
Define
(u,v>G = <u,O v>.
This gives an invariant Hermitian form on
(cfr.
fs(M
the proof of Corollary 5.5, Vogan [1984]).
q.e.d.
51
Definition 2.6.6.
If
Z E A(g,K)
and
6 E K,
write
Z(6) = Hom K(V 6 ,Z).
Then
(2.6.7)
0
Z
Z(6) 0 V .
6EK
If we
fix a positive definite form on
inherits a Hermitian form.
non-zero Hermitian form
z(6)),
Z(6)
for
where
Write
<
Suppose
,
>.
Write
the multiplicity of
(
>
,
sgn(< , >IZ(6))
V6
is positive
the signature of
<
Z
V6 ,
Z(6)
is equipped with a
p(b)
(resp.
q(6),
in the subspace of
(resp. negative or zero).
,
>
on
Z(6)
as
= (p(b),q(6),z(6)).
Then write,
formally
sgn(<
,
>)
=
(p(6),q(6),z(6)).
6EK
We will prove in the next chapters
the following
result.
Theorem 2.6.7.
and
X E A(g,K)
Let
G = SL(n,R),
SU(p,q)
or
SP(n,R)
irreducible, endowed with a non-zero
52
<
invariant Hermitian form
>
,
and regular integral
infinitesimal character.
If
are a
X
= A
,(X'),
for any
L
a)
X
and
9
X
is
'.
an
Then there
(L,LfK)-module
such that
i = 1,2
6
(LfK)-types
and
q = t + u,
0-stable parabolic
and
XL
q'
the unique irreducible submodule of
A
(XL)'
occurs only once as a composition factor of
(XL).
h
b)
is endowed with a Hermitian form
XL
Write
(pL'qLzL)
L
>L
0.
Then
signature.
x0
q
and
2
0.
A+(k) = A+(Lfk) U A(unk).
Choose
,
L
L
L)
c)
for its
<
L
has highest weight
= p
L1, P
L+
+ 2p(n)
6.
Then, if
A (k)
is
dominant.
Chapters 3,
result.
Assume this for
Using
X.
on
this
the moment.
We need to check that
< , >
the proof of
this result, we want to prove non-unitarity of
induced on
X;
that for
the
L
nK
the corresponding
the signature of
same as that of
the Hermitian form
<
< , >G
by Proposition 2.6.5 is a multiple of
q(XL)h
Theorem 2.6.7,
that
4, 5 will be devoted to
,
the form on
L
>
on the
types satisfying c) of
K
types occur
in
X
K-types
is
the
these
L
6.
1*
and
53
Theorem 2.6.8.
Suppose
X E d(9.K)
< .
a non-zero Hermitian form
stable and
unique
XL
an
(L,LK)
(LfK)-type of
LL
L + 2p(ulp)
XL
in
>L
X(6) A 0
>IX(6)]
,
t
if
6
A(unk)
X
and
Q L
L
with highest weight
p,
Sign (
q(XL),
Ots(X)
is dominant for
highest weight
=
u
+
module such that
<
non-zero Hermitian form
q
Let
>.
irreducible submodule of
once as composition factor
is irreducible and has
X
is
0the
occurs only
X
L
E (LfK)
AL
be
has a
is an
such that
then if
6 E K
p
has
and
= Sgn<
,
>L
L
Proof.
Applying the appropriate definitions and results
K
q
and
n
k
we have maps
91A
:
A(Lk,LfK)
vi
:
A(Lk,LK)
Qflk
A(k,K)
A
A(k,K).
-
qfk
If
Y E A(L,LfK)
there are natural maps
-
pro Y
indk
qflk
Y
pro qnk
ind9Y.
Q
to
54
These induce
(k,K)-maps
i
r
i
qYq
4
kY
Se y
.jY
Then, the following diagram is commutative
h
2s-i
[i Y]h
r
It h
i
Yh
h
912s-i
_ ____
qnk
The
isomorphisms across are Theorem 5.3 in Vogan
[1984] for
and
(G,q,)
Arguing as
(K,qk), respectively.
in the Proof
of 2.6.5
(for
K)
maps
qnk
*
.
k
Y
pro
k
h
Y
qflk
:~kink Yn_
qflk
tk .2s
-nu
Y
- gs
qnk
Yh
and we have the following commutative diagram
we have
55
G
s
es Y
h
sY) h
Ih
(2.6.10)
Ir
K
ps
K
s
s
Yh
(Qnk~h
nk
And we have a Hermitian form on
Is
h
;fnk
(Y)
flk
<xy>K
0K = ro Got,
Since
[1984],
L
_
K
and by Proposition 6.10
in Vogan
is a unitary map,
<x,y>K = (Lxy>G
(2.6.11)
(2.6.12)
Write
sign(<
,
> ) = (PK'qK"zK)
sign(<
,
>G
=(G'q G"zG)
PK' q,
-[
ZK
-4IN
'G
G', qG,
N
and again
sign(<
,
> ) = (PL q LzL);
By 2.6.11,
zL
(LfK)
>
IN
56
ZG(6 ) > zK( 6 )
The main ingredient
is
in the proof of
the following result due
to T.
Proposition 2.6.8
Enright.
Proposition 2.6.13 (Enright [1984]).
Let
q = L + a
0-
stable parabolic.
Let
pL
+
6L C (LnK)
L
with highest weight
Set
p =
2p(unp).
a)
If
4
is not
A(u 1k)-dominant,
2
b)
If
w
is
representation of
0.flk
A(unk)
K
with
Y(6) = 0.
dominant, write
highest weight
L
qK
then
=
L( 6
zK06) = ZL(
6
)
L
6 E K
p.
Then
for the
57
For a proof of
Lemma 2.6.14.
S
is
result see Vogan
this
Suppose
V
is a module of
[1984] 6.5-6.8.
finite length and
irreducible.
Assume
a)
of
S C V
factor
V.
b)
Any non-zero
c)
S
,
>
W C V
contains
to scalars,
Vh
has a unique Hermitian form
and
S = Vh/rad(<
The proof of this
Theorem 2.6.8.
(2.6.15)
and
S.
is equipped with a Hermitian form.
Then, up
<
occurs exactly once as a composition
,
>1).
lemma is standard.
We can now prove
By Proposition 2.6.13 and 2.6.11
pG(
L( 6 )
qG(6)
L(L )
zG(6)
zL56 L
Apply Lemma 2.6.14 to
V = 9
(XL)
and
S = X.
58
We know that a)
part of
<
,
-
c) hold in this Lemma since they are
our assumptions on
>G X 0
X.
We also know that
by 2.6.15.
Hence, we have the following result:
Proposition 2.6.16.
In
<
X
So
(
,
>GIX
the setting of Theorem 2.6.8
,
>GX = c<
]h /rad(<
=- [p(X
>
,
,
>G
is nondegenerate and has signature
sgn(<
,
>)
= (pG'qG )
q.e.d.
It
is now straightforward to prove Theorem
1.3.
Theorem 2.6.7, proved in chapters 3-5 for our groups
question, we have that
the hypotheses
true and by 2.6.15
p(G 6
and
qG(6 2
) > 0
> 0
Using
in
in Theorem 2.6.8 are
59
and
<
the form
,
>
on
X
is indefinite
too.
q.e.d.
2.7.
Methods to detect non-unitarity.
To prove Theorem 2.6.7, we will need a few
that we will discuss here.
techniques
Fix a positive root system
A+(k).
Lemma 2.7.1
(Parthasarathy's Dirac operator
See Borel-Wallach [1980] 11.6.1.1.)
unitary representation of
G
and
Let
9K
inequality.
(r,A)
be a
its Harish-Chandra
module.
Fix a positive
A+(k)
and a
weight
p E
t-root system
k-type
t'c.
6
occurring in
Pn = p(A+(A))
g
acting
dominant
on
for
with highest
E (C )*
PC= p(A(k)) E
C
WK
compatible with
Write
p = p(A+ (9))
Let
A+(g)
be
IK'
(tc)*
= P - PC
the eigenvalue of
and
A +(k).
w E W(k,t)
Then
(c)*.
the Casimir operator of
making
w(P-pn
60
<W(A-pn)
Lemma 2.7.2.
K-type
<
1)
6,
There
(00Ln) + Pc>
Suppose the Dirac
>.
some choice of
for
is a
inequality fails
A+ (p).
Then
V6
occurring in
ri
k-type
<P,P>.
> C0 +
with a non-zero, invariant
X E A(g,K)
Let
Hermitian form
on a
+ Pc'
® p
such
that
S,>V6
is
indefinite.
2)
Suppose
is Hermitian symmetric with a one-
G/K
compact center,
dimensional
with the property
that
decomposition of
g
z,
so that we can choose
g = k @ P+
into
the eigenspaces
ik 0
the
is
0,
+1,
-1
of
respectively.
Set
p
on
6
fails
V6 ® P
n
= p(A(p±)).
for
p-,
Then, if
there
is a
the Dirac
k-type
such that
>I(V
is
-
z E
indefinite.
6
V F)
rq
inequality
occurring in
61
Proof.
Recall
definition of
from Borel-Wallach [1980],
V
defined over
<
positive definite inner product
the unitary structure on
also,
(Y,H)
=
v E V
x,
the
,
>.
R,
finite
with a
Write
(
,
>
for
such that
-
<x,((V)y>
y E S(V).
the definition of the Dirac operator
D(ves)
H 0 S
H 0 S
:
(go 0 ko)-module and
a unitary
(2.7.3)
S(V)
(T(v)x,y>
D
for
§6,
the space of spinors of a
(7,S(V)),
dimensional vector space
Recall
II
=
S = S(PO).
r(Xa)v ® i(X-a)s.
aEA(k)
Since
m-V
A+
(where
then
m = 2 dim ac/2])
is
o(p-pn)
occurring in
V
+(k)
(cfr. Borel-Wallach [1984] II §6)
the highest weight of a
@ V n C H ®
S.
k-representation
62
f
Let
=
v ® s
Write also
product
on
be a weight vector for
,
<
H @ S;
D
for
the
G(p-p
tensor product
).
inner
then the proof of Lemma 2.7.1
shows
that
0 > <DE,Df>DD
(<W(p-pn)
So
Df
Df
0
+ Pc 'W(WPn)+Pc> -
c
0
PD'
-
and
=
(Xa)v ® r(X-a)s E p.V 6
S C H ® S.
aEA(p)
This gives a non-zero map
p ® V.
So
Hom k(l
V6, H)
X 0.
pov 6.
a
Let
positive definite this means
E = Im a.
that
(
Since
>
is
<
>5
is
indefinite on
V 6 @ E.
This
proves a)
For b)
p(p+)
and
of
the
simply observe
+
is a
lemma.
that
p -
p~n = p + pn;
representation of
A(k)
<p
,P>
= 0.
k.
pn
Hence if
1 E
63
So
V +
is one-dimensional.
Since
p
+ a
is not a
pn
weight of
S,
for
a E A(p+)'
V+
is killed by
r(X )
pn
and
(2.7.3) becomes, for
[ E V6
0 V
_
Pn
Df
=
r(Xa)v ®
(X-a )s
aEA(p )
so
(p
Df E
).V
@ S C H @ S.
Similarly for
p~.
q.e.d.
Lemma 2.7.4.
group.
Let
G
be a connected, reductive
linear Lie
Assume that
rank G = rank K.
Then, any representation with real
has a Hermitian
Proof.
G
of a principal
character
form.
By Proposition 2.6.3 it
lemma for
infinitesimal
is enough
quasisplit and a Langlands
series
I(6@v)
Hs
=T
the
subrepresentation
6 ® v
with
a maximally split Cartan subgroup
to prove
ss
A
a character of
64
Since
G
is equal
{ay, ... ,ak}
of
that ,
Hs
G,
since
then
B
Hence
strongly orthogonal
ref 1ections
simple real
roots such
is the maximally split Cartan subgroup of
spans
s
w
,
= Lie(A s
a
w = sa
if
B =
rank there is a subset
...
sa
is
acts by
the product of simple
-1
on
As
and by the
LL.
1
T .
identity on
Recall
nA(2,R)
--
from Definition 2.4.3 the maps
+
a
ino.
Consider
Sa
:
0a
the exponentiated map
SL(2,R)
>
Ma
set
1
m
a
=
0
0
-1
(cfr. Vogan [1981] page 172).
connected,
Ts
w E M'/M = W,
But
m
Ra
=m
then there is
Recall
-a
Then,
is generated by
O-m
=m.
a
=
M.
E
a
a
TM U
0
E M'I
aa U -1
since
G
is
(ma
a
real}.
such
that
w-a
a
the elements
6 E
M
and
v E A.
Then
Let
65
(w6)(ma) = 6(w-ma) = 6(ma
W-6
and
Hence
w5 = 6.
Since
I(60v)
infinitesimal character,
Also since
This
is
IT 0
o A = -1
V
=
6.
is assumed
to have real
is real.
then
-Ov = -V
= -V.
the condition of Proposition 2.6.3 for
the
existence of a Hermitian form.
q.e.d.
66
G = SL(n,IR)
Chapter 3.
Preliminary Notation.
3.1.
G = SL(2n,R);
To fix notation consider
is
the odd case
similar.
G = {g E GL(2n,R)
The maximal compact
I
subgroup
det g =
K = SO(2n,R) = {g E G
The
ko =
0
is
I
G
of
is
gtg
corresponding Lie algebras are
go =
If
K
1}.
{X E gL(2n,R)
I
trace
X = 0}
X + tX = 0} = oa(2n,IR).
X E
the Cartan involution defined by
then
p0
=
{X E 90
The Cartan subgroup of
K
I X _ tX}.
is
O(X) = -tX,
67
rr(e1)
Tc
10
E
IR;
r(n)
sin
cosO.
-sin
."
cos 0.
0.
1
l
and if
ri
ry
A
=
r. E [R det g = 1
1
<
r
n
rn
then
Hc =TcAC
is a maximally compact Cartan of
G.
0
-06
0
o
02
C
0
0. E Rl>
2
0
-0
n
en
0
68
a1
a1
a2
C
a0
a2
2 a. = 0
1
x
=
a.
a
E llR
n
a
n
h 0 = t0 + aC
a0
Complexifying:
g = ot(2n,C)
and
k = oo(2n,C).
Define
e. E i(t
j
0
-0
j
0 )*
=
1,2,. .. ,n,
by
0
0
0
e.
0
2g.
=
0
0
-0
Then
the
roots of
tc
in
k,
A
n
and
n
0
g
are,
ly:
A (k, t')
= f±( i ±ek)
1
jo..
1 < j < k < n}
respective-
69
A(t,tc) = {±2 ee;
(e
A(,,tc) = {±2 ee;
can be
+ e
A +(k,tc)
=
(a 1 a 2
. .. .
an)
Computation of
Ii =
Let
of a LKT of
(a,a
X.
n
as a
root
1 < j
in
j< k (
1 1
< k < n}.
is
2.
Then
n}.
K
a1
a2
>...> a n-1
>
IaI|}.
LV(X) for a Harish-Chandra module X.
2 ...
an) E i
be
Fix a positive root
a1
as
ek
1 < j < k < n}
e < n;
I1
± ek
{e
< n;
identified with the set
{p =
3.2.
1 < e
|
±(e ±ek)
The multiplicity of
Choose
±ek)
a2 2
the highest weight
system
A +(k)
so that
|a ni
...
in 3.1.
To obtain
LY(X) = LV(p)
2pc =
Let
p + 2pc
phism of
tion of
A +(qc)
dominant.
K
in 2.4.8 we need:
(2n-2,2n-4,...,2,0).
be a
8-stable positive system making
After conjugating by an outer automor-
we may assume that
A+ (,hc)
as
to
tc
is
an
0.
Then the restric-
70
A+ (g t )
Write
t.
{e.;
k' 2e±|
e
j < k}.
1 < j,k,O < n;
for the set of simple roots restricted
O(g.,tc)
to
Then
*(gtc)
=
{e1
-
e 2 ; e2 - e 3 ;...;
en-1
-
en;
2e
I.
Let
p + 2pc =
(x 1 x 2
.
n.).
We can form an array with the coordinates of
p + 2pc
by
grouping
them into maximal blocks of elements decreasing by
2.
is,
That
if
(3.2.1)
(a
,..
,a.1 ,a 2 ,...,a 2
r 1 times
where
a1
times
...
,a
rt
times
0 ...
R
0)
times
> a2 > ...> at > 0.
Then
since
the array would
(3.2.2)
r2
a
in1
m 1 -2
the coordinates of
2pc
decrease by two,
look like
...
m 1 -2p1 +1
m2
mi2 -2
...
m 2 -2 p2 +
2R-2,...,2,01.
.
71
Proposition 3.2.3.
K
of a representation of
weight
p + 2pc
a)
Suppose that
= x
has
E
and
it 0
is
the highest
that
the picture for
Then
is as in 3.2.2.
Xv(p)
p
the form
'
2
r1
2 .. x t ...xt ,0... 0)
'
rt
r2
R
where
X. = a. .
.1
i)
ii)
So
b)
t
=
if
> x 2 >..>xt > 0.
x
0
if
[yp()l d
oA (r
xt\ > 0
at=
xt
Proof.
1.
is isomorphic
,1 ) (D oA (r 2,C)
to,
(
(D...@
either
A (r tC)
(D
oA (2R.R),
or
A0....
(r 1C)
if
1
a
t(rt
,
) ( ot(2(R+rt
'
'
S0.
Notice that
2pc -
p =
(-1.-,...,-1).
Define
72
{
}I = {2e
.
I <p
{1n-}+=1{2en-j+l
XV(p) = p + 2p
Then
(3.2.4),
orthogonal
{e 1 -e
2
Moreover the
to
Xv(p)
; 2 e3 'e'
- 2 c P.
2
i J
- p +
-
and the conditions a)
are obvious.
> = -c.
+ 2p -p,2e
has
< 0}.
the
form
e) of Proposition 2.4.7
subset of simple roots
is
r -1~r
1
} U
{er
er1+1'..er +r2-1
r +r2
U...U {...;e n-1 en;e2d.
This spans
(Ar
r 1
since
A
the root system
)
r1
D (A r
r2
the roots
-
e
A
e
2-
) (..
A
(Ar -l @A r
) CD A 2 k'
rs 1
s1
2k
involved are
complex roots and therefore occur
restrictions of
twice
in
A(g,Lc).
Now
the proposition is clear.
q.e.d.
3.3.
Lowest
K-types of
the modules
We will give some criteria
representation of
modules
A
(X).
K
is
A
(M).
to determine when a
the LKT of one of the
(g,K)
73
q = t + a
parabolic subalgebra
6-stable
to construct a
Recall from 2.3 that
x E it.
we need a weight
Suppose
x = (x 1 . . .
(3.3.1)
x
x
.
r 1 times
where
x1
q = q(x) =
Write
x
as
x,
...
x
times
r2
t
L(x) + a( x)
for
0)
R times
times
rt
> x2
0,.
x
t...,i
>0
the parabolic defined by
in 2.3.
Clearly
(
2p(ank)
=
(s 1s
,0..
sy
..
r2
r
s. = 2(n-r 1-..
with
J
R
rt
-
.-r j.)
-
.0)
r.
-
1
(3.3.2)
2p(ulp)
and
=
(u
.u
...
r
I
u
r
= 2(n- r
u
...
u ,0 ...0)
R
r
2z
-..-
t
r+)1
r
+
74
Proposition 3.3.3.
ji
Let
be as
(3.2.1) and suppose it
in
is
the highest weight of a representation of
is
the LKT of a
(g,K)-module
a)
a
-
b)
at
> 2R + rt
ai+ 1
r
(X)
A
V
Then
K.
if and only if
+ r i+1
and
X + 2p(unp)
orthogonal
t C-roots
in the
tc
the roots of
to
u.
in
,
X = (X, . . . .
a
at
a. - a.
t
SJ+l1
a
+
= X
2(n-r1
x
-..
is
is positive
t
rt
.-
rt-)
r
r
.'
R
> 0.
t
..
2
+ 2(n-r -. .-
j
X
.'.'I'
,0 ,...,0)
2' ' '2'
x1
Then
=
That is,
r2
and
and it
t
in
Hence
and b).
satisfying 2.5.1 a)
L
character of
p
Then
(X).
is the weight of a one-dimensional
X
and
A
is the LKT of an
V
Suppose
Proof.
+ 1.
-rt
r. + 1
)
-
+
1 > 2R + rt
l)-
+
1
=
x j - x j+1
-
r
j
+
1
+
2r
,j
+
r.
j
-
1
>
-
r. + r j+ .
3
J+
75
4
Conversely, suppose
b)
is a weight
and
satisfying a)
then we can define
and
q = q(p)
A. = a. J 31
p
Then
2(n-r
-r.
will be the LKT of
) + r.
-
1.
A (A).
q.e.d.
3.4.
G = SL(nR).
Proof of Theorem 1.6.7 for
Suppose
X E d4(g;K)
infinitesimal character
is as
Y E
of a LKT of
highest weight
in Theorem 2.6.7 with
(/c)*
X.
Write
IL
Considering what the weights in
1.
2R + r
2.
at
+ 1
® t
the
in 3.3.1.
look like, we
> a.t
> 2R + rt + 1.
By the conditions given in 3.3,
A
then
(A),
Therefore,
in case
Case
V
as
(itc)*
study 2 cases:
will
an
p E
and
1
X
p
if
is
the LKT of
is in case 2.
the first
thing we must do
is not unitary:
1.
We will use the
V
following result.
is verify
that
76
Lemma 3.4.1.
+
r
1.
Le t
inequality fails
p=
be as
that
Suppose
Proof.
1
a.
1
for
s=
The hypotheses on
a i+
-
Then Dirac operator
1.
(n,n-l,...,.1).
p
imply that
(x+t-l,..x+t-l, x+t-2,...,x+t-2.
that
1 < x < 2R +
R + 1
(4)
..x...,x,0.. .0)
r2
rN
Note
=
11
- x
rt
implies
R
< 2R +
at
in 3.3.1 and suppose
and
R
that
R + r
t-
-
x
> -R.
Now,
Pn =
(n,n-l,...,R+rtR+rt-1,...,R+1,R,R-1....,.1),
so
pn
(n-x-t+ln-x-t....,R+rt-x R+rt x-1,...,R+1-x,R,R-1,...,1).
By
(*),
the sequence of
R + rt
- x,
integers
R + rt
-
x -
1,....R +
1
-
x
77
overlaps
the sequence
R,R-1,R-2,...,-R+1,-R.
Clearly, the first
n-R
coordinates of
pn
-
decrease by steps of at most one.
So
if
W E WK
is such that
then the coordinates of
w(P-p
o(p-pn)
in the
is dominant,
will be a sequence of
integers decreasing by at most one,
and
)
ending in
latter case, there must be
0
or
repetitions
+1;
in the
sequence.
Since
P
it
=
(n-l,n-2,...,R+1,R,R-1,...,2,1,0)
follows that
W(4-P n.P
' c
WA-P )(-n
Hence
<w((-pn) + pc
<
n' Pc>
n><<
n)+Pc
n' Pn
<
'.
Pn+p C'n+pC> =
Pp>.
q.e.d.
Now to prove nonunitarity
the minimal
1
for all
integer
i > i0 '
in
for case 1,
{l,2,...,t}
take
such that
i0
a.
1
to be
-
a i+
+
=
78
K = r 0+1 + r 0+2 +...+
Let
and
S=
ot(r 1+r 2 +...+ri
L C L
Then
rt
+ R
) @ ot(2K,R) = L@
and by Proposition 2.4.15,
Langlands quotient of
1
E
(I6
OvVV))
QVL (6
if
L2*
X
is
the
and if we set
XL = Snf[ILv(6V vY)),
then
X
is
the Langlands quotient of
L
gg(X
and a)
V))
of Theorem 2.6.7 holds.
Also, by Corollary
2.6.4,
h
XL
has a Hermitian form
< L
Write
pL
=
p
-
2p(unp).
Then
jL
is a LKT of
By 3.3.2,
2p(unp) =
(2n-(r 1 +.. .+ri )+l, ... ,2n-(r +..
r +r2+...+r
So
SL(2K,R)
.+r
)+1,0,...,0).
K
0
=
ISL(2K,R)
2
XL
79
and by Lemma 3.4.1,
the Dirac inequality
Lemma 2.7.2, there
is a
that makes the Hermitian form
The roots
A(L 2 lp)
in
V 2
K-type
<
V 2
in
>L
,
fails on
By
.
2
indefinite.
are
{(0.... 1,0... 0 +1 0.. .0),(0... 0,+2,0...0)}.
K
It
is
clear that
A(L 2 lp)
if
is also dominant for
j
there
=
a
+ P
is dominant
ai +1
-
A(nlk).
for
f3 E
some
p +
K-type
the
2
Hence
note that if
1...,.t,
Theorem 2.6.7
follows
a,
j+1
> r. +j+1
an
for
-
the LKT of
an
A (X)
and
to prove.
So, assume that
a. 0 - ai + 1
a.
then we hav e
is nothing
Set
2)
=
1.
For case 2,
all
C2
r
then, since
for case
K
there is
< r
+ r0 +1'
K = r
+ r2 +...+
rt
0
t
such that
and
t = at(K,C) @ oL(2R,R).
Note
that
o L(2R,R),
at
t ;
> 2.
LV
Hence
and
XL=
= OL(r 1 ,C)
and we can find
Theorem 2.6.7 holds.
UL
L
t
In fact,
IL ( 6 V®VV),
as
if
XL
G... (D
s.t.
oO = t n
in Definition
ot(rtC)
(a)
V
=
in
v + t n
2.4.17,
we
can
80
choose
to be
XL
XL
since
=
L
then, by Proposition 2.4.15
(XL
where
A(a) = A(v)\A(t)
and
L + U.
q =
By Corollary 2.6.4,
form
(
h
L
admits a non-zero Hermitian
>L
Write
XL2
X
where
XL
as the exterior
L
is an
XL.
tensor product
(Li,L fnK)
0
XL
L = XL
module,
1
L1
=
oc~(K,O)
By Theorem 6.1
p -
inequality fails on
2p(unp)
L2 = SL(2R,R).
in Enright [1979], and especially its
proof (pp. 518-523), if
Dirac
and
XL
the
is not an
lowest
and
1 =
M= 4
L 1
L1
A
,(X')
K-type.
then
Write
L =
81
By Lemma 2.7.2 there
V
(L 1 fK)-type
is an
with
1
r1
=
Ti
1 + P
for
(3.4.1)
Then
p + P
If
P E A(L 1 lPL).
a. -
for all
a i+1
+
i X
ri+1
0
2.
is dominant.
Otherwise
take
K'
r
2
=
with
iEB
B =
{i = 1,...,t
Then apply Enright's result
to
|
3.4.1 holds}.
the rest.
q.e.d.
82
G = SU(p,q)
Chapter 4.
Preliminary Notation
4.1.
n = p + q.
Let
m
for the conjugate transpose of
q,p
I0
K = {g E G
X E od.t(n,C)
=
A
0]
B;
0
BJ
[A
I
X
C
I .
is
G
A E U(p),
B E U(q)}.
0
p
0
E
= 0
+
0
B
=
q
notation:
the usual
with
g0
|I g =
of
K
Then the maximal compact subgroup
0
=[I
0
G = {g E SL(n, C)
Also,
the
We define
A.
matrix
A
and
GL(a,C),
for the identity matrix
I
Write
ot(nC)
I
-I
A E u(p).
q
D E
L(q)
D
skewhermitian;
-B *=
C
A
k0
=
E go
X
0
I
A E u(p),
D E u(q)}.
in
83
If
8
is
the Cartan
involution defined by
8(X)
=
-X
and
{X E
0
go
0(X)
= -X
then
L
I B
X
0=
B
.B
arbitrary
G
1o
Tc =
matrix .
0.
The compact Cartan Subgroup of
H c=
p x q
is
n
e
,e n)
g = diag(e
0. = 0
I
J
j=1
Then,
t
=
diag(i(
1
..... n))
E ,
8
|I
j=1
Complexifying everything we have:
= 0
84
9 = o L(n,C)
0"
k = o(gL(p)
gL(q))
= {X =
A E gL(p),
0
Di
D E g-L(q)
nn
t = {diag(z
,...,zn
zi = 0}.
j=1
K
can be
identified with the space
{1 =
(al , . . . ,ap
I ap+1 , . .
a p+)
correspond to
in
> an
j
ej E IR
If we denote by
the dual basis
...
Rn
.
then
,an) I ai
= 0
a
1, .r.. nf
=
aP;
a. -
aE
Z.
the elements of
the roots of
t
in
g
the set
A(9)
= A(qt) = {e, - ej
A(k)
= A(k,t) =
I
t 9 J;
-, j
1
n}.
Also
{eL - e
i
1
L,
J
p} U {ek
-
em
I
p < k, m
n}
85
the compact imaginary roots of
= A(,t)
A(l)
=
(+ (e -e p+j)
t
and
I
1<
e
p; . 1
t
in
g-.
the noncompact imaginary roots of
LY(X)
Computation of
4.2.
i
weight of a
A+(k)
so
=
(a 1 ,a2 ,. . .
aplb1 ,b 2 , ...
q}
for a Harish-
,b )
be
the highest
Fix the positive system
X.
K-type of
lowest
that
a
1
> ... > a ; b
>
... > b .
to obtain an explicit expression of
We want
parameters
XY(X),
and
j
X.
Chandra module
Let
g.
and
XV
L
attached
to
pi
the
by 2.4.7 and
2.4.8.
2pc = (p-l,p-3,....-p+lq-lq-2,...,-q+l).
Let
p +
2
pC
=
(xlx
2
'.
.
xp I3 1 'Y2''''Yq)
We can form an array of two rows with
the coordinates
of
86
p + 2pc
left
to
the
y.
so that
right as
follows:
the
are in the second;
right in
the array.
j
S >
xi
x
are
in the
first rows;
and terms decrease from
j+1
xi+3
>i+2
>
k
j+2
k+1
j+3..
'''
Xk
Xk+l
look like:
xi+l
i+2 ~~-*
y.y
i+3
yj+2
+1
This array gives a choice of positive roots
A+(g,t),
left to
For example, if we have
>i+l
the array would
...
they are aligned in decreasing order from
compatible with
A+(k).
roots are given by the arrows.
That is,
the
j+3
A+
=
simple
In the preceding example,
they would be
... ei
- e
;+j;
ep+j
-
i+1;
e i+2
...
ek
~ ep+j+
2
;p+j+2
e 1+3
e i+l
"p+j+1
e
''.
e k+l ;ek+1
-e i+2;
87
Because
the terms
in each row decrease by at
entire array is a union of blocks of
least 2,
the
the following five
types.
1.
2.
r
r-2
r
r-2
r
r-2k
...
r-2
...
r-1
3.
r-2k
r-3
r
r-2k
r-2k-l
...
r-2
r-l
4.
r-2
r-3
r-2k+l
r-2
r
r
Zr+1
r-2k
...
r-l
...
r-2k+1
I
5.
r
r+l
...
r-1
From now on we will drop
since
r-2k
r-2
the ordering of
r-2k+l
the arrows
the roots
is clear
r -2k-
in the pictures,
from the
88
arrangement of
agree on choosing
also
refer
the order prescribed
to them as |\
Example.
p + 2pc,
the coordinates of
p = 7,
Let
_
q = 8,
provided we
in block 1.
We will
etc.
and
= (2,2,2,-2,-2,-2,-311,1,0,0,0,-2,-3,-3)
then,
2pc = (6,4,2,0,-2,-4,-617,5,3,1,-1,-3,-5,-7)
p + 2p c = (8,6,4,-2,-4,-6,-98,6,3,1,-1,-5,-8,-10).
We obtain
8
6
8
6
the
following picture
4
p
(g
.g
1
r 1 times
where
r.
number of
is
as
-9
-10
8
p + 2p c
the picture of
coordinates of
=
-5
3
Using
-6/
4
-
we can split
the
follows
...
...
rt
g
times
the number of
f
f
f
' t
s1times
st times
p-coordinates and
q-coordinates making up
the i-th
f)t
t
..
s.
block
the
of
the
89
array of
p + 2p c,
and
gt
r. > 0,
s.
1
1
9t
9
f 2>
f1
i = 1.
0,
.
..
2
-
t.
Write
X
-2= V1)
in Proposition 2.4.7
as
Proposition 4.2.1.
1.
=
XV(X)
in chapter 2.
The expression
for
has
XV(p)
the
form
V
I
times
r 1 times
rt
times
t
s
times
where
X1
2.
Let
> X2
t*
t
sttimes
t
90
A(Ly)
=
A(Lv, tc)
=
{a E A
I XV, a
(@EA(L
xa)
=
0}.
Then
tv(X)
V)
=
Example.
r1
Note
r3
Let
@ u(r t,sd)
...
In our current example we have
,
(2,2, 2
=
o(u(r 1 ,s1 ) (
c
r2
-2,-2, -2, -3
r4
r5
,
| 1,1, 0 ,
r6
s1
s2
s3
0
,
s
-2,
s5
-3,-3)
s6
0.
=
X\
= p + 2pc
=
-
p
(1,1,1,-l,-2,-2,-3|1 2,2,1,0,-i, -2.,-3, -3).
Then
1
V2
r
and
tv
(X)
r2
-1,
r4
-2,-2,
-3
r5
r6
~Y2'2'
|I
,0
s1
1
s2
, 0 , -1,
s3
s
-2,
-3,-3
s5
s6
91
Proof.
Take a block of the
form 1.
Then coordinatewise we
have
I...
p + 2pc
=
p = (...
,s,s-2,.. .,s-2k,... I...,s-l,s-3,...,s-2k-1,...)
(...,r,r-2,..., r-2k,...
x V = p + 2pc
=
-
,r,r-2,...,r-2k,...)
p
(. ..,r-s,r-s,...,r-s... |
... ,r-s+l,r-s+1,...,r-s+1,...)
Let
e m-e p+;ep+em+1
be the set of
..
;"m+k e p++k}
simple roots making up this block.
I i = 1......k+1}.
-- e
.
{03i = e m m+1-1p+e+i-1
Then
-c.
1
=
->
V
1
=
-1
.
Choose
92
Also
<,6
> = 6
P
j
and
c.
V
Let
{(1.....
13
noncompact
of
be
0.
i
the
subset of imaginary
r
positive roots chosen this way from all blocks
the form 1.
subset of
}
k.'...'Pk
1
i
2
For blocks 2 we can choose the
simple roots.
same kind of
In this case
V
c. = -IV
,.
=
0
since coordinatewise, we have
p + 2p
=
(...,r,r-2,...,r-2k,...|
... ,r-l,r-3,...,r-2k+1,..
p = (. . .,s,s-2....,.s-2k,.. .I ...
Vy =
A
..
similar
,r-s,r-s,...,.r-s,
...
,.s-1,s-3,...
|...
situation occurs in cases
s-2k+1
,...)
,r-s,r-s,
3-5.
Choose all
the
93
simple roots
For
of
the form
cases 4 and 5 choose all
For all
clear
is
is
a set
involved in blocks 3.
those of
the
form
ep+b
of
that
= -<X
(,9i
the union
>
IT
of
strongly orthogonal
V
these 5 kinds of subsets
simple imaginary noncompact
+
2
V
c.fp..
1 1
Pi Ell
-
d)
of
Proposition 2.4.7 are clear.
Let
A+
=
a E A+
ja,p
> = 0}
A+ \{eiem;em-eji;e
A
i-e
;e +,-e.}
im m
i p+e
p~ep+e
1 =
0
then
t
ea.
= 0.
Def ine
roots.
a)
-
these subsets
c
It
ep+b
-
ea
1
{(x
can be
. . .
0
identified with
+x
x p Ix p+1'
the set
n ) E t 0 IXm =Xp+L
p
==
94
So if
i
1
=
= A Ig
it
a +b
=
(a1.,
To verify e)
. . ., . 2
a m+b
2
,apb..,...,
,
,..,b).
of proposition 2.4.7 in chapter 2 we use the
following lemmas.
Lemma 4.2.2 (Schmid [1975])
1
+
PC = p(A (k
1
(see Vogan [1981] p.
1
, t )),
A +A (g 1 ,#L 1 ) = A (g) n
13
and
+
11
p 1 = p(A+(g.
, h ))
then
1
C
So
1
=
(2p
)
'
247).
If
95
1
~"1
1
1
x=
2pc
+
1
-
p
V it1
=
(Vogan [1981],
Lemma 4.2.3
e)
So 2.4.7,
Now it
4.3.
1
-c
2 11
+
p
p. 249).
1 + 2p 1
is dominant
c
).
A + (q ,
for
V
is
is clear.
straightforward to verify Proposition 4.2.1.
The Lowest
K-types of
the Modules
A
(N).
In this section we obtain necessary and sufficient
conditions
of
the g-modules
If
parabolic
replacing
it
for a representation of
is of
x =
(x
to be
the LKT of
one
A (N).
, . . . ,oxn) E i(t
subalgebra
x
K
q(x) =
)*
L(x) + u(x)
by a conjugate under
the form
we obtain a
as
W(KT),
0-stabl e
in 2.3.
A f ter
we may ass ume
96
x = (x
1 ,...,x
,x2'''
p1
x 2'
Pt
P2
x
,
x
..
,
x
..
t''
qq
Et
such that
x1
> x2
>'
Pi ' q j
t ';
>
and
Pi = p
q
and
-0
= q.
Then
L(x) =
o(u(pj,qj)
2q
( u(p
2
)
.. E
D t't
Clearly
2p(aflt) = (rl,...,rl,r 2
2
S.
s
,...,s
q1
and
,1s
,r
.....
,rt ... ,r
...
1t
,...,'s
q2
,
,...s , ...,s
q t
)
'
t
97
2p(ank) =
(sl,...,s
,s2....
s 2,.
. ,st,...,st
p2
rr,...,r
,r2 , .
q
where
-
+q2+..
=
S
= -(pl+p
q n
Wri te
pi,
and
their
2
p E L(t 0c)
represen tation of
.t,...,.rt).
+.. .+p
be
qt
.+q_)
_)
+ q+1
+
+ pj+
+
1
S+
qt
+ Pt.
the highest w eight of
a
K.
k = (qfk)(W).
= p + 2p(ufk).
p.'
r2,.
q2
r
Let
Pt
co mpact parts
2p(unk)
as above,
(Note that
for
Q(W) A 0'(W')
but
coincide.)
Wri te
' = (z,...,Z ,Z 2 ,...,z
k
2
,
. . . . . Za,...
k2
za
ka
Z ,
..
Z
aa
..
,.
,...,.
)
............................
Proposition 4.3.1.
e..
n
n.
1
Then
+1
+
n..
i+1~
p
In the above
is the LKT of an
setting,
A (N)
iff
let
z.
n.
1
= k. +
1
i+1-
98
t c-roots
the
is positive on
t
the roots of
to
is orthogonal
X
b).
satisfying 2.5.1 a),
L,
dimensional character of
Then
in
L
in
and it
a.
By Proposition 2.5.6 and its proof we may assume
p
q' n
determines
and if
a(p')
k
and
p'
determines
satisfies 2.5.1,
X
X = (X 1 ,
, 2'
''
k
..,
,X
1'
C'.
So
then
x 2'''''
a'''''
k2
1
a
ka
..
a
xa
.2'
aX,
a
2
with
x
p
that
Suppose
z.
= N
=
+
X
+
1x2
>..>xt'*
2p(aup),
p'
then
(-n -n2-,..-n
)
X
=
+ ni+1 +
+ 2p(u);
...
+ nt
and
z.
I
-z
i+1
L + a
the weight of a unitary admissible one-
(tc *,
X E
and
a =
some
for
w = X + 2p(ulp)
Suppose that
Proof.
=
i
i+1
n.
1
+ n.
i+1
n
i
+ n
i+1*
that
'
=
99
On
the other hand i f
then let
o'
X.= z
p
+ n1 +
satisfies
z. - z
...
-
1+1
I
+ n,_,
+ n
> n.
-
I
and
1+. .. +nt)
(n+
= q(p'),
i.e.
'=
{a E A(g.,tc)
I <p',a>
(a E A(,
= A(U')
Then
I
tc)
> O} U
<p',a> = 0}
U A(L').
if
,
9x
tI
.t9.
k
1
1
t
> 0
<X,A(L )> = 0
and
p = p'
-
2p(ua'k) = X + 2p(u'np).
q. e.d.
1+1'
100
4.4.
XV(A (X))
The parameters
Aq (X) = g
By definition
since
By definition 2.5.1,
highest
weight
ji
= X
+
By Lemma 2.4.23,
first
X
(M))
s = dim a
where
is perpendicular to
The LKT of
A
n
k.
the
has
(X)
2p(unfk).
GL
X v(A ())
LV(CX) + p(a).
=
Assume
that
N = (0.. .010.. .0)
(p-1,. ..- p+l|q-l,....-q+1)
X + 2p c
picture (say
p
which gives a
q)
p-3.
p-i
L = g
and
q
p
then
tv (A
(C.),
X E [center(L)]J.
L,
roots of
and
q-1
q-3
...
-q+1
q-1
q-3
...
-q+1
p+
(if
(4.4.1)
p = q+2k)
or else:
p-l
/
p-3
...
q-4
q-2
q
q-1
q-3
...
-q
-p+
...
-q+1
(if
p = q+2k+l)
(4.4.1)
101
Then
p-q-1 p-q-3
2
2
1 0
2
-p+q+1
1
2'''.'.2
0
0 .q
q
(p
XV
(C )
= q (mod
2))
=
p-
0..
-
0
0 ...O ;
2
q+1
q
(p
So
X =
if
give
(a,...,ala,...,a)
and
t
E q+1
= g,
(mod
2))
will
X + 2pc
the same picture and
Xv(ex)
=
[a +
2
,a
+
+
a +
2
a.. .a,
q+6
1
...a
S ..
p
6
q+e
= 1,0.
q
Now suppose
a1
a2 2
..
a a taI
t =
a) . .(G
...
(
t'q
t))
102
and
X =
(a 1.
...
,a1 ,a 2 ,. . . .,a 2 ,. . .
p1
. at,...,ati
p2
Pt
al,... ,a ,a2,.
. a2
...
,at,...,ad
q2
1
-
qt
Clearly
2
k
=
(pl-1,....-p.
+,p2
'p t-1 ,
'
,-p t+1|
q -1....,-ql+l,... ,q t-1
So on the
(p
,q )-coordinates we have a similar picture
and
xv c)
I.
.a.+
..
-
p.q.i-1
1+6.11
1+6.
1
1
Il
,...a.+
,a....a,a...
2
2
.. .. a
. ,
1 - 1 a -1 2 2
a
q
-
(p.q.i-1)
1
1
2
+61
1
P.
...
|I...
a .... a .i... I
qi
if say,
p.
picture for
= q.
+ 2k + e.,
1
X + 2 pLflk
6.
1
= 0,1,
k > 0.
Also
the
will have pictures like (4.4.1) or,
103
if
q1
p
. . .
z-1
.
y
z- 2p 1 + 1
z-2
z-2p-
..
- 2 p-
...
3
..
(4.4.2)
y y-2
...
z
z-2
...
z-2p
+1
z
z-2
...
z-2p
+1
A...
Now
p(a)
=
(u
,...,u
,...,Ut,...,ut|u
..
,..
Pt
,u 1
.ut,....u
t
qt
.11
So
Sd.
(u(1)
LY(Aq(X)) = (N)1
d.
1
,s)
x u(r
X
((1))
]
t
C
'.
1=1
((U(Pitq i))
where
r
( F
=
1
if
p=
q
= min(p
,q
+ 1 (mod 2)
+
1)
and
.
p
> qq 1;
= 0
104
otherwise)
s. = min(p.,q.)
+
(6. = 1
if
1
p.
q. +
1
1
2d
4.5.
i
+
1
and
r.
I
+
1
Proof of Theorem 2.6.7.
Suppose
X E d(g,K)
infinitesimal character
=
1
coordinates of
=
i
p.
'
2
Y1 >
2>
p E
(Cc)
X.
'
' '
...
t'
.. xt
Pt
y
,y 2
q1
> x2
and let
than that of Section 4.2:
(x 1. . . . . x 1 , x 2 ''
x1
1
slightly different splitting of the
y9 ...
that
q..
G = SU(p,q).
),
p2
so
+
1
otherwise)
1
in Theorem 2.6.7, with
r E (c)
the highest weight of a LKT of
Let's consider a
1
for
is as
6. = 0
> p ;
q.
s.
6.
' .y 2 . . .
. . . ..
q2
>
t
s
'ys'
ys)
qs
be
105
p , q
but here
necessarily
It
of
p
is
compatibl'e with
convenient
splitting is not
this
that is,
the blocks
given by
to draw a picture of
with the same blocks obtained from
In
the
example
11
But now,
of
section 4.2
(2 2 2
-2 -2 -2
-3
Pi
p2
p3
p + 2p C
means
p
is
|
1 1
0 0 0
q1
q2
-2
-3 -3)
q3
4
first
W.
We may assume
that either
x
+ p-1
> y1
+ q-1
or
x
+ p-1 = y 1 + q-1
otherwise we can interchange
-3
-3
to study what happens around the
coordinates of
coordinates
the
-2/
the splitting of
We are going
2pC'
-2
-
0
0
this
p +
-3
A
2
[
p =
> 0,
and
p
and
pl
q.
qi,
p1
106
If
for
< p
p1
we can have
the following configurations
p
p1
1.
x
x 1**x
x
Y1 ...Y
y q -1
2
x2
'.'
'
'
Y2
p1
or
x
x1
x2
Y
11
1
2.
xl
-- - y_1
where
y.i-i
3.
...
xl
*...
with
x 1
x
...
y1
i-i
...
> y.1 > z.
x1
x1
_
.. y i-i
y.1i
z.
y.i-i > y.1 >
-
...
x1
x2
...
y.1
z
''
'
1
...
Y
x
x2
y.
z
107
4.
xl
x
yi
y1
x
5.
>
1
x
x.
...
x.
z
y
...
y
y2
x
...
> z
j
. x
..
x.
z
...
y1
y
> x
x
x
1
..
y2
y
> z.
The blocks
in these pictures are simple factors
of
LvX).Note
in case
that
1.
ji
if
is
the LKT of an
So we have to check
reductive subgroup
L C G,
tion of
A
A
(X),
Dirac
the representation of
the indefiniteness of
that
o
must be
X
as
the Langlands
in the case of an
the He rmitian form is preserved
the derived functor, and in the case when we don't
have an
V L
we
coming from a representa-
but in such a way that,
the signature of
(N),
under
L;
(X),
we can find a
that
and embed
submodule of a Zuckerman module
A
(Lfl)
7L
inequali ty fails on
L,
and
the form
the
on
with highest weigh SIL
+ 2p(unp)
is dominant
p
L
,
the LKT of
(LfK)-types involved in
L,
and occurring
= AL +
P
in
will be such
108
On the other hand,
non-unitarity.
in
In each case a group
1, making sure
All
for cases 2.
that a)
this will
-
c)
-
5. we need
L
will be
to prove
found as
of Theorem 2.6.7 hold.
reduce the problem to
the case
p1
=
p.
In this case we have two configurations
6.
x
x
1
1
yI
Yi
...
x
...-
y
***
x1
''
k
yk'
x
Y
Case 6.
x
z
...
> yk > Z.
x
7.
.***
x 1...
...
can be
y
x
...
ys
x
.
x
Ys
included in either 2.
or 3.
and case 7. will
be dealt with similarly.
Note
that as soon as we have shown that a)
2.6.7 holds,
then by Lemma 2.7.4
of Theorem
the representation of
in question, as well as its Hermitian dual,
L
have a
Hermitian form.
For
is
1,
either
Then
=V
-
Lv
+
t = o(u(pj,q a)
let
q,
or
L D L
L n av'
a'
a)),
here
qa
0.
and by Proposition 2.4.15, if
= L
n
109
DE[
q
Assume
that
XL
Od0
=A9, (X
(X ))
QV
Ly
~ ®(XL
)
)
LV
A1 (N).
Define
Q2
by
a2
a + au
1
%2 ~ 41 + a
a1
+
t
=
a.
+
Then
(A
O
1d
ck2
Proposition 2.4.15 again.
To
prove
see
that
that
9 2(C.)
(X,a> > 0
is a module
a E A(a
for all
A2(ex)
2
).
we need to
But, by Lemma
2.4.22,
xV
-
X + p(a)
+ P afL
=
X
+
+
p(a)
+
Hence
By
A (XL)
Proposition 2.4.16,
is
N + P,
and it
the
p
-
res
res
infinitesimal character of
is regular.
110
Hence,
is dominant
X\
res
+ pLre,a> > 0
KX+p,a> = (X
particular,
a E A(u
if
<XV+P Lres , a>
2
=
X + p
4-*
for all
is dominant
a
positive.
In
)
0.
<presa>
a> +
(X
Now
2p(nlp)
=
(q-a**q-q a'-q............,l-,......P-P1 'P
l '
1
Qap''1
'''9a
p-p 1
Pi
P-Pi
qa
1
9
9Pl)
l
-a
Set
p=L
i-2p(unp) =
(U
,...,u
,...ut,...,utivi'''''
Pi
with
u1
> ui+1'
By c)
LKT of
j
Hence, by
p1
< p
q1
s,.
, s)
qs
>vj+
of Lemma 2.4.23
the module
Since
Pt
1'''''
pL
is
the highest weight
of a
XL.
then
L A G
induction, Theorem 1.3
and
dim L
implies
< dim G.
that
there exists
111
an
L
n
K-type
in
V L
V L
11
77
o
V L
(Lfp)
such that,
on
1~1
77
the Hermitian form is indefinite.
L
p
are
p-
U {+(0... 0 0.. .0 1
if
0.. .0 0.. .0)
qa
1
0.. .0
q
qa
qa =
4a
10.. .0 0.. .0 -1 0...0)
P-Pi
Pi
the roots
0.. .010.. .0 -1
±(0.. .0 1 0...0
A(Llp) =
n
L
the weights in
Now,
a
1
or
A(Lflp) =
(0.. .0 0.. .0
P-Pi
p1
if
1 0.. .010.. .0 -1 0.. .0)
q
= 0.
q
A highest weight of an
then of
the form
candidates
pL +
for
L
n
K-type in
some
C
V L
t
A(tnp.).
for highest weights are the weights
o (fl)
So
the
is
112
L
.,u
77~ (
3
+1,u.,...
u....
.
.V
p.- 1 .1k
p
x p1
k q
V
-k
V k-
... )
q -
qk X qa
or
L
TL =
(u 1 +1,u
...
p1i
u
.Iva ...v a
1',..
v -1...).
a
qa -1
p1 -1
So
T7 =
(x
1 .
.. x 1 ,...
x .+1 x
x..
yk ..
k Yk 1
'..
p .-1
or
Ti = (x
+1
x.
...
p-
,x1,x
2 .'.
1
lyl.
.
.Yi yl -1 ... )
1 -
qa
are dominant.
This
completes the proof of Theorem 2.6.7
for
this
case.
To prove the
following
lemmas.
result for cases 2.
5.
we need the
113
Lemma 4.5.1.
G = U(m,m)
If
the Dirac operator inequality fails
then
pPo = (,i.t.e.
Proof.
Write
And
=
pI
(
1
a
as
W
1
,...,
|
..
pc
+ As
1 '
,..
m
W(p -p
s
n
)
+
PC
C(p~O-pn)
=
is a
with
,
ps
E
)
d =
(x...xjx...x)
=
m
m
m
(0,-i,...,.-m+ljm-1,m-2,...,.1,0)
+ p
+ c
(OI-P+)+p >=
n
c
<c'
X
for
(Cfr. 2.7.1.)
I,
E (center 9)*
Sc
If
p = (a+1,..
and
c
+
(g,K)-module with
s
-n~c'
n
C'
infinitesimal character
then
(-,-(>
And
(
s
+)+p
nP
> <P'c4 c> +
W
P)+p>
(p, p>
<
(p,p>.
q. e.d.
114
Lemma 4.5.2.
If
w
G = U(m+1,m),
=
(b+1,...,b+1lb...b).
m+1
Then Dirac operator
inequality
m
+
fails
-m-1
c
as
p =
m
Write
4-1
p s s=F-:M.
1'
2
and
for
m -m-1
m+1
Proof.
m
=
m+1
b + 2m+1
s + 4
''
m+1
2m+ 1
b+ m+1
'''''2m+
1
as
m+1
2m+l'.'.'
b+
-p
+)
(
same argument
to show
-P )+p > < (Pp>
but
pc
~
m m-2
2 2'
-m
M-1
m+1
2m+1
By the
in the preceding Lemma we only need
(
b+
M+1
2m+1''
in Lemma 4.5.1.
m+1
2m+l
-m+1
.
p = (m,m- 1 ,...,-m
2
)
that
115
P
P+
n +
c
11-s
.,-m+1
(1,0,-i,
=
-m+1
-(m+1)
2m+1 '
+
.. 2m+1
lemma.
the
this proves
m,m-1,..1)
q.e.d.
for each case 2 -
Now,
so
5 we will
choose a subgroup
cases discussed in
the problem to the
that we can reduce
L
Lemmas 4.5.1, 4.5.2.
2.
the
Remember that we have
xi ,.x
1
following picture:
x 1...x
x2
y. i
y.1-1
yy.
Z
-y
s
r
where
y
_,
t
Choose
L
obviously
=
Let
= UV -
a' = u V n L
UI
L = Normalizer of
Let
=
_ )@u(r,r)Ou(p-pl,s))
o(u(pl-r,ql+...
C t.
U
L
> z.
> y
Q =
and
q
in
G,
t +
U.
then
S(U(p 1 -r,ql+...+q_ )xU(r,r)xU(p-pl,s)).
116
By Proposition 2.4.15 there
of
such that
L
is
X
XL
the Langlands subrepresentation of
Adim
the standard module
is a representation
Q
unk
(XL).
Now,
2p(tLnp)
=
(a,....a b,...,b
r
p-r
By Lemma 2.4.23
LKT of
C,....,C |I
p-p1
e,...,e f,...,f).
d,...,d
ql+...+q
pLL = p-2p(anp)
is
_-1
r
the highest weight of a
XL.
We claim now that
pL U(r,r)
=
(x+l,...,x+1|x...,x).
r
In fact
s
r
117
PLIU(rr) -
-ql-...-q.
= (x 1 +q 1 +...+q
_+sjp-p1 +r...
_-s,...,
x 1 +ql+...+q._-sly-p+2p,-r,...,y.-p+2p,-r).
We know from the picture for
+ p -
x
2p 1
A
+ q -
= y
that
2(q 1 +...+q
y. + s -
So
x
+
q, +...+
,_,
-
s -
y
+ p -
(q
+.. .+q
2p,
V L
D v L
indefinite.
Now if
L+
necessarily
P = (0.. .0 -1 1 1 0.. .0).
3
j
+ 1.
1.
p E
>L
such that the Hermitian form
is
1 =
_1+r)
+ r =
Hence, by Lemma 4.5.1 and 2 of Lemma 2.7.2,
A(u(r,r)fpl)
+
_1 +r)
on
is dominant,
118
Also
A + P =
is dominant
(x,
for
. . . .,x 1 ,x 1 -
1,x 2 '...
1y1 '''yi- 1 yi+1,y....)
A(unk).
This proves Theorem 2.6 .7 for case 2.
For
3,
the picture
that we had is
p-p
yi-1
x1
x1
-x
yi
yi''
.y
d
Let
x =
that
d
p-p 1
q
in
G,
s
if
that
q-d-r
S(X)
in 2.3,
and
L =
--U(p1- r-1,d) (D u(r+1,r) (D u(p-pl,q-d-r).
t
/
Lv.
= dim u
(q ,H
Y
V
r
c) E it*
0
then
However, Proposition 8.2.15 of Vogan
[1981] p. 545 gives us that
where
I a.. .a b.. .b C...
0-stable parabolic subalgebra as
t
Note
q-d-r
r+1
Normalizer of
> yi > > z.
i-1
z
b.. .b c.. .c
pl-r-1
define a
x2
r
(a.. .a
1
then
fl k.
X
V
V
v7 )
Vf
LS
(XL)) =
In fact, all we need
is
the
to check is
0-stable data attached to
V~
L 3 Hv
s,
[
Y =
and that
q
contains some Borel
V
subalgebra of
This
qV.
is clear by
infinitesimal
the picture
character of
Y
is
+ 2p
of
-T
=
(X
.
Then the
v) E hV.
V
V.
Since
119
r + 07 =
for
(2 Xv,0),
As before,
a E A(u).
2p(unpl)
pL = p -
that if
it is straightforward to verify
(a+l,. .. a+1Ia.. .a).
r+1
Lemma 4.5.2 and 2) of
,
<
>
,a> > 0
then
pL U(r+1,r) =
Hermitian form
<X
enough to show that
it is
is
r
Proposition 2.7.2
indefinite on
imply th at
V L
( L +P
the
with
P = (0,...,0 -111,0,...,0) E A(u(r+l,r)fpl).
p + P
Also,
is again dominant
for
A(unk).
Hence Theorem
this case.
2.6.7 also holds for
4 and 5 are solved in exactly th e same way as 2 and 3,
p
using
So
+
and
reduced the problem to
I have
the case
pi = p.
6 can be
included in either 2 or 3.
For 7 write
the picture for
W =
W
is
(a a.. .alb1 .. .b ,b2..b2...bt...bt
120
r1
q 1 +6
a.. .a
a.. .a
b
with
e. = 0,
J
r2
qt+st
a.. .a
.
.b
rt+ 1
a..
t...a
t.
.a
t
1
t+1
t
r. +
p=
q.
6
+
.
1
t
qj-
1
Then
=
(a.. .ala+s
...
a+s
, a+s
1 q
2
..
.a+s . .. a+s
...
a+s
qt
p
sk = -(r 1 +.. .+rk)
+ (r k+1+. ..+r t+1)
-
In fact,
(
1+62+..
.+F k1) +
from the above picture
for
p,
(ek+1 +. .. +6
td.
we know that:
)
121
a + p-2(r +...+rk)
1
2(q 1
-
+61+
+ 6 k-1)-1
= bk + q-2(q 1 +...+qk-1)-1+Ek
k
Sbk = a + p-q -
k-i
2
r
1
= a -
6
2
j -
(rl+...+rk) + (rk+I +...+r
t+1l
(61+...+6k-1) +
Note
that
assume
if
q
= 0
Fk
1
for all
i > 1
(Fk+ 1 +...+Etd
this
.
is case 1.
So
L
to which I
can
q
+
t > 2.
As before,
I want
to find a group
apply some reduction argument.
(*)
Suppose
Then
let
> r1
r t+1
L = U(s,q
set
s =
r1 +
1) x U(p-s),q-q1 ).
U(r1 ) x U(q 1 +el,q 1 ) x U(r
x...x U(qt+Ft'qt
2 )
So
verify
L 3 LVs
a)
61
Note
)
+
r
that
LV
x U(rt+1)-
and again, we can use Lemma 2.4.15
to
of Theorem 2.6.7.
2p(alp) = (q-q1 ,..q-q1 ,-q 1
s
.. .
p-s
-q
Ip-s ... p-s,-s...-s
qi
q-q
1
122
Y + (N
By Lemma 2.4.22 and 2.4.23 if
infinitesimal character of
is
the infinitesimal character of
In fact, by definition of
(X Ia> > 0
(4.5.3)
Write
= U(s,q),
L1
L L1
For
=
before;
restriction of
However, this
r2 .
=
V
XV(p)
C A(uv).
regular
a+sl-p+s).
it could be possible
the minimal value of
the
the Cartan of
Lr
integral.
is not possible for all
involve all
values of
instead.
v
of
then
(x,x-l,...,x-r 1+1,w,w,...,w,x-r - 1,x-r -2.. .z
r_1
to
inequality as we have done
to the split part of
Therefore, we need to
If
x
r 1, r2
by simply using
V
a E A(u)
,a-q+qlja+s1 -p+s ...
of Dirac
-TLIL
that makes
A(uv),
for all
some values of
that is,
XL'
then
(a-q+q , ...
the failure
prove
(Nv-P(L),v)
L
then
9 (XL),
is the
,V)
q 1+e1
x-r 1....x-r
q1
...
)
r
,1
123
s
where
x = a
Hs
If
and
=
D
+
+ (n-1)
=
=
2
p
1
+
r1
1+6
-
1-1
2
r
.
Cartan subgroup of
1
1.
v
v
2*
1
(Xv, D)
1
0..
q
regular
2
0 v 1 . ..
2
0.. .0 |
62 +r
2
1
-1)
1
3
2
1
1
,
-q2
a+s +6 -1+r1
2
Ir
=> V 1
2
a +
a-(s
s
2
V1
1
si
> max[2i
+e1 -1)
2
= -r 1 + r 2 +...+
By (+*),
...
integral we need:
6
s
Lv
then
(0...0
0
+ v
-1
2
s
= a
2
-v
a +
+6
+
2
61 +r
To make
a
is a maximally split
s
vyEA
(n-1)
-
= a -
TsAs
=
1
s1
w = a +
z
-
+ p
r
+
rt+1
0.
S0
62 +..
1)
.+
> si +
Et*
611
+
r 11.
r
-1
1
, s +72+
r
124
vl=~ 1
Let
+
6L-
V.
Now,
p (U)
+
2
+ r1
=
(XV-P(U))jL
+ j
J1
n-s-q 1
n-s-q
1
2
'
IL
V
2'
'
'
n-s-q
2
n-s-q
'''''2
2
=
s1
a+ 2
-1
s+q
2- Ir+
1
-n
2
s1
-
,..a+T
2
1+
s+q2-n
2
'
r1
s
s+q 1 -n
a+2
2
s
s+q
2
1
-n
2
q1
s
+
6
-1
2
s+q -n
2
s1
,aT
s+q
-n
2
s
1.
.
+
s+q
n
2
-2
1 +r2
We would
like
to prove that
-n
s
...... ,a+-
s+q 1 -n
s1
a +-2+
s+q
the
K-type with Highest weight
1
2
2
125
=
(a+1,a,...,a
p-
a+s
1
a+s 1, a+s
q,
,...
-1,a+s 2 ... a+s
q
in
...
a+s
the representation
X
and
that
the Hermitian
indefinite on
form is
V
It
t
24
q1 -l
occurs
a+s
is enough
to prove the
11
@ V.
77
failure of Dirac operator
inequality on
p1
L
1
and
p
(t
) = p(Lfk) -
2
''
s
q
So
L
IL
P
1
q1
a+ql-q+2,
Since
q1
s
s
. . . , a+ql-q+-2 a+s +s-p-,...,a+sl+s-p-
)
126
s-1
=
s-3
'''
L
IL
A
(
1
1
-s+ 1 ql-1
2 ' 2
L nk
2
q 1 +s-1
,...,
2
a+ql-q+
and
-
(-da))
2
s-q
a+s-p
Then if
X =
+1
2
( v-PG
)) IL1
,
6-
1S
s+q1 -n
2
.
2+
y = a +
+11
q 1 -s+1
s+q.-1
a+s 1 -p+
2
Let
1
2
=
) + Pc(L)
a+q 1 -q+
-q
2
-
+
1
6
2-+r1 ,y+
-1
6-1
r 1 ,..y
2
r1
y Y+
y ....
q1
If
w
E WK
6
6
-1
2
2
61
such that
-1
65...
y+
2
2
r2
wY
is dominant,
then
y .
y)
127
-1
6 -1
- 2 ir 1 + +ql-1... y+s 1+
2
1 +5,
6
+
(0'r, = + y
1
6-16y+ 2
-1-1 11
'r1 ,. . ..
1,y+
2+
2
2
6
1,...,y+
-1
2
2
r2
r
y-s - 2
r 1 -6,y-s1 -
r -6-q
2
+1J
Also
4 L
) + p+ flk
i +qI,..,y+
y+ 2
2
r'' +1,
1
6
y+
-2
-1
6
r
,
+
-1
6
-1
6
2
r
r2
2
To prove what we want
-1
,+
-
it
is enough
-
to prove
2
that
2
128
(~EA4)
< l' W
r1
< IL i
>
p
+
)
L
Pn
ILi
But
this
P
k'
+ Pt
)
1
is equivalent to
Ity+si
j=1
+ 2 1+
1
+-+2
r 1+6+ji-l1
2
[
Using that
Ic I > |d I
-1
- 2
6-
6
+
Lys
2
-
6
2 1'r 1+j
(b+c)2 + (b-c) 2 -
(b+d)2
-
we conclude that (**) holds
6-1
s1
+
2
6
ir
1
+6+j-l
-2
r 1-j
(b-d)2
> 0
-1
>
then
j > 0.
6
6
1
> 0
+ j -
Now
and
s1
> 0 * s1
+
2 t
1
if
.
+j
H ence
-1
+
2
+ry
1
+ j
6 +
e 1
-1
2
> 0.
if
6
> 0
Since
j -
> 0.
ink >
>
21
2
1
'
-qt)
1.
if
r.
> r
t+1,
L = U(r t+q t+e t+rt+1'qt)
and repeat
the
we choose
x U(p-(r t
same argument for
this
t+F t+rt+1
case.
+
129
Note
6tI
that,
since
s
+ r
= r
...
2 +
0
or some
So
r.
1 <
> 0;
this reduces
j, i -
rt+1
= r t+1
r1
then (**) will also hold if
0t
+
+
62
+...+
and some
6.
>
t.
to the case
r
ql+e 1
a.. .a
a.. .a
1
q2
r
a.. .a
b
b
2
...
b
a.. .a
q,
q2 > 0
2
q1
But
conclude
But
by symmetry, using
that
the case
r 1 > r t+1'
we can
e 1= 0.
then we have
r
a...a
q
r
a.. .a
a.. .a
b.. .b
q
With which we have dealt before.
the same way as case 1. for
p1
This is
solved in
< p.
This proves Theorem 2.6.7. for
G = SU(p,q).
q.e.d.
130
G = SP(n,IR)
Chapter 5.
Preliminary Notation
5.1.
Im
Let
GL(a,C).
in
identity matrix
be the
We
def ine
G = SP(n,R)
tg
{g E SL(2n, IR)
=
.- I n
The maximal compact
K
subgroup
K = SP(n,[R)
n
of
=.
0
nI
I0J
0
11
n
1n}
0
is
G
U(2n) 2 U(n).
Also
I
9'
that
= oP(n,IR)
X E oC(2n,R) It
Kn
.-- I n
nj
+0
=
0
n
is
g'0
and
=
if
=
X
0
is
=
A
B
A
C
tA
A,B,C
E gt(n,IR),
B,
C symmetric
the Cartan involution defined by
(x)
=
-tX,
0
131
=
{X C o p(n,R)
A
A = -tA
A0
X
=
.
,
go
1
8(X)
A
B"
B
t A.
{X
=
n)
tB
B
= U(n)
A
-B
d(O
tX = X}.
B]
{x
=
If
| -
=
-X
|I A, B symmetric}.
then
=
the compact
n
n
Cartan subgroup of
cos
8
1.
0
H
G
is
0
n
8
1.
81
hc
0
=
0
n
8. E lRl
= Tc
0
cos
++ [R
d(
. ..
X
8
)
0
-d( 81 ...
C
sin 8
its Lie algebra is
0
t
0
cos 81
-sin
and
0
cos 6
=
-sin
sin 81
0n
R
=
0
nl
The complexification
of these Lie Algebras gives
tc
0
132
Ii
g = op(n,C)
= {X E oL(2n,C)
n
tX
I
+
0
0]
-I
}
0n X
n
A
x =
=
B =
tB,
C =
tC
C
k 2 gt(n,C)
d(z
0
h
x
=
_
z. E C
d(z1 .
K
,
n
0
can be identified with the
{p
.. .z )
= (a 1 .. .an) I a 1
space
a2
. .
a n;
a.1 E Z}.
x1
0
c
n
If
-N 1
0
n
j
Define, for
of
t
in
g
A(g)=
=
=
1,2,...,n,
e.(X) = ix..
3
3
Then,
the roots
are
A(,,tc)
{±e
e k;
-2e
I J,k,e
= 1,2,...,n; j < k}
133
also
A(k)
= A(k,tC)
the compact imaginary
A(P) = A(p,tc)
=
t(e j-ek)
roots of
= {±(e +ek);
j < k < n},
1
tc
in
± 2 e,
| 1 < j < k ( n;
1 < e < n},
the non-compact
5.2.
imaginary roots of
Computation of
As
A+(k)
for
in
g.
for any module
the preceding cases,
X.
fix a positive root
so that if
A =
then
is
(a 1 1 a 2
Ii +
2
. . . . . an)
a1
a2
pc =
=
an'
(xx2'
(n-1,n-3,...,-n+3,-n+1).
..
''n).
Choosing a positive Weyl Chamber for
corresponds
...
A+(k)-dominant and
2p
Let
LY(X)
tc
to
forming an array of
A(g.,hc)
two rows with the
system
134
absolute value of
in decreasing order as
are aligned
x
,...,x
x2
x1
If
r
the coordinate s of
are
... >
>
X
they
follows:
> xr+1
in the first row
so that
+ 2p
>xn
...
then
_x n _x n,'*-
x r1
en-1' t
r+1
xn'
in
in
left to right in the
the second and they all decrease from
array.
For
example,
X...> -x.
1
-x.jt-1
if we have
> xi+1
> x+
k+1
>i+2
k ~
> ... > -x r1=
'= -x.-
r+1
-rj-2
x r1>
r-
-
x r= 0
r
the array would look like
.
x.
Xk
) Xi+2
xi+1
J-1
x
As
for
the case of
A+
j-2
Xr-1
r+1
the choice of arrows
SU(p,q),
gives a positive root system
..
xk+1
=
A+ (
c),
compatible
A +(k).
with
Again, the entire array is a union of blocks of the
following types.
1.
m
m-2
...
20
m
m-2
...
2 /
0
135
2.
m
m-2
...
1
m
m-2
...
11
m+3 N*/
3.
m+2
...
2
...
m
m-2
m
m-2...
2
m
m+2
or
M+3.m
4.
m+3
...
m
n+2
...
..
/
1
...
m
1
..
or
m+2
m
m
...
...
3
...
m-i
1,
or
...
...
1
..
m+3m
5.
2
m
...
m-1
2/
2
4
...
3
(5 is a particular case of 4.)
6.
All
blocks
of
0
not containing
Again, using
by the blocks
If
that
p + 2pgc
the five
or
types discussed for
1.
the picture ,
p + 2pc
gives
SU(p,q),
split
the coordinates
determines as
follows.
of
136
p1
entries
Pt entries
q
entries
qt entries
5.2.1
with
B
a block of some
=
(a ... a 1
where
m
block B,
Examp e:
.at.. .at
.
is
times
pt
times
p1
c2
CI
'
c 1 c 2 .Cm
btibts
b ...b1 )
...
m entries
q t times
a.
cm.
(2
2p c=
2
pc
(8
2 2
1
1
1
6 4 2 0 -2
= (10
1
1 -1),
-4
-6
-8),
-3
-5
8 6 3
1 -1
6Z
3
3
gives
U
Then
set
if
A +
this
1-5,
times
the total number of coordinates composing the
=
since
type
B
1
1
1
-9)
137
=
(2 2
1
1
2
1
1
1
-1)
as
in 2.4.7.
m
p1
the splitting that we want.
is
Write
=Y11)
-
t
xv (4)
(NJ...
gives
figure 5.2.1
then
=
1
times
p1
yp
p Ei i(c
If
Proposition 5.2.2.
-
x=
t
A2***' 2*'
P2
times
p2
-x t''
t-
-x
1
.
N1 )
m
times
(D
(D...@
2- a (p 1, q 1)
(p)
0.. .0
A t'''. t
(pt'qt)
(D op(m, R)
with
x1 > x2
Proof.
a)
-c i
= (p
E A +(
P
+ 2pc
> 0.
that
Observe
If
t
Se.
-
p. p
tc)
is
such that
+ ek
-
1
<p + 2pC - p + i c
2i
i
then
and
e.
0
P3
ek > < 0,
10
if
138
then
eo
-
should be
ek
included in
the set
{f P}
of
Proposition 2.4.7.
b)
Suppose that
+ p -
a
root system generated by
+ q
1 > -b
1.
-
Then the
the roots involving the
(pi, ql)
coordinates:
n -e 2 ;e+e
2
n-1'
{ei+en
ne+;-
is
isomorphic
Ap.+ql-1.
to
elliptic element then
1i
Ap 1 +q
is
Except
U(p
1
t C
Since
L.
K
Lowest
x
=
Hence the
real
form of
for these extra considerations, the proof of
corresponding result for
Let
centralizes an
,ql).
this proposition is analogous
5.3.
L
(x...x 1.
p
.
the
the one for
SU(p,q).
types of
x E i(tc)
to
We may assume
, x t'
. . . .
A
the modules
that
(N).
it
is of the
.xt 0.. .0
Pt
m
-x t ..'* -
-x"1
qt
x
> xx2 >
q1
'
t
> 0.
form
139
Write
A(L)
A (U)
as
in 2.3.
{a C A(g.,hc)
I
<a,x> =
= (a E A(,,hc)
I
(a,x>
=
>
Then
a)
L
b)
2p(ufnp) = (n-q1 +1.
' u(pl,ql)
19 u
2'2)
2
.
(9...G
1
(D oMp(m,IR);
b t'qt)
.n-ql+1,n-2q1 -q 2 +1
1
..n-2q,-q
2
+1...
P2
(5.3.1)
n- 2 (q
n-2(qi+. . . N t-1
+...+qt
)-qt+1,p-q,...,p-q
m
Pt
-1-n+2(p 1+. .. +pt-1
t. ..- 1-n+2(p. +... .+pt-1t
qt
-1-n+p 1*,
.
q1
1-n+p )
.
140
2p(unk) =
c)
(n-p 1 .. .n-p
1
,n-2p1 -p 2 . . .n-2p 1 -p 2 ...
P1
n-2(pl+. . . pt_
)-pt
p2
,. . . ,n-2(p l+. ..+pt_
)-p ,q-p,. . ..q-p
Pt
-n+2(q 1 +.
.
m
.+q t 1 )+qt.. .- n+2(ql+.
..+q
t-)+qt...,-n+q
-t
Set
n
d)
=
2p(u)
p
n+q
t1
+ q .
= (2n-n +1... 2n-n +1,2n-2n -n
1
1
*
2
+1 ... 2n-2n -n +1
1
P1
2n-2(n1 +n 2.
.
2
P2
.+nt-)-nt+,...,2n-2(n 1 +n
2
.. .+nt
1
)-nt+1,
Pt
0 .. .0,-2n+2(n1 +.. .+n -)+n t-l.. .-2n+2(n +...+nt_,)+nt-1
m
...
t
-2n+n -1...-2n+n -1)
q1
Now suppose
a
that
representation of
4 E
K.
i(tc)*
is
the highest weight
of
141
By
determine a compact parabolic subalgebra
f
n
p
the proof of Proposition 2.5.6 we may use
q
n
to
t n
k =
+
k
k.
2p(ank) = 2p(A(unk)).
Set
Suppose that
p + 2p(alk) =
(a,... a.
r
pi
is
If
hence
-
A
-
a
> nt + 2m + 1.
aii+1
p = X +
...
n
.--
setting,
a,).
si
st
the above
a
X1 = N x
x1
In
the LKT of some
and
Proof.
m
rt
Proposition 5.3. 2.
then
0 ... 0 -at...-at...-a
.. at...at
set
n.
=
r
+
(N)
+ ni+1
2p(W1p),
then
- x ***x t0..
x
> 0
the coordinates of
i
+ 2p(ank) = X + 2p(u)
give
s.
142
X
+ 2n -
2(n 1 +...+ni _)-n+1-(X.+1+2n-2(n +...+n.)-n +1+1)
11+
Ni1
~
+ ni+1
i+1 + n
> n
+ ni+1
and
xt + 2n - 2(n +
.+ nt-)
-
nt
+
Conversely, suppose we are given
conditions of
parabolic
the theorem.
S= a. -
A
+ 2m
+ 1
satisfying the
be the
Set
2n + 2(n +...+n
So
> n
q = t + a
Let
p + 2p(alk).
defined by
1
)
+ n
<X,A(u)> > 0,
<xA(t)>
and
=
A
+
= 0,
2p(unp).
q.e.d.
5.4.
Proof of Theorem 2.6. 7 for
Let
X
character
LKT of
X.
as
G = SP(nR).
in Theorem 2.6.7, with infinitesimal
-r E (h )
Suppose
A E (tc
X
),
the highest weight of a
is not a module
A
(A).
Let
143
Yi) =
LV =
(
'(pl
9q 1 )
U (p
'q
2
p = 1 p
(cfr. 5.2.2) and
t1
) $...
t2 =O
t = L1 @
Define
by
a C uV
Then
q D qV
u
=
S(pt , q
q = 2 q1 .
= u(p,q),
then
2
U
+
,
Set
(m,[R)
2LV
(uvnt).
and by Proposition 2.4.15, a)
of Theorem
2.6.7 holds.
Now let
XL
be an
(L,LfK)-modul e
occurs only as composition factor of
XL
as
the exterior tensor product
that
31 (XL).
XL
=XL
<
>L
X
can
We
0 X
L2
see
with
(t ,LifK)-module.
an
XL.
such
1
That
Xh
L
has a Hermitian form
follows
from
Lemma 2.7.4.
Lemma 5.4.1.
0 C.
L
-
XL
L1
~A
X0
for
some
q0 C L 1 ;
X0
)
A
D
Proof.
By Theorem 2.6.7, b) and c) (proved for
and Theorem 2.6.8, if
(LinK) ,
j = 1,2,
XL/
A (N)
K-types of
XL1
SU(p,q))
then there are
such that
6.
E
144
< ,>
i
4=
IV
L
4
1V 2is
L.
A(L lp)
indefinite.
L
i
=
such that
<
and
2p(ulp).
-
,>L
=
(x
..
,xp+ 1 '
.,x
Then
V 1
that if
there is
is indefinite on
V 1 E
If
Moreover, we know
P E
the sum
.
xp+M+1''
.'. x p+m
'
m)
and
2p(unp) = (n-q+l,...,n-q+1,p-q....,p-q,-l-n+p...-1-n+p)
then,
since
A(L 1 no)
it
is
q
m
p
clear
= ±{(ei+e )
that
+ P
then
-
p+m
if
is dominant
+
< p, p + m <
i
1
is dominant for
for
A (k),
unless
n}
j
A(L fk),
x
p
= x
then that
xp+ 1 '
xp
L
Note
and henc e
Suppose
or
p+m+l*
Suppose
So
p+1
that
p
p2 C
that
2p(u p)IL2
is fine and
=
2p(alp)I
s aL2
XL
((0.. .0);(l... 1,0.
i2
is of
the
th en
form
is a principa l
series.
2
145
2
1.. .1,0.. .0)
=
m-a
a
2
Then for
TI
(0.. .0,-1...-1),
=
m-a
XL2 ,
> 0.
a
with
V 2
is also a LKT of
a
by Frobenius reciprocity.
So
Since
if
77 =
17
A
1
+
T2
+
2p(unp),
is also dominant, it
V
1?
follows
is a LKT o f
xp+m -
that
X.
1
>
p+m+1
Suppose
iI
that
+ P3
is dominant for
3 E
some
If
A(L 1np) n U(pt'q t).
'B = (0.. .0 1 0... 00...0 1 0...0),
P-pt
since
x p+m
- 1>
13 = (0..
p+m+l1
0
...
m
Pt
p +
since
m
= 17L
pL
is dominant.
If
0 0 -1 0.. .0 0.. .0 -1 0..
p
then,
P
q
then
qt
+t
Tj
+
0
q~
is a dominant
candidate.
If
argument
and
Ap2
=(0 ... 0,-1...-1),9
shows
m-a
a
that
x -1
p
If
2 = (0... 0)
xp+m
~ xp+m+l
with
a > 0,
a similar
xp+1*
then both differences
must be strictly positive.
x p - x p+1
This
is
146
clear from the pictures of
p+
2pC
q.e.d.
Lemma
5.4.2.
A o(N
0
)
In the above setting, assume
some
for
q
0
& t1
and
X
true
if we assume
:
C
q
Then, Theorem 2.6.7 is
that
XL
o.
that
> 2
p - x p+1i
(5.4. 3)
and
Proof.
if
.+
P
Suppose
first
x
p+m
+
-
2
that
m+1
m+1
2 ' ' ' ' ' 2 ] ,
2
2
.. 0 -1 0..
V 2
p.
2
Then
1,0.. .0).
+ Pt 2nk> ( (p,P>.
n
is
a
.0 -1),
into a
(0,0.. .0 -2)}
m
m-a
a
making
(1,1,.....
+
n( + pL 2lk'
f3 E {(0.
2.
an easy calculation shows
in Lemma 2.7.2, there
By 2)
p+m+1
space on which
(
is
+j3
indefinite.
Moreover
Similarly
if
A + P
is
A +(k)-dominant,
2
by (5.4.3).
then
m-a
a
147
( E ((1,0...0 1 0.. .0);(2,0...
m-a
Now,
if
=
j2
a
(0... 0)
then the Dirac operator
inequalit y fails for any choice of
unless
r
and,
obvi ously,
Now
of
if
is
XL
p + f
Yit2
is also dominant
=
representation.
L2
and
6
=
V
Hence
I(6,
XL
L2
V
)
m+1
m+1
for
P E A(t+ l 2).
y
=
vIt
(In
fact,
the
is a principal series
1)
trivial;
=
then, the Langlands subquotient
,
the trivial
representation
p(A+(t 2 t)),
p
if
in particular,
2
2
pn
=
L2
.
V
the Lang lands submodule o f
1XL2 )
A(XL
Q(XL2
(XL
is
X = d(A
q
By
0 (X
0
)) ® M (trivial
induction by
stages,
X
contradicting our assumptions on
This proves
representation).
q
the
is an
A (N),
X.
lemma..
q.e.d.
148
To finish the proof of Theorem 2.6.7, suppose now that
x
<
- x
1.
Lemma 5.4.4.
= 1
x
Under
and
x p+m-
the hypothesis of Lemma 5.4.2,
>
Xp+m+
2,
if
x
then Theorem 2.6.7 is
true.
Proof.
that
The assumptions on
the picture of
the coordinates of
ji + 2pc
around
Ii
imply
the coordinates
involved is either
...
m+3
...
m
...
m-2
m-1
m+4
or
M+
...
m+4
...
that
m
m-2
...
Im
m-2
...
is,
p + 2pc =
(...m+5
m+3
m
m-2
-m+1
...
|
-m-4
...
or
p +
2 pc
= (...m+5
m+3
| m
...
-m
|
-m-4
...
)
)
p
-
149
Observe
K-type
that
LV =
that gives
np)
- 2p(a
M-2
rn-i
is
p2 =
p2
=
f ine and that
the fine
the picture
m
is
is
(1,1,....,.1,0.
.0)
...
...
and
the fine
m
m-2
...
m
m-2
...
K-type
that gives
(0...0).
Arguing as
in the proof of Lemma 5.4.2. we
can find,
in both cases
P E
((0.. .0 -1,0..
0 -1);(0...0 -2)}
as we want.
q.e.d.
Assume now
that
0
x
-
xp+1
1I
(5.4.5)
p+m
xx p+m+1
150
We want
to contradict
infinitesimal character
Since we have an
have
the assumption that
is regular and integral.
-r
on
Recall
L = U(p,q) x SP(m,R)
= U(pq),
L
A (X)-module for
some control
that
the
we
-Y.
and
L ;
L
=
t
TI
(U(p.,q.))J
x SP(m,IR).
W e may assume
qt.
pt
By
i=1
the
computation
in
4.3, either
V U(ptqt)
S(X
t+s,X t+s-1...X t+1
X t''*
s
X tX t-l...X t-s
qt+1
It.'X t)
s
qt
or
XV U(pt'qt
1
t+s, ... Xt+
1
t*'
t
t 2
qt
t-st
t
qt
and
V U(pt ,qt )
Inside
= (0...0 v 1.
SP(n,R)
Vt 0... 0 | -v
this gives
1
...-
d
151
(Xt+s,..
. Xt-'
t
(0...0 v1 ... v
If
is
-Y
xt +
regular
t
t
-
xt
s
.'x'
t-s
0
I
-Nt**
x~t)
v0 , .. *v
1*
integral
+
s,
> 0 > -Nt
+
-x t
v
v
,
'*t
11
+
t
-1
>s
qt
S>
t
v
xt
x
>
Claim.
If
Proof.
The picture
can be of
1.
. . .
.--
2.
(5.4.5)
m-1
m+2
...
n+3
types.
m-2
m
m+2
i + 2pc
for
the following
. . .
.
satisfies
p
m
. ..
rn-i
M
m
..
.
m-2
m-2
+ q
at
s
+
qt
then
x
t-
-
s
< 1.
around these coordinates
152
3.
4.
...
m+4
m+
...
m+4
m+2
...
m+4
m+2
m-1
m-3
...
m+4
m+2
m-1
m-3 ...
...
m+3
5.
...
m+4
m-2
/
or
...
m
.
.
...
m-2
1
r+3
m
...
m+3
m
...
...
\rn-i
i+
So we either have
...
(considering that 5 and 2,
and 3 and 1
are symmetric)
p + 2pc =
p =
and
(...m+k+2,m+k
-m-k...)
-m-k+1...)
m+k+2,m+k
(...
or
A
+ 2pC =
(
...
...
with
In both
m+2
m+1
|
|
-m-2...)
cases we get
xV
This proves
=
(..
.1 1
the claim.
0.. .0
-1 -1...).
153
This
2pc
reduces
gives,
...
quasisplit.
= 0.
But
then,
p +
at worst,
m+2
Because if
q
to the case when
q.
= 0,
So,
m+4
p. =
we have
This concludes
M-1
...
m
1,
x p+m
since
-
U(p.,q.)
xp+m+
>
is
2!
the proof of Theorem 2.6.7.
q.e.d.
154
Bibliography
A.
Borel-N. Wallach
[1980] "Continuous cohomology, discrete subgroups and
representations of reductive groups".
Annals of
Mathematics Studies, 94 Princeton University Press.
J. Dixmier
[1977] "Enveloping algebras".
North Holland Publishing
Co., Amsterdam, N. Y., Oxford
T.
J. Enright
[1979] "Relative Lie algebra cohomology and unitary
representations of complex Lie groups".
Duke Math. J.
46, 513-525.
[1984] "Unitary representations for two real forms of a
semisimple Lie algebra:
a theory of comparison", in
Proc. Special Year in Harmonic Analysis, University of
Maryland 1024, 1-29.
Lecture Notes in Math.,
Springer-Verlag, Berlin-Heidelberg-New York-Tokyo.
A.
Knapp-G. Zuckerman
[1976] "Classification theorems for representations of
semisimple Lie groups", in non-commutative Harmonic
Analysis.
Lecture Notes in Math. 587, 138-159,
Springer-Verlag, Berlin-Heidelberg-New York.
W. Schmid
[1975] "On the characters of discrete series (the Hermitian
symmetric case)".
Inventiones Math. 30, 47-144.
B.
Speh
[1981] "Unitary representations of SL(n,IR) and the
cohomology of congruence subgroups".
In noncommutative harmonic analysis and Lie groups 880,
483-505, J. Carmona and M. Vergne (eds.), Lecture
Notes in Math., Springer-Verlag, Berlin-Heidelberg-New
York.
B.
Speh-D. Vogan
[1980] "Reducibility of generalized principal series
representations", Acta Math. 145, 227-229.
155
D.
Vogan
[1981] "Representations of real reductive Lie groups",
Birkhauser, Boston-Basel-Stuttgart.
[1984] "Unitarizability of certain series of
tions", Annals Math. 120, 141-187.
D.
representa-
Vogan-G. Zuckerman
[1984] "Unitary representations with non-zero cohomology",
Compositio Mathematica 53, 51-90.
G.
Zuckerman
[1977] "On construction of representations by derived
functors."
Handwitten Notes.
Download