LIE ALGEBRA COHOMOLOGY AND
THE REPRESENTATIONS OFSEMISIMPLE LIE GROUPS
by
David Vogan
B.A. University of Chicago
(1974)
S.M. University of Chicago
(1974)
SUBMITTED IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE
DEGREE OF
DOCTOR OF PHILOSOPHY
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
September 1976
Signature of Author......... W. ......
-- -' :
--&0.-------.0
Department of Mathematics, August 9,
--
-
-
-
1976
....-...-....--.---.--------------.-.
Certified by...
Tei
.
.
Super-visor
------.-----------.
.. ----. --.
Accepted by................. ......
Chairman, Departmental Committee
on Graduate Students
ARCHIVES
SEP
27 1976
2.
Lie Algebra Cohomology and
the Representations of Semisimple Lie Groups
by David Vogan
Submitted to the Department of Mathematics on August 9, 1976
in partial fulfillment of the requirements for the degree of
Doctor of Philosophy.
ABSTRACT
We obtain a classification of the irreducible quasisimple representations of most semisimple Lie groups, including all the classical ones. We also prove for these
groups the conjecture that every irreducible representation
contains some representation of a maximal compact subgroup
with multiplicity one. The techniques involved are essen-tially algebraic, and are much simpler than those used by
Langlands in his classification ([17]) of the representations for linear groups.
Let G be (say) a finite cover of a classical semisimple
Lie group, and K a maximal compact subgroup. We define an
ordering on K (essentially by the length of the highest
weight.) Then every irreducible representation u of G has
a minimal K-type p c K; this is just the smallest K-type
occurring in T.TK. To each p we associate a cuspidal parabolic subgroup P = MAN of G, and a tempered irreducible
unitary representation 6 e M, in the "limit of the discrete
Then
FV = Ind
(6 0 v 0 1).
series."
For each v c A, set
MANtG
p is the minimal K-type of fV, and occurs with multiplicity
one. Let i denote the uniqe irreducible subquotient of
Then {ffv} is precisely the set
TIV containiRg the K-type p.
oi irreducible representations of G Nith minimal K-type p.
The only difficult part of this program is showing that
there are not too many irreducible representations with
minimal K-type p. This is done as follows. One proves
first that certain other K-types are smaller than p. (This
is done case-by-case, and occupies the bulk of the thesis.)
If 7 has minimal K-type p, it follows that these K-types do
not occur in ff; and one can use spectral sequence techniques
to show that certain cohomology groups do not vanish. KIt is
on the
shown that this greatly restricts the action of U( )
p-primary component of 7, and hence greatly restricts 7T.
Thesis supervisor: Bertram Kostant
Title: Professor of Mathematics
Table of Contents
Page
Abstract (some summary)
2.
Table of contents (some page numbers)
3.
Acknowledgements (some thank-you's)
4.
1.
Introduction (some philosophy)
5.
2.
Preliminary results and notation (some definitions) 8.
3.
Lie algebra cohomology (some new theorems)
21.
4.
The classification problem (some computations)
38.
5.
The subquotient theorem (some facts)
6.
Existence of the representations (some old theorems)166.
108.
Bibliography (some light reading)
184.
Biographical note (some even lighter reading)
187.
4.
Acknowledgements
I would like to thank my advisor, Professor Kostant,
for providing large quantities of advice, encouragement,
and proofs while this thesis was being written.
I worked
very closely with Allan Cooper for several months just
before the main results were obtained, and they owe a great
deal to many long discussions with him.
Floyd Williams
taught me Lie algebra cohomology and Kostant's proof of the
Borel-Weil theorem, and Henryk Hecht was an apparently inexhaustible source of information about the discrete series.
I
hesitate to mention more names, for fear of having to include
the entire mathematics department; but many others have contributed their time and expertise very generously.
Finally, I would like to thank Professor Paul Sally
of the University of Chicago, who undertook the formidable
task of introducing me to representation theory while I was
an undergraduate.
It is not clear whether he will approve
of the absence of analysis in this thesis, but nonetheless
it is dedicated to him.
5.
1.
Introduction
The beautiful simplicity and power of representation
theory for compact Lie groups have inspired many attempts
at generalization.
This thesis concerns one of the
simplest possibilities which has been considered, namely
a classification theory for the irreducible K-finite
representations of a semisimple Lie group.
Such a theory,
for linear groups, is provided by the work of Harish-Chandra,
Langlands, and Knapp and Zuckerman ([17],
[15]).
Unfor-
tunately their theory relies on rather deep analysis on
the group, and the results are less than explicit from an
algebraic perspective: an arbitrary representation is
realized, roughly speaking, as the image of a certain
integral intertwining operator between two induced representations, which are specified in terms of the asymptotic
behavior of matrix coefficients of the representation.
For tne classical groups, and some of the exceptional
groups, this thesis provides an essentially algebraic
classification theory.
The basic notion is that of a
minimal K-type (Definition 4.1.) Theorem 3.12 relates
the existence of a nice minimal K-type for a representation
to the structure of certain Lie algebra cohomology groups.
According to Theorem 3.15, this in turn gives a great deal
_-W
of information about the action of U(o) )
K-type.
on the minimal
In this way the "uniqueness" part of a classifi-
cation theory is reduced to showing that the minimal
K-type of any representation is nice (conjecture 4.2).
This essentially geometric problem is dealt with in
section 4 for a number of examples, including all the
classical groups.
In particular we prove for these examples
the well known conjecture that every irreducible K-finite
representation has a K-type (the minimal one) which occurs
with multiplicity one.
The "existence" part of the
classification theory is proved along fairly standard lines
(Theorem 6.2).
Not surprisingly, it turns out (for purely formal
reasons) that this algebraic classification is very closely
related to the analytic one.
The algebraic translations
of some of Langlands' results provide rather explicit cyclic
vectors in certain induced representations (Corollary 6.7);
this is a partial generalization of results of Kostant and
Helgason for the spherical principal series, and may be of
independent interest.
About two-thirds of this thesis is devoted to the
proofs of Theorems 4.6 and 5.4, which proceed on a
case-by-case basis.
It seems reasonable to hope that these
7.
arguments can be greatly improved before too long, and
they should probably be omitted on a first reading.
Al-
though the results are crucial, the proofs have no bearing
on the rest of the development.
8.
2.
Preliminary results and notation
One or two general comments about the terminology
are in order.
This thesis might better have been called
"Little tiny K-types," and it may require some effort to
keep track of the various notions of little tiny.
The
key concepts are "7[ -minimal in X" (Definition 3.11);
"minimal X -type" (Definition 4.1); and "small
(Definition 5.3).
Ak-type"
Conceptually the distinctions among
these are quite straightforward; keeping them in mind may
make the overall structure of the arguments clearer.
Most
of the proofs below are quite easy, and one only has to
worry about putting them in the right order.
The real dual of a vector space V is written V'; an
asterisk denotes the complex dual.
Especially in section 3,
however, an asterisk may also mean "plunk," as in A*V, the
exterior algebra of V.
X
denotes the centralizer of Y in X,
or the subspace of X annihilated by Y, etc.
Much of the material in the present section may be
found in Warner's book [25], but a number of proofs are
sketched for the convenience of the reader.
Let G be a connected reductive Lie group.
algebra of G will be denoted by
(00
The Lie
00, with complexification
' and universal enveloping algebra U (
Analogous notation for other groups (i.e. H, 10 1,
is followed throughout without further comment.
Cartan involution
bilinear form
0
of
< , >
on
0
position
0f+
Let
Lie algebra.
Choose a
and a non-singular invariant
0 induces a Cartan decom-
y
here A-0
o
0
K
0
U(N))
is a compact
be the connected subgroup of
G
with
Lie algebra 't 0: notice that K need not be compact.
G (or rather the triple (G, Of<
Definition 2.1
>))
is
< , >|e0
is
,
said to satisfy definition 2.1 if
with respect to < ,>
i) P 0=~0
ii)
<
, >|h0
is negative definite, and
positive definite
iii)
K x P0
by
G
+
(x,y)
+
is an analytic
x-exp(y)
diffeomorphism.
Henceforth
G
will be assumed to satisfy Definition 2.1.
(Notice that this is certainly the case if
simple and
< , >
is the Killing form.)
is a closed subgroup.
can write G =
2
1
is semi-
In particular
By passing to a finite cover
(exp 10)(exP 20),
semisimple, and K2 (exp
G
where (K exp
) is central in G.
K
G, we
) is
One easily
deduces the existence of an Iwasawa decomposition of G,
and obvious generalizations of Harish-Chandra's representation
10.
These will be cited as they
theory for semisimple groups.
are needed.
= 0
Put
nected.
k is
Lemma 2.2.
S
+
C
c=6
,
is compact, so all its
0
A=
(tb +
,
and
Since
is abelian.
it remains only to check that
is semisimple or zero, so it will suffice to show
[
[AA] C1$.
[
[I, ]
,
[t~t] = [t, 7] = 0, and
Clearly
= .
c'
Let
in o;
The form
A
_
so
] c
,
+
[[[t,
A
.
i is reductive.
0O; so
0 +
C
i
But
subalgebras are reductive.
0 + i
+
a Cartan subalgebra of 0.
The Lie algebra
Proof.
=t
+Y2
=
is con-
T
be a Cartan subgroup of K:
T g K
Let
Q.E.D.
.
denote the set of non-zero roots of
A c it 0 +
it is easy to check that in fact
< , >
-
induces a nonsingular form (also written
< , >) on A *, which is positive definite on
9
The decomposition
elements of
<(a*,f3),
( Y3O~
as
=
t
(a,f3)
+ (
)
with obvious notation.
If
)
is orthogonal; write
accordingly.
(a',3')> = <a,a'>
it' + (
0
*
Set
+ <f3,f'>
y = (af), y' =
it will occasionally be convenient to write
<y,y'>
11.
instead of
Since
c A, a $ 0.
if (a,1)
t
7, every root has support on
=
O)
A = A
We write
U A
i.e.
the imaginary
and complex roots, where
Ai
= {(a,c)
£ Aj
If
X
= 01, A
=
{(a,1)
c AIS /
01.
is a root vector of weight (a,6), then
root vector of weight
O(a,)
=(a,-).
So
OX
A
is a
is
0--invariant, AiR is precisely the set of 0-fixed roots,
and the elements of
Z $ 0,
If
AC
occur in 0-conjugate pairs.
Cj = -@h 0
the decomposition
invariant under -h.
is
not
For this reason it will be important
to understand the structure of the system of roots
restricted to -,
(a,6) c A, and
X
which we call ft-roots.
is a non-zero root vector of weight
(a,B), as above, so that
of weight
(a,-6).
If
OX
is a non-zero root vector
6 = 0, then the fact that the
OX = cX; and
root spaces are one dimensional forces
c = + 1.
If
c = 1, then
a compact root.
If
c = -
implies
c2 = 1, i.e.
we call
(a,0)
(a,O)
is a noncompact root.
are linearly independent,
But
X + OX
vector for
so
If
X
S
#
with weight a.
0, then
4
Similarly
0 2
1
X Ck, and
1, X ef, and
is neither in
is a non-zero element of
t
Suppose
X
and
ARk
nor
OX
in
, and is a root
X - OX
is a
f .
12.
t-root vector in g , also of weight a.
non-zero
We claim
that these span the space of i -root vectors of weight a.
What must be shown is that if
Replacing
<(a,6),
6
by
-6
which forces
and
X - OX
6
> 0.
(a,6)
3'.
=
3' = + f.
if necessary, we may assume
(aS')>>_<aa
this implies that
(ac') e A, then
-
By abstract root theory,
(a,a') = (0,3-a')
is a root,
X + OX
This proves the claim.
are called the compact and noncompact t(-root
vectors of weight a.
Depending on which of these is under
consideration, we will speak of the compact root a
noncompact root c.
or the
One could make this quite rigorous by
speaking of 1~ -0 roots, i.e. the weights of/t-invariant,
6-invariant subspaces of
<
; but no confusion should arise
as long as the reader is aware of the situation.
If
A,(V)
V
is a t-invariant subspace of
for the corresponding set of
tiplicities; if
V
is also
A(V)
will be
When there is no possibi-
lity of confusion, the subscript dropped to simplify the notation.
we write
t-roots with mul-
A -invariant,
the corresponding set of roots.
o,
At(V)
in
If
V
may be
is 6-invariant,
we may refer to the compact and noncompact f~-roots in
A(V) , in accordance with the preceding paragraph.
this case a I -root is called imaginary or complex
In
13.
depending on whether it corresponds to an imaginary I.-root
or to a 0-conjugate pair of complex .i-roots. This discussion may be summarized as
Proposition 2.3
A+(o
The imaginary elements of
occur
)
with multiplicity one; the complex elements occur with
multiplicity two, namely one compact and one noncompact
-t-root.
Fix once and for all an algebraic ordering
A+ (A) = A (4A) = {a c Ak(AC)t|a
and let
system of compact positive roots.
-
0}
off
it,
be the associated
To every
which
y E it
A+ (k), we associate a positive
is dominant with respect to
root system
A
=
Then
A+
{{a,)
Mj, and
> 0, or <a,y> = 0 and a
e A(%)I<a y>Y
A)
+
Put
Since
p =E
0
+
A
p = (p,O).
the associated Borel subalgebra.
0 =-J0rO
All of these are 0-invariant, so
Ir0
; write
A
be the nilpotent subalgebra corresponding to
Let qf 0 C
A+, and
in
will be regarded as fixed in this
A+
A (.+
is 0-invariant, 0-p = p, i.e. p c
=- n
(\
_ + _
is
WK = W(ik, j)
+
I-onf
a Borel subalgebra of K.
The associated set of positive roots is
Let
-a}.
is a 0-invariant positive root system for
section and the next.
etc.
-
A+ (k).
be the Weyl group of
t in
14.
this will be referred to as the compact Weyl group.
In
the interest of mathematical purity, we would like to
consider
WK
as a subgroup of
compact t -root.
flection
-
a
be a
a
is imaginary, we send the re-
sa e W(4, ,t
)
to
(a,-S)
s (a,0)
W(C , A ).
C
a
If
(a,3), we know
is not a root.
(2a,O) = (a,
i)
Let
If
corresponds to the complex root
(a,3)
A).
W( 9,
There are two cases:
) + (a,-6)
is a root.
Send
s
to
s (2a,O)'
(a,r3) + (a,-3)
ii)
root theory, <(a,3),
is not a root.
Then by abstract
(a,-S)> = 0, i.e. <a,a> =
<,> .
By direct computation, this implies
_
s (arB)s
; their common value on
s (a)
s (a
2<x, a>k
is (x,y) -
(x,y) E:
to this element of
(
W(ojA).
generated by the sa.
2<y, S>
a.,
Let
inspection, these maps satisfy
implies
of
= 1.
$(ca)
$(c)-p
W(,i)
Thus
$(oa) |t
$
s
be the free group
$ : WK
$ : WK
$i(T)I±
= $ (a)-p = 1-p = p.
p
=
$ ().
WK'
+
W(V,4,
By
Suppose
$
But the only element
is the identity, so
actually defines a map
Clearly
Map
p e tj, the preceding observation
Since
which fixes
= a.
WK
We have defined a map
and of course there is a natural map
that
.
$ : WK
-+
$()
= 1.
W(j4A), satisfying
is an embedding; henceforth
WK
15.
will be freely identified as a subgroup of
W(f,9V).
The
problem of finding a more natural way to do this is left
to the reader.
We will be quite interested in a certain special
class of groups.
Definition 2.4
50
is said to be split if the real
semisimple Lie algebra
sense,
i.e.
if
[5
'9001
0]
is split in the usual
1
contains a Cartan sub-
o'0 ].
[
algebra of
0
[V 0'
It will be useful to have another description
of such
V0'
Recall that a subset {a ... an} of the positive roots in
an abstract
a
if
root system is
+ a
said to be strongly orthogonal
is not a root for any
the span of
At
Proposition 2.5
i,
j.
Let
t root
be
r oot
is
,C.
in
0
is
split
if
and only if
spanned by a strongly orthogonal set of (positive) noncompact
imaginary roots.
Proof.
It is enough to assume
*
is semisimple, so that
*
root =
t
.
Extend the abelian algebra
maximal abelian subalgebra
010
of
0
00
90
00.
Clearly
/M
.
f0
to a
Then
is reductive, and
16.
contains Az
/7at
as a Cartan subalgebra; and
A(mz) =A
is an equal rank Lie algebra, i.e.
contains a Cartan subalgebra of
Suppose
subalgebra of
/M 0
tC
is
split.
o
is split.
50,
60
Then
=
AiR
Cl 0
spans
t
is a Cartan
,7,0; in particular
is maximal abelian in
does not meet the center of
A ()
[/7O' fool
[Mlt0 'M 0].
and therefore of
Also
; so
.
411.
110, so
It follows that
-;O is a compact Cartan
subalgebra of the split semisimple equal rank group
[VL0'?n0].
't
By a theorem of Martens (see Schmid [22])
is spanned by a strongly orthogonal set of positive
noncompact roots of
[I7n0'
0 ' i.e. by a strongly ortho-
gonal set of positive noncompact imaginary roots.
For the converse, this argument may be reversed Martens' theorem is a necessary and sufficient condition.
This amounts to writing down the split Cartan subalgebra
corresponding to a strongly orthogonal set of noncompact
imaginary roots, using a Cayley transform.
Let
r 2r
subalgebra of
~
then
Q.E.D.
be an arbitrary 0-invariant parabolic
,
with nil radical 17
=
+
.
and Levi subalgebra
We claim that
complexification of a real subalgebra
.
is the
Let bar
denote conjugation with respect to the real form
90'
17.
we must show
and
A(Y ) = A(,
).
If
(a,3) e A, then a e it;
) , so
ce (
(a,)
(-a,)
=
-
-
Hence
(2.6)
(a, B) E A =>
Since
)
A(
= -
(a, 6)
is stable under
6 (a,
-1
).
and
0, the assertion
is proved.
Let
+c + +
C= t
+
be the center of
g
1
Notice that
~ = (,t+)
c
Clearly
((a , ))(x,y) > 0
(a,f)
(a,-)
e A+
.
By the general theory of
A+
£
a(x) > 0.
(-, )
-
A(
).
A( P) (
A+.
x;
T
so that
Now suppose
Since this set is 0-invariant,
((a,+
Thus
))(xy) > 0, which forces
Hence the corresponding root vector
not commute with
Set
£
(x,y) c C
(a, ) c A+, with equality
whenever
A(I ).
-
to
[ ?,Q].
parabolic subalgebras, we can find
precisely for
< ,
=
Proposition 2.7
Proof.
n()
~=
- root ( t ); thus
corresponds under
so
X(
.
X~'a
This proves
Q.E.D.
L = G t
.
Then
L
is a closed subgroup of G,
does
18.
with Lie algebra
i0.
The decomposition
00
is invariant under conjugation
by exp (Z
L = K
-(exp
P
).
standard fact for
K
present situation.
That
K
G
=
0
0
K-exp
) , so
is connected is a
compact, easily generalized to the
Hence
L
satisfies definition 2.1
with respect to the restriction of
<
,
PutRrdim rVLY , S = dim 0.
-invariant subspace, with
>
If
A = A
0.
and
V coy
is a
(V), write
p(V) = p(A) =12 ce a e
acA
We use the same notation for
in other words, if
V
is an
p(V) E 4*. An unqualified
0
+
p(Tt ) = p(A ).
A
A
p
-invariant subspaces and A;
-invariant subspace,
will always mean
We also define
Finally, we need a little representation theory.
details see Warner [25].
modules for
K, J
,
Finite dimensional irreducible
L n K, T, etc., will be freely iden-
tified with their highest weights (with respect to
an appropriate part of
Y0 (0A.)
V
t
and
The set of all equiva-
lence classes of such modules will be written
Let
For
K, i,
be a complete locally convex space, and ff
a continuous irreducible representation.
V
etc.
G
-+
has a dense
Aut V
19.
subspace
V
of differentiable vectors, on which
acts.
Suppose the center of
(e.g.
if
V
G
acts by scalars on
is a Banach space and
completely irreducible
(TCI)).
=
(
U(J )
)
V
topologically
V
decomposes
00A
V, y c K. Let
Y
denote the center of U(J ) . T is said
to admit a central character
z c
is
'
Then
nicely into K-primary components
(o)
U(o)
acts on
V
X
(OJ )
:
by a scalar
+
C
XT (z)
if every
(e.g. if
7
00
is TCI as above).
In this case each
dimensional, and the
U(cg )
algebraically irreducible.
Definition 2.8
u
module
Even if
V
VK
Y
is finite
yCK
V
is
ff is not irreducible,
is called admissible if
Vw
decomposes
00
into finite dimensional K-primary components
V
=
00
Z V
YCK
In case
the
and
V
U(6) module
is called the Harish-Chandra module of ff.
ff is admissible,
irreducible iff
VK
V ; the
V
VK
is algebraically
is topologically irreducible.
Both
have (possibly infinite) composition series;
subquotients are admissible and irreducible.
Two
such representations are called infinitesimally equivalent
if their Harish-Chandra modules are algebraically equivalent.
(This is the same as Naimark equivalence or, in the unitary
case, unitary equivalence.)
Let
U
be the centralizer
20.
k
of
in
decomposition of
each
ye
This ring preserves the Jo-primary
U(
VK, so
VO
Y
irreducible,
V
y
U
module for
N.
Theorem 2.9 (Harish-Chandra)
of
becomes a
and
V
y
determines .r
If
Tr is admissible and
/ 0, then the
U
module structure
up to infinitesimal equivalence.
of course all of this works for
L
as well as for
21.
3.
Lie algebra cohomology
Most of the basic results on Lie algebra cohomology
and spectral sequences used here may be found in
Cartan-Eilenberg [2].
The proofs of Theorems 3.3 and
3.5 were inspired by a paper of Casselman and Osborne ([4]);
Allan Cooper suggested that their result (a weak form of
Theorem 3.3) could be formulated along the more natural
lines adopted here.
Write
E: U(rg)
U(
) = [U( f) 0 U(g)] $ ? U(g ), and let
U(U) 0 U(gq)
be the corresponding projection.
The idea of the proof of Proposition 2.7 shows that if
u 6 U(?)
, then
quite standard since Z
of
(.
; notice that this is not
E(u) c U(P)
is not a full Cartan subalgebra
But the usual argument does show that
E|U(J)
is a homomorphism, and that
(3.1)
U
Now
R + S = dim (enW
+
U f)
n
) + dim (i c)
is a one dimensional 1 module.
denote the scalar by which
m
+
into
m + $(m)
.
m
For m e
acts on
R+S
= dim97, so A/ 4 1
1,
let
*(m)
AR+97.
Then
defines a Lie algebra homomorphism of
U(), which extends to an algebra autororphism $ of
22.
U(e).
E = $ o (; then
Define
fo
(3.2)
Suppose
X
of
.
with respect to
X
ith Lie algebra cohomology group of
the subalgebra A
H (/,X) for the
Write
is a V-module.
Recall that there is an operator
d : Hom (A /7,X)
+
Hom (Ai+1
X),
the coboundary operator of Lie algebra cohomology, and
that
H (i ,X)
is defined to be the ith cohomology group
then
has the structure of an 4
H (/Z ,X)
*
Hom (AY7 ,X)
can be made into a
H (7 ,X) is a
d
this is just the fact that
One knows that the action of
module.
) module, by
f(
H0
In the case of
(1
is invariant under
X
(i )
a certain homomorphism of g ( )
/,
centralizes '7 ) so that
(because :5(9J )
(p) module.
normalizes
This action clearly
acting only on the second factor.
commutes with
4 c o
If
of the resulting cochain complex.
on
into
X
,X),
p(O).
factors through
)
-
this ob-
servation is the key to the computation of the central
characters of the finite dimensional representations of
The following theorem is a fairly straightforward generalization to arbitrary cohomology; of course the proof is by
"dimension shifting."
23.
Theorem 3.3
If
z E:
z -
w = ((z)
w E H (71,X), then
and
.
Here the action on the left side of the equation is that
just defined, and the action on the right is the usual
--module structure.
Proof.
Proceed by downward induction on i, starting
i = R + S = dim /7
with the case
It is immediate from
.
the definition that
*
R+S
R+S
R+S
{ ,X/ffX) = (A /7 ) 0 X/ftX.
(7,X) = Hom (A
H
both (AR+S) * and X/{ X
HR+S (1[ ,X)
S()
N 0 x E H
z =(z)
The action of
for the corresponding element
x
R+S *
N e (A 77) , x E X, and that
Suppose that
=
As an f module,
(W ,X) is on the X/y7 X factor alone.
x 6 X, write
X//7X.
=
are ) modules.
is their tensor product.
on H
If
of
is invariant under I?. Thus
normalizes /( , ? X
Since
R+S
("y,X).
Then
mod /7U((), and /J
Recall that
m
acts on
N E (AR+S
*,
m - N =
R+S
AR-
by the scalar
$(m)N.
modules,
Also
x ..//X; so
On the other hand, suppose
z - w = N 0 Z(z) - x.
tensor product of
)-
z - w = N 0 z - x.
$ m).
.
m
Since
By the definition of a
24.
(N 0 x)
-
$(m)
Since
$
(3.4)
for all
In
-
(N O x)
=
(m + $(m))
=
(m
=
-$(m) (N 0 x) + N 0 m - x + e (m) (N 0
=
N
N) 0 x + N 0 m - x +
0 mx
is an automorphism of
particular,
y c U( f ),
z
(m) (N 0 x)
- o
=
), this implies
U(
4(y) - (N 0 x) = N 0 y - x
NO
- x
((z)
$(((z))
=
-
(N 0 x)
=
((z)
This establishes the result for i = R + S.
Suppose that the theorem holds for some
i + 1 < R + S.
By choosing a set of generators for X, one can find a free
U( ) module
0
+
F0
-+
F
F
-+
F0
and a submodule
X - 0
is exact.
so that
There is then a long exact
sequence
H 1 (7Z ,F) + H (Mz ,X) +
Now
is a free U(rq ) module, so
U(9)
as a U(VZ) module.
j
< dim rn
Thus
6
=
R + S.
It follows that
Since
is injective.
z ' 6(w) =
Hi+l (ri ,F0
((z)
- 6(w).
with the action of
F
is also free
H (l ,F)
0
=
for
= 0.
i + 1 < R + S, H (IF)
By the inductive hypothesis,
6
It is standard that
UU?), so that E(z)
We have observed that the action of
- 6(w)
(
)
=
commutes
6(g(z) - w).
commutes with
d
ow.
25.
on the cochain level; it follows easily that
z - 6(
= 6(z - w).
So
w) = 6((z) *
6(z
6 is injective, z - w =
-.
*(z)
).
Since
Q.E.D.
Notice that the argument was entirely formal: the
definitions of this section are much deeper than the theorems.
The next result is a little more subtle.
can be made into a
second factor.
H (arnk ,X) is
U(O)
Hom (A (1ak ,X)
module, by acting only on the
This action commutes with d, so that
a U(v )
In many cases H ((nlk ,X)
module.
is quite large, and this action will reflect all of the
U(g )
structure of X; so one cannot hope to compute it as
easily as the
5ig)
action.
The appropriate rabbit is
the following.
*
~
*
*
The decomposition AL = [A ((7z
[A (4ng )]
A )]
)
A
and thus extensions
R
i
Hom (A (gP n ), X) 0 [A (a-in
i+R
+ Hom (A /n, X)
that R = dim 'NR&}.)
extensions intertwine
))]
n
A
induces projections ffi : A +R
(Recall
It is not difficult to see that these
d 0 1
map of cochain complexes.)
and
(i.e. they define a
d
This will be proved in the
course of establishing Theorem 3.14, but the reader can
easily provide a direct argument.
S:H
R
Hi
(gifa,X) 0 [A (MA
We therefore have maps
R+i
*
f)]
+
H
(rt,X)
26.
which preserve
the
module structure.
)?o
*
H (r ,X)
tends to be relatively small, so that
in general has a large kernel.
)
U(
Theorem 3.5 computes the
action modulo this kernel; the rest of section 3
is devoted to finding conditions for
Theorem3.5
P c
R*
u E U(gf)
-then
,
ff. to be non-trivial.
W 6 H ('t
rr ((u -*)
O P)
=
,X),
and
((u)
-
_(W O P)
Proceed by downward induction on i, starting with
the case
i
HS( ,/' n,X)
R+S
If
[A (n ay.)]
Proof.
H
7-
=
dim qlAQ
=
(A
(t ,X) =(A
R+S
S.
=
(,nr/)
())
*
Then
0X
S
X/rrX = (A (a7n))
*
R
*
® (A (ri
))
@ X/X.
With respect to these identifications,
(A (gr ()())
7rS
S
*
(A (,r/n
))
0 X/( 1 ,ak)x 0 (A (/))
R
0 (A (in
*
))
X
/r
lX
is just the obvious
map induced by the projection
X
/(9n 1 1 t)X
action of
,X)
U(<
)
on
H S(aT
is
+
X
/pX. The
on the
X/(G"A)X
factor only.
x s X, write
If
=
Then
Now
x s
Q 0 x E HS (/jl nk ,X),
u
-
o
=
Q 0 U
u =(u) modrtU(
,
X
/(,ok)X, x E
with
so
)cj and
X
Q £ (AS (4r Ql))
r ((u
-
-U(- )
o) 0 P)
Suppose
/X.
=
,
x E X.
(Q 0 P) 0 u - x.
- xc iX;
so
27.
7T
(Q 0 P) 0
((u - w) 0 P)
Q0 P 0
SS
(u)
x
x.
(u)
By
(Q 0 P 0
=4((u))
This proves the result when
i
(3.4),
X)
(u)
=
S.
=
i + 1 < S.
Suppose then that the theorem holds for some
Choose an exact sequence of
with
F
modules
U(g ) module.
a free
0 P))
((W
O
0
F0
-+
F
+
+
X
+
0,
The two long exact sequences
in cohomology give
H (An,,,F)0(A (tap)) -+H (j,nn
H
R+i
(,F)
i < S, H
Since
+
R+i
H
rX) 0(A
R+i
*R11 >H1 i+l (//La9,F )0(A R nn)
(~t)-
62
(TX)
(n,F) = 0;
so
R+i+l
H
->
62
(,FO
is injective.
By the
inductive hypothesis,
Tri+l ((u
0 P)
- 61 (o))
observed that the action of
U
commutes with
cochain level; it is immediate that
The maps
level; so
ff, intertwine
i+1
6
2 i((u - W) 0 P)
0 1)
=
d
=
F(u)
and
is
62
iT((u - w) O P)
injective,
ff
((u
d
61(M)-
u
d 0 1
on the
61 (u - W).
on the cochain
62 ri. Thus
-6
2
(r(w 0 P)).
L
is standard that the action of
that
We have
0 P).
- 7i+l (SoI)
=(u)
commutes with 621
= 62 ((u) -
- W) 0 P)
=
Finally, it
r (w 0 P)).
E(u)
Henceforth it will be assumed that
7 (W 0 P).
X
is a
so
Since 62
Q.E.D.
28.
Harish-Chandra module, i.e. that there is a k -invariant
decomposition
X =
E
X(y).
Here
X(y)
is a finite
YC
dimensional y-primary semisimple A -module.
Recall that
we are identifying A -modules with their highest
be an irreducible
F
y c A, and let
Suppose then that
F
highest weight space of F, i.e. the subspace of
by
/Za
4,-weight
of
X
subspace of
is one dimensional and has
F
y
which is annihilated
Then the dimension of
.
multiplicity of
y
in X.
X2
to be the
X C_ X(y)
Similarly, define
t -weight y.
for the
FY
Write
k -module in the equivalence class y.
annihilated by nz A
1-weights.
XY
is precisely the
U(o ),
is stable under
and the corresponding action completely determines the
action of
for
&CQk
Now rt
on
)
U(
X(y).
We make a similar convention
and
X
and
follows easily that
are semisimple
H (?7,X)
Ai
(
is as well.
modules, and it
The maps
f
may be restricted to the various highest weight spaced
50A : write
with respect to
y-
Since
2p (n a
(3.6)
(3
6)
*y-2P pHnn
AR (rt,))*]Y+
jX
® (AR
[1Hi (,(ermaX)
ARnn
)
IXy
i+R
HP(r),X)
is one dimensional and has
-2 p (4nq7)
.
,-weight
), this may be rewritten as
Tr
H
H (/n-1
qi
(
0 (AR
,X)
(1
X)Y
) * + Hi+R
,X)y-2p (9/ )
29.
To study these maps, we need some information about
i
H (Gn
,X), which (as the reader might guess) requires
If
a c WK, set
(a)
A (A)}.
a little more notation.
A(
++
{a c A (
=
-l-
)a
(Clearly this
refers to t -roots; it should be rather obvious why the
Z
subscript
in
At
has been omitted.)
As is well known,
and easy to check,
2p(A(
(3.7)
=E
a
A)
a = Pc -
a
c
-
p
c
a+
Define
1(a) = IA+(k)
W = {a
Suppose
W
|
(the number of elements of A (Jk)),
c A)},
+
y e k, with
F e y
(Kostant [16])
Theorem 3.8
y = a(y+pc
H3(Ondr ,F)"
space is one dimensional, i.e. the
with multiplicity one in
f
*
A-module.
= 0
unless
In that case the
£nk -type
y
occurs
H3(/Iyil k,F).
Consider the graded
module
*
Hom (Ar,X), with differential d (the coboundary
operator of Lie algebra cohomology.)
{Va }=
that
= j
Y(a)
can now be studied by standard spectral
sequence techniques.
A=
W
an irreducible
for some a E WK(j).
p
The maps
{a
W (j)
of 4//a
Pick a sequence
-invariant subspaces of
1u3)
A (f
so
30.
1)
V 0 = A 0 (ang)
2)
Va
3)
(7 n)ii
C
Ar(a) net); and a
<
Va c<V 0
a'
=> r(a) < r(a')
ia-1>
This is always possible: choosing x c t+ =
(centQ )(I
as in the proof of Lemma 2.7, we see that if E c A*( ,a)
is
kaA
then
x
irreducible, and
acts on
x
(nOk )
E
acts on
E
by the scalar c,
with eigenvalues strictly
*
greater than c.
Using the eigenspaces of
x
in
A
we may therefore arrange (3).
Filter
whenever
qn-r
An-r (nr
)
and
pr
a
A 0 = A, A 1 = {f~f = 0 on A (1,Z
Then
r)
n-r
Ann
An = {f
by
A
Va ,
)},
= 0
for some a' < a}.
etc.
There are
module maps
T
),X) 0 Va
a : An
a - Hom (An-r(a) (47r
For Hom (Anr(a) (\k
),X) @
=
Hom ([An-r(a) (i'trfA)]
@ V ,X);
is just the natural restriction map corresponding
n
n-ra
Then
Va c- An.
)
[A -r (a)(
to the inclusion
and
T
1)
2)
0 for all a' < a; in particular,
f e An iff
n,(f) =
a
a
iff
if f e An , then f 6 A
a+l
a'
If f
1 (df) =
A , then Tn+
a <a.
a
Tat~ (df)
=
0 for a'
<
a.
(d
Tn(f)
a
= 0.
0 l)(T a (f)) ; and
-----------
31.
Statement 1) follows immediately from the definitions.
n
suppose f C Aa
For 2),
a' < a, Q = q0A ..
'''
(An+1elemet)
EA-
n-r(a')
P = pn-r(a')+1
and
Fix
n s Val.
Val
(An element of Va'
actually a linear combination of such terms; since the
argument would be unchanged in general, we take
this form to simplify the notation.)
n+1l
aI
Tn+l (df)
is a map from An+lr(a)
(df)(Q 0 P)
T
df(y 0 A
n
i=O
..
(-1),
A
y
By definition,
V , t'O X;
(.-nn )
n
f(y
0
y. .'
.
A
n)+ E(-1)
y ]Aye...yi.
'f([y
O<i<j<n
df(QAP)
=
n-r (a')
(-1) q -f(qOA...
n
_
qn-r(a')
i
+
i=n-r (a')+1
+
i<j<n-r
iv)
+
(I
(-1
p
(a')
n
f
n-r (a')+1''
il+ f0[q','q']q.
_,)&')4
i''
i''
n
)AP)
A(
A
-rW
A
v)
+
y
indicates that the argument is to
i=O
iii)
..
Thus
be omitted.)
ii)
of
Recall that in general,
= df (QAP) .
(Here the circumflex
i)
P
E (-1)i+i
n-r (a')+1<i< j
.,p]AQAP
'*Pi..
...
A
p. ... ApP)
32.
n
f c Aa, f(X
Since
k > n
-
r(a).
A
Y) = 0
Since
r(a')
and (v) is of this form.
if
X
A (/trvk),
e
< r(a),
for some
every term in lines
(ii)
Line (iv) may be rewritten as
n
E
+ f(qO0i'''i'''n-r
...q..
q n-
,l
A (q.*P))
.
1'
(a')
By assumption (3) in the definition of the V a,
qi-P £ <VO'
0V,..
f(X A Y) = 0
V-1
'
0~
a-i>>.
1 *
a1
*
whenever
Thus line (iv) is zero.
X c A (in nk)
If
lines (i) and (iii) vanish.
ment of claim (2).
are precisely
f e A ,
Since
Y c <v0'''''
a-l >,
a' < a, P e <V0''
a-1
so
This proves the second state-
Finally, if
(d0
and
a = a',
lines (i) and (iii)
1)(Ta (f))(Q 0 P).
This proves (2).
From (1) and (2) one deduces immediately
(3) dA
_ Aa
(4) T
induces a map from
A /Aa+
Hom (An-r(a) (/n/I ),X)
0 V*
phism of differential
Q(1A
to
which is an isomormodules.
Aa
The spectral sequence of the filtration
Theorem 3.9
Eab
Ha+b-r(a) (iQ
differential
map.
ab
There is a spectral sequence Et
dt
,X) 0
now gives
a+b
=> H
; here Va- cr(a)rt(I
has bidegree (t,l-t) and is an
(rz,X).
.
I'k
The
module
33.
Corollary 3.10
H (t,X) is a Harish-Chandra module for
The corollary follows rather easily from Kostant's
theorem 3.8; in fact
already has the
Eab
ink
structure
of a Harish-Chandra module.
An apology may be in order for the proof of the preceding theorem. A slightly stronger version (i.e. smaller
ab
E aterms) can be read off from the famous Hochschild-Serre
spectral sequence ([10]).
However, I know of no applications
of the improvement, and the grubbier approach adopted above
may indicate what's going on a little more clearly.
Fix a I -type
Definition 3.11
J > 1, and
yp of
y
F e y
X
with
X(y)
4 0.
is n -minimal in X if whenever y sil,
is an irreducible A
* y-2 p(nng)
(H J-1 (nk,F) 0 [AR-J
$
module, then
0 implies X(y) = 0.
Thus the condition is that certain k -types
occur in X.
y
should not
A more computable formulation is given by
Proposition 4.8.
Theorem 3.12
TS
Suppose
: H(a(71IRX) ,X)0
y
is n -minimal in X.
[AR
HR
*+
Then
X)-2p 04p)
is injective.
Proof.
Let
r(N) = R.
N
Then
be the index with
VN
=
ARf
c
); of coirse
AN+1 = 0, and it is easy to see that
34.
ANR+n. is precisely the image of 'n.
n
*
vn:Horn (An('^r nk ),X) O V
is the inverse ofopen earlier.
R+n+1
TTN
R+n
TN .
+
In fact,
Horn (AR+nt,X)
(We can now settle a point left
Recall that we proved
R+n
. 1)(TN
NN
(df) = (d
It follows that
Ae Hom (A (rt
d(fT (A
R+n
N = image of
forfcA0
nn'
= rn+l (d 0 1(f)) for
k ),X) 0 VN; this was used in the proof
of Theorem 3.5.)
Thus the assertion of the present theorem
is that the term (ENR-N)y-2p(/qnt)
of the spectral sequence
in Theorem 3.9 "persists to E. *"
dt
is
0
Since the bidegre of
(t,l-t), it suffices to check that (Eab)y-2p(/ap)
whenever
a+b = R-1.
(Eab)y-2p(nV)
1
In this case,
(HR-l-r(a) (gykl, X) ® V )-2p
a
=
CE
(H
(MX
(y))
R-J
[-2p
*
(1)
YE:k
here we put
J = R-r(a).
This vanishes by Definition 3.14.
Q.E.D.
Because of the symmetry of
to W
H (
,X) with respect
( cf. Theorem 3.8) one can actually get considerably
more information about H (r(,X) from the assumption that y
is 7'-minimal.
This is probably important for relating
H ('7,X) to the global structure of X; but that problem will
35.
not be pursued here.
We want to study the action of
case
p
on VA7
X
is n -minimal.
U(9 )
on
X(y)
in
By definition of the action of
U
-cohomology, this is the same as studying
= H 0 (17nz<,X)E
as a
U
module.
Combining Theorems 3.5 and 3.12, we see that this
U
action factors through the homomorphism (.
be improved a little, using Corollary 3.10.
This can
Let
P c U(2A)
be the kernel of the representation p, and put
I
= IP (
) = U(c )
[U (o ),
(~
91
].
Then we have the follow-
ing more precise version of Theorem 2.9.
Proposition 3.13
action of U
I
on Y
is a two-sided ideal in U
factors through U
Harish-Chandra module Y.
+
U /I
,
and the
for any
This establishes a one to one
correspondence between irreducible Harish-Chandra modules Y
Y(y ) /
with
0 and irreducible U(j) /I
modules.
For a proof, see Lepowsky-McCollum [20].
this result to
We will apply
, and the Harish-Chandra module H (7,X).
Let
(3.14)
P : U(f)
be the composition of
+
U(f )
(
/I
with the quotient map.
Notice
own
I
-
_
- -
'-"
36.
that
2p(ftop )
is the weight of a one dimensional
4
ii
module, so that
integral.
Q/1k dominant
is
By 3.5, 3.12, and 3.13, we have proved
Theorem 3.15
action of
p-2p(nop)
Suppose
U()
on
y
is Y7 -minimal in X.
X1
Then the
.
factors through
The consequences of this theorem are dealt with in
detail in section 4, but the following example indicates
its usefulness and limitations quite clearly.
the reader may consult Schmid [23].
For details,
Suppose that
o0
semisimple and equal rank (so that i = k) and that
finite center.
Let
A c it0
-
entials of characters of T.
Let
(A+p)'
G
An element
<X,a> / 0
A
of
A+p
for every
is
a e A.
be the set of nonsingular elements of
A+p.
Harish-Chandra has shown that the discrete series of
parametrized by the F1R(1 fA
Let
A
has
denote the lattice of differ-
0
said to be non-singular if
is
G
is
dominant weights in (A+p)'.
be such a weight, and
X
the Harish-Chandra module
of the corresponding discrete series representation.
Let
A+ be the positive root system associated to A (defined in §§2)
and let
J
=
fr.
Put
y = A-pc + p(il0 ng ) (which is
the
highest weight of a K-type).
Theorem 3.16
(Schmid) With notation as in the preceding
paragraph, y is
f10
minimal in X, and occurs with multiplicity
37.
one.
Suppose
p
is ro minimal in the irreducible
Harish-Chandra module Y; then Y is equivalent to x.
Proof.
Schmid has proved that
X(y) / 0.)
particular that
j t=
${&;
so
is 7W minimal in X (in
In the present case,
)
U(
y
=U (t).
Iy-2p
-2 (gI
gp)
is just the kernel of the character yp-2p(i1V')
Thus
)
U(k
(
/J
is a character of
)
U(o
)
=
C, and
=
dp-2p ('trnf)
: U(t)
u(g)
+
+
0.
C
By Theorem 3.15 and
.
Proposition 3.13, X' and YP are irreducible modules for
U(c
) /ker
C so obviously they are equivalent and
=;
one dimensional.
So
y
occurs with multiplicity one in X;
and the second part of Proposition 3.13 gives
X = Y.
Q.E.D.
Notice that the proof actually exhibits the action of
U
on X1
rather explicitly.
This character has also been
computed (in a slightly different form) by Enright and
Varadarajan ([7]).
Schmid's proof that
y
is
7r4 minimal in X, while
basically analytic, invokes a much stronger form of the
preceding result.
With a little more work, however, the
results of this section can be applied in his argument
(cf. the proof of Lemma 6.3), giving a complete proof of
Theorem 3.16 independent of his (rather difficult) algebraic
computations: we omit the details.
In any case one doe not
get his estimates for the multiplicities of other K types inx.
38.
4.
The Classification Problem
We would like to produce a classification theory for
Harish-Chandra modules, based on the results of section 3.
Recall the algebraic ordering
<
ordering
<y
+ 2pc
1 +
2 pc
<yg +
1 y4 y22.
by
||yff
X
=
Define a second
< Y2if
y
> 2<
+
+ 2p+2p,
pc
Put
Let
i0
on
of il
-4
2pc'
2 + 2pc>
or
y2 + 2p.>
<Y
cr Y2
and
c
<y + 2p , y +2p'>
c
c
be an irreducible Harish-Chandra module for o.
To avoid some minor complications, assume that
whenever
X(y)
k-module.
$
A will
0, i.e. that
X
y e it0
is a unitarizable
henceforth refer only to
y
of this sort.
(In case X is the Harish-Chandra module of a representation
and
is compact, this condition is automatic.)
K
irreducibility of
X
implies that
{yIX(y)
$ 0}
The
is
contained in a certain translate of the lattice of f-roots
in
it
.
Hence the minimum of the positive real .numbers
{ilyl| IX(y) / 0}
of y.
is attained on a finite non-empty set
The following definition therefore makes sense.
Definition 4.1
The minimal
unique minimal element of
k-type
y
of
X
is the
{yJX(y) $ 01 with respect to <.
39.
y
||y|
is essentially the K -type with
|1Jl
The main reason for using
Casimir"
<p+p, ,
+pc>
Schmid's "lowest type"
minimal.
instead of the "compact
is that Definition 4.1 produces
([23])
for discrete series
representations; and one can (with some effort) produce
examples in which the
+p c
definition does not.
To solve the classification problem, it is clearly
enought to describe the set of Harish-Chandra modules with
a fixed minimal k-type p.
To apply the machinery of
Let
section 3, we need some 0-invariant parabolics.
p+2 pc; recall
be the positive root system associated to
that if
a e At(c
and a
-a.
),
a c A+ iff <a,y+ 2 p >
> 0 or <a,-+2p
;>=
Thus we have a 0-invariant Borel subalgebra
as in section 2, with nil radical r0 ; A( 0
0
A+
It remains to define the parabolic
0r.
=
+
It is not clear
how to do this in general, but examples (cf. Theorem 4.6
and Corollary 4.16) support
Conjecture 4.2
0
0
jr
The 0-invariant subalgebra
may
be chosen in such a way that
i)
Chandra
whenever
y
is the minimal k-type of the Harish-
factors through the homomorphism
ii)
U(
module X, then the action of
the Levi. subalgebra
(Definition 2.4)
V
('
of
)
on
X
(Definition 3.14)
el
is split
0
40.
iii)
the
aQOk -type of highest weight V-2p(litnp ) is
small
iv)
to
if
p- 2 p(t en
v)
)
y-.2p (fi-n
),
and
y+2p c-p
is
a small
-type associated
yp' is dominant for A ,then
y
y
is dominant with respect to the positive
imaginary roots supported on
For the definitions of "small" and "associated",
see section 5.
Conditions iv) and v) are needed only to
prove the existence of the representations described by
Theorem 4.5.
The philosophy is this.
the center of
algebra of
Set ci
C.
=c01(y) =
' c [ , ] CP0
Let
(10 0 I
+
+rz;
then
G
with the reductive group
realizations of Section 6).
L = G
.
t
+ C7
V+ is
be a Cartan sub-
(which exists because
of & , and hence also of V.
concentrating on
Recall that
is
0
is split).
a Cartan subalgebra
Harish-Chandra often works
(which will appear in the
The present development is
According to, for instance,
Kostant's quantization theory, the really fundamental
objects are the Cartan subgroups
G
;C+a
.
The problem with dealing directly with these is that (in
Kostant's language) the associated polarizations are
41.
usually neither real (the split principal series case) nor
purely complex (the discrete series case.)
In Harish-
Chandra's point of view, the discrete series are taken as
fundamental building blocks, and one works essentially with
Here
real polarizations (i.e. induced representations).
we start with the principal series for split groups, and
use purely complex polarizations (i.e. 7 -cohomology).
When Conjecture 4.2 holds, the "uniqueness" part of
the classification problem reduces to a study of the
representations of a split group containing a small
K-type (see Theorem 4.5 below).
These representations can
be described almost completely, using Harish-Chandra's
subquotient theorem (Theorem 5.2).
The result is
Let G = KAN be an Iwasawa decomposition of a
Theorem 4.3.
split reductive group, and let y E K
is an analytic family
G
tions of
i)
ii)
ffv
Then there
of admissible representa-
(namely a certain principal series) so that
ffr contains
K-types in
{V}V}v6A
be small.
fr
y
exactly once; and the other small
are precisely those associated to
is irreducible for almost all
finite composition series in general.
p.
v, and has a
-_
__
a
-
42.
iii)
and
VI have equivalent composition series
V = a - v'
iff
W
fV
of the Weyl group of
By i),
ff
A
in
G.
has a unique irreducible subquotient
containing the K-type p.
iv)
in a certain subgroup
a
for some
fw
Then
The representations
{In-)}
y
exhaust the infinitesimal
equivalence classes of admissible irreducible representations
of
G
containing
p.
Since the only proof available for i) is very long,
this theorem will be proved in section 5.
Corollary 4.4
An irreducible Harishf-Chandra module for a
split group contains A-types from at most one family of
associate small .- types.
minimal
A-type
If such a family exists, the
belongs to it, and each element occurs with
multiplicity one or zero.
Suppose that Conjecture 4.2 holds for the h-type p;
choose dr
accordingly.
Recall that 070
0
X(h.
is a maximal abelian subalgebra of
Theorem 4.3 to the split reductive group
knjl
type
p -2p(nn
).
Suppose
v
0
0'~ 0
p(y) = A(p).
L
+ 00
0
Apply
L, and the small
We may choose the subgroup
from the Iwasawa decomposition of
O7(y)0.
0
A (p)
to have Lie algebra
By 4.3
(i),
the highest
43.
weight space of the
type y-2p(mnap ) in
f
one dimensional module for
x
Let
: R
+
Recall the map
: U(o5)
+C
Theorem 4.5
C
a
(?).
/I
R =-U(Y)
is
17
be the corresponding homomorphism.
E
: U(5)
E
by
Let
X
R
+
=XV
(3.14); define characters
Then
P.
*E
be an irreducible Harish-Chandra
module with minimal k-type y, and suppose that Conjecture 4.2
(i-iii) holds for p.
With notation as in the preceding
*
paragraph, there is a
U(5)
y
on
X
v E 0O(y)
so that the action of
E ,
is given by the character
In particular,
occurs with multiplicity one.
Proof.
By 4.2 (i) and Proposition 3.13, X11
irreducible module for
FP(U)
image of the center of
U(J)
U()
+
Chandra homomorphism,
extension.
&(I (ig))
integral extension; so
for
R .
XP
-+
Z
is an integral ring
is commutative, and Z
y
Hence
+
R
'
)
C
R P(Uis an
lifts to an irreducible module
R
in the P-2p(7n)-type of some
irreducible Harish-Chandra module for i?.
PJ
be the
By Proposition 3.13 again, this lifted module is
just the action of
R
Z c R
is essentially a Harish-
By Corollary 5.5, R
is an integral extension.
Let
under the quotient map
)
P(I
R ; since
cR .
is an
By Theorem 4.3 (iv),
must therefore act on XP by the scalars X
some
44.
*
v e 6T(p)
.
But this is precisely the conclusion of the
theorem.
Q.E.D.
It should be remarked that the ring theory here is all
avoidable.
We will prove Conjecture 4.2 (i) by exhibiting
explicitly certain cohomology groups on which
P
acts;
so the necessary liftings appear automatically (although not
uniquely, just as above).
In some cases the construction
of the cohomology is a little subtle, however; and it seemed
best to make Conjecture 4.2 as simple as possible, despite
the resulting uglinees of the preceding proof.
The existence theorem corresponding to 4.5 is
Theorem 6.2.
By a straightforward reduction, it suffices to prove
G
Conjecture 4.2 for
Lie algebra.
ridiculous.
Theorem 4.6
simply connected and
g0
a simple
A case-by-case attack is therefore not utterly
Let
G
denote the universal covering group of G.
Conjecture 4.2 holds in the following cases
has only one conjugacy class of Cartan subgroups
a)
G
b)
G = SL(n,R)
f)
G = SO (2n)
c)
G = SU(pq)
g)
G = SO(p,q)
d)
G = SP(pq)
h)
G = split form of G2
e)
G = SP(nR)
i)
G = rank one form of F 4
45.
For every case but g), we will actually prove the following
stronger versions of parts of 4.2;
4.2 (i)':
Whenever p is
the minimal a-type of the Harish-
Chandra module X, then y is /7-minimal in X. (Definition 3.14)
4.2 (iii-iv) '
P-2p (flo ) is principal series minimal
(Definition 5.3)
That 4.2 (i)' => 4.2 (i) is precisely Theorem 3.15.
That
4.2 (iii-iv)' => 4.2 (iii) and 4.2 (iv) is trivial from the
We will in general say little
definitions(see section 5).
about 4.2 (v) - the reader may observe that in each case
much stronger conditions will be proved in the course of
establishing 4.2 (i).
We will say that the .k-type p is ?I -minimal if whenever p is the minimal a-type of an irreducible Harish-Chandra
module X, then p is /Z -minimal in X.
The first step toward
proving Theorem 4.6 is to give some computable conditions
0
r
for p to be 1 -minimal, for a fixed 0-invariant Yr
Definition 4.7
q
The weight
t
is said to give rise to
J-cohomology larger than p if there is a A -type
that if
then
F
is an irreducible
H (rF)
Write
A
y > p
module of highest weight y,
/ 0.
A(lnp) = {
,.-
-,
so
R }' A
{a
00
. a }
(the noncompact and compact ,t-roots in In, respectively).
46.
Proposition 4.8
Suppose that for any non-empty subset
-.-.- . } of A(fng),
{i
1
the weight
-
does not
-.
JJ
give rise to J-1 cohomology larger than y .
Then p is
H -minimal.
Proof.
Suppose that
F
is an irreducible
A
module of
highest weight y, and that
I *) p2 (')
(Pns,F) 0 {A R-(I)
(H
0.
By the definition of 7-minimal, together with
Definitions 4.1 and 3.11, what we must show is that
y < p.
By a standard fact about highest weights in a
tensor product, there are weights
HJl(fn,F)
$+$
0,
= p-2p(uiW).
$ is
Thus
) =
=
E:A (qnn)
$= y-6. - --.
,
and
$
such that
a weight of [AR-J (ngp)],
= -(61 +...+3, R-J)
$
R-J element subset of A(,tAno).
R
2 p(vnu
$
E
and
for some
Recall that
..
It is immediate that
{6.
}
is the complement of
H
(-nleF)
i=l1
where
'1JJ
in
{S}
A(fop ).
Now
/
0; but by hypothesis,
$ does not give rise to J-1 cohomology larger than y.
Definition 4.7, this forces y < p.
Suppose then that
y-.
..
1J
.-6
J > 1, { S
By
Q.E.D.
J
C A(rtn6),
}
j j=l
and that
gives rise to J-1 cohomology larger than p.
47.
Let
y
be the corresponding k-type.
{ . },
restrictions on the
IliiI-I IYI ; since
To get some
we want to compute
y > y, this number must be non-positive.
By Kostant's theorem 3.8, there is a
(4.9)
y
-
3
..
.
= a(y+p
10*
a 6 W (J-1)
with
- C'
c
Thus
y+2p
a -l
Since
IIyII
= a
[(y+2pc
~
...- . -(p cP
i
1
*)]
J
is an isometry, it follows that
= <(y+ 2 pc )-u
-(p c-ap c),
(y+2pc
By a short computation, this may be written as
yut 1 -IIy|I =2<p+2pc ,Z .
By
(3.7), pc-a-pc=2p(A
+p c~a
.+pc - c
J
a
So Jy
I--|JY|=2<y+2pc U i.
X = X(y)
A(AA),
i.
J
J
y j~-||yj|=2<p+2p c~4' U
inition 4.10
'
c
+pc
J
Since a 6 W (J-l),
(W
is a J-1 element subset of
A+
Def
c
i.
J
<
J
.
say {a.
J
J
J
...a
}.
J
+<2p-(Ui +Ea
),Ei +,O. >
J
J
J
J
= y + 2pc -~
'
48.
For an equal rank semisimple group, (4.10) is the
equation relating Harish-Chandra's discrete series
parameter A and the discrete series' "lowest" h-type y
This observation, which came out of some
(Schmid [23]).
very helpful discussions with H. Hecht, was the original
motivation for Definition 4.1.
{B
a
Set
. .
'1
} U {. 1....
.a
A
J
'1
J-1
().
The
last equation may now be written as
|
y
(4.11)
|
2<X (y) ,
E i
+
E
J
J
+ <2p(B ),
)
A
Bc
Here
-
2p(B)>.
B, so that p (Bc) + p (B)
= p.
Suppose we can show that the right side of (4.11) is always
positive
(Or zero, with y
< y)
. Then Proposition 4.8 will
0
77 c f7 , and
0 (by the
X+p = p+2p c is dominant with respect to /'
0
So <X+p, 3 > and <X+p,a > are non-negative
0).
T
of
choice
is
y
imply that
T{ -minimal.
Now
for all i; it follows that the first term of 4.11 is never
very negative.
The second term is easy to handle, as we
shall see.
If
(4.12)
T 6
W(jA), set A+ 6
2pj +
+ a
={
P-T*p
A+
a e Al}.
Then
49.
Because it preserves A, it is easy to see that the Cartan
involution 0 defines an automorphism
We claim that
T
=
T
<=> Tp
AT ()
For suppose that
A+ (o) is O-invariant.
c
is 0 invariant.
+
2p(A+(g)) is 0-invariant, i.e. is in t
*
since p s )t, this implies
If
6(T-p)
T~1(OTe)-p =
p.
Suppose
Hence P-T-p 6
.
p
A ()
*
then
i.e.
W(f,A)
Finally, if
B < A , Bc
> 0.
Then clearly
(eTO)-p = T-p,
T = OTO.
with 0, it is clear that
<2p(B),2p(Bc)>
*-
The only element of
is the identity, so
Lemma 4.13
.
T-p
since Op = p,
= T-p;
OTO of W(,A-1).
+
T
fixing
T
p
commutes
is 0-invariant.
_
+ - B.
Then
Equality holds iff B = A ()
for
some T s W(t,A) which commutes with 0.
Proof. (Kostant)
By a short computation,
<2p(B),2p(Bc)>
=
<pp>
- <p-2p(B),p-2p(B)>
Consider the finite dimensional irreducible representa'
tion of
to
A) are
with highest weight p.
Its weights (with respect
precisely the various p-2p (B),
B
C
A+.
By the
Cartan-Weyl theory, each weight of a finite dimensional
representation is at most as long as the highest one, with
equality precisely for the extremal weights.
<pp>-
Thus
<p-2p(B),p-2p(B)> > 0, with equality iff
50.
p-2p(B) = T-p
for some
p-2p(B) =
iff
T-p
T
e W(g, A).
One knows that
B = AT (+).
*
p
E;
,
<p, p> = <p,p>h.
so
so <p-2p(B),p- 2p.(B)
p-2p.(B) is a sum of roots,
> 0, with equality iff p-2p(B) E
Thus <p-2p(B),p-2p(B)> > <p-2p(B),p-2p(B)
,
with equality
*
iff
p-2p(B) e
Assembling these inequalities,
.
<2p(B),2p(Bc)>
=
<p,p
<p-2p(B),p-2p(B)>
-
B = A+ (),
with equality iff
and
T'p
F-
> 0,
Q.E.D.
.
A noncompact imaginary positive root 6
Definition 4.14
is bad if there is a
a 6 WK
such that
+K
a)
A (k)
b)
X is singular with respect to 6 and each
consists entirely of complex roots
a E ACY()
c)
if the noncompact imaginary roots in A ( ) are
($
1 +...+$-6)
then a
ordering f
Proposition 4.15
i)
ii)
iii)
X((p)
of
0
it0 .)
Suppose
X(p)
satisfies
is dominant with respect to la
<a, A(p)>
A.(Rng)
> 0 for a e A
(7za)
contains no bad roots.
.
(in the
51.
Then
y
Proof.
is
.7 -minimal.
By Lemma 4.13 and hypothesis i), both terms in 4.11
are non-negative.
If one is positive we are done; so
suppose both terms are zero (so that
Lemma 4.13, there is a 0-invariant
B = A ().
y
y
-r e W(1,,A)
f.)
By
such that
Since the first term of (4.11) is zero,
0
=
a c B.
whenever
a1 ...a.
a.
li
'J-i
follows that
it
By hypothesis (ii),
A (r114).
C
are all in
Since
B
is 0-invariant, it must contain the J-1 corresponding
the discussion preceding
noncompact complex roots (cf.
Proposition 2.3).
So we may assume
a.
1-
J
j
=
S.
The remaining
1...J-1.
=
.
1.
for
J
is necessarily 0-fixed,
JJ
By
e: A. (4T OP).
i.e.
(4.9),
J
a1
+
= crJ-l
j=l ~
-1
By (3.7), a
So
y
=
a
Since
y+2 pC -p
=
y
+PC)-p
Za.
E-
< ca>
)-p
=
a
-l
(-PC+G-P +pC)-P
=
c
0.
8.
= 0
a (y+2pc -p).
(4.12), we get
+P
C
c
for all
a e A(k ), a*A
A; i.e.
Rearranging , and applying (3.7) and
52.
(*) y-a-y = p-a-p-2(p -OP
Now
A ()
C)
A ( )
contains
2p(A ())
=
-
and the corresponding non-
compact complex roots, together with (say)
Clearly
2p(A
a
(
= 4p(A
4p(A +
a (p))
+ $
..
N
$ .. 4Y.0
Thus (*)
.
becomes
(**)
P - a-P = $l + ...
Every compact root in
follows that the
*
is
already in A ();
it
a
are all noncompact imaginary roots.
a
(J)
By (**)
y
= a
1
-
y
a
1
-I
= y +
1J
=P + a
What we need for tP -minimality
Iy|| = 111j11, this amounts to
is not a bad root, y-y = - a~
Corollary 4.16
i)
ii)
iii)
X(y)
Suppose
is
7
A(p)
P > Y; since
y-Y
($ +.. +$~-
Since
)
6-1
J
Q.E.D.
satisfies
dominant
A(-y) is non-singular with respect to Ai
A.R (n0'
0QP) contains no bad roots (Definition 4.14;
in particular if X(p)is non-singular with respect
to
A.
0
53.
y
Then
is
such P, with
Proof.
7Z
0
minimal.
.Q=f,
Thus Conjecture 4.2 holds for
and 6Y)
0.
0
0 F0
i =A is
By Proposition 4.15, y is 7f0 minimal.
abelian, and therefore split, with every
iAk type principal
Q.E.D.
series minimal.
It is fairly clear that the conditions of 4.16 hold
for "most" y , i.e. when V is sufficiently far from every
root wall in it 0*
In that case Enright ([6])
has in fact
proved Theorem 4.5 (that the representations with minimal
type y are more or less parametrized by
(5))
using a much
stronger non-singularity condition on yp. The condition
given here seems to be more nearly precise; this should
be clear from the considerations of section 6.
An example may be enlightening.
Let
G = SL(3,R).
The
A -roots
of
V are pictured
below.
Here
~
is one dimensional
(the vertical
direction) and Y
is one dimensional
(the horizontal).
54.
The complex roots (i.e. those with support on pt) are
{+a,+a};
is imaginary and noncompact.
There is one
compact positive root, namely the compact -t-root
corresponding to the complex root a; we call this a also,
Then the noncompact t -roots are a and
as a t-root.
=
2a.
(All of this is easily checked by finding the
eigenvectors of
in s2(3,R).)
respect to
2r e z.
T
=
A+(.)
=
y
ditions for
= p+2p
0
-sin 0
cos 0
0
0
0
l
{a}
y =
system is necessarily
1
sin O
c K = SO(3)
A ;-weight ra is dominant integral with
So let
p = -(a+6a+
cos O
to be
) =
=
(=
iff
a;
r
the associated positive root
{a,6a,)
n70
is non-negative and
= A(77 0 ).
= 2a as a
/t-root.
n-2 )a.
= a, and
2p
minimal.
If
We want con-
So
n > 2, A
is
0
dominant and non-singular with respect to ? . By
It
0
minimal; since dim P0 = 1, there
Corollary 4.15, y is 7
is at most a one (complex) parameter family of representations of G with minimal k-type p.
(It is not hard to see
that these may be realized essentially as a principal
series associated to a maximal cuspidal parabolic subgroup
of G.)
If n = 0, 1, or 2 (i.e. p is the 1, 2, or 3 dimen-
sional representation of SO(3)) then yp is in fact principal
55.
series minimal (see Table 5.8).
Theorem 4.3 describes the
2 parameter family of representations of G with minimali-type p in those cases; in particular, there are too
many representations for yp to have been
minimal.
7t
Let
us see why Corollary 4.16 does not apply to the case
=
Here
a (i.e. n = 2, the 3 dimensional representation of K.)
n-2
0
2
X
AiR~rdt)
a= 0, so
X
is dominant with respect to I'{
is empty, since there are no imaginary compact
roots; so the second condition of (4.16) is vacuously
satisfied.
Hence the third condition must fail, which
means that
must be a bad root in the sense of
Definition 4.14.
Set
a = -1
This can be seen directly as follows.
WK
complex root; and
A+()
Clearly
A (k)
A (G) = A+
=
{a}, and a is indeed a
{a,6a,}.
As t-roots,
therefore consists precisely of A+ (k),
noncompact root, and 6 = $
bad root.
= 0, and
so a- (
is a
Finally, the problem can be seen from the per-
spective of Proposition 4.8.
roots a and 6 = 2a.
=
the corresponding
-3a
a-3a
Consider the two noncompact
Then
=
-2a
=
a(a+pc
-P
c c
=~~
atc)
~c
Thus the k-type y has 1 cohomology of weight y-a-8, which
prevents y from being '71
0
minimal.
So far we have classified the irreducible admissible
representations of a rank two group, explained a mysterious
56.
definition, and illustrated the structure theory of
section 2 in one fell swoop.
Still more mileage can be
gotten out of this example, however.
One might hope that
all discussion of higher cohomology is irrelevant.
There
are a number of results about the uniqueness of the discrete
series (see for example [23]) which assert that if a
representation contains a A -type y, but does not contain
p-f for any noncompact positive root 6,
then the representa-
tion is equivalent to a certain discrete series representation; or in general that there is at most a (dim Vt) paraAll of these results
meter family of such representations.
require strong non-singularity hypotheses on yp, however.
In the present case(G = SL(3,R),-p,=a) the minimality of y
in
X
y-S
= p-2a = -a
-a
implies that
=
0
cannot occur in X; and
is not even dominant.
Yet there are a
great many representations with minimal k-type y (more
than a dim
P
= 1 parameter family.)
one must consider the fact that y-a-
To explain this,
gives rise to
1-cohomology which can occur in X.
Proof of 4.6(a).
Since
of Cartan subalgebras,
subalgebra of
%0.
and the roots of
roots of 4A in
.
j0
b
has only one conjugacy class
must be a maximal abelian
.
0j
Thus (
)
in
are obviously the imaginary
So
A(
t' nJ
Now
is empty.
Since
57.
AC(C
I
q
)
( 1NC(
by Proposition 2.3, it follows
that every t -root is equal to a compact 1t-root. In
0
particular yp is dominant with respect to fP . To prove
that the conditions of Corollary 4.16 hold, it is there2pc p
fore enough to show that
to
is dominant with respect
n
('fO
70 and non-singular with respect to A.
= p(A
AiR (I
(n'/14))
by the preceding remarks.
is a root system in its own right, with Ai1R014)
as positive roots.
It follows immediately that 2p -p is
dominant and non-singular with respect to A
(a,6) be a simple positive root of A.
<2pc
>
-
proved; so assume
0.
If
34 0.
Then
We claim that
is not a root.
Let
X
weight (a,6); put x
=
x k+
XL
,X
E;,
(7VtA ) .
Let
We want to show
6 = 0, this has just been
is not a root.
xK 6
).
(a,)
(a,f)
(a,-=)
(O,2 )
-
+ (a,-8)
(2a,O)
be a non-zero root vector of
X + OX, xs = X = 2X, and x
-
OX.
x,
Then
If
20X.
=
(2a,O) were a root, [X,OX] would be a non-zero root vector
for (2a,O).
But [X,OX] =1
+,
=
-
[x ,
No such
so (2a,0) would be an imaginary noncompact root.
roots exist, which proves the claim.
theory, <(at),(a,-F)> = 0, i.e. <a,a>
E
By abstract root
<,>
,
or
58.
=
1=
Since
(a,)
is
simple,
'Pd>
) >
(ac,
2 <p
<aa>.
=<aa,
i.e. <p,ca>
==<a"a>
a is necessarily a positive compact root, <2 pc a
so < 2 p ,a>
c
2p-p
-
k
<p,a>
k
> 0.
of SO(n).
Here
> <a,
This completes the proof that
0
is /Y -dominant, and
Proof of 4.6(b)
Since
4.6(a) follows.
K = SO(n), the universal cover
We omit the famous case n = 2 (which is of
course well understood; in any case it will appear
later as
SU(1,1)).
cover of SO(n).
Thus K = Spin (n),
a two-sheeted
For definiteness assume n is even; hence-
forth it will be written as 2n.
On the SO(2n) level, a
torus is
cos
sin el
1
-sin 6
Take the
as coordi-
cos 61
1
.
nates in t;
1
in the dual
coordinates,
cos 0
n
-sin 0n
identified with
Rn.
usual inner product on
sin 0n
cos 0
After normalization,
R
0 is then
<
,
>
is
the
59.
A straightforward computation produces the weight
vectors of t
e
basis
q
=
are
p
iLJ
of
. ..
++
+ e.
_=
s9 (2n,R). With respect to the usual
in
n , the compact roots are
1 <i< j < n.
+
+
+ :E,
j
1 .
and
p
=
1
The noncompact roots
+
2s..
-
has
1
dimension n-1.
Clearly the complex roots are the
q.
for according to Proposition 2.3, a
and p:
;
- root occurs with multiplicity two iff it is complex.
The imaginary roots are the
Order
itr
p+
lexicographically.
=
compact roots are the
q.
,
2pc = (2(n-1),2(n-2) , ... 2,0)
weight
(a1 .. .an)
decreasing,
all
c Z +
a
i.e.
V
y+2 pc =
.
of
q
++
so that
by an easy computation.
is dominant integral if f
.
Suppose
(yl...
is dominant integral.
Then
1+ 2 (n-1) ,...yn- 1 +2 yn).
Clearly
Pn)
(a )
Recall the definition
++
<p--, y+ 2 p > = <q.-, y+2p > > 0; so
ij
c
13
c
(y )
is decreasing and y n-1n
i < n.
Thus
<p+,y+ 2 pc> > 0 for i < n.
for
or
is a k-type,
0+U>1
c A(00) . Since
> 0
A
is
a. e 2
an-l + a > 0, and either all
n1
n- y =
0++
Then the
If
60.
gnT
in
0, <pn
and
p+
P
n
cally.)
sgn
0, <p,
P
n
> =
+2p
+pC
(0) = + ; then we have shown that
0(=
A
Aa{pip,
sgn yi
++
Pi
p = 2 c + (1,1 ...l,sgn vI ),
One computes immediately that
so that
y+ 2 pc-P
(4.17)
If
2
' ' ' ' ]n-l~
s = n; otherwise let
y n = 1, set
fI
the largest index with
> 1. Let
' n-sgn yp
n)*
s
A
denote
be the
0-invariant subalgebra corresponding to the t -roots
perpendicular to
A ()=
c . .s
}i,j
, pT,
{q-
p-
> s}
be the span of the root vectors corresponding
Let p
to the remaining positive roots;
A (q-) =,
(4.18)
Then
&
The span
clearly
=
J @ 7
<s+l ...
++
Z_ r-,
,troot (P )
0
is ordered lexicographi-
it0
(recall that
Define
If
> 0.
,+2p>
c
++
p
sgn yi
, p.
|i < s}.
is a 0-invariant parabolic.
of the . -roots of
n>; and
{p
Ai > s}
S
/ in I;
is
is
a strongly
orthogonal spanning set of noncompact imaginary roots for
61.
*
is split; in fact
By Proposition 2.5,
).
root (
0 is
it is easy to see that the semisimple ideal of
s9 (2 (n-s) ,R).
isomorphic to
We check first that p-2p(40fl2) is principal series
minimal for /
.
=A
s =n, then
If
there is nothing to prove.
is abelian, and
Suppose then that
s < n.
From (4.18), we deduce
~~
f++-
~
A
Thus
2p(r op)
and
V-2p(Innp)
of s, Ip+1
<
1
I
vanishes on the last n-s coordinates,
=
(.
s+1'''n)
1; since
y
By the definition
is dominant integral, it
follows that the only possibilities for
(p s+1'''in)
are (1,...l,0,...O) (with at least one zero),
1
1
and (1
Consulting Table 5.8,
(.__).
( . ,
we see that these are all principal series minimal for
the semisimple ideal
immediate that
for
s9(2(n-s),R)
y-2p(fa(Fp)
0 ; and it is
is principal series minimal
i .
To show that
tion 4.15.
Since
y
/t-minimal, we apply Proposi-
is
Ag (,k)
is empty, condition ii) will
always be vacuously satisfied.
0
and
of
y
> 1; thus
1- =
-.
Suppose first that
Since
(Vi)
s = n
is decreasing,
62.
y 1
for all i.
It is clear from (4.17) that
dominant with respect to 70.
positive imaginary root
i t
Suppose
a e WK
WK
is bad.
Rn
it0
A (
that
)
{$1 ... $2}.
then
contains on even number of
A (9o),
say
0.
One knows
say
-1 +
a pr
A
p
If
Then
p+
r
($1 +...+ N2-p
is
It follows easily
There are now two cases.
pr
pi
by permutation and
changing an even number of signs.
a
Notice that since
is as in Definition 4.14.
operates on
is
Suppose a noncompact
is ordered lexicographically, every
0
that
p
y
2+..+
Every
is a
a'$
negative noncompact imaginary root, i.e. some
p..
Thus their sum is <0, contradicting c) of Definition 4.14.
-1 +
and
1(-pr+
A+ ' then a-l
If, on the other hand, a pr
-1
is a negative noncompact imaginary root;
every a $
+
-1
(r+.
a
so again
contradicting 4.14 c).
<r
So there are no bad roots in this case; by Proposition 4.15,
y
is 770 minimal.
ps
In every other case,
i < s.
By (4.17),
sgn X(p)=
A(p).
sgn y..
/
>
1,
' so that
fly
>
so
X (p)
1 for
0 for i < s, and
We have seen that
sgn y.
AiR(17op)
=
{pi
ji < s}
(see (4.18)),
is
63.
dominant and non-singular with respect to A
In particular there are no bad roots.
X(p) is 7{ dominant.
p ).
(tR
We claim that
By (4.18), it remains only to show
{p-Ii < s}; recall that p-.
this for the roots
=
iJ
as t -roots.
y
. Since i < j < n,
p
Consider for example
> 1.
|y
is positive; and we have seen
ij
> 1.
Thus
Also
= <e.+E.,+>
<p i,
iLJ
If
p. < 0, <p.
J
J
,A>
Suppose then that
If
y.
=
1i
c Z, then y.
=
y
y.+y.
1
= y.+y .-1-sgn yj.
1
J
J
> 0 since y is dominant.
> 0, so that <p..,X> = i.+pj.-2.
iJ
3
1
6 Z + T, then yi > _,
y. >
,and
> 0.
97
j
so y.+y.-2 > 0.
C Z +
and also P
In every case <p
A similar argument can be given for p
indeed
1
> 1 implies p. > 2;
If y
y +p -2
J
.
1 ; so
,X>
Thus
>
y
0.
is
dominant, and so rf minimal by Proposition 4.15.
This proves 4.6 (b) for
SL(2n,R).
SL(2n+l,R)
is
similar but considerably less messy; so the details are
left to the reader.
To continue, we need a refinement of 4.11.
Let
v 6 i I; then by 4.11, with notation as in that situation,
WIMM
64.
| fy [-j
(4.19)
Lemma 4.20
Suppose
1
v
=2<X(y)+
v =
E
C
+
> +<2p(B )-v,2p(B)>
J
J
c6, 0 < c
1.
<
6cA-
<2p (Bc)-v,2p(B)>
Then
i)
B =
ii)
Li( )
>
Equality holds iff
0.
for some
e W(gA) commuting with 0.
T
0 for all 6 e A+ ()
<6,p-T - p
with c.
/
0.
By taking appropriate convex combinations, it is
Proof.
enough to do this in case c6 = 0 or 1; so that v = E6
for some subset
e A+O)6
C0 =
C
A+ ( ) .
{6 } of
- {6 i<2p(B),6 >
Define
6i, any i, and <2p(B),6>
> 0}.
k
Considering the positive root system for
C = CO
2p(B), it is easy to see that
some
W(T,x) commuting with 0.
T1
() <2p( +
Since 97
Put
> 0}
l
} - C 1 ; so 2p(C
C 2 = {1.
T
(C), for
By Lemma 4.13,
))-2p (C0)-2p (C ) ,2p (CO) +2 p(C )>
is invariant under
defined by
=0.
, <p(n), 6 >O =0 for all 6 ElA+;A
2)
these observations, we compute
+ 2p(C 1 ) = v.
Applying
65.
-
+<2p (C 0 )-2p (
<2p-2p (B) -2p (C) -2p (C),2p (B)
= <2p-2p (B
COV Cl) ,2p (B U CO
+ <2p(C 0 )-2p(C 2 ), 2p(B)>
write
-2p (C 2 ),2p(B)>
= <2p-2p(B)-2p(C)
<2p(Bc )-v,2p (B)>
.
-U
<2p (B c) -v,, 2p (B) >= (I)
(B)>
<2p-2p (B U CU Cl) ,2p (C00 Ci)
Call the first term (I), and
+ 2p(A+(k )).
2p = 2p(71)
2 ),2p
Then
-<2p (7Z) ,2p (C0
C l )k+<2p (B) ,12p (C00 Cl) >
-<2p (A + (k) )-2p (CO0) -2p (C1) ,2p (CO0)+2p (C )> +<2p (B), 2p (C0)-2p (C2
= (I)
+ <2p(B),4p(CO)+2p(C
remarks following it.
by
)-2p(C 2 )>
(*)
and the
By Lemma 4.13, (I) > 0; and the
second term is non-negative by the definition of Co' Cl,
This proves the desired inequality.
and C 2 .
equality, necessarily
and
C
C1
0and
E>
-<2p(B),2p(C2)> = -<2p(B),H
In case of
are empty, (I) = 0,
0.
=
By Lemma 4.13,
this is equivalent to the conditions stated (recall that
p-T-p =
2 p(A(
By choosing
)).)
Q.E.D.
v = Ec 6 6
be possible to arrange for
appropriately, it will always
X(p) +
be dominant with respect to 7
0
v
=
=
to
X((p,v,)
in the cases we consider.
(The reader may easily see how this would be done for
st(n,R).)
This is the strong form of conjecture 4.2(v)
L_
66.
which was alluded to after the statement of Theorem 4.5.
It is easy to check whether
for let
6 i.
A
is n
0 '
dominant:
be the simple roots in
770.
Since
<6 i . p 2
___7
<6.1 ,6.>
1
<y+2p +
c 2
=
1,
X
>>
will be dominant if f
for all
2
i.
X
Furthermore
is
singular precisely with respect to those simple roots for
which equality holds.
To illustrate the methods, let
and put
G = SU(p,q)
y = 0, the trivial representation of K.
(p < q)
K
is
S(U(p) x U(q)), and has a torus consisting of the diagonal
matrices
i01
e
iG
T=
e
re.
P
,1
i4~
1
0
e
Ia
+ 4.
3
= 0
67.
t0 may then be identified with the subspace
.. 9),($...)]46e.
{[(
if 0
similarly
+
$
0}
R
of
Rq;
is identified with the same space, with
the usual inner product for
rank group, so
=
0.
<.
G
is
Identifying
an equal
with
0)C
=
±
sP(p+q,C), one checks easily that the roots of
in
are the following (with respect to the basis
{6 , E21
i
1<
< p
< i
j
compact:
1
e1
i,i
< q} of
i
noncompact:
g++
g-i3
2
12
-
i3
1 < i < p,
J
We will sometimes refer to
s.
as the j-coordinates.
lexicographically.
Then
e.
Order
i
34 it
< q, j
j
< p,
i'
1 < j,
1_2
+
( Rg)
-- i
"i
f2
R
1 < j < q.
as the i-coordinates and
c.R
it
0J-
A+
{e
Ni
C Rq
J,J
and one computes easily that
2p c
(q-1, ...q-2j+1,...-(q-1))]
lp-3,...p-2.141,...-(p-1)),
'[p
Since
y = 0, y + 2p
A(rfn
)
= 2pc.
By the definition of
IF ,
=
{gt.p-2i+l > q-2j+l} U {g jp-2i+l < q-2j+l}
=
+
{g. .j1)
1
i > -(q-p)}
-
{g
j
-;
|i
1
< 1~-p
-)
68.
Assume for definiteness that
is a non-negative integer.
q-p
is even, so that
The simple roots of
2
are
A
precisely those which cannot be expressed as a sum of
two other positive roots.
Using this, one can check that
the simple roots are
i2
f
< qJ
<
qj Vilg
g
9 ,---+1
jPpj.
22
g i,9
-
1
l<i<p}
Q-0- be the (e-invariant) parabolic corresponding
Let
to the simple noncompact roots
{g+
}.
Since these
are strongly orthogonal, the semisimple ideal of
0 is
a product of p copies of sZ(2,R); so
TO is split. We
0
leave to the reader the straightforward verification
that
Set
+2p
yp - 2p(-no(q)
v=
c
is principal series minimal for
Then by direct computation
E g
.
i=l
,~'2
+ 1
1
(p -'...p-2i+
q -2( q-~p +1
1
q-2(
/O'
3
3
2,...-(p- 2 )),(q-l,...q- 2 (q-p
2P)+1,
1
1)1.
From this one computes
1e-(S+)
.q-2j+ 2...,q-2
69.
j+1' y+2pc+$--v>
<f j,
2
iu+2p
3
v
+
2
2 '
<g
yrp+2pc+v
= p-2i+
-
(q-2(-i)
+
1
'2
= 1
1l< i<p
13
=q,Pi+2P c+-v>
2( PP-2i.
2 k4)"--(-i~
<g
1
i'2
2 < i < p
= 1
<,j2 g
q-p
'2
7
+2p
=
) = 3
(p+)-(p- 1(P-
Since all roots have length 2, the remarks following
0
Lemma 4.20 imply that X will be 71 dominant if
1
> 1 for each simple root 6; this has just
<V+2pc +v,6>
been verified.
With the usual notation, suppose now that
J
gives rise to J-1 cohomology larger than y;
-
j=1
j
recall that
and
are the corresponding compact roots,
{a. i
j j=1
B = {a. I
1.
U {.
i.
J
}.
By
(4.19) and Lemma 4.20, there
+
is a
T e W(J,h)
Since
y = 0, it
with
B = AT (
follows that
); also
IIl
=
y.
y = 0; so by (4.9) and (3.7)
70.
:i,
1. 3
J
a..
=
so
o(Y+p )-p
=
c
J
=
p-T-p
since
=
c
E
*-pc
.
1. 3
c=
= -Ea
F. it
+
i
follows
1i
that
every coordinate of
(*)
p-T-p
is even.
Lemma 4.20 and (4.19) also imply that every root in A (
is a sum of the simple roots annihilating X, namely
2 < i < p}.
1< i < p} U {g
{+
.q-p
'2
W(TA)
2
is the permutation group on the coordinates, it
is not difficult to deduce that
acts only on the
T
i coordinates, and the j coordinates between
q+p
2
p = [p
Write
1
1
1
2
2
'
)l''''q)H;
,p2''''.Pp
-+1
Since
are simple roots of length 2,
and
2
'2
=
and
1
say p = n.
(Actually, n = 2p21, but this is unimportant.)
1
Since
q-p.
..
1
<p,g
22
~
i -
P2
:
> =
2
1
2
i
9
p.-p
i,
and
1
Pi+1
2
We deduce immediately that
for
1 < i
< p.
p=
n-2i+2, p2
+T1
= n-2i+1
(Henceforth such computations will often
be left to the reader.)
By (*),
T
preserves the
71.
i coordinates and the j coordinates separately; say
=
T
[aa
2]
with obvious notation.
1
for
We claim that
1a< i < p.
If not, a
finite set argument shows that necessarily
a (i)
+ i) 12 > a2(qLE.
2
<T
-1
(g
+
)
,>
for some
2
=<g
+
2
i.
qpT-P> = p
Then
1
p
2
-1
So
+
- (g
T
)
a negative root, i.e.
_
S(i)
phisms of
det T =
that
li)
A()
c A(n ).
1
But
So
We deduce that, as automordet [ail] = det [1,a2 ]
it ,
(det [a1
£
{ ; a contradiction.
is a root of
- a2 (9
_
g
would be
l
i,q
+
< 0
"2
by the computation of p.
g
2
1])
Now if
so that
w 6 W(A), one knows
det w = (-1 ) (w); so it follows that
On the other hand,
Q(T)
contradiction.
y
So
=
B|
=
k(T)
is even.
J+J-l = 2J-1; a
cannot exist, and
y
is 77-minimal,
proving Conjecture 4.2 in this case.
By Theorem 4.5, we have proved that
SU(p,q)
has
at most a p-parameter family of spherical representations.
This is of course less than earthshaking; the point was to
indicate the idea of the following computations in a
72.
relatively simple situation.
The reader may consider
himself warned.
Proof of 4.6
1
p,...
(c)
This is
the case
1
P=)
*2
2
-..- q)]
su(p,q), p <
=
an arbitrary A-type; we
retain the notation of the preceding example. Since y
1
2
is dominant integral, both sequences
(11) and (P2) decrease
1
2
2
by integers. Set y+ 2 Pc = a = [(a1
1 ,...ap), (ai...aq)];
then the positive root system defined by
A(7710pp
='{g. .a.-a.>0}L)
13
i j1
2
Define
01
- a
1 < i < p, define j(i),
a
2
a
1
2
1
a .
if
13ij
=
(formally).
-
For
0 < j(i) < q, so that
> aji)+1.
transitional
satisfies
2<0}.
{g.J.a.-a
2
a
The index
>
1-1. --
a
2
3 (i)
,
and
i
is called upper
lower
transitional
2
1
It is called upper critical if
+1.
a2(j)+ 1 > a i+
i
2
1
1
2
- a < 1, and lower critical if 0 < a 1__ -aj ()+
0 < a2
if
j(i)
i
The simple roots in
1
i
(*) {fj
.
fe i+
c){g . .
___
_
A+
<1.
_M_+-
are then
is not lower transitional and i+1 is not upper transitional}U
1
j
isnotj(i)
for any i}U{g'
li is lower transitionall.
)i
is upper transitional}
73.
l2
By the integrality of y , a -a
e e [0,1).
1
iff . a 22
- a
fixed
j(i)
a
1
j
2
aj(i)+1
= E.
coordinates,
1
a.
1
pc
1
a.
2, and
+l >j
It follows that
1
-
,
and
i
Within the
mod
~
i
for some
is upper critical
is lower critical iff
i
coordinates or the
necessarily decreases by 1; so
2
2
a.
- a+1
2.
One deduces easily
that every upper (respectively lower) critical index is
also upper (lower) transitional.
We can therefore let
r
be the parabolic corresponding to the simple roots
g
(i)i is upper critical) U {g , j(i)+ 1
critical].
|i
is lower
We claim that these are strongly orthogonal;
since they are simple, it suffices to show that the sum
of two of them is never a root.
For pairs of the same
sign this is obvious; so suppose
is a root.
either
gi
.
+ g+
By the description of the root system,
i = i', or
j(i)
= j(i')+l.
If
i = i', i
is
both upper and lower critical, so that
a2
a
1 ))
--a2
aj i)+1 = (a2
(a()-ab
which is impossible since
Similarly one can rule out
claim.
a1 a2
166=1
(a+-aj(i)+
1 ) =-see=1,
(a )
J
decreases by at least 2.
j(i) = j(i') + 1, proving the
Thus the semisimple ideal of
40
is a product of
copies of sk(2,R), and is therefore split.
Since
AR+S fl
74.
is
a one dimensional
2p(1'ing) + 2p(gutm)j,
9 -module of weight
y
-
2 P((7
minimal iff y + 2p(Trk) is.
2p (n7~k)
= 2p c;
4p)
is principal series
Here .?
on
so we must show that
principal series minimal for
O4k, so
=
a
=
y+2pc
is
This is immediate from
.
the characterization of critical indices, and the
description of principal series minimal types in Table 5.8;
details are left to the reader.
yp is
7
It remains to show that
minimal.
+
E-g..
v =
Set
+(-e)g.
i upper
critical
We claim that
=
i lower
critical
y + 2p c + 1
fv
0 dominant.
7-doiat
is
-P
All roots have length 2; so according to the method
outlined after Lemma 4.20, we must show that
<p + 2p
+ 1v, 6>
Consider first a simple root
i
that
-
E
<
i+l
is not upper
Using the definition of v, this implies that
<VE.>
> < 1-s;
<v,
<
so
- 1+1
-1
<P + 2pc +
We know from (*)
ei i+1.
is not lower critical, and
critical.
is a simple root.
6
whenever
> 1
ye
>
=
<p +
2
pc + -v 1
+,
2
ic~
2
13
T2~ 3
-
~
e
>i+
75.
Similarly, if
f lj+1
is a simple root,
<y + 2p c + 2-2v f3,3i+1l >
simple root not in
a
so
2
2
a.~)
- a
1
non-zero only if
and
Then
i
Now
i
<v,F 1.> and <v,e 2 .
3j(i)
is lower critical or
is lower critical; in any case
0,<
<v c.>
<
1-E,
O > <v,
.
1
a
(g )
> 2
+
-
Suppose that equality holds; then
1
1
j(i') + 1,
=
a ,-
1
2
2
a (i)-
p
with a 1 , -
i
i'
=2,so that
aj (i)+1
= 2.
21
2
j(i)+l
I
2
a.
j (i)
-e
=
= s;
= e.
Since
A( P).
P
and
v
2
-
E;
These imply
a = p + 2pc , it follows that
e.
and fj(i) ,j(i)+1'
,(i)+f
so v is
= V
To
1, then j(i-1)+l =
1
and the noncompact roots
in
=
and
= -(1-6)
v
1-1 = l- = v.,
1
j (i)
singular with respect to these compact roots.
v.
summarize: if <p+2p
(1-))
1.
a.
1
-
-
= i-l1, and that
is singular with respect to
Also
Thus
.
2
j (i)
a. - a .
is lower critical, so
j(i)
a.
can be
1
(1-e)
-
>
j (i) = j (i') + 1
-
2
,
i
is not upper critical,
i'
<yP+2p+
i
gi,3i)
e; by integrality, this forces
-
e.
-
i
Suppose
2
A( 0 ) .
1
1
- a. > 2
J1(i)
>-
g.
and
gi,j i)+1
are
are singular with respect to the
(i),
76.
compact roots
e
f.
and
-j.
gij(i)+1
Suppose next that
in
e = 0, we still get
identical results. If
I +
<p+ 2 pc +$v,gj i)+1
1.
Suppose that equality holds.
g
We cannot deduce that
, say, is a root of
<p+2pc
but suppose
g
is a simple root not
e 3 O, we argue exactly as above, and get
If
.
(i)+1
1.
A(
(i) is necessarily in
.9);
for otherwise the
preceding paragraph would imply that
root of 0
In this case
gij(i)+1 is a
which we have assumed is not the case.
,
Exactly as before, we deduce that
y
and
singular with respect to the compact root
Similarly,
j(i+1)
and
v
=
if
j(i)
1<y+2pc+Vygi+1,j(i)+1
+ 1, g 1
,j(i)+
1
A = 1,
v
fj(i),j(i)+1'
then
is a root of f
are singular with respect to
are
,
and
y
ei,i+1'
Finally, it is very easy to check that
c+v,6>=
This proves that
1
X
is a simple root in A(f).
6
if
is
iy
-dominant.
J
Suppose then that
cohomology greater than
y -
y,
f3.
j=1 j
E
gives rise to
J-1
with all notation as usual.
77
By (4.19) and Lemma 4.20,
1ph = llylf, all a.
i.
and S.
i.
are sums of simple roots annihilating X, and B = A (U)
for some 0-invariant T 6 W(oj,%).
Let
0
DQ
be the
subalgebra of
90 corresponding to the roots which
annihilate X.
Then the simple roots in A(Q 0 ) are certain
noncompact simple roots
g..
which are described above
1J
(namely those whose inner product with y+2 pc+2v is 1.)
There is a natural notion of adjacency among the
simple roots of
sk(n,C); two simple roots 61162 are
=
adjacent iff 6 +62 is a root.
Call the sequence 61...6r
of simple roots contiguous if the 6. are distinct, and
is adjacent to 6i+
for 1 < i < r--1.
Then 61 + ...
i
+ 6r
is a root iff (6 ) is a contiguous sequence, and these
are all the roots.
If the 6
is a compact root iff
of
in 9k
roots of f.
r
are noncompact, 6
+
+ ar
Thus the simple roots
is even.
are the 61+62' where 6162 are adjacent simple
Such a pair is clearly of the form
(gifj(i)' girj(i)+1)
or
, gi+1,.j(i+1)).B
(g
is
the remarks in the proof that
X
one root of such a pair is in
A(Q).
compact roots are f
,
dominant, exactly
The corresponding
i or eiti+1 respectively;
78.
we saw that y and v are necessarily singular with respect
to such roots. Since A () c A(I flh), it follows that
-1
-1
v = a v, y = a y.1; and since by definition the roots
rV-
in A(f) annihilate
Now
, a
-1
-
-
()
=
X.
decomposes as a direct sum of simple sub-
.
algebras
S, (plus center) corresponding to
maximal contiguous sets of simple roots in
call these components blocks.
A(V); we
Assume the blocks are
ordered in accordance with the lexicographic ordering
I+
-u
+
"V
of i~
: if g-- .
A(u
), u < v,
o
i ll
i
22
< i
i
then
The first simple root in a fixed block
(type I) or g
g.
is either
The key
(type II).
observation is that, since Aa () c AC(i4.), and
-
) C A(t), both
A+(
T
y=a
=
1
(y+2pc+-v-p)-a
~Y(M -a~
-1
=X-1
-1
-(p-E
))-2p c
c
)-2p
J
~1
Pu
)-PC; thus
a~1 (-Ui+p
1b37
c21.
J
C
=
in
1
(pc+2v-P+i. )-pc
J
+ v(p-ES.
(Ea
y+2pc+-=
y
(PC~ GOPc+
2
a
A
for the part of
(4.9) that
Recall
-l
u
We write
sition.
and a respect the block decompo-
T
1(
---(za i~ +ES 1J ))-2p
(by3.7)
79.
-l
1 -1
(4.21)
(by 4.12)
v)-p+1 (T-p)
y = y+- (v-a
y = P+1(v-aV -(Y-
1 T) -P)
(Here we used only the fact that the roots in A+
a
annihilate X; accordingly we may apply 4.21 in later
computations.)
In the present case, of course, v =
so the middle term may be ignored.
each u, y = y on
tu,
v,
We claim that for
and length (T|u) is even.
The
claim is proved by induction on u; suppose it is known
for u'< u.
Consider first the type II case, so that the
From the proof
first simple root in A(ku) is g.
00'3
that
X
0
770 dominant, we deduce that g
is
u ae
and that the simple roots of Pu
are g. O'
. .i)
A()
-+
+.
, g 1i0+. .
g 1i0 +1,j . (1)+1r
0
+1, 3 (10 )+2f
i.e. each
O'i0)+1
etc.; every second root,
is in A().
g
0
= 1 for
<p,6 >
Since
each simple root 6, it follows that the ku part of # is
j(i0 )place
(i0 place)
[(...
n
-
1,
Suppose that
n
-
y /
3,
y
n -
5
...
) (...
n
,
n
-2
on the i coordinates of
n
4
u
By
induction, y = y on the earlier i coordinates; since
i..)]
,
80.
y
p
in the lexicographic ordering, we deduce that
there is an
y. > y
11
1
k,
in
i
1>0
y
Since
.
so that
for
i < il, and
is annihilated by the compact roots
the sequence (p
similarly for (p).
yi = p.
is constant on each block;
On the other hand, y is dominant
so (y.) is decreasing; it follows that
leading i coordinate of
iu.
= i
i
Recall that
is the
W(QuJu)
acts
by permuting the coordinates. We deduce from (4.21)
-l
that the a T(i 0 ) coordinate of p is greater than the
iO coordinate of p, which is possible only if
aT(i0)
block,
(i0 ).
=
T(i)
=
Suppose that for some
j0
A()
)-gi,j(i)'
c A(); but
< i, j(i)'
P
2
1
--
6
-PT
=
by our description of p.
a contradiction.
So
-
n
p
a
>
T(j 3 )
+c
(i)
fTp
4- T
j(io
(j
(n-k) <
+
This proves
Consider the last j coordinate
Clearly
in the
is a simple root of
i,j(i)
and thus cannot occur in
i
T0
j3
p. -1; since
3
(i))
0
for some
2)
of the block
y
(
g
2
2
u
is dominant,. (4.21)
- -j
81.
implies that
> p. -
p
G
u-l
a
In particular,
-l
0)
T(i 0
a
T(j(i 0
a~
(j0)
Setting
j4
=
since
j coordinates.
j
for all
j
y
a
in the block
a
i0 .
=
T(j(i'))
again)
a E WK
or
-1
j2 , we get
=
0
j(i
))
we must have
thus (by the dominance of
for all j.
i
T(j)
T(j)
i0 , and
i(j)
=
i(j 2)
T0
Since no i coordinate can equal a
Au.
the i coordinates of
Jju .
each i coordinate in
y on
=
By (4.21), p. = p
for
a
It
T(i)
ar1 TI
follows that
preserves the i coordinates, and hence also the
Every such permutation is in
a E WK'
are
u
{a. iCF
} =
again, y
=
2)
which preserves the i and
j coordinate, this is a contradiction; thus y
nates.
Since
e WK( u).
+(k);
y on
);
The compact roots in
so in fact
A2\u.
WK
T
= a
on
j coordi-
since
A
Au
(
)
By (4.21)
u
The proof that length (TIAu)
even proceeds just as in the spherical case.
is
The argument
for the type I case proceeds along the same lines; the
assumption that
y / y
on the i coordinates in
almost immediately to a contradiction.
u
eads
We omit the details.
Since length (TIu) is even for all u, length (T) is
even.
(Actually, one can see fairly easily that we have
deduced that
} =
If{a. 1{J
}|]
This will be needed later
82.
on for the program described in the following paragraph.)
But length (T) = 2J-1; a contradiction.
exist, and y is indeed
'( minimal.
So no such y can
This completes the
proof of 4.6 (d).
Although it would be difficult to formulate a general
lemma, we will need the arguments above repeatedly.
What
happens is that the more complicated algebras contain
It will
assorted subalgebras isomorphic to su(p,q).
often be possible to deal with these separately; this
will generally be done with a reference to the "su(pq)
We have tried in such cases to arrange the
theory."
notation so as to make the argument clear, and details
will be left to the reader.
The next case illustrates
the technique.
Proof of 4.6 (d)
for
T
Here
0= sp(pq), p < q.
the usual Cartan subgroup of
K = SP(p) x SP(q) C SP(pq);
Rq
consists of diagonal
it
with respect to the usual inner product.
we refer to
and
T
This gives an identification of
matrices.
RP ( Rq
We take
Rp
Again
as the "i coordinates," with basis {e },
as the "j coordinates" with basis {F }
equal rank, so
with
=
0.
The compact roots are
O is
83.
ei
f
.
= + F_ + e ,
1-
-
-JJ
(i<i'), c
1
2
2
+ e .,
J J
+
-
j')
(j <
E.
-
noncompact roots are
g..
13
p qj
Order
= R
it
R
1
+
+
=
2
+
and d. = +
The
.
J
1 2
+
12
+(e.-.), h. =+(E+E).
1
J
1)
1
)
J
-
lexicographically.
The
corresponding positive compact roots are
e.,
11
,
+
1
d.
and
f.., ,
J)
J
j
i < i',
(recall
j')
<
One checks easily that the simple compact roots are
e..,
1,1+1
d , and that
q
c p , f.
jj+1 ,and
2p c = [(2p, 2p-2,... 2),(2q, 2q-2, ... 2)].
y =
[(y ,...P
HP,--2 )].
)(
1
y-, > y2 > .. . > P
integral, y .
the j coordinates.
Define
l
1
y
Since
is dominant
1
similarly for
> 0;
Set
2
2
1
1
y+2pc = a = [(a 1 ,. ..a) (a ,...a2) I.
1
2
a0 = a0
(formally)
Fix a k-type
1
, ap+1
2
=aq+
=0; andfor
1 < i < P, define j(i),
0 < j(i) < q, so that
2
1
2
a.
> a > aj.).
3(i)+10
i 3 (i)
Call the index
transitional
a
2
if
a .
1-1
>
a
3 (i),
and
1
; upper critical if
and lower critical if
upper
loe
j~~oe
> a0
i
0 < a
2
- a.
0'< a.
I
j(i)+
_rastonli
<
-
1-
ifiJ
17-
trnstina
'M,
21
If
0
1
is
84.
associated to a, then A(7oag))
-
U{9
a -a
roots of
1
< 01.
A+
(gi,j(i)+
of
a,
13
2
|a1-a
1
13
>
3 -
0
It can be deduced that the simple
not lower transitional and i+l is not upper transitional}u
j # j (i) for any i} U {c, or d according as a < a
p
q
p
q
i is upper transitionall
> a }i
orp
{h.} U {g.
are
{ei, i+l1 is
{f . +1J
=
i is lower transitionall.
1
i is upper critical iff
critical iff
a
1
2
- a. i)+1
= 0.
By the integrality
2
1
- a.
3(i)i
a.
=
Just as for
every critical index is transitional.
1, and lower
su(p,q),
Let
be the
parabolic corresponding to the simple roots
j i is upper critical} u{gi j(i)+l
{
As for
and
su(p,q),
is a product of copies of sl(2,R),
y-2p(f1fn6)
to check that
is principal series minimal.
y
It remains
is rf-minimal.
E
g
.
.
i lower
critical
1
0
X = y+2pc p+2v is n dominant.
Set
i lower criticall.
v=
We claim that
For the various simple
roots of length 2 this is proved as for su(p,q).
Assume
for definiteness that the simple root of length 4 is d .
q
()
2 = 1,so
c q
a2 =
q
(y+2p ) 2> 2.
c q -
No coordinate of 1
2
is
85.
less than
1
-
<y +
so
-,
1
2pc +
+
v,
l
= 2(p + 2p
to
d
.
q
+
+
q
q
is actually nonsingular with respect
%
j ;
Let
.
2
3 > 2 =
> 2(2-)=
X
2c >
+ -v)
c
It follows that
2
= <p + 2pc + i'
d q>
be the subalgebra corresponding to
the roots annihilating X; since the simple root of length 4
is not in 2 , one can now apply the
y
deduce that
is 1--minimal.
su(p,q)
theory to
This completes the proof
of 4.6 (d).
Proof of 4.6 (e)
description of
-
identified with
<
.
0
Here
= +
i0
=
n.
/
it
is
Let
n
e
=g
1
{E }
be
lexicographi-
E.,
J
which
i < j. The noncompact roots are
and d. = + 2E:..
(s. + E.,
Since
is decreasing, and
0 < c < 1.
= 0.
We order
Then the compact roots are
y= (yl..n)
We use the
Rn, with the usual inner product for
is equal rank, so
is positive for
g..
sp(n,R).
given in section 5; thus
the usual basis of
cally.
=
y.
y
Fix a
k-type
(y )
is dominant integral,
=e mod I for some fixed
One computes easily that
E;
say
86.
2p
(n-1,n-3,...-(n-1)).
=
0
0
Let
a =
Set
(a,...an).
+2p=
be the Borel subalgebra associated to a; then
sgn a.
sgn(a.+a.)
A U"n
=
{g
We want to describe the simple
(as usual sgn 0 = +).
> 0
a
so that
r
Define
roots of A+.
}
U {d
}1
1
i < r,
for
a. < 0 for j > r; we call the coordinates up to r the
J
i coordinates, and those after r the j coordinates.
Define
an+l = -
'.
For
< a. <
a
1 < i < r, define
so that
-a
r < j(i) < n.
(Clearly we want
j(i)
With this restriction
- ar+l > a .
j(i) may not exist, since we may have
This will cause no difficulties, however, so we will not
is called upper
lower transitional if
- a (i) < a._l
transitional if
i
The index
correct the notation.)
aj(i) > a 1 ; upper critical if 0 < - a
if
2c < 1, and 6 = 2c -
1
< 1.
+ a
0 < a
and lower critical if
if
2e > 1.
by the integrality of a, one sees that
critical iff
- a. - a.
1
iff
a1 + a.(~
=
6.
=
1 -
J(i)+1
Just as for
- a j(i)+1
Set
Then
i
6 = 2c
0 < 6 < 1;
is upper
6, and lower critical
su(p,qi), a critical
87.
index is necessarily transitional.
Define the permutation
a
I>
|aw2)i
>
f
of
> ja
...
{l,.. .n}
so that
Jail =
If
then (since (a.) is strictly decreasing)
and we list the positive element first.
In terms of
0;
7r,
are
, and the various
d7(n)
fr(n) = r
r+l, according as
ar > - ar+1
ar < - ar+1, or
are {e. i.
or
(1 < i < n-1).
i+1) I(i+1)
[sgn a
-
7i
[sgn a
Notice that
A
j
i
a. = - a.
it is easy to see that the simple roots of A
sgn a(n)
i =
|a.,
Thus the simple roots of
ll < i < r, i is not lower transitional,
i+1 is not upper transitional})
{e.
+Ir
<
j
< n,
j
is
not
j(i)
for any iJU
{gi)ji is lower transitional} U
gi,J(i)
{i is upper transitional}U {d
Next we must define
possible so that
Q.
0 < ar(s)
Choose
s
sgn a (n)
}.
as small as
< n-s+l, or 0 < -a,(5 ) < n-s+l.
If these are not satisfied for any s, s is undefined, and
the corresponding set of simple roots in I
is empty.
Let
&, 2 1 0
(to be defined)
correspond to the simple roots
88.
sgn a
{n
U
,
s i< i < n, and d 'r(a) Tr (n)
(if s is defined)>}
i
i, j(i)+li is upper critical}U {g
is lower critical}.
(The first set is just the simple roots supported on the
coordinates
f(s) ...r (n).)
simple, so that
Suppose
that
qk
A+ (Q) be
is upper critical; say gi.()
is supported on
this forces
c < a
(1-6)
< n - s + e.
< -a.
-
1 + 6;
3(i)+l
0 < e, 6 < 1,
0 < -aj
i(s)...7r(n).
Now
k > s, so
i
a
is integral,
is upper critical,
so
< n
s
-
-
+ c +
(1-6).
c + 1 - 6 > 0, and s - 1
i)+1 < n
-
(s-1) + 1 +
< n
-
(s-1) + 1.
may be given for simple roots
shown that the simple roots of
g
-
6 < 0;
thus
(e-1-6)
This contradicts the definition of s.
two subsets:
:n.
=
If not, then
0 < ai < n - s + 1; since
ai = -aj(i)+ 1
e +
Since
.
1,J(i)+1
k+l > s; then we claim that actually
k+1 = s, and so
i.e.
i
g
Let
A similar argument
.i
A ().
We have
1, J (i)
may be divided into
P1 , consisting of those supported on
'r(l)..r(s-l); and P 2 , consisting of those supported on
Tr(s)...n(n).
Clearly these two sets are strongly orthogonal
89.
to each other.
Arguing as for
su(p,q), one sees that
the first set is pairwise strongly orthogonal; the
corresponding ideal of
0
is a product of copies of
sk(2,R), and is therefore split. The ideal corresponding
2
to P
is clearly just sp(n-s+l,IR), which is of course
split; the strongly orthogonal spanning set is
{d
sgn a() n
}i=s
Thus
is split.
?
complicated nature of
(The slightly
reflects the slightly complicated
set of conjugacy classes of Cartan subalgebras of
Next we must show that
series minimal for
.
y-2p(h Af )
sp(n,R).)
is principal
On the P1 ideal this proceeds
as for
su(p,q); we omit the details. The P2 ideal requires
a little more work. Since A R+Sg is a one dimensional
Q-module of weight 2p(n
consider the weight
y+2p(/i[k) = y+2pc
say.
D) + 2p(fl nk),
yp-2p ('ny ) + 2p (nrn
-
2p(
I-/1On.rl)
= a -
it
is
) + 2p (//rQK) =
2p(IA/Qnk)
To simplify the notation we may assume
=
2p(JDr700)
2 pc
=
(n-1,...-(n-1)).
equivalent to
Since
= b,
s = 1; thus
a
is
decreasing, 7(l) = 1 or n; assume for definiteness that
7(l) = n.
So
b
n
=
By the definition of s, 0 < -a < n-l+1 = n.
n
a
n
+ (n-1) > -
Tr(1) = n, -a > |a.f
ni
1.
On the other hand, since
for all i; in particular,
90.
n > -an > ja1l, so that
b=
a1 -
(n-1) < n -
general theory that
hK =k
(n-1) = 1.
b
; so since
(a ...
a, < n.
p- 2 p(170
)
b
We know from the
is dominant integral for
1 > b1 , bn> -1,
b
1 > a > 0.
-l,...-), with
we see that
Thus
is of the form
From Table 5.8,
is principal series minimal.
Thus
is principal series minimal.
It remains to check that
y
is
'q minimal.
It is
convenient to treat several cases separately.
Case I
s
undefined;
i.e.
I n - i + 1, all i.
-a
E
i lower
(1-6)gi
)
+
ij(i)+l
su(p,q)
theory,
all the simple roots
av (n) < - 1.
or
Set
, A = a - p+
6g
i upper
critical
critical
By the
aT(i)> n-i +1
X
q..
is dominant with respect to
Necessarily
For definiteness we assume
a (n) > 1, or
a
n
> 1;
the
other case is similar.
a)
fact that
a (n-1) > 0.
a
< T)n-1
By the definition of fr, and the
decreases by at least 2, this forces
' a>
=
a 7T(n--L)
=
a(1Tr(n)-
91.
nn-1
Thus
is zero; and the
,na-p+
n-1rX
Hence
<d
+
7T(n)
> > 2 - 1 +
-2
<n-l
,X>
<d
=
1
+
2
TT(n) ,a-p+-v>
Q2 Q
dominant.
Let
singular.
Just as for
in-1 / A(M); so
X
is
be the subalgebra on which
X
is
But we have shown that
.
and those supported on
sp(l,R)
Then
-a
a
< 0, Tn
theory; one sees that
g~
1
),~n
7T
is upper critical.
(n-1) > n - (n-1) + 1 = 2, and so by integrality,
-a
>
=
n-1
-
,ff(n-1)
3
-
£
-a
.
Also
> 3-s.
(n)
x~-)-
a (n)
theory, and the
su(p,q)
su(1,l)
=
The
Tf(n).
q1-minimal.
is in fact
b)
2 =.
decomposes into the roots supported
A( )
(ff(l)... f(n-1)),
second by the
>.2 -
su(p,q), the n-minimality of p
first piece is handled by the
y
> 0
So
reduces to a problem on
on
2
<qn-1'p> = 1, since qn-l has length 2.)
(we have used
Also
in
1
coordinate is at most
fr(n-1)
absolute value.
<
coordinate of v
Tr(n)
is not critical, so the
-
(1-6)
=
2
-
(n-1)
-
e + 6.
2V
ff(n) coordinate of
'r(n)
1.
v
-6,
so
In the f(n-1),fr(n)
2S g I(n),7r(n-1)2
X = a - p +
=1
(=- -6-
-vis 6
coodintes
coodiate
a
is
- 6);
2
so the
92.
a
a (n)
P
r(n)
=a
2v (n) = a (n)
+-_v
(Here we have used p
Hence
X
s
2
>
2
=
(1-6) +
-
+
6
-
> 0
1
-<d + n '
2(n7T(n)I
)
1
-
2xd7+n)
(n)'
<d
+
d
7T(n)
Tr(n)
>
is dominant and non-singular with respect to
. If we let 1 correspond to the roots annihilated
Tr(n)
by ~, then the su(p,q) theory applies to Q , and y is
d
a-minimal.
c)
A
is zero, and it is immediate
v
Then the ff(n) coordinate of
that
is not upper critical.
g (n) , T(n-1
n-1
0'
a (n-1)
is dominant with respect to
If
d+
is
it
non-singular, we can apply the su(p,q) theory as in case b).
= a(n)~
If it
is singular, then
0 = A
i.e.
an (n) = l.
a (n-l)
Since
i.e.
s
Hence
is undefined,
-an (n-1) > 3.
f (n) = a
-
is a negative integer.
(n-1) + 1 = 2,
-a (n1 ) > n -
The 7 (n-1) coordinate of
v
is at
most 1 in absolute value, so
<n n-l'
<gT (n-1) , 7 (n)
> -aT (n-l)
>
3
-
11
-
3
n)
a
1
>
-
0.
1
1
1;
93.
So
X
is nonsingular with respect to
n,
and we can
argue as in case a).
Case II
v1
s is defined.
=
Set
g+
2(1l-6)
+
+
1
0
gi,j(i),
on coordinates
{ 2(p-a)
1
-
gi,j(i)+1l
v2
E 6gi
, 1r(i)+1 +gij(i)
i lower critical
i upper critical
on
7(s)...
()...r(s-1)
Tr(n).
It is not difficult to check that, since
p - 2p(t1op)
is principal series minimal, v 2 is of the form required
by Lemma 4.20.
Put
su(p,q) theory,
l'''Es-2
X
v = v 1 +v 2 ,
=
a-p
X = 0
on
coordinates, X is dominant with respect to
and
<
*
ff(s)... rr(n)
s'...n-l'
Suppose that in fact
,1s'X
> > 0.
<s-l'
Let
=
on
.
By the
is dominant with respect to
; and since obviously
sgn ar(n)
d T(n)
+
I
correspond to the roots annihilated by X; then
fl + fwith
r(l)...rr(s-1),
'(s)...rr(n).
corresponding to roots supported
and
,2
to roots supported on
The su(p,q) theory applies to
e;
and
94.
It follows
2 c i, so 7z-minimality is trivial there.
that
y
So it is enough to show (*).
9?-minimal.
is
a
Assume for definiteness that
a
definition of s,
s-1) > n -
integrality, a
already seen that
's-1
s-i)
=
-
p
(s-1)
=
n -
We have
is critical; since
ns-2
(s-2 ) - a
-a
-(s)
Hence
(s-1) + 1.
+
1
a(s)-p2v
Co
.1
>
or
6=2c
a
(s-i) = n -
(since
1
- 6 =
(n-(s-1)+1)+ -(n-(s-1)+)
11
2c-1;SOE-
(s-
2
$
(s-1) + 1 + e; and if 6
0, v ( 1 )/
a (s-
1)
=
(n -
(s-1) +
1)
+
C +
1
=
(n
(s-2)
1)
+
E
-
6
)
-
+16
+
-
Then necessarily
so that
-
(5
s0)
Hence
!6 > 0.
Suppose equality holds.
<ns-l' X> > 0.
16,
Finally, it is easy to see that
6.
Ss-1 (s-1)
Now
So
is not a critical root.
al s-1) > 0, this forces
v
By the
(s-1) + 1; by
(s-1) + 1 + c.
0, necessarily
VnT(s-1)
if
) > n -
1
0.
>
-
6
0,
95.
Because of the equality, we also have
(n -
=
1) -
(s-2) +
-6 < n -
2
= 0.
s-1)
2
6; so
(s-2) + 1,
-
which contradicts the definition of s.
6 = 0, e = 0, and v
c =
So in fact
(If we had considered
instead a (s-1 ) < 0, we would have gotten <ns-'X> > 0
in general.)
We must now investigate this last case, i.e.
and that
n s-2
easily that
= n -
We have seen that as
<s-l'> = 0.
is not a critical root.
(s-1) + 1,
It follows
<ns-2'a> > 2, and hence that
X
is
non-singular with respect to ns-2. Thus we can define
^1
~12
=
.+1 as usual, and treat I"'1 by the su(p,q)
theory; and we are left with
roots supported on
(s-l)...n(n).
notation, we may assume
y =
A+r07)
If(cl+c.) 12
Y, a,
Let
T,
A+ ()
a
{
.
To simplify the
Then
a,(l) = n
forces
to see that
not difficult
i < nIL'~el
<
{.1. }, {a.1. }
J
that
s = 1.
It is
(li...lU...O).
=
j , which corresponds to
} cA
-
be as usual.
Recall
J
=
{(e-E.)}.
1
J9~
Now the
J
compact Weyl group acts by permutation on the
is easy to deduce that
i.
It
96.
a(b.*.bn) = (bJb *.b J1 bJ+...b n);
A+A) = ({E
1 -s.12 < i < J}.
and in this case
other hand, by
(4.9),
(recall
pc-pc
c
= (yg~
(*) y-E6
=
(F1+2)
yl+l = y2
Ea.
)
c
- - -y**YJr yJ+1'''
n)+(-(J-1) ,1,
(YJ-(J-l),yl+1...y _ +1yJ+1'
.
J
Hence
-
On the
.
.10.
n)
(c); here c = 0 or 1 according as
-
is or is not a 6
J = 1, this fails; but
(If
.
J
an easy special argument shows
yJ = 0, -1, or -2.
this implies that
y
Since
y=
||yll
0
= ||y'|,
y < y, and hence
y
or
-1).
Hence
and y, = 1,
cannot occur.
So
is qminimal in every case; which completes the proof
of 4.6 (e).
4.6 (f) is quite similar, but much easier: the
root system of SO*(2n)
is just that of SP(n,R), except
that there are no long roots.
One can copy the preceding
argument, omitting all reference to "2,,
We have now dealt with all the classical groups
except the
SO(p,q).
These are by far the most complicated.
To begin with, we need a more delicate technique for
.0)
97.
proving Conjecture 4.2.
4.2 (ii-v) hold.
Suppose
k/ is chosen so that
Suppose furthermore that there is a
e-invariant parabolic 5 2 fr
,
.
=
+
'L, with the following
properties.
(4.22)
a)
the k-type
y is 4-minimal.
b)
the knk -type
p-2 p(//Znc)
in I
;
is uniformly minimal
i.e. it is the minimal QOA'.
type of
every irreducible Harish-Chandra module for
2
c)
yp-
in which it occurs.
2
is
p (71()
7
-minimal in
Q.
Then we claim that 4.2 (i) holds, i.e. that the action
of
U()
on
factors through the homomorphism E'.
X
Essentially this is left to the reader, modulo the following
hints.
c),
Using b),
can deduce that
Theorem 3.15, and Corollary 5.5, one
y-2p (n
g)
occurs with multiplicity zero
or one in each irreducible Harish-Chandra module for Q.
By results of Lepowsky [18], it follows that
U()
/I p-2p
(r(O
)
is abelian.
The rest of the
argument is quite formal: essentially one uses
Theorem 3.15 to study
H* (7,X).
H*(rtnQ,H*(NX)) instead of
98.
Proof of 4.6 (g)
We give details only for the
simplest of the various possibilities of parity, namely
50 = so( 2 p+l,2q+l), p < q.
G
p = 0, then
If
has only
one conjugacy class of Cartan subgroups, and we are done;
so assume
k is identified with
p > 1.
so(2p+l) xeso(2q+l)
in the usual way; choosing a maximal torus in each factor
as usual (i.e. with
cos O
-sin
SO( 2 p+l) level) we get
sin 6i
e.
1
cos 0.
1
it0 =
@ R.
lexicographically; as a basis, choose
as usual.
Since rank ()
necessarily dim
1.
=
=
blocks on the
[2 p+
This we order
{e }
ii=1
2q+
+
and {Es}
jj=1
p + q +,
One computes (more or less
easily, depending on one's point of view) that the
1
2
2
1
compact roots are + E 1 + .,
, + e +
., , + C., and + E:2
2
the noncompact roots are + c1 +
E-,
the complex roots are just the
2
+ s1 and + e.;
since all
roots of
so(2p+2q+2,C)
are
c
1
1
+ <a=a>
1
2
+ e , , Ec, C. +
and + cs.
So
have the same length, these have
complex length two (i.e. if a e A, af
<aa>= <aca
12
+ E,
2).;
2
,,
= +
e , then
The positive compact roots
2
(i < i' , j < j').
2p c = (2 p+l, 2p-1, ...1)(2q+l, 2q-1,...1).
Hence
99.
a~-tp
Fix a Fik-type
a = J+2pc.
~= [(y 1 ...yP),
1
y=
y
Since
2
(I2 ...yP2)],
and set
is dominant, the coordinates of
are non-negative and decreasing on each block.
is integral, all
y
similarly for the
Since
are integers, or all are
2
y.
The coordinates of
y
y
- mod Z;
a
are thus
half integers > 1, decreasing by at least two on each block.
We now arrange (a) in decreasing order: define a map
1 2
S: {l,2,.. .,p+q} +
I
so that (with obvious
k < m => a (M)
notation)
only if
9(m; equality should hold
a
n(k) is an i coordinate and 7(m) is a j coordinate.
One computes easily that
the simple roots are
6 c [0,1)
Let
all i, j; of course
= p + q -
p r(
cTr M-
k + 1, and that
e7 (Z+i)' E(p+q) *
be defined so that
6 = 0 or
.
a
1
2
- a2
6 mod Z,
We leave the definition
of critical roots to the reader.
s0
Let
be the largest integer such that
(aT(p+q-2s 0 +1)
It is not hard
+ a(p+g-2s 0 +2) < 4s0
to verify that (*) must also hold with
s0 replaced by any k < s0.
On the other hand, a
and it follows easily from the value of
a rp+q-2k)
>
22
~ 1
+ 1, all k > 0.
2 pc
If the root
that
>T(p+q)
_1
100.
> 2 R+ 3, and
a T(p+q-2P.+1)
a
+a
Tr (p+q-2 (Q+l)+1)+a
So for
compact, then
is
'(p+q-2Z)
Tr(p+q-2k--I)
k < s0' C (p+q-2k-1) ~
0'
fp+q-2kl)p~
By induction,
f(p+q-2k)
Furthermore, condition (*) forces
s = s
1
(p+q-s0 +1)
s = s0 -l.
1,
-
and
1
' (p+q-s
unless
0
is noncompact.
(p2+q-2
r{p+q-2,+1, p+q-29+2} = {
We define
so kA-1 > s
>4(k+l);
p+q-2 (Z+1)+2)
}(
C2
+
<
2
0 +1)
=E
p-s0
> 1; in this case
2
Set
= {critical roots supported on f{l ....
.p+q-2s 0
P
P2 = {roots supported on f{p+q-2s+1,.
.
.p+q}}
We let &r be the parabolic corresponding to the simple
roots in
.1
P 10P2; then
2
A(P) = Pl U P2.
~= 11 +
1
2 , according to
2
= so(2s+l, 2s+l), and
Clearly
a product of copies of sl(2,R); so
su(p,q) theory, y -2p(770g )1,t
On the f 2
is split.
is
By the
is principal series minimal.
piece, we must check that
Conjecture 4.2 (ii-iv).
Since
y-2p('7I/lW)
satisfies
2 2
KC\ [ ,F I
so(2s+l)xso(2s+1)
is centerless, and 2p(Pfy ) is the weight of a one-dimensional
{ fK-module, this amounts to considering
101.
1
[P-sp....
1
2
' q)].
We have seen that
2
1; so on these coordinates
y
p9+
[(0...
2
( q-)
s
0)
(1...
1
1
[(1,...1),
1,,0
...
),
1
1
(1...)],
[(}..
. -)
(0
...
0)],
y
is
[(0...
or [(1...1, 0...0)].
0)(f..}]
By Table 5.8,
the first four are principal series minimal, and hence
satisfy 4.2 (ii-iv).
The last is a small I(-type, and
the only associate one is [(O... 0)(1,1.. .1,0 ... 0)].
Since
2
by the definition of d, yl-s-1
< 1, [this is not dominant
for
; so 4.2 (iii-iv) are satisfied.
Henceforth we assume
s
is defined, i.e.
s < p+q;
the other case is rather easy and is left to the reader.
It should be pointed out that, although
'r(p+q-2s+l)
ETr(p+q-2s)
not included in
Set
v
may be a critical root, it is
.
E
=
(1-6) (e
i lower
i
critical
-1
Tr
(i)<p+q-2s
2
Just as for
i)1)
+
E
61 E
-
)
i upper
i
critical
-l
(i)<p+qj-2s
0
on
7r(1)...r(p+q-2s)
2(p-a)
on
n(p+q- 2 s+l) ... ff(p+q).
sp(n,R), one can easily check that
sum of positive roots in P2, with coefficients
v2
is a
0 < c<
1.
102.
v = v
Set
A = 0
on
a- p +!v
2
+ v2'
1 2
=
A +
2v.
Then clearly
f(p+q-2s+1)... ff(p+q); by the su(p,q) theory,
A is dominant with respect to (e
-e
).
'rr(1)
Tr(2)
dominant, we must show that
(recall
> 0
Tr(p+q-2s)
and a
Pi(p+q-2s) = 2s + 1,
1
2h (p+q-2s)
and
0
a
T (p+q-2s)
X7(p+q-2s+l)
+
is
A
To show that
e(p+q-2s))
(f(p+q-2s-1)
v
VT(p+q-2s)
2
= 0).
Now
(p+q-2s) =P(p+q-2s) + 2s + 1;
unless
s
T(p+q-2s-1)
is a critical root, in which case it is >
-
E
Tr (p+q-2s)
-T.
Recalling
the definition of critical, we see that the desired
inequality holds, unless
1
Pff(p+q-2s-1) = 0 or
y (p+q-2s) = 0, and
; also w{p+q-2s,p+q-2s-l} must
consist of one i coordinate and one j coordinate.
But
it is easy to see that this contradicts the definition
of s.
So
A
4
Let
is dominant.
correspond to the simple roots annihilated
We want to split & into two pieces, roughly
corresponding to P 1and P 2 ; so suppose X is zero on
by
A.
7(r)...1T(p+q),
Then
=
T(r)...T(r-1),
and A
,
r-1) 3 0,
(of course r < p+q-2s+l).
corresponding to roots supported on
and 7T(r)... r(p+q) respectively.
Suppose
103.
y, a,
T,
}, and f.
{a
are as usual.
J
su(p,q) theory, y = y
}
{.
and
.
}
{.
J
in
and
there are equally many
P ; so we need only consider
J
2
suppose
2
r = p + q - 2s + i, then
If
7(riZ = 0, and
x
By the
J
J =fs
}|
r < p + q - 2s + 1.
(p+q-2s)
=
0;
2
2=
0, a contradiction.
=
So
This means that
so by the remarks of the preceding
y rr(p+q-2s) + vrr(p+q-2s) = 0.
paragraph
, so that
p Tr(p+q-2s)
'-and
7Tr(pq-2s) >
is
Since
a non-negative
half-integer, there are very few possibilities:
1
=r(p+q-2s)
(p+q2s) =
I)
by the definition of critical root,
and
{f(p+q-2s-l), frf(p+q-2s)}
and one j coordinate.
II)
P
= 2s + 2.
0,
this forces
Suppose that in fact
follows easily that
In this case,
y'r(p+q-2s-1)
2'
This contradicts the definition of s.
by the definition of v, v Tr(p+q-2s-1)
-
1.
consists of one i coordinate
i(p+q-2s) = 0, V Tr(p+q-2s)
r(p+q-2s-1)
-
= 0.
- 0.
In this case,
Since
a r(p+q-2s-1)
A.
it (p+q-2s-l)-
_Tr(p+q-2s-1)
r(p+q-2s)
i(p+q-2s-1)
r
0.
It
104.
noncompact; so that
(2p )
c 7 (p+q-
2
and
So
pi
So
p'(p+q-2s-1)
v
p=-
= 2s + 1,
s-1)
'r(p+q-2s-l)
7 (p+q-2s-1) *F
(p+q-2s-l)
331
or 2.
One easily rules out.; for in that case 6=
21'
so that v 7r(p+q- 2 s-l)= 0 or - 1.
4
This contradicts the definition of s unless
is an i--coordinate, and
w(p+q-2s)
(Note that in this case
c7 (p+q-2S-l)
not a critical root.)
1.
'r(p+q-2s-1)
is a j-coordinate.
S(p+q-2s)
is
Also we must have (again from the
2
definition of s)
=
q-s-2-
> 1; it follows easily that
X
u(p+q-2s-2)
To summarize, II) breaks into two cases:
a)
X
b)
X T(p+q-2s-1)
'i
(p+q-2s-1)
> 0,
=
0.
and
In this case
Xw(p+q>2s2)
0,
y 7 (p+q-2s-l) (an i coordinate) is 1, and yp
7r (p+q-2s)
( a j coordinate) is 0.
in
Consider first a).
Since we ere interested only
"2
/ , we may as well assume p+q
2s-l; thus
... 0)]
go = so(2(s-1)+1,2s+l), y1 = [(c,...c,0 . )
1
0, i, or 1) and ftis the so (2(S-1)+1,2 (S-1)+1)
(c
which excludes the first j-coordinate.
(We may have
105.
interchanged the i and j coordinates here, but the reader
may convince himself that this is irrelevant.)
A ()
c A(77), it
the first
is
immediate that
so(2(s-l)+l)
is
a
factor of
trivial on
Hence
.
Since
a-y =y
so by (4.9)
-l
-l
c
1.
Y
J
-l
=
(.)
-
c
-1CC
a
a(P
+
-p
)
J
(**) y
(.-Ea.)
EJ
&
=
Now
by
(3.7).
J
A(r7)
consists of those positive roots with
2
support on the first j coordinate el.
Clearly
'r(2s-l) = £ 2 , so these are
2
Fl
1
(1
<.p + q).
- E
1 in the e
=
J
There are
2
2
and c1+
J
6
gg
7T
and
J
.
coordinate.
a .; so
J
J-1
Now the Weyl
J
group of
so(2s+l)
acts by sign changes and permutation
2
c.; it is easy to deduce that
on the
J
2
2
(b ..
Since
Ea.
i.
=
cpcl
J
(k,
2
2
2
(+ b ,Ib I....bl, br+*b
= p -c-p
J
Ea
2
b )=
2
).
, we have
-1, ... -1,
0.. .0)
on the j coordinates.
- -
=__
MM.-ME
106.
Hence
ES.
i 1.
J
-
1.
J
and
-l
a
J
y
(-l,
=
(1, 1...1, 0...0)
E. )
Za.
1..1.
(1,...
=
J
..
-1, + 1, 0... 0)
2
This forces
2
r = 1, a(b ...b)
2(s-l)+l
2
2
coordinates.
2
b ,b2..b
=
elements.
j
on the
y = (1,0...0) on the j-coordinates.
which has
0...0); so by (**),
:1
+
on the j coordinates,
),
Also A () =
J = 2(s-l)+2; but
So
IA(n ny) I = 2 (s-1)+1, a contradiction.
So
y
is
77-minimal
in this case.
For b),
we must apply (4.22).
0 = so(2s+1,2s+l),
=
is the so(2s-1,2s-1)
In this case
omitting the first i-coordinate
and the first j-coordinate.
Let
f
be the so(2s-l,2s+l)
omitting the first i-coordinate, with 0'3
(maximal) parabolic.
twice)
y
is
the obvious
By the argument just given (used
r -minimal,
and
y
is
7/Af()
minimal in
One can use the subquotient theorem to show that
uniformly minimal in
and
[(il.. ., 0... 0)(0... 0)],
Q
y
.
is
; or one can simply observe that
the preceding paragraph never uses the minimality of | | y | i:
y cannot exist for purely algebraic reasons.
In either
case, this completes the argument for 4.6 (g); as promised,
we will spare the reader a detailed argument for
107.
so(2p+-,2q)
and
so(2p,2q).
Any reader who has actually followed the computations
to this point could undoubtedly treat the exceptional
groups mentioned in .(4.6) in his sleep: no machinery of
any kind is needed
other than a knowledge of the root
systems in a convenient basis.
details.
Q.E.D.
We therefore omit the
101MMEMMS00-
108.
5.
The Subquotient Theorem
This section is clevoted to the proof of Theorem 4.3.
Choose a maximal abelian subalgebra
O
of
an associated system of positive roots, so that
is an Iwasawa decomposition of G.
centralizer of
of
G, and
N
A
in
K; then
is normal in
MAN
MAN.
Let
M
,
and
G = KAN
denote the
is a closed subgroup
Let
a c M
be a
(necessarily finite dimensional) irreducible representation of M, and
Then
6 0 v
V e A
defines a representation of
and hence of MAN.
I
of G.
Ind
MANtG
=
a non-unitary character of A.
MA
M / N,
Define
(6 0 v),
a principal series representation
This representation is non-unitary in general;
appropriate p's are introduced in the definition of
induced representation so that Theorem 5.1 holds as stated
below. Let M' denote the normalizer of A in K:
M'/M is a finite group W, the Weyl group of A (in
G).
W acts on
A
and
M; e.g. if w e M', w e W, then
a representation of
M
on the representation space of
defined by
(w-6) (m) = 6(w
w-6
mw).
is
6
109.
Theorem 5.1
(Bruhat, Harish-Chandra)
I
is an
admissible representation with a finite composition series.
I
and I6'O'
(6',v')
have equivalent composition series iff
(a-6,a-v)
=
for some
a c W.
For each 6, 16
is irreducible for almost all v.
Theorem 5.2
(Harish-Chandra's subquotient theorem)
Every admissible irreducible representation of
G
is
infinitesimally equivalent to a composition factor
("subquotient") of some
I60V
'
It should be remarked that Lepowsky has given a purely
algebraic proof of this result ([18]); in fact even his
arguments can be substantially simplified for the cases
we will need.
From the Iwasawa decomposition it is clear that
I
|
nd 6.
=
K
By general facts about induced represen-
MtK
tations, Ind 6
MtK
is equivalent to
Ind a-6
MLt X
for any
a c W.
By Frobenius reciprocity, the multiplicity with which any
K-type
y
occurs in
Ind 6
is just the multiplicity of
MtK
6
in
plM.
Suppose that
G
collection of K-types
is
y
split.
Let
such that
A (6)
6
c K be the
occurs in
pfM, and
110.
that j|||
is minimal with respect to this property.
Definition 5.3
6 e M.
The K-type
y e A(6),
yp is small if
some
It is principal series minimal if it is the minimal
element of
A(6)
with respect to
are associate if
y2 c A(6),
y,
Two K-types p1, P2
.
6 c M.
some
Another characterization of small K-types is given
by Proposition 5.11.
Theorem 5.4
y
Suppose
is small, say
y
£
A(6).
Then
y M1 is the sum of the M representations in the W orbit of
in
M, each occurring with multiplicity one.
(It
is
known by Bernstein,
very likely that this resultis
Gelfand, and Gelfand ([1]);
but it was obtained independently
for this thesis.)
Theorem 5.4 for a moment.
Assume
Proof of Theorem 4.3.
Set
for some 6.
y
6
in
one.
claim
fo = I
Since
.
y
For i), the multiplicity of
fX
is just the multiplicity of
11
Suppose
T
y
6' = G-6
y
in
y, which is
contains the small K-type y' c A(6'); we
6 must occur in p'|M, since y occurs
A(6) = A(6').
in Ind 6.
MtK
is small, y E A(6)
Since 1' is small, Theorem 5.4 implies that
for some
a
£
W.
By the remarks following
111.
Theorem 5.2, Ind 6 = Ind 6'; taking the "small" elements
MtK
MtK
of each side, A(6) = A(6').
Statements ii) and iii) are
just Theorem 5.1; the subgroup
is the stabilizer of
in
6
mentioned in iii)
W
For iv),
W.
suppose the
occurs in an admissible irreducible represen-
K-type
y
tation
Tr of
G.
is infinitesimally
f
By Theorem 5.2,
6'
particular,
for some
16 1®V
a c W.
By Theorem 5.3, 6' = a-5
pvlM.
occurs in
in
I6'®v,
equivalent to a composition factor of some
By Theorem 5.1, 7
P
and
-Il_=
=
~60G 1V
Thus
have equivalent composition series.
fr is
infinitesimally equivalent to some subquotient of
-l
-l
p
of
nav.But the only irreducible subquotient
-l
0. E. D.
a
containing the K-type p is
I
Recall the ideal
k
C U ; set
yi-
R
p
=
U
/I.
p
Another
consequence of Theorem 5.4 is
Corollary 5.5
be the stabilizer of
action of
W
y c A(6)
Suppose
on
M.)
R c-U(OZ)
p
in
6
W
is small.
Let
(with respect to the
Then there is an injection
- U (Ot).-
_
W
C W
112.
S(0)
The image o:f
This follows easily from results of
Proof .
Lepowsky
is precisely _U(c'()_
[18].
WA
It is known (cf.
polynomial ring.
W
that
G
admits a factorization
W
Knapp and Stein
[13])
= W(A')-R; here
is a normal subgroup generated by reflections,
W(A')
R
and
need not be a
U(OT)
It should be noted that
is a product of
Z 2 's,
(Knapp and Stein assume
is linear, but the non-linear cases can easily be
checked froin the proof of Theorem 5.4.)
IRI
that we may arrange
=
IA(6)j; this would imply that
W
is a polynomial ring when
U(OX )
be shown that
R
and
It seems likely
U(G( )
|A(6)
1 = 1.
It can
have the same field of
fractions, so that their integral closures are isomorphic.
It would be nice if the map given by the corollary were
an isomorphism; but if
W6 / W, this is probably very
hard to prove.
Proof of Theorem 5.4.
covering group of G.
of
for
AG
M'.
in
K~
G
Thus
Since
Let
A
G
denote the universal
is connected, the centralizer
is just the preimage of
W
=
4
in
G; similarly
W~ , so it is enough to prove the
G
113.
theorem for
G.
G
Now
is a direct product of simply
connected groups with simple Lie algebras, and abelian
groups; by standard facts about the representation theory
of direct products, it suffices to prove the theorem for
each factor.
For the abelian factors, M = M' = K, and
the result is trivial.
It remains to investigate the
universal covers of the simple split groups: up to
coverings, these are SP(n,R)
(n > 4),
SO(n,n)
of
SO(n+l,n)
G 2 , F4 , E 6 , E 7 , and E8.
(n > 3),
and the split forms
These we check one by one.
K
is compact and
all
SP(n,R),
Except for the universal covering group of
have finite center; so
(n > 3),
SL(n,R)
(n > 1),
M
is finite.
Accordingly we make heavy use of simple arguments from
representation theory for finite groups, notably the
following ones.
space of
6.
If
{61...
has dimension
=
M, and the representation
d., then
Ed2 = |MI.
(This
is usually used to show that some set of representations
of
M
exhausts
M.)
If a representation of
M'
(for
example a K-type restricted to M') contains an M-type 6,
then it contains all the M-types in the W orbit of
in
M, each with the same multiplicity.
6
If some M-type
occurs with multiplicity one, and the representation is
irreducible under M', then it is the sum of the M types
in the W orbit of 6, each occurring with multiplicity one.
114.
In every case the argument will run along roughly
similar lines.
First we compute M.
Next we exhibit a
collection of representations of K, and show that, after
restriction to M, these exhaust the representations of M;
and that the representations of
one.
M
occur with multiplicity
The proof that the K-types listed are actually the
(Details are
small ones is often left to the reader.
SP(n,R), which is one of the most tedious cases.)
given for
The final step is to show that the K-types listed are
irreducible under M'.
pute
M'
Thus it is never necessary to com-
completely, but only to exhibit enough elements
to act irreducibly on the K-types in question.
Some attempt has been made to keep notation consistent
with Helgason ([9]),
H C- G, we write
K
H
H
for the preimage of
in
G.
K
is the universal covering group of K, so that
i.e.
the representations of
K
A
=
.
G = SP(n,R) = {g c GL(2n,R) t
(Helgason ([9])
Thus
are indexed precisely
by the dominant integral weights for
Case 1
If
especially Chapter IX, section 4.
p. 342; recall that
Jn
=
(.)
0
I
n
.
115.
0
>0
n -Ix
-l
K consists of the
i
x-l
n
such that
matrices
-YL
X + iY c U(n),
and this
X
K
defines an isomorphism of
onto
U(n).
Since
A
contains elements with all diagonal entries distinct,
M = KA
consists of the diagonal elements in
n
M
S=
=
+
11.
In
the
K, i.e.
U(n)
picture,
n
M is therefore the diagonal matrices with all entries + 1.
(
Clearly
P
o
M' C
0
O
so in
U(n),
M'
)
P
isa permutation matrix ;
includes all the permutation matrices.
(The other generators of
M',
as elements of
U(n),
are
116.
+ 1
the diagona l matrices with all entries
It follows that the index of
M
in
M'
or
is
is indeed the order of the Weyl group of
+ i.
2 n -n!,
which
sp(n).)
i61
e
Choose T to be the diagonal matrices
- iG
e n
in
U(n).
M c T,
so
M C T.
We use the
e.
as
coordinates in ;; after normalization, the restriction
of the invariant form
il 0
inner product.
(a1 *..an)
S;so
<
,
>
is the negative of the usual
may be identified with n-tuples
is the usual inner product on
>
a
=
En.
The
(0,...0,+l,0,...0,l ,0,...0)
i < j; of course the non-zero entries are in the
ith and jth places.
weight
if f
,
of real numbers in the dual coordinates to
compact roots are
for
<
a
=
(a ..
2<a,a> c Z
2- (a.-a.)
i
1+ 1
=
a.
-
an
By the Cartan-Weyl theory, a
gives rise to a character of
for every compact root a.
a. c 7,
i.e.
j
a. E a. mod 2
i
3
T
This says
for all i,j.
T may be identified as t 0 modulo the kernel A of
the exponential map for G.
Clearly A is just the lattice
Zn with respect to the coordinates e .
A' = (2Z)n is the
117.
preimage of M, so
lattice
A0
/A.
A
contains a smaller
(the dual of the integral weights in i±)
0
~ o/A10
T =
such that
Mi =
~ A'/AO.
M =
Clearly
Using these
identifications, it is essentially obvious that two
iff
in
T, and in fact act on
normalize
M'
the restrictions of (a.) and (as)
SP(n,[R)
same W orbit in M. (For
by
it
(a.),
is a permutation of
(a.)
Thus if
permutation.
The permutation matrices
for all i.
mod 2
a.
1
a.
1
M
have the same restriction to
a'
and
a
weights
to
M
lie in the
WK
the groups
and
W
are closely related; but they should not be confused.)
(a .
lexicographically, i.e.
it
Order
an)
(al ...
n
=
a 1 ...a. = a.,
The corresponding set of positive (compact)
ai+l < ai+l
roots is
a
iff
{a
(a ...an)
A weight
.1.
13
100
n
is dominant if its
inner product with every positive root is non-negative.
a. - a. > 0
This says
D-
when
i < j, i.e.
(a1 .. .an)
is decreasing.
The sum of the positive roots is easily
computed to be
2o
(2pc) i = n - 2i + 1.
and let
y
=
(n-l,n-3,... -(n-1));
Let
6
so
be a fixed element of M,
be a small K-type containing 6.
p
has
118.
(p ,...yn);
highest weight
b
=
(b
..bn)
y
of
is dominant integral,
must transform according to
1. We know that
A(6)
so, replacing
by a W-translate
the
p
a non-negative integer for i < j. Some weight
is
p-y.
since
b
6
and assume
6
under
depends only on the W orbit of 6;
b
a-6, we nay permute
is dominant integral (every weight
of a representation is integral.)
Since the permutation
matrices in
T,
M'
normalize all of
weight of the K-type p.
a
a = y;
equality iff
> <a+p c ,a+p c>
the same argument shows
y = a.
witn equality iff
is still a
It is well known that whenever
p, <p+p c ,p+p >
is a weight of
b
Suppose
P / b.
with
||p|
>
||ail,
By the
preceding observation, the K-type b (i.e. the K-type with
highest weight b) is strictly smaller than
of section 4; but the K-type
the weight
of
p.
biM
in the sense
still contains
M
6, since
This contradicts the minimality
p = b; i.e. the highest weight of
Thus
forms under
is a-6.
blM
y
p
trans-
according to an element of the W orbit
of 6.
So among all dominant weights whose restriction
to
is
M
a-6, p
has the smallest possible norm
|P I.
Obviously some element of the W orbit of 6 is the
restriction of a T-weight of the form (
here the first r' terms are
, and we may assume 0 < 3 < 2.
119.
If
...3-2); here
3 > 1, replace this by ( -,...3-1,I-2
the first n-r' terms are
weight to
M
-l.
The restriction of this
is in the same W-orbit as y.
So in any
case, there is a dominant weight (e,....,e-1,...e-l),
with the first r terms = e,
0 < E < 1, which restricts
If s = 0, there
to an element of N in the W orbit of y.
is also (1,...1,0,...0), with the first n-r terms = 1.
We know that if
the
b.
r
of
are = c mod 2Z, and the rest are E e-l mod 2Z,
then the restriction of
elements of this orbit.
K
b
to
M is in the W orbit of 6.
(the binomial coefficient)
Clearly this produces
tation of
is such that
(bi,...bn)
On the other hand, the represen-
with highest weight (e,...£,£-1,...E-l)
(or (1,...1,0...0)) has dimension
(n);
this is well
known, or may be checked using Weyl's dimension formula.
So the W orbit of
6
has at most
(n) elements, so we
rl
have them all.
K-type to
Furthermore the restriction of this
M must be the sum of one copy of each element
of the orbit.
To prove Theorem 5.4 in this case, it
remains only to check that p = (s,...£,£-1,...e-l) e A(8)
(and (1,...1,0...0) e A(6) if E = 0).
several steps.
This we do in
In each step, it is shown that if p did
not have some property, there would be a K-type p' with
120.
PM, but |Ip'|| <
in the same W orbit as
y'IM
I|l|i.
By the minimality of y, it follows that y must have the
stated property.
y 1-- pi+
i)
it follows that
Since
pi > yi+1 + 2
yp'
-
lip'll
= <p+ 2 pc'y+
=
(y
is decreasing,
J
for some i. If
= y
2 pc>
<y'+2p cp'+2pc
(y-2)
+(n-2i+1) )2 -
y i~-
y' < p, a contradiction.
(n-2(i+l)+l) < y
+
+
(n-2i+l))
> 0.
+ (n-2i+l) < 1, then
p
If
(n-2i+l)
one can use the same argument with
ii)
j.
for i /
but
= 4(p.+(n-2i+l)-l)
So
i.
y
y '|M = PM,
is dominant, and
|i|
all
= yi- 2 , P.'
+ (n-2i+l) > 1, set yp.'
Then
+ 1,
pi+
Suppose not.
Proof.
y
i=
or
-
4 < -3
< -1,
and
pi+l = 1i+l + 2.
The sequence (p.) contains no subsequence
...
Proof.
x+1,x...
x,x-l...).
Suppose that such a subsequence occurs, with
the last x+1 in the ith place and the first x-l in the
jth place
ithpplace
th
th
j
place.
//
Setting
y'=
..
x
, x
...
x
X-1,
X-1,
..
2
121.
we see that
and
pIM
are in the same W orbit
y'|M
(P' is obtained from p by transposing the ith and (j-l)th
places, and subtracting 2 from the (j-l)th place.)
Also
|| || - ||p ' I |
-
= <p+2pc ,p+2p c>
<p'+2p c,'+2pc >
= (x+l+(n-2i+l)) 2- (x+(n-2i+l) 2
+ (x+(n-2(j-l)+l)) 2
(x-l+(n-2(j-l)+l)) 2
= 4((n-i-j+2)+x)
which is positive if
y" =
(...
than
P
of
y
x > -
x+1, x+1, x,
x < -
Similarly,
jth place
uith place
if
(n-i-j+2).
... x,
x, x-l ...)
is smaller
(n-i-j); so in any case the minimality
is contradicted.
//
It follows from i) and ii) that
V
is of the form
(t,3.--,3-1,...I3-1); it remains to show that
0 <
Say the first r' terms are S.
iii)
Proof.
0 <
< 2
Suppose not, e.g.
6 > 2.
Set
y'= (f-2,...S-2 ,I-3,...r-3), with the first r' terms
-2.
Then pIM = y' IM, but
< 1.
122.
r'
I ||- ||lv '
[( +(n-2i+l))2- (-2+(n-2i+1))2
i=1
n2
2
+ E
[(-l+(n-2i+l)) -(-3+(n-2i+l))
i=r'+1
=
4A(S-l+(n-2i+l)) +
i=1l
= 4
n
E
4(-2+(n-2i+l))
i=r'+1
E (f-l+(n-2i+1)) + 4
= 4n(f-l)
-
4(n-r')
(-1)
E
i=r'+1
E (n-2i+l)
since
=0
i=1
> 4n -
4(n-r)
> 0.
We get equality only if
that case we may rewrite
iv)
0 <
Proof.
S=2
and
-l.
In
=0.
yp = (0',...1'), with
f3'
=
1.
< 1.
Suppose not, i.e.
1 <
< 2.
P' = (f-1,...j-1,r-2,...f-2); here the first
are
r'
Then
y' M
n-r'
(S-l+n-2i+l)
-
2
-
|M,
and
n
2
E
(r-2+n-2i+1)
i=n-r'+1
-
2
E ((-1)+n-2i+l)
i=1
n
terms
n
(-l+n-2i+1)
E
i=r'1+1
i=1
n
=
n-r'
is in the same W orbit as y
E ( +n-2i+l) 2 +
I yvI -I lvII
Set
+
r'
E
[2(
-l+(n-2i+1))+1]
i= 1
n
2
E (2U3-l+(n-2i+l))-l].
(6-l+(n-2i+l)) +
i=n-r' +1
123.
Changing the last index,
r'
{ [2 (-1+ (n-2i+l) ) +1]
+
[2 (a-1- (n-2i+l))-l1] }
i=1
4r' (6-1) > 0.
=
r' = 0.
Equality holds iff
(
r = n, e =
defining
In that case, y =
1, p is in the desired form.
-
This completes Case 1.
G = SL(2n,R),
Case 2
Here of course,
n > 2.
0
i > 0,
A =0
=
X
11 ,
and
K is SO(2n).
n
of
K
that the fundamental group
Chevalley [5])
One knows (cf..
G
has order 2; so
and K = Spin (2n).
is a two-sheeted cover of
G,
Clearly
0
1e.
M=
0
)
n
=
1
+1, Hs.
-
=
1 , and
M'
1
of the permutation matrices of determinant 1.
IMI = 2 2n-1, and
|MI
=
2 2n.
consists
Since
M
So
is abelian,
M
has at least 22n-1 one dimensional representations.
K = Spin (2n) has two representations of dimension
2 n-
124.
We will eventually show
which we call spin+ and spin~.
that these are irreducible and inequivalent as represen2n-1
2
n-1 2
2n
) + (2n-l )2 ,
+ (2
-(1)
=2
Mj = 2
tations of M. Since
these exhaust the representations of M.
Leaving to the
reader the easy fact that spin+ and spin
are actually
the small K-types of the corresponding representations
of M, this will prove Theorem 5.3 for the M-types
So we consider the one dimensional represen-
spin-fM.
tations of
M
Embed
M
O(n).
in
6
S
e D
by
D
first.
in the full group
2n
If
has order 22.
.
6
S
=
.
S
D
of diagonal matrices
S C {1,2... 2n}, define
s..
e
1
This exhausts
65 IM exhaust M. Since
so the
£
M
D,
iff
n
65 |M
HE
= 1,
If
a s W, and
that
M'
=
65 ,jM
a
S = (S')
s = s' or
iff
is the corresponding permutation (recall
consists of permutation matrices)
It follows that the orbits of
exactly one representation
Consider first the case
W
in
Then
6,.
ST
=
M each contain
6r = 6{1,2 .
r < n.
a-6
r}
W-6r
for
0 < r < n.
is clearly in
125.
1-1 correspondence with the r element subsets of
2n
{l,...,2n};
so |W-6 | = ( ).
Let p r be the
r
r 2n
if
[A (R2n
representation of SO(2n) on
e1 . . . e 2 n
is the usual basis of R 2n,
clearly transforms according to
6r
prIM.
occurs in
2n
rr ) =
dim y
r
element of
W-6r
6r
then e1 A
under
.
M.
On the other hand,
-W2r|,
r'sso
p|M
PrI
exactly once.
must contain every
(In particular
is irreducible, which is of course well known.)
argument along the lines of that given for
shows that
taining
yr
6 r'
A er
Thus
yr
An
SP(n,R)
is in fact the unique small K-type con-
which proves Theorem 5.3 for the 6r, r < n.
To consider
6n
and the spin representations, we
need a little more structure.
Let
cos 01 sin 01
-sin 01 cos 01
cos
n sin
)
n
n
n
0
-sin 0 cos 0
n
n
be the usual torus in K.
in %,
again identifying
We use the
i
0
with
0.
Rn
as coordinates
(with
<
>
the
126.
usual inner product).
The Weyl group of
t
in
K, WK'
acts by permuting the coordinates and changing an even
number of the signs of the coordinates.
We order
it0
The positive
lexicographically in these coordinates.
(0,...1,0...,+1,...0); so
(compact) roots are then
2pc =
A weight (a1 ...an)
(2n,2n-2,...2,0).
is dominant
an-l + an > 0; it is integral
if it is decreasing and
a
is an integer or if every
is an
if every
a
integer.
All this may be checked just as we checked the
Consider now
6 n.
the W orbit of
in
6n
1 2n
) elements.
has 2-(
tation of
The pairs
K
M.
It follows that this orbit
2n
The weights of the (R )C represen-
(0,...0,+1,0...0),
are
of n-element
{S,S c}
are in 1-1 correspondence with
{l,2.. .2n}
subsets of
above.
U(n)
corresponding statements for
-
a
+ 1
in the
ith place corresponding to the weight vector e2i- T i 2i
1
in the usual basis.
one.
are
These weights occur with multiplicity
Hence the weights of the
(e1 ... en);
here
r
(An.2n C representation
of the
and the rest are zero; furthermore
e
are 1,
n-(r+s)
s
are -1,
is necessa-
rily even, since +1 weights and -l weights can cancel
only in pairs.
In particular the dominant weights
127.
yn
=
and
(l,...l)
y~(1,...l,-l)
occur, and
1y + a
does not occur for any compact positive root a.
follows that
2 n)
(An
It
contains the two irreducible
+C
K-types y n.
-
dim y=--(
1
By weyl's dimension formula,
2n
1
n 2n
)=-dim (A R
np2n
)
, so (A
are
e A ... Ae
1
n
6n
contains each element of
1 2n
|W-6n 1
n ),
+
-
@Pyn
2n
twice (the corresponding vectors
e
and
=y
it is clear that An
Using the usual basis of R2n
contains the M-type
)
1
2n+1
W-6
'
..
n
).
A 2 )n
An2n
Hence
Hnc
at least twice; since
each element occurs exactly twice.
It
follows that the conclusion of Theorem 5.4 holds for
either
y
or
yn.
IPy|
= | |I
yn'
One can compute that |
and it is easily verified that
A(6nl
~
It remains to show that spin+ and spin
irreducible and inequivalent under M.
are
By Schur's lemma,
++
it is enough to show that T = (spin @ spin~) 0 (spin @ spin )
contains only two copies of the trivial M type.
n
well known that
Since
i 2n
A ER
=
A
=
(A R2 n ),
2n-i 2n
R
But it is
the exterior algebra of R 2n
as a K-module, our previous
results show that the only M invariants in ff are
AR
2n
e A 2n
2 n.
This completes Case 2.
The case of
128.
G = SL(2n+lIR)
is considerable simpler for more or less
obvious reasons, so we leave it to the reader.
G = SO(2p+1,2p), p > 2.
Case 3
G
is the subgroup of
SL(4p+l)
consisting of matrices preserving the quadratic
2
+
2
-x 2
(cf.. Helgason ([9])
1 -x
2p+1 2p+ 2 .
4p+1
form
{e. .}
be the usual basis of the
JJ
4p+l x 4p+1 matrices. As a basis for the Lie
algebra
p. 340).
Let
of A, we may use {ei2p+l+i + e2p+l+ifi
X
K
is
(
X
0
0
Y
i
7/
) 1 x.E SO(2p+l)
0/0
1 1 < i < 2p}.
, Y E SO(2p)}.
A matrix
Y
commutes with
if
iff
01
j, i,j < p, and X
X.
=
Y..
= 0
for
1,
=Y
(X,Y) c SO(2p+l) x SO(2p),
for
i < p.
For
this implies
0
E
0
e2p
X
0
0
Y)
1
0
E.
P is
P
0
S1
0
0
2p
= +1, HIe.
=
1.
If
a 2p x 2p permutation matrix, det P = 1, then
0
0P
1
M';
' and if
D.
is diagonal, det D. = 1, then
129.
D
0
0
D2
(
M'.
These assertions may be checked by direct
calculation.
Clearly
M
has order
2
2 p-1
On the other hand,
K = Spin (2p+l) x Spin (2p) is a 4 sheeted cover of K.
|M| = 2p
so
G
Spin (2 p+l,2p).
has a 2 sheeted linear cover
For a linear group, M is abelian
(cf. Warner [25], theorem 1.4.1.5, or the proof of
Lemma 5.5 below),
So
M
has at least
2 2p
one dimensional
K
onto Spin (2p).
representations.
Let
Then
72 (M)
"M'
denote the projection of
T2
for
is
SL(2p);"
remarks on the
of
"M for SL(2p)," and
K, and
R2
r2 (M')
contains
this follows from the corresponding
SO(2p+1,2p) level.
If
H
is a subgroup
is a representation of Spin (2p), then
the representation
1 0 R2
(outer tensor product) of K,
restricted to H, is trivial on ker
the representation
R2
V2 , and corresponds to
restricted to
immediately that the various
1 O un, are irreducible under
72 (H).
It follows
1 0 pr' 0 < r < n, and
M', and correspond to
22p-1 distinct one dimensional representations of M.
Also the two representations 1 0 spin
under
M
are irreducible
and inequivalent; each has dimension
2 p.
130.
M'
Next, notice that
for SL(2p);"
ment for
SO(2p+1,2p) level.
on the
M'
"M
(Here "spin" is the 2
SO(2p+l).
sional spin representation of
Theorem 5.4 for G = SL(2p+l,R)
for
"M
irreducible under
SL(2p,R).)
exactly as for
that spin 0 spin
It follows that
is irreducible under M'; it has
dimension 2P.21 = 22p-1.
K
x
again this follows from the corresponding state-
K-type spin 0 spin
of
"M for SL(2p+1)"
contains
dimen-
The proof of
shows that spin is
SL(2p+l,IR);"
this proceeds
It is not difficult to verify
actually lives on the two-sheeted cover
in the linear cover of G, but not on K.
level is abelian, so spin 0 spin~ provides
2
2 p- 1
M at this
more one
dimensional elements of M - we need only show that they
are distinct.
Let
H
This can be done in the following way.
denote the standard
SO(2p) c SO(2p+l), h C-4
0)
(0
Let
M
C H
denote
"M for SL(2p)", and let
corresponding subgroup of
'r 2 (K) = SO(2p).
M2
Then
l
be the
M
= M 2,
so it makes sense to speak of the diagonal subgroup
A C M1 x M2
A is just M.
We will show that
spin 0 spin IA'
22p-1 distinct components.
= spin+ @ spin~.
spini
H
Thus M
2
A(M
x
M22)
A'
M1 .
already consists of
First, one knows that
So as a representation of
131.
A' = M1 , spin 0 spin
(here
of
0
spin-)
means inner tensor product.)
0 spin
As a representation
A(H x H) = H = Spin (2p), this is known to be
p-1
( E Pr
r=O
2
is simply (spin +
2 p-
n], which we know (from Case 2) consists of
distinct N
= A'-types.
Finally, consider the K-type spin 0 1.
denote the projection of
for 1 0 Spin,
K
on
Spin (2p+l).
Let
1
Arguing as
we see that spin 0 1 is equivalent as an
M representation to the representation of f 1 (M) on
spin
|
rr(M).
Now
' 1 (M)
=
MC-
H (since this holds on
$ spin~.
the SO(2p+1,2p) level) and spinip = spin
So
spin 0 1 splits into two irreducible M components of
dimension
2 P;
they are inequivalent.
We need to show
that spin 0 l.is M' irreducible; but fr(M') 2 "M for
SL(2p+l)," which acts irreducibly on spin.
In this way (i.e. by hook and by crook) we get
+ 2
2
four
one dimensional representations of M, and
dimensional representations.
2
2p-1
2p+1
|MI = 2
2
+
2p-1
2
+ 4-(2
p-1 2
),
Since
this exhausts M.
Leaving to the reader the proof that the K types exhibited
are the small ones, we have proved Theorem 5.4 in Case 3.
132.
The various other split orthogonal groups are similar and
easier.
For the exceptional groups, the following lemmas
are helpful.
Lemma 5.6.
of
K.
Let
T : K
Choose
+
a .l. an
be the V
SO(Y 0 )
representation
a strongly orthogonal set of
noncompact imaginary roots, and let
X
. Xn
be
non-zero elements of the corrresponding weight spaces
P
of
= (60O C.
element
a.
Suppose that for each
in the normalizer of
2
= -a ., and f(a. ) = 1.
aui-a.
a~
by
{X
of
.
+
Also
subalgebra
Proof.
0 of
in
there is
=
(X
Ca.
spanned
is an abelian subalgebra
so that
0(=
for some abelian
O
That C( is abelian follows immediately from
the strong orthogonality of the ai; notice that
commutes with
an
K, such that
Then the space 07
a.
)-(a.)X
} and
01=OZ,
T
i
X
since
a.
J
is an imaginary root.
Since
ti
, it suffices to see that
+ ff(a )X
C.
) = C
(X
+ 7(a)X
i)
for each i; here
For simplicity of notation, we drop the subscript i.
133.
7(a)Xax clearly has weight -a.
-ea = -a
a
since
is imaginary.
Xa
are one dimensional,
sides; since
Thus
C X; so (Xa+7(a)Xa)
group.
Since the weight spaces
Ca7 (a)Xa.
=
Apply
ff is a real representation,
with conjugation.
Lemma 5.7.
By (2.6), Xa has weight
ff(a)Xa
Suppose that
=
commutes
2
T(la
)Xa
C
Cx ((a)XX+Xa).
=
QED.
is a split semisimple matrix
is abelian, and
M
Then
G
to both
7(a)
CaTr(a)Tr(a)X
=
Xa + C X0
=
'r(a)
IMI <
2 rank
G
By hypothesis, G c GL(n,R), so that
Proof.
C gk(n,R),
and
Vc gk(n,C).
connected subgroup of
Let
GL(n,C)
G
denote the
J'.
with Lie algebra
bar denotes the natural conjugation in GL(n,C),
is just the identity component of
If
then G c G
{g c G |g = g}.
The center of a connected Lie group consists precisely
of the elements which act trivially in the adjoint
representation.
center of
G
=
Since
it follows that the
(%)C'
is contained in the center of
GC.
A complex
semisimple Lie group has finite center, so the center of G
is finite, and therefore
closed and discrete,
Let
AC
=
exp 0.
M
K
is compact.
Since
M C K
is finite.
One knows (or it follows from the
results of section 2) that
A
=
G ; hence
Mc A..
The
is
-A
134.
kernel of expi
is a lattice
A C ic', and exp ig0
is isomorphic to a product of dim C
Suppose
Since
M
m c M; choose xy c 0£0 so that exp (x+iy) = m.
is finite, mN = 1
this implies
that
= rank G circles.
for some positive integer N;
Since
Nx + Niy e A.
x = 0, so
m e exp iOl
A c ic(0 , it follows
On the other hand, since
m c G, m = m = exp (iy) = exp
(-iy) = m
-l
, so m
2
=
1.
A product of r circles has exactly 2r elements of order 2,
rank G
.
QED.
so IMI < 2
One can give a much shorter proof using known results
for instance the fact that
-
M c exp (io0) is a special
case of a theorem of Osborne and Rader (cf. Warner [25],
proposition 1.4.1.3).
But is seems likely that anyone
familiar with these results knows the preceding proposition.
Case 4.
G = simply connected split form of G 2 .
Here
a2+|2
K = SU(2) x SU(2) =
2+6j2=l}.
We let the first factor here correspond to the long compact
root in the diagram below.
Each
SU(2)
acts on polynomials
in two variables (say (x,y) and (z,w) respectively) so that
K acts on polynomials in x, y, z, and w.
representations of
K
degree n in
and
(x,y)
The irreducible
are the actions on polynomials of
m
in
(z,w); we call this
135.
representation (n+l) 0 (m+1), these numbers being dimensions.
Weight vectors with respect to the usual Cartan subgroup T
of K (consisting of pairs of diagonal matrices) are the
The p
monomials.
representation is isomorphic to 2 0 4.
Such an isomorphism is indicated in the following root
G
diagram for
xz 3
with respect to
xz 2w
xzw 2
T.
2D
strongly orthogonal
yz
yz w
a e N(T)
Let
yzw
yw
noncompact roots.
be the element
(0
1)
a2 = (-I,-I); aince -I acts by -1
0)
-l
0)
-l
in every even dimensional irreducible representation of
((0 1)
SU(2),
xzw
2
a2 acts by (-1)
and
xz
3
(-1)
=
1 in
-.
The root vectors
correspond to strongly orthogonal roots,
so we may apply Lemma 5.6. Since a-(x,y,z,w) = (-y,x,-wz),
2
2 =
-)-~
2
3
3
3
a-(xzw ) = (-y)(-w)z = yz w, and a-(xz ) = (-y)(-w) = yw3;
2
2
3
3.
as a basis
so we may take xzw + yz w and xz + yw
for
07
(of course
O is equal rank, so
= 0 ).
It
136.
is easy to check that the eight elements
e 3iO
0
/
0
ie
(jO
e
ei
'0
0
e-3i)
-3iO
0
We claim that
group.
G
and (-I,I)
act by
e4i
1
U
e
0
e3ie
(-e-_ie
0)
(e 4iO
fix O(
is a two-sheeted cover of its adjoint
The center of
in the center of K.
0))
G
The center of
-l
Y
is just the centralizer of
on
),
K
is
(+I,+I); (I,-I)
so the center of
G
is
{(IFI),
(-I,-I)}.
dlaim.
The adjoint representation is faithful for the
This has order two, which proves the
adjoint group, so Lemma 5.7 applies to it:
"M for the adjoint group" hasat most 22 = 4 elements.
Thus IMI < 8, so the 8 elements listed above exhaust M.
It also follows that
representations.
has at least 4 one dimensional
M
It is easy to check that
1 0 3|M
consists of 3 inequivalent non-trivial M-types, spanned
by
z2 +w 2,
M-type.
z2-w , and zw.
We claim that
recalling that
0
10 1 is of course the trivial
1 0 2
is irreducible under M:
denotes internal tensor product, we
compute
1
2*
(1 0 2) 0 (1 0 2)
=
1 ® (2 0 2)
=
(1 0 1) @ (1 0 3)
137.
which contains only one copy of the trivial M-type by the
By Schur's lemma, 1 0 2 is M-irreducible.
preceding remarks.
3 + 1 + 22 = 8, this shows that 1 0 1, 1 0 2, and
Since
1 0 3 exhaust M, each M type occurring with multiplicity
Leaving to the reader the verification that these
one.
are small, it remains to check that they are M'-irreducible.
Of course 1 0 1 and 1 0 2 are already M-irreducible.
2
2 = i; then one checks easily
e C so that
Choose
E-3
that
0
0
Pl=
and
(0
P
2
(3)
Q=
(
d/
1
(0
~
, --
1
2
are in M'. The
1
1,
2
2
z , zw, and w of
three weight spaces
according to distinct scalars under
2
2
(z) =
It follows that
irreducibly on 1 0 3.
K = SU(2)
transform
and
1
2
(-(z-w)) , which has a non-zero component in
each weight space.
Case 5
P
1 0 3
P1
and
P2
act
This completes Case 4.
G = simply connected split real form of Fg.
x
SP(3).
of diagonal matrices
A torus for
K
Here
consists of the pairs
138.
e
(0
0
0
2
00
e
i0,
i03
eie 3
-ie
e
0
2
e
-i0 3
e
We use the
01
identified with
dual basis
t0 ; then
as coordinates in
IR4
is
in the dual coordinates, with the
{e0 'e1 'e2 le3 }; after normalization, the
invariant inner product is the usual one.
roots are +2e 0
(the roots of SU(2))
1 < i, j < 3, i / j
for
it 0
and +2e , +e.+e.
(the roots of SP(3)).
+e 0 +ej,
noncompact roots are
The compact
1 < i < 3, and
The
+e0+e +e
(This is the dual of the usual presentation of the roots
of
F4 ; cfi, Humphreys ([11])
,(
Let a =
L-l
a2
=
[
0
c K. Then
0O
-I
= (e
(e
,e
,e
e
7),
which
We claim that
2
acts trivially on z
is K-irreducible, and
a2
is central in K, it
is central in K.
since
p. 43.)
is enough to check this on one weight vector of
, say
;
139.
(1,1,1,1).
(x
0
[e
Now
,...e
'lX2'x 3 ) by the scalar e
(1,1,1,1)
e
by
=
acts on a weight
']
ixi
a2
; so
acts on
1.
As a strongly orthogonal system of noncompact roots,
one can choose
with an odd number of
(1, +1, +1, +1),
Apply
minus signs; clearly there are four such roots.
Lemma 5.6 with
a
= a
for all
i.
get a maximal abelian subalgebra
M nT.
0,
670
= 0, we
with a e M.
3
0
We want to compute
P
Since
[ei0 ...e
]
is in
M
T
iff it acts trivially on each of the strongly orthogonal
weights (and their negatives.)
00-81+62+63
0+61-02+e3,
being elements of 277.
easily that
and
in
and
00+O +62 63, and 60-61-62-63
Using this condition, one checks
[+1, +1, +1, +1]
[+i, +i, +i, +i]
This amounts to
(even number of minus signs)
(odd number of minus signs)
Mg T, for 16 elements in all.
Mn1T
Also
G
a e M; clearly
generate a group of order 32.
hand, a computation like that given for
are
On the other
G2
shows that
is a two-sheeted cover of its adjoint group, so that
IMI < 2.24 = 32.
M
So this exhausts M.
has a direct product decomposition
a
140.
M=
<
[-7ii
, [1 -
= P x Q x R
-i
-,-1]
, y> X <l~,1 - ] > X < l -,11]>
(here < > denotes "group generated by".)
This is quite easy to check, using the fact that if
t c T, ata
=
t~1.
P
is the non-abelialn group of
order 8, which was studied as
of course isomorphic to
Z2.
"M for G 2 ."
R
and
Q
are
Thus the representations of
M come in sets of four, each set corresponding to an
R.
and parametrized by the action of
P
element of
We continue to denote representations of
their dimensions; also write
Q
and
by
SU(2)
"l" and "6" for the trivial
and six dimensional representations of
respectively.
SP(3)
Then the small K-types are 1 0 1, 2 0 1, 3 0 1, 1 0 6,
and 2 0 6.
1 @ 1 is obviously the trivial M type.
2 0 1 and 3 0 1 are trivial on Q x R, since this group
projects to I in the first factor (i.e. if (T1 ,T2 ) E Q x R,
then T1 = I e SU(2)).
The projection of
P
onto the first
factor is "M for G 211, so 2 0 1 is the 2-dimensional
representation of P, and 3 0 1 is the 3 non-trivial one
dimensional representations of P.
[-1,-1,-1,-1]
P
acts by -1 on 1 0 6, so this must split into three copies
of the two-dimensional representation of P: if
xlIx2'
3'
l'
2'
3
is the usual basis of
C6
as an
SP(3) module, the corresponding components are <xi,yi>,
141.
as is easy to check.
Q x R
acts on each of these pieces
by a different character, the three non-trivial characters
of Q x R.
Finally, the 2 dimensional representation of P
tensored with itself is the sum of the 4 one dimensional
representations; from our description of 2 0 1 and 1 0 6,
this implies that 2 0 6 consists of the 12 one dimensional
representations of M which are non-trivial on Q x R.
Thus all M types occur in the given list, each with
multiplicity one.
Leaving the verification that these
are small to the reader, it remains to show M' irreducibility.
This is of course obvious for 1 0 1 and 2 0 1.
To compute some elements of M',
M
.
observe first that
For it is easy to see that the only noncompact
roots on which
MA T
acts trivially are the given
strongly orthogonal set and their negatives; and
6Z
is
precisely the space of a-invariants in the span of those
8 root vectors.
If
g c K, it follows that
so that M' is the normalizer of M in K.
Then tt
1
iff t2 c M.
= t(at 1 a
g-O=
g9"
Suppose t c T.
)a = t-t-a = t 2 a, which is in M
Since T normalizes MC T, t e M' iff t 2
M.
Using this, it is easy to check that each of the weight
spaces in any one of the representations 3 0 1, 1 0 6,
and 2 0 6 transforms according to a different character
142.
of
M'n T.
To prove irreducibility, it is therefore
enough to show that any of these weight spaces is cyclic
under M'.
2 0 6.
For definiteness, we restrict attention to
Let
(x0 'y0 )
SU(2) module.
be the usual basis of
as an
Considering the action of a, it is enough
to show that one of the subspaces V
W = <y0
C2
= <x0 0 x
y0
O
0 y i>, i = 1, 2, 3 has components in
i
all the others under the action of M'.
Let
$
be a
3 x 3 permutation matrix such that $2 = I; then $ = $so
y
i>'
= (I,
commutes with
=-
a, and it follows
0$
easily that
(y )-W
M'-(EV )
=
W
y
e M'.
.
Clearly
So it is enough to show that
has a component in EW.
Q
g=
can be checked that
g
-1
ag =
g e M';
g
/
)
g (x0
has a component in EW.
completes Case 5.
-1
l) =1
-i
-iI
commutes with
[i,-i,-i,-i], and g
and
For this we use the
I
11
element
= V
(y )-V
K. It
Q
and
R,
[i,-i,-i,-i]g = -a.
0iy0
(x1 -iy1
So
,
which
So 2 0 6 is M' irreducible, which
143.
Case 6
G = simply connected split form of E
K = SP(4).
We use the coordinates
6.
Here
in the Lie
I
algebra of the torus
e
i1
i64
T
=
-io
[e
i4
,...e
]}.
e
With respect to the usual basis
the compact roots are +2e.,
e1 .. .e 4
+e+e
of
(i / j).
42 dimensional subrepresentation of
i
I-
=
4
R,
2 is a
A4 C , the fourth
exterior power of the standard C8 representation of
SP(4); its weights are +e 1 +e 2 +e 3+e , +ei+e
and two zero weights.
(i / j),
By the general results of section 2,
the imaginary noncompact roots are those which are not
also compact, i.e. +e1 +e 2 +e 3 +e 4.
Notice in particular
that
A4C8
contains any element of
of weight
+e +e2 +e3 +e
4 , since these occur with multiplicity one
48
in A C . Also the center of K, which has order 2, acts
trivially in
; so the center of G has order 2, and
.4
144.
fM| < 2-2
128.
=
0
a s SP(4)
Let
denote the element
Consider the group
Interpret
C
4
(r2
'r,srs}
2
L (D), and take
as
as a basis for
D = Z2 x2
8
4
4
C8 = C @ C .
For
2
x 1 ... ,x4 ,y .. .y4
g c D, let
P
g
denote the 4 x 4 matrix defining the action of g in
0 1
the regular representation: e.g.
P
=
1
0
0
0 1
1
Clearly
P
Pg1 P
91 92
=t-1
P
g
P .
=
g
g1 g92
;
in particular
a c
Define
SP(4)
by
P=
1, so
0
=
a
0
9
0
a
then the
x...x ,
S8;
y-
commute with
.-
Y4
notice that
a.
a-x
= -y,, a-y
X(g)
2
2
2
has weight
If
for
X
is a
g c D;
Set
1-X (rs)
1-x (1 )
X
= x,.
= + 1 or - 1
1-X(g) = 0 or 1 accordingly.
=ca
Now consider
as the usual basis of the SP(4) module
character of D, then
b
P
e , and
yg
48
* s4
-AC.
has weight -es, b
Since
x.
has weight
145.
,, = (x(1),x(r),X(s),x(rs));
{S }
check that
b.
so
It is easy to
is a strongly orthogonal set of real
X XED
noncompact roots, and of course
a 2 = -I
since
a
x
=
b
x
+ a-b
x
fixes the
a -a
=a
g Xg9
, together with
a
;
a
2
as
,
a
and all
a
basis of (X .
are in
g
M.
Obviously
and
2
a -(a
(Ha
By Lemma 5.6,
.
-S
is central in G, we may take
We claim that
a
=
a*3
-X 1 ...
- (aG
x
^a
1-2 (rs)
2
- x 4 ] + a[
1-x(rs)
2
)A .. .^a
- (a
-x 4 )]
(since a and a
commute)
1-X (rs)
=
Since
OXg-1
2
-x g4]
+ a[
]).
a2 = 1, it is easy to see that
1-X (9 )
1-x (g)
a 2
([a
A...A 0
,,
2
1-x (g)
- 2
1-X (gg1 )
2
.
1-x (g-1)
2
- ( [C
x g- I
But the permutation of
(l,
Therefore
(recalling g 2 =1)
1-X (g -rs)
A
Aj
2y
-x g
4
]+ a[
2, 3, 4) defined by the regular
146.
action of
=
g
is necessarily even; so
1-X (g)
2
l-X(1)
2
K 1A
1-x (rs)
2
* AC
x4
+ a[
])
1-X (g)
2
=a
since
That
a
a
X
and
a
of
z A z
x1
Ax
SP(4)
(z1 -z 3 )A (z2 -zg)
Now
a-z
.
= z2'
zi = xiA y ; then
c
A4 C 8 .
a .
9
Z AZ
= 0,
We have seen that
axA
y
to this, one gets
(z1 -z 2 )
(z3 -z4 )
in g ; by dimension, these span(
= -y A x
= z,; so
= x Ay
a
For reasons of symmetry, it is enough to
consider any non-identity
or (z 1
fixed by
is spanned by those combinations
and
=
is
a.
cp ; applying appropriate elements of the
Lie algebra of
fixes
Write
which lie in
4
Thus
is just an explicit computation,
and
z A z.,
=
a.
fix
g
which we only sketch.
z A z
X
is fixed by
X
a
= a
a , say ar.
One checks that
ar (z2) = z1 , ar (z3 ) = z 4 , and ar (z4) =
hence ar((z -z2 ) A (z3 -z4 )) = (z2 -z1 )A (z4 -z3)
(z1 -z 2 )A (z3 -z 4) , and
(z2 -z 4 )A (z1 -z3 )
the proof that
=
a
ar ((z1 -z3 )A (z2 -z 4 )) -
(z1 -z 3 )A (z2 -z4 ).
and the
a
are in
This completes
M.
3'
.
147.
Let
by
a
MW
be the 16 element abelian group generated
a .g Notice that
and the
MW
The 16 element (abelian) subgroup of
normalizes
T
T.
consisting of
[+1, +1, +1, +1]
(even number of minus signs) and
[+i, +i, +i, +i]
(even number of minus signs)
contained in
M; this is checked exactly as for F 4 .
we call this subgroup
MW CM
MT
{+I}; so
=
MT,
MTMW
is therefore all of
M.
Next, we claim that
If
then MW normalizes MT
16.-16
= 128 elements, and
2
has
|MJ
(Recall that we saw
q
=
PM; in fact
For one checks that the only weights of J
under
is also
< 128.)
Q = P0<MTta>
invariant
are + 1X and 0, and the two dimensional space
MT
corresponding to the weights + 13 has only one a invariant.
It follows as before that
of
M
in
M'
is precisely the normalizer
K.
Since
M/+l
is abelian, there are at least 64 one
dimensional representations in M; one may check that each
occurs with multiplicity one in the representation
C8 08
of K.
It follows by Schur's lemma that
irreducible under M; since 128 = 8
A
M.
Now
S2 (C8 )
C8 @ C8
A (C8)
C8
is
+ 64, this exhausts
decomposes as a K representation into
=
S
7 2 0 1; here
S2
is the symmetric
product, 1 is the exterior form defining SP(4)
(atrivial
148.
K-type) and
f2
is a 27 dimensional subrepresentation
of
A2 (C ); these are irreducible.
1,
88
are small; the proof is as usual left
and S
2
We claim that
It remains to check that
to the reader.
f2
and
S2
are M' irreducible.
Let
E
denote the permutation group on the 4 element
set D; recalling
~2
4
C 4 = L (D), an element of Z4 corresponds
to a 4 x 4 matrix P, which in turn gives an element
0)
e SP(4).
D
Z4 , and
is a normal subgroup of
the conditions defining
are invariant under permuta-
MT
tion of coordinates; so it is not hard to check that
E4 C M'.
Also the elements
T
-of
Using these, together with
space of
iT2
character of
or
S2
are in
M'.
MT, one sees that every weight
transforms according to a distinct
M' 0 T.
Hence every M'-irreducible
component of those representations is a sum of weight
spaces.
For definiteness we restrict attention to
Considering the action of
M'-irreducible component of
2
2
four spaces
V1
V3
E
and a, one sees that every
2
S
is a sum of certain of the
<Xi ,y 2 >' V 2 = <xi .,yiy.
i<x
y.
I
2
S
j>' V 4 = <xi
y>
I
149.
To prove irreducibility, we exhibit two more elements of
M'
and
where
X=
the matrix of characters of D.
g2
g
Put
which mix these subspaces.
-
1
-l
1
-1
1
1
-l
-1
One may thing of
as diagonalizing the commutative families
{a }
D
=
respectively.
normalize M, and hence belong to M'.
Thus
V4,
S
Case 7
g1 -V2
meets
g,
and
{l,a}
and
From this perspective, or by
explicit computation, one can check that
meets
is
V 3 , and
g,
Clearly
g2 V
meets
and
g2
g1 -V1
V 2.
is M' irreducible, which completes case 6.
G = simply connected split forms of E7 and 2 .
8
These are entirely analogous to E6; since E8 involves
some rather messy computations in aClifford algebra, we
give details in that case.
So
K = Spin (16).
Using
the usual Cartan subgroup on the SO(16) level to give
an identification of
of
is
1
(,...
representation.
in the center of
1
)
it
with
R8,
the highest weight
is the positive half spin
One knows that two of the four elements
K
act trivially on -9
(this will be
150.
seen explicitly below) so the center of
IMI < 2-2 8
Thus
Define
G
has order 2.
=512.
D =Z 2
Z2 X Z2 , with generators r, s, t.
Writing D
{l,r,s,rs,t,rt,st,rst}, we identify
I 8
2
it
= R
with L (D). A strongly orthogonal set of
weights for 0
is
For each
X
we choose
is a representative
a2 = 1
{u
X
u
a
c
is a basis for CZ .
X
~'7
of weight
X/2.
2If
There
By Lemma 5.6,
M should now be
M n T = MT, a, and a certain copy of
generated by
the normalizer of
T
in
K.
other than direct computation.
IR16 with the usual basis
(cf... Chevalley [5]); then of course
for
i / j.
C
subspace
--
e1
=
<e
e1 6
e .2
C , the regular elements of C.
C2
...
C
There is an exponential map
*
exp
D
So...
Consider the complexified Clifford algebra
e e. = -e.e.
D
i / j>
in
Unfortunately I know of no
easy way to guarantee the existence of this copy of
generated by
e tc .} .
~
of the -l element of WK so that
(as will be seen below.)
+ a-u }
~7
(-'
) ,,
T,,o, ,
I
p
{X Xf D }f = {f (I,
The real
is a Lie algebra, and
151.
For j = 1,... 8,
K = exp C 2 is Spin (16).
x. = e .
2j-1
J
Then
z3 x 3
+ie
.j y.
2j.
f(e.2
=Jee . - 1-e 2j
2
1
z.j
define
=' 1-(e
j_
.
e2
).
.
2
+ie 2j-l 23
e2 .ej2
e2 +ie
1(-e
2j
.j 2 _e2)
2
2 2j,
=e -1I(2j
j
1 j1
=
)
.
.i e 2j-1
-(e 2 j-
i
)
+ie
.
2j
j-1
- 2 (e
The other identities below are similar.
z~x.
=
= -x.x.
(*) x.x.
1 J
J li
z.z.
1
J
= z.z.
J1
We may take
<z.>
exp
(Xz.)
exp fz
=
-,= -y.j
xiy.l =J -y.x.
J l1
yiy.
= -yjyi
z.x. = x.z.
1J
J 1
ziy. = y.z.
1JJ
l
= cos .1 + 2z.
2z.,
i $
and
z
2l
sin
= i-4;
,
one easily computes
in particular,
exp (2Tz.) = 1.
One knows (as was pointed out to me by Kostant)
that
Y = y... -y8
j.
as the Lie algebra of the Cartan
Since
subgroup T C K.
i
z~y
-x.
-
generates a maximal left ideal in C;
so the unique irreducible C-module is CY.
This has a
152.
basis indexed by subsets
S
of
{l,...8}; X-
(
x)Y..
icS
a = e 1 e 3 e 5 '''
Set
e1 3 e 15.
-1
2
= -z .
a2 = 1, and az a-
Then
4
(It is easy to see that
a
=
X5 has weight (+ , +
By (*),
I exp (7e4 j-3e 4j-1)
j=1
,...
K.)
with respect to t ;
+ )
here the minus signs correspond precisely to elements of S.
XS, where
span of the vectors
X e D, let S
representation the
D
So we may take as a model for the
has even order.
S
If
denote the set where X = -1; of course
Is I = 4 or 0. Then
a
+ a-X S }.
has basis {X5
Exactly
X
X
as for the other exceptional groups, one easily checks
M n T
that
H
2z.,
is
1
contains
H
2z.,
is
1
i/SleSleSi/S~~
[ H
l(l+2z.)][ H
1 .
is
(one choice of sign throughout.)
These four elements are
distinct for each X, giving 32 in all.
H
element
2z
=
8
2 -z
l(l+2z.)]
z 2... z
For X = 1, the
is
=1e2e3e4'''e15e16
i/S
the promised central element of
on
P.
Clearly
a
K
which acts trivially
is also in M, and normalizes T.
The
resulting 64 element group already picks out OT: in fact,
set M
=
<a, 24 z1 z2 z3 4 4z2
2
z2 z5 z6 > C M.
153.
This is a 16 element abelian group, which is in some sense
dual to the group generated by D and a (once we show D C M.)
M
We claim that
01
=
Arguin g as for E 6 , it is enough
.
are the only weights annihito show that the various +
-2
lated by M 0 n T . Notice that , for example, 2 z1 z z 3z
2
4
1
acts on the weight
(egOR2
(F( 2 i) (cei)
1 i)C
e1 .. 4
4
2
(64 i)
=
+ 1;
are
(.
234
61E:2 E:3
if
(2- )
E:
4
.
= + 1)
by
Thus an even number of
i s a weight of -5, it
follows also that an even number of
c5'''
8
are +.
Proceeding along these lines, it is easy to see that
)
(6
or
is a character of
-
details are left to the reader.
K
element of
D;
It follows that any
which conjugates the four generators of
M 0 into M, is necessarily in M'.
D
We want to exhibit
W
K
inside M.
The element of
transposing z. and z. has a representative
i.=
i
J
(1 + e 2 i- 1 e 2 j 1 ) (1 + e 2 ie2 j).
Every non-identity
element of D, acting in the regular representation on
2
'
~
8
= R ,
may be represented as a product of
0
four such transpositions, e.g. ar = CY1 2 Y3 4 a 56 a7 8 * We
L (D) = iO
154.
claim that there is a certain choice of this representative
with
a
such that
9
a ge M.
gJ
Clearly each
a, so it is enough to show that if
a -X
= X
X
a-X
or
X
according as
a..
commutes
g e D,
X(g)
= 1 or -1.
X
For purely formal reasons, these equations hold up to
sign; verifying that they hold exactly is the only reason
for considering the Clifford algebra.
{(i 1 jl),... (i4 1j4 )}
Suppose
sign of the permutation
=
Set
w
{l,...8}; let
(1,...
((i,,,j))
8).
=ee
e
be the
Then by (*)
4
a =
I
(E)
.
w1.
By direct computation,
k=1
)
iJ
a. (x
(**)
ij
ij
i
(ixxjij)=Xi
i
xj i j
w (x(xyyy ) = x y y
iy)
j
w
a..i (x y yj)=-xiy.y
a
,w
.
Yk for k / ij.
commute with xk'
a.. and w..
(x y y. )=-x y y.
commute with aki
{ij}
Fix g / 1, and a character X c D.
and w
if
{k,.
Suppose first that
{l,...8} = {(i1,j1)... (ig1j4 )} with
X(g)
=
1.
Write
g-it
=
j ,
and set a
= a
.
9 11J1
a
(Later we will
4J4
155.
put some conditions on the choice of the (i 9,,j
X(g)
= 1,
X(J ,); set L
X(i,)
e
Then for some fixed
XS
xS
IT( x
=
y. y. )(
= 1[
eX
g
y. y. ).
X ',
a,
. - (x. x. y. y. )][
(x. x. y. y. ) H
9,#L
9 k
1 , 9
H
1,L
a. 3- (y. y. )]
k/L
AL X
X
SX
=1
by
I
,j 9, 'Z/L
LX
Thus a
-1}-
=
+ 1, we have
=
x
Since
X(i
{Xi
=
).)
X
t
z
I
(y. y. )
J9 39
=X S
(**) ;
So the desired equation holds for any representative if
L
= {X(i
pairs (i,
=
,
a =
j , ) is at our disposal, we may assume that
. (i 4 ,jg))
4
H w...
9,l
6
Set
Since the order of the various
-l}.
(1,...8) + (ij)..
then
X(i9 ) = -X (j9 )
X(g) = -1, then
X(g) = 1. If
= +1, we have
is an even permutation;
On the other hand, for some
Y
XS
= E2
y. )
x
9eL
.X
X
z
y. y.
I x
/L
9 9
j 9
9
Using the middle equations of (**) just as in the previous
case, one sees that
ag -X5
= a-X .
SSg
g
So
a
E M.
156.
It is fairly clear that the 32 elements
of MT
and the 8 elements of D generate a group of order
32-2-8
512; so they exhaust M.
The subgroup MT is
normal; a acts on it by the inverse map, and the various
Y by permutation.
With this structural information, and
the various explicit formulae given, it is not hard to
work out the representation theory of M: there are 256
one dimensional representations, each occurring with
.
16
multiplicity .16
one in the (R
0 IR ) representation of K.
By Schur's lemma, (R1 6
is M-irreducible; since
2
16 + 256 = 512, this accounts for all of M. Now
16
(R,
2
16
0 RP)
A
(D S
2
=A
2
0
(ff2 e 1); components of
dimension 120, 135, and 1 (the invariant quadratic form
defining SO(16)).
These components are irreducible under
2
K. To complete the proof, we must show that
A and F2
are M'-irreducible. Recall that g c M' iff gM0 g -1 C N,
where
MO
is generated by
a
and 3 elements of MT. As
for F , it follows immediately that x c T, x 2 6 M => x c M.
We now consider
{x ..y.}
as a basis for the (R1 6 C
representation of K, which we denote by
-1
if
a -y.
g c K, g-x.
= y
,
etc.
gxig
.
Then
.
One knows that
a-xi = yi, a -x.
g
1
x
g-i
.
For the differentiated representation,
157.
z.
-x=
z..y. = iy., z. x. = z.-y. = 0
1
1
J
1J
-ix.,
Set
u..
exp ('rz. + nz.).
J
=
1)1
u..
13
u..
1 c M, it follows that
=
2
A .
definiteness
1J
1
31-(I
square roots of the elements
M'A
easy to see that the
i /
j.
e T, and
Si~nce
u.
e M'.
Consider for
.
lJ
together with various
u.
Using the
for
+2z. ) 6 M, it
is
T-primary decomposition of A2
is just the weight space decomposition.
. .
commutes with
normalizing
a..
1J
MT
is in
the permutation
f
Now every
of
D, then
w
of
ao, so any product of
In particular, whenever
M'.
defines an automorphism
8}
{l, ....
expressed as a product of
a..
1J
is in
M'.
D together with Aut D (D acting by the regular action)
is doubly transitive on {l,.
.
It follows that the
.8}.
M'-irreducible components of A2 are sums of the following
subspaces:
<xiA x , yiA y
V
=
V
= <xiA y,
V3
1
V
y.A x.
y1 + ...
= <x./AY.
-
+ x8
i
>
i
j>
y8
x- A y->.
(The zero weight space
V3 $ V4
splits only as indicated
158.
because Aut D' acts transitively on the non-trivial characters of D; one proves the splitting on the "Fourier
transform side" of D.)
Consider
We claim
of
p =
P s M'; one checks that the inner automorphism
induced by
C
(1+e
)(1+e1el
e) (l+ee
(1+e
e2 * e3' e 3
-
fixes all ei, except
P
-e22 '
.'
14
-*e1 5 , e1 5
-e1 4 '
PM0P
C M.
direct computation shows that
Then a
It also
follows that
P-x
=
P- (e +ie 2 )=+e
for the other
3
xi, y .
1
+y )
x2
2)
2
1 A
Thus
A1 17
a)
P (x A x2 )
b)
P- (x A yi) = - (x +yj)/\ (x2 y2
Because of a)
by b),
=
P-V
similar.
Q.E.D.
2
1A
y1
has non-zero components in V2 and V ,
and similar equations for
is not M'-invariant.
: similarly
Thus
x A.1
yi
in general, V 3
A2 is M'-irreducible; ff
2 is
e
TABLE 5.8
Small K-types
(non-spherical) for simple split groups
Associate
Principal series minimal
K-types
SP (n,R)
Small K-types
U(n)
(l...l1.0... 0)
(0 < C < 1)
SL (2n,R)
Spin
Ar (C2 n)
(2n)
(An)-
(#1's
l < r
=
#-l's)
< n
n + 2n
(A ) (C)
2n
spin
SL(2n+l,R)
Spin
(2n+l)
Ar (C2n+l)
l < r
< n
spin
SO (2n, 2n)
Spin (2n) x Spin
(2n)
1 0 Ar(C2n
1 0
< r < n
Ar (C2n
(An)-(C2n)
0 1
n+
C2n
2n
(An+(C
Spin
0 Apin
1 0 spin
9pin
0 1
spin
0 spin
(c=0)
non-spherical principal
series minimal
Associate small
K-types
SO (2n+1, 2n+l)
K-types
Spin (2n+1) x Spin (2n+1) 1 0 Ar (C2n+)
< n
1 < r
A r M 2n+
spin 0 spin
spin 0 1
1 0 Spin
SO(2n, 2n-1)
Spin (2n) x Spin
(2n-1) 1 0 Ar (C2 n-1)
spin
1 < r < n-l
$pin
0 spin
0 spin
1 0 Spin
spin
SO (2n+1,2n)
Spin (2n+1) x Spin (2n) 1 0 Ar
1 0
spin+ 0
0 1
2n
(An)- (2n
spin 0 spin
1 0 Spin
Spin 0 1
<
<
n +
Spin 0 spin
2n
non-spherical principal
Associate small
series minimal
K-types
K-types
x SU(2)
1 0 2,
1 0 3
F4
SU(2) x SP(3)
2 0 1,
3 0 1, 1 0 6, 2 0 6
E6
SP (4)
C8
82
(8
S M8
7T2
2 (C
SU(2)
SU(8)
)
A2
8
A 2 (C 8 )
S2 8
adj oint
C8
*
(C)
(8
Spin (16)
AA2 (C16
(016
Tr
2
(Cl
~S22(16
x2
A2
8C*
2
8 *
162.
[g, 0 ~
Suppose
Remark
] = sS(2,R).
Let
of
it 0 .
Then
be the
in .the ordering
0
noncompact imaginary root which is
a
y' is principal series minimal iff
2<
it is small iff
< 1;
-l
(This is just the case
< 1.
-1 <
above; we mention it
SP(1,R)
separately only for convenience.)
It is worth noting another technical property of
small K-types.
Recall the Iwasawa decomposition G = KAN.
a e A(
Corresponding to every root
0.
sk(2,R) c-
1
0
0
).
and
{Z a
Let
X
e 770
denote the image of
1
0
The image of
A(F)
Definition 5.9
%), there is a map
10
(0
0)
is
z
= Xa + OX a
is an orthogonal basis of
0
AO'
(Bernstein, Gelfand, and Gelfand [1])
The representation y of K is fine if the eigenvalues of
Z
in y all lie in [-1,1].
(If
G
is linear, they are
necessarily integers.)
Theorem 5.10
7T be a finite dimensional irreducible representation
Let
of
6
(Bernstein, Gelfand, and Gelfand [1]).
G
on a vector space V, and
be the representation of
space
V
y
a fine K-type.
Let
M on the (one dimensional)
Then the multiplicity of
y
in
fK
is
I
163.
precisely the multiplicity of
6
in
y M.
Amazingly enough, we have
Proposition 5.11
Proof.
y is
fine <=>
is small
It is enough to assume
G
is simple.
We
then proceed on a case-by-case basis, using the following
criteria for fineness and Table 5.7.
...
Let
*
be a strongly orthogonal spanning set for
rroot
consisting of noncompact imaginary roots.
that
A
S
the
is
is obtained from
We may assume
and a Cayley transform on
Then it is easy to see that the sk.(2,pR) through
one of the "root s P(2,.R.) 's" for
corresponding
Z
is in the
S
A(rt 0 6
direction in
0 ).
The
'.
There
are several possibilities.
i)
All roots of
SO(n,n), E6 , E 7 , E 8 .)
C
have the same length (SL (n,R),
In this case all
Z
are conjugate
under W, so it is enough to verify the fineness condition
for one of them; say in the
direction in
t0.
Computing
for sk(2,R), one sees that the fineness condition is
2<-p,
-1
<
.>
<
1
whenever
y
is a weight of
y.
Since
all weights lie in the convex hull of the extremal ones,
this is equivalent to
164.
-<
ii)
2<y,>
1.. for each noncompact imaginary root 3.
There are two. root lengths,
both occurring among the K
Similar reasoning leads to precisely
(SO(2n,2n-1),G2.
the same condition.
iii)
There are two root lengths, only one occurring among
the
.
(SO(2n+l,2n), SP(n,R), F4 .) The
S.
are
necessarily long roots (by inspection; or the reader can
provide the "general" proof.) One can find two Si, say
1
6 and 2' such that f(o + .2
is a short compact
It is easy to see that the Lie algebra
imaginary root.
corresponding to these roots and their negatives is
sp(2,R); and it also corresponds to certain
isomorphic to
roots of
control on
In this way one can get explicit
in 9% 0.
00
Z
when
a
is a short root.
Computing for
sp(2,R), and reasoning as before, one gets the criterion
for fineness:
-1 < 2<y,1S> < 1 for every long noncompact
imaginary root; and -1 < <Y-a> < 1 for every short compact
imaginary root.
We leave to the reader the straightforward verification
that the small K-types listed in Table 5.8 are the only
dominant integral weights satisfying the conditions given
above.
Q.E.D.
165.
Corollary 5.12
A finitef dimensional irreducible repre-
sentation' of a split group contains exactly one associate
class of small K-types, each with multiplicity one.
Proof
Combine 5.4, 5.10, and 5.11.
It follows that the
A(6)
form a "strong system of
minimal types" for a split group, in the sense of
Lepowsky ([19]).
The considerations of section 6 give
analogous results for arbitrary semisimple groups; this
will be considered in detail in a later paper.
166.
6.
Existence of the Representations.
It will be convenient in this section to assume that
K is compact.
This makes available the theory of the
discrete series in its most standard form.
Essentially
the same results hold without this assumption, however.
Fix a K-type yp. We assume that Conjecture 4.2 holds
for yp;
thus we are given a 0-invariant parabolic Jr
=1+ r(
with the following properties:
i)
is
and
split;
O(O
i.e.
is
I
-
y =
y2-2p(/nA
according to the decompo-
] ()
then
iii)
=
is
))
recall that
;
y2 - 2 p (P1 n
and if
[f
is a Cartan subalgebra
(y ly)
sition t= f++t=
ii)
),
(and 01).
Write
(6.1)
,
a maximal abelian subalgebra of
, then
of
l()center of
+
if
a small [T, ] n
H
)
type, and
(y,
y+2p -p
=- 1([1,1].
type;
is an associate small
(y
,
2)
is
t
)
is
dominant for j,
dominant with respect to the
imaginary roots in
(
)
167.
(6.1)
iv)
the irreducible Harish-Chandra module X has
If
minimal K-type y
on
X
U()
then the action of
K
factors through -.
MAN be a Langlands decomposition of a
Let P
cuspidal parabolic subgroup corresponding to
tit 0 =
((p)
Cf1
0
T
subgroup of G;
of
T+
Then
.
L(OMA= T A
is a Cartan
C K, and the identity component
0.
has Lie algebra t 0
T+
T+
0
is a compact Cartan
subgroup of M.
Suppose Conjecture 4.2 holds for the K-type y;
Theorem 6.2.
choose P
=
Then there is
MAN as above.
unitary representation
6
a certain tempered
c M, in the "limit of the discrete
series, " such that
i)
ii)
If
v
A, Ind 6
PtG
0 v = T
has minimal K-type y;
1'
and
y
occurs exactly once.
If
v
is the unique subquotient of
y
containing yi, then
{Tv}
y
7
y
is precisely the set
of irreducible quasisimple representations of G
with minimal K-type yi, up to infinitesimal
equivalence.
Proof.
Suppose we can prove i).
formal reasons, which we only sketch.
Then ii) follows for
Recall the map
bE
~-~---~
168.
EP: U
+UU )
2p (4
/
The latter ring may be embedded in
(.)I *
way (Corollary 5.5) so we get a map
the
character of
UA
E2 o(evaluation at v),;
U
(2 :U
One can also construct a map
+U(0).
U(0OZ),
+
on the k-type
V
in a natural
U(OT)
y
so that
is just
and E2 agree on
()
(Spec R means the set of ring
and
We therefore have maps
El
homomorphisms of R into C)
Spec U
Spec U(O()
Spec
()
Spec U
E2
so that the diagram commutes.
By 6.1 (iv),
By simple properties of the map Spec U(7)
-+
Im E2
- Im El.
Spec
(g)
(which is essentially a Harish-Chandra homomorphism) and
a little algebraic geometry, it follows that Im E
=
-m
E2'
which is essentially statement (ii) of the theorem.
So it is enough to prove i).
identity component of M.
Let
G
G
denote the
be some linear form of
G, such that there is a covering map
that
M
Let
G
+
G .
(One knows
has a finite dimensional representation which is
faithful on 03.)
Set
F = w~
(T(M) n exp (i(70 )), a
169.
finite group; then
Put
M4 = M F.
F C M A L n K, and
F
We will produce a representation, 6
the limit of the discrete series of MO
6
dimensional representation
6
centralizes M0 '
of
act by the same scalars on
and a finite
'0
F, so that 6
M
6
0 6
;
and we
0 6.
Ind 6
=
M
and
6
Thus
F.
will be a well-defined representation of
will set
in
M tM.
l.
Choose a strongly orthogonal set
of
~.
noncompact positive imaginary roots which span
want to borrow some arguments from [22].
there that
G
transform, one can define a system
in
T+
(It is assumed
is equal rank; but the results we need can
easily be generalized from that case.)
of
M, so that
T
Via some Cayley
T
of positive roots
++
)
corresponds to A A (
the positive imaginary roots supported on
p. 68).
2 pc
Then
so that
We
()-p(Y)
2ppc)
-
=
(2p -p) |
+ p(T) =
+
Schmid [22],
([22], p. 128) ,
We claim that
=
y1 is dominant with respect to every compact root
a s T.
If
a
is compact as an element of
is immediate from the dominance of
noncompact in
root
6
i
A( ,).
±
.{-
-
q
}
A(o(,),
yp. Suppose
Then ([22].,.,p.
68)
a
there is
this
is
a
which is not strongly orthogonal
170.
to
Since
a.
and
since
a
Since
a F A(17),
a -+
.
i , a + Si
a £
are noncompact,
.
e A(k),
and
a' + S.
are compact.
<Pia.> =- <ya> =-(<pt,a+.> +
y
since
so
- A(7),
+
and
Thus
are positive roots.
1
A(N,A);
are roots in
>
0
is dominant with respect to
Schmid has defined certain invariant eigendistributions
0(P,X)
([22],
respect to
T
which for
[8])
X
strictly dominant with
are discrete series characters.
apply the following lemma to
We will
(All of the statements
MO.
are more or less known by now, although it is difficult
to give a specific reference.)
Lemma 6.3
rk G = rk K, P is a system of
Suppose
positive roots in
A(gt), A E i.t
respect to T, X+p lifts
0(-p,X)
unitary representation
to
K,
p
is dominant with
to character of T,
is dominant with respect to
eigendistribution
0
'
A K().
and p = XA2 pc+P
Then the invariant
is the character of a tempered
6 (A).
In the restriction of
6 (A)
occurs with multiplicity one; and every K-type
which occurs is of' the form
positive roots.
p+Q, where
Q
is
a sum of
171.
Proof.
see [8].
For the proof that
Langlands ([17])
O($,,X)
is tempered,
has shown that a tempered
representation is unitarizable.
(Apparently
is irreducible or zero whenever
X
®($,X)
is
is dominant, but I
have been unable to find a reference for this.
If
linear groups it is asserted in [15].
X
For
is non-
singular, the irreducibility is of course part of
Harish-Chandra's description of the discrete series.)
The other assertions are known (see for example
Schmid [23])
if
A
is non-singular.
We pass to the
general case using a technique of: Zuckerman.
Let
y
be the finite dimensional irreducible
-p
representation of lowest weight -p, with character 0
.
-p
Then the character of 6 (A+p) 0 y_
ee(,
=
+p)*-
Z
v . a weight
is
m. 0(iP,X+p+v.)
of y
here
m.
is the multiplicity of
Each
V.
is
-p+fl.,
where
v.
in
y
(cfi.
[8]).
'{ is a sum of positive roots;
Thus
0($,A+p)-
Since
A
is
_P
=
Z
m.
()($,X+
-dominant, it is easy to see that the only
172.
is
term with the same central character as -($,X)
6(X+p) 0 y
identified with the subspace of
and
Q"
is
y
So every K-type of
has highest weight p+p+Q'-p+Q" =,y+Q;
multiplicity at most one.
y
It remains to show that
fact occurs in the piece of
-p
6 X+p' 0
has
p
and by being slightly more careful, one sees that
central character.
Q'
of course
-p+Q";
are sums of positive roots.
6(X) C 6(X+p) 0 Y_
with a
has highest weight
6(X+p)
know that every K-type of
Every weight of
may be
X+p is non-singular; so we
certain central character.
y+p+Q'.
6 (X)
So
which occurs exactly once.
E($, X),
in
with the right
We do this by producing some cohomology.
(Easier and less interesting arguments exist, but it never
hurts to justify one's thesis title.)
Let
X
be the Harish-Chandra module of
the representation space of
y_ .
Borel subalgebra corresponding to
is dominant with respect to
which occurs is
p+p+Q, with
it follows easily that
Also
p+p
T;
Q
Let
T.
-2
Now
it
6 ,+p,
Y
and
be the
(p+p)+
2 pc
X+2p
since every other K-type
a sum of roots in
F,
is the minimal K-type of X.
X(p+p) = (p+p)+2p -P =.X+p
is; dominant and non-
singular with respect to T; so by Proposition 4.15, p+p is
f(-minimal in X.
By Theorem 3.12,
173.
7
0
,X)'+P
H
: H0(.p
*/)
0 (AR
HR (,X)y+p-2p(fin8)
-
is injective.
-p.
weight
contains the K-type of lowest
Y
On the other hand,
By a theorem of Rao ([.21]'),
(y+p)-p = y; i.e.
the K-type of highest weight
H0 (77 nA, X 0 Y)
is non-zero.
Y
-submodule of
space of weight -p.
X@Y 1 )
Y
Let
be the
generated by all the weight vectors
is a one dimensional
Y/Y
except the lowest one; thus
S-* H0 (k,
X 0 Y contains
By the long exact sequence,
H (/Ir,X)
+ H0
0 (Y/Y 1 ) + ..
Using a filtration of Yl, it is easy to see that the first
a
group is zero, so that
is an isomorphism.
it is not hard to prove directly that
(Actually
H (ftr4, X 0 Y1 )
in this way one can eliminate the use of Rao's theorem.)
So there is a commutative diagram (for convenience we put
Q = AR
*,
and y = i-2p(ni
HR ('(, X 0 Y )y
Now
-*
HR (77, X)Y
HR (7X 0 Y)
+
0
))
0 rqd, X 0 Y)
(Y/Y )
0
0
0 Q
t is an isomorphism and
l
0
0
+p(Y/Y
5is injective;
so
0Q
0;
174.
necessarily
1
fTr
is injective. Thus there is a cocycle
0
R
w c Hom (A r(, x 0 Y)., with values in
(X 0 Y) , which does
not vanish on the cohomology level; and
(p+ 2 p
)=
p-2p(R
Theorem 3.5 that
-p)
p
-
Y(O)
X
-
acts on
the central character of
but
=
w
has weight
It follows from
p.
(X 0 Y)
according to
Q.E.D.
O($,X).
It should be pointed out that if
X
+p
X-2p c-
G($,X)
is not k-dominant, then
is Q-dominant,
0.
For
in this case there is a simple compact root a (simple
for
$ A A(k))
such that
2<a,A-2p +p>
<a,a>
2<a,p >
Now
<a,X> > 0,
$ A A(k), and
=
1.
<a,a>
Schmid ([8])
c
<a,a>
=1
2<a,p>
> 1.
<a a> Then
a
since
a
is simple for
So necessarily
is simple for
have proved that if
X
<a,X> = 0, and
$; and Hecht and
is singular with
respect to a compact root which is simple for
$, then
E(),X) = 0.
Actually, one can get an existence theorem like 6.2
using only the existence of a representation with the
K-decomposition specified in the lemma.
This can be
proved very algebraically, without invoking the theory
175.
of the discrete series (see for example Wallach [24]).
But we will use the full strength of Lemma 6.3. to relate
our results to Langlands' (Corollaryl6.7) .
M
Consider now the representation of
with
e(4p, t ). Lemma 6.3 applies; so we let
character
MO
be the irreducible component containing the
0
60
K-type p.
I0
6
Producing the representation
little more work.
requires a
F
fl(1n
Recall that
is a parabolic in A .
oT nri
+
With obvious notation, therefore,
( nn ) + (b0 k)
=
of
+ (Titik),
(*)
U(A) = U(/7tdlk ) @ U(E A k ) @
Let
V
put
V1
denote the irreducible
and
(70k
A-module
of highest weight p;
V
.
irreducible
k'k
module of highest weight p.
roots of
in
n<
of
V
/t
is of the form
every weight of
where
By Kostant's theorem 3.8,
Q
V
are supported on A
(y ,*)
not in
V
Using (*),
V
Since the
every weight
,
one sees that
is of the form
is a (non-empty) sum of roots in
is the
(y1 ,*) + Q,
A(Onk ).
On
the other hand, the proof of Lemma 2.7 provides an element
x s t+, such that
a (x) is positive, zero, or negative
according as a e A (c),
A(
A(),
or A (I). Hence a weight E
of
s(x) = p(x); so
V
is in
V1
iff
V1
consists
176.
precisely of the weights
(y,,*)
a is a compact positive root of
is. precisely the
Recall that
Now
2p(rtng4)
n
M
V.
If
a(x) > 0; so
, then
V
K-highest weight space of weight y
was a small
)
P.-2p(7n
L f K
L
nK
)) , so Theorem 5.4 applies to p:
(L n K)A
on
V1
decomposes into a Weyl
group (of A in L) orbit in ((L n K)A),
each representation
occurring with multiplicity one. Let 61 be one of the
y
A
(L n K) types occurring. Now one knows that +
A
+
(L n K) = T0 F, and
+
T0
is central in
L = G
0
Hence
.
6 IF is still irreducible.
We must check that
1
0
6 and 6
act by the same
Since this group is central in F
0
1
and 6 . Since
and Mo, it does act by scalars in 6
scalars on
F I MO
F fA M
.
. K n Mo, 0necessarily
n M 0.)
F n mO
maximal abelian in K
the definitions that
6
0
10
and 6 .
So
representation of
We claim that
Now
F
1
6
6
F n MO0
TO0
(for
T+
0
is
But now it is obvious from
acts by
yiFn M0
in both
is a well defined irreducible
M
6
is central in
in V.
type.
is the weight of a one dimensional
module (namely AR (T77r
the action of
occurring in
M
Ind 6
0 6
if
is linear; and it follows
G
is irreducible.
177.
easily that
M
is normal in
must show that if
6.
a c M
in general.
6e M/M'
1
0
not equivalent to 6 .
a-6 00
M
1
6 .
- (60
then
So we
61)
is
In fact a little more is true:
For we may of course choose a representative
K of a, which normalizes
T
positive Weyl chamber for M 0~n K.
and preserves the
Then (cf.
Schmid [22])
a determines a non-identity element of the complex Weyl
group of T0 in MO; since
set of compact roots.
a e M
n K,
o-pc
)
So
a
preserves the
(f , and (since
=PC
is
dominant for $) a-(X|
+ p))
A + p(M).
We have seen that 6 0 has minimal M
K-type y ; so
P0
-6- has minimal MO A K-type
X1
n
-
a(X e
+ p($)
(XJ
+ p($))
2
-
-
pc
2pc
X + p($) - 2p (
So
a6
00
60, and
6
Fix a character
f=
Ind 6 0 v
11 PfG 'P
TrV
K
y
Ind
K'M''+
Kp
l'
is irreducible.
v c A.
It remains to show that
has minimal K-type p.
G = KP ,and
Since
(V)
K
P=
[(60
KM
M)
0
K
M ,we
®
]
Set
have
P0 = M*AN.
178.
Suppose
y c K.
plicity Of
y
By Frobenius reciprocity, the multi-
Kis the dimension of
in
01
(((6
Hom
0 6
y) .
It follows easily from the
discussion preceding the definition of
K. Suppose
SK
contains a weight
(
exactly once in
Then
y
0 61.
By Lemma 6.3,
y
of weights
(( ,*).
containing
[
6
.
type.
L .
61
Recall that
Since
Q
is a sum
Consider the subspace
under
F, and thus a weight
[/,/]A Q
y 2-2p(1f ri)
2p(n a
)+2p(71 k)
type
is a small
is the weight
y 2+2p(T n k)I
is also
22
Definition 5.3 now implies (writing jvj2 = <v,v>)
of a one dimensional A
small.
+ Q, where
the highest weight of some
C2
with
occurs.
=(YY2)
This must contain elements
transforming according to
(El,(2),
occurs
which restricts to
-i
of positive roots supported on
of
p
that
M 0 K-type occurring in
the highest weight of some
6
y
Tr
6
I+2p(nnk) !
(**)
equality holds iff
associated to
module,
+
~
+2 p(A
2~
I2
E -2p(i7tp)
p 2 -2 p (17ng ).
2p(nV) I + 2p(A+(9nk)
=
2
P
2~r4
2
is a small [ ,]O
Of course
2 pc I-
so we have
+2p(A+(1~
-type
12;
179.
y+2pc 2
[y
2 +
1+2pc
c
+2pc~ 2
c2
t
2
> y1+Q+2pcit12 + Ii2+2pc
+
S +2pc
2
12
2
y2 +2pc It-I
I y||.
=
The first inequality is obvious; the second is (**); and
the third is just the fact that
+
the definition of A .)
y=
(E
Suppose equality holds.
Q = 0; and C 2 -2p(7ng )
I);
[$,Q] A A type associated to
y
_
y;
so
<Qy+2p c> > 0
y
(recall
Te
Then
is a small
P2 -2p(r/cD ).
is in fact the minimal K-type.
By 6.1 (ii),
This completes
the proof of Theorem 6.2.
We note for future reference a corollary to the proof.
If
X
is an arbitrary Harish-Chandra module, the k-type y
Ily|
of X is called small in X if
A-types
of X.
of
are the various
i')
is minimal among the
Then the proof shows that the small k-types
(y 1,y2 ), where
dominant, and
y 2 -2p(r(p) It-
associated to
y 2 -2p(I a)
I
(pl'y 2 )
is
is a small [1,e]AA -type
;
one need only check that
these actually occur (in fact with multiplicity one), which
is left to the reader.
that the small
Modulo Conjecture 4.2, it follows
k-types of an arbitrary irreducible X
180.
occur with multiplicity one.
least, and
rr
When
G
is linear, at
is tempered, it seems to be possible to
show (using Corollary 6.7 below) that every constituent
of
'rr contains a small K-type; in particular that every
constituent occurs with multiplicity one (which was proved
by Knapp [12].)
I hope to pursue this matter in a later
paper.
Our next goal is to relate the classification of
Theorem 6.2 to that of Langlands ([17]),
describe.
Langlands assumes that
G
which we now
is linear, but this
is almost certainly unnecessary for the results we will
quote.
Let
P = MAN
be a Langlands decomposition of an
arbitrary parabolic subgroup of G.
Let
6 c M
irreducible tempered representation, and let
arbitrary.
I'
= Ind 6 0 v.
60V
PtG
Put
one finds that if
M'A' = MA,
P'
P
I60v
then
be an
v e A
be
By computing characters,
is any other parabolic with
P'
and I
have equivalent com60v
position series.
Definition 6. 4
V
Re <v,a> > 0; it
Theorem 6'. 5
e A is dominant if for every root a.of A in N,
is
strictly dominant if
(Langlands
[17]).
Suppose
Re <v,a> > 0.
G1
is linear,
181.
P is as above, and v c A is strictly dominant.
is a cyclic module.
such a
j
denote the unique
Then every irreducible quasisimple
irreducible quotient.
is infinitesimally equivalent to
G
representation of
Let
Then
J 60 ; the conjugacy class of (mA, 6®v)
Langlands does not state explicitly that
is unique.
J60V
is
cyclic; the observation that this follows immediately
from the proof of Lemma 3.13 of [17] is due to Milicic.
Suppose
is dominant.
see that there is a parabolic
induction from
P'
to
P
The case Re v = 0 amounts
In general, it is easy to
a unitary character.
v
to
v
P'2
P, so that the
is unitary (say
and v' is strictly dominant. Unitary
Ind (60v) = 6'0v')
PtP'
induction preserves temperedness, so 6' is tempered. As
such it splits into finitely many irreducible constituents
6
...
6r
which are tempered.
Using this construction,
one easily deduces the following corollary of Theorem 6.2.
Corollary 6.6.
Suppose Conjecture 4.2 holds for G.
Let T
be an irreducible representation with minimal K-type p.
Then there is a parabolic P = MAN, a tempered irreducible
representation
6 e M, and a strictly dominant character
v c A, so that
I6
has minimal K-type p, which occurs
with multiplicity one; and if we let JP60V denote the
182.
unique subquotient containing yp, then
~P
simally equivalent to Jv
ir
1is infinite-
Combining this with Theorem 6.5, we obtain
Corollary 6.7.
G
Suppose
Conjecture 4.2 holds for G.
above,
i.e.
with
v
is linear, and that
MAN,
Let P
strictly dominant.
Then
the minimal K-type of Ind 6 0 v
PtG
Proof.
6', and v be as
J 60v = J 60v ;
is cyclic.
There are only finitely many (non-conjugate)
P', 6', v'
so that
character as Id6V.
JP'
~P'
J'tv'=
a 6 '0
'.
I' 'v
has the same central
We claim that for each of these,
This is proved by downward induction
(with respect to the ordering of section 4) on the
minimal K-type y ' of
:t
irreducible subquotient of
contain
p'.
'
v
6''.
I
J 6 'Ov,
,
,
is maximal, every
must (by Corollary 6.6)
y'
occurs with
is irreducible; so
In general, suppose that
'
Then the minimal K-type y' of J6'O'
is greater than p '.
=
I6,0v,
p'
By Corollary 6.6 again,
multiplicity one, so
J6
If
Applying Corollary 6.6 to
we see that
J6 '®O'
is infinitesimally
183.
equivalent to
J,
which is
inductive hypothesis.
just
J ,
V,,
by
But this contradicts the uniqueness
statement of Theorem 6.5, and completes the induction.
Corollary 6.7 is of course full of possibilities.
For example, using the fact that every discrete series
representation is a quotient of some principal series
(see [14]), one can get Casselman's theorem that any
representation is a quotient of some principal series
([3]).
Of course this proof uses Conjecture 4.2, is
confined to linear groups, and is much harder than
Casselman's argument; but one gets more specific information about which principal series to choose, as well
as generalizations.
later paper.
All of this will be pursued in a
QED.
184.
Bibliography
i.
I. N. Bernstein, I. M. Gelfand, and S. I. Gelfand,
"Models of representations of compact Lie groups,"
Funk. Anal. Pril. 9 (1975) No. 4, 61-62.
2.
H. Cartan and S. Eilenberg, Homological Algebra,
Princeton University Press (1956).
3.
W. Casselman, "The differential equations satisfied
by matrix coefficients," manuscript.
4.
W. Casselman and M. S. Osborne, "The 77-cohomology of
representations with an infinitesimal character,"
manuscript.
5.
Chevalley, Theory of Lie Groups, Princeton University
Press (1946).
6.
T. H. Enright,' "Irreducibility of the fundamental series
of a semisimple Lie group," forthcoming manuscript.
7.
T. J. Enright and V. S. Varadarajan, "On an infinitesimal
characterization of the discrete series," Ann. of Math.,
V
to appear.
8.
H. Hecht and W. Schmid, "A proof of Blattner' s conjecture," Inv. Math. 31 (1975), 129-154.
9.
S. Helgason, Differential Geometry and Symmetric Spaces,
Academic Press (1962).
10. G. Hochschild and J. P. Serre, "Cohomology of Lie algebras,"
Ann. of Math. 57 (1953), 591-603.
185.
11.
J. Humphreys, Introduction to Lie Algebras and
Representation Theor
12.
Springer-Verlag (1972).
A. W, Knapp, "Commutativity of intertwining
operators II," Bull. Amer. Math. Soc. 82
(1976),
271-273.
13.
A. W. Knapp and E. M. Stein, "Irreducibility theorems
for the principal series," in Conference on Harmonic
Analysis (Lecture Notes in Mathematics 266),
Springer-Verlag (1972).
14.
A. W. Knapp and N. Wallach, "SzegS kernels associated
with discrete series," mimeographed notes, Institute
for Advanced Study.
15.
A. W. Knapp and G. Zuckerman, "Classification of
irreducible tempered representations of semisimple
Lie groups," manuscript.
16.
B. Kostant, "Lie algebra cohomology and the generalized
Borel-Weil theorem," Ann. of Math. 74
17.
(1961), 329-387.
R. Langlands, "On the classification of irreducible
representations of real algebraic groups," mimeographed
notes, Institute for Advanced Study.
18.
J. Lepowsky, "Algebraic results on representations
of semisimple Lie groups," Trans. Amer. Math. Soc.
176
(1973) , 1-44.
186.
19.
J. Lepowsky, "Representations of semisimple Lie groups
and an enveloping algebra decomposition," thesis,
MIT (1970).
20.
J. Lepowsky and G. W. McCollum, "On the cetermination
of irreducible modules by restriction to a subalgebra,"
Trans. Amer. Math. Soc. 176 (1973), 45-57.
21.
K. R. Parthasarathy, R. R. Rao, and V. S. Varadarajan,
"Representations of complex semisimple Lie groups and
Lie algebras," Ann. of Math. (2) 85 (1967), 383-429.
22.
W. Schmid, "On the characters of the discrete series
(the Hermitian symmetric case)," Inv. Math. 30 (1975),
47-144.
23.
W. Schmid, "Some properties of square-integrable
representations of semisimple Lie groups," Ann. of
Math., to appear.
24.
N. Wallach, "On the Enright, Varadarajan modules: a
construction of the discrete series," Ann. Sci. Ecole
Norm. Sup., to appear.
25.
G. Warner, Harmonic Analysis on Semisimple Lie Groups I,
Springer-Verlag (1972).
187.
Biographical Note
David Vogan was born in Mercer, Pennsylvania on
September 8, 1954.
He attended public schools in
Bellefonte, Pennsylvania.
Beginning in 1971 he attended
the University of Chicago, receiving an A.B. with honors
and an S.M. in mathematics in June, 1974.
Since then he
has been a graduate student at MIT, supported by a
National Science Foundation fellowship.
In September,
1976 he will begin a one year appointment as an instructor
at MIT.