C OMPOSITIO M ATHEMATICA
D AVID A. V OGAN
G REGG J. Z UCKERMAN
Unitary representations with non-zero cohomology
Compositio Mathematica, tome 53, n
o1 (1984), p. 51-90
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51
UNITARY REPRESENTATIONS WITH NON-ZERO COHOMOLOGY
David A.
Vogan,
Jr. * andGregg
J. Zuckerman© 1984 Martinus Nijhoff Publishers, Dordrecht. Printed in The Netherlands.
1. Introduction
Let G be a real connected
semisimple
Lie group with finite center.Suppose (03C0, H03C0)
is an irreducibleunitary representation
ofG,
and(p, F )
is an irreducible finite dimensional
representation
of G. Then one canconsider the continuous
cohomology of
G withcoefficients
in ir ~ p,(see [2]
or[5]).
The zerocohomology
groupH0ct(G, H03C0
0F)
consists of the G-invariant vectorsin A’ ,
0F,
and thehigher
groups are the derived functors ofH°
in anappropriate
category. One of the main reasons for the interest in thiscohomology
is its connection with thetheory
ofautomorphic
forms. Thesimplest
aspect of this connection can be described as follows. Let K c G be a maximal compactsubgroup,
and letr c G be a discrete
subgroup.
Assume that0393 B G
is compact, and that racts
freely
onG/K. (Such subgroups
ralways exist.)
Thenis a compact manifold. The action of G
by right
translation on the Hilbert spaceL2(0393BG) decomposes
into a Hilbert space direct sum of irreducibleunitary representations
ofG,
eachoccurring
with finite multi-plicity :
with m03C0 a
non-negative integer.
Matsushima’s formula([2],
page223)
is* Supported in part by NSF grant MCS-8202127.
The
cohomology
groups on the left are theordinary topological
ones forthe manifold X. The numbers m03C0 are
essentially
dimensions of spaces ofautomorphic
forms for r. In order toapply
this formula(and
refine-ments or
generalizations
ofit)
tocompute
the m03C0, we need to understandHci(G, H03C0).
Theproblem
we consider(and
more or lesssolve)
is there-fore this:
describe,
in as much detail aspossible,
the irreducibleunitary representations ( ’TT, Ye,)
such thatfor some finite dimensional
representation
F of G.To understand the solution to this
problem,
we have to know what itmeans to describe a
unitary representation. (The
reader with any back-ground
in this area should nowskip
to Theorem1.4.)
The bestdescrip-
tion is
usually
a realization: we make G act on a vector bundle over ahomogeneous
space(say),
and consider therepresentation
on an ap-propriate
Hilbert space of sections of the bundle.Unfortunately,
therepresentations
withcohomology rarely
have such realizations.(This
isperhaps
a failure more oftechnology
than of vision.Rawnsley, Schmid,
and Wolf in[21] ]
havesuggested
apossible
realization which is a verysophisticated
version of "sections of a bundle" butthey
have been able toproduce
itonly
inspecial cases.) Instead,
we consider some invariants which anyunitary representation has,
andspecify
whatthey
are in ourrepresentations.
Two invariants are needed: theeigenvalue
of the Casimir operator of therepresentation,
and the restriction of therepresentation
toK.
To describe the Casimir operator, we need to get a
representation
ofthe Lie
algebra
g0 of G out of aunitary representation (03C0, H03C0).
There is adense
subspace H~03C0 ~ H03C0,
the smooth vectors ofH03C0.
If x e g0 andv
~ H~03C0,
then we definethe limit
exists,
andbelongs to H~03C0.
ThereforeThe Casimir operator is a certain element
03C0-(03A9)
in thealgebra generated by
the operators03C0(X);
we will describe it moreexplicitly
in a moment. Itcommutes with all of the operators
03C0(X).
Since(03C0, H03C0)
isirreducible,
this suggests that03C0(03A9)
should be a scalar operator.(Schur’s
lemma doesnot
immediately apply,
sinceH03C0
is infinitedimensional,
and03C0-(03A9)
is definedonly
on a densesubspace.) Nevertheless,
I.E.Segal
has shownthat
The
(real)
constant c03C0 is our firstgeneral
invariant of v. To definev(Q ),
recall the
Killing
formon g0. It is
non-degenerate
since g0 issemisimple.
Fix a basis{Xt}
of g,and let
{XJ}
be the dual basis:Then
It is easy to check that this is
independent
of the choice of basis.If 03C0 is realized in a space of functions on
G/K (which
is a Riemannianmanifold in a natural
way),
then c’1T may beinterpreted
as theeigenvalue
of the
Laplace-Beltrami operator
on that space of functions. NowHodge theory suggests
that it is the zeroeigenspace
of theLaplacian
whichshould be related to
cohomology;
and this turns out to be the case.PROPOSITION 1.2
([2], Proposition IL3.1): Suppose (7r, A’,)
is an irreduci-ble
unitary representation of G,
and(p, F)
is an irreduciblefinite
dimen-sional
representation of
G. Write c03C0,cp for
therespective eigenvalues of
theCasimir operator
( see (1.1)).
ThenHci(G, H03C0,
(9F) =1=
0only if
c03C0 =cP .
Inparticular, H*ct(G,H03C0) ~
0only if
c03C0 = 0.We therefore know the value of this first invariant in a
representation
with non-zero
cohomology.
Next,
recall the maximal compactsubgroup
K of G.Any unitary représentation (y, H03B3) of
Kdecomposes
as a Hilbert space direct sum ofcopies
of the various irreduciblerepresentations
ofK,
which are finite dimensional. If wewrite k
for the set of irreduciblerepresentations
of K,we can write
symbolically
Here
m(03B4, y )
is a cardinalnumber,
themultiplicity of
8 in y. This meansthat
(if
8 actson
(a
Hilbert space directsum),
with theisomorphism respecting
the actionsof K. In
particular,
if(03C0, H03C0)
is an irreducibleunitary representation
ofG, then
here we have written
m(03B4, 03C0)
for themultiplicity
of 8 in the restriction ofv to K. A theorem of Harish-Chandra says that all the cardinal numbers
m(03B4, 03C0)
arefinite;
thatis, they
arenon-negative integers.
The secondgeneral
invariant of 77 which we have in mind is the set ofintegers m(03B4, 03C0).
Inpractice,
oneusually
uses much weaker information. Onemight
knowonly (for
some fixedv)
that aparticular m(03B40, v)
isnon-zero. If v is realized in a space of functions on the
symmetric
spaceG/K,
forexample,
thenm(trivial, 77-)
must be non-zero: thatis,
v mustcontain the trivial
representation
of K.Matsushima’s formula shows that the
representations having
non-zerocohomology
are connected with thecohomology
oflocally symmetric
spaces; so
analogy
with the DeRham theoremsuggests
that such repre- sentations should be realized on the space of sections of the form bundleon
G/K.
This isessentially
correct; and one concludes thatthey
mustcontain certain
particular representations
of K.PROPOSITION 1.3
([2], Proposition Il.3.1): Suppose (03C0, 03C0)
is an irreduci-ble
unitary representation of G,
and( p, F)
is afinite
dimensional represen- tationof
G.Write P for
thecomplexified
tangent spaceof G/K
at theorigin. Suppose HJct ( G, 03C0
0F) =1=
0. Then there is a 8 EK
such that 8occurs in both ’lT and
Propositions
1.2 and 1.3 show how toget
some information about our twogeneral
invariants fromknowing
that thecohomology
of 17 isnon-zero. Our results
imply
that this very weak informationactually
determines the
representation.
THEOREM 1.4:
Suppose ( p, F)
is an irreduciblefinite
dimensional represen- tationof G,
and 8 Ek
occurs inThen there is at most one irreducible
unitary representation (03C0, H03C0) of
Gwith the
following properties:
(a)
c03C0 =cp (notation (1.1))
(b) 03B4
occurs in 03C0.Assume that
(7r, H03C0)
exists. Then thefollowing things
may becomputed explicitly from
8:(1) H*ct(G, H03C0 ~ F ) ( together
with itsHodge
structure,if G/K
is Hermi-tian
symmetric)
(2)
theposition of
03C0 in theLanglands classification of
irreducible repre- sentationsof
G(3)
the characterof 77
on afundamental
Cartansubgroup (4)
themultiplicity of
anyrepresentation of
K in 77.This summarizes
Propositions 6.1, 6.4,
and6.19,
and Theorems 5.5 and6.16. The theorem can be
rephrased
as follows.Suppose
v isunitary
andv 0 F has non-zero
cohomology.
If we can determine asingle K
type of v which lies inHomC(F, 039B’p),
then we can compute all of the otherthings
mentioned.
For
simplicity
ofexposition,
we treat the case of untwisted coefficients( F = Ç)
first. The first step is to exhibit a certain collection of represen- tations of G(Theorem 2.5).
These were first constructedby Parthasarathy
in
[12],
but we need a characterization of them different from the one hegives. Next,
wecompute
thecohomology
of theserepresentations (Theo-
rem
3.3). Finally,
we show(confirming
aconjecture
ofZuckerman)
thatany irreducible
unitary representation
of Ghaving
non-zero continuouscohomology belongs
to our collection(Theorem 4.1).
The method is due toParthasarathy [13].
Our results
rely heavily
on work of Kumaresan in[10] (the
cases À = 0of
Propositions
5.7 and5.16).
For someapplications,
such asvanishing theorems, they
do notimprove
on[10]. (We
havecompleted
theexplicit
calculation of Kumaresan’s
vanishing
theorem in Section8.)
When G iscomplex,
our results deduce to those ofEnright [4],
withessentially
thesame
proof.
WhenG/K
is Hermitiansymmetric
andH03C0
is ahighest weight representation,
our results are those ofParthasarthy [13].
When Gis
SL( n, R), sharper
results than ours have beengiven by Speh
in[15], [16].
There is-one
annoying
gap in our results: we do not know how to prove that all of therepresentations
which we construct are in factunitary;
so we have notactually
classified theunitary representations
with non-zero
cohomology.
This isunimportant
for the obviousapplica-
tions to
automorphic forms;
but it is of greatimportance
in thestudy
ofthe
unitary
dual of G. Somepartial
results aregiven
in Section 6(Propositions
6.3 and6.5).
Finally,
our results have somebearing
on thetheory
of Dirac opera- tors onlocally symmetric
spaces. This is discussed in Section 7.This paper has
deliberately
been written at two levels. The results should be of interest outside ofrepresentation theory,
sothey
have beenformulated in as
elementary
a way aspossible.
Theproofs
are notreally
very
deep,
butthey
are a littleconvoluted;
and it is difficult toimagine
that
they
will appear natural orenlightening
to a non-expert.They
areaccordingly
addressed to a much smaller audience. From Section 5 on, infact, many proofs
are omitted or sketched on thegrounds
that thisaudience could
easily supply
the details.It is a
pleasure
to thank J. Arthur and D. Barbasch forhelpful
discussions.
2. The
représentations A a
Recall that G is a connected real
semisimple
Lie group with finite center.Write g o for the Lie
algebra
ofG,
and g =(g0)C
for itscomplexification.
Analogous
notation is used for other groups. Let K c G be a maximal compactsubgroup, 0
the Cartaninvolution,
andthe
corresponding
Cartandecomposition.
Wewrite (, )
for theKilling
form on g o and its various natural
complexifications, restrictions,
and so on.Let
(03C0, Yê)
be an irreducibleunitary representation
of G on a Hilbertspace P,
and £00 thesubspace
of smooth vectors. Then H~ is a densesubspace
of H invariant underG,
and there is a natural action of g on H~(see
theintroduction).
Definehere (v(K)v)
denotes the linear span of all the vectors of the form’1T ( k ) v,
with k E K.PROPOSITION 2.1
(Harish-Chandra [6]): £K
is stable under the actionsof
K and
g on H~. As a g module,£K is irreducible,
and determines ’1T up tounitary equivalence.
We call
ye K
the Harish-Chandra moduleof
’1T. If x ~ g and v~ HK,
wewill use the module notation x - v instead of
03C0(x)
v.What we will
actually
describe are modulesAn
for g. The maintheorem will assert that if
(03C0, H)
has non-zero continuouscohomology,
then
£K
isisomorphic
tosome A q
as a g module. Because of the last assertion ofProposition 2.1, Aq
then determines v.(The problem
dis-cussed in the introduction is
that, given Aq,
we do not know how to finda
unitary representation (03C0, H)
with4,K
~Aq.)
The firstproblem
is todescribe the parameter q.
Fix an element x E
i f o;
here i =1 .
Since K is compact, the linear transformationad(x)
of g isdiagonalizable,
with realeigenvalues;
andcomplex conjugation interchanges
thepositive
andnegative eigenspaces.
Define
q = sum of
non-negative eigenspaces
ofad( x )
u = sum of
positive eigenspaces
ofad( x )
1 = sum of zero
eigenspaces
ofad( x )
= centralizer of x.(2.2)
Then q is a
parabolic subalgebra
of g, andis a Levi
decomposition. Furthermore,
1 is thecomplexification
oflo = q ~
g o . Since 03B8x = x,q, l,
and u are all invariantunder 0,
soand so on. In
particular,
q n f is aparabolic subalgebra
off,
with Levidecomposition
We call the
subalgebras
q obtained in this way 0-stableparabolic subalge-
bras
of
g.(Since
not everyparabolic subalgebra preserved by 03B8
is of thisform,
theterminology
isunfortunate.)
With notation as in the
preceding paragraph,
choose a Cartansubalge-
bra
to
off0 containing
ix(as
ispossible).
Then t isautomatically
contained in 1 E f.
Let f
c q be anysubspace
stable underad(t).
Thenthere are a subset
( 03B11, ... , 03B1r}
of t *(the
dual oft),
andsubspaces f03B1,
off ,
such that
if y E
t and v~ f03B1 ,
thenWrite
the
weights
or roots of tin f.
Often we assume that0394(f)
is a set withmultiplicities, with 03B1, having multiplicity dim f03B1l.
Then ifwe have
Fix a system
0394 + ( ~ f )
ofpositive
roots in the rootsystem A(
~ f,t).
(Of
course0394(
~f, t)
as defined above includes zero, and so it is notreally
a root system; but we will overlook such abuses ofterminology.)
Then
is a
positive
root system for t in f.Recall that
(given 0394+())
the irreduciblerepresentations
of the com- pact group K may beparametrized by
theirhighest weights,
which areelements of t *. Define
03BC = 03BC
( q ) = representation
of K withhighest weight 2 p ( u ~ p).
(2.4)
Since not every element of t * is the
highest weight
of arepresentation,
the existence of IL is not
quite
obvious. We will postpone thisproblem
until Section
3,
where IL will be exhibited as asubrepresentation
of theexterior
algebra 039B* p.
We can now describe therepresentation Aq.
THEOREM 2.5 : Let q = 1 + u be a 0-stable
parabolic subalgebra of
g(see (2.2)). Then
there is aunique
irreducible moduleAq for
g with thefollowing properties:
(a)
The restrictionof Aq
to f contains the irreduciblerepresentation 03BC(q) ( see (2.4)).
(b)
The centerof
the universalenveloping algebra of
g acts inAq by
thesame scalars as in the trivial
representation of
g.(c) If
therepresentation of f of highest weight
8 occurs inA q
restricted tof,
then 8 must beof
theform
( notation (2.3)),
with ne anon-negative integer.
If 1 c f
(and
in factonly then), Aq
is a discrete seriesrepresentation.
More
generally, A q is
a fundamental seriesrepresentation
if andonly
if[, ] ~ f. If [, ] f,
thenAq
is nottempered;
thatis,
it does not appearin Harish-Chandra’s Plancherel formula for G. If u n p = 0
(for example,
if q =
g),
thenAq
is the trivialrepresentation
of g. The otherAq
are lessfamiliar
representations.
ForSU(n, 1)
andSO(n, 1), they
are all therepresentations having
the same infinitesimal character as the trivialrepresentation; they
occur at theendpoints
of certaincomplementary
series. In Section
6,
asimpler
characterizationof Aq is given (Proposition
6.1,
with 03BB =0),
as well as a formula for itsglobal
character on the fundamental Cartansubgroup (Proposition 6.4)
and otherproperties.
The existence of
Aq
isproved by Parthasarathy
in[12]. Alternatively,
the results of
§6.3
of[19]
show that thecohomologically
induced repre- sentationRsq() (with
S = dim u nf)
satisfies(a) - (c);
soAq
may betaken to be an
appropriate
irreduciblesubquotient
of it.(Actually,
it isnot hard to show that
RSq(C)
isirreducible,
and so coincides withAq.
This construction of
Aq
is due toZuckerman.)
Still another construction is that ofSpeh-Vogan [17]: by
theirresults, X( q, Ç, 03BC(q)) (notation [17],
p.
247)
satisfies(a), (b)
and aweakening
of(c): 0394(u~p)
isreplaced by 0394(u). (This
weaker version of(c)
suffices for ourapplications
in thispaper.)
The
uniqueness of A,
is moredifficult, particularly
because we need toestablish it under a much weaker condition than Theorem
2.5(c).
Choosea
system 0394+()
ofpositive
roots for t in1, containing
thepositive
system0394+(
r r1f )
chosen above. Thenin a system of
positive
roots for t in g.(It
should be admitted that t is not a Cartansubalgebra
of 1 or g ingeneral;
but0394()/{0} and 0394(g)/{0)
can be shown to be root systems, so the discussion of
"positive systems"
is
justified.)
SetWe will be interested in
representations
of fhaving
ahighest weight
ofthe form
with (J E
wi, and Bo
a subset of0394(u ~ p).
We assume that either (J’ =1=1,
or
Bo =1= j).
LEMMA 2.7: No
weight
8 E t * cansatisfy
both(2.5) ( c)
and(2.6).
PROOF:
Suppose (2.6)
holds. Weput B
=B, U AI (l
np);
thenTherefore
(2.6)
may be rewritten asAs is well known
(e.g., [20], 2.5.2.4),
the first term is a sum of roots in- 0394(u ~ f). By [19],
Lemma5.4.5,
each of the next two terms is of the formTherefore
Here
C1 ~ 0394(u ~ f); C2
andC3
are containedin A’(P). Suppose
nowthat 8 also satisfies
(2.5) (c). By inspection,
thisimplies
that alln03B2
arezero, and
Ci
isempty.
Thus 8 =2p ( u ~ p)). Since C1
isempty,
so a = 1. Therefore
(2.6)
becomesThis forces
Bo
to be empty,contrary
to thehypotheses
in(2.6).
0 Another
proof
of Lemma 2.7 can begiven
based on Lemma 4.7. Here is the result weactually
need.PROPOSITION 2.8: Theorem 2.5 remains true with
(c) replaced by (c)’:
Norepresentation of
f whosehighest weight
isof
theform (2.6)
occurs inAq.
PROOF.: Existence.
By
Lemma2.7,
condition( c) implies
condition( c)’;
so the constructions
given
after Theorem 2.5apply.
Uniqueness.
Let X be an irreducible g modulesatisfying
the conditionsof
Proposition
2.8. The calculation in[18],
before(5.3),
shows that therepresentation
p of f isstrongly
u-minimal in X([18],
Definition3.13).
Write 03BCL for the trivial
representation
of 1 ~ f.By
Theorem 3.14 of[18],
I’L occurs in
HR(u , X) ( R
= dim u~ p) exactly
as often as Il occurs in X. Let Y be an irreduciblesubquotient
ofHR(u, X) (as
an 1module) containing
03BCL. Since 03BCL is onedimensional,
the centralizerU()~f
oft r1 f in the
enveloping algebra
of 1 actsby
scalars on the 1n f-type
IL L of Y. Writefor the
corresponding homorphism.
Recall from[18], (3.2)
the homomor-phism
Define
By
Theorem 3.5 of[18], U(g)f
acts on the t type it of Xby
thehomomorphism
cp. Thus X is determinedby
Y. Tocomplete
theunique-
ness
proof,
we will show that Y must be the trivialrepresentation of 1;
this is where we use
hypothesis (b)
on X.Choose a
maximally split
0-stable Cartansubalgebra b o
of10;
writeWrite W =
W(g, b)
for theWeyl
groupof b
in g, and3(9)
for thecenter of
U( g ).
Recall from[9]
the Harish-Chandraisomorphism
If 03BB ~ b *,
thencomposition of X
with evaluation at Àgives
with X. = X s if and
only
if À e W·03B4.Fix an Iwasawa
decomposition
of lo (with
a o asabove),
and let m0 be the centralizer of a o in10 n to.
Choose a
positive
rootsystem 0394+(,b) compatible
with thisdecomposi- tion ;
thatis,
for some choice
of 0394+(m, t+).
DefineLet
WA
denote the "littleWeyl group"
of a0 in10’
The irreduciblerepresentations
of 1 whose restrictions to 1 ~ f contain 03BCL are parame- trizedby WA
orbits in a*;
write v for the parameter of Y. The parameter of the trivialrepresentation
is03C1(0394+(l, b))|03B1;
so what we aretrying
toprove is
Now
3(1)
acts in Yby ~L03B3 (defined
inanalogy
with(2.9)),
withSince Y occurs in
H*(u, X),
the Casselman-Osborne theorem([3])
says that8(g)
acts in Xby
X03BB,, withOn the other
hand, hypothesis (b)
ofProposition
2.8 says that8(g)
actsby
X,. Therefore there is an element w E W such thatSince u is
0-stable, 03C1(u) ~ (t+)*;
so we getfinally
By [20], 2.5.2.4,
the left side isWe now consider the inner
product
of both sides withp(u).
Since03C1(u) ~ (t+)*,
we mayignore
the v and the pA . Sincep ( u )
isorthogonal
to the roots
of b in 1,
it isorthogonal
to 03C1T+ . SoSo each root in the sum must
belong to 0394+();
so w EW(,b),
and wfixes
p(u).
Now(2.120
becomesSince what we want to prove is
(2.11),
we maymodify v by WA.
Assume this has been
done,
in such a way thatwhenever « E
0394(n,b);
thatis,
that v is dominant for the restricted rootsof a in 1. We claim that when w =
1;
this willcomplete
theproof.
If a is asimple
root of t + in m, thenso a is also
simple
inw(0394+(r, b)).
PutIt follows that
W(m, t+)
permutes0394+w.
Letw0 ~ W(, t+)
be theelement
taking 0394+(m, t+) to -0394=(m, t+).
If03B1 ~ 03B4+2,
thenSince « E
0394+w
and this set is stable under wo, the last term ispositive. By (2.14), 03B1 ~0394(a, b);
soObviously
this forces w =1,
as we wished to show.~
The
uniqueness
part of Theorem 2.5 follows fromProposition 2.8, by
Lemma 2.7.
3. The
cohomology
ofA ci
DEFINITION 3.1: Let X be any module for g.
Identify
the exterioralgebra 039B*p
with thequotient
of039B*g by
the idealgenerated by f;
thusHom(A*p, X)
is identified with asubspace
ofHom(A*g, X).
In thisidentification,
the d map of gcohomology
preserves thesubspace Homf(039B*,p, , X),
which therefore becomes acomplex.
Itscohomology
groups are called
H * ( g, f, X),
the relative Liealgebra cohomology
groups.For more about this
definition,
see[2]
or[5].
PROPOSITION 3.2
(see [2]): Suppose (03C0, H)
is an irreducibleunitary representation of
G,and HK
is its Harish-Chandra module.(a) H*ct(G, H) ~ H*(g, f, HK)
(b) H*(g, f , HK) is
zero unless the Casimir operator actsby
zero inHK.
(c) If the
Casimir operator actsby
zero in.)Ié?K,
thenStatements
(b)
and(c)
make sense for any g module inplace of ye K.
Inthat
generality, (b)
is true when the Casimir operator actsby
ascalar;
and
(c)
is false ingeneral (even
for(g, K ) modules).
Because of thisproposition,
we can compute the continuouscohomology
of theunitary representation
whichought
to be attached toAq,
in terms ofAq
alone.The
result,
due toZuckerman,
is this.THEOREM 3.3: Let q = 1 + u be a 0-stable
parabolic subalgebra of
g(see (2.2)), and put R
=dim( u ~ p).
ThenZuckerman’s
original proof
of this theorem was verysimple:
sinceAq
=RSq(¢) ([19],
Definition6.3.1), and R1q(¢)
= 0 for i ~S,
thespectral
sequence of
[19], Corollary
6.3.4collapses
to theisomorphism
we want.However,
a much moreelementary
argument can also begiven.
Since it is of some interest for thelight
it sheds on the structure ofA* p,
we willgive
it here.
Write
recall that R is the dimension of u ~ p . For each i, 1 i r, choose a non-zero
element X,
of u ~ p ofweight 03B2l,
and a non-zeroelement Yl ouf p
of
weight - 03B2l.
Choose also a basis{Z1,...,Zm}
of 1 ~ p,consisting
ofweight
vectors for t. If A and B are subsetsof {1,...,R},
and C is asubset of
{1,..., ml,
putThese elements form a basis of 039B* p
consisting
ofweight
vectors for t.LEMMA 3.4: With notation as
above,
let À E t * be theweight of XA
AYB
AZ..
ThenEquality
holdsif
andonly if
A= (1, .... RI
and B=)1.
PROOF. If a is a root of t in u, then
and if
/3
is a root of t inf,
thenThe lemma is now obvious.
n
LEMMA 3.5:
Suppose
x E039B*p
is aweight vector for
t,of weight 2p( u ~ P)
+
8,
with 8 aweight of 039B*( ~ p).
ThenPROOF: If U E U n f is a
weight
vector ofweight
a, thenad(U)x
hasweight 2p( u ~ p)
+ 8 + a. SinceLemma 3.4 says that
039B* p
has no vectors of thisweight.
PROPOSITION 3.6: Let
(7rL, FL)
be an irreduciblerepresentation of
~f, of highest weight 8, occurring
in039B*( ~ p);
and let(03C0, F)
be the irreduciblerepresentation of t of highest weight
8 +2 p ( u ~ p) ( if
itexists).
Then thereis an
isomorphism
PROOF: Denote
by V
the one dimensional spaceAR( u ~ p);
andby F °
the
subspace
of F annihilatedby
u ~ f. Thenas
representations
of 1 ~ f.By
theCartan-Weyl highest weight theory,
Every weight
ofFL ~
V is of the form203C1(u ~ p)
+8,
with 8 aweight
of039B*( ~ p). By
Lemma3.4,
these occur inA* p only
inside V ~A* ( ~ p);
so
In the
preceding proof,
we were a little careless about thequestion
of theexistence of 7r. There is no
difficulty,
however. As theproof shows, (03C0, F)
can be taken to be the f submodule of039B*p generated by V
times acopy(03C0L, FL)
in039B*(~p).
COROLLARY 3.7:
PROOF: Let 8 be the
highest weight
of arepresentation
of foccurring
inAQ’ By
Theorem2.5(c),
with
n03B2
0.By
Lemma3.4, 8
cannot occur inA* p
unless all né are zero.This proves the first
isomorphism.
The second is aspecial
case ofProposition
3.6.~ To
complete
theproof
of Theorem3.3,
we must show that the differential in thecomplex Homf(039B*p, Aq)
is zero. If we knew thatAn
arose from a
unitary representation,
this would follow fromProposition 3.2(c);
and since this is theonly
case wereally
careabout,
the reader may wish to omit the rest of this section. We will in any casesimply
use thestandard
proof
in theunitary
case.LEMMA 3.8:
Aq
admits anon-degenerate
invariant Hermitianform ~ , >;
that
is, if
v, w EAq,
andX E g o,
thenPROOF: Let Y be the space of all
conjugate-linear,
f-finite functionals onAq:
Make Y into a g0 module
by
and then into a q module
by complexification.
We claim thatY ~ Aq;
wewill prove this
using
the characterization of Theorem 2.5.By general
arguments, Y isirreducible,
andY|f ~ Aq|f.
So(a)
and(c)
of thecharacterization are satisfied. Let
be the
antiautomorphism
determinedby
for
X ~ g0 ,
À eÇ.
LetI ~ 8(g)
be the maximal idealannihilating A,,;
then ~(I)
annihilates Y.By (b)
of Theorem2.5,
1 annihilates the trivialrepresentation
of g. Since thatrepresentation
admits an invariant Hermi-tian
form, ~(I)
= I. So I annihilates Y,establishing (b)
of the characteri- zation.Choose an
isomorphism 41: Aq ~ Y,
and putObviously
this is anon-degenerate
invariant form on Y, linear in v andconjugate
linear in w. The restrictionof ~ , ~
to thef-type IL( q)
must stillbe
non-degenerate;
so if v, is ahighest weight
vector of03BC(q),
then~v0, v0~ ~
0.Modifying 1/; by
ascalar,
we mayassume ~
vo,va)
= 1. Nowwe claim
that (, )
is Hermitian. To seethis,
defineWe want to show
that 4, == .
SinceAq
are Y areirreducible, and
is ag-module
map, we havefor some À E
Ç.
Butso 03BB = 1.
A f-invariant form on an irreducible
representation
of K isnecessarily definite;
so we may assumethat (, )
ispositive
definite on thef-type IL
of
Aq.
TheKilling
formgives
a naturalpositive
definite Hermitian formon
039B*p;
so we get a natural Hermitian form onSince
only IL
contributes to this Hom(Corollary 3.7),
this form ispositive
definite. That d is zero is now
proved just
as in[2], Proposition
11.3.1.Theorem 3.3 then follows from
Corollary
3.7.4.
Unitary representations
withcohomology
THEOREM 4.1: Let ir be an irreducible
unitary representation of G,
and Xthe Harish-Chandra module
of
7r.Suppose
thatHci( G, 7r)
=1= 0. Then there isa 0-stable
parabolic subalgebra
q = 1 + uof
g( cf . (2.2))
such that X ~An (cf.
Theorem2.5).
The main
ingredient
in theproof
is thefollowing
result ofParthasarathy.
Recall from Section 2 our Cartansubalgebra
t cf,
and thepositive
rootsystem à
+( f ).
LEMMA 4.2:
( Parthasarathy’s
Dirac operatorinequality - [2],
LemmaIL6.11,
and[13], (2.26)).
Let 03C0 be an irreducibleunitary representation of G,
and X the Harish-Chandra moduleof
7r. Fix arepresentation of
foccurring
inX, of highest weight
X E t* ;
and apositive
rootsystem 03B4+ ( g ) for
t in g. WriteFix an element w E
WK
such thatw(~ - Pn)
is dominantfor 0394+().
Let codenote the
eigenvalue of
the Casimir operatorof
g in X. ThenRoughly speaking,
this says that thelength
of thehighest weight
of arepresentation
of foccurring
in aunitary representation
must be at leastthe
eigenvalue
of the Casimir operator. Forexample,
in aunitary spherical
seriesrepresentation,
theeigenvalue
of the Casimir isnon-posi- tive,
since the trivialrepresentation
of f occurs.Here is a sketch of the
proof
of Theorem 4.1.By Proposition 3.2,
Xcontains some
f-type
fromA*p,
ofhighest weight
IL; and the Casimiroperator has
eigenvalue
zero in X. Kumaresan shows in[10]
that anyf-type
ofA* p satisfying
theinequality
of Lemma 4.2 with co =0,
must be of the form03BC(q) (cf. (2.4))
for some 0-stableparabolic subalgebra
q 9 g.To show that
X ~ Aq ,
we use the criterion ofProposition
2.8:roughly speaking,
the badf-types
of(2.6)
do notsatisfy
theinequality
of Lemma4.2. To carry this out, we need some calculations with roots and
weights,
drawn
largely
from[10].
LEMMA 4.3:
Suppose
y, 8 E t * are dominantintegral for 0394+(f),
anda E
Wx .
ChooseQ’,
a" EWK
so that a y - 8 is dominantfor Q’( 0 + (f)),
andy - 8 is
dominantfor 03C3"(
0394+(f )).
ThenEquality
holdsif
andonly if
03C303B3 - 8 EWA. - ( y - 03B4).
PROOF: The "if"
part
isobvious;
we prove theinequality
and"only
if"by
induction on thelength
of a, as in[10],
Lemma 2.2. If a =1,
there isnothing
to prove; so suppose a >1,
and the result is known for shorterWeyl
group elements. Choose a reflection Sa( a ~ 0394+(f))
so thatChoose T E
Wx
so thats03B103C303B3 - 03B4
is dominant for03C4(0394+(f)). By induction,
if suffices to show that
with
equality only
ifay - 8
isconjugate
tos03B103C303B3 - 03B4. By [9],
Lemma13.4(C)
andProposition 21.3,
it isenough
to show thats03B103B3 - 03B4
is aweight
of the finite dimensional
representation F
off,
of extremalweight
a y - 8.Now
(4.4) implies
that(since
y isdominant);
andsince 8 is dominant. So the
a-string
ofweights
of Fthrough 03C303B3 2013 03B4
iswhich contains
LEMMA 4.5
( Kumaresan ’s lemma):
Under thehypotheses of
Lemma 4.2,assume that the Casimir operator
of
g actsby
zero in X. Assume that thereare elements cy, T E
WK,
and a subset A c à +( g),
such that(a)
Qx -p" is
dominantfor 03C4(0394+(f))
(b)
ax - p"+ Tp,
= p -2 p ( A ) ( notation after (2.3)).
Then there is a 0-stable
parabolic subalgebra
q = f + uof
g, such that~=203C1(u~p).
PROOF: Extend t to a Cartan
subalgebra b
of g. Let W be theWeyl
groupof b
in g. The elementsof 0394+(g)
are the restriction to t of apositive
rootsystem 0394+( g, f) for b
in g; and thecorresponding
half sum of roots isjust
p,extended by
zero on theorthogonal complement
of t inb.
Choosea subset A
~ 0394+(g, b)
whose restriction to t is A. Then p -203C1(A) ~ b*
isa
weight
of the finite dimensionalrepresentation
of g ofhighest weight
p;so
with
equality only
iffor some w E W. On the other
hand,
so
Equality
holds if andonly
if there is a w asabove,
and wp is zero on theorthogonal complement
of tin b.
This is the same asrequiring
that wcommute with 0
([18],
Lemma5.5).
In that case,03C9(0394+(g, b))|t is
anotherpositive
system for the restricted roots0394(g, t);
we write it asw(0394+(g)).
We may also
regard w
asacting
on talone,
since it commutes with 0.Thus
with
equality
if andonly
ifNow choose
03C4’ ~ WK
so that~ - 03C1n is
dominant for03C4’(0394+(f)). By
Lemma