C OMPOSITIO M ATHEMATICA
A. W. K NAPP
Commutativity of intertwining operators for semisimple groups
Compositio Mathematica, tome 46, n
o1 (1982), p. 33-84
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33
COMMUTATIVITY OF INTERTWINING OPERATORS FOR SEMISIMPLE GROUPS
A.W.
Knapp*
© 1982 Martinus Nijhoff Publishers
Printed in the Netherlands
To Susan
In an earlier paper
[13],
E.M. Stein and the authordeveloped
atheory
ofintertwining operators
forrepresentations
of real reductivegroups G induced from
parabolic subgroups.
Inpart
of that paper, we dealtspecifically
with themajor unitary representations
that con-tribute to the Plancherel formula for G and determined their irre-
ducibility/reducibility
in terms of a finite group known as the R group.The
present
paper makes a detailedstudy
of the R group, conclud-ing
with a structure theorem for thecommuting algebra
for each of thesemajor unitary representations.
The theorem shows inparticular
that each such
representation splits
into the direct sum of in-equivalent
irreduciblerepresentations. However,
our results will be obtainedonly
for a more limited class of groups G than are thesubject
of[13].
The limited class includes all linear connectedsemisimple
groups. Each group G in the limited class is assumed to have a faithful matrixrepresentation
and to have some other proper- ties that restrict the disconnectedness ofG ; precise
axioms aregiven
in
§ 1.
Anexample
due toVogan [16],
discussed furtherbelow,
shows that some such limitation is necessary.We introduce a minimum of notation needed to
give
aprecise
statement of the main theorem. Let G be a linear reductive group
satisfying
the axioms of§1,
let K be a maximalcompact subgroup,
and let MAN be the
Langlands decomposition
of aparabolic
sub-group of G such that M has a discrete series. To each discrete series
* Supported by a grant from the National Science Foundation. AMS(MOS) subject classifications (1980). Primary 22E45, 20G20.
0010-437X/8204033-52$00.20/0
representation e
of M andeach imaginary-valued
linear functional Aon the Lie
algebra
ofA,
we associate the inducedrepresentation
of Ggiven by
Let
W03BE,039B
be thesubgroup
of theWeyl
group of A of elements that fix A and the class ofe.
To each element w ofW03BE,039B, §8
of[13]
associatesa
unitary self-intertwining operator 03BE(w)A(w, 03BE, A)
forU(03BE, 039B, .).
These
operators multiply according
to the group law ofW03BE,039B, except
for scalar factors. LetR03BE,039B
be thesubgroup
ofW03BE,039B
defined in§6
belowor
in § 13
of[13].
Then theoperators
associatedjust
with the elements ofR03BE,039B
form a linear basis of thecommuting algebra
ofU(03BE, 039B, ), by
Theorem 13.4 of
[13].
MAIN THEOREM : The group
Re,A
is the direct sumof
a number rof copies of Z2
with r --5 dimA,
and theoperators 03BE(w)A(w, 03BE, 039B)
asso-ciated to the elements
of ReA
commute with one another. Inparticular,
the
commuting algebra of U(e, A, .)
is commutative and its dimension is a powerof
two ;therefore, U (03BE, A, .) decomposes
into the direct sumof inequivalent
irreduciblerepresentations.
This theorem has two
parts - the
f act that R= 03A3 Z2 (given
below asTheorem
6.1)
and thecommutativity
of theoperators (given
below asTheorem
7.1). Vogan’s example
shows that the second conclusion is not automatic from thefirst,
even for linear groups: Let G be the semidirectproduct
ofSL(2, R) ~ SL(2, R) by
theeight-element
qua- ternion groupHg (of
standard basis elements of the realquaternions, together
with theirnegatives),
where ioperates
on the firstSL(2, R) by conjugation by (A -) and j operates
on the secondSL(2, R) by conjugation by (i 0-1.
For the minimalparabolic subgroup,
M isZ2 EDZ2 ~ Hs,
and we use A = 0and e
= sgnQ9
sgnQ9 0’
with 0’ anirreducible two-dimensional
representation
ofHg.
ThenR03BE,039B = Z2 0 Z2,
but the
commuting algebra
ofU(03BE, 039B, ·)
isisomorphic
with a full2-by-2
matrixalgebra.
The trouble encountered in
Vogan’s example
is that the discon-nectedness of G is too wild.
In §1
we introduce axioms for ahereditary
class of linear reductive groups G whose disconnectedness is more limited. Akey property
of such groups, not sharedby Vogan’s example,
isgiven
in Lemma 4.4 and is usedcritically
inLemma 7.6.
The main
difficulty
inproving
that R = 1Z2
is inunderstanding
thesubgroup
of theWeyl
group that fixese.
In§§2-3
we make thenecessary detailed
analysis
of how theWeyl
group of A acts on acompact
Cartansubalgebra
of the Liealgebra
ofM ;
thisanalysis
is ofindependent
interest and centers about Theorem 3.7. Theproof
thatR
= 03A3 Z2 begins
in earnest in§4.
Once one knows that R
= 03A3 Z2,
it is not too difficult to deduce thecommutativity
of theoperators.
Theproof
that wegive
here ofTheorem 7.1
departs
from the announcedproof ([9]
and[10])
and isshorter than the announced
proof.
It uses theconcept
of’superor- thogonality
of roots’ introducedby Gregg
Zuckerman and the authorin
joint
work.In the case that MAN is a minimal
parabolic subgroup, parts
of thepaper
simplify considerably:
The results of§2
reduce to easy facts in[11], part
of§4
is notneeded,
and§5
is almostcompletely
un-necessary.
The results of this paper were announced in
[10].
In the case thatMAN is minimal
parabolic, they
had been obtainedearlier,
ahd brief sketches ofproofs
had beengiven
in[8]
and[9].
The press of other matters hasdelayed publication
ofcomplete proofs
until now.§1. Assumptions
on GThe groups G in this paper are real Lie groups
of
matricessatisfy- ing
thefollowing
axioms:(i)
Theidentity component Go
of G has a reductive Liealgebra
g.(ii) Go
hascompact
center.(iii)
G hasfinitely
manycomponents.
(iv)
IfG’
denotes theanalytic
group of matrices with Liealgebra
the
complexification gC
and ifZ(G)
denotes the centralizer of G in the totalgeneral
linear group ofmatrices,
then G cG" Z(G).
From
(iv)
it follows that eachAd(g),
for g inG,
is inAd(G’);
thislatter statement is the 4th axiom in
[13].
Thus thepresent
axioms area
specialization
of those in[13],
which in turn are aspecialization
ofthose of Harish-Chandra in
[4].
All finite groupssatisfy
the axioms of[13],
whereas theonly
finite groups thatsatisfy (iv)
above are theabelian ones.
Since G satisfies the
axioms
of[4]
or[13],
all of the basic notationof
[13]
and groupdecompositions
of[4]
make sense. The Cartandecomposition g
= tEB
p is defined relative to a Cartaninvolution 0,
and
mp (6 ap (D np
andMpApNp
refer to minimalparabolics
con-structed in the standard way from the Cartan
decomposition.
If0
denotes a maximal abelian
subspace
of mp, thenap EB bo
is a Cartansubalgebra
of g. Roots relative to this Cartansubalgebra
are real onap
+1 $o.
Thea p -roots (roots
relative toa p )
are the restrictions toa p
of such roots. If a is anap-root,
the root space for awith g
is denotedga. We
assumegiven
anAd(G)-invariant, 0-invariant, nondegenerate, symmetric
bilinear form B on g x g such thatBe(X, Y) = -B(X, 8Y)
ispositive definite,
and from B we can construct in the standard wayan inner
product (-, .)
on the dual space(ap
+ib0)’.
As a consequence of the axioms of
[4]
or[13],
we have G =GoMp, i.e.,
everycomponent
of G meets thecompact
groupMp.
Furtherproperties
and notation aregiven
in[13].
LEMMA 1.1:
If
z is inZG(GO),
the centralizerof Go
inG,
then z is inZG,
the centerof
G.PROOF: Let g be in G and write g
= x03B6, by (iv),
with x inGC and 03B6
in
Z(G).
Then z commutes with x sinceAd(z)
=1,
and z commuteswith e since e
is inZ(G).
Hence z commutes with g.Define
The group F is
compact
andabelian,
often finite.LEMMA 1.2:
Mp
=(Mp)oF,
where(Mp)o
is theidentity component of Mp.
PROOF: Let m in
Mp
begiven.
Thenb0
andAd(m)b0
are twomaximal abelian
subspaces
of map and areconjugate by Ad((Mp)0).
Adjusting
rnby
a member of(Mp)o
andchanging notation,
we may assume thatAd(m)b0
=b0.
Since theWeyl
group of(Mp)o
is transitiveon the
Weyl chambers,
we may, afterintroducing
anordering,
assumeAd(m)
preserves the set ofpositive
roots of(Mp)o.
NowAd(m)
leaves
ap pointwise fixed,
and we may assume that(ap
+ib0)’
isordered with
ap
before(ib0)’.
ThenAd(m)
normalizesap + ib0
andpreserves the set of
positive
roots. Write m = gzby (iv)
with g inG’,
z in
Z(G).
ThenAd(g)
=Ad(m)
normalizesap
+ib0
and preserves theset of
positive
roots, so that g = exp H with H in(ap
+ib0)C.
Since mis in
Mp, Ad(g)
has to beunitary
ongC,
and thus ad H hasonly imaginary eigenvalues. Consequently
H=Hi+H2
withHl
inb0
andH2
ini ap.
Here expHl
is in(Mp)o.
Thus m, after a leftmultiplication by
a member of(Mp)o,
may be assumed to be of the form(exp H2)z
withH2
ini aP
and z inZ(G).
This m is inMp, clearly
commutes withMp,
and is exhibited as inZ(G)
expi ap ;
hence it is in F. The lemma follows.Let MAN be a
parabolic subgroup containing
a minimalparabolic subgroup
built fromMpAp.
The Liealgebra
is written tnE9 a (9 n,
andwe have M ~
Mp
and a Cap.
M satisfies the axioms of[13].
We writeap
= aE9
aM,orthogonal
sum. Then aMplays
the same role for M thatap
does forG,
and it is shownin §1
of[ 13]
that theMp
group for M isthe same as the
Mp
group for G.LEMMA 1.3: M
satisfies
thepresent axioms,
and M =MoF,
whereMo is
theidentity component of
M.REMARK: F is central in
Mp
but often not central in M. This circumstance isresponsible
for certaincomplications
in§4
below.PROOF: Since M is linear and satisfies the axioms of
[13], only
axiom(iv)
is at issue. Since M satisfies the axioms of[13]
and itsMp
group is the same as for
G,
we have M= MoMp.
Then Lemma 1.2gives
M =
MoMp
=M0(Mp)0F
=MoF.
Now
Mo
is contained inMC,
and F satisfiesTherefore M satisfies axiom
(iv).
Roots relative to a are restrictions to a of
ap-roots.
The root space for an a-root e is denoted 9E. IfHE
is the member of a dual to e, thenis a reductive Lie
algebra,
and one can show that the groupGI"I =
MG&E)
satisfies the axioms. An a-root E is called reduced if cE is notan a-root for 0 c 1.
We define
Weyl
groupsW(ap), W(a),
andW(aM)
as the obviousnormalizers-in-K-divided-by-centralizers.
Members ofW(a) always
have extensions to members of
W(ap);
cf. Lemma 8 of[11].
Now suppose, as will henceforth be the case in this paper, that m
has a
compact
Cartansubalgebra (which
can be assumed to be in f flm).
We shall construct acompact
Cartansubalgebra
6 of m withthe
property
that F normalizes 6(see Proposition 4.5).
First weremark that a + û will have to be a Cartan
subalgebra
of g, and roots relative to it will be real on a + ib. Since any two Cartansubalgebras
of
q’
areconjugate
viaG’,
we will be able to find a member c ofAd(Gc)
withIt is this map c, the
Cayley transform,
that we construct.By
Lemma 4 of
[11],
we can find anorthogonal system 51, ..., 5n
of rootsof
(mC, (aM + ib0)C)
that vanish onb0
and spanaM ;
such asystem
may be constructed so as to bestrongly orthogonal (no 5¡:t Si
is aroot).
For each
Si
fix a root vectorXj
in m so that[Xj, 03B8Xj] = -2|03B4j|-2H03B4j.
SetCj = exp
03C0i 4(Xj - OXi).
The various Cj commute, and then c =II;
cj has therequired properties.
Note that c leaves a andb0 pointwise
fixed.§2.
Odd and even roots of aForm the
decomposition
intoMpApNp
of a minimalparabolic
sub-group of
G, compatibility
with the Cartandecomposition
of g. Fix aparabolic subgroup
MANcontaining MpApNp,
and let notations be asin §1.
We shall assume that MAN iscuspidal, i.e.,
that m has acompact
Cartansubalgebra.
Let b be thecompact
Cartansubalgebra
of m constructed in
§ 1,
and letc: ap + ib0 ~ a + ib
be theCayley
transform.
We have
a p
= aE9
aM, and a dominates aM in theordering.
If a isan
ap-root,
we shall often write a = aR + a, as thedecomposition
intothe
projections
on a’ anda’
As in§3
of[11],
thehypothesis
thatMAN is
cuspidal implies
that ci = aR - aI is also anap-root.
Follow-ing [11],
we say that a is auseful ap-root
if2(a, 03B1~/|03B1|2 ~
+ 1.Whenever
Proposition
lOb of[11] applies,
this notiondepends only
on aR
(i.e.,
it holds for all a, or else fornone);
thus we canspeak unambiguously
ofuseful
a-roots. In theremaining
cases, we say aR isuseful
if it is the restriction to a of some usefulap-root.
Proposition
lObapplies except when g
has a factor that is someform of the
exceptional
groupG2.
A groupG2
can arise as acomplex
group, with minimal
parabolics
as theonly cuspidal
ones, or as anR-split
group, with allparabolics cuspidal.
All of these cases have allap-roots
usefulexcept
for the two maximalparabolic subgroups
in theR-split G2,
in which case a is built from one root and can be examineddirectly.
When a is built from thelong
root and aR is the reduced a-root, then thepositive
a-roots are aR and2aR,
both of which areuseful; however,
there are two not usefulap-roots
that restrict to aR.When a is built from the short root and aR is the reduced a-root, then the
positive
a-roots are aR and2aR
and3aR,
with aR and2aR
useful and3aR
notuseful;
theonly
not usefulpositive ap-roots
are thoserestricting
to3aR.
Proposition
10c of[11]
saysthat, apart
fromG2,
if aR isuseful,
so is cap for every c ~ 0. Moreover theonly possible positive multiples
of a reduced a-root that can be an a-root are
{1, 2}, according
to theCorollary
toProposition
12. These statements remain valid forG2,
with theexception
of the two cases noted above.By Proposition
12 of[11],
the useful a-roots form a rootsystem
A in asubspace
of a, and the Main Theorem of[11]
says that the groupW(a)
isjust
theWeyl
group of A. We shall use these resultsstarting
in
§3.
Butfirst,
we relateusefulness, multiplicities,
andproperties
ofroots relative to a + ib.
If a = aR + aI is the
decomposition
of anap-root according
toa
EB
aM, recall à is defined as aR - aI- It is shown in theproof
ofLemma 9 of
[11]
that thisconjugation
isimplemented by
a member ofW(ap). Consequently
if a is anap-root, ga
and g03B1 have the same dimension.LEMMA 2.1: The
following
conditions on an a-root aR areequivalent :
(i)
Themultiplicity of
aR as an a-root is odd.(ii)
Themultiplicity of
aR as anap-root (when
extendedby
0 onaM) is
odd.(iii) aR is
a rootof a
+ ib when extendedby
0 on ib.PROOF: The root space for aR as an a-root is the sum
and the remarks above show that the sum
03A303B11>0(-)
on theright
iseven-dimensional. Hence
(i)
and(ii)
areequivalent. According
to theremarks after Lemma 3 of
[11], (ii)
holds if andonly
if aR is a root ofap
+1b0
when extendedby
0 on1 $o. Applying
theCayley
transform c, we see that(ii)
and(iii)
areequivalent.
We shall call an a-root aR odd or even
according
as the dimensionof 03A3c>0 ac03B1R
is odd oreven.’
LEMMA 2.2: The
following
three conditions on an a-root aR areequivalent : (i)
aR is odd.(ii)
Somemultiple of
aR is a rootof
a + ib when extendedby
0 oni6.
(iii)
The reductive Liealgebra g(aR)
has acompact
Cartan subal-gebra.
PROOF: If
(i) holds,
then dimgc003B1R
is odd for some c0 >0,
and then Lemma 2.1applied
to COAR shows that(ii)
holds.Conversely
if(ii) holds,
then COAR is a root for some c0 > 0.Moreover,
caR is not a root forpositive
c other than co, since a nontrivialmultiple
of a rootcannot be a root.
Applying
Lemma 2.1 to caR for each c, we obtain(i).
For theequivalence
of(ii)
and(iii),
it is well known thatg(03B1R)
has acompact
Cartansubalgebra
if andonly
if there exists a root of a + ibthat takes on
only
realvalues, i.e.,
vanishes on b. Such a root must bea
multiple
of gR, and theequivalence
follows.LEMMA 2.3:
If
aR ± al arenot-useful
rootsof a p,
then aR + al + 03B3 is not a rootof ap
+1$o for
any03B3 ~
0 in(ib0)’. Moreover, 2aI
is a rootof ap
+ib0
but2a,
+y is
not a rootof ap
+1$o for
any03B3 ~
0 in(ib0)’.
PROOF: First suppose that the roots in
question
are not in anyG2
factor. If aR + aI + y is a root,then aR
+03B1I|2
=|03B3|2 by
Lemma 2 of[11],
and03B1R + 03B1I + 03B3 and
aR - al - y are both roots because con-jugation
relative to a carries roots to roots. Then’ The nomenclature inessential and essential was used in [11] but will be abandoned
now since it is misleading.
with the second
equality holding
since aR + aI is notuseful;
hencewhich is not an
integer,
contradiction. Thus aR + a, + y cannot be aroot of
a p
+ib0.
Since aR + a, is a root of
ap,
theonly remaining possibility
is thataR + ai itself is a root of
ap
+ib0. Similarly
aR - ai is a root, and theirdifférence
2a,
must be a root ofa p
+ib0.
If also2a,
+ y is a root, then the sumis a root, contradiction. Thus
2a,
+ y is not a root ofap
+1$o.
If the roots in
question
are in aG2 factor,
theG2
must besplit
overR. No roots in the factor then have a
component
in(ib0)’.
If aR ± alare not
useful,
then2a,
is anap-root,
hence a root ofap
+ib0,
and therest of the lemma is vacuous.
LEMMA 2.4: Let
03B1R ± 03B2
be rootsof
a + i b with03B1R ~ 0, 03B2 ~ 0,
and2aR
not a rootof a
+ ib. Then aR is auseful
a-rootif
andonly 2j3
isnot a
root of
a + ib.PROOF: For both directions of the
proof,
form the roots ofap
+1$o given by
If aR is not useful as an a-root, then aR ± a, are not useful as
ap-roots.
Hence Lemma 2.3 shows that y = 0 and that
2ci,
is a root ofap
+ib0.
Thus
203B2
=c(2ai)
is a root of a + ib.Conversely
suppose that aR is useful and that20
is indeed a root.In the
expression
a, + y =c-1(/3),
we cannot have ai =0, by
Lemma 1 of[11].
Since 03B1R + 03B1I has to be useful(even
inG2,
under ourhypotheses),
theonly possibility
is thatBy assumption
the rootstring 120,
aR +03B2}
does not extend to2aR,
and thereforeJaR
+03B2|2 ~ |203B2|2,
fromwhich
it follows that|03B1R|2 ~ 3|03B2|2.
Applying (2.1),
we obtainBy
Lemma 1 of[11] applied
to2(03B1I
+y ), 4 y
is not a root ofa p
+ib0,
and thus
Inequality (2.2)
shows that theright
side is>0,
and we have a contradiction.LEMMA 2.5:
If 03B1R ± 03B2
are rootsof
a + ib with aRuseful
and with03B1R ~ 0, 03B2 ~ 0,
and2aR
not a rootof
a +ib,
then~03B1R + 03B2, 03B1R - 03B2~ = 0
and
consequently |03B1R|2
=|03B2|2.
PROOF: The différence
20
of aR+ 03B2
and aR -j3
is not a root of a + ibby
Lemma2.4,
and the sum2aR
is not a rootby assumption.
Hence aR +
j3
and aR -j3
areorthogonal. Since aR
+03B2|2
=1 aR 03B2|2,
itfollows that
|03B1R|2
=|03B2|2.
LEMMA 2.6:
If
aR± 03B2
are rootsof
a + i6 with 03B1R ~ 0 andif 2aR
isan a-root, then
2aR
is a rootof
a + ib when extendedby
0by
ib.PROOF: Assume on the
contrary
that2aR
is not a root of a + ib.Proposition
10c of[11], together
with an examination of the variousG2
cases, shows that aR isuseful,
and hence we conclude fromLemma 2.5 that
IURI2
=Ip 12.
Choose03B3 ~
0 in(ib)’
so that2aR
+ y is aroot of a + ib. Since
4aR
is not an a-root(corollary
toProposition
12of
[11]),
Lemma 2.5gives |203B1R|2 = |03B3|2.
Form the innerproduct
and without loss of
generality
choose thesign of 0
so thatBy
the Schwarzinequality
and thus
|~03B2, 03B3~|
~2|03B1R|2,
withequality
if andonly
if03B2
= cy. From(2.4),
we then haveand we can conclude from
(2.3)
thatis a root of a + ib.
Again
we canapply
Lemma2.5,
with the result that|03B1R|2
=I_y _ p 12
or else y =0.
In the former case,whence
2~03B2, 03B3~ = l03B3l2= 4BaRB2
andequality
holds in the Schwarz in-equality (2.5).
Thus03B2
= cy in both cases.If c ~
1,
then|03B3|2
=2~03B2, y)
and03B2
= cy says that c =1/2
and y =203B2.
That
is,
aR+ 03B2
and2(aR
+0)
are roots of a +ib,
in contradiction to the f act that twice a root is not a root.If c =
1,
then aR must be a root of a + ibby (2.6);
thus03B2
is a root, and203B1R = (203B1R + 03B2) - 03B2
is a root, contradiction. Thiscompletes
theproof.
LEMMA 2.7: Let aR be a reduced a-root.
(a) If
aR and2aR
are both a-roots, then aR has evenmultiplicity
as ana-root,
2aR
has oddmultiplicity
as an a-root,2aR
is a rootof
a + ib when extended
by
0 onib,
aR and2aR
are bothuseful,
andaR is
odd.If
also3aR
is an a-root, it has evenmultiplicity
and isnot
useful.
(b) If aR is
odd and2aR is
not an a-root, then aR isuseful and,
whenextended
by
0 onib,
is a rootof
a + ib.(c) If
aR is notuseful,
then aR is even.(d) W(a)
carries odd roots to odd roots and even roots to even roots.PROOF: In
(a),
Lemma 2.6 shows that2aR
is a root of a + ib. Since twice a root is not a root, aR cannot be a root. The conclusions aboutmultiplicities
of aR and2aR
then follow from Lemma 2.1. If3aR
is notan a-root, then the rest of
(a)
followsimmediately.
If3aR
is an a-root, the roots inquestion
lie in asplit G2
factor with a built from a short root, and(a)
followsby looking
at this casedirectly.
For
(b),
Lemma 2.1 says aR extends to be a root of a+ i6,
and thisextension exhibits aR as useful. Conclusion
(c)
follows from(a)
and(b). Finally
in(d),
if wrepresents
a member ofW(a),
thenAd(w)
carries the root space for aR to the root space for waR, and the conclusion follows.LEMMA 2.8:
If
aR andaR
arenonorthogonal useful
rootsof
thesame
length,
then aR andaR
are either both even or both odd.PROOF: We may assume that aR and
aR
arelinearly independent.
Then the
hypotheses imply
that PaRPakaR =aR,
with the indicated root reflectionsexisting
inW(a)
since aR andaR
are useful. The result therefore follows from Lemma 2.7d.§3.
Action ofW(a)
oncompact
Cartansubalgebra
of iriWe continue with the notation of
§2.
In order to understand the action ofW(a)
on discrete series ofM,
we shall first introduce in Theorem 3.7 an action ofW(a)
on thecompact
Cartansubalgebra
b ofm. For this purpose let us recall the main theorem of
[11] - that W(a)
isexactly
theWeyl
group of thesystem A
of useful roots of a.Ultimately
we shalldecompose W(a)
into a semidirectproduct,
inorder to
analyze
the actionon b,
and the semidirectproduct
decom-position
and action willdepend
upon choices oforderings.
Thus wesuppose that a’ and
(ib)’
have been orderedlexicographically
in somefashion. Let
ne
=f those simple
roots of A that areeven}
We
=subgroup
ofW(a)(= W(0394)) generated by
reflections in the members ofne.
Notice that
simplicity
has been defined relative to the set A ofuseful
roots of a, not to the set of all roots of a. The
key
result behind theaction of
W(a)
on 6 is thef ollowing imbedding
theorem.PROPOSITION 3.1: It is
possible
to chooseJ3
in(i b)’ corresponding
toeach aR in
IIe
so that aR +(3
is a rootof a
+ib,
so that thereflection
P03B2 preserves the setof positive
rootsof (m,.b),
and so that the linear extensionof
themapping given by 03B1R ~ J(03B1R) = J3
is anisometry of aé
into
(ib)’.
Some
explanation
isappropriate. Suppose q
iscomplex semisimple
and the
parabolic subgroup
is minimal. Then trtp =i ap,
and the actionof
W(ap)
on mp isjust
the action onap transported
to mp via i. On the otherhand, if g
is realsplit
and if theparabolic subgroup
isminimal,
then mp = 0 and
W(ap)
actstrivially.
Thegeneral
case behaves like amixture of these two, with a distinction made
according
as whetherroots are even or odd. Reflections in even roots are
analogous
tothose in the
complex
case, and reflections in odd roots areanalogous
to those in real
split
groups. Tocapture
thisaction,
we first imbed theeven
simple
roots into(ib)’ by
ageneralization
of themultiplication- by-i
map of thecomplex semisimple
case.LEMMA 3.2:
(a)
Let aR be an odd a-root. Then there is arepresentative
w inKo of
thereflection
paR on a such that w is in theanalytic subgroup corresponding
toq(aR)
and such thatAd(w)
is theidentity
on m.(b)
Let aR be an evenuseful
a-root, and let aR ±13
be rootsof
a + iûrestricting
to aR. Then there is arepresentative
w inKo of
thereflection
PaR on a such that w is in the
analytic subgroup corresponding
tog(03B1R)
and such that
Ad(w)
is -1 onIRHf3
and is + 1 on theorthocomple-
ment
of Hf3 in
ib.PROOF:
(a)
Since aR isodd,
Lemma 2.2 shows thatq(03B1R)
has acompact
Cartansubalgebra. By
Lemma 4 of[11],
there exists w, inthe
analytic subgroup
with Liealgebra g(03B1R)
~ such thatAd(wl)
is 1on
g(03B1R) ~ l
and -1 onq(aR)
~ p.Applying
Lemma 4 of[ 11]
to m, we obtain W2 in theanalytic subgroup
with Liealgebra
m ~ l such thatAd(w2)
is 1 on m n t and -1 onm n p.
Then w = w1w2 has therequired properties.
(b)
Let a = aR +j3,
and letXa
be a root vector ingC.
With con-jugation
defined relative to g,Xa is
then a root vector for â = aR -13,
since roots areimaginary
on b. Thecomplexification
of the Cartan involution 0 is 1 on i b and - 1 on a, and henceThus
aXa
is a root vector for - a. NowB(X, - 03B8X)
is > 0 for all X ~ 0 ingC
and inparticular
forXa. Multiply Xa by
a constant so thatB(X03B1, - 03B8X03B1)
=2/|03B1|2,
set H =2Ial-2Ha,
and setX-03B1 = -03B8X03B1. Then
and it follows from the
theory
ofSL(2, C)
thatrepresents
theWeyl
group reflection for a in(a
+ib)’. Similarly
represents
theWeyl
group reflection for â. The roots a and a arestrongly orthogonal by
Lemmas 2.5 and2.4,
from which it follows thatConsequently w+w-
is exhibited as in the(real) analytic subgroup corresponding
toq(03B1R). Evidently Ad(w’w-)
is -1 onCH03B1 + CH03B1
and is + 1 on theorthocomplement
in(a
+ib)C.
HenceAd(w+w-)
is -1 onRH03B1R + RH03B2
and is + 1 on theorthocomplement
in a + ib. Thus w =w+w-
has therequired properties.
LEMMA 3.3: Let a, and a2 be distinct members
of ne
such that03B11 + 03B12 is
an a-root. Then2(03B11 + 03B12) is
not an a-root.PROOF: An
exceptional G2
factor has no even useful roots. Thuswe may
disregard
these cases in the lemma. If2(a, + a2)
is an a-root, then al + a2 and2(ai
+a2)
are useful(hence
inA) by Proposition
10cof
[11].
Without loss ofgenerality
letla112 ~ |03B12|2
and form the a,-string
in Athrough 2(a,
+a2).
Thensince ai and a2 are distinct and
simple
within A and since|03B11|2 ~ la212.
Therefore
is a member of
A,
and a2 must be oddby
Lemma2.7a,
contradiction.LEMMA 3.4: Let a
= 03B11 + 03B21
and a’ =a2+02
be rootsof a + ib
suchthat 03B1
1 and
a2 areuseful
even a-roots, and suppose 03B1 + a’ is a rootof
a + ib but
03B11 + 03B12 is
notuseful.
Then|03B11| = |03B12|, = + 1,
and 03B1 isorthogonal
to a’.PROOF: Without loss of
generality,
let|03B12| ~ |03B11|.
Since the restric-tion to a of
is not
useful,
Lemma 2.4 shows that2(p,
+03B22)
is a root of a + ib.Moreover,
Expanding
theright
side andusing
theequality la B2
=2|03B11|2 given by
Lemma
2.5,
we obtainThat
is,
Since a, and a2 are in à and
|03B12| ~ lall,
theright
side is aninteger.
Theleft side is
3/2
of the ratio of thelength squared
oftwô
roots. Thusboth sides ’of
(3.2)
are 3:Let
la212
=nia 112
with 1 ~ n :5 4. Then(3.4) gives 2~03B11, a2)/la 112
= 2 - n,which is
impossible
for n = 3 if a2 islong
and a, is short. If n =4,
then a2
= -2a,;
hence a, + a2 = -a,is useful, contrary
tohypothesis.
Suppose
n= 2,
so that|03B12|2
=21a 112. By
Lemma2.5, la ’12
=21a 12.
Thus
(3.1)
and(3.3)
lead to the conclusion thatwhich contradicts the fact that the sum of a short root and a
long
rootis
necessarily
short.Thus n =
1,
so that|03B12|2
=Ball2
and2~03B11, (2)/lalI2
= + 1.By
Lemma2.5, la ’12
=la 12 .
Thus(3.1)
and(3.3)
lead to the conclusion thatExpanding
the leftside,
we seeimmediately
that(a, a’)
= 0.LEMMA 3.5: Let
03B11 + 03B21
and03B12 + 03B22
be rootsof
a + ib that restricton n to distinct a, and a2 in
ne. If
a, 1 and a2 areorthogonal,
then soare
03B11 + 03B21
anda2:t 132, If
a, 1 and a2 are notorthogonal,
thenexactly
one
of (a,
+01,
a2 +03B22)
and(ail + 03B21, 03B12- 03B22~
is 0. Thefirst
one is 0if
and
only if ~03B21, 03B22~
is>0,
and the second one is 0if
andonly if
~03B21, 03B22~ is
0.PROOF: First suppose a, and a2 are
orthogonal.
Since a, and a2 aresimple
forA,
al - a2 is not a useful a-root. But nor can 03B11 2013 a2 be an n-root that is notuseful, by
Lemma 3.4. Hence(a,
+131) - (a2 ± 132)
arenot roots. Also a, - a2 not in 0 and a,
orthogonal
to a2imply
03B11 + 03B12 is not inA,
and a, + a2 cannot be an a-root that is notuseful, again by
Lemma 3.4. Hence
(a,
+131)
+(a2
±03B22)
are not roots. Thus ci +131
isorthogonal
to a2 ±132.
Now suppose that ai and a2 are not
orthogonal.
Sincethey
aresimple
forA,
we must have(al, a2)
0. Then one of the two innerproducts ~03B11 + 03B21, 03B12 ± 03B22)
is ~ 0.Say ~03B11 + 03B21, 03B12 + 03B22~ ~ 0.
We shallshow that this inner
product
is 0. In thecontrary
case,(03B11 + 03B21) - (03B12 + 03B22)
is a root and theorthogonality
conclusion of Lemma 3.4 shows that a, - a2 isuseful, contradicting
the fact that a, 1 and a2 aresimple
in thesystem
0 of useful roots of a. Thus(a,
+03B21,
a2 +132)
is0.
It follows that the sum
is a root of a + ib.
By
Lemma3.3, 2(a,
+a2)
is not an a-root, and we know 03B11+03B12 is useful since~03B11, 03B12) 0
and a, and a2 are in A. Weshall show