C OMPOSITIO M ATHEMATICA
G. V AN D IJK
Smooth and admissible representations of p-adic unipotent groups
Compositio Mathematica, tome 37, n
o1 (1978), p. 77-101
<http://www.numdam.org/item?id=CM_1978__37_1_77_0>
© Foundation Compositio Mathematica, 1978, tous droits réservés.
L’accès aux archives de la revue « Compositio Mathematica » (http:
//http://www.compositio.nl/) implique l’accord avec les conditions géné- rales d’utilisation (http://www.numdam.org/conditions). Toute utilisation commerciale ou impression systématique est constitutive d’une infrac- tion pénale. Toute copie ou impression de ce fichier doit contenir la présente mention de copyright.
Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
77
SMOOTH AND ADMISSIBLE REPRESENTATIONS OF P-ADIC UNIPOTENT GROUPS
G. van
Dijk
Sijthoff & Noordhoff International Publishers-Alphen aan den Rijn Printed in the Netherlands
§ 1.
IntroductionA
representation
1T of atotally
disconnected group G on acomplex
vector space V is said to be smooth if for each v E V the
mapping
is
locally
constant. 1T is called admissible if in addition thefollowing
condition is satisfied: For any open
subgroup K
ofG,
the space of vectors v E V left fixedby 7r(K)
is finite-dimensional. An admissiblerepresentation
is said to bepre-unitary
if V carries a1T (G )-invariant
scalar
product.
These
representations play
animportant
role in the harmonicanalysis
on reductivep-adic
groups[6].
The aim of this paper is toemphasize
theirimportance
in harmonicanalysis
onunipotent p-adic
groups. Let ,f2 be a
p-adic
field of characteristic zero. G will denote aconnected
unipotent algebraic
group, defined over f2 and G itssubgroup
of J1-rationalpoints.
Let W be the Liealgebra
of G and W itssubalgebra
ofQ-points.
G is atotally
disconnected group. We show:(i)
any irreducible smoothrepresentation
of G isadmissible, (ii)
any irreducible admissiblerepresentation
of G ispre-unitary.
Jacquet [7]
has shown that(i)
holds for reductivep-adic
groups G.Actually,
we make use of a remarkable lemma from[7].
The main tool for theproof
of(i)
and(ii)
is the interference of so-called super-cuspidal representations,
which are known toplay
a decisive role inthe
representation theory
of reductive groups[6].
Weapply
someresults of Casselman
concerning
theserepresentations [3],
whichoriginally
wereonly
stated forGL(2).
For theproof,
which isby
induction on dim
G,
one has to go to the three-dimensionalp-adic Heisenberg
group. A new version of von Neumann’s theorem([11],
Ch.
2)
is needed tocomplete
the induction. All this is to be found in sections2, 3, 4
and 5.Section 6 is concerned with the Kirillov construction of irreducible
unitary representations
ofG,
which is standard now. In the nextsection we discuss the character
formula, following Pukanszky [12].
As a
byproduct
we obtain ahomogeneity
property for the dis-tribution,
definedby
a G-orbit 0 in W’: if dim 0 =2m,
thenfor all
t E a,
t:,-é 0. Similar results are true fornilpotent
orbits ofreductive G in W
[2];
therethey
form a substantialhelp
inproving
that the formai
degrees
ofsupercuspidal representations
areintegers, provided
Haar measures aresuitably
normalized. Let Z denote thecenter of G.
Section 8 deals with
square-integrable representations
mod Z of G.Moore and Wolf
[10]
have discussed them for realunipotent
groups.The main results still hold for
p-adic
groups.Let 1T be an irreducible
square-integrable representation
mod Z ofG. For any open compact
subgroup
K ofG,
letm (1T, 1)
denote themultiplicity
of the trivialrepresentation
of K in the restriction of 1T to K. Normalize Haar measures on G and Z in such a way thatvol(K)
=vol(K nZ)
= 1. Choose Haar measure onG/Z accordingly.
Then, according
to ageneral
theorem([5],
Theorem2)
one has:where
d (1T)
is the formaldegree
of 1T.Now assume in addition K to be a lattice
subgroup
of G : L =log
K isa lattice in G.
Moreover,
letm (ir, 1) > 0.
Then we haveequality:
This is
proved
in section 9.In section 10 we relate our results to earlier work of C.C. Moore
[9]
on these
multiplicities, involving
numbers of K-orbits. We conclude with anexample
in section 11.§2.
Smoothrepresentations
We call a Hausdorff space X a
totally
disconnected(t.d.)
space if it satisfies thefollowing
condition: Given apoint
x E X and aneigh-
borhood U of x in
X,
there exists an open and compact subset to of X such that x OE w C U.Clearly
a t.d. space islocally
compact.Let X be a t.d. space and S a set. A
mapping f:
X - S is said to besmooth if it is
locally
constant. Let V be acomplex
vector space. We writeC°°(X, V)
for the space of all smooth functionsf :
X-> V andC’ (X, V)
for thesubspace
of thosef
which have compact support. If V = C wesimply
writeCOO(X)
andC’ (X) respectively.
One canidentify C’ (X, V)
withC’ (X) (&
Vby
means of themapping
i :
C-,(X) Q9
V --->C’ (X, V)
defined as follows: Iff
EC’ (X)
and v EV,
theni ( f Q9 v )
is the f unctio n x- f(x)v (x E X )
from X to V.Let G be a t.d. group, i.e. a
topological
group whoseunderlying
space is a t.d. space. It is known that G has
arbitrarily
small open compactsubgroups. By
arepresentation
of G onV,
we mean a map1T:
G --> End(V)
such that7r(l)=
1 and1T(XY) = 1T(X)1T(Y) (x,
y EG).
Avector v E V is called 1T-smooth if the
mapping x H ->r (x ) v
of G into V is smooth.Let V 00 be the
subspace
of all 1T-smooth vectors. Then V 00 isn(G)-stable.
Let 1T 00 denote the restriction of 7r on V 00. 1T is said to bea smooth
representation
if V = Voe. Of course n00 isalways
smooth.We call a smooth
representation
1T on V irreducible if V has nonon-trivial
ir(G)-invariant subspaces.
Let 1T be a
representation
of G on thecomplex
vector space V. 7r is called admissible if(i)
7r issmooth,
(ii)
for any opensubgroup
K ofG,
the space of vectors v E V which are left fixedby ir(K),
is finite-dimensional.An admissible
representation
7r of G on V is calledpre-unitary
if Vcarries a
7r(G)-invariant
scalarproduct.
Let Y be thecompletion
of Vwith respect to the norm, defined
by
the scalarproduct.
Then 7rextends to a continuous
unitary representation
p of G on Y such that V = Y. and 7r = poc. It is well-known that 7r is irreducible if andonly
if p is
topologically
irreducible. Note that V is dense in H.Let 1T be a smooth
representation
of G on V and V’ the(algebraic)
dual of V. Then the dual
representation
7r’ of G on V’ isgiven by
Put
V
=(V’).
and if =(1Tf)oo. Then if
is a smoothrepresentation
whichis called
contragredient
to 7r. It iseasily
checked that 1T is admissible if andonly
if if is.Let H be a closed
subgroup
of G and a a smoothrepresentation
ofH on W. Then we define a smooth
representation 1T = ind H tG u
asfollows: Let V denote the space of all smooth functions
f :
G ---> W such that(2) Supp f
iscompact
mod H.Then 1T is the
representation
of G on Vgiven by
Let 1T¡, lr2 be two smooth
representations
of G onVI
andV2 respectively.
We say that 7ri isequivalent
to 7r2 if there is a linearbijection
T :VI ---> V2
such that1T2(x)T
=T1T1(X)
for all x E G.§3.
Smooth and admissiblereprésentations
of the three-dimensionalp-adic Heisenberg
groupLet a be a
p-adic field,
i.e. alocally
compact non-discrete field witha discrete valuation. There is an absolute value
on il, denoted 1-1,
which we assume to be normalized in the
following
way. Let dx be anadditive Haar measure on a. Then
d(ax) = lai dx (a
Ef2*).
Let 6 bethe
ring
ofintegers:
6 =lx
El f2:lx 1::; 1}; (9
is a localring
withunique
maximal ideal
P, given by
P ={je
E J1 :Ixl 1}.
The residue-class fieldCIP
hasfinitely
many, say q, elements. P is aprincipal
ideal with generator w. So P =mû, lwl
=q-1.
Put P n = ’ü1ney(n
EZ).
Since P n is a compact
subgroup
of the additive group of f2 and a =Un P n,
any additive character of f2 isunitary.
Let G =H3
be the 3-dimensionalHeisenberg
group over f?:G is a t.d. group. The group
multiplication
isgiven by:
THEOREM 1:
(1)
Each irreducible smoothrepresentation
7Tof H3
isadmissible; (2)
Each irreducible admissiblerepresentation
1Tof H3
ispre-unitary.
We make use of the
following
result ofJacquet [7].
LEMMA 1: Let H be a group and p an
(algebraically)
irreduciblerepresentation of
H on acomplex
vector space Vof
at mostdenumerable dimensions. Then every operator A which commutes with
p(H) is a
scalar.Let V be the
space of
7T. Let v EV,
v # 0 and K= {g E G: 1T(g)V
=v}.
Then K is open and
G/K
is denumerable. Since V =spanl7r(g)v: g
EGIK},
the lemmaapplies.
Z =1[0, 0, z] :
z Ea}
is thecenter of G.
Therefore,
there exists an additive(unitary)
character«P7T
of D such that1T([0, 0, z]) = t/!7T(z)1 (z
Ef2),
where I is theidentity
inEnd(V).
We have two cases:(a) «P7T
= 1. Then 7ractually
is arepresentation
ofGIZ = a2
whichis
(again by
thelemma)
one-dimensionaland,
as observedabove, unitary.
(b) «P7T#
1. Fix w EV,
w # 0. For any v EV, put cv(g) = (1T(g)V, w) (g
EG).
Themapping
v H c, is a linearinjection
of V into the space of smooth functionsf
onG, satisfying
Let K be a
(small)
open compactsubgroup
of G such thatif(k)w
= wfor all k E K. Call
VK
=Iv
E V :ir(k)v
= v for all k EK}.
Thenf
= cvsatisfies
for all v E
VK.
Write g
=[x,
y,0], k
=[x’, y’, 0].
ThenHence
Therefore for all
x, y (E f2
andx’, y’
small
(only depending
onK,
not on theparticular
choice of v EVK).
Moreover:
for
x’, y’
as above. Sincet/17T =1- 1, f ([x,
y,0])
= 0 for x or ylarge enough (only depending on K,
not on theparticular
choice of v EVK).
Sincef ([x,
y,z]) = t/17T(Z )f([x,
y,0]), f
iscompletely
determinedby
the valuesf([x,y,01), (x,yEf2). Consequently, dim VK = dim(cv : v OE VK) OJ.
Part
(1)
of the theorem is now evident. To prove part(2)
it suffices to take thefollowing
scalarproduct
on V:REMARK: It is clear that the same observations remain true for the
higher
dimensionalp-adic Heisenberg
groups.§4. Supercuspidal representations
G is a t.d. group and 7r a smooth
representation
of G on V.By
a matrix coefficient of 7r, we mean a function on G of the formwhere v and 15 are fixed elements in V and
V respectively.
Let Zdenote the center of G. We call 1T a
supercuspidal representation
ifeach matrix coefficient of 1T has compact support modulo Z. The
proof
of Theorem 1emphasizes
thesignificance
of this kind ofrepresentations. Actually,
one has thefollowing
lemma.LEMMA 2: Let 1T be a smooth
representation of H3
such that71’([0,0, zl) = «/J’TT(z)I (z
Ef2) for
some non-trivial additive character«/J’TT of
J1. Then 1T is asupercuspidal representation.
Assume,
from now on, G tosatisfy
the second axiom ofcountability.
Let 1T be an irreducible smooth
representation
of G on V. Thenby
Lemma
1,
there is a characterAn,
of Z such that1T(Z) = À’TT(z)I (z
EZ).
LEMMA 3: Let 71’ be an
irreducible,
admissible andsupercuspidal
representation of
G on V. Assume À’TTunitary.
Then 7r ispre-unitary
and one has the
following orthogonality
relations: There exists apositive
constantd7T (the formal degree of 1T), only depending
on thechoice
of
Haar measuredg
onG/Z
such thatf or
allTo make 1T
pre-unitary,
choose anyw E V,
w # 0 and define thefollowing
G-invariant scalarproduct
on V:1T extends to an irreducible
unitary representation
on thecompletion
H of V such that H00. = V. The
orthogonality
relations now followeasily
from those for irreducibleunitary supercuspidal
represen-tations
([5],
Theorem1).
The
following
theorem is due to Casselman([3],
Theorem1.6).
THEOREM 2: Let
p be
anirreducible,
admissible andsupercuspidal representation of
G on W such thatp(z) = À (z)I (z
EZ),
where À is aunitary
characterof
Z. Let 1T be any smoothrepresentation of
G on Vsuch that
zr(z) = À (z)I (z
EZ).
Given aG-morphism f:54-’
0from
1T to p, there exists aG-morphism splitting f.
PROOF: Let
S"(G)
denote the space of smooth functions h on G with compactsupport
mod Z such thath (xz) = h (x)À (z-’) (x
EG,
z E
Z). S,(G)
isa G-module,
Gacting by
left translation. FixWo E W,
WO #
0. Themapping
F: W --->S, (G),
definedby
is a
G-morphism.
ChooseWo E W
andvo Cz V
such that(wo, wo) = dp, f(vo)
= wo.By
P we denote theG-morphism
fromSA(G)
to Vgiven by
Then pop is the
G-morphism, splitting f:
Hence
f o
p 0F(w)
= w for all w E W.Let us now turn back to
H3.
The irreducibleunitary representations
of
H3
are well-known(cf. [ 11 ]).
Their restrictions to the space of smooth vectors are admissible.Keeping
in mind Theorem1,
we have therefore thefollowing
list of irreducible admissiblerepresentations
of
H3.
Let xo denote any non-trivial additive character of a. Then:(a)
One-dimensionalrepresentations
PJL,V(IL, v E a),
trivial onZ;
PJL,V
([X,
y,Z]) = x0(>X
+vy).
(b) Supercuspidal representations
pA(À
EJ1 *),
non-trivial onZ,
on the spaceC;(a);
We have the
following analogue
of the famous theorem of von Neumann forH3 ([11],
Ch.2).
THEOREM 3: Let 1T be a smooth
representation of H3
such thatir([O, 0, z])
=Xo(Àz)I (z e f2) for
some À # 0. Then 7r is the(algebraic)
direct sum
of
irreduciblerepresentations equivalent
to p,.PROOF: Let V be the space of ir. Due to Theorem
1,
every irreduciblesubrepresentation
of 1T isequivalent
to pA.By
Lemma2,
n is asupercuspidal representation.
We shall prove thefollowing:
Givenany G-invariant
subspace W
ofV,
W #V,
there exists an irreduciblesubspace
U of V such thatU n W = (0).
An easyapplication
ofZorn’s Lemma then
yields
the theorem.Let W be a proper G-invariant
subspace
of V. Put V =V/
W. V is aG-module;
the action of G is a smooth andsupercuspidal
represen- tation of G. Let uo EV,
Xo # 0. The G-moduleVo generated by
uo contains a maximal proper G-module. ThereforeVo
has an irreduciblequotient,
which is alsosupercuspidal,
and admissibleby
Theorem 1.By
Theorem2, Vo
and henceV,
even has an irreduciblesubspace,
sayVI,
on which G acts as anadmissible, supercuspidal representation.
Let
V
+ W be itspre-image
in V. ThenVi
+ W is a G-invariantsubspace
of V and the canonical map from V to V induces anon-zero
G-morphism
fromVI +
W toVI. Again
Theorem 2implies
the existence of an irreducible
subspace
U of V such that U n W =(0),
U + W =VI
+ W. This concludes theproof
of Theorem 3.§5.
Smooth and admissiblereprésentations
ofunipotent p-adic
groups Let f2 be ap-adic
field of characteristic zero.By
G we mean aconnected
algebraic
group, defined overf2, consisting
ofunipotent elements,
with Liealgebra
19. LetG, %
be the sets offi-points
ofG, W respectively.
We have thef2-isomorphism
ofalgebraic
varietiesexp: % - G,
whichmap Cf}
onto G. Let’log’
denote its inverse. We shall call G aunipotent p-adic
group and say that (9 is its Liealgebra.
Let Z be the center of
G,
its Liealgebra
#. One hasexp
= Z.More
generally:
theexponential
of asubalgebra
of 19 is aunipotent p-adic subgroup
ofG,
theexponential
of an ideal in W is a normalsubgroup
of G.Let G be a
unipotent p-adic
group.THEOREM 4: Each irreducible smooth
representation
1Tof
G isadmissible and
pre-unitary.
PROOF: We use induction on dim G. Lemma 1 is the main source to prove the theorem in case dim G = 1. Assume dim G > 1. Fix any non-trivial character yo of f2.
By
Lemma 1 there exists a(unitary)
character A, of Z such that
1T(Z) = À7T(z)I
for all z E Z.A, -
exp is anadditive character
of 1,
hence À7T 0 exp = Xo -f
forsome f
E #’.Ker(f )
isa
subalgebra
of#, exp(Ker f )
=Ker(,k,)
therefore aunipotent p-adic subgroup
of Z of codimension at most one. If dim Z > 1 or dim Z = 1 and À7T =1,
-ractually
reduces to an irreduciblerepresentation
iro ofGo
=G/Ker
À,. But dimGo
dim G. The theorem follows from the inductionhypotheses.
It remains to consider the case: dim Z = 1 and A, 0 1. We will first show the existence of a
unipotent p-adic subgroup Gi of
codimensionone in G and an irreducible smooth
representation
1TI ofGi
such that1T is
equivalent
toindg, i G
1TI.Let
YDEW
be suchthat, [Yo,WICI, Yo§Éit.
PutCf}l = (U : [ U, Yo]
=01. Cfjl
is an ideal in 19 of codimension 1. ChooseXo §É %i
and defineZo
=[Xo, Yo].
ObserveZo G #, Zo --,-é
0. Then{Xo, Yo, Zo}
is abasis for a 3-dimensional
subalgebra
of Wisomorphic
to the Liealgebra
ofH3.
Let S denote thesubgroup
of Gcorresponding
to thissubalgebra
andwrite,
asusual,
We can choose A
E ,fl,
À =1 0 with thefollowing
property:Let us assume, for the moment, that 1T is an irreducible smooth
representation
of G on V.By
Theorem3,
the restriction of 1T to S is adirect sum of irreducible
representations
ofS,
allequivalent
to therepresentation
pA of S inC;(a) given by
So V =
(D i,,Ej VA
for some index-setI,
eachVf being isomorphic
toC;(a).
We mayregard
I as a t.d. space in the obvious way. Then wehave
where W =
C’ (I). Moreover,
with theseidentifications,
Let
G1
denote theunipotent p-adic subgroup
of G with Liealgebra G1,. G,
is a closed normalsubgroup
of G and G =G1. (exp tXo)tEn (semi-direct product).
SinceYo
is in the center ofWi, 7r(GI)
and-rr(exp yYo) (y E n)
commute. RecallOur aim now is to prove the
following
lemma.LEMMA 4: For each t E
f2,
there exists a smoothrepresentation
gl H
7T(gl, t) of G,
on W such thatObviously,
this lemmaimplies -7r--indG,JG’7ri
where 1T1 isgiven by
1T1(gl) = 1T(g, 0) (g, E G,).
Theirreducibility
of 1Tyields
the ir-reducibility
of n1, .To prove the
lemma,
we start with a linear map A :C;(J1, W) -->
C’ (f2, W), commuting
with all operators7r(exp y Yo) (y E ,fl ).
Thus:for
all t,
y E Q andf
EC’ (n, W).
Since
C’ (f2)
is closed under Fouriertransformation,
we caneasily
establish the
following: Given 0
EC’(f2)
and an open compact subset K offi,
there exists aninteger m
>0, À 1, ...,
Àm E C and y 1, ..., ym Ei f2 such thatFor cP E COO(J1)
letLb
denote the linear mapC’ (É2, W) --.> C-,(n, W) given by Let>f(t) = f>(t)f(t) (f E C;(J1, W». Then, putting
K =Supp f
USupp A f,
we obtain:(t
Ea, f E C;(J1, W )).
HenceALb
=Lq,A
forevery 0
ECoo(J1).
Inparticular
we have:1T(gl)L/>
=L/>1T(gl)
for all 91 EGI, 0
EC’(f?).
LetI/1n
denote the characteristic function of P n. Inaddition,
putL,O(s) = 0(s - t) (s, t
EJ1, cP
any function onf2).
Define:Here,
asusual, LigIn Q9
w is identified with the functions - L,tp, (s) -
w(s E !1). 7r (g 1, t)
is well-defined :assuming n’:5 n,
we obtainBut this
equals, by
the aboveresult,
Let us show now that
ir(gl, t)
satisfies condition(a)
of Lemma 4. Fixf E C;(fl, W)
and determineintegers m, n > 0,
tl,..., tm E a andW1, ..., Wm E W such that
Then
Condition
(b)
is also fulfilled.Indeed,
Furthermore,
Hence,
Finally,
it iseasily checked,
that condition(a)
forces gi Hn(gi, t)
(g 1 E G )
to be a smoothrepresentation
ofG
1 for each t E D. This concludes theproof
of Lemma 4.COROLLARY: Each irreducible smooth
representation of
G ismonomial.
Let us continue the
proof
of Theorem 4.By
induction we assume that1T1 is admissible and
pre-unitary.
Hence 1T =indal
ta 1T1 ispre-unitary.
Let K be an open
subgroup
of G and letVK
denote the space of allf
ECc(f2)
such that7r (g)f
=f
for all g E K. Letf
EVK.
Sincethere exists an
integer
n >0, only depending
onK,
such thatf
isconstant on cosets of P n.
The relation
implies
thatSupp f C
Pm for someinteger
m> 0, only depending
onK. Assume m n. Then P m =
lÙ/=i (t,
+P n )
for some th..., tk E f2- Now consider themapping
of
VK
intoW k.
Thismapping
is linear andinjective.
Sincewe obtain that
f (ti)
is fixedby
expt;Xo - (K n G 1) exp(-t;Xo), being
anopen
subgroup
ofG (
=1, 2,..., k). Therefore,
eachf (ti) stays
in a finite-dimensionalsubspace
of W.Consequently
dimVK
00.We have shown that 7r is admissible. This concludes the
proof
ofTheorem 4.
REMARK: Similar to the
proof
of Theorem 4 one caneasily
showthat the restriction of an irreducible
unitary representation
of G to itssubspace
of smooth vectors is an admissiblerepresentation
of G.§6.
Kirillov’stheory
Let G be as in
§5.
What remains is to describe the irreducibleunitary representations
of G. This is doneby
Kirillov[8]
for the real groups Gand,
as observedby
Moore[9],
the wholemachinery
worksin the
p-adic
case as well. Forcompleteness
and for later purposes,we
give
the result.Given
f
E(9’,
putBf(X, Y) = f ([X, Y]) (X,
YEl (9). Bf
is an alter-nating
bilinear form on G. Asubalgebra J
of W which is at the sametime a maximal
totally isotropic subspace
forB f
is called apolariza-
tion at
f.
Polarizations atf
exist([4], 1.12.10). They
coincide with thesubalgebra’s J
C CO which are maximal with respect to the propertythat J
is atotally isotropic subspace
forB f (cf. [8],
Lemma5.2,
which carries over to thep-adic
case withabsolutely
nochange). Let J
beany
subalgebra
of W which is atotally isotropic subspace
forBf : f e,] =
0. Put H =exp Sj.
We may define a character Xf of Hby
the formula:
Let
p(f, Sj, G)
denote theunitary representation
of G inducedby
Xf.THEOREM 5
([8], [9]):
(i) p(f, J, G) is
irreducibleif
andonly if Sj
is apolarization
atf, (ii)
each irreducibleunitary representation of
G isof
theform
p (f, f), G),
(iii) ?(/!, t. G)
andp(f2, f)2, G)
areunitarily equivalent if
andonly if fj
and/2
are in the same G-orbit in G.§7.
The character formulaThe main reference for this section is
[12].
G acts on Wby
Ad andhence on %’
by
thecontragredient representation.
It is well-known(and
can beproved
similar to the realcase)
that all G-orbits in G’ are closed.Let us fix a non-trivial
(unitary)
character yo of the additive group of n.We shall choose a Haar measure
dg
on G and a translation invariant measure dX on W such thatdg
=exp(dX).
Let
f E fjf, f)
apolarization
atf
and 0 the orbit off
in G’. Put1T =
p(f, Sj, G). Given gi
EC?(G),
we know that1T( «f)
is an operator of finite rank(§5, Remark).
Putt/J1(X) = t/J (exp X) (X
EG). Then ip,
EC?(%).
The Fourier transform oft/J1
is definedby:
’ Here Xo is (as usual) a fixed non-trivial additive character of n.
Observe that
«fr1
EC?(%’).
THEOREM 6: There exists a
unique positive
G-invariant measure dvon 0 such that
for all tP
EC?(G):
Note that the
right-hand
side isfinite,
because dv is also a measure onCfjf,
since 0 is closed in G’.Pukanszky’s proof
of([12],
Lemma2),2
goes over to our situation with no substantialchange.
Observe thateach tp
EC’ (G)
is a linear combination of functions of theform 0 * P (0
EC-,(G» where Ô
isgiven by (g) =,O(g-’) (g
EG).
Thealgorithm
to determine du(given dg
and dX such thatdg = exp(dX))
is similar to thatgiven by Pukanszky:
(i)
PutK = exp;j, r = KBG.
Choose invariant measures dk anddy
on K and rrespectively
such thatdg
= dkdy.
(ii)
Choose a translation invariant measure dHon J
such thatdk =
exp(dH).
(iii)
Let dX’ and dH’ denote the dual measures of dX and dHrespectively.
(iv)
Let= {Xf E %’:(§,X’)= 01.
TakedH’
on;j1-
such thatdX’ =
dHf dH1-.
(v)
Let S be the stabilizer off
in G. Then S C K. Choose dÀ onSBK
such that dÀ is theinverse-image
ofdH 1-
under thebijection
of
SBK
ontof
+H1-.
(vi) Finally,
put dv =image
of dkdy
under thebijective mapping Sg - g-l . f (g E G )
ofS B G
onto O.The invariant measure dv
depends
on the choice of the character Xo.Taking
instead of xo the character x ->Xo(tx)
for some t En,
t #0,
weobtain, by applying
the abovealgorithm,
thefollowing homogeneity
2 Part (d) of his proof has to be omitted here.
property for d v :
COROLLARY: Let 0 be a G-orbit in G’
of
dimension 2m. Thenfor all -0
EC?(%’)
and all t EJ1,
t # 0.Observe that we may choose in the
corollary
dv to be any G-invariantpositive
measure on O.Let 0 be as above. 0 carries a canonical measure IL, which is constructed as follows. For any p E
0,
define ap : G ---> 0by ap(a) =
a . p
(a E G).
The kernel of the differentialBp
of ap,{3p:
G->Tp ( Tp =
tangent space to 0 inp )
coincides with the radical of thealternating
bilinear formBp
on G. LetStabG(p)
be the stabilizer of p in G. Thenalso, Ker (3p
= Liealgebra
ofStabG(p).
HenceBp
inducesa
non-degenerate alternating
bilinear form wp onTp.
In this way a2-form w is defined on 0. One
easily
checks that w is G-invariant(cf.
[12]
for the realcase).
Let d = 2m be the dimension of 0. Assume d > 0.Then li
isgiven by IL = 1(1/2mm !)A mwl.
THEOREM 7: Let us
fix
the character yoof
f2 in such a way that xo = 1on 0, yo 0
1 onP-1.
Let 0 be any G-orbit in G’of positive
dimension. Then the invariant measure dv and the canonical measure IL on 0 coincide.
The
proof
isessentially
the same as in the real case([12], Theorem).
§8. Square-integrable représentations
mod ZLet G and Z be as in
§5.
An irreducibleunitary representation
1T ofG on Y is called
square-integrable
mod Z if there existEi X - (0)
such thatSuch
representations
areextensively
discussedby
C.C. Moore and J.Wolf for real
unipotent
groups[10].
Forp-adic unipotent
groups, see[13]:
the restriction of 1T to the space Y. of ir-smooth vectors is asupercuspidal representation.
Our maingoal
is to find a closed formula for themultiplicity
of the trivialrepresentation
of well-chosen open and
compact subgroups K
of G in the restriction of 1T to K.Let
f
E G’.By O f
we denote the G-orbit off
in Cfjf andby irf
anirreducible
unitary representation
ofG, corresponding
tof (more precisely:
toOf) by
Kirillov’stheory (§6).
Letlltf
denote the space of1Tf. Then we
have,
similar to([10],
Theorem1):
THEOREM 8: The
following four
statements areequivalent:
(i)
irf issquare-integrable
modZ, (ii)
dimOf
= dimG/Z,
(iv) Bf
is anon-degenerate
bilinearform
onNow assume 1Tf to be
square-integrable
mod Z. The orbitOf
carriesthe canonical measure tt. We shall define another G-invariant
measure v on
Of.
Let us fix a G-invariant differential form W on%1#
of maximal
degree.
Let cr denote theadjoint representation
of G on 19and let p be the
representation
of Gcontragredient
to a. Fix p EOf.
We have
StabG (p )
= Z and g -p(g)h
is anisomorphism3
ofG/Z
ontoOf.
CallI3p
the differential of this map at e;8p: %/# - Th.
DefineIn this way we
get
a n-form w’ onOf.
We claim that w’ is G-invariant:Call v the measure on
Of corresponding
tow’; v
isuniquely
deter- minedby
the choice of the volume form w onG/L.
LetIP(f)1
denote3 Here isomorphism is meant in the sense of algebraic geometry.
the constant
relating IL
and P: li =/P (f)/V.4
The volume form wfixes,
on the other
hand,
a Haar measuredg
onG/Z.
It is obvious that v is theimage
ofdg
under themapping g F--> p(g)f
ofG/Z
ontoOf.
Fromthe definition of v we see that the same is true for the
mapping
g ->
p(g)h
ofG/Z
ontoOf,
for any hE Of.
Let us denote
by d( 1Tf)
theformal degree
of 1Tf:THEOREM 9:
d(wf) is a positive
realnumber,
whichsatisfies the following identity : d(wf) = (P (f)[ .
This is
proved exactly
the same way as in the real case([10],
Theorem4).
§9. Multiplicities
Let G be as
usual, f
E %’ such that 1Tf issquare-integrable
mod Z.Let K be an open and compact
subgroup
of G. We shall call K alattice
subgr-oup
if L =log K
is a lattice inCfJ,
i.e. an open and compact, O-submodule of G.THEOREM 10: Let K be a lattice
subgroup of G, L = log K.
Normalize Haar measures
dg
on G and dz on Z suchthat f K dg
=fKnzdz
= 1. Choose a Haar measuredg
onG/Z
such thatdg
= dzdg.
Then the trivial
representation of
K occurs in the restrictionof
1Tf to Kif
andonly if f(L n#)c 6;
moreover, itsmultiplicity m(wf, 1) is l/d(1Tf).
The
proof
of Theorem 10 is ratherlong
andproceeds by
a careful induction on dim G. The theorem is obvious if dim G = 1. So assumedim G = n > 1. Put
iÉ° = Ker f n#
andZ° = exp #°.
We have twocases :
1.
dim #° # 0. Replace % by %liX°
and Gby G/Z°.
The center ofG/Z°
isZ/Z° (cf [13], proof
ofTheorem, (i)). Replace
also Kby K°
=KZ°/Z°. K°
is a latticesubgroup
ofG/Z°: log K°
=L/L n #°.
Letf°, w)
be thepull
down off,
1Tf to%/#°
andG/Z° respectively.
It iswell-known that
w)
isequivalent
to wp. Hencem(wf, 1) = m(wp, 1).
4 P(f ) actually is the Pfaffian of the canonical differential form, defining IL, relative to CJ)
([ 1 ], §5, no. 2).
Furthermore, f (Lnz)=fo(LOnell-îo). Normalizing
the Haarmeasures on
G/Zo, ZIZO
andG/Z°/Z/Z°
asprescribed
in thetheorem,
one obtains
d(irf)
=d(irfo).
The assertion for G now follows im-mediately
from the result forG/Z°,
which is of smaller dimension.2. dim 1 = 1 and
f#
0 on 1. L ni is a lattice of rank one. Let Z bea generator of L ne. Choose
X e 1
such that[X, G
C L. Putego = {U: [ U, Y] = 01. ego
is an ideal in W of codimension one with centerIo = 1 + (-Y) (cf [13],
p.149). -,to n L
is a lattice of rank two;zo n Lli
n L is a lattice of rank one. We may assume that X is chosen in such a way that Xmod(e n L)
generatesio n Lli
n L. Then ob-viously,
Since
L/L fl %o
is a lattice of rank one, we can choose Y EL, Y e Go
such that L = 6y + L nego.
PutGo
= expego, G,
=(exp s Y)sEn.
ThenG =
Go- G1
andGo n Gi = {e}.
Now choose a basis
Z, X,
e,, ..., en-3 ofego
such that L nego
= 6z + 0% +6ei
+... +Oen-3
and such that e,, ..., e,,-3 is asupplemen-
tary basis ofLo
in the sense ofPukanszky ([12],
section3).
Oneeasily
checks that this is
possible.
GivenXo E ego,
writeand choose
(z, t,
tl, ...,tn-3)
as coordinates of the second kind onGo.
Then
dgo
= dz dtdt, ... dtn-3
is a Haar measure onGo
and dsdgo
is aHaar measure on G.
Moreover,
ifZo
= expIo, Ko
= K nGo,
we now have:Let
fo
denote the restriction off
toGo.
It is part of the Kirillovtheory
that 1Tf is
equivalent
toindexe
1Tfo.Moreover,
1Tfo issquare-integrable mod Zo ([13],
p.149).
We need a relation betweend( 1Tf)
andd( 1Tfo).
The Haar measures on
G/Z
andGo/Zo
should be chosen asprescribed
in the theorem. The
following
lemma isproved by computations,
similar to those
given
in([13],
Section5).
LEMMA 5: Let
r = f [X, Y]. Furthermore,
putfor
any sE a,
fs (Xo)
=f (Ad (exp -
sY)Xo) (Xo
EGo)
and 1Ts = 7Tf!>. Then ns is square- 5 We take dz and dz dt as Haar measures on Z and Zo respectively.integrable
modZo
andfor
all s E a.PROOF: The space
Hf
of n f may be identified withL2(a, Hf,,).
Fix asmooth vector v E
Hf0
v # o.Choose «/1 E CllJ(Q), ##0
andput glv(x) = #(x)v (x
En).
Then
tp,
E¡¡et. Furthermore,
thecomputations
in([13],
Section5),
showMoreover,
Hence, w,
issquare-integrable
modZo
andd( 1Ts)
=d( 1To)
for all s E J1.In addition:
This
completes
theproof
of the lemma.Let
cP, CPo
denote the characteristic functions ofK, Ko respectively.
Given «/J
EL 2(l2, ’Jefo)’
we haveHence, by
ap-adic analogue
of Mercer’stheorem,
So,
we obtain thefollowing
relation:Equivalently:
LEMMA 6: m
(1Tf, 1)
=fil
m(1Ts, 1)
ds.Now assume
m(1Tf, 1) >
0. Thenm(1Ts, 1) >
0 for some s E il.By induction, fs (Lo nL0.) C fJ,
whereLo
= Ln Cfjo.
HenceConversily,
assumef (L ni) c C.
Let s e f2. Thenfs(Lo n
c 6 ifand
only
iff, (,Y)
C 6. We have:Hence, by induction, m(1Ts, 1)
> 0 if andonly
if s E(1 / r)(-f (,Y)
+0).
Moreover, again by induction, applying
Lemma 5 and6,
This