C OMPOSITIO M ATHEMATICA
H. J ACQUET
K. F. L AI
A relative trace formula
Compositio Mathematica, tome 54, n
o2 (1985), p. 243-310
<http://www.numdam.org/item?id=CM_1985__54_2_243_0>
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243
A RELATIVE TRACE FORMULA
H.
Jacquet
and K.F. Lai© 1985 Martinus Nijhoff Publishers, Dordrecht. Printed in The Netherlands.
§1.
Introduction and notations(1.1)
Let F be a numberfield, E
aquadratic
extension ofF,
M’ a divisionalgebra
of center F and rank 4. We will make thefollowing assumptions:
(1)
thealgebra M’( E )
= M’ ~ E is a divisionalgebra;
(2)
let D be the set ofplaces v
of F such thatMl,
is a divisionalgebra;
then everyplace v
in Dsplits
in E.We
regard
themultiplicative
group of M’ as analgebraic
group G’defined over F. Thus:
Moreover
G’( FA )
is a closed subset ofG’( EA ).
We will denoteby
Z’ thecenter of G’. If ~ is a continuous function on the
quotient
we will set
and we will say that an
automorphic
irreduciblerepresentation
03C0’ ofZ’(EA)BG’(EA)
isdistinguished
if there is a smooth function ~ in the space of 1T’ such thatB’(~) ~
0.(1.2)
Instead ofM’,
we may also consider thealgebra
M =M(2, F)
oftwo
by
two matrices andregard
itsmultiplicative
group G =GL(2)
as analgebraic
group defined over F. It ~ is a continuous bounded function onthe
quotient
we will set
Let ’17 be an irreducible
automorphic cuspidal representation
ofZ(EA)BG(EA);
we will say that qr isdistinguished
if there is a smooth function ~ in the space of 03C0 such thatB(~) ~
0.(1.3)
Our main result is thefollowing
theorem:THEOREM:
Suppose
03C0’ is an irreducibleautomorphic representation of Z’(EA)BG’(EA);
suppose 03C0’ isinfinite
dimensional and let 77 be thecorresponding
irreducibleautomorphic cuspidal representation of Z(EA)BG(EA).
Then 03C0’ isdistirguished if
andonly if
1T is.The motivations for this result can be found in the work of
Harder, Langlands
andRapoport
onalgebraic cycles
of certain Shimura varieties.This is not the
place
for a discussion of their work. Suffices to say that itconcerns
poles
of certain Hasse-Weil zeta functions attached to a Shimura surface. These zeta functions can becomputed
in terms ofautomorphic
L-functions which have been studied
directly by
Asai(Cf. [S.A.])
Thequestion
arises then ofdeciding
when the L-function attached to anautomorphic cuspidal representation
qr has apole
at s = 1. Now the L-function has anintegral representation.
Indeed consider anintegral
where T belongs
to the space of 03C0 andE(x, s )
is an Eisenstein series.This
integral
has apole
at s = 1 with residueB(~).
On the other hand it isequal
to the L-function times anelementary factor;
inparticular
theL-function has a
pole
at s = 1 if andonly if,
for at least one cp, theintegral
has apole
at s =1,
that is if andonly
if B is non zero on the space of 1T, in otherwords,
03C0 isdistinguished.
Intrying
to extend theresult of
Harder, Langlands, Rapoport
to the case of acompact
Shimura surface one has toreplace
the group Gby
the group G’ and use the above theorem.Although
this will notplay
a role in thepresent
paper, we recall the characterization of thedistinguished representations:
arepresentation
isdistinguished
if andonly
if it is the " BaseChange"
of anautomorphic cuspidal representation
ofG ( FA )
whose central character is thequadratic
idele class character of F attached to E. The
proof
isrelatively
com-plicated
since it involves the "BaseChange"
and thetheory
of the Asai L-function. It would beinteresting
to see if this result could be estab-lished
by using
anappropriate
relative traceformula,
similar to the onewe are
using
here.(1.4).
In order to prove our theorem we need a "relative trace formula":it is stated in section 2. In section 3 we derive our theorem from this trace
formula. The
remaining
sections are devoted to theproof
of our traceformula.
(1.5) Finally
we thank R.Langlands
forsuggesting
thisproblem
to us,for his
encouragement
and his advice. We thank A.Selberg
for his adviceon classical results on L-functions. We also thank L. Clozel for
making
his results available to us;
although
the results are not used in thepresent
paper,they
werequite suggestive.
(1.6)
We summarize ourprincipal
notations.(1)
The group G is the groupGL(2),
P is the group of uppertriangular matrices,
A thesubgroup
ofdiagonal matrices,
N the group ofunipotent
matrices in P. We also denoteby
R thealgebra
of uppertriangular
matrices.(2)
We writeMat[ p, q,
r,s ]
for the matrix with rows( p, q )
and( r, s ).
We also set:(3)
So far as the number field F is concerned we follow standard notations. Inparticular FA
is thering
of adeles andFz
the group of ideles. We denoteby Ç
a non trivial character of thequotient FA/F.
Wedenote
by
dx the self dual Haar measure onFA
so thatvol(FA/F)
= 1and
d(ax) = lalFdx where |a|F
is the module of the idele a; wedrop
thesubscript
F when this does not create confusion. We write dx =1-Idx,, 03C8
=II Bfi v
wheredxv
is the Haar measure onFv,
self dual withrespect
toBfIv.
We denoteby dXxv,
theTamagawa
measure onF" :
Then we denote
by d"x
theTamagawa
measure onF::
We denote
by F1
the group of ideles of norm one andby F+~
the group of ideles whose finitecomponents
are one and whose infinitecomponents
are all
equal
to somepositive number,
the same for all infiniteplaces.
Then
FÂ
is theproduct
ofF and F;;
and theTamagawa
measure is theproduct
of a measure onF
1 and the measured t/t
where t= 1 x 1.
Recallthat
vol(F1/Fx)
= 1. Ingeneral algebraic
groups over F areprovided
with the
Tamagawa
measure.(4)
Let v be aplace
of F. We denoteby K,,
the standard maximalcompact subgroup
of G. ThusKt,
=GL(2, Ru)
whereRv
is thering
ofintegers
if v is finite. IfFv = R,
thenKu
=0(2, R).
IfF,, = C,
thenKu
=U(2, C).
We denoteby KF
theproduct
of theK,,.
The groupG(FA)
andZ( FA ) being provided
with theTamagawa
measures and thequotient Z(FA)BB(FA)
with thecorresponding
measure, wehave,
for afunction
f
on thequotient:
where d k is a certain Haar measure on
KF.
Note thatvol(KF) ~
1.Having
chosen an invariant differential form of maximumdegree
w on Gwe have for each
place v,
themeasure |03C9v| and
theTamagawa
measureL(1, 1v)|03C9v|
onGv.
If wegive
tq thequotient Z,,NG,,
the measurequotient
of theTamagawa
measures then we have for a functionf
on thequotient:
where
dkv
is a certain Haar measure onK v. Again vol(Kv) ~
1.(5)
We have also thequadratic
extension E ofF,
with Galois group{1, 03C3}.
Whenever convenient we write E = F[].
We denoteby Bfi E
nontrivial character of
EA/E. Usually
we assume that is trivial onFA .
If u isa
place
of E we denoteby Ku
the standard maximalcompact subgroup
of
Gu, by KE
orsimply
K theproduct
of theKu.
§2.
A relative trace formula(2.1)
We will denoteby S
a finite set ofplaces
of Fcontaining
D and allinfinite
places;
the set S will beenlarged
as need dictates. For each set T ofplaces
of F we will denoteby
T ~ the set ofplaces
of E which areabove a
place
of T. For eachplace v
of F not in D we choose anisomorphism M’v ~ M(2, Fv);
let a be a basis of M’ over F and au theRu-module generated by a
inM,;
we may assumethat,
for allplaces v
ofF not
in S,
ourisomorphism
takes av toM(2, R,,). Extending
thescalars,
wehave,
for eachplace v
of E not inD~,
anisomorphism M’(Ev) ~ M(Ev).
Fromthese,
weget
for eachplace v
of E(resp. F )
notin D
(resp. D~)
anisomorphism G’v
=G,;
we use it toidentify
the twogroups. For each
place v
of E(resp. F )
we denoteby Kv
the standard maximalcompact subgroup
ofGv.
Wechoose,
in the usual way, two invariant F differential forms of maximaldegree
w and w’ on G andG’;
over an
algebraic
closure of Fthey
are the same.Thus,
for eachplace v
of F the groups
G,
andG’v
comeequipped
withforms w
andwl;
if v isnot in D we may assume the
isomorphism
ofG’v
ontoGv takes |03C9’v|.
Similar remarks
apply
to E.We denote
by f
a smooth function ofcompact support
onZ(EA)BG(EA);
we assume thatf
is aproduct
of localcomponents fv.
For all v the local
component fv
issmooth,
ofcompact support,
bi-Kv-finite;
for almost allfinite v,
it is the characteristic function ofZ,Kv. Similarly,
we consider a smooth function ofcompact support f’
on
Z’(EA)BG’(EA).
It is also assume to be aproduct
of local factorsf’v.
We make the
following assumptions
onf
andf ’ : (1)
for each v not inD - , fv
=f,;
(2)
let v be aplace
in D andvl,
v 2 the twoplaces
of E above v ; wehave
isomorphisms
we demand that the convolution
products
on
G’v and Gv respectively
have the sameregular
orbitalintegrals.
In other
words,
we demand that thehyperbolic
orbitalintegrals
ofh v vanish;
on the otherhand,
if7§
andT,
areisomorphic Fv-subalgebras
ofrank 2 of
M’v
andM, respectively
and ift’~T’v-Fv corresponds
tot~Tv-Fv,
then we demand thatIn this formula the Haar measures on
G,
andG’v
are theTamagawa
measures attached to LVv and
wv respectively;
on the other hand themeasures on
T"
andT’f
are theTamagawa
measures attached todifferential forms TJv and
qi
whichcorrespond
to one another under theisomorphism Tv ~ T’v. Finally
thequotients
aregiven
thequotient
mea-sures.
(2.2)
Theoperator p ( f )
definedby f
on the spaceis defined
by
a kernel K. We call P theorthogonal projection
on thespace of
cuspidal
elements. ThenP03C1(f)P
isrepresented by
a kernelKcusp;
it is a smooth bounded function.Similarly
theoperator 03C1(f’)
defined
by f’
on the spaceis
represented by
a kernelK’;
we call P’ theorthogonal projection
on theorthogonal complement
of the spacespanned by
the functions of the formxodet,
where X is aquadratic
character ofExBExA.
ThenP’03C1(f’)P’
is
represented by
a kernelKcusp.
It is also smooth and bounded.(2.3)
PROPOSITION:Suppose
that the aboveassumptions,
inparticular (2.1.1)
and(2.1.2),
aresatisfied.
Then:In this
formula Tamagawa
measures.The
proof
will occupy section 4 and thefollowing
sections.§3.
Demonstration of the theorem(3.1)
We will assumeProposition (2.2)
and derive Theorem(1.3).
Lettherefore 03C0’ and 03C0 be as in
(1.3).
We will prove that if qr isdistinguished
then 1T’ is
distinguished.
Theproof
of the converse assertion is similar and left to the reader. So from now on we assume 1T isdistinguished.
Wefirst recall a remark of
Langlands. Suppose v
is aplace
of F whichsplits
into v1 and v 2 in the extension
E;
then we haveisomorphisms Gv1 ~ Gv2 ~ Gv.
The restriction of B to the space of 03C0 is a non zero
G( FA )-invariant form; imbedding
the space of 7TvI 0 lrv2 into the space of 1Tappropriately,
we find a non-zero linear form which is invariant under the group
Gv1
XGv2.
Therefore 1TvI and 03C0v2 arecontragredient
to one another. Sincethey
are trivial on the center, we see that thefollowing
condition is satisfied:(1)
if vsplits
into vl andv2,
then 1Tvl ~ 03C0v2.Of course the
representation
03C0’ satisfies theanaloguous
condition. We set 03C0v = 03C0v1 = 03C0v2 and03C0’v
=03C0’v1
=03C0’v2.
(3.2)
Recall that S is a finite set ofplaces
of Fcontaining
D and theinfinite
places.
Let us setWe may choose S so
large
that II contains the unitrepresentation
ofKS - .
We thantake fv
for v OE S ~ to be bi-invariant underKv.
Ingeneral
if a is an irreducible
representation
ofZ(EA)BG(EA) containing
theunit
representation
ofKS~,
we will denoteby e03C3
thecorresponding
character of the Hecke
algebra
of the groupG S - .
Inparticular
if a is anautomorphic cuspidal representation
we will denoteby V(03C3)
the space of forms in the space of a which are invariant underKS .
Theorthogonal projection
onto the spaceV(a)
will be notedPo.
The spaceV(a)
isinvariant under
GS~
and thecorresponding representation
will be noted 1TS - . Thusfor (p
inV(a)
we have:Accordingly
we may writeKc.,p
as the sum of thefollowing uniformly convergent
series:where
Ko
is the kernel attached to theoperator Po’17(fs - )Po
and the sumis over all
cuspidal representation
acontaining
the unitrepresentation
ofKS - .
In turn, if~j
is an orthonormal basis ofV(03C3),
thenKa
isequal
tothe finite sum
Therefore
We will choose now
fs -
in such a way that a03C0 is non zero and thereexists a function
f’ satisfying
the conditions(2.1.1)
and(2.1.2).
To thatend let us observe that 03C0S ~ is
equivalent
to the tensorproduct
represen-tation
We choose therefore an
isomorphism
we have set
and v,
is the space of ’17V. At the cost ofenlarging S
we may assume thereis a
unitary
vector u in the spaceVT ~
such that the restriction of B tou 0
VT ~
is non zero. We choosefT ~
in such a way that’TTT-(fT-)
is theorthogonal projection
on u. Thenif v
is any basis ofVD ~
we have:For each v in D let v1 and v2 be the two
places
of E above v. LetCl,
bean invariant linear form on the tensor
product V:Jl
0V:2.
Thenhn -
is thetensor
product
of the spacesVv1 ~ V:2
and the restriction of B tou 0
hD - is,
up to a constantfactor,
the tensorproduct
of theCv.
Foreach v in D let us choose bases of
Vv1
andVv2
dual to oneanother,
a,and b.
say.Then
the aboveexpression is,
up to a constantfactor,
theproduct
over all v in D of the sums:Since this sum reduces to
where
(3.3)
At thispoint
we need a lemma:LEMMA: We can choose
fv1
andfv2
in such a way that tr03C0v(hv)
= 1 andtr
a( h v)
= 0for
everyinfinite
dimensional irreduciblerepresentation
aof Gv
which is notequivalent
to 03C0v.PROOF OF LEMMA: In any case 03C0v is in the discrete series. Thus if v is finite there is a function
hv
with therequired
traces and it is trivial that it is a convolutionproduct.
So we are done in this case. If v is infinite thenv is real. So we may
identify
E to R andGv
toGL(2, R).
LetKo
be thegroup
SO(2, R)
ofKu and,
for eachn E Z,
X n the characterSay
that ’17v is therepresentation
of the discrete series ofhighest weight
n > 0.
Since ’17v
is trivial on the center n isnecessarly
even. Choosesmooth functions of
compact support a
and b onZvBGv.
We will take then withsupport
in the set of matrices withpositive
determinant so thatwe may think of them as
being
functions onSL(2, R ).
The function b will be taken bi-invariant underKo.
As for a it will be assumed totransform under
x n
on both sides:We will take
h v
to be(a-b)*(a+b).
Then if a is any irreduciblerepresentation
ofGL(2, R)
we have:If furthemore a is in the discrete series then the second term is zero; the first term is zero as well unless the
highest weight m
of a is less than orequal
to n, that is unless abelongs
to a certain finite set. The linearindependence
of charactersimplies
then that we may choose a such that tr03C0v(a)
= 1 and tr03C3(a)
= 0 if a is anotherrepresentation
of the discrete series. Thus all we have to show is that we can choose b in such a way thatif a is a
representation
of theprincipal
series trivial on the center.Unraveling
this we see that this relation isequivalent
toF,
=Fh,
wherewe have set, for x >
0,
and
Fb
is definedsimilarly.
SinceFa
is smooth ofcompact support
onR++
and invariant under the substitutionx ~ x-1,
there is a function bwith the
required properly ("Paley
Wiener Theorem forSL(2, R)",
cf.for instance
[S.L.],
Th.3, V, $2,
p.71).
This concludes theproof
of thelemma.
(3.4). Coming
back to the situation of(3.2)
andchoosing h v
as in thelemma we have
Hence a’7T
= C ~ 0 as desired.Now we choose a function
f’
onG’(EA)
so that(2.1.1)
and(2.1.2)
aresatisfied: so for a
place v
of E not in D - wetake f,,’ = ft,.
For v inD,
the Schur
orthogonality
relations show there is a convolutionproduct h §,
on
G’v
so that tr03C0’v(h’v)= -1
and tr03C32(h) =
0 if a’ is notequivalent
to03C0’v.
Since tr’17v(hv)
= 1 and tr03C3(hv)
= 0 if Q is notequivalent
to 77,, we see thathu
andh’v
have the same orbitalintegrals.
We therefore takef’v1
andf’v2
so thatand we can
apply Prop. (2.2).
The kernelK’cusp
has adecomposition analoguous
to(3.2.2).
Inparticular
we findwhere
By Prop. (2.2) expressions (3.2.3)
and(3)
areequal.
We mayregard
themas infinite linear combinations of characters of the Hecke
algebra
ofGS~.
Since theprinciple
of linearindependence
of charactersapply
tothese infinite sums
([R.L.] $11) and e03C0
= e03C0, we see that a03C0, is non zero.Since
K’03C0(x, y )
is a sum of terms of theform ~(x)~’(y) where ~
and~’
belong
to the space of u’ we see that the restriction of B’ to the space of 77’ is non zero. This concludes theproof
of Theorem(1.3).
§4.
Double cosets(4.1)
In order to prove our trace formula we need tostudy
the doublecosets of
G’(F)Z’(E)
inG’( E )
and the double cosets ofG(F)Z(E)
inG(E).
An element ofG’(F)Z’(E)
orG(F)Z(E)
will be termedsingu-
lar. An element not in this set will be termed
regular.
We start with G. We let
Q
be a set ofrepresentatives
for theG(F) conjugacy
classes in the set of subfields ofM(2, F )
of rank 2 over F. Welet R be the
subalgebra
ofM(2, F)
formed of the uppertriangular
matrices and P the
subgroup
ofG(F)
formed of the uppertriangular
matrices.
(1)
LEMMA:If T is in Q
then anysingular
elementof (T
0E)
is inT .
EX. Similarly
anysingular
elementof P(E)
is inP(F)Z(E).
PROOF OF
(1): Suppose x
issingular
in(T ~ E) .
Then there is c in E =Z(E)
such that x = cywith y
inG(F). Then y
is in the intersec-tion of the
algebras
T 0 E andM(2, F),
that is in T. Since it isinvertible,
it is in T’ and we are done. Theproof
of the second assertion issimilar, using
thealgebra
R instead of thealgebra
T.Recall that the Galois group of E over F is
(1, cr 1.
(2)
LEMMA:Suppose
x is inG(E).
Then x isregular ( in
oursense) if
andonly if x 0 x -
1 isregular
in the usual sense.PROOF OF
(2): Suppose
x issingular
in our sense. Then x = gc with g inG(F)
and c in Ex. Thenx03C3x-1
=co-l
issingular. Conversely,
supposex03C3x-1
= c where c is in EX. Then c03C3c = 1 so c =z03C3-1
with z in Ex. Thisgives (xz-1)03C3 = xz-1.
Hencexz-1
is inG(F)
and x issingular
in oursense.
(3)
LEMMA:If
x is inG(E)
then either x issingular,
or x is in a doublecoset
G(F)pG(F)
with p inP(E) - P(F),
or x is in a double cosetG(F)tG(F),
where t is in(T 0 E)X-
TX. EX and T is inQ.
Moreover thethree
possibilities
are exclusiveof
one another.PROOF OF
(3):
Let us prove the first assertion. Let E be the extensionF[~].
Then any x inM(2, E)
can be writtenuniquely
in the form:with a and b in
M(2, F).
Suppose
x is invertible and notsingular;
recall that in thepresent
context this means that x is not in
Z(E)G(F).
Then a is not zero, otherwise x would be inG(F)EX,
hence would besingular. Suppose a
isnon zero but non invertible. We claim there is an element c of E such that x’ = xc has the form
Indeed a has the
following
form:Let us write also:
We claim that for some
appropriate
z in F the matrix za + b has a non zero determinant.Suppose
not. Thenfor all z in F. This
implies s
= 0 andthen rq
= 0.Say r
= 0. Then thematrix x has the form
and is thus not
invertible,
a contradiction. Let therefore z be an element of F such that za + b is invertible. Set c = z +1/.
Then c is in EXand xc has the form a’ +
b’/Ù
where a’ is invertible. Thus we mayassume that a is invertible.
Replacing
xby xa -1
we may even assumethat a is 1. If b is not invertible or if it is invertible but not
elliptic (in
theusual
sense)
inG ( F ),
there is a g inG ( F )
such that b =gpg-1
where p is inR(F).
Then:Since q
is inR ( E )
we are done in this case. If b is anelliptic
element ofG ( F )
then there is a T inQ
and a t in T’ suchthat b = -gtg - 1.
Then:Since y
is in T 0 E we areagain
done.It remains to prove the second assertion of the lemma
(3).
Because oflemma
(1),
all we have to prove is that for Tin Q
an element x of(T ~ E)X-
T . EX cannot be of the form gl pg2, with p inP(E)
and gl and g2 inG(F). Suppose
it is. After asimple computation
we find:Suppose
the Falgebra
T is notisomorphic
to E. Then T ~ E is afield,
in fact aquadratic
extension of E andx03C3x-1
is an element of that field which is not anelliptic
element ofG ( E ) (in
the usualsense).
Thusx03C3x-
1is
singular
in the usual sense andby
lemma(4.1.2) x
issingular
in oursense, a contradiction.
Suppose
now that T isisomorphic
toE;
inpassing
note that there isexactly
one elementof Q
with thatproperty.
Then T 0 E is not a
field;
it is a sum of twocopies
of E. Inparticular
theminimal
polynomial
of x over F hasdegree
one or two and the same istrue of q. This is
possible only
if the twoeigenvalues
of g are in F. Sincethey
are elements of E whose F norm is1, they
are + 1 or -1. In anycase:
This
gives a2
= 1 and a03C3a = 1. SInce x isregular a ~
± 1(lemma (4.1.2)).
Let us
identify
T ~ E with the sum of twocopies
ofE;
thenFrom a2 =
1 and a ~ ± 1we get a = (1, -1) or ( -1, 1).
Then aOa = -1a contradiction.
We will say that an element of
G(E)
iselliptic
if it is not in the setG(F)P(E)G(F).
(4.2)
We will nowclassify
allelliptic
elements.(1)
LEMMA:Suppose
T and T’ are inQ. Suppose
that x and x’ areregular
elements
of
T 0 E and T’ ~ Erespectively
but have the same double class moduloZ(E)G(F).
Then T = T’.PROOF OF
(1) :
Since E is contained in(T 0 E ) X
and(T ~ E)
we may as well assume that x and x’ have the same double class moduloG(F) :
with c and c’ in
G ( F ).
Then
x 1 -
=CX,0-lC-1.
Moreverx03C3-1
is aregular
element of(T ~ E) ’
in the usual sense
(lemma (4.1.2)).
Hencec (T 0 E)c-1
= T’ 0 E.Taking
the intersections with
M(2, F)
we obtainCTC-l
=T’,
hence T = T’.(2)
LEMMA:Suppose
that c and c’ are inG(F),
T is inQ,
x and x’ areregular
elementsof (T 0 E) ’
and z is in EX.Suppose
thatThen c’ and c normalize T and cc’ is in T.
PROOF OF
(2): Again
we havex03C3x-1 = cx’03C3x’-1z03C3-1c-1
andx03C3-1
is aregular
element in the usual sense. It follows that c normalizesT ~ E,
hence also T.Finally
cc’ =cx’ - lc - lz -’x
is in T 0E,
hence in T.(4.3)
We now deal with theremaining
elements ofG(E).
(1)
LEMMA:Every
element xof P(E)
is in the same double class moduloG(F)Z(E)
as oneof
thefollowing
element:PROOF OF
(1):
Let us write x in the formIf
alc
is in F then aftermultiplying by
an element ofZ(E)A(F)
wemay assume that a = c = 1. If b is in F then x is in
N( F )
and we are incase
(i).
If b is not in F then aftermultiplying by
a matrix inN( F )
wemay assume that b =
u~
with u in F. Thenis the matrix
(ii).
Ifa/c
is not in Fthen,
aftermultiplying by
an elementof
Z(E),
we may assume c = 1. Since a is not in F we may write b in the formwith u and v in F.
Then the matrix
has the form
(iii).
This concludes theproof
of the lemma.(2)
LEMMA: In lemma(1)
cases(i), (ii), (iii)
are exclusiveof
one another.PROOF oF LEMMA
(2):
In view of lemma(4.1.1),
it suffices to show thatcase
(ii)
and(iii)
are notcompatible. Suppose they
were. Then therewould be g and g in
G(F)
and a matrix d inA(E),
where A is thegroup of
diagonal matrices,
such that:Then we would find:
Since
n(2~)
isunipotent
and d03C3-j is adiagonal
matrix this is a contradiction.(3)
LEMMA:Suppose
c and c’ are inG(F),
x and x’ areregular
elementsof A(E)
and z is in E.Suppose
thatThen c and c’ normalize A and cc’ is in A
(F).
The
proof
is the same as theproof
of lemma(4.2.2).
(4)
LEMMA: Let n =n(h). Suppose
g andg’
are inG(F),
z in E.Suppose
thatThen z is in
F, gg’ =
z and g andg’
are inN(F)Z(E),
where N denotesthe group
of triangular
matrices with unitdiagonal.
PROOF oF
(4):
Once more we findgn03C3-1g-1
=z03C3-1n03C3-1.
Sincen03C3-
1 is aregular unipotent element,
thisimplies
thatz’-’ =
1 and g centralizesno-1;
in turn thisimplies
that g is inZ(E)N(E)
hence in fact inZ( F ) N( F ) and
z is in F.Similarly g’ is in Z( F ) N( F ).
SinceZ( E ) N( E )
is a commutative group
gg’
= z and we are done.(4.4) Altough
not necessary for thedescription
of the double classes the twofollowing
lemmas will be useful latter on.(1)
LEMMA:Suppose 1 is in G(E)
and03BE03C303BE-1 is in P(E). Then 1 is in G(F)P(E).
PROOF oF
(1):
Set p= 03BE03C303BE-1. By assumption
p is inP(E).
On the other handp°p
= 1. We claim there is a q inP(E)
such that p =q°ql.
Thiswill prove the lemma because then
eq-1
is invariant under a hence inG ( F ).
To prove our claim we writeThe condition that
p’p
= 1gives:
The condition
q03C3q-1
= pgives:
Because of
(2.i)
and(2.ii)
it ispossible
to solve(3.i)
and(3.ii).
Nextregard (3.iii)
as asystem
of 2 linearequations
for r and s where z = r +s~.
The condition z03C3z = 1 means that the determinant of thesystem
iszero and the condition
(3.iii)
is then thecompatibility
condition for the twoequations.
Our claim follows.Say
that an element ofG ( E )
issemi-simple regular
if it iselliptic
or inthe double class modulo
G(F)Z(E)
of an elementdiag( a, 1)
with a OE F.Say
it isunipotent regular
if it is in the double class of an element of the formn(y) with y
not in F.(4)
LEMMA: An element xof G(E)
issingular ( resp. semi-simple regular,
resp.
unipotent regular) if
andonly if x03C3-1
issingular (resp. semisimple,
resp.
unipotent regular)
in the usual sense.REMARK: In this lemma an element of the form
zn(y)
with z in Z isregarded
asunipotent.
PROOF oF
(4):
We mayalready
assume xregular (lemma ((4.1.2)).
Thenwe know that x
= gyg’
where g andg’
are inG(F) and y
is either in(T ~ E) -
TXEx or of the formdiag( a, b )
witha/b~F,
or of the formzn(u)
with u e F and z inZ(E).
Thenx03C3x-1 = gy03C3y-1g-1.
In the firstcase
y03C3y-1
is an element of T ~ E hence asemi-simple
one. In thesecond case
y03C3y-1
ishyperbolic
and in the last case it isunipotent.
Inany case it is
regular (loc. cit.).
(4.5)
Theprevious
resultsapply
mutatis mutandis to G’.Every
element iseither
singular
orelliptic.
Theset Q
isreplaced by
a setQ’
ofconjugacy
classes for the
subalgebras
of M’ of rank 2. Of course, any element ofQ’
is
isomorphic
toexactly
one element ofQ.
§5.
Terms attached to theelliptic
andsingular
elements(5.1)
To prove our trace forrnula we now write downKcusp
moreexplicitly:
where:
A formula for the two other kernels will be recalled below. In turn we
write
(2)
as the sum of three more kernels:where:
Similarly,
we writeK’cusp
aswhere
We write
(8)
as the sum of two other kernels:where
In this section we will prove
that,
under theassumptions
ofProposition (2.2),
the kernelsKe
andK’e, Ks
andKs, KSp
andKsp respectively,
havethe same
integrals.
(5.2)
We first deal withKe
and prove thatKe
is indeedintegrable
overthe
product
of thequotient Z(FA)G(F)BG(FA) by
itself. To that endwe first prove a lemma. We let
KE
be the standard maximalcompact subgroup
ofg( EA )
and for g inG(EA)
define theheight H(g)
of g to be:We first recall without
proof
a standard lemma(Cf. [J.A.]):
(2)
LEMMA : Let Q be a compact subsetof G(EA).
Then there is a number d > 0 such that the relations:imply e
is inP(E).
We derive from this lemma the
following
result:(3)
LEMMA:Suppose
Q is a compact subsetof G(E).
Then there is anumber d > 0 such that the relations
imply that 1 is in G(F)P(E).
PROOF oF
(3):
We have alsoh-03C303BE-03C3~03A9-03C3.
Aftermultiplication
weget h-03C303BE-03C303BEh
E 03A9-03C303A9.Applying
theprevious
lemma we see that if d is sufficientlarge
andH( h )
> d then03BE-03C303BE~ P(E).
The conclusion follows from lemma(4.4.1).
Recall that
If x
and y
are inG ( FA )
then theprevious
lemma shows that the relationf(x-103BEy)=0 implies
first that xand y
are in setscompact
moduloG(F)Z(EA);
in turn thisimplies that e stays
in a set finite modulo EX.Hence
Ke
is indeedintegrable. Furthemore, by taking f positive
in whatfollows we can
justify
our formalmanipulations.
(5.3)
The results of the sub-section(4.2)
allow us to write:we have noted
N( T )
the normalizer of T inG ( F ).
Since T has index 2in
N(T )
this can also be written as:If we
integrate formally
over theproduct
of thequotient G(F)Z(EA)BG(FA) by
itself we find:the sum is over all T
in Q
andthen 1 regular
inT B(T ~ E)x/Ex;
theintegral
in x is overG(FA)/Z(FA)
and theintegral in y
over thequotient TÂ B G ( FA ); finally
vT is the volume of thequotient FÂ T"B T A.
In turn, each
integral
in(3), apart
from a constantfactor,
can be writtenas a
product
over allplaces v
of F of local ones.If v is a
place
of F which does notsplits
in E and u is thecorresponding place
of E then E is aquadratic
extension of F and G isa
subgroup
of G. Thecorresponding
local factor isnothing
but:The convergence of this
integral
can be established as follows: if theintegrand
is not zero thenxi y belongs
to a setZuQ
where 03A9 iscompact.
Then