C OMPOSITIO M ATHEMATICA
S. J. P ATTERSON
The laplacian operator on a Riemann surface, III
Compositio Mathematica, tome 33, n
o3 (1976), p. 227-259
<http://www.numdam.org/item?id=CM_1976__33_3_227_0>
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227
THE LAPLACIAN OPERATOR ON A RIEMANN
SURFACE,
IIIS. J. Patterson Noordhoff International Publishing
Printed in the Netherlands
1. Introduction
In the first paper of this series
[9]
we sketched a method ofobtaining
the
spectral decomposition
of thefamily
ofLaplace operators 0394k acting
onautomorphic
forms ofweight k
andmultiplier X belonging
tosome Fuchsian group G. In the second paper
[10]
we introduced the Eisenstein seriesE03B6(z, s )
and demonstrated variousproperties
of thesefunctions. On
Re (s) = 1/2 they
may be considered asgeneralised eigenfunctions.
Our firstobject
will be to make this statement exact, and to prove in our case a theorem of Elstrodt[2] (in
his case it isonly
valid when
03B4(G) 1/2).
The chief
object
of this paper is to prove acompleteness
theorem-to the effect that the functions
above, together
with a discrete set ofeigenfunctions,
suffices for thespectral
resolution ofàk. Unfortunately
our
proof
of this fact isonly complete
when k EZ,
and we shall limit ourselves to the case k = 0 in our discussion of thecompleteness
theorem. This is no loss of
generality,
thanks to thetheory
of Maal3 operators,([3], [13]).
In the casek ~
Z we show that the whole of the continuousspectrum of - L1k
is contained in[1/4, ~[,
and that theeigenvalues (which
lie in[03B4(G )(1- 03B4(G)), 1/4[,
cf.[9])
are, to some extent,regularly
distributed.When 03B4(G) 1/2
thecompleteness
theorem was
proved
in[9].
Theplan
of theproofs
of the facts outlined above was sketched in[9].
As a consequence it will follow thatE03B6(z, s )
and
03A6k(s, X)
haveanalytic
continuations tomeromorphic
functions onC,
with thepossible exception
of one or threepoints.
The notations and conventions of this work are as in
[10]. Mostly they
shall be used without any furtherexplanation.
The Fuchsian groups considered here will all befinitely generated
and contain noparabolic
elements.2. Estimates
In
[10; §6]
we deduced estimates for the functionQ-k(03B8, X, s ), and, by implication,
also for itspartial
derivatives. These estimates are not uniform in s in unbounded sets, or, moreimportantly,
in 0 in theneighbourhood
of 0.However,
at the cost of notobtaining
theoptimal
behaviour in X we can obtain uniform bounds of the same
type.
Let us choose 03B11, a2 >
0,
ai + «2 7T.Suppose
that 0 E]0, 03B11].
Recall theintegral representation
forQ-k(03B8, X, s) ([10; Eq. (59)])
Suppose X >
0. We move the line ofintegration
to Im(z) = -
a2 andwrite z = x -
i«2.
Then wesplit
theintegral
into twoparts, correspond- ing
to x > 0 and x 0.If,
in thelatter,
wereplace
xby -
x we obtainAs ex lies on the line
[1, ~[
it follows afterstraightforward
estimatesthat and
Let s = u + it. Then
substituting
these estimates in theequation above,
we
obtain,
for cr >0,
where
We
clearly
obtain the same result when X 0 and thenby continuity
we have that
(2)
is valid for all X.Next we note that the same method can be
applied
toand we see that this is
bounded,
in o- >0, by
for a suitable
C2(U, 03B1). Integrating
this from 0 to0,
we seethat,
for03C3 > 0,
Next, using [10; (35), (37)]
we haveThus, writing
T = Min(03C3,
1-03C3),
andassuming
that 0 03C31,
andIs - 1/2|
~1/3
we find from(3), (4)
andStirling’s
formulaC2(U, a ), C3(U, 03B1)
can be taken to be continuous in (r. Theexpression
onthe left-hand side of
(5)
is entirein |s - 1/2| ~ 1/3
and so,by
themaximum modulus
principle (5)
holds also in the interior of the circle(and
so in thestrip
0 03C31).
Note that inderiving
theseinequalities
we have used the arbitrariness of a2
given
ai to absorb powers of|s|,
but
unfortunately
at the cost ofexplicit
forms forC2(U, 03B1), C3(U, a ).
Observe that
(3), (5)
exhibit the nature of thesingularity
ofQ~k(03B8, X, s )
as 03B8 ~ 0.Using
this with[10;
Th.5]
we obtain anintegral representation
ofR~~k(03B8, s, t, X)
valid inRe (s) > 0, Re(t) > 0.
Theformulae are somewhat
lengthy,
but notcomplicated,
and we shall notwrite them down.
However,
we do need theasymptotic
behaviour ofR~~k(03B8, s, t, X)
which can be deduced from this and(3), (5). Letting
5(+)
=1, 03B4(-)
= 0 we obtainWe shall close this section with some
special
results in the casek = 0. Note
that,
from(1),
This is an Euler
beta-function;
we haveNext,
as in[9],
setand define the
point-pair
invariantLet
hR (s )
be theSelberg
transform of this kernel([9], [14]).
From the formulaegiven by Selberg
we see thatwhere x is the
positive
solution ofSuppose
now that r > 0. We deform thepath
ofintegration
to thepolygonal path -x, -x + i/4,
x+ i /4, x.
Forlarge
r the contribution from the line - x +i /4, x
+i /4
isO(e-r/4), unif ormly
for x incompact
subsets. As u ~ xand,
as u - - x,uniformly
for xlying
in compact sets. It now followsby
Watson’slemma that
and this estimate holds
uniformly
for xlying
incompact
subsets of]0, ~[.
If we let Il =[21r, 4w] then,
for each r > 1 there exists x E Il so thatUsing
theargument
of[ 10; §7]
with the kernelKR
where x =x (R )
EIl
it follows that there exists a cp sothat,
if Re(s)
=1/2, then,
forz E D7T/2,
This makes
explicit
an estimate in[10; §7].
The fact thatKR
is not continuous does not matter; in fact it can beapproximated by
smoothfunctions in the L
1 sense and,
sinceE03B6(z, s )
is bounded on thecompact
set Dl’ this suffices.
3. The inner
product
formulaConsider a function
f (s, 03B6), taking
values inV,
and defined for s with Re(s )
=1/2, and e
E03A9(G).
We shall also assume that,E (C, TJ)
was defined in[9].
Hencef (s, C)
has a Fourierexpansion
of theform
(03B6
E03A903B1(j))
We shall assume that
only
a finite set of thefa (s ) (a
EZ*)
are notidentically
zero as functions of s and that all thefa (s )
are smooth withcompact
support on{Re (s )
=1/2, s 0 1/2}.
These conditions are very restrictive but shall be weakened later. Let L ={Re (s )
=1/2}
andWe shall show
that vf
is squareintegrable
and calculate its norm. Let(·,·)v
be the Hermitian form on V. ThenSince
E03B6(z, s )
is smoothin C
and z([10])
it follows that there is nodifficulty
ininterchanging
orders ofintegration
and summation. The inner sum can be evaluatedby [10;
Th.4].
The result isIn section 2 we found
approximations
tovalid when cp is
small,
with error termsO(~).
It followsby (6),
the conditionsimposed
onf,
and the estimate on theUal3 (s)
in[10;
Th. 8(+ Th. 6)]
that the error termsgive
a contribution0 (cp)
to theintegral
above. Hence
where
By
the estimates for03C303B103B2(s)
mentioned aboveIa.j(s, t)
is a smooth function ofcompact support
on(LB{1/2}) (LB{1/2}). Moreover,
the functionalequation [10; Eq. (40)] gives
and
where 03C3*03B103B2(s)
is defined in[10; §4].
We consider the first and fourth terms in the above
integral together.
We have to evaluate
We introduce the new variables
and
setting
which are smooth functions of
compact support
on R xL,
we see thatwe must evaluate the
integral
By
theRiemann-Lebesgue
lemma the first term is zero. The secondterm is a Dirichlet
integral
and thisgives
the
equality following
from the evaluation ofI1,j(s, s ), I4.j(s, s)
above.By
ananalogous argument
the second and third termsgive
rise toIn
particular,
if thesupport
off (s, C),
as a function of s lies in L+ ={Re (s )
=1/2,
Im(s )
>01
then this term vanishes.We form the Hilbert space
DE
of functionsf(s, 03BE)
on L+ x03A9(G) satisfying (7)
and with the normFurthermore,
as in[9],
letL (G, V)
be the Hilbert space of square-integrable automorphic
forms ofweight k
andmultiplier X,
and normThen the
f (s, 03B6)
which we haveconsidered,
and which havesupport
on L+belong
toDE, and
we have shown that for these functionsLet
L (G, V)c
be the closure inL (G, V) of v f
withf
as above. Then thefollowing
theorem followsby
standardHilbert-space
methods:THEOREM 1: The map
is
defined
on a densesubspace of DE
and takes values inL (G, V).
Thismap extends to an
isometry of ;jE
onto a closedsubspace L (G, V)c of
L (G, V).
This theorem was first
proved
for(not necessarily finitely generated)
G
with 03B4(G) 1/2 by
Elstrodt[2].
Hisproof
is rather more direct thanours
although
the crucialpoint
in bothproofs
is the appearance of aDirichlet
integral.
4. Bounds on the Eisenstein séries and
eigenfunctions
Assume for the moment that k = 0. Let
If Re
(s )
=1/2
then(9), [10;
Th.1] and Cauchy’s inequality
showthat,
for some cp
However
D+0(03C0/2, 03B1, s) = D-0(03C0/2,
and soby (7)
andStirling’s
formula
Again Stirling’s
formula with the first of these twogives
Thus,
from[10;
Th.6]
followsIn the same way we also obtain that
Next we
require
some information aboutA~(s)
itself. Leth (s )
be a functionsatisfying (i), (ii), (iii)
of[9; §3],
and letqh (z, w )
be the associatedpoint-pair
invariant. We shall assume that h ispositive
onthe
’spectrum’
of -do, which
Elstrodt[3]
has shown to be contained in{s :s(1 - s) > 0}.
Now letwhich therefore is the kernel of a
positive integral operator
onL (G, V).
It follows now from§3
and the definition ofh(s)
that the contribution of the spaceL (G, V)c
to the kernel isMercer’s theorem
(or
at least theargument leading
to it in[1])
showsthat this
integral
isabsolutely convergent
and thatHowever,
the seriesdefining Qh(z, z)
convergesuniformly
oncompact
sets
([11], [12])
and henceTaking
thespecial
caseh (s )
=(a
+s(1 - s ))-b (a
>0, b
>1)
thisgives, f or any b > 1,
Cauchy’s inequality
showsthat,
forany b
> 1Let us now consider
eigenfunctions f03BC(z)
EL(G, V)
withparameter s03BC(Re (s03BC)
~1/2).
k can bearbitrary.
From[10; §6]
it follows thatRe(s03BC) > 1/2,
and the Fourierexpansion of f03BC
has the formThe set of all SIL lies in the
compact
interval[1/2, 1].
It follows from[10;
§6]
that for s in this interval and E >0,
Let
Suppose
’P =(~1,
...,’Pp),
cpj 7T/2.
Take 02 = 1T/2,
01 = cpj and onevaluating
the norm off03BC integrated
over the subset of DP definedby
1 s
rj(z) e03BA(j),
03B81 ~0j(z) *
03B82 we seethat,
if E > 0 there is c6 > 0 sothat
From
(3)
we see thatThis shows that
Considering,
asabove,
the contribution of theff, 1
to the kernelQh (z, w )
we obtain thatLet
ko
be theunique
number so thatnoting,
however thatko might
not exist. The1/2
andko (if
itexists)
arethe
only possible
accumulationpoints
of{s03BC}.
Furtherare
convergent,
unlessko
=1/2,
whenconverges.
We now show that there is no continuous
spectrum
in]1/2, 5(G)].
Suppose
I~]1/2, 03B4(G)], k0 ~ I, I compact
and that there exists afamily
of
generalised eigenfunctions e (z, s )
withrespect
to a measure /£. il isa measure
supported
on I so that03BC({s})
= 0. The mapis an
isometry
ofL 2(I, 03BC)
onto a closedsubspace
ofL (G, V). As A,
isan
essentially self-adjoint elliptic operator
thee (z, s )
are, for03BC-almost
all s,
C’-eigenfunctions
ofL1k.
All these facts can begleaned
from[7;
Ch. XVIII
§§8, 9,
Ch.XI §1], [13; §§3, 5].
In
particular, e (z, s )
has a Fourierexpansion
of the formWe have
also, by
an unnumberedequation
between[7;
Ch. XVIII§9, Eqs. (9.2)
and(9.3)],
that ifApplying
the method ofpoint-pair
invariantsagain
showsthat,
if ~ isgiven
there iscp’
so thatTaking
two suitable values of cp andusing [10; §6]
showsthat,
if E > 0 isgiven
there isbe
EL 2(I, 03BC)
so thatIt follows now from
(3) that, if ~ L 2(I, g)
is
square-integrable
onDj
={z :
1 sri (z) e03BA(j),
0Oi (z) - 7r /21.
Fromthe definition of
e(z, s)
it follows that so isThis function
is, however, by
the estimates above and(3)
square-integrable
onD*j = {z : 1 ~ rj(z) e03BA(j), 03C0/2 ~ 03B8j(z) 03C0}.
Thus it issquare-integrable
onDj
UD*j,
which is a fundamental domain forG,,,).
It is then an
automorphic
form ofweight k, multiplier X
whosespectral support
withrespect to - 0394k
lies in[0,1/4].
The spectrum lies in[1/4, ~[ ([3], [9])
and thus we have a contradiction unlessc+03BC(s) =
0. Butthen
e(z, s)
is asquare-integrable eigenfunction.
As03BC({s})
= 0 and asthere are at most a countable set of
eigenfunctions
it follows thate(z, s)
does not exist. Hence:THEOREM 2:
Suppose
karbitrary. If 5(0) 1/2
thenIf
5(G) ? 1/2,
the continuous spectrumof - L1k
lies in[1/4, ~[,
and thepoint spectrum
lies in[03B4(G)(1 - 03B4(G)), 1/4[. If
theeigenvalues
are03BB03BC
=s03BC(1 - s03BC),
SIL >1/2,
andif k0 ~ [1/2, 03B4(G)], k0 - |k| E Z
then(The
second term is to be omittedif ko
does notexist.)
The
only outstanding
statement, thefirst,
appears in[9].
COROLLARY:
Ei (z, s)
can be continued as ameromorphic function
ins to
CB{1/2, ko,
1-k0},
smooth in z andC.
Theonly poles
in Re(s) ~ 1/2
are at the
points
s03BC and these areexactly first
order.PROOF: The method of
[10; §8]
shows that if so E]1/2, 03B4(G)],
and ifRe
(sn)
~0,
sn - so then there is asubsequence s’m,
and a sequence Wm so thatconverges,
uniformly
oncompact subsets,
to a smootheigenfunction
of
àk
in z, smooth also inC. Furthermore, w-. 1E!.(z, s 2)
converges, forinfinitely
many a EZ*,
to a non-zero function. If wm ~ 00 then this isan
L 2
function(from
the Fourierexpansion). Thus,
whens0 ~ {k0, s03BC},
wm ~ and we can take wm to be constant.
Also,
the limit ofE03B6(z, s ),
ass ~ so
(~{k0, s, 1)
must beunique.
Thisgives, by
the reflectionprinci- ple,
theanalytic
continuation toRe(s) > 1/2, s ~ {k0, s03BC}.
Next in
[10; Eq. (51)]
we set s = t = 03C3 +ir,
sum over a and obtainwhere, if 03B6 ~ 03A903B1(j),
as in[10; §7],
Let
A, (s)
be as defined at thebeginning
of this section but now forarbitrary
k. The method ofpoint-pair
invariants shows that the estimates(13), (14), (15)
of this section hold forarbitrary k
when s isrestricted to a
compact
set.Thus,
if r ~0, 1 ri
1 we show that the firstthree terms on the
right-hand
side above areO(A~(s) + 1), uniformly,
when lies in
compact
subintervals of]1/2, 1]. By [10;
Th.5]
Thus,
for 03C3lying
in acompact
subintervalof ]1/2, 1]
uniformly.
In other wordsWe now
bring
in theelementary
lemma:LEMMA:
Suppose f (z)
isanalytic
in a deletedneighbourhood of 0,
andthat, if IRe (z)| ~ 1, Ilm (z)1
~1,
Then
f (z)
can be continued to ameromorphic function
in thefull neighbourhood of 0,
with at most afirst-order pole
at 0.PROOF: Let
Ci
be the square~/2 - 3 ~i, ~/2 + 3~i,
2E+ 3~i, 2E - 3 Ei,
and let L be the linesegment
E +iy (Iy 1::; ~).
L lies insideCi.
is bounded
by (3/2)2 + 62E
7E on the verticaledges
ofel,
andby ((3/2)2+ 32)1/2~/(3~) 7~
on the horizontaledges. Thus, by
the maximum-modulusprinciple,
for z E LHowever,
for z E LHence
From this it follows that
on any square with vertices ± E ± Ei.
Hence, by
Riemann’s theorem onremovable
singularities
this function can be continued to an entire function at 0. This proves the lemma.From the estimates above it follows that
E03B6(z, s)
satisfies the conditions of the lemma and can therefore be continued to each of thepoints s03BC
with at most a first-orderpole
there.We show
finally
thatE03B6(z, s )
has apole
at s = SIL. Letf03BC(z)
be thecorresponding eigenfunction,
and letbe the Fourier
expansion. Then,
as in[10; §5]
Suppose E03B6(z, s )
to beregular
at s = SjL. Then the left-hand side and the second term on theright-hand
side of the aboveequation
vanishidentically ([10;
Cor. to Th.5]).
ButR-+k(~i(03B1)),
SjL, sw,03B1) ~
0(ibid.).
Thuswe have a contradiction when there is an « so that
c(03B1, 03BC) ~
0(and
there is
always one).
Thiscompletes
theproof
in Re(s) ~ 1/2.
The rest followsusing
the functionalequation.
Note
that,
it follows that03A6k(s, X)
can be continued toRe(s) ~ 1/2, s ~ {k0, s03BC}.
One caneasily
check that the residue ofE03B6(z, s )
is a finite linear combination ofeigenvectors
and that therefore03A6k(s)
has apole
of order at most the dimension of theeigenspace
with parameters03BC 2013these
facts we shall not need.This now
completes
theinvestigation
in the case ofarbitrary k
andhenceforth we shall
only
deal with the case k = 0. We shall however prove some of theintermediary
results for all k but it isonly by
meansof the estimates obtained at the
beginning
of this section that we canuse these
effectively.
It is worth
noting
that theeigenvalue k0(1 2013 ko)
of -L1k
can be ofinfinite
multiplicity (cf. [3 ; § 12]).
Whether or not1/2
orko
can be an accumulationpoint
ofthe s,,
remains an openquestion.
5. The local trace formula
We recall the
following Maaß-Selberg
relation([10;
Th.4, §7])
From
[10;
Th.8]
it follows that this is valid forall s, t
inCB[l - 03B4(G), 03B4(G)] (C
if5(G) 1/2),
with the naturalinterpretation
at thepoles.
Thevanishing
of theright-hand
side when t = 1 - sis,
as wenoted in
[10], essentially
the functionalequation
for(03C303B103B2(s)).
We shallnow calculate
as a
limiting
version of(19).
Recall that iffn
is a sequence ofanalytic
functionsconverging uniformly
oncompact
subsets to afunction f,
then
f n
converges in the same sense tof’.
Thisjustifies
the passage to the limit in(19).
Unfortunately
we must now introduce some morespecial
functionsin order to express our results. Define
We shall also
write,
if a =(03B11, ..., an),
and
Finally,
the matrix over W([10; §4])
with entriesB~~(R, s, 03B1)03B403B103B2 (resp.
Mk(~
s,03B1)03B403B103B2)
will be denotedby L~~k(~, s) (resp. Mk(~, s)).
With these notations the passage to the limit
gives
As
long
as we do notseparate
each of the infinite summations this isuniformly convergent
oncompact
sets. Let so be aregular point
ofEc(z, s).
Then there is acompact neighbourhood
K of so and aconstant c >
0,
so that(03B8 given
inadvance),
for s E K. IfK1 ~~ K
there is a constantc =
c1(K1, 03B8)
so that(by Cauchy’s theorem)
for s E
Ki.
This suffices to allow us tosplit
the terms. It is alsopossible, using
the estimates of§2
to find bounds on theL~~k(~, s, a )
which also prove that the series above can besplit apart.
Allproblems
of convergence
arising
here can be treated in this way and we shall not deal with them further.The formula above can be very
considerably simplified by
means ofthe functional
equation
and[10;
Th.6],
i.e.In the calculations it is
advantageous
to use matrices over W. LetIk (s)
be the matrix with entries
qk(s, a )5a¡:l.
With the notationsabove,
andthose in
[10],
we obtainThe functional
equation
can be written in the form(using [10; Eq. (41), (36)]).
Thus we haveFrom
[10;
Th.3,
Cor. to Th.3]
we haveHence
Using
now(20), (21)
and[10;
Th.5 + Cor.]
we findUsing
these identities we obtainThe
reorganisation
of the terms can beeasily justified. Let,
as in[10],
From
this,
and the functionalequation
Expanding
andapplying
the functionalequation again
Define now
Then we
finally
obtainIf
then
A
simple
calculation shows thatand hence
03A6’(1 2013 s)/03A6(1 2013 s) 2013 03C81(1 2013 s )
is invariant under s H 1 - s.If we let
E’)(z, s)
be thecorresponding
Eisenstein series forG03B1(j)
andThen one
finds,
injust
the same way as aboveThe terms in
(24) offering
thegreatest difficulty
are thoseinvolving Mk(~, s )
and we shall nowinvestigate
these. From theintegral representation
forRk--(O, s, t, a)/(s(1 2013 s) - t(1 2013 t))
used in§2
we obtain also anintegral representation
forMk(~, s).
Thus thelimiting
case of
(6) gives
estimates forMk(~, s).
This caneasily
be verifieddirectly.
Wefind,
when Re(s )
=1/2,
Let
h(s)
be a continuous function on L =f Re (s )
=1/2}.
From nowon assume that k = 0. We shall consider
We estimate first the ’error term’. The
diagonal
matrix elements ofIk(s)-203A30(s)
are, up to a constantfactor,
by (15).
ChooseThen we see that the error term is
If h satisfies the conditions that
as s ~ ~, and
as s ~
1/2.
Then it follows from(17)
that theintegral
converges and hence the error term isO(~) =
o(1).
The
principal
termis,
up to a constantfactor,
By (14),
and
this,
with(26), (27)
and theRiemann-Lebesgue
lemma show that this limit exists and is 0.When Re
(s )
=1/2 A~ (s )
= A(s, ’P)
and then(26)
and(27)
show thatexists. It then follows from
(24)
thatconverges
absolutely.
Define
Hence
This formula is the
object
of this section and from it we can deduce thecompleteness
theorem without much additional work. If03B4(G) 1/2
thenA, (s)
is bounded on L and we can therefore omit the condition(27).
6.
Completeness
Let
h(s)
be a functionsatisfying
thefollowing
conditions:can be extended to an
analytic
function inin the
strip
defined in(i),
Here e > 0 is a fixed constant. That such functions exist can
easily
beshown
by examples. Form,
as in[9],
the associatedpoint-pair
invariantqh (z, w),
andLet
which
represents
theprojection
ofQh(z, w )
to anoperator
onL (G, V)c. By (i), (ii), (iii)
it follows(cf. [9])
thatQh
is the kernel of an L 2operator,
and theh (s ) give
thediagonal
elements in a commondiagonalisation
of all suchoperators (which
form acommuting
set ofnormal
operators).
Theoperator
has finite trace-as we shall show
presently.
It is alsopositive
andconsequently
it has a purepoint spectrum (which
will then be thatappearing
in theorem2).
Let
From
[9], [12]
it follows thatwhere « >
1, and 03C3(z, w )
is as before(cf. §2).
It follows withoutdifficulty that,
for z EDj,
Hence
converges
absolutely.
On the otherhand,
from[9]
we have thespectral
resolution for
hyperbolic
groups whichgives
Hence,
from(28)
we obtainand the
right-hand
side converges. Thusconverges. This proves our assertion. Note also that
converges but that this is weaker than theorem 2 since
(ii), (iv), (vi)
imply
that h has a zero of order 4 at1/2.
THEOREM 3:
L(G, V)c
spans the continuous spectrumof - 03940.
This theorem is the chief result of this paper. However we note that
I2
hasalready
beencalculated, Il
can be calculatedby
the transforma- tions ofSelberg [14],
andIn
fact,
note that7i, I,
exist if wedrop
condition(vi) above,
and thatQh - Qch
is still trace-classby
theorem 2.Thus,
forh (s ) satisfying (i),..., (v) (such
that h we calladmissible ),
We now calculate
11,
but to avoid inessential difhculties we shallassume that G has no
elliptic
elements.Let
{G}
be a set ofrepresentatives
ofconjugacy
classes inG,
and letNg
be the normaliserof g
in G. Let{G}0
be the subset of Gconsisting
of those elements not
conjugate
to any element of anyG03B1(j).
Letand
Then it follows that
Let
D(g)
be a fundamental domain forNg.
ThenThe calculation of the
integrals
offer no difhculties and we obtainwhere
H = set of elements
representing
theprimitive hyperbolic conjugacy classes,
N ( y )
=multiplier
of y, andb(03B3) =
1 if y is notconjugate
to a generator of aG03B1(j),
= 1/2
otherwise.Thus
combining
our results we see that forh (s ) satisfying (i),..., (v)
the
following
version of theSelberg
trace formula holdsIt is not difficult to see that condition
(v)
can ingeneral
be removed.This follows
by
afairly simple
and standardapproximation argument
but we shall not discuss it here.If 5(G) 1/2 then,
as remarked at the end of§5,
condition(iv)
can bedropped.
In this case it also follows that the left-hand side is zero as there are noeigenfunctions (cf. [9]).
As in the case of groups of the first
kind, u (D 7T 12)
is atopological
invariant. Infact,
if G isgenerated by A,, B1, ..., Ag, Bg, El,..., Er, H1, ..., Hp,
03C01, ..., 7Tqsatisfying
and
so that
(a)
anyelliptic
element of G isconjugate
to a power of someEj,
(b) H;
isconjugate
to agenerator
ofG03B1(j),
and(c)
anyparabolic
element of G isconjugate
to some power of some 1Tj.Then an
application
of the Gauss-Bonnet theorem shows thatLet us now consider the
case 03B4(G ) 1/2.
We define theSelberg
zeta-function in theregion
Re(s )
> 03B4(G),
Then, using
a well-knownidentity,
Summing
over kgives
From this it follows that
Putting
this into theSelberg
trace formulagives
Since we can find a sequence of admissible h
approximating 5ro
+03B4-r0 (5x
is the Dirac 5- ’function’ atx)
it follows that theexpression
inbrackets,
which is even, is zero. That isIf k = 0 it follows that
This,
with[10; Eq. (37)] gives
Then a
simple
calculation shows thatAs
we have
It is worth
noting
that in this casewhere a
= (03B11, ..., an ).
This formula showsquite clearly
theanalytic
structure of
03C8(s).
Now note that the residues at the
poles
ofare
integral,
and so we defineexists as a
meromorphic function,
with apole
of order 2m + 1 at s =-m, m~ Z, m
> 0.(30)
is valid at firstonly
for Re(s )
=1/2,
but as the left-hand side isanalytic
and definedin 03B4(G) Re(s) 1 - 03B4(G)
it follows that it is also valid there.Integrating,
andusing (31)
we obtain anequation
which can be written in the two formsvalid for all s. The second
gives
theanalytic
continuation and theanalytic properties
ofZ(s ).
The first can beregarded
as anexplicit
formula
for 03A6(s),
useful in03B4(G)
Re(s ) 1 - & (G) -
7. A
special
caseThe case when
5(G)
=1/2
exhibits somespecial features, especially
in
respect
to the behaviour at1/2.
Theobject
of this section is toexamine this case.
So,
unless otherwisestated, 03B4(G)
=1/2
butk, X
areallowed to be
arbitrary. Suppose firstly
that k -1/2 ~ Z.
The
starting point
isagain [10;
Th.4].
Fix ~ and letand
With these notations
[10; Eq. (51)] gives
Now,
as remarkedpreviously,
inRe(s) > 1/2
has
negative
entries.Clearly
I(s, a )
was mentioned in[10; §6],
where it was noted thatand so, as
D+k(03B8,
a,s)
andD-k(03B8, 03B1, s)
arelinearly independent
ats =
1/2, I(s, a )
ispositive
definite. ThusUsing (6)
it follows thatThe
technique
of the lemma of§6
shows now thatwith fixed a,
(3,
is bounded in aneighbourhood
of1/2
in Re(s ) > 1/2.
The method of
[10; §§7, 8]
shows that the same is true of(s - 1/2)E03B6(z, s).
Again by
the method of[10; §7]
there is a sequence Sm ~1/2 along
which(sm - 1/2)E03B6(z, Sm)
convergesuniformly
oncompact
subsets(of
z and
C). Suppose
thatLet
s’m
be another such sequence withcorresponding t’03B103B2.
Taking
the limit in[10; Eq. (51)] along (sm)
showsthat,
for a certain functionfl3 (z),
Taking
now the limitalong (s 2)
Interchanging
the roles of(sm )
and(s 2)
andusing [10;
Th.6]
shows thatHence
E,(z, s )(s - 1/2)
has a continuous extension to Re(s)
=1/2.
Hence so does
D-k(03B8,
a,s)03C303B103B2(s)(s - 1/2),
for fixed0, uniformly
in a,.8.
Consequently, by [10; Eqs. (62), (64)],
and the fact thatJk(1/2)-1 = 0,
03A6(s) = 03A6k(s, X)
has also a continuous extension to Re(s)
=1/2 (cf.
[5]).
Howevershows
that 03A6(1/2) ~ 0. Thus 03A6(s) ~ 0
in aneighbourhood
of1/2
in Re(s) - 1/2. Hence 03A6(s)
is bounded(by
the functionalequation)
in aneighbourhood
of1/2. By
Riemann’s isolatedsingularities
theorem0(s)
can be continued to aregular
function at1/2. By [10;
Th.8]
E03B6(z, s)
has nosingularities
at anypoints
in Re(s) ~ 1/2 except perhaps
at s =1/2. As 03A6(s)
~ 0 in aneighbourhood
of1/2
it followsfrom the functional
equation
thathas a continuous extension to
Re(s) = 1/2
inRe(s) ~ 1/2.
ThusE03B6(z, s)(s - 1/2)
has a continuous extension to1/2. Again, by
Riemann’s theorem on removable
singularities
it follows thatEc(z, s)(s - 1/2)
can be continued to aregular
function at s =1/2.
Now suppose that k -
1/2
E Z. Theargument
above fails at thepoint
where we showed that
(s - 1/2)E03B6(z, s)
has aunique
limit at1/2,
sinceR+-k(~, 1/2, 1/2, a ) =
0.However,
since(from (6))
the sameargument
shows that tal3 = 0. Thus(s - 1/2)E,(z, s)
has a
unique
limit 0 at s =1/2.
The rest of the argument goesthrough
asbefore, remembering
that nowJk(s)
isregular
and non-zero at s =1/2.
(There
is oneexception
to this last statement, for if k -1/2
E Z thenq’(s, 0)
= 0 but this caneasily
be dealtwith.)
ThusTHEOREM 4:
Suppose
that5(0)
=1/2.
ThenEc(z, s)(s - 1/2)
and03A6k(S, X)
areregular
in aneighbourhood of
s =1/2,
and hencemeromorphic
in thes-plane.
Now let
This has abscissa of
convergence 03B4(G)
andessentially
the samebehaviour as s
~ 03B4(G)
asE03B6(z, s ), X
= Id([11]). By
thecorollary
totheorem 2 and theorem 4 we see that
if 03B4(G ) ~ 1/2
thenE03B6(z, s )
ismeromorphic
in aneighbourhood
of03B4(G),
whereas if03B4(G ) 1/2
thiswas
proved
in[10; §4]. By
Landau’stheorem,
since theE,(z, s )
areDirichlet series with
positive
termswhen X
=Id,
thatE03B6(z, s )
has asingularity
at s= 03B4(G).
From the results of[11]
and theorem 4 thispole
is first-order if
03B4(G) ~ 1/2. Consequently
THEOREM 5:
Suppose
k =0,
X = Id. ThenE03B6(z, s)
has apole
ats = 03B4(G), of
thefirst
orderif 03B4(G) ~ 1/2.
Furthermoref (z,
w ;s)(s - 03B4(G)) is
bounded belowby
apositive quantity
as s ~5(G).
This shows that not
only
is5(G)
the abscissa of convergence off(z,
w ;s )
but alsof(z,
w ;s ) diverges
at s= 03B4(G). Using
thespectral decomposition
off (z, w ; s )
it is notpossible
to obtain much finerinformation,
but we shall not deal with thistopic
here.8.
Concluding
remarksThe results of this paper are far from
complete.
The most obviousoutstanding problem arising
from it is to extend theorem 3 toarbitrary
k. This seems to be technical in so far that
good
lower bounds forQk(e, X, s)
are needed.A much more fundamental
problem,
however is that ofdeciding
whether or not the set
Is, 1 appearing
in theorem 2 canreally
be infinite.This seems to
require
more than the verygeneral
methods which areused here.
It is worth
remarking
that the results here have agreat
deal incommon with
scattering theory,
and thetheory
of theSchrôdinger equation (cf. [16]).
Thetechniques
used there toproduce generalised eigenfunctions
do not seem toapply
in our case,although
very similiar methods can be used in thetheory
of groups of the first kind withparabolic
elements([4], [6], [8], [15]).
Could such a method befound,
itwould, presumably,
shedquite
a lot oflight
on theanalytic
nature
of 03A6(s),
which one would like to exhibit as some sort of Fredholm determinant of a resolvent kernel.The case of groups of the first kind can be
regarded
as alimiting
caseof the
theory
described in this series. This appearsformally
onletting
the
G03B1(j)
’shrink’ toparabolic.
This is rathersuggestive
and leads oneto
hope
that if thetheory
here could be extended to moregeneral
Liegroups than
SL (2, R),
that one could infersomething
about thespectral
resolution for these groups.However,
even thesimplest
case,that of Kleinian groups in
SL (2, C) (also
of rank1)
offersformidable,
but very
interesting
diinculties. In this case thespectral
resolution isknown
(Selberg).
It is also worthnoting
that the spaceW,
which classifies the continuousspectrum,
is the space of L 2 functions from theboundary
of the Riemann surfaceG BH,
with values in V. Thisphenomenon
appears also in Euclidean spaces and it would be interest-ing
to knowjust
howgeneral
it is.Notes added in
proof
1. Since this paper was written 1 have received a
preprint
from Prof.John D.
Fay,
"Fourier Coefficients of the Resolvent for a FuchsianGroup",
where the sameproblems
are treated in a rather different way.In
particular,
some of the openquestions
mentioned above are nolonger
open.
2. In
[9]
there are somemisprints
and unclarities. In§3
theargument
is rathersketchy;
this is morefully
dealt with in aforthcoming
paper,"Spectral Theory
and FuchsianGroups",
to appear in Math. Proc.Camb. Phil. Soc.
Also,
in§3,
conditions(i), (ii), (iii)
shouldalways
bereplaced by (i’), (ii’), (iii’).
The most seriousmisprints
are listed below. 1am indebted to J. Elstrodt for a set of very detailed comments on
[9]
from which these are taken.