On Advanced Analytic Number Theory

241 Pages·01.362 MB·English

by  C. L. Siegel

On Advanced Analytic Number Theory
By
C.L. Siegel
TataInstituteofFundamentalResearch
Mumbai,India
i
FOREWORD
“AdvancedAnalyticNumberTheory”wasfirstpublishedbytheTataInsti-
tuteofFundamentalResearchintheirLectureNotesseriesin1961. Itisnow
being made available in book form with an appendix–an English translation
of Siegel’s paper “Berechnung von Zetafunktionen an ganzzahligen Stellen”
whichappearedintheNachrichtenderAkademiederWissenschafteninGo¨ttingen,
Math-Phy. Klasse. (1969),pp.87-102.
We are thankful to Professor Siegel and to the Go¨ttingen Academy for
according us permission to translate and publish this important paper. We
also thank Professor S. Raghavan, who originally wrote the notes of Profes-
sorSiegel’slectures,formakingavailableatranslationofSiegel’spaper.
K.G.Ramanathan
ii
PREFACE
During the winter semester 1959/60, I delivered at the Tata Institute of
FundamentalResearchaseriesoflecturesonAnalyticNumberTheory. Itwas
mayaimtointroducemyhearerstosomeoftheimportantandbeautifulideas
whichweredevelopedbyL.KroneckerandE.Hecke.
Mr.S.Raghavanwasverycarefulintakingthenotesoftheselecturesand
inpreparingthemanuscript. Ithankhimforhishelp.
CarlLudwigSiegel
Contents
1 Kronecker’sLimitFormulas 1
1 Thefirstlimitformula. . . . . . . . . . . . . . . . . . . . . . 1
2 TheDedekindη-function . . . . . . . . . . . . . . . . . . . . 13
3 ThesecondlimitformulaofKronecker. . . . . . . . . . . . . 20
4 Theelliptictheta-functionϑ (w,z) . . . . . . . . . . . . . . . 29
1
5 TheEpsteinZeta-function . . . . . . . . . . . . . . . . . . . 41
2 ApplicationsofKronecker’sLimitFormulastoAlgebraicNumber
Theory 52
1 Kronecker’ssolutionofPell’sequation . . . . . . . . . . . . . 52
2 ClassnumberoftheabsoluteclassfieldofQ(√d)(d <0) . . . 69
3 TheKroneckerLimitFormulaforrealquadraticfieldsandits
applications . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
4 RayclassfieldsoverQ(√d),d <0 . . . . . . . . . . . . . . . 89
5 RayclassfieldsoverQ(√D),D>0. . . . . . . . . . . . . . . 102
6 SomeExamples . . . . . . . . . . . . . . . . . . . . . . . . . 133
3 ModularFunctionsandAlgebraicNumberTheory 149
1 Abelianfunctionsandcomplexmultiplications . . . . . . . . 149
2 FundamentaldomainfortheHilbertmodulargroup . . . . . . 165
3 Hilbertmodularfunctions . . . . . . . . . . . . . . . . . . . . 184
1 EllipticModularForms . . . . . . . . . . . . . . . . . . . . . 220
2 ModularFormsofHecke . . . . . . . . . . . . . . . . . . . . 225
3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
iii
Chapter 1
Kronecker’s Limit Formulas
1 The first limit formula
1
Let s = σ+it beacomplexvariable, σandt beingreal. TheRiemannzeta-
functionζ(s)isdefinedforσ>1by
∞
ζ(s)= k s,
−
Xk=1
here k s stands for e slogk, with the real value of logk. The series converges
− −
absolutelyforσ > 1anduniformlyinevery s-half-planedefinedbyσ 1+
≥
ǫ(ǫ > 0). ItfollowsfromatheoremofWeierstrassthatthesum-functionζ(s)
is a regular function of s for σ > 1. Riemann proved that ζ(s) possesses
an analytic continuation into the whole s-plane which is regular except for a
simplepoleats=1andsatisfiesthewell-knownfunctionalequation
s 1 s
π−s/2Γ ζ(s)=π−((1−s)/2)Γ − ζ(1 s),
(cid:18)2(cid:19) 2 ! −
Γ(s)beingEuler’sgamma-function.
Weshallnowprove
Proposition 1. The functionζ(s)can becontinued analytically intothehalf-
planeσ>0andthecontinuationisregularforσ>0,exceptforasimplepole
ats=1withresidue1. Further,ats=1,ζ(s)hastheexpansion
1
ζ(s) =C+a (s 1)+a (s 1)2+
− s 1 1 − 2 − ···
−
1
Kronecker’sLimitFormulas 2
CbeingEuler’sconstant.
We first need to prove the simplest form of a summation formula due to
Euler,namely.
Lemma.If f(x)isa complex-valued function having a continuous derivative
f (x)intheinterval1 x n,then
′
≤ ≤
1 n−1 1 f(1)+ f(n)
f (x+k) x dx+
Z  ′ − 2! 2
0 Xk=1 n n
= f(k) f(x)dx. (1)
−Z
Xk=1 1
2
Proof. In face, if a complex-valued function g(x) defined in the interval 0
≤
x 1, has a continuous derivative g(x), then we have from integration by
′
≤
parts,
1 1 1 1
g(x) x= dx= (g(0)+g(1)) g(x)dx. (2)
′
Z 2! 2 −Z
0 0
(Formula(2)hasasimplegeometricinterpretationinthattherighthandsideof
(2)representstheareaoftheportionofthe(x,y)-planeboundedbythecurve
y=g(x)andthestraightlinesy g(0)=(g(1) g(0))x,x=g(0)andx=(g1)).
− −
In(2),wenowsetg(x)= f(x+k),successivelyfork=1,2,...,n 1. Wethen
−
havefork=1,2,...,n 1,
−
1 1 1 k+1
f (x+k) x dx= (f(k)+ f(k+1)) f(x)dx.
′
Z − 2! 2 −Z
0 k
Addingup,weobtainformula(1). (cid:3)
ThisisEuler’ssummationformulainitssimplestform,withtheremainder
terminvolvingonlythefirstderivativeof f(x).
It is interesting to notice that if one uses the fact that the Fourier series
e2πinx
(n = 0, omitted) converges uniformly to x 1/2 in any closed in-
− n 2πin −
terPval (ǫ,1 ǫ)(0 < ǫ < 1) one obtains from (2) the following useful result,
−
namely,
If f(x)isperiodicinxwithperiod1andhasacontinuousderivative,then
1
f(x)= ∞ e2πinx f(x)e 2πinxdx.
−
Z
nX= 0
−∞
Kronecker’sLimitFormulas 3
ProofofProposition1.Letusnowspecialize f(x)tobethefunction f(x) =
x−s =e−xlogx (logxtakingrealvaluesfor x>0). Thefunctionx−s isevidently 3
continuously differentiable in the interval 1 x n; and applying (1) to the
≤ ≤
functionx s,weget
−
sn−1 1(x+k) s 1 x 1 dx+ 1(1+n s)
− − −
− Z − 2! 2
Xk=1 0
n n
= k s x sdx
− −
−Z
Xk=1 1
= nk=1k−s− 1−sn11−s(ifs,1) (3)
Pn k 1 log−n(ifs=1).
 k=1 − −
LetusnowobservePthattheright-handsideof(3)isanentirefunctionofs.
We suppose now that σ > 1. When n tends to infinity nx sdx tends to
1 −
n R
1/(s 1). Further, as n tends to infinity, k s converges to ζ(s). Thus for
−
− k=1
σ > 1, the right hand side of (3) convergePs to ζ(s) 1/(s 1) as n tends to
− −
infinity.
Ontheotherhand,lettheleft-handsideof (3)bedenotedbyϕ (s). Then
n
ϕ (s)isanentirefuncitonofsandfurther,forσ ǫ >0,
n
≥
1 n 1+n ǫ
ϕ (s) s k 1 ǫ + −
| n |≤ 2| | − − 2
Xk=1
1 s ∞ k 1 ǫ +1.
− −
≤ 2| |
Xk=1
Thus, as n tends to infinity, ϕ (s) converges to a regular function of s in the
n
half-planeσ>0;thisprovidestheanalyticcontinuationofζ(s) 1/(s 1)for
− −
σ>0.
Nowtheconstanta inthepower-seriesexpansionats=1of
0
1
ζ(s) =a +a (s 1)+a (s 1)2+
− s 1 0 1 − 2 − ···
−
isnothingbut lim ϕ (1). Inotherwords,
n
n
→∞
1 1
a lim 1+ + + logn
0
n 2 ··· n − !
→∞
Kronecker’sLimitFormulas 4
=C(Euler’sconstant).
Proposition1isthusproved. 4
TheconstantC liesbetween0and1. Itisnotknownwhetheritisrational
orirrational;veryprobably,itisirrational.
Onecoulddeterminetheconstantsa ,a ,...alsoexplicitlybutthisismore
1 2
complicated.
We shall consider now an analogous problem leading to the Kronecker
limitformula(KroneckerscheGrenzformle).
Insteadofthesimplelinearfunction x, weconsiderapositive-definitebi-
nary quadratic form Q(u,v) = au2 +2buv+cv2 in the real variables u and v
and with real coefficients a, b, c (we then have a > 0 and ac b2 = d > 0).
−
AssociatedwithQ(u,v),letusdefine
∞
ζ (s)= (Q(m,n)) s, (4)
Q −
m,Xn=
−∞
where denotessummationoverallpairsofintegers(m,n),except(0,0).
′
Q(uP,v)beingpositive-definite,thereexistsarealnumberλ > 0,suchthat
Q(u,v) λ(u2 + v2) and it is an immediate consequence that the series (4)
≥
converges absolutely for σ > 1 and uniformly in every half-plane defined by
σ 1+ǫ(ǫ >0). Thusζ (s)isaregularfunctionofs,forσ>1.
Q
≥
As in the case of ζ(s) above, we shall obtain an analytic continuation of
ζ (s)regularinthehalf-planeσ>1/2,exceptforasimplepoleat s=1. The
Q
constanta intheexpansionats=1ofζ (s),viz.
0 Q
a
ζQ(s)= s −11 +a0+a1(s−1)+···
−
isgiven preciselybytheKronecker limitformula. Theconstanta whichis
1
theresidueofζ (s)ats=1wasfoundbyDirichletinhisinvestigat−ionsonthe
Q
class-numberofquadraticfields.
Letusfirsteffectsomesimplifications. 5
SinceQ(u,v)= Q( u, v),weseethat
− −
ζ (s)=2 ∞ (Q(m,0)) s+2 ∞ ∞ (Q(m,n)) s. (5)
Q − −
Xm=1 Xn=1mX=
−∞
Further
b 2 b2
Q(u,v)=a u+ v + c v2
a ! − a !
Kronecker’sLimitFormulas 5
b 2 d b+ √ d b √ d
=a u+ v + v2 =a u+ − v u+ − − v
a ! a  a  a 
=a(u+zv)(u+zv)   
wherearg(√ d)=π/2andz=(b+ √ d)/a= x+iywithy>0.
− −
Also,wecould,withoutlossofgenerality,supposethatd =1;ifQ (u,v)=
1
(1/√d)(au2+2buv+cv2) = a u2+2b uv+c v2,thenζ (s) = ds/2ζ (s)and
1 1 1 Q1 Q
a c b2 = 1. Thefunction ds/2 (choosing afixed branch) isasimpleentire
1 1 − 1
function of s and therefore, to study ζ (s), it is enough to consider ζ (s).
Q Q1
Thenwehavea=y 1andQ(u,v)=y 1(u+zv)(u+zv)=y 1u+zv2. Now(5)
− − −
| |
becomes
∞ ∞ ∞
ζ (s)=2ys m 2s+2ys m+nz 2s
Q − −
| |
Xm=1 Xn=1mX=
−∞
∞ ∞
=2ysζ(2s)+2ys m+nz 2s (6)
−
| |
Xn=1mX=
−∞
WeknowfromProposition1thatζ(2s)hasananalyticcontinuationinthe
half-plane σ > 0, regular except for a simple pole at s = 1/2. To obtain
an analytic continuation for ζ (s), we have, therefore, only to investigate the
Q
natureofthesecondtermontherighthandsideof (6),asafunctionof s. For
thispurpose,weneedthePoissonsummationformula.
Proposition 2. Let f(x) be continuous in ( , ) and let ∞ f(x+m) be
−∞ ∞ m=
uniformlyconvergentin0 x 1. Then P−∞
≤ ≤
∞
∞ f(x+m)= ∞ e 2πikx f(ξ)e2πikξdξ.
−
Z
mX= kX=
−∞ −∞ −∞
6
Proof. The function ϕ(x) = ∞ f(x+m) is continuous in ( , ) and pe-
m= −∞ ∞
riodicin x, ofperiod1. Ifa =P−∞1ϕ(ξ)e2πikξdξ then, byFe´je´r’sTheorem, the
k 0
R
(C,1)sumof ∞ a e 2πikxisequaltoϕ(x).Inparticular,if ∞ a converges,
k − k
k= k= | |
thenforxin( P−∞, ) P−∞
−∞ ∞
1
∞ f(x+m)=ϕ(x)= ∞ e 2πikx ϕ(ξ)e2πikξdξ.
−
Z
mX= kX= 0
−∞ −∞
Kronecker’sLimitFormulas 6
Inviewoftheuniformconvergenceof ∞ f(x+m)in0 x 1,
m= ≤ ≤
P−∞
1 1
∞ f(ξ+m)e2πikξdξ = ∞ f(ξ+m)e2πikξdξ
Z Z
0 mX= mX= 0
−∞ −∞
= ∞ f(ξ)e2πikξdξ.
Z
−∞
Thus
∞ f(x+m)= ∞ e 2πikx ∞ f(ξ)e2πikξdξ, (7)
−
Z
mX=−∞ kX=−∞ −∞
whichisthePoissonsummationformula.
Now we set f(x) = x + iy 2s = z 2s for x in ( , ). Then we see
− −
| | | | −∞ ∞
thattheseries ∞ m+z 2sconvergesabsolutely,uniformlyineveryinterval
−
m= | |
N x N,foPr−∞σ > 1/2. Forthispurpose,itclearlysufficestoconsiderthe
− ≤ ≤
interval0 x 1,sincetheseriesremainsunchangedwhen xisreplacesby
≤ ≤
x+1. For0 x 1,
≤ ≤
∞ m+z 2s ∞ m+z 2σ
(cid:12) − (cid:12) −
(cid:12)(cid:12)mX= | | (cid:12)(cid:12)≤mX= | |
(cid:12)(cid:12) −∞ (cid:12)(cid:12) −∞
(cid:12)(cid:12) z 2σ+ z 1(cid:12)(cid:12) 2σ+ ∞ (m 2σ+m 2σ)
− − − −
≤| | | − |
Xm=1
<2y 2σ+2 ∞ m 2σ σ> 1 .
− −
2!
Xm=1
(cid:3)
7
Thus,by(7),forσ>1/2,
∞ m+z 2s = ∞ e 2πikx ∞ ξ+iy 2se2πikξdξ
− − −
| | Z | |
mX=−∞ kX=−∞ −∞
= ∞ e 2πikx ∞(ξ2+y2) se2πikξdξ
− −
Z
kX=−∞ −∞
=y1 2s ∞ e 2πikx ∞(1+ξ2) se2πikξdξ, (8)
− − −
Z
kX=−∞ −∞
whenever the series (8) converges. Actually, we shall prove that it converges
absolutely for σ > 1/2. For this purpose, let us consider for k > 0, ∞(1+
R−∞