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and solutions of polynomial congruences
Dirk Segers
To cite this version:
Dirk Segers. Smallest poles of Igusa’s and topological zeta functions and solutions of polynomial
congruences. Mathematics [math]. Université Catholique de Louvain, 2004. English. �tel-00006134�
b
FaculteitWetenschappen
Departement Wiskunde
Smallest poles of Igusa's and
topological zeta functions and
solutions of polynomial
congruences
Dirk Segers
Promotor:
Prof.dr.WillemVeys
Proefschriftingediend tot het behalenvan degraadvan Doctorin deWetenschappen
Allereerstwilikmijnpromotor,Prof. dr. WillemVeys,bedankenomdat hij steeds wel tijd vond als ik vragen had. Hij heeft me een enorme vrijheid gegeven in mijn onderzoek en tegelijkertijd slaagde hij erin me waardevolleideeenaantereiken. Zijninteresseindevorderingenvanhet onderzoekhebbenmebovendiengemotiveerdomsteedsverdertezoeken. Wim,bedanktvoorde professioneleen aangename begeleiding.
Prof.dr.JanDenefwilikbedankenvoorzijnhulpenvoorzijnenorme interesse in mijn werk. Ik dank ook hartelijk de andere leden van de jury Prof. dr. Paul Igodt, Prof. dr. Pierrette Cassou-Nogues en Prof. dr.EnriqueArtalBartolo voorhetbeoordelenvande thesis.
Verder wil ik de collega-assistenten bedanken voor de ontspannende middagpauzes, de spelletjesavonden, de voetbal, enz. Ook wil ik hen bedankenvoordeinteressantegesprekkenendiscussiesoverhetonderzoek enwiskundeinhetalgemeenenvoordehulpdieikkreegalsdecomputer vervelend deed.
De collega's van Gent en Geel waarmee we ons onderzoeksseminarie organiseren verdienenookeen woordjevan dank. Onze samenwerking is echt eenmeerwaarde voorde afdeling.
Tenslotte wil ik nog alle andere mensen bedanken die me nauw aan hethart liggen, inhet bijzondermijnouders voor de morele steun,mijn familie,Katrienenmijnvrienden.
Dankwoord iii
Introduction vii
1 Preliminaries on zeta functions 1
1.1 Igusa'sp-adiczeta function . . . 1
1.2 Topologicaland motivic zetafunction . . . 13
1.3 Historicalnote . . . 19
2 The topological zeta function for curves and surfaces 21 2.1 Introduction. . . 21
2.2 Curves . . . 23
2.3 Surfaces . . . 32
2.3.1 On candidatepoles which arenotpoles . . . 33
2.3.2 Multiplicitytwo . . . 35
2.3.3 Multiplicitylargerthantwo . . . 37
2.4 Arbitrarydimension . . . 42
2.5 Otherzeta functions . . . 43
3 Igusa's p-adic zeta function 45 3.1 Introduction. . . 45
3.2 The toolforourvanishingresults . . . 48
3.3 The vanishingresults. . . 64
3.3.1 Curves. . . 64
3.3.2 Surfaces . . . 66
3.4 Determinationofthe smallestpoles . . . 71
3.4.1 Curves. . . 71
3.4.2 Surfaces . . . 77
4 Vanishing results with character 83 4.1 Charactersums . . . 83 4.2 The vanishingresult . . . 91
5 Congruences and lower bound in arbitrary dimension 103 5.1 Introduction. . . 103 5.2 A theoremon thenumberofsolutions of congruences. . . 106 5.3 The smallestpoles of Igusa'szetafunctionand congruences108
Bibliography 113
Nederlandse Samenvatting N1
N.1 Basisbegrippen en-eigenschappen. . . N1 N.1.1 Igusa's p-adische zetafunctie. . . N1 N.1.2 De lokale topologische zetafunctie. . . N3 N.2 Studiemet behulpvan ingebedderesoluties . . . N4 N.2.1 De lokale topologische zetafunctie. . . N4 N.2.2 Igusa's p-adische zetafunctie. . . N5 N.3 Studiemet behulpvan congruenties . . . N6
(0.1) Number theory and geometry are historically the rst elds of mathematical interest and have always fascinated a lot of people. My research ofthelastfewyearsledto newpropertiesinthesetwo domains.
(0.2)Oneof thebasisconcepts ofthisthesis ispolynomialcongruences. Let f 2 Z[x
1 ;:::;x
n
] be a polynomial over the integers in n variables. Put x = (x
1 ;:::;x
n
). We want to study the number of solutions of f(x) 0modm in (Z=mZ)
n
for an arbitrary positive integer m. The Chinese remainder theorem reduces this problem to the case that m is a power of a prime. Let p be a xed prime and let M
i
, i 2 Z 0
, be thenumberof solutionsof thecongruence f(x)0 modp
i in (Z=p i Z) n . If f has no singular point in Z
n p
, then the behaviour of the M i
is well understood forlarge ibecause of Hensels lemma. Here Z
p
is thering of p-adic integers. The behaviour of the M
i
is very complicated if f hasa singularpoint in Z
n p
. Igusa's p-adiczeta function allows usto study the singularcase.
(0.3)Forzinthe eldofp-adicnumbersQ p
,letordz2Z[f+1gdenote thevaluationofz,jzj=p
ordz
theabsolutevalueofzandacz=zp ordz the angular component of z. Let be a character of Z
p , i.e., a group homomorphism :Z p ;:!C
;:with niteimage. Letfbeapolynomial over Zormoregenerally overZ
p orQ p . Let X=Z n p
. Igusa'sp-adic zeta functionZ
f;
(s) off and is de nedbythep-adicintegral
Z f; (s):= Z X (acf(x))jf(x)j s jdxj
contin-Igusa'sp-adiczetafunctioncan be de nedina moregeneralcontext. LetK be ap-adic eld,i.e., anextensionofQ
p
of nitedegree. LetR be thevaluationringofK,P themaximalidealofR ,a xeduniformizing parameterforRandqthecardinalityoftheresidue eldR =P. Forz2K, letordz2Z[f+1gdenotethevaluationofz,jzj=q
ordz
theabsolute value of z and acz = z
ordz
the angular component of z. Let be a character of R
. LetX bean open andcompact subsetof K n
and let f be aK-analytic functiononX. NowIgusa'sp-adiczetafunctionZ
f; (s) of f and is de ned in the same way as before. Note that we have in our special case above that K = Q
p , R = Z p , P = pZ p , = p, q = p, X=Z n p and f 2Q p [x 1 ;:::;x n ].
Letf beapolynomialoverRor,moregenerally,anarbitraryrigid K-analytic functionon X =R
n
de nedover R , i.e., a K-analytic function on R
n
which is given by a power series over R which converges on R n
. Igusa's p-adic zeta function of such an f has an important connection with congruences. Fori2 Z
0
and u2 R =P i
, let M i
(u) be thenumber ofsolutions of f(x)umodP i in(R =P i ) n . PutM i :=M i (0). Let ebe the conductor of , i.e., the smallest a 2 Z
>0
such that is trivial on 1+P a . ThentheM i+e ( i u), u2(R =P e ) ,describe Z f; (s)throughthe relation Z f; (s)= 1 X i=0 X u2(R=P e ) (u)M i+e ( i u)q n(i+e) q is :
If is the trivial character, all the M i
's describe and are described by Z
f;
(s) throughtherelation
Z f; (s)=P(q s ) P(q s ) 1 q s ;
wherethePoincareseriesP(t)of f is de nedby
P(t)= 1 X i=0 M i (q n t) i :
Using this connection, we can obtain properties of Igusa's p-adic zeta functionfromresultsonthenumberofsolutionsofcongruencesand vice versa.
ThesecondmethodinourstudyofZ f;
(s)usestheexistenceofan em-beddedresolutionofsingularitiesoff,whichisacompositionof blowing-ups. Wewillcalculatethede ningintegralofIgusa'sp-adiczetafunction
(0.4) The study of Igusa's p-adic zeta function by using an embedded resolution of singularities led to the introduction of a new geometric in-variant.
Let f be thegerm of a holomorphicfunction on a neighbourhoodof the origin 0 in C
n
which satis es f(0)= 0 and which is not identically zero. Let g : V ! U C
n
be an embedded resolution of singularities of a representative of f 1 f0g. We denote by E i , i2 T, the irreducible components of g 1 (f 1 f0g), and by N i and i 1 the multiplicitiesof f Æg and g (dx 1 ^^dx n ) along E i . The (N i ; i ), i 2 T, are called the numerical data of the embedded resolution. For I T denote also E I :=\ i2I E i and Æ E I :=E I n([ j2I= E j
). Note thatV isthedisjoint union
ofthe Æ E I ,I T. Letd2Z >0 .
The localtopologicalzetafunctionof f and d isde ned as
Z (d) f (s):= X IT 8i2I:djN i ( Æ E I \g 1 f0g) Y i2I 1 i +sN i :
Here s is a complex variable and () denotes the topological Euler-Poincare characteristic. The candidate poles of this rational function are the i =N i , with i 2 T satisfying d j N i
. It is striking that most candidatepoles areactuallynotpoles.
In ourresearch,we willalways take an embeddedresolution g which isacompositiong 1 ÆÆg t ofblowing-upsg i
withanappropriatecentre. Theexceptional varietyofg
i
and thestricttransforms ofthisvarietyare denoted by E
i
. We will use the same notation if we have an embedded resolutioninthe context of Igusa'sp-adiczeta function.
(0.5) The rst chapter is an introduction to these zeta functions. It is writtenforpeoplewhoarenotfamiliarwiththissubject,butwhohavea mathematicalbackground. Theotherchapterscontainmyresearch. They areall self contained,except Chapter 4. Chapters2,3 and 5 correspond to respectively[SV], [Se1] and[Se2].
(0.6)Chapter2containsourstudyofthelocaltopologicalzetafunction. We will seethat it is easy to prove that forevery n2Z
2
,all the poles ofa local topologicalzetafunctionare intheinterval[ n+1;0[.
form 1=2 1=iwithi2Z >1
,andall thesevaluesarethepoleofalocal topologicalzetafunctionwithd=1. Moreover,ifthemultiplicityoff at theorigin02C
2
isat leastfour, wewillsee that 1is theonlypossible pole lessthan 1=2. Our resultimpliesthewellknown factthatthe log canonical thresholdoff is never in]5=6;1[.
If n=3 and d=1,wedetermine allvalueslessthen 1which occur as a pole of a local topological zeta function. They are 1 1=i with i2 Z
>1
. In particular, there are no poles less than 3=2. Moreover, if themultiplicityof f is at least three,we willseethat there are no poles lessthan 1. ThelastpartwillbethemostdiÆcultbecausewewillhave to prove thatsome candidate polesare notpoles.
(0.7) In Chapter 3 we study Igusa's p-adic zeta function by using the methodof embeddedresolutionof singularities. We obtainresultswhich areanalogousto theones ofChapter 2,butnow forthereal partsof the poles of Igusa'sp-adiczeta function.
We willhave to deal with an extra diÆcultyin theproof of the fact thatsome candidate poles of expected order one arenotpoles.
Forthelocaltopologicalzetafunction,wecouldusetherelationsbetween thenumericaldataofVeys. Ifs
0
isacandidatepoleofZ (d) f
(s)ofexpected order one and ifE
i
isan exceptional varietywithcandidate poles 0
,one can write down easily the contribution of E
i
to the residue of Z (d) f
(s) at the candidate pole s
0
. The relations of Veys allow you to prove that thiscontributioniszeroforsomeintersectioncon gurationsonE
i at the stage whenit iscreated.
For Igusa's p-adic zeta function, we also want to be able to calculate the contribution of an exceptional variety E
i
with a candidate pole s 0 of expected order one to the residue of Z
f;
(s) at s 0
by using only our information when E i is created. Let s 0 be a candidate pole of Z f; (s) of expected order one and let E
i
be an exceptional variety with candi-date pole s
0
. Then it is generally known that the contribution of E i to theresidue ofZ f; (s) at s 0
isequal to theprincipalvalue integralof the Poincare residue of (acf Æg)jf Ægj
s0 g
dx on E i
. We deduce this for-mulaagainwithoutusingtheterminologyofprincipalvalueintegralsand Poincare residues. Our deduction is elementary. We will push forward this formula to the stage when E
i
is created. In the specialized termi-nology,our resultis thatthe contribution of E
i
to theresidueof Z f;
(s) at s is equal to the principal value integral of the Poincare residue of
(acfÆg 1 ÆÆg i )jfÆg 1 ÆÆg i j s 0 (g 1 ÆÆg i ) dx onE i .
Chapter 4 contains the calculations which are needed to obtain our result for Igusa's p-adic zeta functions with (non-trivial) character and n=3.
(0.8) Chapter 5 contains results in general dimension n. Let f be a polynomial over R ora rigid K-analytic functionon R
n
de ned overR . Asin(0.3), let M
i
(u)be thenumberof solutionsof f(x)u modP i in (R =P i ) n and putM i := M i
(0). Let l be thesmallest real part of a pole ofIgusa'sp-adic zetafunctionof f withtrivialcharacter. We prove that there exists an integer a which is independent of i such that M
i is an integer multiple of q p(n+l )i aq for all i 2 Z 0
. We explained this result also inChapter3.
TogetherwithourresultsofChapter3indimensiontwoandthree,we obtain some nice divisibilityproperties. The statements of these prop-erties are so easy that we tried to nd an elementary proof, and with success. Itgeneralizedeasilyto arbitrarydimensionandtothemore gen-eral class of numbers M
i
(u). More precisely, we willdeduce that M i (u) isdivisiblebyq p(n=2)(i 1)q foralli2Z >0 and all u2R =P i .
We will also prove the converse of the result two paragraphs ago: if thereexistsanintegerasuchthatM
i isanintegermultipleofq p(n+l 0 )i aq for all i 2 Z 0 , then l 0
l. This statement has an analogue if we are dealingwitha character.
Let beanarbitrarycharacterofR
. Thelasttwoparagraphsimply thatZ
f;
(s)hasnopoleswithrealpartlessthan n=2. Thisisthemain result of thelast chapter. We willsee that it holdsfor an arbitrary K-analyticfunctiononanopenandcompactsubsetofK
n
,andnotonlyfor rigid K-analytic functions on R
n
which are de ned over R . Moreover, thisresultimpliesthat alocaltopologicalzetafunctionhasnopoles less than n=2. We also give an example which shows that this bound is optimal.
Preliminaries on zeta
functions
Abstract
Wegiveanintroductiontothreezetafunctions: Igusa'sp-adic zeta function, the topological zeta function and the motivic zetafunction. The rstone belongsto thedomainofnumber theory, the others are geometric invariants. We also try to explaintheconnectionbetweenthese zetafunctions.
1.1 Igusa's p-adic zeta function
(1.1.1) Let K be a p-adic eld,i.e., an extension of Q p
of nitedegree. Let R be the valuation ring of K, P the maximalideal of R , a xed uniformizing parameter for R and q the cardinality of the residue eld R =P. For z 2 K, let ordz 2 Z[f+1g denote the valuation of z and jzj = q
ordz
the absolute value of z. Let f 2 R [x 1 ;:::;x n ] and put x=(x 1 ;:::;x n ). (1.1.2)Let M i ,i2Z 0
,be thenumberof solutionsof f(x)0 mod P i in (R =P i ) n . Note that M 0 = 1. If K = Q p and f 2 Z[x 1 ;:::;x n ] or, more generally, f 2 Z p [x 1 ;:::;x n
], it is just the number of solutions of
The Poincare seriesP(t)of f is theformalpowerseries P(t)= 1 X i=0 M i (q n t) i :
Ifweconsidert asa complexvariable,it isabsolutelyconvergent on the openunit discftjjtj<1gC because M
i q ni . (1.1.3)AsubsetXofK n
iscalledopenifforeveryx2X,thereexistsan i2Z 0 suchthatx+(P i ) n X. Inthisway,K n becomesa topological space. Forn=1,it isthetopology onK inducedbytheabsolute value, and forn>1,it isthe producttopologyon K
n
=KK. With this topology K
n
;+ is a locally compact group. Therefore, thereexistsauniquetranslationinvariantmeasure=jdxjonthe Borel--algebra of K
n
for which (R n
) = 1. This measure is called the Haar measure on K
n
. Integration of a complex valued function on K n
with respectto theHaarmeasure is calledp-adicintegration.
Let i 2 Z >0
. Then R n
is the disjoint union of q ni cosets of (P i ) n . Because the Haar measure is translation invariant, the measure of each of these cosets is the same. Hence, we obtain ((P
i ) n ) = q ni because (R n
)=1 and because isadditive.
(1.1.4) Igusa's p-adic zeta function Z f
(s) of f is de ned by a p-adic integralforcomplex numberss satisfyingRe (s)>0:
Z f (s)= Z R n jf(x)j s jdxj:
Theintegrandisthesthpoweroftheabsolute valueof f(x). Wewillsee thatZ
f
(s) isanalytic and thatit extends to a meromorphicfunctionon C,butuntilthereZ
f
(s) isjustacomplexfunctionontherighthandside ofthe complexplane.
(1.1.5) Because fx2R n jordf(x)=ig =fx2R n jf(x)0modP i gnfx2R n jf(x)0modP i+1 g = [ a+(P i ) n 2(R=P i ) n f(a)0modP i a+(P i ) n n [ a+(P i+1 ) n 2(R=P i+1 ) n f(a)0modP i+1 a+(P i+1 ) n ;
whereevery union isa unionof disjointsets, we obtainthat
(fx2R n jordf(x)=ig)=Mq ni M q n(i+1) :
Moreover, itfollowsfromthiscalculationthatjf(x)j isalocallyconstant function outside the zero locus of f. Consequently, jf(x)j
s
is a locally constant function, and thus continuous, outside the zero locus of f for every complex number s. One checks easily that jf(x)j
s
is continuous everywhere ifRe (s)>0.
Sincethede ningintegralofIgusa'sp-adiczetafunctionisanintegral ofacontinuousfunctiononacompactset,itismeaningful. Actually,this isalso clearbylookingat thefollowingexplicitcalculation of Z
f (s): Z f (s) = Z R n jf(x)j s jdxj = 1 X i=0 (fx2R n jordf(x)=ig)q is = 1 X i=0 (M i q ni M i+1 q n(i+1) )q is = 1 X i=0 M i q ni q is P 1 i=0 M i+1 q n(i+1) q (i+1)s q s = P(q s ) P(q s ) 1 q s : (1.1) Note that q s
is an analyticfunctionof s which maps a point of the right hand side of the complex plane to a point of the open unit disc. Moreover, P(t) isan analytic functionon theopen unitdisc. Therefore, weobtainthatZ
f
(s)isananalyticfunctionon therighthandsideofthe complexplane.
It is clear from equation (1.1) that Z f (s) depends only on q s , i.e., Z f (s 1 )=Z f (s 2 )ifq s 1 =q s 2
. Onecanobservethisfactalsodirectlyby lookingat thede ningintegral. We have thatq
s 1 =q s 2 ifand onlyif s 1 =s 2 +2k p
1=logq forsomek 2Z. Consequently,Z f
(s)isperiodic withperiod2
p
1=logq.
Lett beanothercomplex variable. Because themap':fsjRe (s)> 0g ! ft j 0 < jtj < 1g : s 7! t = q
s
is surjective and because Z f
(s) depends only on q
s
, there exists a unique function Z f (t) on ft j 0 < jtj<1gsuchthat Z f (s)=Z f
(t)Æ'. It follows fromequation (1.1) that
Z f
(t)=P(t)
P(t) 1 :
Consequently, also Z f
(t) is an analytic function. Because the constant termofP(t)is1,wecanextendZ
f
(t)toananalyticfunctiononftjjtj< 1g. Ifwe solvethe previousequation to P(t), we obtain
P(t)= 1 tZ f (t) 1 t : (1.1.6) Fix a = (a 1 ;:::;a n ) 2 R n . Let M 0 i
be the number of solutions of f(x) 0modP i in (R =P i ) n
which are equal to a modulo P n = P P. PutQ(t)= P 1 i=1 M 0 i (q n t) i
. Oneobtainsanalogously asin (1.1.5) that Z a+P n jf(x)j s jdxj=Q(t) Q(t) q n t t ; whereof courset=q s .
Ournextgoalistocalculatethisintegralundertheconditionsf(a) 0modP and(@f=@x
j
)(a)60 mod P forat leastone j. We willdo this bycalculatingthenumbersM
0 i
. Inthenextparagraph,wewillshowthat M
0 i
=q
(i 1)(n 1)
. This isactuallyHensel's lemmainseveral variables. Fix a j 2 f1;:::;ng for which (@f=@x
j
)(a) 6 0 mod P. Fix also b 1 ;:::; b j 1 ;b j+1 ;:::;b n 2 R satisfying a l b l
modP for every l 2 f1;:::;j 1;j+1;:::;ng. We want to count the number of solutions y2R =P i 1 of g(y):=f(b 1 ;:::;b j 1 ;a j +y;b j+1 ;:::;b n )0 mod P i :
Theconstanttermofgasapolynomialinyisf(b 1 ;:::;b j 1 ;a j ;b j+1 ;:::; b n
)andthisiscongruenttof(a)andthusto0modP. ThecoeÆcientofy ingisequalto(@f=@x j )(b 1 ;:::;b j 1 ;a j ;b j+1 ;:::;b n ). Thisis0 mod P and di erent from 0 mod P
2 because it is (@f=@x j )(a)modP 2 . The coeÆcient of every y k is of coursein P k
. Puth(y)=g(y)=. We divide the previous congruence by , so we obtain that we have to count the numberof solutionsy 2R =P i 1 of h(y)0 modP i 1 :
Note that h is a polynomial in which the coeÆcient of y is not in P and in which the coeÆcients in terms of higher degree are in P. First, we determine the solutions of h(y) 0modP in R =P. Because this is actually a linear congruence, it has one solution c +P. Next, we
determinethesolutionsofh(y)0 mod P 2
inR =P 2
. Therefore,wehave to determine the solutions of h(c
1 +y 2 ) 0 modP 2 in R =P, which is actually a linear congruence after we have divided by . So we obtain that there is one solution c
2 +P 2 of h(y) 0 mod P 2 . Thereafter, we determinethesolutionsofh(y)0 mod P
3
inR =P 3
. Therefore,wehave to determinethe solutionsof h(c
2 + 2 y 3 )0 mod P 3 inR =P,which is actuallya linear congruence afterwe have dividedby
2
. Ifwe continue in this way, we obtain nally that h(y) 0 mod P
i 1
, and thus also g(y) 0 modP
i
, has one solution in R =P i 1 . Because we have q i 1 choices modulo P i for every b l , with l 2 f1;:::;j 1;j+1;:::;ng, we obtainthatM 0 i =q (i 1)(n 1) . Because we know theM
0 i
,we areable to calculateQ(t):
Q(t) = 1 X i=1 M 0 i q ni t i = q n+1 1 X i=1 q i t i = q n+1 1 1 q 1 t 1 = q n t 1 q 1 t : Consequently, Z a+P n jf(x)j s jdxj = Q(t) Q(t) q n t t = q n t 1 q 1 t q n 1 t 1 q 1 t = q n q 1 q t 1 q 1 t :
(1.1.7)Wehaveobtainedeverythingtowritedownthep-adicstationary phaseformula[Ig7 , Theorem10.2.1]. Itis actuallythetrivialequality
Z R n jf(x)j s jdxj= X a+P n 2(R=P) n Z a+P n jf(x)j s jdxj;
inwhich we replacetheintegrals thatwe know by now. Theseare Z n jf(x)j s jdxj=q n
iff(a)60modP and Z a+P n jf(x)j s jdxj=q n q 1 q t 1 q 1 t
if f(a) 0 mod P and (@f=@x j
)(a) 6 0modP for at least one j 2 f1;:::;ng. If f(a) 0 mod P and (@f=@x
j
)(a) 0 mod P for all j, thentheintegralstays unchangedinthe formula.
The reductionof f moduloP,whichis denoted byf,is obtainedby reducingallthecoeÆcientsoff moduloP and isthusa polynomialover the nite eld F
q
. Note that f(a)0 mod P ifand only iff(a) =0 in F
q
and that (@f=@x j
)(a) 0 mod P if and only if (@f=@x j )(a ) = 0 in F q . Here, a2F n q
is obtainedby reducing every component of amodulo P. Note alsothata singularpointofa polynomialisbyde nitionazero of thepolynomialand of all its partialderivatives ofthe rst order. For example, every pointis asingularpointof thezero polynomial.
The p-adic stationary phase formula can sometimes be used to cal-culate Igusa's p-adic zeta function. I illustrate this with the polyno-mial f = x 1 x 2 +x 2 3
. We rst calculate the number of solutions of f = x 1 x 2 +x 2 3
= 0 over the nite eld F q . If x 2 6= 0, this is a linear equation in x 1 for every x 3 2 F q
. Such an equation has one solution. There are thus q(q 1) solutions for which x
2 6= 0. If x 2 = 0, then x 3 =0 and x 1 is an arbitrary element of F q
. Thereare thus q solutions for which x 2 =0. We obtain that x 1 x 2 +x 2 3 =0 has q(q 1)+q = q 2 solutionsoverF q
. Thereisonlyone singularpoint ofx 1 x 2 +x 2 3 =0 over F q
. Indeed,thesystemofequationsf =x 1 x 2 +x 2 3 =0,@f=@x 1 =x 2 =0, @f=@x 2 = x 1 = 0 and @f=@x 3 = 2x 3
= 0 only has (0;0;0) as solution overF
q
. Notethatthisisalsothecaseifweworkincharacteristic 2. The p-adicstationary phaseformulastates that
Z R 3 jx 1 x 2 +x 2 3 j s jdxj = (q 3 q 2 )q 3 +(q 2 1)q 3 q 1 q t 1 q 1 t + Z P 3 jx 1 x 2 +x 2 3 j s jdxj:
If we apply the coordinate transformation R 3 ! P 3 : (x 1 ;x 2 ;x 3 ) 7! (x 1 ;x 2 ;x 3
) to thelastintegral, we obtain Z P 3 jx 1 x 2 +x 2 3 j s jdxj = Z R 3 jx 1 x 2 + 2 x 2 3 j s q 3 jdxj = q 3 t 2 Z 3 jx 1 x 2 +x 2 3 j s jdxj:
The integralwewant to calculate appearsin thelastexpression. Conse-quently,we geta functionalequation. Solvingthisgivesus
Z R 3 jx 1 x 2 +x 2 3 j s jdxj= q 1 q 1 q 3 t (1 q 1 t)(1 q 3 t 2 ) :
In more diÆcult examples, one applies the p-adic stationary phase formulaseveraltimesuntilonegetsasystemofequationsinseveral unde-terminedfunctionsfromwhichIgusa'sp-adiczetafunctioncanbesolved. It isan open problemwhether thisisalwayspossible.
(1.1.8) Now I explain how Hoornaert obtained in her Ph.D. thesis (see also[DH ])aformulaforIgusa'sp-adiczetafunctionofpolynomialswhich are non-degenerated over F
q
with respect to their Newton polyhedron. Thisis insome sense avery large classof polynomialsand consequently an interesting class for checking conjectures. It was also useful in my research: at some pointsitmade meclearwhat Icould expectand what certainly not. Indeed, theformula isinterms of theNewtonpolyhedron of f and is therefore very interesting to construct examples, which can be calculatedexplicitlybyusingtheformulaorthecomputerprogramof Hoornaertand Loots[HL]based on it.
Letf(x)= P !2Z n 0 a ! x ! = P !2Z n 0 a ! x !1 1 x ! n n beanon-zero poly-nomial over R with f(0) =0. Here Z
n 0
is theabbreviated notation for (Z 0 ) n . Let supp(f) = f! 2 Z n 0 j a !
6= 0g. The Newton polyhedron (f)off isde nedastheconvexhullinR
n 0 oftheset[ !2supp(f) !+R n 0 . Theintersectionof (f) witha supportinghyperplaneiscalled afaceof (f) and a facet is a face of dimension n 1. If is a face of (f), we de ne f (x) = P !2 a ! x !
. We say that f is non-degenerated over F
q
if not any of the polynomials f and f
, with a face of (f), has a singularityin(F q ) n .
Let be a face of (f). Let 1
;:::; i
be the facets of (f) which contain . Let v 1 ;:::;v i be the vectors in Z n 0 nf(0;:::;0)g of minimal length which are perpendicular to respectively
1 ;:::; i . Then = f 1 v 1 ++ i v i j i 2 R >0
g is called the cone associated to . One provesthat R
n 0
nf(0;:::;0)g is thedisjointunion ofthe . For x =(x 1 ;:::;x n ) 2 R n , we denote (ordx 1 ;:::;ordx n ) simply by
Z f (s) = Z R n jf(x)j s jdxj = X k2Z n 0 Z x2R n ordx=k jf(x)j s jdxj = Z (R ) n jf(x)j s jdxj+ X X k2Z n 0 \ Z x2R n ordx=k jf(x)j s jdxj: Let M (f) (resp. M
) be the numberof solutions of f (resp. f ) in (F q ) n .
Becausef isnon-degeneratedoverF q
wehave thatf hasno singular-ities in(F q ) n ,and consequently Z (R ) n jf(x)j s jdxj = ((q 1) n M (f) )q n +M (f) q n q 1 q t 1 q 1 t = q 1 q n M (f) q n 1 t 1 q 1 t :
Let beafaceof (f). Letk2Z n 0
\
. Oneprovesthatthefunction : (f) ! R : x = (x 1 ;:::;x n ) 7! k x = k 1 x 1 ++k n x n reaches it minimum exactly on the points of . This minimum is denoted by m(k). Put(k) =k
1
++k n
. Weapplythecoordinatetransformation ' : (u 1 ;:::;u n ) 7! ( k 1 u 1 ;:::; k n u n ). Then ' jdxj = q (k) jduj and ' 1 (fx 2 R n j ordx = kg) = (R ) n . Because ' (x ! ) = k! u ! and because reaches its minimum m(k) exactly on the points of ,we can write' (f(x))= m(k) (f (u)+f ;k (u)),wheref ;k isapolynomialover R . Becausef isnon-degenerated overF
q
we have thatf
,and thusalso f +f ;k ,hasno singularitiesin(F q ) n . Therefore, Z x2R n ordx=k jf(x)j s jdxj = Z (R ) n j m(k) (f (u)+f ;k (u))j s q (k) jduj = q (k) t m(k) Z (R ) n jf (u)+f ;k (u)j s jduj = q (k) t m(k) q 1 q n M q n 1 t 1 q 1 t :
The part between the brackets does not depend on the choice of k 2 Z
n 0
\
. We have thusthat X k2Z n \ Z x2R n ordx=k jf(x)j s jdxj
isequal to 0 @ X k2Z n 0 \ q (k) t m(k) 1 A q 1 q n M q n 1 t 1 q 1 t :
Onecan provethat X k2Z n 0 \ q (k) t m(k) = A(t) (1 q (v 1 ) t m(v 1 ) )(1 q (v i ) t m(v i ) ) ;
whereA(t)is apolynomialwhich can be calculatedexplicitlyand where v
1 ;:::;v
i
arethevectorsinZ n 0
nf(0;:::;0)gofminimallengthwhichare perpendicularto thefacets thatcontain .
The formula we obtained has important consequences. If f is non-degenerated overF
q
, it follows from theformula that Z f
(s) has a mero-morphiccontinuation to C. Moreover, we can read the candidate poles ofthismeromorphiccontinuationfromtheNewtonpolyhedronoff. The real part of a candidate pole is 1 or of the form (v)=m(v) with v a vector which is perpendicular to a facet of (f). One checks directly that (v)=m(v) = 1=t
0
,wheret 0
istheintersection ofthe supporting hyperplaneof withthediagonal f(t;:::;t)jt2Rg of R
n .
(1.1.9) AllIgusa's p-adiczeta functions we have met sofar arerational functions of t. This is true ingeneral. We now give the ideabehind the proof. Formore details, see[Ig7 ] orChapter 3.
To calculate integrals, one often tries to nd a coordinate transfor-mationsuch thattheintegral canbe calculatedeasilyinthe new coordi-nates. Here, the appropriate coordinates lie on an embedded resolution of (f;dx), which always exists by Hironaka's theorem [Hi]. Hironaka ac-tuallyproved that there existsa compositionof blowing-upswhich isan embeddedresolution of(f;dx).
Let g : Y ! R n
be an embeddedresolution of (f;dx). Note that Y is a K-analytic manifold. Let b be an arbitrary point of Y. Then there exist analytic coordinates y = (y
1 ;:::;y
n
) on a neighbourhood V of b, non-vanishingK-analyticfunctions"; onV andintegersN
i 0,
i 1 forevery i2f1;:::;ng such thaton V
f Æg=" n Y y N i i and g dx= n Y y i 1 i dy:
Because j"j and jj are locally constant functions, we can nd a neigh-bourhoodof bon which theintegral
Z Y jfÆgj s jg dxj isof theform j"(b)j s j(b)j Z (P k ) n n Y i=1 jy i j N i s+ i 1 jdyj: Note that Z f
(s) is a nite sum of such integrals because Y is compact. Becausethevariablesareseparated,theintegralistheproductofintegrals inone variablewhichcan becalculated withoutaproblem:
Z P k jy i j N i s+ i 1 jdy i j= q 1 q q k(Nis+i) 1 q N i s i :
We have provedthat Z f
(t)is arationalfunctionof tof theform
Z f (t)= A(t) Q j2J (1 q j t N j ) ;
whereA(t)is apolynomial. Indeed, apossiblepowerof tinthe denomi-natorcanbecancelledbecausewesawin(1.1.5)thatZ
f
(t)isananalytic functionon the open unit disc. Because P(t)=(1 tZ
f
(t))=(1 t), we obtainthatalso P(t)isa rationalfunctionof t.
SinceZ f
(t)isarationalfunctionoft,we have thatZ f
(s)hasa mero-morphic continuation to C. From now on, Igusa's p-adic zeta function Z
f
(s)off isthemeromorphiccontinuationofthefunctionweconsidered before. Note that some facts of (1.1.5) such as the periodicityof Z
f (s) and the relation between Z
f
(t) and P(t) extend to this larger context. Thecandidate poles ofZ
f
(s)are oftheform
j N j + 2k p 1 N j logq ;
withk 2Zandj2J. Someofthesecandidatepolesarepolesandothers aren'tbecause theycancel withA(q
s ).
(1.1.10) Let us now formulate a theorem of Denef [De1]. Let F be a number eld. LetO betheringofintegersinF. Letf 2O [x ;:::;x ].
We considerf as a polynomialfunction on X = (F algcl ) n . Let g :Y XP m (F algcl
)!X bean algebraic embeddedresolution of f which is de nedoverF.
Ap-adiccompletionofF isacompletionwithrespecttoaprimeideal of O
F
. If K is an extension of nitedegree of a p-adiccompletion, then the embedded resolution g induces an embedded resolution g
K : Y K K n P m (K)!K n
of(f;dx)byusingthesameequationsforY K
andby usingthesame ruleforg
K .
For almost all p-adiccompletions ofF we have the following. LetK be an extension of nite degree of this p-adic completion. Let R be the valuation ring of K, P themaximal ideal of R and K
= F q the residue eld R =P. Let g K : Y K ! K n
be the embedded resolution induced by g. LetE
i ,i2T
K
,be theK-irreduciblecomponentsof g 1 K (f 1 f0g). Let N i be themultiplicityof E i inthedivisorof fÆg K on Y K and let i 1 be themultiplicityofE i inthedivisorofg K dxonY K . LetY K (resp. E i , i2T) denote thereductionmoduloP ofY
K (resp. E i ). Then Z f (s)=q n X IT K C I Y i2I q 1 q i +sN i 1 ; whereC I =cardfa2Y K (K)ja2E i
ifand onlyifi2Ig. Remark that Z
f
(s) dependson K.
Denef actually proves that this holds for a p-adic completion with respect to the prime ideal if g : Y ! X has good reduction modulo . Remark also that the formula can be formulated directly interms of g :Y ! X by using the F-irreducible components of g
1 (f
1
f0g) and by usingreductionmodulo.
(1.1.11)We willalso considersome generalizationsofthe previous de -nitionof Igusa'sp-adiczeta function.
Let f 2 K[x 1 ;:::;x n ]. Let be a character of R , i.e., a group homomorphism :R ;:!C
;:with niteimage. Putformally (0)= 0. For z2K,let acz=z
ordz
bethe angularcomponent ofz. Igusa's p-adiczeta functionZ
f;
(s) off and isde ned by thep-adicintegral
Z f; (s)= Z R n (acf(x))jf(x)j s jdxj
Moreover, if f is an arbitrary K-analytic function on an open and compactsubset X ofK
n
,thenthemeromorphic continuationof
Z f; (s)= Z X (acf(x))jf(x)j s jdxj;
whereRe(s)>0,isalso calledIgusa's p-adiczeta functionoff and .
(1.1.12)Themonodromyconjectureisoneofthemostimportantreasons for thestudy of the poles of Igusa's p-adic zeta function. For a polyno-mialf overa number eld,this conjectureof Igusaassertsa connection betweenthepolesofIgusa'sp-adiczeta functionandthetopologyof the map f : (C
n ;f
1
f0g) ! (C;0). Moreover, the striking fact that most candidate poles of Z
f;
(s) are actually not poles would be elucidated if thismonodromyconjecture istrue. We now state theconjecture.
LetF bea number eld. Letf 2F[x 1
;:::;x n
]. Foralmostall p-adic completionsK ofF wehave thefollowingforanycharacter ofR
. Ifs isa poleof Z f; (s), thenexp(2 p
1Re(s)) isan eigenvalueof thelocal monodromyof f at some complexpoint of f
1 f0g.
(1.1.13)Finally,I wantto enumeratesome di erent methodswhichcan be usedto studyIgusa'sp-adiczeta function. Some ofthem can be used toobtainageneralstatement,othersstudyaspecialclassoffunctionsf.
1. One can use the relationbetween Z f;
(s) and thenumber of solu-tionsofcongruences. Resultsforthenumberofsolutionsof congru-encescan have implicationsforZ
f; (s).
2. One can usean embedded resolution of (f;dx). Especially embed-ded resolutions which are a composition of blowing-ups with well chosen centra can be very useful.
3. Onecan usetheformulaofDenef to obtainresultsforpolynomials withgood reductionmoduloP.
4. One can use the formula of Hoornaert to obtain results for poly-nomials which are non-degenerated over F
q
with respect to their Newtonpolyhedron.
5. Onecanstudytheclassofpolynomialswhicharearelativeinvariant ofa prehomogeneous vector space.
6. Onecan usep-adiccelldecomposition. Denef usedthismethodfor exampleto obtainanotherproof oftherationalityof Z (t).
1.2 Topological and motivic zeta function
(1.2.1) Let f :(C n
;0) ! (C;0 ) be the germ of a holomorphicfunction (resp. f 2C[ x
1 ;:::;x
n
]). We willalways assumethatf isdi erentfrom 0. Let g :Y ! X C
n
(resp. g :Y ! C n
) be an embeddedresolution ofsingularities ofa representative f :X !C (resp. of f :C
n
! C) . We denote by E
i
,i2T,the irreduciblecomponents of g 1 (f 1 f0g), and by N i and i
1themultiplicitiesof fÆg and g dxalong E i . The (N i ; i ), i 2 T, are called the numerical data of the embedded resolution. For I T, we denote E I :=\ i2I E i and Æ E I :=E I n([ j2I= E j ). Notethat Y is
thedisjointunion ofthe Æ E I with I T. (1.2.2) To the germ f :(C n ;0) ! (C;0) of a holomorphic function we associate thelocaltopologicalzetafunction
Z top ;0;f (s):= X IT ( Æ E I \g 1 f0g) Y i2I 1 i +sN i : Toapolynomialf 2C[x 1 ;:::;x n
]we associate theglobaltopological zetafunction Z top ;gl ;f (s):= X IT ( Æ E I ) Y i2I 1 i +sN i :
Inbothcases,sisacomplexvariableand ()denotesthetopological Euler-Poincare characteristic. Of course, one has to prove that the ex-pressionson the right handsidedo notdependon thechosen embedded resolution. Thisremarkablefactwas rstprovedasfollowsbyDenefand Loeserin[DL1 ]. Forapolynomialf overtheringofintegersO
F
ofa num-ber eld F, take a completion K of F such that the formulain (1.1.10) holdsforall extensionsofthiscompletion. Usingp-adicinterpolation,we can considerthe global topological zeta function as the limit for e ! 0 ofIgusa's p-adiczeta functionsof f overcertain non-rami edextensions ofdegree eoverK. Onehasto useacohomologicalinterpretationof the numbersC
I
,whichisprovidedbyGrothendieck'straceformula. Totreat theother cases,one has to domore.
Notethatonecan alsode neatwistedversionofthelocalandglobal topologicalzetafunctionforevery d2Z
2
. The de nitionis exactlythe same asabove, exceptforthefactthat thesumruns onlyover subsetsI ofT forwhich djN for every i2I.
(1.2.3) We are interested inthe poles of these zeta functions. The can-didate poles are the
i =N
i
, i 2 T. Some of them are poles, others are not. We are going to discuss the case n = 2. Let E
i
, i 2 T, be an exceptional curve. We have thus E
i = P 1 . Let E j , j 2 J, be the irreducible components of g 1 (f 1
f0g) which are di erent from E i and which intersect E i . Suppose that j =N j 6= i =N i ,which isequivalent to j := j ( i =N i )N j
6=0, forevery j 2J. Actually,we assume thus thatthesumofthetermsinthede nitionofthetopologicalzetafunction forwhichi2I hasa candidate pole
i =N
i
of expectedorder one. The contribution R of E
i
to the residue at the candidate pole i
=N i
is by de nitiontheresidueofthissumat
i =N
i
. NotethatR=0ifandonly ifthissumhasnot apolein
i =N
i .
Everything we write down from now on is for the local topological zeta functionZ
top ;0;f
(s) associatedto the germ f :(C 2
;0)! (C;0 ) of a holomorphicfunction. Oneobtains easilythat
R= 1 N i 0 @ ( Æ E fig )+ X j2J 1 j 1 A : (1.2)
Wewillprovethatthisisequalto zeroifJ containsone ortwo elements. Therefore, we needarelationbetweenthe
j
,j2J.
WegivehereadeductionofVeyswhichusesintersectiontheory. Note thatwe mayassumethatf isapolynomialbecausethere existsadegree from which the termsdo notin uence theembedded resolution and the numerical data. Because div(f Æg) =
P j2T N j E j is a principal divisor, we obtain that ( P j2T N j E j )E i =0. Because E j E i = 1 ifj 2J and E j E i =0 ifj62J[fig, we get X j2J N j = N i E 2 i : (1.3) Here,E 2 i
istheselfintersectionofE i onY. Becausediv(g dx)= P j2T ( j 1)E j
is the canonical divisor, we obtain from the adjunction formula that( P j2T ( j 1)E j )E i = 2 E 2 i . Hence, we get X j2J ( j 1)+2= i E 2 i : (1.4)
the j : X j2J ( j 1)+2=0: (1.5)
WearenowabletoprovethatR=0ifJ containsoneortwoelements. If J =f1g containsone element, then it follows from relation(1.5) that
1
= 1. We obtainfrom (1.2) that
R= 1 N i 1+ 1 1 =0:
IfJ =f1;2g contains two elements, thenit follows from(1.5) that 1
+
2
=0. We obtainfrom (1.2) that
R= 1 N i 0+ 1 1 + 1 2 = 1 N i 2 + 1 1 2 =0:
We have treated the easy part of the following theorem. The other partis more diÆcultand isproved in[Ve5 ].
Theorem. We have that s 0 is a pole of Z top ;0;f (s) if and only if s 0 = i =N i
for some exceptional curve E i
intersecting at least three times other components, ors
0
= 1=N j
for some irreduciblecomponent E j
of thestrict transformof f
1 f0g.
(1.2.4)Thetopologicalzetafunctioninvolvesthetopological Euler-Poin-carecharacteristic. Thisinvariantisquiterough: sometimesonewantsto detect more. Now we introducetheuniversalEulercharacteristic,which is the nest invariant with the following properties of the topological Euler-Poincare characteristic: (X)= (Y) ifX is isomorphicto Y and (XnY)= (X) (Y)ifY is aclosedsubvarietyofX. Thisuniversal EulercharacteristicistheclassofanalgebraicvarietyintheGrothendieck ring of algebraic varieties over C. We rst recall this notion, which we willgiverelative toan arbitrary eldof characteristic zero.
Let k be a eld of characteristic zero. An algebraic variety over k is in this paragraph a reduced separated scheme of nite type over k. An algebraic variety is thus not necessarily irreducible. Let G be the free abelian group generated by the symbols [V], where V is an alge-braic variety over k. Let N be its subgroup generated by f[V] [V
0 ] j V is isomorphicto V
0
over kg[f[V] [V nW] [W] j W is a closed k-subvariety of Vg. The Grothendieck ring K
0 (Var
k
vari-by [V][W] := [V W]. We set L := [A 1 k ] and we denote by M k the localization ofK 0 (Var k ) withrespectto L. (1.2.5)Iff :(C n
;0)!(C;0 ) isthegerm of aholomorphicfunction, the localmotivic zeta functionoff isthefollowingelement ofM
C [[t]]: Z mot;0;f (t)=L n X IT [ Æ E I \g 1 f0g] Y i2I (L 1)L i t N i 1 L i t N i : If f 2 C[ x 1 ;:::;x n
] is a polynomial over C, the global motivic zeta functionof f isthefollowingelement of M
C [[t]]: Z mot;gl ;f (t)=L n X IT [ Æ E I ] Y i2I (L 1)L i t Ni 1 L i t N i :
Thesemotiviczetafunctionsarede nedintrinsicallybyDenefandLoeser in[DL2 ]. Inthispaper,theyobtainedthepreviousexpressionswhichare interms ofan embeddedresolution.
Note that 1 L i t Ni is a unit in M C
[[t]] and that its inverse is 1=(1 L i t N i ) =1+L i t N i +L 2 i t 2N i
+ . We can consider these motiviczetafunctionsasanelement ofthelocalizationofthepolynomial ring M
C
[t] with respect to the set f1 L i t N i j i 2 Tg or to the set f1 L a t b ja;b2Z >0
g. TheselocalizationsaresubringsofM C
[[t]].
(1.2.6) The topologicalzeta functions are specializationsof themotivic zetafunctions. Thisgivesusanotherproofofthefactthattheexpressions inthede nitionsofthetopologicalzetafunctionsareindependent ofthe chosen embeddedresolution.
We now explain how Z top ;gl ;f
(s) can be seen as a specialization of Z
mot;gl ;f
(t). First we specialize Z mot;gl ;f
(t) to the Hodge zeta function Z Hod ;gl ;f (t): Z Hod;gl ;f (t) = (uv) n X IT H( Æ E I ) Y i2I (uv 1)(uv) i t N i 1 (uv) i t N i :
Here, H(V) 2 Z[u;v] represents the Hodge polynomial of the algebraic variety V. This polynomial codes important information of the mixed Hodge theoryofV. Itis anEulercharacteristic at anintermediatelevel:
and one can specialize these further to the topological Euler-Poincare characteristic byputtingu=v=1. Lets2Z 0 . Then Z Hod ;gl ;f ((uv) s ) = (uv) n X IT H( Æ E I ) Y i2I (uv 1)(uv) i sNi 1 (uv) i sN i = (uv) n X IT H( Æ E I ) Y i2I uv 1 (uv) i +sN i 1 = (uv) n X IT H( Æ E I ) Y i2I 1 1+uv++(uv) i +sN i 1
We applythetopologicalEuler-Poincarecharacteristicto this expres-sion. So we have to replace H(
Æ E I ) by ( Æ E I ) and H(L) =uv by 1. We obtainthefunction
Z 0 !C :s7! X IT ( Æ E I ) Y i2I 1 i +sN i : (1.6)
Thisfunction is independentof theembeddedresolution and can be ex-tendeduniquely to arationalfunction onC, which is
Z top ;gl ;f (s)= X IT ( Æ E I ) Y i2I 1 i +sN i :
We used Hodge polynomials to deduce that (1.6) is independent of the embeddedresolution. I try to clarifywhythis is crucialbygiving a slightly di erent deduction. We want to considerZ
mot;gl ;f (L
s
), but we can notgive senseto thisinM
C
. We deal withthis problemas follows. Supposethatwe have two embeddedresolutions. Then
L n X IT [ Æ E I ] Y i2I (L 1)L i t N i 1 L i t N i =L n X I 0 T 0 [ Æ E I 0 ] Y j2I 0 (L 1)L j t N j 1 L j t N j :
We bring the denominators to the other side, so we get a polynomial identity in which we make the substitution t = L
s
for s 2 Z 0
. If we applynowthetopologicalEuler-Poincarecharacteristic,wegetthetrivial identity0=0 because (L 1)
jTj+jT 0
j
isafactor ofeach sideandbecause (L 1) = 0. We have to cancel (L 1)
jTj+jT 0
j
we do not know that L 1 is not a zero divisor in M C
. However, if we specialize rst to the Hodge polynomials,we can cancel H(L 1) = uv 1 because Z[u;v]
uv
isan integral domain. Thereafter, we applythe topologicalEuler-Poincarecharacteristic andwebringevery factorto its originalsideto obtain(1.6).
(1.2.7) If we work relative to a number eld, there exists a connection betweenthemotivic zetafunction andIgusa's p-adiczetafunction.
Let F bea number eld. Letf 2O F [x 1 ;:::;x n ] be a non-zero poly-nomial. Letg:Y !C n bean embeddedresolutionof f :C n !C which isde nedover F. We denotebyE
i
,i2T,theF-irreduciblecomponents ofg 1 (f 1 f0g),and byN i and i
1themultiplicitiesof fÆg andg dx along E i . For I T, we denote E I := \ i2I E i and Æ E I := E I n([ j2I= E j ). Theglobalmotivic zetafunctionZ
mot;gl ;f
(t)isinthiscontextan element ofM
F
[[t]], and isgiven by thesame expressionas in(1.2.4). Let be a prime ideal of O
F
such that the statement in (1.1.10) holds for the p-adic completion of F with respect to . Let K be an extension of nite degree of this p-adic completion. Let R be the v al-uation ring of K and let P be the maximal ideal of R . Let q be the number of elements of the residue eld O
F = and let e 2 Z 1 be such that R =P = F q e
. For an algebraic variety X over F, let X be the re-duction modulo and let X(F
q e
) be the F q
e
-rational points of X. Be-cause card(X(F q e ))=card(Y(F q e )) ifX =F Y and card(XnY(F q e ))= card(X(F q e )) card(Y(F q e
)) if Y is a closed F-subvariety of X,we can de ne card( (F q e )) for 2K 0 (Var F
) in an obvious way. We have that card(( )(F q e )) = card( (F q e ))card( (F q e )) for ; 2 K 0 (Var F ) and that it induces a notion of cardinality on M
F
. Because of (1.1.10), we obtainIgusa'sp-adiczeta functionbyspecialization.
(1.2.8)Theweakfactorizationconjectureisprovedrecently(see[AKMW ] or [W l]) and can be used to obtain the independence of the embedded resolutionof theglobal topologicaland theglobalmotivic zetafunction. The weak factorization theorem states the following. Let f : X 99KY be a proper birational map between non-singular algebraic varieties X andY overan algebraicallyclosed eldofcharacteristic zero,whichisan isomorphismover anopen setU. Then f can befactored as
X =X f0 99KX f1 99K f n 1 99KX =Y;
whereeach X i
is a non-singularalgebraic varietyand f i
is a blowing-up or blowing-down at a smooth centre which is an isomorphism over U. Moreover, if XnU and Y nU are divisors with normal crossings, then eachD
i :=X
i
nU isadivisorwithnormalcrossingsandf i
isablowing-up orblowing-downatasmoothcentrewhichhasnormalcrossingswiththe componentsofD i . (1.2.9) Example. Let f = x 5 z 2 +x 8 y 2 +x 11
. We used the computer program of Hoornaert and Loots [HL ] to compute Igusa's p-adic zeta functionof f and theglobaltopologicalzeta functionoff.
Let K = Q p
. If p 3mod4, we obtained that 1=2 is a pole of Z
f
(s). Ifp 1mod4 we obtained that 1=2 is nota pole of Z f
(s). In bothcases, there arepoles withreal part 1=2 and with imaginarypart between0 and2
p
1=logp.
Because 1=2 is a pole of in nitelymany p-adic elds Q p
, we have thatZ
mot;gl ;f
(t)hasapolein 1=2. However,weobtainedthatZ top ;gl ;f
(s) has not a pole in 1=2. For the de nition of a pole of the motivic zeta function, see[RV2].
1.3 Historical note
The study of these zeta functions has a rich history. Igusa's p-adic zeta functionwasintroducedin1965byWeil[We ]becausehewantedtoprove the convergence of the Eisenstein-Siegel series. The basis properties of this zeta function were rst studied by Igusa ([Ig1 ], [Ig2 ], [Ig3 ]). He also usedZ
f
(s) to solve a conjecture ofBorevichand Shafarevich. They introduced the Poincare series P(t) in the 1950's and they conjectured the rationality of it [BS]. Igusa proved the rationality of P(t) rst in [Ig2 ]and in[Ig3 ] hegavetheproofofthisfactwhichwehave seenbefore andwhichismucheasier. See[De2 ]formore informationandforfurther developmentson Igusa'sp-adic zetafunction.
In theearly 1990's Denefand Loeser introduced thetopological zeta function[DL1 ]. They obtainedthisinvariantas a limitof Igusa's p-adic zeta functions. In 1998 they de ned the motivic zeta function [DL2 ]. This invariant specializes to thetopological zeta functionand is de ned intrinsically.
The topological zeta
function for curves and
surfaces
Abstract
We study the local topological zeta functionassociated to a complexfunctionthatis holomorphicat theoriginof C
2 (re-spectivelyC
3
). Wedetermineallpossiblepoleslessthan 1=2 (respectively 1). On C
2
ourresult isa generalizationof the factthatthelogcanonical thresholdis neverin]5=6;1[. Sim-ilarstatementsare trueforthemotivic zeta function.
2.1 Introduction
(2.1.1) Let f be the germ of a holomorphic function on a neighbour-hood of the origin 0 in C
n
which satis es f(0) = 0 and which is not identically zero. Let g : V ! U C
n be an embedded resolution of a representativeoff 1 f0g. WedenotebyE i
,i2T,theirreducible compo-nentsofg 1 (f 1 f0g), andbyN i and i
1themultiplicitiesof fÆgand g (dx 1 ^^dx n ) alongE i . The (N i ; i
),i2T,arecalledthenumerical data of the resolution (V;g). For I T denote also E
I := \ i2I E i and Æ E I :=E I n([ j2I= E j ).
The set of germs of holomorphic functions on a neighbourhood of 02C n willbe denoted byO n .
(2.1.2) To f one associatesthelocaltopologicalzeta function
Z f (s)=Z top ;0;f (s):= X IT ( Æ E I \g 1 f0g) Y i2I 1 i +sN i :
Here s is a complex variable and () denotes the topological Euler-Poincare characteristic. The remarkablefactthatZ
f
(s)doesnotdepend on the chosen resolution was rst proved in [DL1] by expressingit asa limitof Igusa'sp-adiczeta functions.
(2.1.3) Thelog canonicalthresholdc 0
(f)of f at 02C n
is byde nition the supremum of the set fc 2Q j the pair (C
n
;cdiv f) is log canonical inaneighbourhoodof0g. Wecandescribeit(see[Ko2 ,Proposition8.5]) in terms of the embedded resolution (V;g) as c
0 (f) = minf i =N i j i 2 Tg. Inparticular,thisminimumis independentof thechosen resolution. Consequently, c
0
(f) is the largest candidate pole of Z f
(s). The log canonicalthresholdhasalreadybeenstudiedinvariouspapersofAlexeev, Ein,Kollar,Kuwata,Mustata,Prokhorov,Reid,Shokurovand others;in particular,thesets
T n :=fc 0 (f)jf 2O n g; withn2Z >0
,arethesubjectof interestingconjectures. Itisnaturaltoinvestigatewhethermorequotients i
=N i
,i2T,yield invariantsofthegermoff at0. Ofcourse,thewholesetf
i =N
i
ji2Tg dependsonthechosen resolution(forn=2,however,one couldconsider suchasetassociatedto theminimalresolution),butitssubsetconsisting ofthepolesof Z
f
(s) isaninvariant off. Philosophically,these poles are inducedby`important' componentsE
i
,which occur inevery resolution. Forn2Z
>0
,we de netheset P n by P n :=fs 0 j9f 2O n : Z f (s) hasa polein s 0 g:
Thecasen=1istrivial: T 1 =f1=iji2Z >0 gandP 1 =f 1=iji2Z >0 g.
(2.1.4)When n=2,itisknownthatT 2
\]5=6;1[= ;(see [Ku1]or[Re]). Because itfollows from [Ve5 ] that c
0
(f) is apole(and thusthelargest pole) of Z (s), thestatement P \] 1; 5=6[=; would bea remarkable
generalization. It isinfact nothardto provethisgeneralization. Inthis article,wewillprovemore:
P 2 \] 1; 1=2[ = f 1=2 1=iji2Z >1 g (2.1) = f 1; 5=6; 3=4; 7=10;:::g:
We willalso seethat 1 istheonlypossiblepolelessthan 1=2 iff has multiplicityat leastfour.
(2.1.5) Kollarproved in [Ko1] that T 3
\]41=42;1[= ;. It turns out that there isnoanalogousresult forP
3
. Infact, we willgive examplesof zeta functionswithpolesin] 1; 41=42[whichare,moreover,arbitrarilynear to 1. Ontheotherhand,weprove theanalogueof(2.1),whichappears to be P 3 \] 1; 1[ = f 1 1=iji2Z >1 g: (2.2)
In particular,there are no poles lessthan 3=2. Moreover, ifthe multi-plicityof f isat least three,we willsee thatthere areno poles lessthan 1. ThelastpartwillbethemostdiÆcultbecausewewillhaveto prove thatsome candidate polesare notpoles.
(2.1.6) Ingeneral, we expectforn2that
P n
\] 1; (n 1)=2[=f (n 1)=2 1=iji2Z >1
g:
We willseethat forn2, itis easy to prove that
P n \] 1; n+1[=;: 2.2 Curves (2.2.1) We will determine P 2
\] 1; 1=2[. Let f be the germ of a holomorphic function on a neighbourhood of the origin 0 in C
2
which satis es f(0) = 0 and which is not identically zero. Let (V;g) be the minimal embedded resolution of f
1 f0g. Write g = g 1 ÆÆ g t as a composition of blowing-ups g i , i 2 T e := f1;:::;tg. The exceptional curve of g i
and also the strict transforms of this curve are denoted by E
i
. Theirreduciblecomponentsoff 1
f0gandtheirstricttransformsare denoted byE ,j2T .
(2.2.2) Thedual(minimal) embeddedresolution graphof f 1
f0g is ob-tainedasfollows. Oneassociatesavertextoeachexceptionalcurveinthe minimalembeddedresolution(representedbyadot), andto eachbranch ofthestricttransformoff
1
f0g(representedbyacircle). Onealso asso-ciatestoeachintersectionanedge,connectingthecorrespondingvertices. Thefact thatE
i
hasnumericaldata(N i ; i ) is denotedbyE i (N i ; i ). (2.2.3)Let E i
bean exceptional curveand letE j
,j2J,bethe compo-nents that intersect E
i in V. Set j = j ( i =N i )N j forj 2 J. Then we have therelation
X
j2J (
j
1)+2=0; (2.3)
whichwas rstprovedbyLoeserin[Lo1 ],andlatermoreconceptuallyby thesecond author in[Ve2 ].
Suppose that j 6= 0, which is equivalent to i =N i 6= j =N j , for all j 2 J. Then one computes easily that the contribution R of E
i
to the residueof Z
f
(s) at thecandidate pole i =N i is 1 N i 0 @ ( Æ E fig )+ X j2J 1 j 1 A : (2.4)
From (2.3) and (2.4) it follows that R = 0 if J contains one or two elements. Thisis theeasy partof thefollowingtheorem. The otherpart ismore diÆcultand is proved in[Ve5 ].
(2.2.4) Theorem. We have that s 0 is a pole of Z f (s) if and only if s 0 = i =N i
for some exceptional curve E i
intersecting at least three times other components, ors
0
= 1=N j
forsome irreduciblecomponent E
j
of thestrict transformoff 1
f0g.
Thefollowing lemmais obtainedbyelementarycalculations.
(2.2.5) Lemma. Suppose that we have blown up k times but we do not yet have an embedded resolution. Let P be a point of the strict transform of f
1
f0g with multiplicity in which we do not yet have normalcrossings. Letg be theblowing-upat P.
(a)Supposethat two exceptional curvesE i
and E j
containP. Then thenew candidatepole
k+1 =N k+1 = ( i + j )=(N i +N j +) islarger thanminf i =N i ; j =N j g.
(b)SupposethatexactlyoneexceptionalcurveE i
containsP andthat 2. ThenE
k+1
hasnumericaldata(N i + ; i + 1)and ( i + 1)=(N i + ) isbetween 1= and i =N i .
(c)SupposethatexactlyoneexceptionalcurveE i
containsP andthat =1. Notethatthetwocurvesaretangent atP becausewedonothave normalcrossingsatP. Letg
k+2 betheblowing-upatE i \E k+1 . Because thestricttransformoff
1
f0gdoesnotintersectE k+1
afterthis blowing-up,wedonothavetoblowupatapointofE
k+1
anymore. BecauseE k+1 is intersected once, it follows from (2.2.3) that the contributionof E
k+1 to theresidue at the candidatepole
k+1 =N
k+1
is zero. The numerical dataof E k+2 are(2N i +2;2 i +1), and (2 i +1)=(2N i +2) isbetween 1=2 and i =N i .
(2.2.6) Suppose that after some blowing-ups, we do not have normal crossingsat a point P. Supposealsothat thecandidate poles associated totheexceptional curvesthroughP arealllargerthanorequalto 1=2. Then it follows from the above lemma that the components above P in the nalresolution do notgivea contributionto a polelessthan 1=2.
Corollary. Zetafunctionsofsingularitiesof multiplicityat leastfourdo nothave apolein] 1; 1=2[nf 1g.
Indeed, every exceptional curve in the minimal embedded resolution of f
1
f0g lies above a point of E 1
(considered in the stage when it is cre-ated),whichhasa candidatepolelarger thanorequalto 1=2.
(2.2.7)Todealwithmultiplicitytwo andthree,we willstudyan`easier' element of O
2
with thesame zetafunction. We mention two methodsto obtainasimplefunctionwith thisproperty.
Method1. (See[Ku2])Letf 2O n
havemultiplicityd andletf d
bethe homogeneous part of degree d in the Taylor series of f. Let N 2 Z
>d . Take a maximalset V of homogeneouspolynomialsof degree largerthan d and at most N which are linearly independent in the quotient vector spaceO n =(@f d =@x 1 ;:::;@f d =@x n
). Then f isholomorphicallyequivalent to f d + P u i 2V a i u i +',forsome a i 2C andsome '2O n whichsatis es
Remark. (i) The similar statement in [Ku2, Lemma 3.2] is not correct; some homogeneityconditionis needed.
(ii)Letf 1
f0ghave anisolatedsingularityattheoriginand suppose that we have an embedded resolution which is an isomorphism outside thissingularity. Then 'doesnotin uencetheembedded resolutionand the numerical data if N is big enough. Consequently, to calculate the local topological zeta function, we can omit' if we take N big enough. Notethatwhenn=2,f
1
f0g hasanisolatedsingularityat0ifandonly iff is reducedand mult(f)2.
Method 2. (Weierstrass Preparation Theorem) If f(z 1 ;:::;z n 1 ;w) = f(z;w)2O n
isnotidenticallyzero on thew-axis,thenf can be written uniquely as f = (w e +a 1 (z)w e 1 ++a e (z))h, where a i (z) 2 O n 1 satis esa i (0)=0and h2O n satis es h(0)6=0.
Becauseh(0)6=0,theresolutionsand thelocaltopologicalzetafunctions off andw e +a 1 (z)w e 1 ++a e
(z)arethesame. Afteran appropriate coordinate transformation, the desired form will appear. For example, thecoordinate transformation(z;w)7!(z;w a
1
(z)=e) cancelstheterm a 1 (z)w e 1 . (2.2.8) Example. Let f 2O 2
have multiplicitythree and let f 3 =y 3 + xy 2 = y 2
(y +x). First we illustrate method 1. Because @f 3 =@x = y 2 and@f 3 =@y=3y 2 +2xy,we get(@f 3 =@x;@f 3 =@y) =(y 2 ;xy). Therefore, we set V = fx 4 ;x 5 ;:::;x N
g, and we obtain that f is holomorphically equivalentto afunctionoftheformy
3 +xy 2 +a 4 x 4 ++a N x N +g(x;y), withmult(g(x;y))>N. Ifall a
i
can betaken zeroforeveryN,thenf is holomorphicallyequivalenttoy 3 +xy 2 . Ifthereexistsana i di erentfrom zero, whichis thereducedcase, theformabovewillallowusto calculate thelocal topologicalzetafunction off.
Now we illustrate method 2. By the Weierstrass Preparation Theo-rem,wemayworkwithafunctionoftheformy
3 + a 1 (x)y 2 + a 2 (x)y+ a 3 (x), with mult(a 1 (x)) = 1, mult(a 2 (x)) 3 and mult(a 3 (x)) 4. One can checkthatthere existsacoordinatetransformation(x;y)7!(x;y k(x)) such that the function becomes of the form y
3 +b 1 (x)y 2 +b 3 (x), with mult(b 1 (x)) =1and mult(b 3
(x))4. Afteranothercoordinate
(2.2.9) Theorem. We have P 2 \ 1; 1 2 = 1 2 1 i i2Z >1
and every local topological zeta function has at most one pole in the interval] 1; 1=2].
Proof. (a)Supposethatmult(f),themultiplicityoff attheoriginofC 2
, isequalto two. Thenf isholomorphicallyequivalenttoy
2 ory 2 +x k for some k 2 Z >1 . If itis y 2
, the only pole of Z f
(s) is 1=2. If k = 2, the onlypoleof Z
f
(s) is 1. Ifk is odd, writek =2r+1. After r blowing-ups, the strict transform of f
1
f0g is non-singular and tangent to E r
. ThenumericaldataofE
i
,i=1;:::;r,are(2i;i+1). Togettheminimal embeddedresolution,wenowblowuptwice. Thedualresolutiongraphis
::: s s s s s s c E 1 E 2 E 3 E r E r+2 E r+1
andthenumericaldataareE 1 (2;2),E 2 (4;3),E 3 (6;4),:::,E r (2r;r+1), E r+1 (2r+1;r+2) andE r+2 (4r+2;2r+3).
If k is even and larger than 2, write k = 2r. By easy calculations we obtainthedualresolution graph
::: s s s s s H H H c c E 1 E 2 E 3 E r 1 E r
and the numerical data E 1 (2;2), E 2 (4;3), E 3 (6;4), ::: ,E r 1 (2r 2;r) and E r (2r;r+1). Because (2r+3)=(4r+2) = 1=2 1=(2r+1) and (r+1)=(2r) = 1=2 1=(2r),it followsfrom (2.2.4) that
fs 0 j9f 2O 2 :mult(f)=2 and Z f (s) hasa pole ins 0 g = 1 2 1 i i2Z >1 [ 1 2 :
Note that Newton polyhedra could also be used to deal with (a), see [DL1 ].
(b)Supposethat mult(f) =3. Up to an aÆne coordinate trans-formation,there are threecases forf .
(b.1)Casef 3
=xy(x+y). Afterone blowing-upwegetanembedded resolution. Thepoles ofZ
f (s) are 1and 2=3= 1=2 1=6. (b.2) Casef 3 =y 2
(y+x). According to example2.2.8, we may sup-pose that f =y
3 +xy
2
+g(x), where g(x) is a holomorphicfunction in the variable x of multiplicityk 4. If g(x) =0, the poles of Z
f
(s) are 1 and 1=2. Now consider thecase when k is odd. Write k =2r+1. After r blowing-upswe getan embedded resolution withdual resolution graph ::: c s s s s H H H c c E 1 E 2 E r 1 E r
and with numerical data E 1 (3;2), E 2 (5;3), ::: , E r 1 (2r 1;r) and E r (2r+1;r+1).
Ifk iseven, writek=2r. After r+1 blowing-upsweget
::: s s s s s s c c E 1 E 2 E 3 E r 1 E r+1 E r and E 1 (3;2), E 2 (5;3), E 3 (7;4), ::: , E r 1 (2r 1;r), E r (2r;r +1) and E r+1 (4r;2r+1).
Thepolesappearingin(b.2)areinthedesiredsetbecause (r+1)=(2r+ 1)= 1=2 1=(4r+2) and (2r+1)=(4r)= 1=2 1=(4r).
(b.3) Casef 3
=y 3
. Wemaysupposethat f isof theform
y 3 +a 4 x 4 +b 3 yx 3 +a 5 x 5 +b 4 yx 4 +a 6 x 6 +b 5 yx 5 +; where a i ;b i 2 C. If f = f 3 = y 3
then the only pole of Z f
(s) is 1=3. Otherwise there is an integer r 1 such that after blowing up r times and always taking the charts determined by g
i (x;y) = (x;xy), we get (g 1 ÆÆg r ) dx^dy=x r dx^dyandfÆg 1 ÆÆg r =x 3r (y 3 +a 3r+1 x+ b 2r+1 yx+a 3r+2 x 2 +b 2r+2 yx 2 +a 3r+3 x 3 + ),witha 3r+1 ;b 2r+1 ;a 3r+2 ;b 2r+2 and a 3r+3
not all zero. The equation of E r
in this chart is x = 0 and thenumerical dataofE
r
are(3r;r+1). The zerolocusof y 3 +a 3r+1 x+ b 2r+1 yx+a 3r+2 x 2 +b 2r+2 yx 2 +a 3r+3 x 3
+ is the strict transform of f
1
(b.3.i)Ifa 3r+1
6=0,weobtainafterblowingupthreemoretimesthedual resolutiongraph ::: s s s s s c E 1 E r E r+3 E r+2 E r+1
andthenumericaldateE r (3r;r+1),E r+1 (3r+1;r+2),E r+2 (6r+2;2r+3) and E r+3
(9r+3;3r+4). The pole (3r+4)=(9r+3) is inthe interval ] 1; 1=2] if and only if r = 1, and in this case the pole is equal to
1=2 1=12. (b.3.ii)If a 3r+1 =0and b 2r+1 6=0, calculationsgiveus ::: s s s s c c E 1 E r E r+2 E r+1
and thenumericaldateE r (3r;r+1), E r+1 (3r+2;r+2) and E r+2 (6r+ 3;2r+3). The pole (2r+3)=(6r+3) is inthe interval] 1; 1=2] if and onlyifr=1,and inthiscase thepoleis equalto 1=2 1=18.
(b.3.iii) If a 3r+1 = b 2r+1 = 0 and a 3r+2
6= 0, we get the dual resolution graph ::: s s s s s c E 1 E r E r+2 E r+3 E r+1
andthenumericaldataE r (3r;r+1),E r+1 (3r+2;r+2),E r+2 (6r+3;2r+3) and E r+3
(9r+6;3r+5). The pole (3r+5)=(9r+6) is inthe interval ] 1; 1=2] if and only if r = 1, and in this case the pole is equal to
1=2 1=30.
(b.3.iv) The last case is a 3r+1 = b 2r+1 = a 3r+2 = 0 and (b 2r+2 6= 0 or a 3r+3 6=0). Ify 3 +b 2r+2 yx 2 +a 3r+3 x 3
isaproductofthree distinctlinearfactors, we getanembeddedresolution afterone blowing-up. The numericaldataof E r+1 are (3r+3;r+2) and (r+2)=(3r+3)2]= 1; 1=2[. Ify 3 +b 2r+2 yx 2 +a 3r+3 x 3
isnotaproductofthreedistinctlinearfactors, thenitisequaltoy
3 +xy
2
afteranaÆnecoordinate transformationthat does notchange the equation x = 0 of E . Let g be the blowing-up
at the origin of the chart we consider. The strict transform of f 1
f0g only intersects the exceptional curve E
r+1
, which has numerical data (3r+3;r+2). Because (r+2)=(3r+3) 1=2forallr,itfollowsfrom (2.2.4) and (2.2.6) that Z
f
(s) has no pole in] 1; 1=2[di erent from 1.
(c)Supposethatmult(f)4. We explainedin(2.2.6) that Z f
(s) hasno polein] 1; 1=2[di erent from 1.
(2.2.10) We now present a similar result for the following generalized zeta functions [DL1 ]. The case d = 2 is used in the next section. To f 2O
n
and d2Z >0
one associates thelocaltopologicalzeta function
Z (d) f (s)=Z (d) top ;0;f (s):= X IT 8i2I:djN i ( Æ E I \g 1 f0g) Y i2I 1 i +sN i : Forn;d2Z >0 ,we set P (d) n :=fs 0 j9f 2O n : Z (d) f (s) hasa pole ins 0 g: Consequently, Z f (s)=Z (1) f (s) and P n =P (1) n . (2.2.11)Let E i
be an exceptional curve and letE j
, j 2J, be the com-ponentsthatintersectE
i inV. Then X j2J N j 0 (modN i ); (2.5)
see, e.g., [Lo1 ] or [Ve3]. Fix d 2 Z >0
and suppose that d j N i
. Let J
d
J be thesubset of indicesj satisfyingd jN j . Suppose that j := j ( i =N i )N j
isdi erentfrom 0forall j2J d
. ThenthecontributionR ofE
i
to theresidue ofZ (d) f
(s) at thecandidate pole i =N i is 1 N i 0 @ ( Æ E fig )+ X j2J d 1 j 1 A : (2.6)
This contribution is zero if J contains one or two indices. Indeed, if J contains one element, relation(2.5) impliesthat J =J
d
R=0. If J contains two elements, relation(2.5) impliesthatJ d =J or J d =;. IfJ d
=J,we obtainR=0 analogously as intheprevious case. If J
d
= ;, we get R = 0 because the Euler-Poincare characteristic of a projective lineminustwopoints iszero.
(2.2.12) Theorem. Letf 2O 2 andd 2Z >1 . Then P (d) 2 \ 1; 1 2 = 1 2 1 i i2Z >2 and djlcm(2;i) :
Proof. We use the notation and the calculations of (2.2.9). First we considerthe candidate pole
j =N j , j 2 T s . Because E j has numerical data (N j
;1), the candidate pole j
=N j
isless than 1=2 ifand onlyif N
j
=1, andinthiscase d-N j
.
(a)Supposethatmult(f)=2. So f is holomorphicallyequivalent to (i)y 2 or(ii)y 2 +x k forsome k 2Z >1
. Incase (i)withd=2theonly pole of Z
(d) f
(s) is 1=2. In case (i) withd >2 and incase (ii)fork =2, we obtainthatZ
(d) f
(s) isidenticallyzero.
Ifk=2r+1,weobtainfrom(2.2.11) andtheconsiderationaboveabout E
j ,j2T
s
,thatonlythecandidatepoleassociatedto E r+2
canbeapole. Sowehavetocomputetheresidueatthecandidatepole 1=2 1=(2r+1). Ifd -N r+2 ,then R=0. IfdjN r , djN r+1 and djN r+2 ,thend =1 and we have a contradiction. If djN r , djN r+2 and d -N r+1 (which is equivalent to d = 2), then the contribution R is r=(2r +1) and this is not zero because r 1. If djN r+1 , djN r+2 and d - N r , then R = 1=(4r+2). If djN r+2 , d - N r and d - N r+1
, then R= 1=(4r+2). We conclude that 1=2 1=(2r+1) isa poleof Z
(d) f
(s) ifand onlyifdj4r+2. Ifk =2r, r2, onlythecandidate pole associated to E
r can be a pole. Ifd-N r ,then R=0. If djN r and djN r 1 ,thenR=(r 1)=(2r)6=0. If djN r and d-N r 1 ,thenR= 1=(2r). Consequently, 1=2 1=(2r) isa pole ofZ (d) f (s) ifand onlyif dj2r. Remark thatwe have proved that
fs 0 j9f 2O 2 :mult(f)=2 and Z (d) f (s) hasa poleins 0 gnf 1=2g = 1 2 1 i i2Z >2 and djlcm(2;i) :
(b)Supposethat mult(f)=3. (b.1) Casef
3
=xy(x+y). We getthat 1=2 1=6 is a pole ifand only if dj3 (which is equivalent to d = 3). This is consistent with the claim because 3jlcm(2;6). (b.2)Casef 3 =y 2 (x+y). Ifg(x)=0,thenZ (d) f (s) isidenticallyzero (foreveryd2).
If k = 2r+1, only the candidate pole associated to E r can be a pole. If djN r , then d - N r 1
because d > 1 and because N r
and N r 1
are odd numbers with di erence 2. Consequently, if dj2r +1, then R =
1=(2r+1)6=0.
Ifk =2r,theonlycandidate polewhichcan beapoleis r+1 =N r+1 . If djN r+1 and djN r 1
, thend =1, which is a contradiction. So we have to considertwo cases. If djN
r+1 , d - N r and d -N r 1 , then R = 1=(4r). If djN r+1 ,djN r and d -N r 1
,then R=1=(4r). We obtainthat 1=2 1=(4r) is apole ifand onlyifdj4r.
(b.3)Casef 3
=y 3
. Wegetanalogouslythatifwehaveapole, itisof the desired form. Remark that we onlyhave to consider thecase r =1 in(b.3.i),(b.3.ii) and (b.3.iii).
(c) Supposethat mult(f) 4. As before we get that Z (d) f
(s) has no pole lessthan 1=2.
2.3 Surfaces
Inthissection,weprove thefollowingtheorem.
(2.3.0) Theorem. We have P 3 \] 1; 1[= 1 1 i i2Z >1 : Moreover, if f 2 O 3
has multiplicitythree or more, then Z f
(s) has no pole lessthan 1.
Remark. (i) It is a priori not obvious that the smallest value of P 3
is 3=2. This is in contrast to the fact that it easily follows from Lemma 2.2.5 thatthe smallestvalue ofP
2 is 1. (ii)In(2.3.3.9)wegivefunctionsf
k 2O
3
ofarbitrarymultiplicitysuch that Z f k (s) has a pole in s k , where (s k ) k
is a sequence of real numbers largerthan 1andconvergingto 1. Inparticular,P
3
\] 1; 41=42[6=;, which isincontrast to T \]41=42;1[=;.
