Conjectures Ab out Discriminan ts of Hec Algebras of P - PDF document

ConjecturesAboutDiscriminantsofHeckeAlgebrasofPrimeLevelFrankCalegari???1andWilliamA.Steiny21HarvardUniversityfcale@math.harvard.eduhttp://www.math.harvard.edu/~fcale2HarvardUniversity,was@math.harvard.eduhttp://modular.fas.harvard.edu/Abstract.Inthispaper,westudyp-divisibilityofdiscriminantsofHeckealgebrasassociatedtospacesofcuspformsofprimelevel.Byconsider-ingcuspformsofweightbiggerthan2,weareareledtomakeapreciseconjectureaboutindexesofHeckealgebrasintheirnormalisationwhichimplies(iftrue)thesurprisingconjecturethattherearenomodpcon-gruencesbetweennon-conjugatenewformsinS2(¡0(p)),buttherearealmostalwaysmanysuchcongruenceswhentheweightisbiggerthan2.1BasicDe¯nitionsWe¯rstrecallsomecommutativealgebrarelatedtodiscriminants,thenintroduceHeckealgebrasofspacesofcuspforms.1.1CommutativeAlgebraInthissectionwerecallthede¯nitionofdiscriminantofa¯nitealgebraandnotethatthediscriminantisnonzeroifandonlyifnobaseextensionofthealgebracontainsnilpotents.LetRbearingandletAbeanR-algebrathatisfreeof¯niterankasanR-module.Thetraceofx2Aisthetrace,inthesenseoflinearalgebra,ofleftmultiplicationbyx.De¯nition1(Discriminant).Let!1;:::;!nbeanR-basisforA.Thenthediscriminantdisc(A)ofAisthedeterminantofthen£nmatrix(tr(!i!j)).???SupportedinpartbytheAmericanInstituteofMathematicsySupportedinpartbyaNationalScienceFoundationPostdoctoralFellowship Thediscriminantisonlywell-de¯nedmodulosquaresofunitsinR.WhenR=Zthediscriminantiswellde¯ned,sincetheonlyunitsare§1.WesaythatAisseparableoverRifforeveryextensionR0ofR,theringA­R0containsnonilpotents.Proposition1.SupposeRisa¯eld.ThenAhasnonzerodiscriminantifandonlyifAisseparableoverR.Proof.Fortheconvenienceofthereader,wesummarizetheproofin[Mat86,x26].IfAcontainsanilpotentthenthatnilpotentisinthekernelofthetracepairing,sothediscriminantis0.Conversely,ifAissepa-rablethenwemayassumethatRisalgebraicallyclosed.ThenAisanArtinianreducedring,henceisomorphicasaringtoa¯niteproductofcopiesofR,sinceRisalgebraicallyclosed.ThusthetraceformonAisnondegenerate.TheDiscriminantValuationWenextintroduceHeckealgebrasattachedtocertainspacesofcuspformsofprimelevelp,de¯nethediscriminantvaluationastheexponentofthelargestpowerofpthatdividesthediscriminant,andobservethatthereareeigenformcongruencesmodulopexactlywhenthediscriminantvaluationispositive.Wethenpresentanexampletoillustratethede¯nitions.Let¡beacongruencesubgroupofSL2(Z).Inthispaper,wewillonlyconsider¡=¡0(p)forpprime.Foranypositiveintegerk,letSk(¡)denotethespaceofholomorphicweightkcuspformsfor¡.LetT=Z[:::;Tn;:::]½End(Sk(¡))betheassociatedHeckealgebra,whichisgeneratedbyHeckeoperatorsTnforallintegersn,includingn=p(wewillsometimeswriteUpforTp).ThenTisacommutativeringthatisfreeasamoduleoverZofrankequaltodimSk(¡).WewillalsosometimesconsidertheimageTnewofTinEnd(Sk(¡)new).De¯nition2(DiscriminantValuation).Letpbeaprime,kapositiveinteger,andsupposethat¡=¡0(p).LetTbethecorrespondingHeckealgebra.Thenthediscriminantvaluationof¡inweightkisdk(¡)=ordp(disc(T)): Weexpectthatdk(¡)is¯niteforthefollowingreason.TheHeckeoperatorsTn,withnnotdivisiblebyp,arediagonalizablesincetheyareselfadjointwithrespecttothePeterssoninnerproduct.Whenk=2oneknowsthatUpisdiagonalizablesincethelevelissquarefree,andwhenk�2oneexpectsthis(see[CE98]).IfTcontainsnonilpotents,Proposition1impliesthatthediscriminantofTisnonzero.Thusdk(¡)is¯nitewhenk=2andconjecturedtobe¯nitewhenk�2.Letpbeaprimeandsupposethat¡=¡0(p).Anormalisedeigenformisanelementf=Panqn2Sk(¡)thatisaneigenvectorforallHeckeoperatorsT`,includingthosethatdividep,normalisedsothata1=1.Thequantitydk(¡)isofinterestbecauseitmeasuresmodpcongruencesbetweennormalisedeigenformsinSk(¡).Proposition2.Assumethatdk(¡)is¯nite.Thediscriminantvaluationdk(¡)ispositive(i.e.,thediscriminantisdivisiblebyp)ifandonlyifthereisacongruenceincharacteristicpbetweentwonormalizedeigen-formsinSk(¡).(ThetwocongruenteigenformsmightbeGaloisconju-gate.)oof.ItfollowsfromProposition1thatdk(¡)�0ifandonlyifT­Fpisnotseparable.TheArtinianringT­FpisnotseparableifandonlyifthenumberofringhomomorphismsT­Fp!FpislessthandimFpT­Fp=dimCSk(¡):Sincedk(¡)is¯nite,thenumberofringhomomorphismsT­Qp!QpequalsdimCSk(¡).ThepropositionfollowsfromthefactthatforanyringR,thereisabijectionbetweenringhomomorphismsT!Randnormalisedeigenformswithq-expansioninR.Thesameproofalsoshowsthataprime`dividesthediscriminantofTifandonlyifthereisacongruencemod`betweentwonormalizedeigenformsinSk(¡)Example1.If¡=¡0(389)andk=2,thendimCS2(¡)=32.LetfbethecharacteristicpolynomialofT2.Onecancheckthatfissquarefreeand389exactlydividesthediscriminantoff.Thisimpliesthatd2(¡)=1andthatT2generatesT­Z389asanalgebraoverZ389.(IfT2onlygeneratedasubringofT­Z389of¯niteindex�1,thenthediscriminantoffwouldbedivisibleby3892.) Modulo389thecharacteristicpolynomialfiscongruentto(x+2)(x+56)(x+135)(x+158)(x+175)2(x+315)(x+342)(x2+387)(x2+97x+164)(x2+231x+64)(x2+286x+63)(x5+88x4+196x3+113x2+168x+349)(x11+276x10+182x9+13x8+298x7+316x6+213x5+248x4+108x3+283x2+x+101)Thefactor(x+175)2indicatesthatT­F389isnotseparableoverF389sincetheimageof(f=(x+175))(T2)inT­F389isnilpotent(itisnonzerobutitssquareis0).Thereare32eigenformsoverQ2butonly31mod389eigenforms,sotheremustbeacongruence.ThereisanewformFinS2(¡0(389);Z389)whosea2termisarootofx2+(¡39+190¢389+96¢3892+¢¢¢)x+(¡106+43¢389+19¢3892+¢¢¢):ThereisacongruencebetweenFanditsGal(Q389=Q389)-conjugate.2ComputingDiscriminantsInthissectionwesketchthealgorithmthatweuseforcomputingthediscriminantsmentionedinthispaper.ThisalgorithmwasinspiredbyadiscussionofthesecondauthorwithHendrikLenstra.Weleavethedetailsofconvertingthedescriptionbelowintostandardmatrixoperationstothereader.Also,themodularsymbolsalgorithmsneededtocomputeHeckeoperatorsarequiteinvolved.Let¡=¡0(p),andletk¸2beaninteger.ThefollowingsketchesanalgorithmforcomputingthediscriminantoftheHeckealgebraTactingonSk(¡).1.Foranygivenn,wecanexplicitlycomputeamatrixthatrepresentstheactionofHeckeoperatorsTnonSk(¡)usingmodularsymbols.Weusethesecondauthor'sMAGMA[BCP97]packagesforcomput-ingwithmodularsymbols,whichbuildsonworkofmanypeople(in-cluding[Cre97]and[Mer94]).2.UsingtheSturmbound,asdescribedintheappendixto[LS02],¯ndanintegerbsuchthatT1;:::;TbgenerateTasaZ-module.(Theintegerbisd(k=12)¢[SL2(Z):¡]e.)3.FindasubsetBoftheTithatformaQ-basisforT­ZQ.(ThisusesGausselimination.)4.ViewTasaringofmatricesactingonQd,whered=dim(Sk(¡))andtryrandomsparsevectorsv2Qduntilwe¯ndonesuchthatthesetofvectorsC=fT(v):T2Bgarelinearlyindependent. 5.WriteeachofT1(v);:::;Tb(v)asQ-linearcombinationsoftheele-mentsofC.6.FindaZ-basisDfortheZ-spanoftheseQ-linearcombinationsofelementsofC.(ThisbasisDcorrespondstoaZ-basisforT,butismucheasierto¯ndthatdirectlylookingforaZ-basisinthespaceofd£dmatricesthatTisnaturallycomputedin.)7.Unwindingwhatwehavedoneintheprevioussteps,¯ndthetracepairingontheelementsofD,anddeducethediscriminantofTbycomputingthedeterminantofthetracepairingmatrix.Averytime-consumingstep,atleastinourimplementation,iscom-putingDfromT1(v);:::;Tb(v)expressedintermsofC,andthisexplainswhyweembedTinQdinsteadofviewingtheelementsofTasvectorsinQd£d.AnimplementationbythesecondauthoroftheabovealgorithmisincludedwiththeMAGMAcomputeralgebrasystem.Therelevantsourcecodeisinthe¯leGeometry/ModSym/linalg.minthepackagedirectory(oraskthesecondauthoroftheappertosendyouacopylinalg.m).WeillustratetheuseofMAGMAtocomputediscriminantsbelow,whichwererununderMAGMAV2.10-21forLinuxonacomputerwithanAthlon2800MPprocessor(2.1Ghz).�M:=ModularSymbols(389,2,+1);�S:=CuspidalSubspace(M);�timeD:=DiscriminantOfHeckeAlgebra(S);Time:0.750�D;629670054720061882880174736321392595498204931550235108311\Factorisation(D);[,00;53,,00;4,,00;6,1,0;2,7,0;1,Ή,;1,...]Ή,;M:=ModularSymbols(997,2,+1);S:=CuspidalSubspace(M);Ή,;timeD:=DiscriminantOfHeckeAlgebra(S);Time:55.600Thereasonforthe+1intheconstructionofmodularsymbolsissothatwecomputeonaspacethatisisomorphicasaT-moduletoonecopyofS2(¡0(p)),insteadoftwocopies.3DataAboutDiscriminantValuationsInthissectionwereportonourextensivecomputationsofdk(¡0(p)).We¯rstnotethatthereisonlyonep50000suchthatd2(¡0(p))�0. Nextwegiveatableofvaluesofd4(¡0(p)),whichseemstoexhibitanicepattern.WeightTwoTheorem1.Theonlyprimep60000suchthatd2(¡0(p))�0isp=389,withthepossibleexceptionof50923and51437.Computationsinthisdirectionbythesecondauthorhavebeencitedin[Rib99],[MS01],[OW02],and[MO02].Forexample,Theorem1isusedforp1000in[MS01]asacrucialstepinprovingthatifEisanellipticcurveoverQ(¹p),with17·p1000,thennotallelementsofE(Q)[p]arerationaloverQ(¹p).Proof.Thisistheresultofalargecomputercomputation.Therestofthisproofdescribeshowwedidthecomputation,sothereaderhassomeideahowtoreplicateorextendthecomputation.Thecomputationde-scribedbelowtookaboutoneweekusingaclusterequippedwith10Athlon2000MPprocessors.Thecomputationsarenontrivial;wecomputespacesofmodularsymbols,supersingularpoints,andHeckeoperatorsonspacesofdimensionsupto5000.TheaimistodeterminewhetherornotpdividesthediscriminantoftheHeckealgebraoflevelpforeachp60000.IfTisanoperatorwithintegralcharacteristicpolynomial,wewritedisc(T)fordisc(charpoly(T)),whichalsoequalsdisc(Z[T]).Wewilloftenusethatdisc(T)modp=disc(charpoly(T)modp):Weruledoutthepossibilitythatdk(¡0(p))�0formostlevelsp60000bycomputingcharacteristicpolynomialsofHeckeoperatorsus-inganalgorithmthatthesecondauthorandD.KohelimplementedinMAGMA([BCP97]),whichisbasedontheMestre-Oesterlemethodofgraphs[Mes86](orcontactthesecondauthorforanEnglishtranslation).Ourimplementationisavailableasthe\ModuleofSupersingularPoints"packagethatcomeswithMAGMA.Wecomputeddisc(Tq)modulopforseveralsmallprimesq,andinmostcasesfoundaprimeqsuchthatthisdiscriminantisnonzero.Thefollowingtablesummariseshowoftenweusedeachprimeq(notethatthereare6057primesupto60000): qnumberofp60000whereqsmallests.t.disc(Tq)6=0modp25809times3161(largest:59471)543(largest:57793)715(largest:58699)1115(thesmallestis307;thelargest50971)132(theyare577and5417)173(theyare17209,24533,and47387)191(itis15661)Thenumbersintherightcolumnsumto6049,so8levelsaremissing.Theseare389;487;2341;7057;15641;28279;50923;and51437:(Thelasttwoarestillbeingprocessed.51437hasthepropertythatdisc(Tq)=0forq=2;3;:::;17.)Wedeterminedthesituationwiththeremaining6levelsusingHeckeoperatorsTnwithncomposite.pHowwerulelevelpout,ifpossible389pdoesdividediscriminant487usingcharpoly(T12)2341usingcharpoly(T6)7057usingcharpoly(T18)15641usingcharpoly(T6)28279usingcharpoly(T34)ComputingTnwithncompositeisverytimeconsumingwhenpislarge,soitisimportanttochoosetherightTnquickly.Forp=28279,hereisatrickweusedtoquickly¯ndannsuchthatdisc(Tn)isnotdivisiblebyp.Thistrickmightbeusedtospeedupthecomputationforsomeotherlevels.Thekeyideaistoe±cientlydiscoverwhichTntocompute.ComputingTnonthefullspaceofmodularsymbolsisdi±-cult,butusingprojectionswecancomputeTnonsubspacesofmodularsymbolswithsmalldimensionmorequickly(see,e.g.,[Ste00,x3.5.2]).LetMbethespaceofmodpmodularsymbolsoflevelp=28279,andletf=gcd(charpoly(T2);deriv(charpoly(T2))).LetVbethekerneloff(T2)(thistakes7minutestocompute).IfV=0,wewouldbedone,sincethendisc(T2)6=02Fp.Infact,Vhasdimension7.We¯ndthe¯rstfewintegersnsothatthecharpolyofTnonVhasdistinctroots,andtheyaren=34,47,53,and89.Wethencomputedcharpoly(T34)directlyonthewholespaceandfoundthatithasdistinctrootsmodulop. 3.2SomeDataAboutWeight4Thefollowingarethevaluationsd=d4(¡0(p))atpofthediscriminantoftheHeckealgebrasassociatedtoS4(¡0(p))forp500.Thisdatasuggestsapattern,whichmotivatesConjecture1below.p235711131719232931374143475359d00000222244666688p6167717379838997101103107109113127131137139d1010101212121416161616181820202224p149151157163167173179181191193197199211223227229233d2424262626282830303232323436363838p239241251257263269271277281283293307311313317331337d3840404242444446464648505052525456p347349353359367373379383389397401409419421431433439d5658585860626262656666686870707272p443449457461463467479487491499d727476767676788080824SpeculationsMotivatedbythepromiseofapatternsuggestedbythetableinSec-tion3.2,wecomputeddk(¡0(p))formanyvaluesofkandp.Ourobser-vationsledustothefollowingresultsandconjectures.Theorem2.Supposepisaprimeandk¸4isaneveninteger.Thendk(¡0(p))�0unless(p;k)2f(2;4);(2;6);(2;8);(2;10);(3;4);(3;6);(3;8);(5;4);(5;6);(7;4);(11;4)g;inwhichcasedk(¡0(p))=0.Proof.From[Rib91],modpeigenformson¡0(p)ofweightkariseexactlyfrommodpeigenformson¡0(1)ofweight(k=2)(p+1).Moreover,thereisanequalityofdimensionsofvectorspaces:dimS(k=2)(p+1)(¡0(1))+dimS(k=2)(p+1)¡(p¡1)(¡0(1))=dimSk(¡0(p)):ThusthedimensionofSk(¡0(p))isbiggerthanthenumberofmodpeigenformswheneverdimS(k=2)(p+1)¡(p¡1)(¡0(1))isnon-zero.Thecasesofdimensionzerocorrespondexactlytothe¯nitelistofexceptionsabove,forwhichonecanexplicitlycalculatethatdk(¡0(p))=0. Notethatfork=2,however,thereisacanonicalidenti¯cationofspacesS(p+1)(¡0(1);Fp)'S2(¡0(p);Fp);describedgeometricallyin[Gro90].Fork=4,thedatasuggeststhatthediscriminantsd4(¡0(p))aresigni¯cantlylargerthanzeroforlargep,andthetableabovesuggestsaformulaoftheform2¢bp=12c(Notentirelyco-incidentally,thisisthedi®erenceindimensionofthespacesS4(¡0(p))andS2(p+1)(¡0(1))).Thisexactformulaisnotcorrect,however,asevidencedbythecasewhenp=139.IfweconsidertheHeckealgebraT4forp=139inmoredetail,however,weobservethatT4­Q139isrami¯edat139,andinparticularcontainstwocopiesofthe¯eldQ139(p139).Justasinthecasewhenk=2andp=389,thereisa\selfcongruence"betweentheassociatedrami¯edeigenformsandtheirGaloisconjugates.Forallotherpintherangeofthetable,thereisnorami¯cation,andallcongruencestakeplacebetweendistincteigenforms.SuchcongruencesaremeasuredbytheindexoftheHeckealgebra,whichisde¯nedtobetheindexofTinitsnormalisationeT.Ifweareonlyinterestedinmodpcongruences(ratherthanmod`congruencesfor`6=p),onecanrestricttotheindexofT­Zpinsideitsnormalisation.Thereisadirectrelationbetweenthediscriminantandtheindex.SupposethatT­Qp=QKiforcertain¯eldsKi=Qp(WemayassumeherethatTisnotnilpotent,forotherwiseboththediscriminantandindexarein¯nite).Thenifip(¡)=ordp([T;eT]),thendp(¡)=2ip(¡)+Xordp(¢(Ki=Qp)):Ifwenowreturntotheexamplek=4andp=139,weseethatthediscrepancyfromthediscriminantdp(¡0(139))=24totheestimate2b139=12c=22isexactlyaccountedforbythetwoeigenformswithcoe±-cientsinQ139(p139),whichcontribute2totheaboveformula.Thisleadsustopredictthattheindexisexactlygivenbytheformulabp=12c.NotethatforprimespthisisexactlythedimensionofSp+3(¡0(1)).Similarcomputationsleadtothefollowingmoregeneralconjecture.Letk=2mbeanevenintegerandpaprime.LetTbetheHeckealgebraassociatedtoSk(¡0(p))andleteTbetheintegralclosureofTinT­Q(whichisaproductofnumber¯elds).Conjecture1.Supposep¸k¡1.Thenordp([eT:T])=¹p12º¢Ãm2!+a(p;m); wherea(p;m)=80ifp´1(mod12),3¢Ãdm3e2!ifp´5(mod12),2¢Ãdm2e2!ifp´7(mod12),a(5;m)+a(7;m)ifp´11(mod12).Here¡xy¢isthebinomialcoe±cient\xchoosey",and°oorandceilingareasusual.TheconjectureisveryfalseifkÀp.Whenk=2,theconjecturespecializestotheassertionthat[eT:T]isnotdivisiblebyp.Apossiblymorefamiliarconcreteconsequenceoftheconjectureisthefollowingconjectureaboutellipticcurves.ThemodulardegreeofanellipticcurveEisthesmallestdegreeofasurjectivemorphismX0(N)!E,whereNistheconductorofE.Conjecture2.SupposeEisanellipticcurveofprimeconductorp.ThenpdoesnotdividethemodulardegreemEofE.Usingthealgorithmin[Wat02],M.Watkinshascomputedmodularde-greesofahugenumberofellipticcurvesofprimeconductorp107,andnotfoundacounterexample.Lookingatsmallerdata,thereisonlyoneellipticcurveEofprimeconductorp20000suchthatthemodulardegreeofEisevenasbigastheconductorofE,andthatisacurveofconductor13723.Thiscurvehasequation[1;1;1;¡10481;408636],modu-lardegreemE=16176=24¢3¢337.Themodulardegreecanbedivisiblebylargeprimes.Forexample,thereisaNeumann-Setzerellipticcurveofprimeconductor90687593whosemodulardegreeis1280092043,whichisover14timesasbigas90687593.Ingeneral,foranellipticcurveofconductorN,onehastheestimatemEÀN7=6¡²(see[Wat04]).5ConjecturesInspiredbyConjecture1First,somenotation.Letpbeanoddprime.Let¡=¡0(p),andletSk(R):=Sk(¡)new­R:ThespacesSkcarryanactionoftheHeckealgebraTnew,andaFrickeinvolutionwp.If122R,thespaceSkcanbedecomposedinto+and¡eigenspacesforwp.WecalltheresultingspacesS+kandS¡krespectively. Similarly,letM+kandM¡kbethe+1and¡1eigenspacesforwponthefullspacesofnewmodularformsofweightkfor¡0(p).Itfollowsfrom[AL70,Lem.7](whichisanexplicitformulaforthetracetolowerlevel)andthefactthatUpandwpbothpreservethenewsubspace,thattheactionoftheHeckeoperatorUponSkisgivenbytheformulaUp=¡p(k¡2)=2wp:ThisgivesrisetotwoquotientsoftheHeckealgebra:T+=Tnew=(Up+p(k¡2)=2)andT¡=Tnew=(Up¡p(k¡2)=2):whereT+andT¡actonS+andS¡,respectively.RecallthateTisthenormalization(integralclosure)ofTinT­Q.LeteTnewdenotetheintegralclosureofTnewinTnew­Q.Lemma1.ThereareinjectionsTnew,!T+©T¡,!eTnew:WenowbeginstatingsomeconjecturesregardingtheringsT§.Conjecture3.Letkp¡1.ThenT+andT¡areintegrallyclosed.Equivalently,allcongruencesbetweendistincteigenformsinSk(Zp)takeplacebetween+and¡eigenforms.Notethatfork=2,therecannotbeanycongruencesbetween+and¡formsbecausethiswouldforce1´¡1modp,whichisfalse,becausepisodd.Thuswerecovertheconjecturethatp-[eT:T]whenk=2.OurfurtherconjecturesgoontodescribeexplicitlythecongruencesbetweenformsinS+kandS¡k.LetE2bethenon-holomorphicEisensteinseriesofweight2.Theq-expansionofE2isgivenexplicitlybyE2=1¡241Xn=1qn0Xdjnd1:Moreover,thefunctionE¤2=E2(¿)¡pE2(p¿)isholomorphicofweight2andlevel¡0(p),andmoreoveronq-expansions,E¤2´E2modp.Lemma2.Letp�3.Letf2Mk(¡0(p);Fp)beaHeckeeigenform.ThenµfisaneigenforminsideSk+2(¡0(p);Fp). Proof.Oneknowsthat@f=µf¡kE2f=12isofweightk+2.Onq-expansions,E2´E¤2modp,andthusforp�3,µf´@f+kE¤2f=12(modp)isthereductionofaweightk+2formoflevelp.ItiseasytoseethatµfisacuspidalHeckeeigenform.LetusnowassumeConjecture3andconsidertheimplicationsfork=4inmoredetail.ThespaceofmodularformsM2(¡0(p);Fp)consistspreciselyofS2andtheEisensteinseriesE¤2.Themapµde¯nedaboveinducesmaps:µ:S+2(Fp)!S4(Fp);µ:M¡2(Fp)!S4(Fp):Theimagesaredistinct,sinceµf=µgimplies(withsomecareaboutap)thatf=g.Conjecture4.Letf2S2(Zp)andg2S4(Zp)betwoeigenformssuchthatµf´gmodp.Thentheeigenvalueofwponfandghaveoppositesigns.Assumingthis,wegetinclusions:µS+2(Fp),!S¡4(Fp);µM¡2(Fp),!S+4(Fp):Nowwearereadytostateourmainconjecture:Conjecture5.ThereisanHeckeequivariantexactsequence:0-µS+2(Fp)-S¡4(Fp)-S+4(Fp)-µM¡2(Fp)-0:Moreover,themapS¡4(Fp)!S+4(Fp)hereisthelargestsuchequivariantmapbetweenthesespaces.Equivalently,aresidualeigenformofweight4andlevelpoccursinboththe+and¡spacesifandonlyifitisnotintheimageofµ.LetusgivesomeconsequencesofourconjecturesfortheindexofTnewinsideitsnormalisation.Fixaresidualrepresentation½:Gal(Q=Q)!GL2(Fq)andconsidertheassociatedmaximalidealminsideT4.If½liesintheimageofµthenourconjectureimpliesthatitisnotcongruenttoanyothereigenform.If½isnotintheimageofµ,thenitshouldariseexactlyfromapairofeigenforms,oneinsideS+4(Qp)andoneinsideS¡4(Qp).Supposethatq=pr.Ifthereisnorami¯cationinT­Qover p(thisisoftentrue),thenthe+and¡eigenformswillbothbede¯nedovertheringW(Fq)ofWittvectorsofFq.SinceUp=ponS¡4and¡ponS+4,theseformscanbeatmostcongruentmodulop.ThusthecompletedHeckealgebra(T4)misexactlyf(a;b)2W(Fq)©W(Fq);ja´bmodpg:Oneseesthatthishasindexq=prinsideitsnormalisation.Thusthe(logofthe)totalindexisequaltoPrioveralleigenformsthatoccurinsideS+4andS¡4,whichfromourexactsequenceweseeisequaltodimS¡4¡dimS+2:Conjecture1whenk=4,wouldthenfollowfromtheequalityofdimen-sions:dimS¡4(Fp)¡dimS+2(Fp)=¹p12º:Weexpectthatsomethingsimilar,butalittlemorecomplicated,shouldhappeningeneral.Inweight2k,therearemodpk¡rcongruencesexactlybetweenformsintheimageofµr¡1butnotofµr.5.1ExamplesWewritesmalls'sandm'sfordimensionsbelow.Letp=101.Thens+=1,m¡2=7+1=8,s¡4=9,s+=16.Wepredicttheindexshouldbe9¡1=8=b101=12c.Inthetablebelow,weshowthecharacteristicpolynomialsofT2onS¡4andS+4,andforweight2,wetakethecharacteristicpolynomialofµT2(orthesame,takingF(x=2)whereF(x)isthecharacteristicpolynomialofT2).NotethatwehavetoaddtheEisensteinseries,whichhascharacteristicpolynomialx¡1¡2,whichbecomesx¡6´x+95mod101underµ.FactorsoftheCharacteristicPolynomialofT2forp=101.µS+2(F101)S¡4(F101)S+4(F101)µM¡2(F101)(x)(x)(x+46)(x+95)(x+46)(x+95)(x2+90x+78)(x2+58x+100)(x2+58x+100)(x2+96x+36)(x5+2x4+27x3(x2+90x+78)(x3+16x2+49x2+7x+65)(x2+96x+36)+35x+72)(x3+16x2+35x+72)(x5+2x4+27x3+49x2+7x+65) Herearesomefurtherconjectureswhenk�4.Conjecture6.Letpandkbesuchthat4kp¡1.ThereisanHeckeequivariantexactsequence:0-µS+k¡2(Fp)-S¡k(Fp)-S+k(Fp)-µS¡k¡2(Fp)-0:Moreover,allformsnotintheimageofµcontributemaximallytotheindex(afactorofp(k¡2)=2).Thusthetotalindexshouldbeequalto(k¡2)2(dimS+k¡dimS¡k¡2)+theindexatlevelpandweightk¡2.ThisisthesumkXn=2(2n¡2)2(s+n¡s¡2n¡2):Whenk=4,weneedtoaddtheEisensteinseriestoS¡2inourpreviousconjecture.Notethats+¡s¡k¡2=s¡k¡s+k¡2fork�4(andwiths¡replacedbym¡2whenk=2).Thisfollowsfromourconjectures,butcaneasilybeproveddirectly.Asanexample,whenp=101,wehaves+=1,s¡4=9,s+=17,s¡8=26,s+=34,s¡12=42,s+=51,andsowewouldpredicttheindexesIktobeasgiveninthefollowingtable:kIk?2048=8+0624=24+0851=48+31083=80+312123=120+314177=168+9Thisagreeswithourconjecturalformula,whichsaysthattheindexshouldbeequalinthiscaseto8Ãk=22!+3Ãdk=6e2!:italsoagreeswithcomputation.ReferencesA.O.L.AtkinandJ.Lehner,Heckeoperatorson¡0(m),Math.Ann.185(1970),134{160. 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