Du al Singularities Igor Burban In tro duction consider quotien singularities G where - PDF document

DuValSingularitiesIgorBurban1IntroductionWeconsiderquotientsingularitiesC2=G,whereGSL2(C)isanitesubgroup.SincewehaveaniteringextensionC[[x;y]]GC[[x;y]],theKrulldimensionofC[[x;y]]Gis2.NotethatC[[x;y]]2=C[[(x+y;xy)]]C[[x;y]].Inordertogetakindofbijectionbetweennitesubgroupsandquotientsingularitiesweneedthefollowingdenition.Denition1.1LetGGLn(C)beanitesubgroup.Anelementg2Giscalledpseudo-re\rectionifgisconjugatedtodiag(1;1;:::;1;),where6=1.AgroupGiscalledsmallifitcontainsnopseudo-re\rections.Proposition1.21.LetGGLn(C)beanitesubgroup.ThenC[[x1;x2;:::;xn]]G=C[[x1;x2;:::;xn]]G0;wherethegroupG0isacertainsmallsubgroupofGLn(C).2.LetG0;G00GLn(C)betwosmallsubgroups.ThenC[[x1;x2;:::;xn]]G0=C[[x1;x2;:::;xn]]G00ifandonlyifG0andG00toareconjugated.3.LetGGLn(C)beasmallnitesubgroup.ThenC[[x1;x2;:::;xn]]GisalwaysCohen-Macaulay.4.LetGGLn(C)beasmallnitesubgroup.ThenC[[x1;x2;:::;xn]]GisGorensteiniGSLn(C).Remark1.3NotethateverysubgroupofSLn(C)issmall.Nowwewanttoanswerthefollowingquestion:whatarenitesubgroupsofSL2(C)moduloconjugation?1 2FinitesubgroupsofSL2(C)Lemma2.1EverynitesubgroupofSLn(C)(GLn(C))isconjugatedtoasub-groupofSU(n)(U(n)).Proof.Let(;)beahermitianinnerproductonCn.Denehu;vi:=1jGjXg2G(gu;gv):(NotethatincaseGU(n)itholdsh;i=(;)).Thenh;iisanewhermitianinnerproductonCn:1.hu;ui0.2.hu;ui=0impliesu=0.3.hu;vi=hv;ui.Itholdsmoreoverhhu;hvi=1jGjXg2G(ghu;ghv)=hu;vi:HenceGisunitarywithrespecttoh;i.Moreover,h;ihasanorthonor-malbasisandletS:(Cn;h;i)!(Cn;(;))beamapsendingvectorsofthechoosenorthonormalbasisoftheleftspacetovectorsofthecanonicalbasisoftherightone.Itholdshu;vi=(Su;Sv).Weknowthathgu;gvi=hu;viforallg2G,u;v2Cn.Henceweget(Su;Sv)=(Sgu;Sgv),or,settingS1uandS1vinsteadofuandv(u;v)=(SgS1u;SgS1v)forallg2Gandu;v2Cn.NowwewanttodescribeallnitesubgroupsofSU(2).RecallthatSU(2)=fA2SL2(C)jA1=Ag=fjjj2+jj2=1g:So,fromthetopologicalpointofviewSU(2)S3.Theorem2.2Thereisanexactsequenceofgrouphomomorphisms1!Z2!SU(2)!SO(3)!1:Moreprecisely,ker()=f1001g:TopologicallyisthemapS32:1!RP3.2 FromthistheoremfollowsthatthereisaconnectionbetweennitesubgroupsofSU(2)andSO(3).TheclassicationofniteisometrygroupsofR3isaclassicalresultofF.Klein(actuallyofPlaton).TherearethefollowingnitesubgroupsofSO(3;R):1.AcyclicsubgroupZn,generated,forinstance,by0cos(2n)sin(2n)0sin(2n)cos(2n)000112.DihedralgroupDn,jDnj=2n.Itistheautomorphismgroupofaprisma.Itisgeneratedbyarotationaandare\rectionbwhichsatisfytherelationsan=e;b2=e;(ab)2=e.Notethatthelastrelationcanbechoosenasba=an1borbab1=a1.3.GroupofautomorphismsofaregulartetrahedronT=A4.NotethatjTj=12.4.GroupofautomorphismsofaregularoctahedronO=S4,jOj=24.5.GroupofautomorphismsofaregularicosahedronI=A5,jIj=60.Observethatallnon-cyclicsubgroupsofSO(3)haveevenorder.Lemma2.31001istheonlyelementofSU(2)ofdegree2.Proof.Theproofisasimplecomputation.2=2jj2+2jj2=1001implies2=2or=.Hence=x;x2Ror=ix;x2R.Moreover6=0sinceotherwisemustholdjj2=1.Incase2Rwehave+=2hence=0and=1.Ifa=ixispurelyimaginarythen2jj2=jxj2jj20:LetGSU(2)beanitesubgroup,:SU(2)!SO(3)the2:1surjec-tion.Considertwocases.1.jGjisodd.ThenG\Z2=feg(noelementsoforder2inG).Soker()\G=fegandG!(G)isanisomorphism.HenceGiscyclic.2.jGjiseven.TheduetotheSylow'stheoremGcontainsasubgroupoforder2kandhencecontainsanelementoforder2.ButthereisexactlyoneelementofthesecondorderinSU(2)(seethelemmaabove).Henceker()GandG=1((G)).So,inthiscaseGisthepreimageofanitesubgroupofSO(3).3 FromwhatwassaidwegetthefullclassicationofnitesubgroupsofSL2(C)moduloconjugation.1.AcyclicsubgroupZk.Letgbeitsgenerator.gk=eimpliesg"00"1;where"issomeprimitiverootof1ofk-thorder.2.BinarydihedralgroupDn,jDnj=4n.TondthegeneratorsofDnwehavetoknowtheexplicitformofthemap:SU(2)!SO(3).Skippingalldetailswejustwritedowntheanswer.Dn=ha;biwithrelations8an=b2b4=ebab1=a1Tobeconcrete,a="00"1;"=exp(in);b=0110:3.BinarytetrahedralgroupT,jTj=24.T=h;;i;where=i00i;=0110;=1p2"7"7"5";"=exp(2i8):4.BinaryoctahedralgroupO,jOj=48.Thisgroupisgeneratedby;;asinthecaseofTandby="00"7:5.FinallywehavethebinaryicosahedralsubgroupI,jIj=120.I=h;i,where="300"2;=1p5"+"4)"2"3"2"3""4;"=exp(2i5):Remark2.4TheproblemofclassicationofnitesubgroupsofGL2(C)ismuchmorecomplicated.Indeed,everynitesubgroupGGL2(C)canbeembeddedintoSL3(C)viathegroupmonomorphismGL2(C)!SL3(C)g7!g001det(g):FinitesubgroupsofGL2(C)givemainseriesofnitesubgroupsofSL3(C).Nowwecancomputethecorrespondinginvariantsubrings.4 3DescriptionofDuValsingularities1.AcyclicsubgroupZn=hgi,g:x7!"x;y7!"1y;"=exp(2in).ItisnotdiculttoseethatX=xn;Y=ynandZ=xygeneratethewholeringofinvariants.C[[x;y]]Zn=C[[X;Y;Z]]=(XYZn)C[[x;y;z]]=(x2+y2+zn):ItisanequationofAn1-singularity.2.BinarydihedralgroupDn.:x7!"xy7!"1y:x7!yy7!x;where"=exp(in).ThesetofinvariantmonomialsisF=x2n+y2n;H=xy(x2ny2n);I=x2y2:TheysatisfytherelationH2=x2y2(x4n+y4n2x2ny2n)=IF24I2(n+1):ThestandartcomputationsshowthatC[[x;y]]DnC[[x;y;z]]=(x2+yz2+zn+1):ItisanequationofDn+2-singularity.3.ItcanbecheckedthatforthegroupsT;O;Iwegetsingularities(a)E6:x2+y3+z4=0,(b)E7:x2+y3+yz3=0,(c)E8:x2+y3+z5=0.Wegetthefollowingtheorem:Theorem3.1DuValsingularitiesarepreciselysimplehypersurfacesingular-itiesADE:Wewantnowtoanswerournextquestion:whatareminimalresolutionsanddualgraphsofDuValsingularities?4A1-singularityConsiderthegermofA1-singularityS=V(x2+y2+z2)A3.Considertheblow-upofthissingularity.~A3=f((x;y;z);(u:v:w))2A3P2jxv=yu;xw=zu;yw=zug:5 Takethechartu6=0,i.e.u=1.Weget8x=xy=xvz=xwWhatis~S=1(Snf0g)?Considerrstx6=0(itmeansthatwearelookingfor1(Snf0g)).x2+x2v2+x2w2=0;x6=0;or1+v2+w2=0;x6=0:Inordertoget~Sweshouldallowxtobearbitrary.Inthischart~SisacylinderV(1+v2+w2)A3.Whatis1(0)?Obviouslyitistheintersectionof~Swiththeexceptionalplane((0;0;0);(u:v:w)).Inthischartwejusthavetosetx=0inadditiontotheequationofthesurface~S.1(0)=1+v2+w2=0x=0:WeseethatE=1(0)isrationalandsinceall3chartsof~Saresymmetric,weconcludethatEissmooth,soE=P1.NowwehavetocomputetheselntersectionnumberE2.Wedoitusingthefollowingtrick.Let:~X!Xbetheminimalresolution.Itinducesanisomorphismofeldsofrationalfunctions:C(X)!C(~X).Letf2C(~X)bearationalfunction.Thenitholds:(f):E=degE(OE\nO~X(f))=degE(OE)=0:Inparticularitholdsforf2mOX;0:(f):E=0:Considerthefunctiony2OX;0.Inthechartu6=0wegety=xu.Whatisthevanishingsetofy?1.x=0isanequationofE.2.u=0impliesv2+1=0orv=i.Hence(f)=E+C1+C2andwehavethefollowingpicture:6 C2C1ESo,(f):E=E2+C1:E+C2:E=E2+2=0.FromitfollowsE2=2.5E6-singularityInthissectionwewanttocomputeaminimalresolutionanddualgraphofE6-singularityX=V(x2+y3+z4)A3.Let~X!Xbeaminimalresolution,E=[Ei=1(0)theexceptionaldivisor.InordertocomputeselntersectionnumbersE2iwehavetoconsiderthemap:C(X)!C(~X).Letf2mXOX,thenequalities(f):Ei=0willimplytheselntersectionnumbersofEi.Firststep.LetX=V(x2+y3+z4)A3,f=x.Considertheblow-upofA3:~A3=f((x;y;z);(u:v:w))2A3P2jxv=yu;xw=zu;yw=zug:Takerstthechartv6=0(i.e.v=1).Wegetequations8x=yuy=yz=yw:Togettheequationofthestricttransformof~X1weassumethaty6=0andy2u2+y3+y4w4=0;oru2+y+y2w4=0:Inthischart~X1issmooth:Jacobicriteriumimplies8u=01+2yw4=0w3y2=0:Itiseasytoseethatthissystemhasnosolutions.Inthechartu=1thestricttransform~X1isagainsmooth.Considernallythechartw=1.8x=zuy=zvz=z:7 Thestricttransformisz2u2+z3v3+z4=0;oru2+zv3+z2=0:Jacobicriteriumimpliesthatthissurfacehasauniquesingularpointu=0;v=0;z=0,orintheglobalcoordinates((0;0;0);(0:0:1)).Weseethatthispointindeedliesonlyinoneofthreeanechartsof~X1.Wenowneedanequationofexceptionalbre.Theexceptionalbreisbythedenitiontheintersection~X1\f((0;0;0);(u:v:w))g.Togetitslocalequationinthechartw=1wejusthavetosetz=0intheequationof~X1.z=0impliesu=0.HencewegetE0=f((0;0;0);(0:v:1))gA1:GoingtotheotherchartsshowsthatE0P1.Finally,thefunctionfinthischartgetstheformf=zu.Agreement.Sincethenumberofindicesdependsexponentiallyonthenumberofblowing-ups,weshalldenotethelocalcoordinatesofallchartsofallblowing-ups~Xibytheletters(x;y;z).Secondstep.Wehavethefollowingsituation:8surfacex2+zy3+z2=0functionf=xzexceptionaldivisorE0x=0;z=0:Consideragaintheblowing-upofthissurface.Itiseasytoseethattheonlyinterestingchartis8x=yuy=yz=yv:Wegetthestricttransformy2u2+yvy3+y2v2=0;y6=0;oru2+y2v+v2=0:Againy=0;u=0;v=0istheonlysingularityoftheblown-upsurface.Theexceptionalbreofthisblowing-uphastwoirreduciblecomponents:y=0impliesuiv=0(wecallthiscomponentsE01andE002).Whatisthepreimage(underpreimagewemeanitsstricttransform)ofE0?x=0;z=0impliesu=0;v=0.Thefunctionf=xzgetsinthischarttheformf=y2uv.8 Thirdstep.Wehavethefollowingsituation:8surfacex2+y2z+z2=0functionf=xy2zexceptionaldivisorE0x=0;z=0exceptionaldivisorE1y=0;xiz=0:EEE110'''Letusconsiderthenextblowing-up.8x=yuy=yz=yv:Thestricttransformisy2u2+y2yv+y2v2=0;y6=0;oru2+yv+v2=0:ItisanequationofA1-singularity(anditmeansthatwearealmostdone).TheexceptionalbreconsistsagainoftwoirreduciblecomponentsE02;E002.Theylocalequationsarey=0;uiv=0.Thefunctionfisuy4v.ItiseasytoseethatthepreimageofE0isu=0;v=0.WhataboutthepreimageofE1?OursurfaceliesintheanechartA3em-beddedintoA3P2viathemap(y;u;v)7!((yu;y;yv);(u:1:v)).Butthentheconditiony=0wouldimplythatthepreimageofE1liesintheexceptionalplane((0;0;0)(u:1:v)).Butitcannotbetrue!ThesolutionofthisparadoxisthatthepreimageofofE1liesinanothercoordinatechart.Consider8x=xy=xuz=xv;9 Thestricttransformisx2+x2u2xv+x2v2=0;x6=0or1+xu2v+v2=0:TheequationsoftheexceptionalbreE2inthischartarex=0,whatimpliesv=i.ThepreimageofE1xiz=0y=0isgivenbyxixv=0xu=0;x6=0;hence8v=iu=0xarbitraryInthepictureitlookslike:E'E"E'E"122(u=0, v=-i, x=0)(u=0,v=i, x=0)1Itiseasytoseethatallintersectionsaretransversal.Fourthstep.Wehavethefollowingsituation:therearetwocoordinatecharts8surfacex2+yz+z2=0functionf=xy4zexceptionaldivisorE0x=0;z=0exceptionaldivisorE2y=0;xiz=0:8surface1+xy2z+z2=0functionf=x4y2zexceptionaldivisorE1y=0;z=iexceptionaldivisorE2x=0;z=i:10 E2'E2''E0E1'E1''Ournextstepistheblowing-upatthepoint(0;0;0)intherstcoordinatechart.Again,inordertogetequationsofthepreimagesofE0andE2wehavetoconsidertwocoordinatecharts.8x=yuy=yz=yw:Thestricttransformisacylinderu2+v+v2=0:ThepreimageofE0isgivenbyequationsu=0;v=0,theexceptionalbreE3isgivenbyu2+v+v2=0;y=0,ourfunctionf=uy6v.Inanotherchartwehave8x=xy=xuz=xv:Thestricttransformisgivenbyx2+x2uv+x2v2=0;x6=0or1+uv+v2=0:TheexceptionalbreE3isgivenby1+uv+v2=0;x=0,thepreimagesofE02andE002aregivenbyu=0;v=i,f=x6u4v.HenceourexceptionalbreEisgivenbythefollowingcongurationofprojectivelines:11 E0E'2E'1E"1E"2E3ThedualgraphofthiscongurationisFifthstep.Wehavetotakeintoaccountthreecoordinatechartsofaminimalresolution.~X:x2+z+z2=0f:xy4zE0:x=0;z=0E3:y=0;x2+yz+z2=08~X:1+yz+z2=0f:x6y4zE3:x=0;x2+yz+z2=0E2:y=0;z=i8~X:1+xy2z+z2=0f:x6y2zE1:y=0;z=iE2:x=0;z=i:Nowwehavetocomputethedivisor(f).LetXA3beanormalsurface,YXaclosedcurve,f2C(X)arationalfunction.SupposethatpC[X]istheprimeidealcorrespondingtoY.ThenC[X]pisadiscretevaluationringandmultY(f)=valC[X]p(f):1.Considertherstchart.Letf=xy6z=0.x=0impliesz=0orz=1.x=0,z=0isanequationofE0,x=0;z=1isthestricttransformCofthecurvex=0inV(x2+y3+z4)A3.WhatisthemultiplicityofE0?Thegeneratorofthemaximalidealofthering(C[x;y;z]=(x2+z+z2))pisxandx2z.ThereforemultE0(f)=3.y=0givesanequationofE3.ItiseasytoseethatmultE3(f)=6.NotethatthecurveChastransversalintersectionwithE3atthepointx=0;y=0;z=1.2.Considerthesecondchart.Inthischartholdsf=x6y4z=0:z=0isimpossible,x=0cutouttheequationofE3andy=0equationsofE02andE002.ThesamecomputationasaboveshowsthatmultE3(f)=6(whatisnotsurpriseandmakesussurethatwedidnotmakeamistakeincomputations)andmultE02(f)=multE002(f)=4.12 3.InthesamewayweobtainthatmultE01(f)=multE001(f)=2:Thereforeweobtain:(f)=6E3+4(E02+E002)+2(E01+E001)+3E0+C:WehaveC:E3=1,allotherintersectionnumbersofCwithirreduciblecomponentsofEarezero.Intersectionnumbersofirreduciblecomponentsarecodedinthedualgraph(whichisE6,seethepictureabove).Thewholejobwasdoneinordertocomputeself-intersections.(f):E0=6+3E20=0=)E20=2:Inthesamewayweconcludethattheotherselntersectionnumbersare2.Remark5.1LetXbeanormalsurfacesingularity,:~X!Xitsminimalresolution,E=nSi=1Ei=1(0)theexceptionaldivisor.Supposethat~Xisagoodresolution,allEiP1andE2i=2.ThenXisasimplehypersurfacesingularity.Indeedweknowthattheintersectionmatrix(Ei:Ej)n=1isnegativelyde-nite.LetbethedualgraphofX.Thenthequadraticformgivenbyintersec-tionmatrixcoinsidewiththeTitsformofthedualgraph:Q(x1;x2;:::;xn)=2(nXi=1x2X1=i=naijxixj);whereaijisthenumberofarrowsconnectingverticesiandj.FromthetheoremofGabrielweknowthatQisnegativelydenite(andquiverisrepresentationnite)ifandonlyif=ADE.Sinceoursingularityisrational,itistautanduniquelydeterminedbyitsdualgraph.62-dimensionalMcKaycorrespondenceRecallthatwedenedDuValsingularitiesasquotientsingularitiesC[[x;y]]G,whereGSU(2)issomenitesubgroup.Anaturalquestionis:arethereanyconnectionsbetweentherepresentationtheoryofGandgeometryoftheminimalresolutionofasingularity?Letusrecallsomestandartfactsaboutrepresentationsofnitegroups.Theorem6.1(Mashke)LetGbeanitegroup.ThenthecategoryofC[G]-modulesissemi-simple.ThistheoremmeansthatanyexactsequenceofC[G]-modulessplits.Inpartic-ular,everynite-dimensionalC[G]-moduleisinjectiveandprojective.Butan13 indecomposableprojectivemodulebyatheoremofKrull-Schmidtisisomorphictoadirectsummandoftheregularmodule.LetC[G]sMi=1niibeadirectsumdecompositionofC[G].Then1;2;:::;sthewholelistofindecomposableC[G]-modules.Lemma6.2LetC[G]=sLi=1niibeadecompositionoftheregularmoduleintoadirectsumofindecomposableones.ThenitholdsdimC(i)=ni:Inparticular,thefollowingidentityistrue:sXi=1n2=jGj:Denition6.3LetGbeagroup,(;V),:G!End(V)itsrepresentation.Thecharacterofarepresentationisthefunction:G!Cdenedbytherule(g)=Tr((g)).Remark6.41.Itiseasytoseethatthecharacterdoesnotdependonthechoiceofarepresentativefromtheisomorphismclassofarepresentation:Tr((g))=Tr(S1(g)S):2.Itholds:\n = ; =+ :Inotherwords,denesaringshomomorphismfromtheGrothendieckringofC[G]toC.3.Itholds:(h1gh)=Tr((h1gh))=Tr((h)1(g)(h))=Tr((g))=(g):Itmeansthatisacentralfunction,i.e.afunctionwhichisisconstantonconjugacyclassesofG.Theorem6.5AnitedimensionalrepresentationofanitegroupGisuniquelydeterminedbyitscharacter.14 Ideaoftheproof.Let'; betwocentralfunctionsonG.Seth'; i:=1jGjXg2G(g) (h):ItdenesanhermitianinnerproductonthespaceofallcentralfunctionsonG.Thetheoremfollowsfromthefactthat1;2;:::;sisanorthonormalbasisofthisvectorspace.Indeed,letbeanynite-dimensionalrepresentationofG.Thenweknowthatmii:Thenitobviouslyholdsmi=h;ii.Corollary6.6ThenumberofindecomposablerepresentationsofanitegroupGisequaltothenumberofitsconjugacyclasses.Denition6.7(McKayquiver)LetGSU(2)beanitesubgroup,0;1;:::;sallindecomposablerepresentationsofG.Let0bethetrivialrepresentation,natthenaturalrepresentation(i.e.therepresentationgivenbytheinclusionGSU(2)).DenetheMcKaygraphofGasthefollowing:1.Verticesareindexedby1;:::;s(weskip0).2.Leti\nnat=sMj=0aijj:(or,thesameinat=sXi=0aijj:Thenweconnectverticesiandjbyaijvertices.Itiseasytoseethat0\nnat=nat.Remark6.8Itholdsaij=aji.Indeed,aij=hinat;ji=1jGjXg2Gi(g)nat(g)j(g)=Xg2Gi(g)nat(g)j(g1)(hereweusethatgn=1andhence(g)n=id.Fromthisfollows(g)diag("1;"2;:::;"k)and(g1)diag("11;"12;:::;"1k)=diag("1;"2;:::;"k):)Sincenatisthenaturalrepresentation,allnat(g)2SU(2);g2G:LetA2SU(2).IfAdiag(a;b)thenA1diag(b;a)(ab=1.)Thereforewehavenat(g)=nat(g1).Thenwecancontinueourequality:Xg2Gi(g)nat(g)j(g1)=Xg2Gi(g)nat(g1)j(g1)=hi;natji=aji:15 Example6.9LetG=D3beabinarydihedralgroup.Aswealreadyknow,jD3j=12.ThegroupD3hastwogeneratorsa;bwhichsatisfythefollowingrelations:8a3=b2b4=eaba=b1:ThegroupD3has41-dimensionalrepresentationsa=1;b=1;a=1;b=1;a=1;b=ianda=1;b=i.Thenaturalrepresentationisalsoknown:itisjusta="00"1;b=0110:where"=exp(i6)=12p32i:Thereisalsoanotheroneirreducible2-dimensionalrepresentation:a=cos23isin23isin23cos23;b=0110:WehavefoundallindecomposablerepresentationsofG:1+1+1+1+4+4=12=jD3j.Wecansumuptheobtainedinformationintothecharactertable.(a)(b)dim0111trivial111121i131i14102natural5102FromthistablewecanderivethewholestructureoftheGrothendieckringofjD3j.2canbeonly0+1+5:Inthesameway5nat=2+3+4.WegettheMcKaygraphofjD3j:1221521043ObservethatweobtainedthedualgraphoftheD5-singularity.NotethatthefundamentalcycleoftheD5-singularityisZfund=E1+2E4+2E5+E2+E3:ButthecoetientsofthisdecompositionarethesameasthedimensionsoftherepresentationscorrespondingtotheverticesoftheMcKayquiver.16 Theorem6.10(McKayobservation)LetGSU(2)beanitesubgroup,C[[x;y]]Gthecorrespondinginvariantsubring.ThentheMcKayquiverofGcoinsidewiththedualgraphofC[[x;y]]G,dimensionsoftherepresentationcor-respondingtoavertexofMcKayquiverisequaltothemultiplicityofthecorre-spondingcomponentoftheexceptionalbreinthefundamentalcycle.17