The
Ekedahl
invariants
for
finite
groups
Ivan Martino1
Departmentof Mathematics,UniversityofFribourg,1700Fribourg,Switzerland
In2009Ekedahlintroducedcertaincohomologicalinvariantsoffinitegroupswhich are naturally related to the Noether Problem. We show that these invariants aretrivialfor every finite groupinGL3(k) and for thefifthdiscreteHeisenberg group H5. Moreover in the case of finite linear groups with abelian projective reduction,theseinvariants satisfya recurrencerelation inacertainGrothendieck groupforabeliangroups.
LetV beafinitedimensionfaithfullinearrepresentationofafinitegroupG overafieldk ofcharacteristic primeto theorderofG.InspiredbyaworkofBergström[1],Ekedahlin[4]and [5]investigatedamotivic versionofpointcountingoverfinitefields.OneapplicationofEkedahl’sresultsistostudywhentheequality
{GL(V )/G} = {GL(V )} (1)
holdsintheKontsevichvalueringK0(Vark) ofalgebraick-varieties.
Alltheknowncases wherethisequalityfailsarecounterexamplestotheNoetherProblem.Inthe begin-ningofthelastcentury,Noether[13]wonderedabouttherationalityofthefieldextensionk(V )G/k forany
finitegroupG andany fieldk,where k(V )G aretheinvariants ofthefieldofrationalfunctionsk(V ) over theregularrepresentationV ofG.(TheNoetherProblemcanbe statedforany arbitraryfield,butwewill notneedthefullgenerality.)
Thefirst counterexample,Q(V )Z/47Z/Q,was givenby Swanin[18] and it appearedduring1969. Inthe
1980s morecounterexamples werefound: for everyprime p Saltman[16] and Bogomolov [3] showed that there exists a group of order p9 and, respectively, of order p6 such that the extension C(V )G/C is not
rational.
Saltman used the second unramified cohomology group of the field C(V )G, H2
nr(C(V )G,Q/Z), as a
cohomological obstruction to rationality. Later, Bogomolov found a group cohomology expression for H2nr(C(V )G,Q/Z) whichnowbearshisnameand isdenotedbyB
0(G).
E-mailaddress:ivan.martino@unifr.ch.
1 PartialsupportbytheSwissNationalScienceFoundation:GrantNo. PP00P2_150552/1.
Published in -RXUQDORI3XUHDQG$SSOLHG$OJHEUD ± which should be cited to refer to this work.
RecentlyHoshi,KangandKunyavskiiinvestigated thecasewhere|G|= p5.TheyshowedthatB
0(G)= 0
ifandonlyifG belongsto theisoclinismfamilyφ10;see[8].
In2009,Ekedahl[5]defines,foreveryintegerk, acohomological map
Hk: K
0(Vark)→ L0(Ab),
where L0(Ab) isthegroupgenerated bytheisomorphismclasses{G} offinitelygenerated abeliangroups
G under therelation{A⊕ B}={A}+{B}.
Let Li be theclass of theaffine space Ai
k in K0(Vark). Inparticular,L0 is theclass of apoint {∗}=
{Spec(k)}. We observe that Li is invertible in K
0(Vark). To define Hk on K0(Vark) is enough to set
Hk({X}/Lm) = {Hk+2m(X;Z)} for everysmooth and proper k-variety X (for more details see Section 3
in[11]).
The class {BG} ofthe classifyingstack of G isan element of K0(Vark) (seeProposition 2.5.bin[11])
andso onecan define:
Definition1.2.Foreveryintegeri,thei-thEkedahlinvariant ei(G) ofthegroupG isH−i({BG}) inL0(Ab).
WesaythattheEkedahlinvariantsofG aretrivial ifei(G) = 0 forallintegers i= 0.
InProposition2.5.aof[11],theauthor rephrasestheequality(1)intermsof algebraicstacks, usingthe expression
{BG} = {GL(V )/G}
{GL(V )} ∈ K0(Vark).
Since {GL(V )} is invertible inK0(Vark),the equation(1) holds ifand onlyif{BG}={∗} and, ifthis is
thecase,thentheEkedahlinvariantsofG aretrivial, becauseH0({∗})={Z} andHk({∗})= 0 for k= 0.
These new invariants seema naturalgeneralization of the Bogomolov multiplierB0(G) because of the
following result.
Theorem 1.3. (SeeThm. 5.1of [4].) Assume char(k)= 0. If G isa finite group, thenei(G) = 0 forevery
i < 0, e0(G) = {Z}, e1(G) = 0 and e2(G) = {B0(G)∨}, where B0(G)∨ is the Pontryagin dual of the
Bogomolov multiplierof thegroupG.
Moreover, for i > 0, the invariant ei(G) is an integer linear combination of classes of finite abelian
groups.
Using that e2(G) = {B0(G)∨}, one finds some groups with non-trivial Ekedahl invariants (and so
{BG} = {∗}).
Corollary. If G is one of the group of order p9 defined in [16], of order p6 defined in [3] or of order
p5 belonging to the isoclinism family φ
10 (see [8]), then the second Ekedahl invariant is non-zero and so
{BG}= {∗}.
ItisnotclearifhigherEkedahlinvariantsareobstructions totherationalityoftheextensionk(V )G/k.
InCorollary5.8of[4],itisalsoprovedthat{BZ/47Z}= {∗}∈ K0(VarQ),butitisunknownif{BG}= {∗}
impliesanegativeanswertotheNoetherproblem.
Totheauthor’sknowledge,therearenoexamplesoffinitegroupG suchthatB0(G)= 0 (i.e.e2(G) = 0)
ande3(G)= 0.ItisworthmentioningthatPeyre[14]showedaclassoffinitegroupshavingB0(G)= 0,but
non-trivialthirdunramifiedcohomology:itisnotcleartotheauthorifthisisconnectedtothenon-triviality ofhigherEkedahlinvariants.
TherearealotofclassesoffinitegroupsthathavetrivialEkedahlinvariants(seeSection4of[11]).We wantto mentiononeexamplerelevantforthis work:ifG⊂ GL1(k),then {BG}={∗}∈ K0(Vark), under
the condition thatk contains a primitive root of unity of degree the exponent of G [4, Proposition 3.2]. Since, everyabeliangroupA is adirectproduct ofcyclic groups,{Bμn}={∗} impliesthat{BA}={∗}:
thiswillbe usefulintheproofofTheorem 3.2.
Inthis paperwefirstgeneralizethejustmentionresult.
Theorem2.4.Assumechar(k)= 0 andassumethatk containsaprimitiverootofunityofdegreetheexponent
ofG. If G isafinite subgroup ofGL3(k),then {BG}={∗} inK0(Vark) andtheEkedahl invariantsofG
aretrivial.
ThecasewhenG isafinitesubgroupofGL4(k) ismorecomplicatedbecauseitinvolvesthestudyofthe
classinK0(Vark) ofresolutionofsingularitiesoftheaffinevarietiesA3/A,forafinite groupA⊂ GL3(k).
Nevertheless,wecan stilltacklethisproblemifweassumethegroundfieldtobe thecomplexfield(and weusethetoricvarietiestools).Indeed,inhigherdimensionweareabletoprovethattheEkedahlinvariants satisfyakindofrecurrenceequationinL0(Ab).
Theorem 3.2. Let G be a finite subgroup of GLn(C) and let H be the image of G under the canonical
projectionGLn(C) → PGLn(C).
IfH isabelianandifPn−1
C /Hhasonlyzerodimensionalsingularities,theneachsingularityistoroidaland
foreveryinteger k
ek(G) + ek+2(G) +· · · + ek+2(n−1)(G) ={H−k(X;Z)},
whereX →Pn−1
/H isasmoothandproper resolutionwithnormal crossing toric exceptionaldivisors.
Wefocus,then,onthep-discreteHeisenberggroupHp,whereweonlydealwithcyclicquotient
singular-ities.This is aninterestingcandidate for thestudy of theEkedahl invariants,because B0(Hp)= 0 (using
Lemma4.9in[3])andsothefirstunknownEkedahlinvariantise3(Hp).
Theorem4.4. TheEkedahl invariantsofthefifthdiscrete HeisenberggroupH5 are trivial.
Kanghasproved thattheNoetherproblem fortheHeisenberg groupHp hasapositiveanswer.Indeed
usingTheorem1.9 of[9]oneshowsthattheextension C(xg,g∈ Hp)Hp/C isrational.
We show a general approach for the study of the Ekedahl invariants of Hp, butwe narrow down our
investigationtop= 5 becauseofthedifficultiestoextendthetechnicalresultinTheorem 4.7.Therefore,it isnaturalto conjecturethat:
Conjecture.TheEkedahl invariantsof theHeisenberggroupHp of order p3 are trivial.
Afterapreliminarysectionwherewereview thetheory oftheEkedahlinvariants,inSection2weprove thatthesearetrivialforallfinitesubgroupsofGL3(k).Then,inSection3,westudythefinitelineargroups
withabelianprojectivequotientand inthelastsectionwedealwiththefifthHeisenberg group.
Notation.Alongthewholework,weconsiderfinitelineargroups.InSection2wealsoassumethatk contains aprimitiveroot ofunityofdegreetheexponentofG.InSection3andSection4thegroundfieldisC.
Weset ∗= Spec(k).
Finally,ifX is ascheme with aG-action, then wedenote by X/G theschematic quotient andby [X/G]
thestackquotient.
1. Preliminaries
InthissectionwereviewthebackgroundandthedefinitionoftheEkedahlinvariants.Amorecomplete and self contained introduction to these new invariants can be found in [11]. In this section we consider finite lineargroupandweworkoverafieldk ofcharacteristic primetotheorder ofG.
TheGrothendieckringofalgebraicvarietiesK0(Vark) isthegroupgeneratedbytheisomorphismclasses
{X} ofalgebraic k-varietiesX,subjectto therelation{X}={Z}+{X \ Z}, forallclosedsubvarietiesZ
of X.ThegroupK0(Vark) hasaringstructure givenby{X}· {Y }={X × Y }.LetL betheclass ofthe
affineline.ThecompletionofK0(Vark)[L−1] withrespectto thedimensionfiltration
FilnK0(Vark)[L−1]
={{X}/Li: dim X− i ≤ n}
iscalled theKontsevichvalueringanddenotedbyK0(Vark).
In[5],EkedahlintroducedtheconceptofGrothendieckgroupofalgebraicstacks.
Definition 1.1.Wedenote byK0(Stackk) theGrothendieckgroupof algebraick-stacks.This isthegroup
generatedbytheisomorphismclasses{X} ofalgebraick-stacksX offinitetypeallofwhoseautomorphism groupscheme areaffine(algebraic k-stacks offinite typewithaffine stabilizers,inshort).Theelements of thisgroupfulfillthefollowingrelations:
1. for eachclosed substackY ofX,{X}={Y }+{Z}, whereZ is thecomplementofY inX;
2. for eachvectorbundle E ofconstantrankn overX,{E}={X × An}.
Similarly to K0(Vark), K0(Stackk) has a ring structure. In Theorem 4.1 of [5] it is proved that
K0(Stackk)= K0(Vark)[L−1,(Ln− 1)−1,∀n∈ N].Moreoverweobserve inLemma 2.2of[11]the
comple-tionmapK0(Vark)[L−1]→ K0(Vark) factorsthrough
K0(Vark)[L−1]→ K0(Stackk)→ K0(Vark).
Theclassifyingstackof thegroupG isusuallydefinedasthestack quotientBG= [∗/G] and,viathismap,
oneseestheclassoftheclassifyingstack{BG} asanelement ofK0(Vark) (seeProposition2.5.bin[11]).
Using theBittner presentation(see[2]),given an integerk, Ekedahldefines in[5]acohomological map
fortheKontsevichvaluering,sending{X}/Lmto{Hk+2m(X;Z)},foreverysmoothandproperk-varietyX:
Hk: K
0(Vark)→ L0(Ab)
{X}/Lm→ {Hk+2m(X;Z)}.
Ifk = C,itisnaturaltoassigntoeverysmoothandproperk-varietyX theclassofitsintegralcohomology
group Hk(X;Z). If instead k isdifferent from C, then we send {X} to the class {Hk(X;Z)} defined as dim Hk(X;Z) {Z}+p{tor Hk(X;Zp)}.Thismapcan beextendedto K0(Vark).InSection3of[11], we
provethatthismap iswelldefined.
Notation.Everycohomologygroup(ifnotexplicitlyexpresseddifferently)isthesingularcohomologygroup withintegercoefficients,thatisHk(−) = Hk(−; Z).
Definition1.2.Foreveryintegeri,thei-thEkedahlinvariant ei(G) ofthegroupG isH−i({BG}) inL0(Ab).
WesaythattheEkedahlinvariantsofG aretrivial ifei(G) = 0 forallintegersi= 0.
2. ThefinitesubgroupsofGL3(k)
Inthissectionweassumethatchar(k) isprimewiththeorderofG andthatk containsaprimitiveroot ofunityofdegreetheexponentofG.LetV beann-dimensionalk-vectorspace.WeassumeG tobeafinite subgroupGLn(k) andH beitsquotientinPGLn(k):
0 K G H 0
0 Gm GLn(k) PGLn(k) 0.
WesometimesuseforsimplicityPn−1 fortheprojectivespaceP(V ).
Togetinformationabout{BG},westudy theclassofthestackquotient{[P(V )/G]}.
Lemma2.1. (SeeLemma2.4of [11].){BG}(1 +L + · · · + Ln) ={[P(V )/G]} in K
0(Vark).
Bydefinition H acts on P(V ) and P(V )/G ∼= P(V )/H. Their stack quotients are not isomorphic, butthe
classesof theirstack quotientsareequalinK0(Vark).
Proposition 2.2. Let L → X be a G-equivariant line bundle over a smooth scheme X, where the group
G acts faithfully on L. Assume that the kernel K of such action on X is cyclic and set H =G/K. Then
{[X/G]}={[X/H]}∈ K
0(Vark).
As a consequence, if G is a subgroup of GL(V ) and H is its quotient in PGL(V ), then {[P(V )/G]} =
{[P(V )/H]}∈ K0(Vark).
Proof. Thenaturalmap[L/G]→ [X/G] isak∗-torsorandusingProposition1.1ii in[5]oneobtains{[L/G]}=
(L− 1){[X/G]}. Similarlyfrom[(L/K)/H]→ [X/H], one gets{[(L/K)/H]}= (L− 1){[X/H]}.The first partof
thestatementfollowsfrom[(L/K)/H]= [L/G].
ThesecondpartofthestatementfollowsbysettingL= V \{O},X =P(V ) withthenaturaltautological bundle. 2
Wenowset upnotations, definitionsandremarks regardingthequotient ofalgebraic varietiesbyfinite groups.(Thisisnecessary tointroducenextlemma.)
LetY beasmoothcomplexquasi-projectivealgebraicvarietyandletA beafinitegroupofautomorphisms ofY .LetY →Y/Abethequotientmap andy be¯ theimageofy.
A nontrivial element of G ⊂ GL(V ) is a pseudo-reflection if it fixes pointwise a codimension one hy-perplaneinV . (Intheliteraturepseudo-reflectionsaresometimescalledreflections.)Thepseudoreflection subgroup, Pseudo (G), of G ⊂ GL(V ) is its subgroup generated by pseudo-reflections. The Chevalley– Shephard–ToddTheorem saysthatthequotientV/G issmoothifandonlyifG= Pseudo (G).
ThewellknownCartan’s Lemmasaysthatforallthepointsy ofY ,theactionofthestabilizerStaby(A)
of y on Y induces an action of Staby(A) on the tangent space on y, TyY . Moreover the analytic germ
(Y/A,y) is¯ isomorphicto (TyY/A,O),¯ where O is¯ theimage ofthe originO∈ TyY under thequotient map
TyY → TyY/A. Aneasy consequence is thatfor allthe points y ofY , Staby(A)⊆ GLdim(Y ) and onealso
provesthatp isasingularpoint ofV/G,p∈ Sing (V/G),ifand onlyifPseudo (Stab
p(G))= Stabp(G).
Comparingthe classes {[P(V )/H]} and {P(V )/H}, we are going to prove thatthe Ekedahl invariants for
everyfinitesubgroup G inGL3(k) aretrivial.Weproveitbyinductiononn,where G isafinitesubgroup
ofGLn(k), andn= 1,2,3.Thebasecase,n= 1,iscoveredbyProposition3.2in[4].
Proposition 2.3. If G is a finite subgroup of GL2(k) then {BG} = {∗} in K0(Vark) and the Ekedahl
invariantsof G are trivial.
Proof. LetU be thenon-empty opensubsetofP1where H actsfreely.Weset I =P1\ U,thefiniteset of
non-trivial stabilizerpoints. Usingproperty1)inDefinition 1.1,wewrite
{[P1 /H]} = {U/H} + p∈I/H {[p/Stab p(H)]} = {U/H} + p∈I/H {B Stabp(H)}.
Weobservethatthesumistakenovertheorbitsofthenon-trivial stabilizerpoints.Similarly,intheclassical quotient, {P1 /H} = {U/H} + p∈I/H {∗}.
Therefore,comparing thelatterexpressions:
{[P1
/H]} = {P1/H} +
p
({B Stabp(H)} − {∗}).
Using (inorder) that{[P1/G]}= (1 +L) {BG},Proposition 2.2,theprevious formulaand P1/
H∼=P1,one
has
{BG}(1 + L) = {P1} +
p
({B Stabp(H)} − {∗}).
Using Cartan’s Lemma, Stabp(H) is a subgroup of GL1 and, hence, for Proposition 3.2 in [4],
{B Stabp(H)}={∗} for everyp ∈I/H. Hence,{BG}(1+L)= {P1} andthis implies {BG} = 1,because
Ln− 1 isinvertible inK
0(Vark),L2− 1= (L− 1)(L+ 1) andso 1+L isinvertible too. 2
Notethatweactuallyprovedthat{[P1
/H]}={P1} inK0(Vark) andinasimilarwaywetacklethenext
case. We are goingto useresolution of singularities and for this we requirethe characteristic of thebase fieldtobezero.
Theorem 2.4. Assume char(k)= 0. If G is a finite subgroup of GL3(k) then{BG} = 1 in K0(Vark) and
theEkedahl invariantsofG are trivial.
Proof. Using equation {[P2/G]} = 1 +L + L2{BG} and Proposition 2.2, we know that {BG}{P2} =
{[P2
/H]}.Since{P2} isinvertibleinK0(Vark),itissufficienttoprovethat{[P2/H]}={P2}.
LetU betheopensubsetofP2whereH actsfreelyandletC bethecomplementofU inP2.Wedenoteby
C0andC1respectivelythedimensionzeroandthedimensiononeclosedsubsetsofC sothatC = C0 C1.
Oneobservesthat[C0/H] isthedisjointunionofafinitenumberofquotientstacks[Oi/H] whereOiarethe
orbitsof Pi ∈ C0 undertheaction ofH.We note that[Oi/H]= [Pi/StabPi(H)]=B StabPi(H).ByCartan’s
Lemma,StabPi(H) isasubgroupofGL2(k) and then,byusingProposition 2.3,
{[Oi/H]} = {B StabPi(H)} = {Oi/H} = {∗}.
Therefore{[C0/H]}={C0/H}.
Weobservethat{[S/H]}={S/H} holdsforeveryfinite stablesubsetS ofP2 withthesameargument.
ThesetC1 istheunionofafinitenumberoflinesLi.WedenotebyI theunionofpairwiseintersections
Li∩ Lj,fori= j.WealsodenotebyC1∗ thecomplementof I inC1.
LetL be aline inC1 andSL = StabL(H).Byachange ofcoordinate we can assumeL to be the line
at infinity of P2 and so S
L ⊂ GL2. Another way to rephrase this is to observe that StabL(PGL3(k)) ∼=
GL2(k) k2.UsingSL⊂ GL2(k) andProposition 2.3,onegets{[L/SL]}={L/SL}.
We set L = L∩ C1∗. Then [L/SL] = [L
/SL]∪ [L\L
/SL]. By whatwe said in the zero-dimensional case
{[L\L /SL]}={L\L /SL} and so {[L /SL]}={L
/SL}. We call Oj theorbit of Lj under H. Since C1∗ is the
disjointunionofafinite numberof orbitsOj,then
{[C1/H]} = {[C1∗/H]} + {[I/H]} = j {[Oj/H]} + {[I/H]} = j {[Lj/S Lj]} + {[I/H]} = j {Lj/H} + {I/H} = {C1/H}.
Summarizingtheprovenfacts,onehas
{[P2
/H]} = {[U/H]} + {[C0/H]} + {[C1/H]} = {U/H} + {C0/H} + {C1/H} = {P2/H}.
Therefore there remains to provethat {P2
/H}= {P2}. For this purpose let X be aresolution of the
sin-gularities of P2
/H, π : X → P2/H. Castelnuovo’s theorem implies every unirational surface is rational in
characteristiczero(see Chapter Vof [7])andonecan construct abirational morphism, π : X → P2.The
quotientsingularitiesofP2
/H are rationalsingularitiesand theexceptionaldivisorDy ofy ∈ SingP2/His
atreeof P1 (seefor instance [17]).This impliesthatD
y =∪ ny
j=1P1,where ny isthenumberof irreducible
componentsofDy.Then{Dy}= ny{P1}−{∗}.(Thesumrunsovertheintersectionpointsofthecopies
ofP1.)
Since the graphof the resolution is atree, then there are exactly ny− 1 intersection points in {∗}.
Hence{Dy}= ny{P1}− (ny− 1){∗}= nyL+{∗}.Then, {P2 /H} = {X} − y ({Dy} − {y}) = {X} − y (nyL + {∗} − {∗}) ={X} − L y ny={X} − Ln,
where n = yny is the number of irreducible components in the full exceptional divisor D = ∪yDy.
Similarly,onegets{P2}={X}−Lm,wherem isthenumberofirreduciblecomponentsinthefullexceptional divisorE oftheresolution X−→ Pπ 2.
Weshallprovethatm= n.Letusconsider thefollowing spectralsequencefromthemap π : X→P2
/H:
E2i,j= HiP2/
H; Rjπ.QX
⇒ Hi+j(X;Q) .
Since the map is an isomorphism away from a finite number of points, Rjπ.Q
X is zero elsewhere
but those points. If y is one of those points, then HiP2/H; (Rjπ.QX)y
equals Hiπ−1(y);Q and so
H0π−1(y);Q=Qn and for i > 0, Hi
π−1(y);Q = 0. The spectral sequence degenerates and then we obtain
0→ Q → H2(X;Q) → Qn → 0.
This implies H2(X;Q) = Qn+1. Similarly, for π : X → P2, one gets H2
(X;Q) = Qm+1 and, thus, the
equalitym= n. 2
ItisnaturaltoaskiftheprevioustheoremremainsvalidifGL3(k) isreplacedbyGLn(k),forn≥ 4.For
largeenoughn,itwillfail.Forn= 4 onecanmodifythefirstpartoftheproofofTheorem 2.4toshowthat
{[P3/
H]}={P3/H}∈ K0(Vark). We donot know whetheror not{P3/H}= {P3} or whetherTheorem 2.4
remains valid for n = 4. Indeed, proving that{P3
/H}= {P3} involves the study of the resolution of the
quotientsA3
k/AforcertainA⊂ GL3(k).
3. Finitegroupswith abelianprojectivequotient
InthissectionthegroundfieldisassumetobeC.Inwhatfollowswearegoingtousethefollowingfact. Lemma 3.1. Let Y be a smooth quasi-projective complex algebraic variety and let A be a finite group of
automorphisms ofY .LetY →Y/A bethecanonicalquotientmapandy be¯ theimageof y.
Lety¯∈Y/A.Thegerm(Y/A,y) is¯ asimplicialtoroidalsingularity(i.e.locallyisomorphic,intheanalytic
topology,totheorigininasimplicialtoricaffine variety)ifandonlyifthequotientStaby(A)/Pseudo(Staby(A))
in TyY isabelian.
Aproof canbe found,forinstance,inthelemmainSection1.3of[15].
AsintheprevioussectionG isasubgroupofGLn(C) andH isitsquotientinPGLn(C).IfH isabelian
and ifthesingularitiesofPn−1
/H arezerodimensional,then,for thepreviouslemma, suchsingularitiesare
toroidal.Underthisconditions,theEkedahlinvariantssatisfyarecursiveequation. Theorem3.2.If H isabelian andifPn−1
C /H hasonlyzero dimensionalsingularities,thenforeveryintegerk
ek(G) + ek+2(G) +· · · + ek+2(n−1)(G) ={H−k(X;Z)}, (2)
where X→Pn−1
/H isasmoothandproperresolutionwith normal crossingtoric exceptionaldivisors.
Wefirstshowatechnicallemma.WedenotebypX(t)=i≥0βi(X)tithevirtualPoincarépolynomialof
acomplexalgebraicscheme X,where βi(X)= dim(Hi(X;Q)) isthei-th BettinumberofX (herewe use
thenotationofSection4.5of[6]).ForeverysmoothprojectivetoricvarietyY ,pY(t) isanevenpolynomial
(seeSection5.2of[6]). WeobservethatH∗Pn−1
/H;Q= H∗Pn−1;QHbecauseH isafinitegrouponthesmoothschemePn−1
andthecardinalityofH isinvertibleinQ.WeremarkthatH∗Pn−1;Q= H∗Pn−1;Z⊗ Q.Anyelement
of H∗Pn−1;Q can be written as ni=1−1aihi where h is the first Chern class of OPn−1(1) and ai ∈ Q.
Now,weobserve theactionistrivialonthecoefficientsai anditcomes fromalinearactioninGLn andso
H∗Pn−1;QH
= H∗Pn−1;Q. Hence,ifG isafinite subgroupofGL
n,then pPn−1/H(t)= pPn−1(t).
Lemma 3.3.LetG, H, Pn−1
/H andX satisfythehypothesis ofTheorem 3.2.Then:
i) {BG}(1+L+· · · + Ln−1)={Pn−1
/H} and,in particular,
ek(G) + ek+2(G) +· · · + ek+2(n−1)(G) =H−k{Pn−1/H}. (3)
ii) Every singularityof Pn−1
/H isatoroidalsingularityand
{Pn−1 /H} = {X} − y ({Dy} − {y}) , (4)
http://doc.rero.ch
where the sum runs over y ∈ SingPn−1
/H; Dy = π−1(y) is the exceptional toric divisor of
the resolution of y with irreducible components decomposition Dy = D1y ∪ · · · ∪ Dyr; {Dy} =
q≥1(−1)q+1i1<···<iq{D i1 y ∩ · · · ∩ D iq y }.
iii) Ifk is non-zero andeven,onehas
1 = βk(X)− y q≥1 (−1)q+1 i1<···<iq βk(Di1 y ∩ · · · ∩ Diyq) and,fork = 0, 1 = β0(X)− y q≥1 (−1)q+1 i1<···<iq β0(Di1 y ∩ · · · ∩ Diyq)− 1 . iv) βodd(X)= 0.
Proof. AsintheproofofProposition 2.3andofTheorem 2.4,weconsiderallthesubvarietiesofPn−1globally fixedbycertainsubgroupA ofH.ThissubgroupisabelianandbyProposition3.2in[4]{BA}={∗}.Thus,
{[Pn−1
/H]}={Pn−1/H}.Using inorder,{BG}(1 +L + · · · + Ln) ={[P(V )/G]} inK0(Vark),Proposition 2.2
andthelatterequality,weobtainthefirstpartof i).
Forthesecondpartof i),wenote thatapplying thecohomologicalmapH−k onthelefthandside, one has:
H−k{BG}(1 + · · · + Ln−1)=H−k({BG}) + · · · + H−k{BG}Ln−1
=H−k({BG}) + · · · + H−k−2(n−1)({BG})
= ek(G) +· · · + ek+2(n−1)(G) .
Regardingitem ii):EverystabilizergroupofH isabelianandsoitisforthequotientofStabx(H) modulo
Pseudo (Stabx(H)) in TxX. Then, by Lemma 3.1, each singularity of {Pn−1/H} is an isolated simplicial
toroidal singularity. One produces a toric resolution X with normal crossing toric exceptional divisors (see Section2.6 of [6]). This means that each intersectionDi1
y ∩ · · · ∩ D
iq
y is asmooth toric variety.The
resolution X → Pn−1
/H restricted to the pull back of Pn−1/H\ SingPn−1/H is an isomorphism and then
{X}−y{Dy}={P
n−1
/H}−y{y}.
Therefore,pDy(t)=
q≥1(−1)q+1i1,...,iqpDi1y ∩···∩Dyiq(t) and theodddegree coefficientsof pDy(t) are
zero.
Finally,wewanttocomputethevirtualPoincarépolynomialofX via formula(4)andusingpPn−1/H(t)=
pPn−1(t),
pPn−1(t) = pX(t)−
y
(pDy(t)− 1).
Comparing,degreebydegree,thepolynomialinthelefthandsideandintherighthandside,onegetsthe Bettinumbersequalities anditem iv). 2
ProofofTheorem 3.2. Fromthefirst itemofthe previouslemma, weknowthat(3) holds.We shallshow thatH−k{Pn−1
/H}={H−k(X;Z)}.Using theprevioustechnicallemmawe express{Pn−1/H} in (4)as a
sumof smoothandpropervarietiesandsmoothtoricvarieties{Di1
y ∩ · · · ∩ D
iq
y}.
Ifk > 0 or k <−2(n− 2), H−k({Dy} − {y}) = 0 fordimensional reasonand so the recurrenceholds.
Thesame holds, if k is oddintegerbetween 0 and −2(n− 2), because the cohomology of asmooth toric varietyistorsion freebySection5.2in[6].
It remains to consider thecase 0≤ k = 2j ≤ −2(n− 2). For these values,inthe left handside of(3), therearecertainEkedahlinvariantswithnegativeindices(sozero),e0(G) andsomepositiveevenEkedahl
invariants e2(G) +· · · + e2j+2(n−1)(G) that are an integer linear combination of classes of finite abelian
groups(weusethesecond partofTheorem 1.3).
Ontherighthandsideof(3)theonlypossibletorsionpartis{tor H−k(X;Z)},becausethecohomologies of a smooth toric varietyis torsion free.Hence, what remains to proveis that the free partscancel each other: thisisequivalentto item iii) ofthepreviouslemma. 2
4. Thediscrete HeisenberggroupHp
Letp beanodd.Thep-discreteHeisenberg groupHpisthefollowing subgroupofGL3(Fp):
Hp= ⎧ ⎪ ⎨ ⎪ ⎩M (a, b, c) notation = ⎛ ⎜ ⎝ 1 a b 0 1 c 0 0 1 ⎞ ⎟ ⎠ : a, b, c ∈ Fp ⎫ ⎪ ⎬ ⎪ ⎭.
WeobservethatHpisgeneratedbyX = M(1,0,0),Y = M(0,0,1) andZ = M(0,1,0) modulotherelations
ZYX = X Y,Zp=Xp =Yp= M (0,0,0),ZX = X Z andZY = YZ.ThecenterofH
p,Z Hp,iscyclicand
generated byZ. Wedenote byAp the group quotient Hp/Z Hp =∼ Z/pZ×Z/pZ. Moreover, Hp is the central
extension ofZ/pZbyZ/pZ×Z/pZ:
1→Z/pZ→ Hp φ
−→Z/pZ×Z/pZ→ 1. (5)
UsingLemma 4.9in[3], oneproves thattheBogomolovmultiplierB0(Hp) iszeroforeveryprimep.
We alsoremark thatthediscrete Heisenberg grouphas p2+ p− 1 irreduciblecomplexrepresentations:
p2of themareonedimensionalandtheremainingp− 1 are faithfulandp-dimensional.
Let V bea faithfulirreducible p-dimensional complexrepresentation ofHp, Hp ρ
−→ GL(V ).There is a
naturalactionofHponV anditinducesanactiononthecomplex(p− 1)-dimensional spacePp−1.Oneso
defines thequotient Pp−1
/Hp.
Since Z belongsto the center,ρ(Z) = e2πip Id, for some0< i < p,where Id is theidentity element of
GL(V ). Hence,thecenteractstriviallyonPp−1 andPp−1
/Hp∼=P
p−1
/Ap.From Lemma 3.1,we knowthatif
Pp−1
/Aphassingularities,thentheyaretoroidal.WestudythesesingularitiesandsowefocusonStabx(Ap).
Proposition 4.1.Letx∈ Pp−1.If theaction ofAp atx isnotfree, then|Stabx(Ap)|= p.
Proof. Call Wx the one dimensional subvector-space of V corresponding to x. By assumption, the
sta-bilizer of x is a nontrivial subgroup of Ap and, by Lagrange’s Theorem, it could have order p or p2. If
Stabx(Ap) = Ap,thenforeveryg∈ Ap,gWx= WxandApWx= Wx.ThenHpWx= Wx.Therefore,Wx is
a onedimensionalirreducibleHp-subrepresentationofV contradictingthefactthatHpactsirreducibly. 2
Thereareexactlyp+ 1 subgroupsoforderp inAp.LetB beoneofthem.WedefineB to beasubgroup
ofHp suchthatthesurjectionin(5)restrictedto it,φ| B,isagroupisomorphism:B ∼=Z/pZ⊂ φ−1(B). WerestricttherepresentationHp
ρ
−→ GLp(C) tothesubgroupB = Z/pZandwewriteV asadirectsumof
onedimensionalirreduciblerepresentations:V =⊕χ∈Z/pZVχ,whereVχ ={v ∈ V : g ·v = χ(g)·v,∀g ∈Z/pZ}.
In other words, B fixes p one dimensional linear subspaces Vχ and so B fixes p points Pχ ∈ Pp−1, with
StabPχ(Ap)= B,thatis (P
p−1)B={P
χ0,. . . ,Pχp−1}.
Proposition 4.2.If B and B are twodistinct p-subgroupsof Ap,then(Pp−1)B∩ (Pp−1)B
=∅.
Proof. Trivially,underthehypothesis,Ap= B⊕ B andifP∈ (Pp−1)B∩ (Pp−1)B
,thenStabP(Ap)= Ap,
contradictingProposition 4.1. 2
WeobservethatAp/Bactsregularlyon(Pp−1)B.Thus,thesepointslieinthesameorbitundertheaction
ofAp/B andthismeansthattheycorrespond toauniquepoint yB inPp−1/Ap.
Theorem4.3. ThequotientPp−1/Ap has p+ 1 simplicial toroidalsingularpoints.
Proof. Thereareexactlyp+ 1 subgroups,B, oforderp in Ap.Eachof themcorresponds toapointyB in
Pp−1
/Ap.ByProposition 4.2,thesepointsaredistinct.
Lety∈ Pp−1 suchthaty = y¯
B. Weconsidertheactionof Staby(Ap) onthetangentspace TyPp−1.The
pseudo-reflection groupPseudo (Staby(Ap)) = {e}, because it is asubgroup of Staby(Ap)∼=Z/pZand, so,
itiseitherthetrivialgrouporStaby(Ap). Thelatterisnotpossible becauseStaby(Ap) stabilizes onlythe
originof the vector space TyPp−1. Thus, Pseudo (Staby(Ap))= Staby(Ap) in TyPp−1 and by Lemma 3.1
thesesingularitiesarealsotoroidalandsimplicial. 2
We now outline a method to calculate the Ekedahl invariants for Hp: we write {Pp−1/Ap} as a sum of
classesof smoothandpropervarietiesandweuseTheorem 3.2. LetXp
f
−→Pp−1
/Apbe theresolutionofthep+ 1 toroidalsingularitiesof P p−1
/Ap.Onehasthefollowing
geometricalpicture: P(V ) π U π|U Xp f Pp−1 /Ap Up f|Up ∼ Up
whereU istheopensubsetofP(V ) whereAp actsfreely;Up=U/Ap.
SinceAp isabelian,using Theorem 3.2onegets
ek(G) + ek+2(G) +· · · + ek+2(p−1)(G) ={H−k(Xp;Z)}.
BecauseofTheorem 1.3,e0(Hp) ={Z} ande1(Hp) = e2(Hp) = 0.Thus,wefocusone3(Hp).Wearegoing
toshowthate3(H5) = e4(H5) = 0.Wesetp= 5 becauseofthedifficultyincomputingtor(H∗(Xp;Z)) for
p> 5.
Claim.tor(H5(X5;Z))= 0.
Usingthisclaim,weprovethemainresult.
Theorem4.4. TheEkedahl invariantsofthefifthdiscreteHeisenberg groupH5 are trivial.
Proof. Byusing Theorem 3.2 for G= H5, n= 5, k =−2· 5+ 5 and X = X5 and also by applying the
secondpartofTheorem 1.3,wehavee3(H5) ={tor
H5(X5;Z) }.ByPoincaréduality, e3(H5) ={tor H5(X5;Z) } = {torH4(X5;Z) }
http://doc.rero.ch
andthis iszerobytheclaim.Similarly,fork =−2· 5+ 6, e4(H5) ={tor H2·5−6(X5;Z) } = {torH4(X5;Z) }.
Moreover, ei(H5) = 0 for i ≥ 5. Indeed, e5(H5) = {tor
H3(X5;Z) } = e2(H5) = 0 and e6(H5) = {torH2(X5;Z)
}= e1(H5) = 0.Inaddition,ei(H5) = 0 for i> 6 fordimensionalreason. 2
Weobserve thatthesameproof wouldfollow forHp forp> 5 if wehad enoughinformation aboutthe
vanishing of the torsion in thecohomology of Xp. This factand the recentproof (byKang in[9]) of the
rationality oftheextensionC(xg,g∈ Hp)Hp/C suggest thefollowingconjecture:
Conjecture.The Ekedahlinvariants oftheHeisenberggroupHp of orderp3 aretrivial.
4.1. Proof oftheclaim
To obtaintheclaimitsuffices toshow thatHodd(U5;Z) = 0 andH5E(X5;Z) = 0, whereE is theunion
ofexceptionaldivisorsoftheresolutionX5−→f P5−1/A5.
Theorem4.5. The cohomologyof thesmoothopensubset Up of P(V )/Ap fork < 2p− 2 is
Hk(Up;Z) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ Z if k = 0; 0 if k is odd; Z ⊕ (Z/pZ)k2+1 if k= 0 and even.
Proof. Since Ap acts freely on U , let us consider the Cartan–Leray spectral sequence (see Section 5 or
Theorem 8bis.9 in[12])relativeto thequotientmapπ : U → U p: Ei,j2 = H i Ap; Hj(U ;Z) ⇒ Hi+j (Up;Z) .
Let Sp = Pp−1\ U. One sees that Hi(U ;Z) ∼= Hi(P(V ); Z) for i < 2p− 3, H2p−3(U ;Z) = Z[Sp]0 and
H2p−3(U ;Z) iszero otherwise.Here Z[Sp] is thegroup freely generated bythe p(p+ 1) pointsin Sp and
Z[Sp]0is thekernel oftheargumentationmap.
Toread theEi,j2 -termsweobserve thatthecohomologyofAp hasaZ-algebrastructure:
H∗(Ap;Z) ∼= Z[x1
, x2, y]
(y2, px
1, px2, py)
,
where deg(x1) = deg(x2) = 2 and deg(y) = 3. Indeed the Z-algebrastructure comes fromthe Bockstein
operatorforH∗(Z/pZ;Fp).(Thereadermayfindadetailedproof inAppendixAof[10].)
Inthisproof, weonlycareaboutthetermsE2i,j withj < 2p− 3 wherethedifferentialdi,j2 iszero. Let h be the first Chern class of OPp−1(1) (hence hk generates H2k
Pp−1). Using this notation one
writesanelementofE2i,2k(for2k < 2p− 3)ashk· α whereα∈ HiA
p; H2k(U ;Z)
.Thiskeepstrackofthe differentialH∗Pp−1;Z-algebrastructure(seeChapter2in[12]):for2k < 2p−3,d
3(hk)= k·hk−1d3(h).In
addition,d3(h) isanon-zero multipleofy inH3(Ap;Z).Thus,thedifferentiald3isadegree3 homomorphism
fromH∗(Ap;Z) toitself:
di,2k3 : H
i
(Ap;Z) → Hi+3(Ap;Z)
1→ αy,
fori,k > 0.Then,onecomputesthatfor0≤ j < 2p− 4,onehas
E4i,j∼= ⎧ ⎪ ⎨ ⎪ ⎩
Hi(Ap;Z) for i even and j = 0;
Z for j≤ 2p − 4 even and i = 0;
0 otherwise
andE40,2p−4 =Z andE 1,2p−4
4 = 0.
Moreover, the spectral sequence degenerates, Ei,j
∞ = E4i,j, for j = 0 with i < 2p− 2, 0< j < 2p− 4
and j = 2p− 4 with i = 0,1. We onlyremark thatfrom the E∞-level, one reads the information about
H∗(Up;Z) viaE∞i,j = gr
Hi+j(Up;Z)
. 2
Thefollowing resultisalsotrueforeveryprimep.
Theorem 4.6. Each singularity of Pp−1
/Ap is locally isomorphic to the origin of the toric affine
vari-ety Ap−1
/Z/pZ. Moreover it is of the type 1p(1,2,. . . ,p − 1), that is Z/pZ acts on Ap−1 diagonally via
j→ diag(ζj,ζ2j,. . . ,ζ(p−1)j) where ζ isaprimitivep-root ofunity.
Proof. InTheorem 4.3 we havealreadyshown thatPp−1
/Ap has p+ 1 zero dimensionalsimplicialtoroidal
singularities.Wekeepusing thenotationinthebeginningofSection2.
LetV decomposeasadirectsumofonedimensionalirreduciblerepresentations,V =⊕χ∈Z/pZVχ.Apoint
Pχ in(Pp−1)B correspondstosomeVχ forsomecharacterχ and soV/Vχ=⊕χ∈ B∨,χ=χVχ.
Theaction of Ap/B on(Pp−1)B isregular and using Cartan’s Lemma, thegerm (Pp−1/Ap,yB) is locally
isomorphicto (TPχPp−1
/StabPχ(Ap),O),¯ where O is¯ the image of theorigin ofTPχPp−1 underthe quotient
map
TPχPp−1→TPχPp−1/StabPχ(Ap).
Without loss of generality letχ = 1. Using those factswe obtain TPχPp−1 =1=χ∈Z/pZCeχ. Thus, the
action of Z/pZ on thetangent space is given by g· v = χ(g)v forany v ∈ Vχ. Therefore Z/pZ acts viathe homomorphismZ/pZ → (C∗)p−1 sending j to(ζj,ζ2j,. . . ,ζ(p−1)j),whereζ isap-rootofunity. 2
Toproceedweneedaresolutionforsuchtoroidalsingularitiesand forthisreasonwehaveto setp= 5. Weconstruct the toric resolution of A4
/Z/5Z using the computeralgebra program Magma.The resolution
is made by adding a suitable number of rays to the single cone of the fan of A4
/Z/5Z (see Exercise on
page 35 of [6]). The new fan is denoted by Δ5 and consists of 10 rays and 21 maximal cones. To each
ray one associated a toric divisor, Di = V (Star(ri)), in the resolution of A4/Z/5Z. There are 6 new rays
thatcorrespond to 6 exceptional divisors Di for1≤ i ≤ 6. LetD be the unionof them.One hasthatif
k > 3,thenDi1∩ · · · ∩ Dik =∅ (thatisthere existnomaximalconesofΔ5,generated onlybyexceptional
divisors).WewillnotwritedowntheraysandtheconesinΔ5;thedetailsofthiscomputationcanbefound
inChapters 6and7of[10].AnexampleoffanofD1isinFig. 1.
Theorem4.7. H5E(X5;Z) = 0.
Proof. LetE = f−1(SingP4
/A5) betheunionofexceptionaldivisorsfromtheresolutionofthesixtoroidal
singularitieslocally isomorphicto A4
/Z/5Z.ThenE =
y∈SingP4/A5E
(y),withE(y)= f−1(y), and eachE(y)
isisomorphictoD.Thus H∗E(X5;Z) = y∈SingP4/A5 H ∗ D(X5;Z) = H∗D(X5;Z)⊕6.
SincewewanttoprovethatH5E(X5;Z) = 0,fromnow onwe focusonH∗D(X5;Z).
Fig. 1. ThefanofthethreedimensionaltoricvarietyD1.Thefaniscompleteandweshowtheassociatedtriangularizationofthe 2-dimensionalsphere(thesimplexr4,r3,r6closesthesphere).Weshowashellingorderofthistriangulationthatisanordering ofthemaximalconeinthefanofD1satisfyingproperty(∗) onpage101of[6].The2-simplexhavingverticesr4,r3andr6isσ13.
Table 1
TheE1-termsandthedifferentialsd1.
0 0 0 9 ⊕ H2(D i1∩ Di2∩ Di3) d3,81 −−−→ ⊕ H4(D i1∩ Di2) d2,81 −−−→ ⊕ H6(D i1) 8 0 0 0 7 ⊕ H0(D i1∩ Di2∩ Di3) d3,61 −−−→ ⊕ H2(D i1∩ Di2) d2,6 1 −−−→ ⊕ H4(D i1) 6 0 0 0 5 0 ⊕ H0(Di1∩ Di2) d2,61 −−−→ ⊕ H2(D i1) 4 0 0 0 3 0 0 ⊕ H0(Di1) 2 0 0 0 1 0 0 0 0 −3 −2 −1 −ki
We denote Di1 ∩ Di2 ∩ · · · ∩ Dik by Di1,i2,...,ik. To compute H
∗
D(X5;Z), we use the second quadrant
spectral sequence E−k,i1 = i1<···<ik HiD i1,i2,...,ik(X5;Z) ⇒ H ∗ D(X5;Z) .
TheE1-termsaredefinedforeveryi andfork > 0.
We first observe that Di1,i2,...,ik is a smooth toric variety corresponding to the star of the cone
ri1,. . . ,rik,so
H∗D
i1,...,ik(X5;Z) = H
∗−2 dim( ri1,...,rik)(Di
1,...,ik;Z) .
Recallingthatthereareat mosttriple intersections,weimmediately havethatE1−k,i= 0 ifk > 3.
Table 1 shows the E1-termsof thespectral sequence and the non-zero differentials. All the indexes of
the direct sums run over the indicies of the exceptional divisors Di. All the cohomologies are integral
cohomologies.
Now,we focusonthehomomorphism
⊕ H2
(Di1,i2)
d2,61
−−−→ ⊕ H4
(Di1) .
Let us assume thatKer(d2,61 ) = 0. Then E 2,6
2 = 0 and so E∞2,6 = E22,6 = 0. We remark that thespectral
sequenceconverges toHk+i+1D (X5;Z),thatis
Ek,i∞ = grHk+i+1D (X5;Z)
.
Table 2
Thematrixofthedifferentiald2,61 .Wedecoratecolumnsandrowswiththebasiselements.
τ25,6 τ22,4 τ22,3 τ32,3 τ21,3 τ21,4 τ21,6 τ21,5 τ31,5 τ21,2 τ31,2 τ41,2 τ51,2 τ3(4) 0 1 0 0 0 1 0 0 0 0 0 0 0 τ3(6) ±1 0 0 0 0 0 1 0 0 0 0 0 0 τ4(3) 0 0 0 1 0 0 0 0 0 0 0 0 0 τ5(3) 0 0 1 0 1 0 0 0 0 0 0 0 0 τ4(5) 0 0 0 0 0 0 0 0 1 0 0 0 0 τ5(5) 1 0 0 0 0 0 0 1 0 0 0 0 0 τ6(2) 0 −2 0 0 0 0 0 0 0 0 0 0 1 τ7(2) 0 −1 1 0 0 0 0 0 0 0 0 1 0 τ8(2) 0 0 0 0 0 0 0 0 0 0 1 0 0 τ9(2) 0 +1 0 1 0 0 0 0 0 1 0 0 0 τ5(1) 0 0 0 0 0 0 −2 1 −2 0 0 0 0 τ6(1) 0 0 0 0 0 0 −3 0 −3 0 0 0 0 τ10(1) 0 0 0 0 0 −2 0 0 0 0 0 0 1 τ11(1) 0 0 0 0 1 −1 2 0 2 0 0 1 0 τ12(1) 0 0 0 0 0 0 4 0 4 0 1 0 0 τ13(1) 0 0 0 0 0 1 3 0 3 1 0 0 0
(Since columns are counted from k = −1, there is a shift by one.) The terms of the spectral sequence involvedingrH5D(X5;Z)
areE∞−1,5,E∞−2,6 andE∞−3,7 andallofthemarezero.Thus,H5D(X5;Z) = 0.
ItremainstoprovethatKer(d2,61 )= 0.Byusing,essentially,thetheoremonpage102of[6],onecomputes
fromthefan Δ5 a basis{τj(i1,i2)} forthe cohomologies H
2(D i1,i2) andabasis {τ (i1) j } for H 4(D i1). Using
thetools of intersection theory fortoric varieties (see Chapter 5in [6]),one constructsthe matrix of the homomorphismd2,61 :thisisinTable 2.Thedetailsofthecomputationofsuchhomomorphismcanbefound
inChapter 7of[10]. Thismatrixhasazerodimensionalkernelover Z. 2 Theorem4.8. tor(H5(X5;Z))= 0.
Proof. Letusconsider thelongexactsequence
· · · → H∗E(X5;Z) → H∗(X5;Z) → H∗(U5;Z) → . . .
FromLemma 3.3iii),weknowthatβodd(X
5)= 0 andhenceHodd(X5;Z) = tor
Hodd(X5;Z)
.Theorem 4.5 showsthatHodd(U5;Z) = 0 andhenceweobtain
· · · → H5
E(X5;Z) → tor(H5(X5;Z)) → 0.
Theorem 4.7saysthatH5E(X5;Z) = 0 andonehastor(H5(X5;Z))= 0. 2
Acknowledgements
I would like to thank to Torsten Ekedahl for suggesting me this problem and for his advice. I thank Anders Björner and Angelo Vistoli for the priceless help and great support.I am very grateful to Fabio Toniniforhissuggestions.
Finally, my deep gratitude goes to Zinovy Reichstein for the comments on the paper. I have really appreciatedthesuggestedimprovementsonSection2.Inparticular,Proposition 2.2hasthisformbecause ofhiscomments.
The author is partially supported by the Swiss National Science Foundation, Grant No. PP00P2_ 150552/1.
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