A note on Mumford-Roitman argument on Chow schemes

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by  Kalyan Banerjee

ANOTE ON MUMFORD-ROITMAN ARGUMENTON CHOWSCHEMES
7 KALYANBANERJEE
1
0
2 ABSTRACT. Inthisnotewearegoingtounderstandtwoquestions.One
n isthefiberofthenaturalmapfromaprojectivealgebraicgroupG to
a G/Γ, where Γ denotes the Γ-equivalence onG. The other one is to
J defineanaturalmapfromHilbertschemeofthegenericfiberofafi-

2 brationX →StotheChowgroupofrelativezerocyclesonX →Sand
1
tounderstandthefibersofthismap.

]
G
A
h. 1. INTRODUCTION
t
a
In the breakthrough paper [M], Mumford had sketched an outline of
m
thefactthatthefibersofthenaturalmapfromthesymmetricpowersofa
[

smoothprojectivevarietyX totheChowgroupofX arecountableunions
1
v ofZariskiclosedsubsetsinsidethesymmetricpowersof X. Inthepaper
1
by[R]Roitmanhasproventhatthefibersareindeedcountableunionof
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4 Zariskiclosedsubsetsinsidethesymmetricpowersofthesmoothprojec-
3
0 tivevarietyX. Thatisthedepartingpointofthisarticle.Weaskthesame
1. questionbutforΓ-equivalenceonprojectivealgebraicgroups.HereΓisa
0 smoothprojectivecurve.Twopointsg,honGaresaidtobeΓ-equivalent,
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1 ifthereexiststwopoints0,∞onΓ,andarationalmap f fromΓtoG such
:
v that
i
X
r
a f(0)=g, f(∞)=h.
Now we consider the natural map θ fromG toG/Γ, where Γ denotes
theΓ-equivalencerelationandaskwhatisthekernelofθorthefiberofθ
overe,theidentityelementofG.Soourmaintheoremofthisarticleisas
follows.
LetG beaprojectivealgebraicgroupoveranuncountable,algebraically
closedgroundfieldk. Letθ denotethenaturalmapfromG toG/Γ. Then
θ−1([e])isacountableunionoftranslatesofanabelianvariety A ofG.
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0MathematicsClassificationNumber:14L40,14L10
0Keywords:Projectivealgebraicgroups,R-equivalence
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ToprovethatwemainlyusedtheRoitman’stechniquetostratifyθ−1([e]),
in terms of quasi-projective schemes and show that the Zariski closure
of each of them is again in θ−1([e]), obtaining that θ−1([e]) is a count-
ableunionofZariskiclosedsubsetsofG. Thentheuncountabilityofthe
groundfieldiscomingintothepicture,givingusthefactthatoneofthese
Zariskiclosedsubsetsisactuallyanabelianvarietyandweobtaintheoth-
ers as the translates of the abelian variety. Over complex numbers the
pictureismuchmoreinteresting,becausewecanusetheanalyticstruc-
tureofG, thatitisa complexcompactmanifoldandθ−1([e])isa locally
compact hausdorff topological subgroup of it. Hence it is a Baire sub-
spaceofG givingusthatonlyfinitenumberoftranslatesgiveusθ−1([e]).
Thisisourfinaltheorem.
LetG beaprojectivealgebraicgroupoverC. Considerthenaturalmap
θfromG toG/Γ. Thenθ−1([e])isafiniteunionoftranslatesofanabelian
subvariety A ofG.Henceθ−1([e])isanalgebraicsubgroupofG.
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Thenextsectionisdevotedtorelativezerocycleswhichwasfirstintro-
ducedbySuslinandVoevodskyin[SV]. WedefinetheChowgroupofrel-
ativezerocycles andproduceanaturalmapfromtheHilbertschemeof
lengthd zerodimensionalsubschemesonthegenericfiberofafibration
X → S, (X,S smooth projective) to the Chow group of relative zero cy-
cleson X →S. Weprovethatthefibersofthismapiscountableunionof
ZariskiclosedsubschemesintheHilbertscheme. Hereweusethetech-
niquescomingfrom[R]toprovethisresult,alsoasketchofthisproofwas
givenbyMumfordin[M].
Acknowledgements: The authorwould like to thankthe ISF-UGC grant for
funding this project and is grateful to the hospitality of Indian Statistical In-
stitute, Bangalore Center for hosting this project. The author also thanks the
anonymousrefereeforpointingoutaninaccuracyaboutR-equivalence inthe
earlierversion ofthepaper. FinallytheauthorisgratefultoVladimir Guletskii
fortellingtheproblemaboutgeneralizationoftheMumford-Roitmanargument
forthecaseofrelativecycles,totheauthor.
2. PRELIMINARIES
Let G be a projective algebraic group over a ground field k. Two k-
pointsonG aresaidtobeΓ-equivalentifthereexistsa chainofrational
maps from Γ connecting them. Precisely, let a,b be two k-points onG.
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They are said to be Γ-equivalent if there exists rational morphisms f :
Γ→G ,0,∞inΓsuchthat
f(0)=a, f(∞)=b.
Let Γ(e) be the class of the identity e of G under the above relation,
that is collection of all k-points of G, Γ-equivalent to e. Then Γ(e) is a
subgroupofG andG/Γ(e)isagroup.
2.1. Mumford-Roitmantechniques. Let us considerthefollowingmap
θ:G→G/Γ(e),defineby
θ(g)=[g]
where [g] denotes the class of g in G/Γ(e). Since the addition law in
G/Γ(e)isdefinedtobe
[g]+[h]=[g+h]
wegetthatθ isahomomorphismofgroups. Weareinterestedtounder-
standwhatisthekernelofθorθ−1([e]).Therewasasimilarsuchquestion
askedforthenaturalmapfromthesymmetricpowerofafixeddegreeof
analgebraicvarietytotheChowgroupofzerocycles. Itwassketchedin
Mumford’sarticle [M] and later proved by Roitman in [R], that the fiber
over zero of a such a naturalmap is a countableunionof Zariski closed
subsetsofthesymmetricpowerofthegivenalgebraicvariety. Inthissec-
tionwearegoingtoadaptthetechniquespresentintheRoitman’sproof
in[R]tooursetuptoderiveatthefactthatθ−1([e])isacountableunion
oftranslatesofanalgebraicvariety.
Proposition 2.2. Let θ be the natural map from G to G/Γ(e) defined as
above. Then θ−1([e]) is a countable union of translates of Zariski closed
subsetsofG.
Proof. Considerθ−1([e]). Supposethatg belongstoθ−1([e]),thatmeans
thatthereexists f :Γ→G suchthat f(0)=g and f(∞)=e. Nowtheidea
intheRoitman’sproofistostratifyθ−1([e])bythedegreeof f. Consider
Td(e)={g ∈G|∃f ∈Homv(Γ,G),f(0)=g,f(∞)=e}.
Here Homd(Γ,G) is the hom-scheme parametrizing the degree d mor-
phisms from Γ to G, it is known to be a quasi-projective subscheme of
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the Hilbert scheme of Γ×G, parametrizingsubvarieties of Γ×G having
Hilbertpolynomiald. Itiseasytoseethat
θ−1([e])=∪d∈NTd(e).
NowweprovethateachTd(e)isaquasi-projectivesubschemeofG. For
thatconsidertheCartesiandiagram
V =Homd(Γ,G)× G // G
d G×G
(cid:15)(cid:15) (cid:15)(cid:15)
Homd(Γ,G) ev // G×G
Wherethemorphismev isgivenby
ev(f)=(f(0),f(∞))
andthemorphismfromG toG×G isgivenbyg 7→(g,e). Thenitiseasy
tocheckthatTd(e)isnothingbutπ(V ), whereπ istheprojectionfrom
d
V to G. Since V is a quasi-projective scheme, we get that π(V ) is a
d d d
quasi-projective subscheme ofG. Therefore Td(e) is a quasi-projective
subschemeofG.
NowweprovethatTd(e)isasubsetofθ−1([e])provingthatθ−1([e])is
acountableunionofZariskiclosedsubsetsofG. Let g belongstoTd(e).
Thenwehavetoprovethatthereexists f :Γ→G suchthat
f(0)=g, f(∞)=e.
LetW beanirreduciblecomponentofTd(e)whoseZariskiclosurecon-
tainsthepointg. LetU beanaffineneighborhoodofg suchthatU∩W
isnon-empty. LetustakeanirreduciblecurveC passingthroughg inU.
LetC¯ betheZariskiclosureofC inW¯ . NowembeddingG inG×G bythe
homomorphismg 7→(g,e),wehavetheregularmorphism
ev :Homd(Γ,G)→G×G
givenby
ev(f)=(f(0),f(∞))
andTd(e)istheimageofthemorphismev.Thenwecanchooseaquasi-
projectivecurveT inHomd(Γ,G)suchthattheclosureofev(T)isC¯. We
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givedetailsoftheconstructionofT. Letusconsiderev−1(C¯). Itisofdi-
mension greater or equal than 1. So it contains a curve. Consider two
distinct points onC, consider their inverse images, then there will be a
curve in ev−1(C) which map to the curve C. Then this curve is our re-
quiredcurveT.
Now let T¯ be the closure of T in PN. Let T be the normalization of
k e
T¯ and let T be the inverse image of T in T. Consider the evaluation
f0 e
morphism
f :T ×Γ→T ×Γ⊂Homd(Γ,G)×Γ→e G
0 f0
where
e:(f,t)7→ f(t).
Thisdefines a rationalmapfrom T ×ΓtoG, sinceT is non-singular,on
e e
each fiber T ×{Q}, f defines a regularmap from T toC¯. So theregular
e 0 e
morphismT →T →C¯ extendstoaregularmorphismT →C¯. LetP bea
f0 e
pointinthefiberofthismorphismoverg.Foranyclosedk-pointQ onΓ,
T ×{Q}mapsontoC¯.Thenwegetthatthereexistsx ,x onΓsuchthat
0 1
f | (P)=g, f | =e.
0 Te×{x0} 0 Te×{x1}
Thisgivesusthatg isΓ-equivalenttoe. Sowegetthatθ−1([e])contains
Td(e). So we can write θ−1([e]) as a countable union of Zariski closed
(cid:3)
subsetsofG.
3. VARIETIES OVER UNCOUNTABLE GROUND FIELDS
Inthissectionweprovethatθ−1([e])isacountableunionoftranslates
ofanabelianvariety,whenthegroundfieldisuncountable.
Firstweprovethefollowingfewlemmas.
Lemma3.1. LetX beaprojectivevarietyoveranuncountablegroundfield
k.ThenX cannotbewrittenasacountableunionofproperZariskiclosed
subsetsofitself.
Proof. Suppose that X can be written as a countable union of proper
Zariskiclosed subsetsof it. By Noether’s normalizationthereexistsa fi-
nitemap from X →Pm where m =dim(X). Since X can bewrittenas a
k
countable union of Zariski closed subset of itself, we can write Pm as a
k
countableunionofZariskiclosedsubsetsofitself,say
Pmk =∪i∈NZi .
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SincethecollectionofZ ’siscountableandthegroundfieldkisuncount-
i
able,wegetthatthereexistsahyperplaneH notcontainedinanyofthe
Z ’s. Sowecanwrite
i
Pmk −1=H =∪i∈N(Zi∩H).
Continuingthisprocess we obtainthatP1 is a countableunionof its k-
k
(cid:3)
points,whichcontradictstheassumptionthatk isuncountable.
Lemma3.2. Letk beuncountable. Let Z =∪i∈NZi beacountable union
ofZariskiclosedsubsetsembeddedinsomePm. Thenwecanwrite Z asa
k
uniqueirredundantcountable unionofirreducibleZariskiclosedsubsets
ofPm,thatis
Z =∪i∈NAi
suchthat A 6⊂A fori 6= j andthisdecompositionisunique.
i j
Proof. WewriteeachZ asafiniteunionofirreduciblecomponentssay,
j
Z =∪ki Z′
i l=1 il
. Thenwegetthat
Z =∪ (∪Z′ ∪···Z′ )
i i1 ik
forsimplicitywewritetheaboveas
∪ B
i i
whereeachB isirreducible.NowordertheB ’sbysetinclusionandonly
i i
consider those B ’s which are maximal with respect to inclusion. Then
i
we get an irredundant decomposaition cup B . Now we have to prove
i i
thatthisdecompositionisunique. Supposethatthereexistsanotherde-
composition∪j∈NAj.ThenobservethateachAj iscontainedinsomeBi,
otherwise,wecanwrite
Aj =∪i∈N(Aj ∩Bi)
where A ∩B isaproperclosedsubsetof A ,whichcontradictsthepre-
j i j
vious lemma 3.1. Similarly B is contained in some A , so we get that
i k
A = A andconsequently A =B . Sowegetthatthedecompositionis
j k j i
(cid:3)
unique.
Proposition3.3. θ−1([e])isacountableunionoftranslatesofanabelian
subvarietyofG.
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Proof. Bytheprevioustwolemmas3.1,3.2wegetthatθ−1([e])isacount-
able union of Zariski closed closed subsets in G such that the union is
irredundant. So let θ−1([e]) is a countable union say ∪i∈NAi, such that
A 6⊂ A for i 6= j. Then we claim that there exists a unique A among
i j 0
these A ’s which passes through e and moreover this A is an abelian
i 0
variety. So suppose that there exists A ,···,A passing throughe, then
0 m
consider A +···+ A . Since θ−1([e]) is a subgroup of G, we get that
0 m
A +···+A isasubsetofθ−1([e]). Sinceitistheimageofthemorphism
0 m
A ×···×A →G givenby
1 m
(a ,···,a )7→a +···+a
1 m 1 m
itisirreducibleandZariskiclosed. Thereforebylemma3.1,itmustland
insidesome A . Alsosincee belongsto A ,···,A ,wegetthat A ⊂ A +
j 0 m i 1
···+A ⊂ A , for all i = 0,···,m. So by the irredundancy we get that
m j
A = A =···= A . So A istheuniqueirreducibleZariskiclosedsubset
0 1 m 0
inthedecomposition∪ A suchthatitpassesthroughe.
i i
Now we claim that A is an abelian variety. For thatsupposethat x ∈
0
A . Thenconsider−x+A ,sincetranslationby−xisahomeomorphism,
0 0
−x+A is Zariski closed and irreducible. Hence by 3.1, it is a subset of
0
someA .Nowebelongsto−x+A andtherepassesauniqueA through
j 0 0
e so we get that A = A and hence −x+A ⊂ A . Now we show that
j 0 0 0
A +A isinside A . Forthatweobservethat A +A istheimageofthe
0 0 0 0 0
regularmorphismfrom A ×A toG givenby
0 0
(a,b)7→a+b.
Thenagainbylemma3.1, A +A isinsidesome A and A isinside A +
0 0 j 0 0
A ,sowegetthat A =A . So A isanabelianvariety.
0 j 0 0
Thereforewecanwrite
θ−1([e])=∪x∈θ−1([e])(x+A0)
wheretheabove unionis disjoint. We provethattheabove unionis ac-
tually countable. So let us consider x+A , since it is Zariski closed ir-
0
reducible, it mustland insidesome A . So we get that A ⊂−x+A , by
j 0 j
similarargumentwegetthat−x+A ⊂ A sowegetthat A = A ,which
j k k 0
inturngivesusthatx+A =A ,sincethereareonlycountablymanyA ’s
0 j j
wegetonlycountablymanyx+A ’sgivingus
0
θ−1([e])=∪i∈Nxi +A0.
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(cid:3)
4. PROJECTIVE ALGEBRAIC GROUPS OVER C
In this section we are going to understand that θ−1([e]) is actually a
finiteunionoftranslatesofanabeliansubvarietyofG,whentheground
fieldiscomplexnumbers.
Proposition4.1. LetGbeaprojectivealgebraicgroupoverC.Thenθ−1([e])
isafiniteunionoftranslatesof A .
0
Before going to the proof of the theorem we recall the definition of a
Bairespace. AtopologicalspaceX iscalledBaire,ifanycountableunion
of closed sets having non-empty interior implies that one of them has
non-empty interior. Any complete metric space or a locally compact
HausdorffspaceisBaire. AlsoifX isanon-emptyBairespace,whichisa
countableunionofclosedsubsets,thenitfollowsthatoneoftheclosed
subsetshasnon-emptyinterior.
Proof. Bytheproposition3.3
ker(θ)=∪i∈N(xi +A0)
since A isanabeliansubvarietyinG itisclosedintheanalytictopology.
0
Also G is a metric space and A is closed, so it is complete under this
0
metric.Nowweclaimthatker(θ)iscompleteunderthismetric.Sotakea
Cauchysequence{y } inker(θ). Weclaimthatthereexistssomen ∈N
n n 0
such that for all n ≥ n , y belongs to one of the x +A . Suppose the
0 n i 0
opposite. Thatis foreach N, thereexistsn,m ≥N such that y belongs
n
toonex +A and y belongstox +A ,where(x +A )∩(x +A )=;.
i 0 m j 0 i 0 j 0
Nowgivenanyǫ>0,thereexistsN ∈Nsuchthatforn,m≥N wehave
d(x ,x )<ǫ.
n m
Now take ǫ to beless than theinfimumof d(y ,a), where y belongsto
n n
x +A anda∈x +A ,where(x +A )∩(x +A )=;. Thenforlargem
i 0 j 0 i 0 j 0
wehave
d(y ,y )<ǫ
n m
butontheotherhand
d(y ,y )≥inf(d(y ,a)),
n m n
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whereavariesinx +A .Sincex +A iscompactintheanalytictopology
j 0 j 0
wehavethat,thereexistsb suchthat
inf(d(y ,a))=d(y ,b).
n n
Nowifd(y ,b)=0thenwehave y =b,but(x +A )∩(x +A )=;. So
n n i 0 j 0
d(y ,b)>0,thereforechoosingǫtobelessthaninf(d(y ,a))wegetthat
n n
d(y ,y )≥ǫ
n m
contradicting the fact that {y } is Cauchy. So there exists N ∈ N such
n n
thatforalln≥N wehave y belongstoonefixedx +A . Sincex+A is
n i 0 0
completeforeachx∈G,wegetthatthesequence{y } convergesinx +
n n i
A . Henceker(θ)iscomplete.SoitisaBairespace. Thereforethereexists
0
one x such that the interior of x +A is non-empty. Since translation
i i 0
by−x isahomeomorphismwegetthattheinteriorof A isnon-empty.
i 0
Now A isatopologicalsubgroupofker(θ)whoseinteriorisnon-empty.
0
So A is open in ker(θ). Therefore each x +A is open in ker(θ). So we
0 i 0
haveanopencoverofker(θ). Sinceker(θ)completeinthegivenmetric,
itisclosedin A .SoitiscompactintheanalytictopologyofG.Therefore
0
wegetthatafinitelymanyx +A coverker(θ). Soker(θ)isafiniteunion
i 0
oftranslatesof A . Sinceeachx +A isZariskiclosedandirreduciblein
0 i 0
(cid:3)
G,wegetthatker(θ)isanalgebraicsubgroupofG.
5. RELATIVE RATIONAL EQUIVALENCE AND THE MUMFORD-ROITMAN
TYPE ARGUMENT
NowwewouldliketogeneralizetheMumford-Roitmanargumentsay-
ing that the natural map from the Chow variety of a smooth projective
variety to the Chow group of the variety itself, has the fibers equal to a
countableunionofZariskiclosedsubsetsoftheChowvariety. Allthisis
happeningover an uncountableground k. Now we supposethat X is a
smooth-projectiveschemeoveranotherNoetherianschemeS. Thenwe
observe that there is a natural map from the k(S)-points of the Hilbert
schemeHilbd(X/S)toCH (X/S). Weprovethatthefibersofthismapis
0
acountableunionofZariskiclosedsubschemesinHilbd(X/S)(η),where
ηisthegenericpointofS.
Firstofallwerecallthedefinitionoftherelativecyclesonthescheme
X/S dueto[SV]. Arelativecycleofrelativedimensionr on X,isanalge-
braiccyclesuchthatallitsprimecomponentsmapstothegenericpoint
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η of S and for any k-point P on S, the pullback with respect to any fat
point corresponding to P coincide. Now observe that any r-cycle on X
whichisflatoverS,thatisitscompositionof Z →X →S isflatisarela-
tivecycle. InviewofthisweconsidertheHilbertschemeHilbd(X/S)and
itsk(S) points,which is nothingbutHilbd(X ), that is thelengthd zero
η
[d]
dimensionalsubschemesofX .WedenoteitbyX ,andwehaveanatu-
η η
[d] Z
ralmapfromX to (X/S)associatingazerodimensionalsubscheme
η 0
oflengthd toitsfundamentalcycle. Forsakeofconvenienceweidentify
[d]
X withitsimageundertheHilbert-Chowmorphismtothesymmetric
η
powerSymdX ,anddenoteitbythesamenotationX[d].
η η
Z
We define the rational equivalence on (X/S) as follows. Let Z ,Z
0 1 2
be two relative cycles of relative dimension 0, they are said to be ratio-
nallyequivalentifthereexistsamorphism f :P1 →SymdX andarelative
S
effectivezerocycleB,suchthatimageof f andsupportofB iscontained
inX[d,d],and f(0,s)=Z +B;,f(∞,s)=Z +B Inthefollowingwedenote
η 1 2
X[d1,···,dn]tobeQ X[d1].
η i η
[d,d]
Proposition5.1. Thenaturalmapθ fromX toCH (X/S)hasfibers
X/S η 0
[d,d]
equaltoacountableunionofZariskiclosedsubschemesof X .
η
[d,d]
Proof. Let W be the subset of X consisting of pairs (A,B), where
d η
u,v
θ (A,B) is relativelyrationallyequivalentto0 onCH (X/S). LetW
X/S 0 d
[d,d]
bethesubsetofX whichconsistsofpairs(A,B),suchthatthereexists
η
f inHomv(P1,X[d+u,d])with f(0,s)=(A+C,C)and f(∞,s)=(B+D,D).
S η
Thenwehave(A,B)relativelyrationallyequivalent.Thenitiseasytosee
u,v
thatW is a subset ofW . On the other hand suppose that (A,B) be-
d d
longstoWd. Thenthereexists f :P1 →X[d,d] andarelativezerocycleC,
S η
[d,d]
suchthatimageof f iscontainedinX andwehave
η
f(0,s)=A+C,f(∞,s)=B+C
thenwecanfindu,v suchthat(A,B)belongstoWd .
u,v
u,v [d,d]
Now we prove that the setsW is a quasiprojectivevariety in X
d η
and its Zariski closure is contained in W . Then we can write W as a
d d
[d,d]
countable union of Zariski closed subsets of X . Consider the mor-
η
phismefromHomv(P1,X[d+u,u])→X[d+u,u,d+u,u],bysendingamorphism
S η η
f tothepair(f (0),f (∞)).TheothermorphismfromX[d,u,d,u]toX[d+u,u,d+u,u]
η η η η
givenby(A,C,B,D)7→(A+C,C,B+D,D). Thenifweconsiderthefiber
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