degeneracy loci of vector bundle maps and ampleness

22 Pages·2002·0.16 MB·English

MATH.SCAND.90(2002),13–34
DEGENERACY LOCI OF VECTOR BUNDLE MAPS
AND AMPLENESS
JØRGENANDERSGEERTSEN
1. Introduction
Let X be an algebraic variety and E,F vector bundles on X, of ranks e,f
respectively. For any vector bundle map φ : E → F and k ∈ N, 0 ≤ k ≤
min{e,f},thek’thdegeneracylocusofφ istheset
Xk(φ):={x ∈X |rankφ(x)≤k}.
Since a linear map has rank ≤ k if and only if all k +1-minors vanish, the
set Xk(φ) can also be described as the zero locus of the section ∧k+1φ ∈
(cid:13)(X,(∧k+1E)∗ ⊗ ∧k+1F). As such, it comes with a natural structure as a
closedsubschemeofX.
Degeneracy loci turn up in a number of geometric constructions. For in-
stance, the common zero locus of a finite set of sections of a vector bundle
is a degeneracy locus. In Brill-Noether Theory, degeneracy loci show up as
the subsets of the Jacobian which parametrizes complete linear systems of a
specificdegreeanddimensionboundedbelow.
For a map φ : E → F, the expected dimension of Xk(φ) is the number
dimX−(e−k)(f −k).Wedenotethisnumberbymk throughoutthispaper.
Thenwehave,foranyφ :E →F,
If Xk(φ)(cid:13)=∅ then dimXk(φ)≥mk.
Forinstance,ifE = OX istrivialandk = 0,thenφ : E → F canbeviewed
asasectionφ ∈(cid:13)(X,F).ThenX (φ)isequaltoZ(φ),thezero-locusofthe
0
sectionφ,andinthiscasetheexpecteddimensionofZ(φ)isdimX−f.This
is just Krull’s Hauptidealsatz: Z(φ) is locally cut out by f equations, so its
codimensioninXshouldbef.
Nowitmightverywellbethecasethatnoφ :E →F hasXk(φ)withthe
expecteddimension.ForexampleonP2,thebundleE =O⊕O(1)hasrank2
sooneexpectsagenericsectiontovanishincodimension2.Butmostsections
ReceivedJanuary4,1999.
14 jørgenandersgeertsen
don’t vanish at all, namely those of the form (λ,σ) where λ ∈ k \{0}, and
the other sections vanish too much, namely on a line, i.e. in codimension 1.
SoevenforverynicegloballygeneratedbundlesitcanhappenthatnoXk(φ)
hasthe’right’dimension.Wedefinethenumbertk =tk(X,E,F)asfollows:
tk :=min{dimXk(φ)|φ :E →F amorphismwith Xk(φ)(cid:13)=∅}.
Notethatsincewecantakeforφ thezero-morphism0 : E → F,whichhas
Xk(0) = X for all k, the number tk is well-defined and tk ≤ dimX. Also,
the inequality mk ≤ tk holds for all k (0 ≤ k ≤ min{e,f} as always). For
instance,intheaboveexample,m =0andt =1.
0 0
We are concerned with the following relative situation. Suppose E,F are
vector bundles on X as above. Let Y be an irreducible algebraic variety and
suppose f : X → Y is a proper surjective map. We say that the bundle
E∗ ⊗F is ample relative to f, written ‘relf’, if it is ample on all fibers of
f,i.e.therestrictionE∗⊗F |f−1(y) isanamplevectorbundleforall(closed)
points y ∈ Y. Our purpose is to describe what happens to the images of the
degeneracylociunderthemapf.Theresultsarethefollowing.
Theorem1. SupposeXisacomplexalgebraicvarietyandE,F arevector
bundlesonX.Letf : X → Y beapropersurjectivemaptoanothervariety
Y, andsupposeE∗ ⊗F isamplerelf.SupposefurthermorethatE∗ ⊗F is
globallygenerated.
Letφ :E →F beanyvectorbundlemap.Then
(1) If tk ≤ dimY, then for any irreducible component Z ⊂ f(Xk(φ)) we
havedimZ ≥tk.
(2) Iftk >dimY,thenf(Xk(φ))=Y.
(3) Furthermore, if mk ≤ dimY, then tk ≤ dimY. If mk = dimY, then
f(Xk(φ))=Y.
Remark1. IncaseXk(φ)isempty,statement(1)isinconclusivesincethere
arenoirreduciblecomponentsZtochoosefrom.However,statement(2)says
that Xk(φ) must necessarily be non-empty if tk > dimY, since its image
under f is equal to Y. Similarly, statement (3) says that all Xk(φ) must be
non-emptyifmk =dimY.Theseresultsaresharp:Ifforinstancetk =dimY
butmk <dimY,onecannotassertthatalldegeneracylociarenon-empty.For
example, take the bundle F := O(1)⊕O(1) on P1 and look at zero-sets of
sectionsofF.Thent =0,m =−1andF hasanowherevanishingsection.
0 0
TherelationsofTheorem1withtheoremsinthelitteratureareasfollows.
TakingY = pt,weareinthesituationofanamplevectorbundleE∗ ⊗F on
a complete variety X. In this case statement (1) is trivial. Statement (2) and
degeneracylociofvectorbundlemapsandampleness 15
(3)howevercombinetotheexistenceresultofFultonandLazarsfeld[4](for
globallygeneratedbundles).FultonandLazarsfeld’sresultsaysthatifmk ≥0,
then for any φ : E → F, we have Xk(φ) (cid:13)= ∅. If mk > 0 then also tk > 0
sothisstatementfollowsfrom(2).Ifmk = 0 = dimY,thestatementfollows
from(3).
Again taking Y = pt, the statement in (3) “mk ≤ dimY ⇒ tk ≤ dimY”
meansinthiscasethatmk ≤ 0 ⇒ tk = 0, sincewealwayshavetk ≥ 0.For
instance, if E = OX and we look at zero-schemes of sections of the bundle
F then m = dimX −rankF and the statement is the following geometric
0
observation:
Corollary2. SupposeXisaprojectivevarietyandF anample,globally
generated vector bundle on X with rankF ≥ dimX.Then there is a section
σ ∈(cid:13)(X,F)suchthatZ(σ)isafinitenon-emptysetofpoints.
This in turn gives an easy proof of a special case of the Bloch-Gieseker
theorem [2] which states that for an ample vector bundle F on a projective
variety X, if rankF ≥ n := dimX, then cn(F) > 0. In fact, letting C =
{σ ∈(cid:13)(X,F)|Z(σ)(cid:13)=∅}betheconeofsectionsthatdovanishsomewhere,
and l the length of the zero subscheme of a generic section in C (which is
well-defined by the corollary) then borrowing some of the ideas behind the
proofofTheorem1yieldsthefollowingformulaforcn(F):
cn(F)=l·degC.
(Here degC is the degree of the cone C). This formula gives a new lower
boundforcn(F)validforallampleandgloballygeneratedvectorbundlesF
(Corollary13).Theformulaalsoimpliesthefunnyresultthatifcn(F)happens
tobeaprimenumber,andrankF >n,thenF hasasectionσ suchthatZ(σ)
consists of exactly one (reduced) point (Corollary 12). In general, if F is
“sufficientlypositive”,oneexpectsl =1-forinstance,thisholdsifF(−1)is
globallygeneratedforsomeveryamplelinebundleO(1)onX(Example7).
IfontheotherhandonetakesY =Xandf =1inTheorem1thenthereis
noamplenessconditiononE∗⊗F andstatement(1)musthold,i.e.dimZ ≥tk
foranyirreduciblecomponentZ ofXk(φ).ThisshowsthatforbundlesE,F
such that E∗ ⊗F is globally generated the number tk could equally well be
defined as the minimum dimension of any irreducible component Z in any
Xk(φ).
Statements (1) and (2) imply that for any φ : E → F, if Z denotes an
irreduciblecomponentoff(Xk(φ)),thendimZ ≥min{dimY,tk}.Sincetk ≥
mk,thisimpliestheweakerstatement
dimZ ≥min{dimY,mk}.
16 jørgenandersgeertsen
ThisistheTheoremofSteffen[11,Th.0.3](forgloballygeneratedbundles).
However, Steffens algebraic methods are completely different from ours. In
fact, our Theorem 1 arose from trying to give a simple geometric proof of
Steffens result. The underlying geometric idea of our approach does not use
globalgeneration,buttogetauniformstatementlikeTheorem1wemusthave
E∗⊗F globallygenerated.Ontheotherhand,weobtainastrongerresultthan
Steffens for this class of bundles, and the statement inTheorem 1 is optimal
(seeExample4).Furthermorewecan(almost)workinarbitrarycharacteristic:
ifoneassumesthatthemapf inTheorem1isgenericallysmooth(whichis
ofcourseguaranteedinchar.0)thentheproofgoesthroughoveranarbitrary
algebraicallyclosedgroundfield.
Acknowledgements.IamindebtedtoR.Lazarsfeld,A.Hirschowitzand
J.Hansenformanysuggestionsandvaluableconversations.Severaloftheideas
insection5developedduringdiscussionswithL.ManivelatInstitutJ.Fourier
in Grenoble, France, and I thank Manivel and the Institut for inviting me to
Grenoble.
2. Ample vector bundles
LetXbeanalgebraicschemeoverthealgebraicallyclosedgroundfieldKand
EavectorbundleonX.Recallthatbydefinition,Eisampleifforallcoherent
sheavesF onXthereexistsanintegerN ∈Nsuchthat
n≥N ⇒ F ⊗SymnE isgloballygenerated.
Denote by P(E) the projective bundle of lines in E. Then E is ample on
X if and only if the tautological line bundle OP(E∗)(1) is ample on P(E∗)
[7]. Next suppose Y is another scheme and f : X → Y is a map. We say
E is ample relative to f (written ‘relf’) if there is an open affine covering
Y = ∪Uα suchthatforallα,thebundleE|f−1(Uα) isanamplevectorbundle.
If we let q : P(E∗) → X denote the projection map, then this condition can
be translated as follows: E is ample relf if and only if OP(E∗)(1) is ample
relf ◦q.
Iff isapropermap,theconditionofrelativeamplenessisequivalenttothe
following property [11, Lemma 1.5]: For all coherent sheaves F on X there
existsN ∈Nsuchthat
n≥N ⇒ Rif∗(F ⊗SymnE)=0, i >0.
IfY isapoint,thisissimplySerreVanishing,sincethenRif∗(F⊗SymnE)=
Hi(X,F ⊗SymnE).
degeneracylociofvectorbundlemapsandampleness 17
OfcourseifE isamplerelativetoamapf,thenE isampleonthefibers
off, thesebeingclosedsubschemesofX.Conversely, iff isapropermap,
thisconditionisalsosufficient:E isamplerelf ifandonlyifforall(closed)
pointsy ∈Y,therestrictionE|f−1(y) isanamplevectorbundle[7,Prop.4.4].
Itisthisveryexplicitdescriptionofrelativeamplenessthatweshalluse.
Example1. Letπ :Bl (P2)→P2 betheblow-upoftheprojectiveplane
pt
atapoint.IfE denotestheexceptionaldivisor,thenOX(−E)isamplerelπ.
BecauseonthefiberE = P1 itisO (1)whichisample, andallotherfibers
P1
are points so there is no condition. For the same reason, OX(E) isn’t ample
relπ.
Example2. Consider
A1×P1
p(cid:1) ❅p
(cid:24)1 (cid:25)2
A1 P1.
Thenp∗O(1)isamplerelp becauseonthefibersP1it’sjustO (1).Sincep
2 1 P1 2
isanaffinemorphism,p∗O(1)isalsoamplerelp .
2 2
ThefollowinglemmawillbeusedintheproofofTheorem1.
Lemma3. Supposef :F(cid:26) →F isabirationalmapofpositivedimensional
completevarietiesandE anamplevectorbundleonF.Then
H0(F(cid:26),f∗E∗)=0.
Proof. Tobegin,notethatthemorphismf isautomaticallyproper.Indeed,
F isavariety,henceseparatedoverK,andthecompositionF(cid:26) →F →K is
proper.
We first prove the lemma in case E = L is a line bundle.Assume σ ∈
H0(F(cid:26),f∗L∗)isanon-zerosection.Thenc (f∗L)∩[F(cid:26)]=−[Z(σ)];hence
1
(cid:1) (cid:1)
c (f∗L)dimF(cid:26) ∩[F(cid:26)]=− c (f∗L)dimF(cid:26)−1∩[Z(σ)]
1 1
F(cid:26) F(cid:26)
(cid:1)
=− c1(L)dimF(cid:26)−1∩f∗[Z(σ)]
F
≤0
bytheprojectionformulaandNakai’scriterionforampleness.Butsincef is
birational, f∗[F(cid:26)] = [F] and dimF(cid:26) = dimF, so again by Nakai’s criterion
18 jørgenandersgeertsen
andtheprojectionformula,
(cid:1) (cid:1)
c (f∗L)dimF(cid:26) ∩[F(cid:26)]= c (L)dimF ∩[F]>0;
1 1
F(cid:26) F
acontradiction.WhatwehavedoneisbasicallytoobservethatsinceL isbig
andnefandthesenotionsarebirationallyinvariant,thepull-backf∗L isalso
bigandnef,hencethedualdoesn’thavesections.
Now let E be an ample vector bundle of arbitrary rank on F. Suppose to
thecontrarythat0 (cid:13)=σ ∈H0(F(cid:26),f∗E∗)isanon-zerosection.Thenthemap
f∗E −σ→∗ OF(cid:26)isnon-zero.ChooseanamplelinebundleL onF –forexample,
thedeterminantofE willdo.Pickn∈NsoL∗⊗SymnE isgeneratedbyits
global sections.Then also f∗L∗ ⊗Symnf∗E is globally generated. Hence
thereisasurjection
(∗) O⊕l −→f∗L∗⊗Symnf∗E −→0.
F(cid:26)
Since the map f∗E −σ→∗ OF(cid:26) is non-zero, also the symmetric power map
Symnf∗E −−S−ym−n−σ→∗ OF(cid:26) is non-zero. Tensoring with f∗L∗ we see that the
map
f∗L∗⊗Symnf∗E −1−⊗−S−ym−n−σ→∗ f∗L∗
isnonzero.Comparingwith(∗),weconcludethatthecomposition
O⊕l −→f∗L∗⊗Symnf∗E −→f∗L∗
F(cid:26)
isnon-zero.ButthismeansthatH0(F(cid:26),f∗L∗)(cid:13)=0,incontradictionwiththe
firstpartoftheproof.ThisconcludestheproofofLemma3.
Remark2. Theconclusionofthelemmaholdsundertheweakerassump-
tionthatdimF = dimF(cid:26) andf isany dominant(=surjective)map,notnec-
cesarily birational.All we have to do in the proof of Lemma 3 is to change
the equality f∗[F(cid:26)] = [F] to f∗[F(cid:26)] = d[F] where d > 0 is some positive
integer.Thentheexactsameproofapplies.
3. Degeneracy loci
Let X be an algebraic scheme and E,F vector bundles on X of ranks e,f
respectively.Supposeφ :E →F isamap.Thek’thdegeneracylocusofφ is
theset
Xk(φ):={x ∈X |rankφ(x)≤k}.
ThisisaclosedsubschemeofX,givenlocallybythevanishingofthek+1-
minors of φ. To analyze the degeneracy loci Xk(φ) we use the following
degeneracylociofvectorbundlemapsandampleness 19
construction. Put k(cid:26) = e − k and consider G := Gk(cid:26)(E), the grassmanian
bundleofk(cid:26)-planesinE withitsnaturalprojectionmapπ :Gk(cid:26)(E)−→X.
OnGwehavethetautologicalexactsequence
0 −→S −→π∗E −→Q −→0
where S,Q are the tautological sub- and quotient bundles respectively.We
denoteapointinGby(x,Hx);Hx isak(cid:26)-planeinthevectorspaceE(x).The
fiberofS overthepoint(x,Hx)∈Gisjustthek(cid:26)-planeHx ⊂E(x),andthe
fiberofQover(x,Hx)isE(x)/Hx.Considerthecomposition:
σ :S (→π∗E −π−∗→φ π∗F.
Then Z(σ) = {(x,Hx)|Hx ⊂ kerφ(x)}. So Z(σ) maps onto Xk(φ) by π,
becausethemapφ(x):E(x)→F(x)hasrank ≤kifandonlyifthekernelis
atleastk(cid:26)-dimensional,i.e.containssomek(cid:26)-plane.Thefiberoftherestriction
π : Z(σ) −→ Xk(φ) over the point x ∈ Xk(φ) is just the grassmanian of
k(cid:26)-planescontainedinkerφ(x).Ifx ∈Xk(φ)\Xk−1(φ),i.e.therankofφ(x)
is exactly k, then dimkerφ(x) = k(cid:26) and π−1(x) = {kerφ(x)} is a point. In
otherwords,incaseXk(φ)\Xk−1(φ)(cid:13)=∅,π isbirational.
Theideaofthegrassmanianconstructionistoreducequestionsongeneral
degeneracy loci to questions on zero-schemes of sections of vector bundles.
Thisisachievedbyconstructingthesectionσ ∈(cid:13)(G,S∗⊗π∗F)whosezero-
schememapsontoXk(φ)andingoodcasesevenisbirationaltoXk(φ).Then
wewouldliketotransporttheinformationwehaveonE∗⊗F toinformation
onS∗⊗π∗F:
G ⊃ Z(σ)
π↓ ↓
X ⊃ X (φ)
k
f↓ ↓
Y ⊃f(X (φ)).
k
InthesettingofTheorem1,weknowthatE∗⊗F isampleonthefibersoff.
SowewouldliketohaveS∗⊗π∗F ampleonthefibersoff ◦π,inparticular
S∗⊗π∗F wouldthenhavetobeampleonthefibersofπ.Butonthesefibers
f
(cid:2) (cid:3)(cid:4) (cid:5)
S∗⊗π∗F| =S∗| ⊕S∗| ⊕...⊕S∗|
π−1(pt) π−1(pt) π−1(pt) π−1(pt)
sowearereallyaskingfortheamplenessofS∗|π−1(pt).However,π−1(pt)isjust
theususalgrassmanianG=Gk(cid:26)(E(pt))ofk(cid:26)-planesinthevectorspaceE(pt)
20 jørgenandersgeertsen
andS∗|π−1(pt)isjustthedualofthetautologicalsubbundleonG.Unfortunately
thisbundleisalmostneverample,asthefollowingproposition(whichsurely
belongstofolklore)shows.
Proposition4. LetV beafinitedimensionalvectorspaceandG:=Gk(V)
thegrassmanianofk-planesinV.LetS bethetautologicalsubbundleofG×V.
ThenS∗ isample⇔k =1.
Proof. Ifk =1,G=P(V)andS∗ =O(1)isample.
Ifk >1,pickak−1-dimensionalsubspaceW ⊂V andput
P:={H ∈G|H ⊃W}∼=P(V/W).
ThenP = Pn−k isaprojectivespace(heren = dimV).Forx ∈ P,denoteby
lx thelineinV/W correspondingtox.ThenitisclearthatS|P(x)=lx⊕W,
henceS| =O (−1)⊕O⊕k−1.SoS∗| =O (1)⊕O⊕k−1whichisn’tample
P P P P P P
fork >1.
However,thekeytoproveTheorem1turnsouttobethefactthatthedual
ofacertainvectorbundledoesnothaveanynon-zeroglobalsections,andfor
amplevectorbundlesthispropertyispreservedunderbirationalmaps(Lemma
3)eventhoughthepropertyofamplitudeitselfisnotpreserved.
SupposenowthatthebundleE∗⊗F isgloballygenerated,onthealgebraic
variety X. Let V ⊂ (cid:13)(X,E∗ ⊗ F) be a finite dimensional vector space of
sectionsthatgenerateE∗⊗F.Considertheprojection:
X×V
p
↓ 1
X.
Thereisatautologicalmapτ :p∗E →p∗F definedby
1 1
τ(x,φ):p∗E(x,φ)−−−→p∗F(x,φ)
1 1
E(x)−−φ(−x→) F(x).
Alternatively,letτt bethetautologicalsectionofp1∗(OX⊗V)onX×V;thenτ
isjusttheimageofτt underthe(surjective)mapp1∗(OX⊗V)→p1∗(E∗⊗F).
DenotebyDk(τ)thek’thdegeneracylocusofτ.ThenbydefinitionDk(τ)=
{(x,φ)|rankφ(x)≤k}={(x,φ)|x ∈Xk(φ)}.
We can apply the grassmanian construction described earlier to the map
τ : p∗E → p∗F ontheschemeX×V.Henceweconsiderthegrassmanian
1 1
degeneracylociofvectorbundlemapsandampleness 21
G = Gk(cid:26)(p1∗E)withitsprojectionmapπ toX×V.Wehavethetautological
exactsequenceonG:
0 −→S −→π∗p∗E −→Q −→0,
1
andthesectionσ ∈(cid:13)(G,S∗⊗π∗p∗F)definedasthecomposition
1
σ :S (→π∗p∗E −π−∗→τ π∗p∗F.
1 1
Thecrucialpointis:
Proposition 5. σ is a regular section of S∗ ⊗π∗p∗F and Z(σ) is an
1
irreduciblevariety,ofdimensiondimX+dimV −(e−k)(f −k).
Proof. Toprovethepropositionweuseanotherdescriptionofσ andthen
employageneralresult(Lemma6)onlocallyfreesheavesandtheirassociated
vectorbundles.FirstnotethatGk(cid:26)(p1∗E)identifieswithGk(cid:26)(E)×V andthat
wehaveacommutativesquare
G (E)×V −−q−1→G (E)
k(cid:26) k(cid:26)
↓ ↓
X×V −−−→ X,
where q is the projection on the first factor. We denote the natural map
1
Gk(cid:26)(E) → X by π(cid:26). Let S(cid:26),Q(cid:26) denote the tautological sub- and quotient
bundles on Gk(cid:26)(E); under q1 these bundles pull back to S and Q. On X we
havetheexactsequence
OX ⊗V −→E∗⊗F −→0,
henceonGk(cid:26)(E)thefollowingexactdiagram
OGk(cid:26)(E)⊗V −−→π(cid:26)∗(E∗⊗F)−−→0
↓
S(cid:26)∗⊗π(cid:26)∗F
↓
0.
DenotethekernelofthecompositionbyK :
0 −→K −→OGk(cid:26)(E)⊗V −→S(cid:26)∗⊗π(cid:26)∗F −→0.
ThenK isalocallyfreesheafofrankdimV−(e−k)f onGk(cid:26)(E).ViewingGas
Gk(cid:26)(E)×V amountstoconsideringitasthetotalspaceofthelocallyfreesheaf
22 jørgenandersgeertsen
OGk(cid:26)(E)⊗V onGk(cid:26)(E),andthetautologicalsectionσt ∈(cid:13)(G,q1∗(OGk(cid:26)(E)⊗V))
issimplythepull-backviaπofthetautologicalsectionτt ∈(cid:13)(X×V,p1∗(OX⊗
V)).Nowσ isdefinedastheimageofπ∗τ in(cid:13)(G,S∗ ⊗π∗p∗F),henceas
1
the image of π∗τt ∈ (cid:13)(G,π∗p1∗(OX ⊗ V)) in (cid:13)(G,S∗ ⊗ π∗p1∗F) via the
composition π∗p1∗(OX ⊗V) → π∗p1∗(E∗ ⊗F) → S∗ ⊗π∗p1∗F.Viewing
π∗τt = σt as a section of π∗p1∗(OX ⊗ V) = q1∗(OGk(cid:26)(E) ⊗ V) and noting
S∗⊗π∗p1∗F =q1∗(S(cid:26)∗⊗π(cid:26)∗F),thismeansthatσ istheimageofσt underthe
mapq1∗(OGk(cid:26)(E)⊗V)→q1∗(S(cid:26)∗⊗π(cid:26)∗F).NowthetotalspaceSpecSymK ∗
of K is a subscheme of SpecSym(OGk(cid:26)(E) ⊗V) = G. The proposition will
certainly follow if we show that SpecSymK ∗ = Z(σ), because any closed
embedding of an affine space bundle in another is always regular, and the
dimensionofSpecSymK ∗ isdimGk(cid:26)(E)+rankK =dimX+(e−k)k+
dimV −(e−k)f =dimX+dimV −(e−k)(f −k).NowSpecSymK ∗ =
Z(σ)followsfromthenextgeneralobservation,suggestedbyAndersThorup:
Lemma6. LetT beaschemeand
0 −→H (cid:26)(cid:26) −→H −→H (cid:26) −→0
an exact sequence of locally free sheaves on T. Let H = SpecSymH ∗ be
the total space associated to H and p : H → T the projection map. Let
σt ∈ (cid:13)(H,p∗H )bethetautologicalsectionofp∗H andσ ∈ (cid:13)(H,p∗H (cid:26))
itsimageglobalsection.Thenasschemes,
Z(σ)=SpecSymH (cid:26)(cid:26)∗.
Proof. We show that the corresponding functor of points are equal, see
[10,Prop.1,Chap.II.6].HenceletQbeascheme.Wemustshowthat
Hom(Q,Z(σ))=Hom(Q,SpecSymH (cid:26)(cid:26)∗).
NowbytheuniversalpropertyofZ(σ),togiveamorphismfromQtoZ(σ)is
thesamethingasgivingamorphismf : Q → H suchthatf∗σ = 0.Butto
giveamorphismf :Q→H isequivalenttogiveamapg =p◦f :Q→T
+asectionofg∗H (namelyf∗σt).Henceweseethat
Hom(Q,Z(σ))=morphismsg :Q→T +asections ofg∗H
suchthattheimageofs is0in(cid:13)(Q,g∗H (cid:26)).
But the kernel of the map (cid:13)(Q,g∗H ) → (cid:13)(Q,g∗H (cid:26)) is (cid:13)(Q,g∗H (cid:26)(cid:26)).
Hence
Hom(Q,Z(σ))=morphismsg :Q→T +asectionofg∗H (cid:26)(cid:26)
=Hom(Q,SpecSymH (cid:26)(cid:26)∗).