Enriques Surfaces and other Non-Pfaffian Subcanonical Subschemes of Codimension 3

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ENRIQUES SURF ACES AND OTHER NON-PF AFFIAN SUBCANONICAL SUBSCHEMES OF CODIMENSION 3DAVID EISENBUD,SORIN POPESCU,AND CHARLES WALTER Abstract.We give examples of subcanonical subvarieties of codimen-sion 3in projective n -space which are not Pfa?an ,i.e.de?ned by the ideal sheaf of submaximal Pfa?ans of an alternating map of vector bun-dles.This gives a negative answer to a question asked by Okonek [29].Walter [36]had previously shown that a very large majority of sub-canonical subschemes of codimension 3in P n are Pfa?an,but he left open the question whether the exceptional non-Pfa?an cases actually occur.We give non-Pfa?an examples of the principal types allowed by his theorem,including (Enriques)surfaces in P 5in characteristic 2and a smooth 4-fold in P 7C .These examples are based on our previous work [14]showing that any strongly subcanonical subscheme of codimension 3of a Noetherian scheme can be realized as a locus of degenerate intersection of a pair of Lagrangian (maximal isotropic)subbundles of a twisted orthogonal bundle.There are many relations between the vector bundles E on a nonsingular algebraic variety X and the subvarieties Z ?X .For example,a globalized form of the Hilbert-Burch theorem allows one to realize any codimension 2locally Cohen-Macaulay subvariety as a degeneracy locus of a map of vector bundles.In addition the Serre construction gives a realization of any subcanonical codimension 2subvariety Z ?X as the zero locus of a section of a rank 2vector bundle on X .The situation in codimension 3is more complicated.In the local setting,Buchsbaum and Eisenbud [5]described the structure of the minimal free resolution of a Gorenstein (i.e.subcanonical)codimension 3quotient ring of a regular local ring.Their construction can be globalized:If φ:E →E ?(L )is an alternating map from a vector bundle E of odd rank 2n +1to a twist

of its dual by a line bundle L ,then the 2n ×2n Pfa?ans of φde?ne a degeneracy locus Z ={x ∈X |dim ker(φ(x ))≥3}.If X is nonsingular and codim(Z )=3,the largest possible value,(or more generally if X is locally Noetherian and grade(Z )=3),then,after twisting E and L appropriately,O Z has a symmetric locally free resolution

0→L p ??→E φ?→E ?(L )p ?→O X →O Z →0(0.1)

2DAVID EISENBUD,SORIN POPESCU,AND CHARLES W ALTER

with p locally the vector of submaximal Pfa?ans of φand with Z sub-canonical with ωZ ～=ωX (L ?1)|

Z .(The reader will ?nd a general discussion

of subcanonical subschemes in the introduction of our paper [14]and in Sec-tion 2below;for the purpose of this introduction it may su?ce to know that a codimension 3subvariety Z of P n is subcanonical if it is (locally)Gorenstein with canonical line bundle ωZ =O Z (d ).)

Okonek [29]called such a Z a Pfa?an subvariety ,and he asked:

Question (Okonek).Is every smooth subcanonical subvariety of codimen-sion 3in a smooth projective variety a Pfa?an subvariety ?

This paper is one of a series which we have written in response to Okonek’s question.In the ?rst paper (Walter [36])one of us found necessary and suf-?cient numerical conditions for a codimension 3subcanonical subscheme X ?P N k to be Pfa?an,but was unable to give subcanonical schemes failing the numerical conditions.In the second paper [14]we found a construc-tion for codimension 3subcanonical subschemes which is more general than that studied by Okonek,and proved that this construction gives all sub-canonical subschemes of codimension 3satisfying a certain necessary lifting condition.(This lifting condition always holds if the ambient space is P n or a Grassmannian of dimension at least 4.)

The construction is as follows:Let E ,F be

a pair of Lagrangian subbundles of a twisted orthogonal bundle with dim k (x ) E (x )∩F (x )

always odd.There is a natural scheme structure on the degeneracy locus Z :={x ∈X |dim k (x )

E (x )∩

F (x ) ≥3}(0.2)which is subcanonical if grade(Z )=3;the structure theorem asserts that

every strongly subcanonical codimension 3subscheme (in the sense of the de?nition in the introduction of our paper [14])arises in this way.

Another more complicated but also more explicit description of the struc-ture theorem may be given as follows (see our paper [14]for more details):If one has a subcanonical subvariety with ωZ ～=L |

Z ,then the isomorphism

O Z ～=ωZ (L ?1)|Z ～=E xt 3O X (O Z ,ωX (L ?1))is a sort of symmetry in the de-rived category.Let L 1:=ωX (L ?1).Then O Z should have a locally free resolution which is symmetric in the derived category.This is the case if O Z has a locally free resolution with a symmetric quasi-isomorphism into the twisted shifted dual complex:

0-E -O X -0

0-G ?(L 1)φ?

?ψ?-E ?(L 1)φ?

?

-O X

NON-PF AFFIAN SUBCANONICAL SUBSCHEMES OF CODIMENSION33 In this paper we give examples of non-Pfa?an codimension3subcanonical subvarieties.Thus we give an answer“No”to Okonek’s question.Because of the numerical conditions of[36],our examples fall into four types:

1.Surfaces S?P5k with char(k)=2and withωS～=O S(2d)such that

h1(O S(d))is odd.

2.Fourfolds Y?P7withωY～=O Y(2d)such that h2(O Y(d))is odd.

3.Fourfolds Z?P7R such thatωZ～=O Z(2d)such that the cup product

pairing on H2(O Z(d))is positive de?nite(or otherwise not hyperbolic).

4.Examples in ambient varieties other than P N.

Our examples include among others:codimension3Schubert subvarieties of orthogonal Grassmannians,non-Pfa?an fourfolds of degree336in P7,and nonclassical Enriques surfaces in P5in characteristic2.

Structure of the paper.In§1we review well known about quadratic forms,orthogonal Grassmannians and Pfa?an line bundles.In§2and§3 we review the machinery and results of our previous papers[14]and[36]. The rest of the paper is devoted to the examples of non-Pfa?an subcanonical subschemes.The?rst example,in§4,is a codimension3Schubert variety in OG2n.In§5we construct a smooth non-Pfa?an subcanonical fourfold of degree336in P7with K=12H.In§6we construct some subcanonical fourfolds in P7R which are not Pfa?an over R but become Pfa?an over C. In§7we analyze Reisner’s example in characteristic2of the union of10 coordinate2-planes in P5.

In the last two sections§§8,9we analyze nonclassical Enriques surfaces in characteristic2and their Fano models.As a consequence we obtain a (partial)moduli description of Fano-polarized unnodal nonclassical Enriques surfaces as a“quotient”of OG20modulo SL6(Corollary8.9).In§9we identify the closure of the locus ofα2-surfaces within OG20by calculating the action of Frobenius on our resolutions.

The philosophy that(skew)-symmetric sheaves should have locally free resolutions that are(skew)-symmetric up to quasi-isomorphism is also pur-sued in[15]and[37].The former deals primarily with methods for construct-ing explicit locally free resolutions for(skew)-symmetric sheaves on P n.The latter investigates the obstructions(in Balmer’s derived Witt groups[1])for the existence of a genuinely(skew)-symmetric resolution. Acknowledgments.We are grateful to Igor Dolgachev,Andr′e Hirschowitz, and N.Shepherd-Barron for many useful discussions or suggestions.

The second and third authors would like to thank the Mathematical Sci-ences Research Institute in Berkeley for its support while part of this paper was being written.The?rst and second authors also thank the University of Nice–Sophia Antipolis for its hospitality.

Many of the diagrams were set using the diagrams.tex package of Paul Taylor.

4DAVID EISENBUD,SORIN POPESCU,AND CHARLES W ALTER

1.Orthogonal Grassmannians

In this section we bring together a number of facts about orthogonal Grassmannians which we will need throughout the paper.All the results are well known,so we omit proofs.

Quadratic forms[23][31].Let V be vector space of even dimension2n over a?eld k.(We impose no restrictions on k;it need not be algebraically closed nor of characteristic=2.)A quadratic form on V is a homogeneous quadratic polynomial in the linear forms on V,i.e.a member q∈S2(V?). The symmetric bilinear form b:V×V→k associated to q is given by the formula

b(x,y):=q(x+y)?q(x)?q(y).

(1.1)

The quadratic form q is nondegenerate if b is a perfect pairing.

A Lagrangian subspace L?(V,q)is a subspace of dimension n such that q|L≡0.There exist such subspaces if and only if the quadratic form q is hyperbolic.By de?nition,this means that there exists a coordinate system on V in which one can write q= n i=1x i x n+https://www.doczj.com/doc/b37905072.html,grangian subspaces L satisfy L=L⊥,and the converse is true if char(k)=2.

If X is an algebraic variety over the?eld k,then the quadratic form q makes V?O X into an orthogonal bundle.A Lagrangian subbundle E?V?O X is a subbundle of rank n such that all?bers E(x)?V are Lagrangian with respect to q.

Orthogonal Grassmannians[27][28][32].Let V be a vector space of even dimension2n with the hyperbolic quadratic form q= n i=1x i x n+i. The Lagrangian subspaces of(V,q)form two disjoint families such that dim(L∩W)≡n(mod2)if L and W are Lagrangian subspaces in the same family,while dim(L∩W)≡n+1(mod2)otherwise.

Each family of Lagrangian subspaces is parametrized by the k-rational points of an orthogonal Grassmannian(or spinor variety)OG2n of dimension n(n?1)/2.Let U(resp.W)be a Lagrangian subspace such that dim(L∩U) is even(resp.dim(L∩W)is odd)for all L in the?rst family.Then OG2n contains the following Schubert varieties:

σ2(U):={L|dim(L∩U)≥2},

(1.2)

σ3(W):={L|dim(L∩W)≥3},

(1.3)

σ5(W):={L|dim(L∩W)≥5}.

(1.4)

We have the following basic results about these subvarieties:

Lemma1.1.(a)The subvarietyσ2(U)?OG2n is of codimension1,and its complement(parametrizing Lagrangian subspaces complementary to U)

is an a?ne space A n(n?1)/2

k .

(b)The subvarietiesσ3(W)andσ5(W)are of codimensions3and10in OG

2n

,respectively.Moreover,σ3(W)is nonsingular outsideσ5(W).

NON-PF AFFIAN SUBCANONICAL SUBSCHEMES OF CODIMENSION35 The reason behind Lemma1.1(a)is that if one?xes a Lagrangian subspace U?complementary to U,then the Lagrangian subspaces complementary to U are the graphs of alternating maps U→U?.This fact also underlies the following lemma.

Lemma1.2.Suppose L?(V,q)is a Lagrangian subspace.Then there is a natural identi?cation T[L](OG2n)～=Λ2L?.Moreover,if dim(L∩W)=3, then there is a natural identi?cation

T[L](σ3(W))～={f∈Λ2L?|f|L∩W≡0}.

Pfa?an line bundles[27][32].The Picard group of OG2n is isomorphic to Z.Its positive generator O(1)is the universal Pfa?an line bundle.The divisorsσ2(U)of Lemma1.1(a)are zero loci of particular sections of O(1). The canonical line bundle of OG2n is O(2?2n).

On the orthogonal Grassmannian OG2n there is a universal short exact sequence of vector bundles

0→S→V?O OG

2n →S?→0

(1.5)

with S the universal Lagrangian subbundle of rank n.The determinant line bundle related to the Pl¨u cker embedding is det(S?)=O(2).

The universal Lagrangian subbundle has the following universal property: If X is an algebraic variety and E?V?O X is a Lagrangian subbundle over X whose?bers lie in the family parametrized by OG2n,then there exists a unique map f:X→OG2n such that the natural exact sequence 0→E→V?O X→E?→0is the pullback along f of the universal exact sequence(1.5).The Pfa?an line bundle of E is then Pf(E):=f? O(?1) . It is a canonically de?ned line bundle on X such that Pf(E)?2～=det(E).

2.The Lagrangian subbundle construction

Let X be a nonsingular algebraic variety over a?eld k,and let Z?X be a codimension3subvariety.In[14]we called Z?X subcanonical if the sheaf O Z has local projective dimension over O X equal to its grade(in

this case3),and the relative canonical sheafωZ/X:=E xt3O

X (O Z,O X)is

isomorphic to the restriction to Z of a line bundle L on X.We called Z strongly subcanonical if in addition the map

Ext3O

X (O Z,L?1)→Ext3O

X

(O X,L?1)=H3(X,L?1).

induced by the surjection O X→O Z is zero.The last condition is immediate if X=P n,for n≥4,as the right cohomology group is always zero.Thus all subcanonical subvarieties of P n are strongly subcanonical.

The method we use to construct subcanonical subvarieties of codimension 3was developed in[14].In fact,all our examples can be constructed using the following result,which is a special case of[14]Theorem3.1.

Theorem2.1.Let V be a vector space of even dimension2n over a?eld k, and q= n i=1x i x n+i a hyperbolic quadratic form on V.Suppose that X is a

6DAVID EISENBUD,SORIN POPESCU,AND CHARLES W ALTER

nonsingular algebraic variety over k ,and that E ?V ?O X is a Lagrangian subbundle with Pfa?an line bundle L :=Pf(E )satisfying L ?2～=det(E ).Let W ?(V,q )be a Lagrangian subspace such that dim k (x )[E (x )∩W ]is odd for all x .If W is su?ciently general,then

Z W ={x ∈X |dim k (x )(E (x )∩W )≥3},

(2.1)is a (strongly )subcanonical subvariety of codimension 3in X with

ωZ W ～=(ωX ?det(E ?))|Z W

and with symmetrically quasi-isomorphic locally free resolutions

0-E (L )-O X

-L ?2

--E xt 3O X (O Z W ,L ?2)～=η?

4(c 1c 2

?2c 3),where c i :=c i (E ?).

To check the smoothness of Z W in characteristic 0we proceed as follows.The universal property of the orthogonal Grassmannian gives us a mor-phism f :X →OG 2n .The group SO 2n acts on OG 2n ,and it translates the Schubert varieties into other Schubert varieties:g ·σi (Λ)=σi (g Λ).Since the characteristic is 0,Kleiman’s theorem on the transversality of the gen-eral translate [22]applies.So there exists a nonempty Zariski open subset U ?SO 2n such that if g ∈U ,then σ5(g Λ)intersects f (X )transversally in codimension 10,and σ3(g Λ)intersects f (X )transversally in codimen-sion 3.Moreover,since SO 2n is a rational variety and k is an in?nite ?eld,U contains k -rational points.So if W :=g Λwith g a k -rational point in U ,then W ?(V,q )is a Lagrangian subspace de?ned over k such that Z W :=f ?1(σ3(W ))is of codimension 3and is smooth outside f ?1(σ5(W )),which is of codimension 10in X .

In characteristic p ,it will be more complicated to prove that Z W is smooth (see Lemma 5.2below).3.Pfaffian subschemes

Theorem 2.1allows us to construct subcanonical subvarieties Z ?X of codimension 3with resolutions which are not symmetric.Such a Z might nevertheless possess other locally free resolutions which are symmetric (i.e.Pfa?an).However,the third author gave in [36]a necessary and su?cient condition for a subcanonical subscheme of codimension 3in projective space to be Pfa?an.

So suppose Z ?P n +3is a subcanonical subscheme of codimension 3and dimension n such that ωZ ～=O Z (?).If n and ?are both even,then Serre

NON-PF AFFIAN SUBCANONICAL SUBSCHEMES OF CODIMENSION 37duality de?nes a nondegenerate bilinear form

H n/2(O Z (?/2))×H n/2(O Z (?/2))∪?→H n (O Z (?))～=k.

(3.1)This bilinear form is symmetric if n ≡0(mod 4),and it is skew-symmetric if n ≡2(mod 4).In characteristic 2it is symmetric in both cases.The result proven was the following.

Theorem 3.1(Walter [36]).Suppose that Z ?P n +3is a subcanonical sub-scheme of codimension 3and dimension n ≥1over the ?eld k such that ωZ ～=O Z (?).Then Z is Pfa?an if and only at least one of the following conditions holds :(i)n or ?is odd,(ii)n ≡2(mod 4)and char(k )=2,or (iii)n and ?are even,and H n/2(O Z (?/2))is of even dimension and con-tains a subspace which is Lagrangian with respect to the cup product pairing of (3.1).

Another way of stating the theorem is that a subcanonical subscheme Z ?P n +3of dimension n with ωZ ～=O Z (?)is not Pfa?an if and only either (a)n ≡0(mod 4),or else

(b)n ≡2(mod 4)and char(k )=2;

and at the same time either

(c)?is even and H n/2(O X (?/2))is odd-dimensional,or else

(d)?is even and H n/2(O X (?/2))is even-dimensional but has no subspace

which is Lagrangian with respect to the cup product pairing of (3.1).The simplest

cases where non-Pfa?an subcanonical subschemes of dimen-sion n in P n +3k could exist are therefore

1.Surfaces S ?P 5k with char(k )=2and with ωS ～=O S (2d )such that

h 1(O S (d ))is odd.

2.Fourfolds Y ?P 7with ωY ～=O Y (2d )such that h 2

(O Y (d ))is odd.

3.Fourfolds Z ?P 7R such that ωZ ～=O Z (2d )such that the cup product

pairing on H 2(O Z (d ))is positive de?nite (or otherwise not hyperbolic).

4.The codimension 3Schubert variety in OG 2n

Our ?rst examples of non-Pfa?an subcanonical subschemes of codimen-sion 3are the codimension 3Schubert varieties of OG 2n .

Theorem 4.1.Let n ≥4,and let σ3(W )?OG

2n be one of the codimension

3Schubert varieties of Lemma 1.1(b).Then Z is (strongly )subcanonical with ωZ ～=O Z (4?2n )but is not Pfa?an.

Proof.We apply

Theorem 2.1with X :=OG 2n and E :=S the universal

Lagrangian subbundle.The degeneracy locus is Z :=σ3(W )and it has

resolutions 0-S (?1)-O

-O (?2) --E xt 3(O Z ,O (?2))

～=η?

8DAVID EISENBUD,SORIN POPESCU,AND CHARLES W ALTER

Moreover,ωZ～=ωOG

2n (2)|Z～=O Z(4?2n)sinceωOG

2n

～=O(2?2n).

It remains to show that Z is not Pfa?an.Let U be a general totally isotropic subspace of dimension n?3,and let i:OG6(U⊥/U)?→OG2n be the natural embedding.If Z were Pfa?an,its symmetric resolution would be of the form

0→O(?2)→E(?1)→E?(?1)→O→O Z→0

(4.2)

(cf.(0.1)in the introduction).It would thus pull back to a symmetric resolution of i?1(Z)on OG6(U⊥/U)of the same form.

Now OG6parametrizes one family of P2’s contained in a smooth hyper-quadric in P5.Thus OG6～=P3(see Jessop[21]or Gri?ths-Harris[18]), and i?1(Z)is a codimension3Schubert subvariety of OG6～=P3and indeed a point Q.If we pull the resolution(4.2)back to P3and twist,we get a resolution

0→O P3(?3)→i?E(?2)→i?E?(?2)→O P3(?1)→O Q(?1)→0. This gives1=χ(O Q(?1))=2χ(i?E(?2)),which is a contradiction.This completes the proof.

The contradiction obtained at the end of the proof above is of the same nature as the obstructions in Theorem3.1(see[36]).The contradiction can also be derived from[14]Theorem7.2.

5.A non-Pfaffian subcanonical4-fold in P7

In this section we use the construction of Theorem2.1to give an example of a smooth non-Pfa?an subcanonical4-fold in P7over an arbitrary in?nite ?eld k.

Let V=H0(O P7(1))?,so that P7is the projective space of lines in V.We ?x an identi?cationΛ8V～=k.The70-dimensional vector spaceΛ4V has the quadratic form q(u)=u(2),the divided square in the exterior algebra.(If char(k)=2,then u(2)=1

ξ∈P7,viewed as a subspace ofΛ4V,is the image over

NON-PF AFFIAN SUBCANONICAL SUBSCHEMES OF CODIMENSION 39The ?bers W ξof ?4P 7(4)are all members of the same family of Lagrangian subspaces of the hyperbolic quadratic space Λ4V .Let W be a general mem-ber of that family,and (to abuse notation)W ?a general member of the opposite family of Lagrangian subspaces of Λ4V .Then Λ4V =W ⊕W ?,and the symmetric bilinear form on Λ4V induces a natural isomorphism between W ?and the dual of W .Finally,let

Z W :={

-O (?20)

ψ-W ??O (?10)--O Z W

0-W ?O (?10)φ??ψ?-?3P 7(?6)φ??-O

(5.3)

Proof.The existence of Z W and the resolutions follow from Theorem

2.1,using the fact that det(?4P 7(4))=O P 7(?20).The subcanonical subvariety Z W is not Pfa?an because ωZ =O Z (12)and h 2(O Z (6))=1(as one sees from (5.3)).Using the resolutions and a computer algebra package,one computes the Hilbert polynomial of Z W as:336 n +34 ?2520 n +23 +9814 n +12

?25571n +49549So the degree of Z W is 336.In characteristic 0the smoothness of the gen-eral Z W follows from Kleiman’s transversality theorem as explained after Theorem 2.1.In characteristic p the smoothness follows from Lemma 5.2below.

The free resolution of R/I Z W is of the form

(5.4)0→R (?14)→R (?13)8→R (?12)28⊕R (?20)

→R (?11)56→R (?10)35→R →R/I Z W →0.

Lemma 5.2.In Theorem 5.1,if k is an in?nite ?eld and W is general,then Z W is smooth.

Proof.Let

Υ:={W ∈OG 70|P 7∩σ5(W )=?,

and P 7∩σ3(W )=Z W is nonsingular of dimension 4}

10DAVID EISENBUD,SORIN POPESCU,AND CHARLES W ALTER

We will show thatΥis nonempty without invoking Kleiman’s theorem on the transversality of a general translate.We will do this by showing that the complement has dimension at most dim OG70?1.

Let n:=35,and let Gr(3,2n)be the Grassmannian parametrizing three-dimensional subspaces ofΛ4V.Let

Y:={(ξ,F,W)∈P7×Gr(3,2n)×OG2n|F?Wξ∩W}.

To choose a point of Y,one?rst choosesξ∈P7,then F in a?ber of a Grassmannian bundle with?ber Gr(3,n),then W in a?ber of an orthogonal Grassmannian bundle with?ber OG2(n?3).We see that Y is nonsingular with dim(Y)=dim OG2n+4.A similar computation shows that

Y1:={(ξ,F,W)∈Y|dim(Wξ∩W)≥5}

is of dimension equal to dim OG2n?3.So the image of Y1under the pro-jectionπ:Y→OG2n is not dominant.Over a point W/∈π(Y1),one has π?1(W)=Z W.Hence it is enough to show that the dimension of Y2:={(ξ,F,W)∈Y|dπ:T(ξ,F,W)Y→T W OG2n is not surjective}

is at most dim OG2n?1.The essential question is:if(ξ,F,W)∈Y?Y1,so thatπ?1(W)=Z W,when is the dimension of TξZ W=ker(dπ)(ξ,F,W)equal to4,and when is it more?

Now TξZ W=T W

ξ

(P7∩σ3(W)).In order to identify this space we must

identify T W

ξ(OG2n)and the two subspaces T W

ξ

P7and T W

ξ

(σ3(W)).

There is a well known isomorphism T W

ξ

(OG2n)～=Λ2W?ξ.Essentially, given a?rst-order deformation of Wξas a Lagrangian subspace ofΛ4V,any vector w∈Wξdeforms within the deforming subspace as a w+εu(w)with u(w)well-de?ned modulo Wξ.This gives a map Wξ→Λ4V/Wξ～=W?ξ. The vector w+εu(w)remains isotropic if and only if u is alternating.To describe u as an alternating bilinear form on Wξ,one looks at

w1,w2+εu(w2) =ε w1,u(w2) .

If W∩Wξis a space F of dimension3,then T W

ξ

(σ3(W))={u∈Λ2W?ξ| u|F×F≡0}according to Lemma1.2.I.e.T W

ξ

(σ3(W))is the kernel of the natural mapΛ2W?ξ→Λ2F?.

Now let Vξ:=V/ ξ .Then there is a natural isomorphism TξP7～=Vξ. There is also a natural isomorphism Wξ=ξ∧Λ3V～=Λ3Vξ.A vector θ∈TξP7=Vξcorresponds to a?rst-order in?nitesimal deformationξ+εθofξ,and hence to the?rst-order in?nitesimal deformation(ξ+εθ)∧Λ3V of Wξ.The corresponding alternating bilinear form on Wξ=ξ∧Λ3V is computed by

ξ∧α,(ξ+εθ)∧β =ε(ξ∧α∧θ∧β)∈ε(Λ8V).

If we use the isomorphism Wξ～=Λ3Vξ,then the corresponding alternating bilinear form onΛ3Vξis therefore

gθ(α,β)=?α∧β∧θ∈Λ7Vξ.

NON-PF AFFIAN SUBCANONICAL SUBSCHEMES OF CODIMENSION311 Hence if we think of F:=W∩Wξas a three-dimensional subspace of

(P7∩σ3(W))?Vξis the orthogonal complement of Λ3Vξ～=Wξ,then T W

ξ

the image ofΛ2F inΛ6Vξ～=V?ξ.As a result(ξ,F,W)∈Y2if and only if the mapΛ2F→Λ6Vξis not injective.

Every vector inΛ2F is of the formα∧β.We therefore look at the set T1(ξ):={(α,β)∈(Λ3Vξ?{0})2|α∧β=0}.

We stratify this locus according to the rank of the map mα:Vξ→Λ4Vξde?ned by multiplication byα.Generically this map is injective,and the dual multiplication map m?α:Λ3Vξ→Λ6Vξis surjective.For(α,β)to be in T1(ξ),βmust be in the codimension-7subspace ker(m?α)?Λ3Vξ.Hence T1(ξ)contains a stratum of dimension2n?7where mαis injective.

The locus T1(ξ)contains two other strata,but they are of smaller dimen-sion.In one of them theαhave the formα′∧α′′withα′∈Vξandα′′∈Λ2Vξindecomposable.Sinceα′is really only well-de?ned up to a scalar multi-ple,andα′′is really well-de?ned only inΛ2(Vξ/ α′ ),these form a locus of dimension21=n?14inΛ3V.For theseαthe rank of mαis6,so this stratum in T1(ξ)has dimension2n?20.

In the?nal stratum theαare of the formα1∧α2∧α3,a locus of dimension 13=n?22.For theseαthe rank of mαis4,so this stratum of T1(ξ)has dimension2n?26.

Thus dim T1(ξ)=2n?7.As a result the locus

T2(ξ):={F0∈Gr(2,Wξ)|F0=Span(α,β)withα∧β=0}

is of dimension2n?11.Consequently the locus

T3(ξ):={F∈Gr(3,Wξ)|F?F0for some F0in T2(ξ)}

is of dimension at most dim T2(ξ)+n?3=3n?14.

For each F∈T3(ξ),to give a W∈OG2n such that F?W∩Wξis equivalent to giving a member of one of the families of Lagrangian subspaces of F⊥/F.Consequently,such W form a family of dimension(n?3)(n?4)/2. As a result,the locus

T4(ξ):={(F,W)∈Gr(3,Wξ)×OG2n|F?W∩Wξand F∈T3(ξ)}

is of dimension at most(3n?14)+(n?3)(n?4)/2=n(n?1)/2?8.Finally

T5:={(ξ,F,W)|(F,W)∈T4(ξ)}

is of dimension at most n(n?1)/2?1.But the part of T5where F= W∩Wξis exactly the part of Y2lying in Y?Y1.So dim Y2

As a result Y2does not dominate OG2n,andΥ?OG2n is a nonempty open subset.As before,Υmust then contain a rational point because the base?eld is in?nite,and we conclude.

12DAVID EISENBUD,SORIN POPESCU,AND CHARLES W ALTER

6.Extension of scalars and non-Pfaffian examples

In this section we prove the following extension of Theorem 5.1.It will allow us to construct subcanonical subschemes of codimension 3in P N k which are not Pfa?an over k but are Pfa?an over a ?nite extension of k .

Theorem 6.1.Let k be an in?nite ?eld,let E be a k -vector space of dimen-sion d ≥1,and let b :E ×E →k be a nondegenerate symmetric bilinear form.There exists a nonsingular subcanonical fourfold X ?P 7k of degree (1000d 3+8d )/3with K X =(20d ?8)H such that the cup product pairing (3.1)on the middle cohomology

H 2(O X (10d ?4))×H 2(O X (10d ?4))→H 4(O X (20d ?8))～=k

may be identi?ed with b :E ×E →k .

Proof.If e 1,...,e d is a basis of E and b ij :=b (e i ,e j ),then the quadratic form on E ?Λ4V given by Q

i e i ?u i := i b ii u (2)i + i

b ij u i ∧u j is nondegenerate and has the property that if L is a Lagrangian subspace of (Λ4V,q ),then E ?L is a Lagrangian subspace of (E ?Λ4V,Q ).Consequently E ??4P 7(4)is a Lagrangian subbundle of (E ?Λ4V ?O P 7,Q ).Also,if we make a standard identi?cation Λ4V ?～=Λ4V ,then the symmetri

c bilinear form associate

d to Q is b ?1:E ?Λ4V →E ??Λ4V .

We now choose a general Lagrangian subspace W ?(E ?Λ4V,Q )and apply the construction of Theorem 2.1.This produces a subcanonical four-fold X ?P 7k with K X =(20d ?8)H .The degree of X can be computed from the formula after Theorem 2.1.If char(k )=0,then X is nonsingular by Theorem 2.1.If k is in?nite of positive characteristic,X is nonsingular by an argument similar to Lemma 5.2whose details we leave to the reader.Let N :=10d ?4.The isomorphism η:O X (N )→ωX (?N )induces an isomorphism

H 2(O X (N ))～?→H 2(ωX (?N ))～=H 2(O X (N ))?

(6.1)which corresponds to the cup product pairing of Serre duality (3.1).Our problem is to identify this map.Now twisting the diagram (2.2)of Theorem 2.1produces a commutative diagram with exact rows (where we write O :=O P 7to keep things within the margins):

f

-E ??4P 7g -O (N )-O (?N ?8) --ωX (?N )

(6.2)

NON-PF AFFIAN SUBCANONICAL SUBSCHEMES OF CODIMENSION 313The rows of the diagram produce natural isomorphisms H 2(O X (N ))～=E and H 2(ωX (?N ))～=E ?.Moreover,the Serre duality pairing

H 2(O X (N ))×H 2(ωX (?N ))→H 4(ωX )tr ?→k

corresponds to the canonical pairing E ×E ?→k since the bottom resolution is dual to the top resolution,so the hypercohomology of the bottom resolu-tion is naturally Serre dual to the hypercohomology of the top resolution.It remains to identify the map H 2(O X (N ))→H 2(ωX (?N ))with the map b :E →E ?

.We do this by using the following commutative diagram with exact rows.

0f

-E ??4P 7g -O (N )-O (?N ?8) --O X (N )

0g ?-W ?O (?4) --ωX (?N )

～=η?

i -E ?Λ4V ?O (?4)

i ′-E ??3P 7π?b ?1?

-E ???3P 7

π′?

14DAVID EISENBUD,SORIN POPESCU,AND CHARLES W ALTER

that the map H 2(O X (N ))→H 2(ωX (?N ))induced by ηmay be identi?ed with b :E →E ?.This completes the proof of the theorem.

As an example,we may apply Theorem 6.1with b a positive de?nite sym-metric bilinear form on R 2.There are no Lagrangian subspaces of (R 2,b ),but there are Lagrangian subspaces of (C 2,b C ).So by the Pfa?anness cri-terion of Theorem 3.1,we get the following corollary.

Corollary 6.2.There exists a nonsingular subcanonical fourfold Y R ?P 7R of degree 2672with K Y =32H which is not Pfa?an over R but whose complexi?cation Y C ?P 7C is Pfa?an over C .

7.Reisner’s example in P 5

Another example of a codimension 3subcanonical subscheme Z ?P 5is provided by the Stanley-Reisner ring of the minimal triangulation of the real projective plane RP 2in characteristic 2.This is Reisner’s original ex-ample of a monomial ideal whose minimal free resolution depends on the characteristic of the base ?eld.See Hochster [20],Reisner [33],Stanley [34].Recall that if ?is a simplicial complex on the vertex set E ={e 0,...e n },the corresponding face variety X (?)?P n k in the sense of Stanley,Hochster and Reisner (see [34]for more details)is the locus de?ned by the ideal I ?generated by monomials corresponding to the simplexes (faces)not con-tained in ?.Geometrically,X (?)is a “projective linear realization”of ?.It is the union of linear subspaces of P n k corresponding to the simplices in ?,where for each m -simplex {e i 0,...,e i m }one includes the m -plane spanned by the vertices P i 0,...,P i m of the standard coordinate system on P n k .

If the topological realization |?|is a manifold,then X (?)is a locally Gorenstein scheme with ω?2X (?)～=O X (?).Moreover,ωX (?)is trivial if and only if |?|is orientable over the ?eld k .The cohomology of X (?)is given by the formula H i (O X (?))=H i (?,k ).The Hilbert polynomial of X (?),which coincides with the Hilbert function for strictly positive values,is completely determined by the combinatorial data (see [34]for more details).

We apply this theory with ?the triangulation of RP 2given in Figure 1,and with k a ?eld of characteristic 2.Then X :=X (?)is a union of ten 2-planes in P 5k corresponding to the ten triangles.Moreover,ωX ～=O X since RP 2is orientable in characteristic 2.The cohomology is given by

h 0(O X )=1,h 1(O X )=1,h 2(O X )=1,

since this is the cohomology of RP 2over a ?eld of characteristic 2.We may now prove

Proposition 7.1(char 2).Let X ?P 5be the face variety corresponding to the triangulation of RP 2given in Figure 1.Then X is subcanonical of codimension 3but not Pfa?an.

Proof.The subvariety X is a locally Gorenstein surface with ωX =O X but h 1(O X )=1.Thus it is subcanonical but not Pfa?an by Theorem 3.1.

NON-PF AFFIAN SUBCANONICAL SUBSCHEMES OF CODIMENSION 3150011001

10011001100001111000011110011000011110011012

3

3

44

55

Figure 1.The Minimal Triangulation of RP 2

In addition,X ?P 5is linearly normal (all X (?)are)and is not contained in a hyperquadric (since all pairs of vertices are joined by an edge in ?).This will allow us to conclude below that X is a degenerate member of a family of nonclassical Enriques surfaces (see Proposition 8.7and (8.5)).Over the complex numbers this is a Type III degeneration in Morrison [26]Corollary 6.2(iii a).See also Symms [35]and Altmann-Christophersen (in preparation)for the deformation theory of this scheme.

8.Enriques surfaces in P 5

In their extension to characteristic p of the Enriques-Castelnuovo-Kodaira classi?cation of surfaces not of general type,Bombieri and Mumford [4]char-acterize Enriques surfaces as nonsingular projective surfaces X with numer-ically trivial canonical class K X ≡0and with χ(O X )=1.In characteristic 2,Enriques surfaces are divided into three types:

h 1(O X )

canonical class 0

—1

injective 10

The map F :H 1(O X )→H 1(O X )is the action of Frobenius on the coho-mology.There is a voluminous literature on Enriques surfaces which we do not cite here.Our Pfa?anness criterion Theorem 3.1yields immediately Proposition 8.1(char 2).If X ?P 5is a nonclassical Enriques surface,then X is subcanonical but not Pfa?an.

A Fano polarization on an Enriques surface X is a numerically e?ective divisor H such that H 2=10and such that H ·F ≥3for every numerically e?ective divisor F with F 2=0.It is well known that every Enriques surface

16DAVID EISENBUD,SORIN POPESCU,AND CHARLES W ALTER

admits a Fano polarization,and that for any Fano polarization H the linear system|H|is base-point-free and de?nes a map?H:X→P5which is birational onto its image.The image?H(X)?P5is called a Fano model of X.A surface Y?P5is a Fano model of an Enriques surface if and only if it is of degree10with at worst rational double points as singularities and hasχ(O Y)=1and numerically trivial canonical class K Y≡0.

An Enriques surface is called nodal if it contains a smooth rational curve C with C2=?2.Fano models of unnodal Enriques surfaces are neces-sarily smooth.The following theorem was proven in characteristic=2by Cossec[8].Igor Dolgachev has informed us that the theorem also holds in characteristic2using arguments similar to Dolgachev-Reider[13]. Theorem8.2.Let H be a Fano polarization of an Enriques surface X. Then the following are equivalent:(a)X is nodal,and(b)one of the Fano models?H(X)or?H+K

X

(X)is contained in a hyperquadric.

We will now give an algebraic method for realizing Fano models of unnodal nonclassical Enriques surfaces as degeneracy loci associated to Lagrangian subbundles of an orthogonal bundle on P5.We will need the following tech-nical lemmas about divided squares.Both lemmas hold for even m in all characteristics.

Lemma8.3.Suppose that R is a commutative algebra over a?eld of char-acteristic2,that m≥3is an odd integer,and that E a free R-module of rank2m.Then there exists a well-de?ned divided square operationΛm E→Λ2m E～=R whose formula,with respect to any basis e1,...,e m,is given by

x i1...i m e i1...i m (2):= i1<···

j1<···

i1

{i k}∩{j?}=?x i

1···i m

x j

1···j m

e.

(8.1)

Proof.It is enough to prove the lemma for R a?eld,since one can reduce to

the case where the coe?cients x i

1...i m are independent indeterminates,and

R is the?eld of rational functions in these indeterminates.

We have to show that(8.1)is invariant under changes of basis in GL(E)=

GL n(R).But since R is now assumed to be a?eld,GL n(R)is generated by

three special kinds of changes of basis:permutations of the basis elements,

operations of the form e i λe i,and operations of the form e j e j+αe i. The formula(8.1)is almost trivially invariant under the?rst two kinds of

operations,and it is invariant under the third kind of operation because the

portions

x ijk

1...k m?2x?

1...?m

and x ik

1...k m?1

x j?

1...?m?1

+x i?

1...?m?1

x jk

1...k m?1

do not change.

Lemma8.4.Suppose that m≥3is an odd integer,that V is a vector space of dimension2m over a?eld of characteristic2,and that Q is a nonzero

NON-PF AFFIAN SUBCANONICAL SUBSCHEMES OF CODIMENSION 317quadratic form on Λm V .Then Q (x ∧w )=0for all x ∈V and w ∈Λm ?1V if and only if Q is

a

constant multiple of the divided-square quadratic form of Lemma 8.3.

Proof.Suppose ?rst that Q is a multiple of the divided-square quadratic form,and that 0=x ∈V and w ∈Λm ?1V .Choosing a basis of V of the form e 1=x,e 2,...,e m and applying formula (8.1),one sees easily that Q (x ∧w )=0.Conversely,suppose that Q (x ∧w )=0for all x ∈V and w ∈Λm ?1V .Write Q ( y I e I ):= I J αIJ y I y J .The hypothesis implies that Q (e I )=αII =0for all I .It also implies that each Q (e iA +e iB )=αiA,iB =0.Hence αIJ =0if the multi-indices I and J are not disjoint.Finally Q ((e i +e j )∧(e A +e B ))=αiA,jB +αiB,jA =0.In other words αIJ =αKL if I and J are disjoint,and K and L are disjoint,and I and K contain exactly m ?1common indices.One then easily deduces that all the αIJ with I and J disjoint must be equal.But then Q is the same as the quadratic form of (8.1)up to a constant.

Now let V =H 0(O P 5(1))?,and consider the exact sequence of vector bundles on P 5.

0→?3P 5(3)→Λ3V ?O P 5→?2P 5(3)→0.

(8.2)As in (5.2),if ξ∈V ,then the ?ber of ?3P 5(3)over

-O P 5(?6)

ψ-W ??O P 5(?3)--O Z W

0-W ?O P 5(?3)φ??ψ?-?2P 5φ??-O P 5

(8.4)

Proof.The general Z W is a subcanonical surface of degree 10with ωZ W ～=O Z W and with the resolutions (8.4)according to Theorem 2.1.It is smooth by an argument similar to Lemma 5.2.Since K Z W =0and h i (O Z W )=1for

18DAVID EISENBUD,SORIN POPESCU,AND CHARLES W ALTER

i=0,1,2,it is a nonclassical Enriques surface by the Bombieri-Mumford classi?cation.Since the degree is10,it is a Fano model.Moreover,according to the resolution,it is not contained in a hyperquadric.Hence it is unnodal by Theorem8.2.

The theorem has the following converse.

Proposition8.7(char2).Let X?P5be a locally Gorenstein subscheme of dimension2and degree10,withωX～=O X,with h i(O X)=1for i= 0,1,2,and which is linearly normal and not contained in a hyperquadric. Let V=H0(O P5(1))?,and let Q be the divided square onΛ3V.Then there exists a unique Lagrangian subspace W?(Λ3V,Q)such that X=Z W(cf. Theorem8.6)with locally free resolutions as in(8.4).

For example,the face variety X=X(?)of Proposition7.1satis?es all the hypotheses of the proposition and is de?ned as X=Z W for the Lagrangian subspace W?Λ3V spanned by the exterior monomials

(8.5)

e013,e014,e023,e025,e045,e124,e125,e135,e234,e345.

Any Z W in(8.3)which is of dimension2satis?es all the hypotheses of Proposition8.7.

The following re?nement of the Castelnuovo-Mumford Lemma is needed for the proof of Proposition8.7.Its proof is left to the reader.

Lemma8.8.Let M be a coherent sheaf on P n with no nonzero skyscraper subsheaves,and let0≤q≤n?1and m be integers.Suppose that M is(m+1)-regular,that H i(M(m?i))=0for i≥n?q,and that the comultiplication maps

H i(M(m?1?i))→V?H i(M(m?i))

are surjective for1≤i

Proof of Proposition8.7.The Hilbert polynomial of X isχ(O X(t))=5t2+1 by Riemann-Roch.The reader may easily verify that the sheaves O X(t)have cohomology as given in the following table:

h0(O X(t))h2(O X(t))

t>05t2+10

1

t<005t2+1

Now according to the table O X is3-regular;the only group obstructing its2-regularity is H2(O X);and the comultiplication H2(O X(?1))→V?H2(O X)is dual to the multiplication V??H0(O X)→H0(O X(1))and thus is an isomorphism.So we may apply Lemma8.8to see that all the generators of H0?(O X)are in degrees≤2,and all the relations are in degrees≤3.But for all t≤2the restriction maps H0(O P5(t))→H0(O X(t))are bijections. So the generators and relations of H0?(O P5)and of H0?(O X)in degrees≤2

NON-PF AFFIAN SUBCANONICAL SUBSCHEMES OF CODIMENSION 319

are the same—one generator in degree 0and no relations.Hence H 0?(O X )has only a single generator in degree 0,and its relations are all in degree 3.In other words H 1?(I X )=0,and the homogeneous ideal I (X )is generated by a 10-dimensional vector space W ?of cubic equations.We now go through the argument in [14]Theorem 6.2which constructs a pair of symmetrically quasi-isomorphic locally free resolutions of a sub-canonical subscheme of codimension 3.We have an exact sequence

0→K →W ??O P 5(?3)→O P 5→O X →0.

There is a natural isomorphism

Ext 1O P 5(K ,O P 5(?6))～=H 4(K )?～=H 2(O X )?～=k,

and a nonzero member of this group induces an extension which we can attach to give a locally free resolution

0→O P 5(?6)→F →W ??O P 5(?3)→O P 5→O X →0.

The intermediate cohomology of F comes from the intermediate cohomology of X and is given by H i ?(F )=0for i =1,2,4and H 3?(F )～=H 1?(O X )=k .According to Horrocks’Theorem [30],this means that F ～=?3P 5⊕

O P 5(a i ).Since F and ?3P 5are both of rank 10,they are isomorphic.This gives the top locally free resolution of (8.4).The rest of diagram (8.4)is now constructed as in the proof of [14]Theorem 6.2.

It remains to identify the quadratic spaces W ?⊕W and Λ3V ,and to show that W is uniquely determined by X .However,(8.4)gives us an exact sequence

0→?3P 5(3) ψφ ???→(W ?⊕W )?O P 5(φ?ψ?)

?????→?2P 5(3)→0.This is the unique nontrivial extension of ?2P 5(3)by ?3P 5(3),and thus co-

incides with the truncated Koszul complex (8.2).There is therefore a nat-ural identi?cation of W ?⊕W with Λ3V which identi?es the hyperbolic quadratic form on W ?⊕W with a quadratic form Q on Λ3V for which ?3P 5(3)?Λ3V ?O P 5is a Lagrangian subbundle.By Corollary 8.5,Q is necessarily the divided square quadratic form.

The identi?cation of the quadratic spaces W ?⊕W and Λ3V is unique up to homothety,once the morphisms ψ,φare chosen.However,the diagram (8.4)is unique up to an alternating homotopy α:W ??O P 5→W ?O P 5.One may check that this alternating homotopy does not change the Lagrangian subspace W ?Λ3V but only the choice of the Lagrangian complement W ?(cf.[14]§5).Thus W is uniquely determined by X .This completes the proof.

Corollary 8.9(char 2).There is a universal family of Fano models in P 5=P (V )of unnodal nonclassical Enriques surfaces parametrized by an SL(V )-invariant Zariski open subset U ?OG 20(Λ3V ).The SL(V )-orbits

20DAVID EISENBUD,SORIN POPESCU,AND CHARLES W ALTER correspond to isomorphism classes of Fano-polarized unnodal nonclassical Enriques surfaces(X,H).

In particular,unnodal nonclassical Enriques surfaces form an irreducible family.We will see below that its general member is aμ2-surface(Proposi-tion9.1).

The universal family of unnodal nonclassical Enriques surfaces has the following universal locally free resolution over G:=OG20(Λ3V).Let S?Λ3V?O G by the universal Lagrangian subbundle on G.According the

(3)and q?S on machinery of[14]Theorem2.1,the two subbundles p??3

P5

P5×G(where p and q are the two projections)de?ne a degeneracy locus Z,which one can verify has the expected codimension3using an incidence variety argument.Therefore Z?P5×G has a locally free resolution (8.6)

0→O(?6,?2)→(p??3P5)(0,?1)→(q?S?)(?3,?1)→O→O Z→0. Remark8.10.(a)Barth and Peters have shown that a generic Enriques surface over C has exactly214·3·17·31distinct Fano polarizations([2] Theorem3.11).

(b)The fact that the Fano model of a unnodal Enriques surface is de?ned by10cubic equations was?rst observed by F.Cossec(cf.[8][9]).

(c)(char2)The following geometrical construction of a general Fano-polarized unnodalμ2-surface is inspired by Mori-Mukai([25]Proposition 3.1).Let E be an elliptic curve with Hasse invariant one,and let i:E→E be the involution given by translating by the unique nontrivial2-torsion point.Letπ:E→E′:=E/ i be the quotient map.Let D′be a divisor of degree3on E′and let D:=π?(D′).The linear system|D|gives an em-bedding E?→P5with image an elliptic normal sextic curve.The involution i on E extends to an involution of P5.As in Mori-Mukai[25],(3.1.1)we may choose three general i-invariant hyperquadrics Q1,Q2,Q3containing E,such that X:=Q1∩Q2∩Q3?P5is a smooth K3surface,such that i|X has no?xed points.Let L be the hyperplane divisor class on X.Then S:=X/ i is an Enriques surface of typeμ2(since it has an′e tale double cover by a K3surface,cf.[4]).The divisor class L+E descends to a Fano polarization H on S,and the elliptic curve E?X descends to E′?S. By[7]or[13]Theorem1,(which Igor Dolgachev informs us also holds in characteristic2),S is unnodal if and only if|H?2E′|=?.But this fol-lows since|L?E|=?because E?P5is nondegenerate.An easy dimension count shows that this construction gives rises to10moduli of Fano polarized unnodalμ2Enriques surfaces,and thus to a general one.

9.Calculating the action of Frobenius

We would like to give a method for identifying when a Fano model of a nonclassical Enriques surface constructed in Theorem8.6is of typeμ2 or typeα2.This involves computing explicitly the action of Frobenius on

another的用法

an other others the other the others等。 1 one...the other翻译为一个...另一个...含义是有范畴的共2个2个部分2个人等。 例题I’ve got two tickets for tonight’s concert. One is for me _________ is for you. A other B the other C others D an other 答案D 2 one/some/several... another翻译为一个/一些/几个...另一个...含义中没有范畴可以无限增加。 例句You have had several cakes. Do you really want an other one Neither of the hats doesn’t look good on my daughter. Would you give me an other one 3 some...others...翻译为一些...其它的指剩余的大部分... 含义中没有范畴因为还有部分未提及。 例句/题A few students are playing soccer while others are watching them. 操场上一般不有只踢球和看球的人可能还有做其它活动的人。We should help_____when they are in trouble. A other B the others C others D the other我们不可能帮助除我们以外的所有人所以答案是C。 4 one...the others...翻译为一个...其余的指剩余的全部... 含义是有范畴的. 例句The minotor will go to the Teachers’ Office the others will stay in the classroom. 5some...the others...翻译为一些...其余的指剩余的全部... 含义是有范畴的. 例句Four of the ten boys are standing and the others are sitting round them.

some和any的区别和用法上课讲义

s o m e和a n y的区别 和用法

精品文档 收集于网络，如有侵权请联系管理员删除 some 和any 的区别和用法 要表示"一些"的意思，可用some, any 。 some 是肯定词（ASSERTIVE WORD 〕，常用于肯定句；any 是非肯定词（NONASSERTIVE WORD 〕，常用于否定句或疑问句。例如： There are some letters for me. There aren''t any letters for me. Are there any letters for me? I seldom get any sleep these days. any 也常用于条件分句以及带有否定含义的句子中： If you have any trouble, please let me know. 如果你有任何麻烦，请让我知道。 I forgot to ask for any change. 我忘了要一些零钱。 当说话人期待肯定回答时，some 也可用于疑问句, 比如当说话人期待来信时，他可以问道： Are there some letters for me? 当购物时向售货员提问或者主人向客人表示款待时，也可在疑问句中用some: Could I have some of these apples? Would you like some chocolate cake? 当some 与单数可数名词搭配时, some 相当于a certain （"某一"〕的含义；而any 与单数可数名词搭配，则相当于 every （"任何一个"〕的含义。例如： Some boy has broken a window. 有个男孩打破了窗户。 Any child could answer that question. 任何一个孩子都可以回答这个问题。 替代词one, ones, that, those 在比较结构中的用法辨析 在比较结构中，为了避免重复，常用替代词替代名词词组或它的中心词。可以这样用的替代词常用的且较难掌握的有one ， ones ，that 和those 。 I ．替代词one 和ones 的用法： 1．one 只能替代单数名词，one 的复数形式ones 只能替代复数名词。例如： My child doesn't like this book ．Show her a more interesting one ． 2．替代词one 或ones 必须带有一个限定或修饰词，它们和所替代的名词中心词所指不一定是同一对象，这是替代词one 或 ones 在用法上的一个重要特征。例如： I don't like this book ．I'd like a more interesting one ． 3．当替代词one 或ones 带有后置修饰语时，它们前面总有定冠词。例如： Our new cassette is more expensive than the one we had before ． 4．当替代词one 或ones 在形容词比较级、最高级以及某些限定词如this ，that ， which 和序数词等之后，可以省略。例如： some 和any 既可以修饰可数名词又可以修饰不可数名词，some 常用在肯定句中，而any 则常用在否定和疑问句中。因此 some 和any 的用法主要 是考虑用在肯定句、疑问句还是否定句中，与名词的可数与否无关。 some 意为“一些”，可作形容词和代词。它常修饰可数名词复数。如： some books 一些书，some boys 一些男孩，也可修饰不可数名词，如：some water 一些水，some tea 一些茶叶，some 常用在肯定句中。any 意为 “任何一些”，它也可修饰可数名词复数或不可数名词，常用于疑问句和否定句。如： --I have some tea here. 我这儿有些茶叶。 --I can’t see any tea. 我没看见茶叶。 --Do you have any friends at school? 你在学校有些朋友吗? --I have some English books, they are my best friends. 我有英语书，它们是我最好的朋友。 但在表示建议，反问，请求的疑问句中，或期望得到肯定回答时，多用some 而不 用any 。如： Would you like some coffee? 你要不要来点咖啡? What about some fruit juice? 来点水果汁如何? 当any 表示“任何”的 意义，起强调作用时，它可以用在肯定句中； Any student can answer this question.任何学生都可以回答这个问题。 选题角度： 辨析 some 和any 的不同用法：some 常用在肯定句中，而any 则常用在否定和疑问句中。在表示建议，反问，请求的疑问句中，或期望得到肯定回答 时，多用some 而不用any 。 一般用于疑问句或否定句中，用于never, hardly, without 等词之后，用于if / whether 之后。 而some 则用于肯定句中，用于建议或请求的 疑问句中，用于预料会作肯定回答的疑问句中，用于表示反问的否定的疑句中。 如：1. I’d been expecting ________ letters the whole morning, but there weren’t ________ for me. (全国卷) A. some; any B. many; a few C. some; one D. a few; none 【分析】答案选A 。前一分句是肯定句，用some ，后 一分句是否定名句，用any 。另外，后一分句的weren’t 显示这是否定句，主语应为复数，排除C 和D ；many 一般不用于肯定句，a few 一般用 于肯定句，排除B 。 2. Let us hope we can settle the matter without ________ more trouble. (全国卷) A. any B. a little C. some D. little 【分析】答案选A 。without 表否定，用any 。 3. Sarah has read lots of stories by American writers. Now she would like to read ________ stories by writers from ________ countries. (全国卷) A. some; any B. some;some C. some; other D. other; other 【分析】答案选C 。肯定句中用some 表示“一些”，other 作定 语，意为“别的”。 4. There’s ________ cooking oil left in the hou se. Would you go to the corner store and get ________ ? (北 京卷) A. little; some B. little; any C. a little; some D. a little; any 【分析】答案选A 。后句是表示请求的问句中，用some ，排 除B 和D ；由后文的意思可知，语气是否定的，排除C 。 5. ―I fee a bit hungry. ―Why don’t you have ________ bread? (全国卷) A. any B. some C. little D. a 【分析】答案选B 。表示建议的疑问句中 用some 。

another 的用法

1. another可视为由“an+other”构成，但总是写成一个词，不能写成an other；其后一般只接单数可数名词，不接复数名词或不可数名词。another有两个基本意义： (1) 外加的，同样的。如： Don’t say another word. 不要再说了。 Let me have another cup of tea. 给我再来一杯茶。 (2) 不同的，另外的。如： That’s another Story. 那是另一码事。 Give me another cup. This one’s cracked. 请给我换个杯子，这个裂了。 If I were you, I should get another lawyer. 如果我是你，我就请别的律师。 这样用的another表泛指。比较： Give me another (one). 另外给我一个。（表泛指） Give me the other (one). 给我另外那个。（表特指） 2. 在通常情况下，another 后不能接复数名词或不可数名词，但是若复数名词之前有few 或数词修饰，或不可数名词之前有piece of 之类的单位词时等，则可以与another 连用。如： I could go on for another two hours. 我再讲两个小时都讲不完。 I need another few days before l can make up my mind. 我还需几天才能决定。

3. one 有时可与another 对照使用。如： One (boy) wanted to read, and another wanted to watch TV. 一个（男孩）想看书，另一个则想看电视。 One day he wanted his lunch early, another day he wanted it late. 他一天要早点吃午饭，另一天又要晚点吃午饭。 4. 习语one after another意为“一个接一个地”“相继地”“依次地”，在句中主要用作状语。如： Planes took off one after another. 飞机陆续起飞。 One after another all his plans have failed. 他的计划都一一失败了。有时也可用作主语或宾语。如： One after another began to choke, and at the end all the women were weeping. 人们一个接一个哽咽起来，到结束的时候妇女们全都哭了。注意，该结构多用于三者或三者以上的“依次”，如指两者“依次”，则通常用one after the other。如： The boy showed me his dirty hands one after the other. 那男孩把他脏兮兮的手依序伸给我看。 5. one another 与each other：两者均表示“彼此”“互相”，原认为one another 用于三者或三者以上，each other 用于两者，但在现代英语中，两者常可换用。值得注意的是，“互相”一词，在中文里给人的感觉好像是副词，但其实它们是代词，因此它们在句不用作状语，若用于不及物动词之后时，要考虑添加适合的介词。如： We don’t always agree with one another [each other]. 我们的意见并

Some和any的用法及练习

S o m e和a n y的用法及练 习 Prepared on 24 November 2020

Some 和any的用法及练习 不定代词：不是指确定的对象，而是仅具有非确定特指含义，不明确指代某事或某人的代词。通常用作替代词。 有all /both/every/both/either/neither/one/none/litt le/few/many/much/other/another/some/any/ no以及由some/any/no/every组合的复合不定代词。 some意为“一些”、“几个”,通常用于肯定句中。 现将其主要用法归纳如下: 一、 some作形容词用时,可以修饰复数可数名词,也可以修饰不可数名词。例如: There are some students in the classroom. 教室里有几个学生。 There is some milk in the glass. 杯子里有一些牛奶。 二、 some作代词用时,可在句中作主语或宾语。例如: Some of the boys are playing games now. 有几个男孩现在正在做游戏。 I have no paper. Please give me some. 我没有纸了,请给我一些吧。 三、 some有时也可以用在疑问句或条件句中,表示请求、建议,并期望得到对方的肯定回答。它也可以用于反问句中。例如:

Can you give me some money 你能给我一些钱吗 Would you like some more rice 再来点米饭好吗 Why don’t you buy some flowers for her 你为什么不给她买些花呢If you want (some), I’ll give you some. 如果你想要(一些),我就给你一些。 四、some 与单数可数名词搭配，表示未知的人或物，相当于a(an) 或 a certain。 There must be some job I could do. 肯定有我能做的事情。 Some book on this topic was published last year. 去年出版了有关这主题方面的书。 注意：some + 可数名词复数中的“some”作“一些”解。 五、some用在基数词和few之前，是副词，表示“在概”或“在约”的意思，等于about。 例：There were some 40 or 50 people there.那里大约有40或50人。 any也有“一些”的意思,但它常用在否定句、否定答语或疑问句中。其具体用法如下: 一、 any作形容词用时,可以修饰可数名词的单数或复数,也可以修饰不可数名词。例如: Do you have any questions to ask 你有什么问题要问吗There isn’t any water in this bottle. 这个瓶子里没有水。二、 any作代词用时,也可用于肯定句中,表示“任何”的意思,起强调作用。例如:

some与any的用法区别教案资料

s o m e与a n y的用法 区别

some与any的用法区别 一、一般说来，some用于肯定句，any用于否定句和疑问句。例如： She wants some chalk. She doesn’t want any chalk. Here are some beautiful flowers for you. Here aren’t any beautiful flowers. 二、any可与not以外其他有否定含义的词连用，表达否定概念。例如： He never had any regular schooling. In no case should any such idea be allowed to spread unchecked. The young accountant seldom (rarely, hardly, scarcely) makes any error in his books. I can answer your questions without any hesitation. 三、any可以用于表达疑问概念的条件句中。例如： If you are looking for any stamps, you can find them in my drawer. If there are any good apples in the shop, bring me two pounds of them. If you have any trouble, please let me know. 四、在下列场合，some也可用于疑问句。 1、说话人认为对方的答复将是肯定的。例如： Are you expecting some visitors this afternoon?（说话人认为下午有人要求，所以用some）

代词another和other的用法及练习

代词三 四、other的用法：other, another, others, the other, the others的用法区别 基本用法 other： other+ 复数名词（ other student s） another： . another +单数名词, “另一个”(数目不清楚) the other： The other +复数名词= the others “其他的人或物”（指确定范围内剩下的全部） others (别人)：其后不能带名词，代换上文中提及的可数名词（复数，泛指） the others：代词，其后不能带名词，代换上文中提及的可数名词（复数，定指）； 考点要求注意两个句式、一个搭配和两个区别 1、两个句式的用法 （1）One … the other … 一个……另一个 ~ 注意：使用该句式时，其前应出现说明具体数量的数词two；如出现的数词大于two，one 可以根据实际情况调整成其它数词；如出现的数词减去one或调整后的数词后仍大于“1”时，the other应变为the others或“the other + 数词”（两个数词相加应等于所给数词）。 There are two apples here. One is for you, the other is for your sister. There are five apples here. Two are for you, the others are for your sister. There are five apples here. Two are for you, the other three are for your sister. （2）Some … others …一些……另一些 注意：使用该句式时，其前应出现说明不定数量的词语；如出现说明具体数量的数词，others前应加上the。 There are many people on the beach. Some are swimming, others are enjoying the sun. Mrs. Smith bought 25 books. Some were for her daughter, the others were for her son. 2、一个搭配：any同other连用时应注意之点： Any others：any同单一的other连用，other应使用others； Any other + 单数名词：any后如还带有名词，用other，名词用单数； 【 Any of结构：any of后的other前应加the，如含名词用other，名词用复数；如不含名词，用others。 Of all the cities in China, Shanghai is bigger than any others. Of all the cities in China, Shanghai is bigger than any other city. Of all the cities in China, Shanghai is bigger than any of the others. Of all the cities in China, Shanghai is bigger than any of the other cities. 3、两个区别：（1）同数词连用时another和more的区别 another用于数词前，more用于数词后。（鞍前马后）

some和any用法和区别-小升初英语语法汇总

some和any用法和区别，小升初英语语法汇总 some 表示一些 any表示任何一个 somesome用在肯定句中，也可以用在表示婉转意思的疑问句中，表询问或建议；any 用于否定句和一般疑问句。 请注意看下列例句后扩号中说明的用法。 There is some water in the glass. （肯定句） There are some flowers in the garden. （肯定句） There aren’t any lamps in the study.（否定句） Are there any maps on the wall?（一般疑问句） Would you like some orange juice? （希望得到肯定回答，表建议） Would you like some lessons in Wangjing Dynamic English World with me this week ? （表询问，希望得到肯定回答） 这周你想和我一起上望京能动英语的几堂课吗？ Do you want to take any photos at the party? （一般疑问句） 练一练：选用some或any填空。 1) There isn’t ______milk in the fridge. 2) I can see______cars, but I can’t see______buses. 3) He has ______ friends in England. 4) Were there ______fruit trees on the farm? 5) Here are ______presents for you. 6) Does Tom want to take ______ photos? 7) Is there______rice in the kitchen? 8) There are______new buildings in our school.

some和any的用法与练习题

some和 any 的用法及练习题( 一) 一、用法： some意思为：一些。可用来修饰可数名词和不可数名词，常常用于肯定句 . any 意思为：任何一些。它可以修饰可数名词和不可数名词，当修饰可数名词 时要用复数形式。常用于否定句和疑问句。 注意： 1、在表示请求和邀请时，some也可以用在疑问句中。 2、表示“任何”或“任何一个”时，也可以用在肯定句中。 3、和后没有名词时，用作代词，也可用作副词。 二、练习题： 1.There are ()newspapers on the table. 2.Is there ( )bread on the plate. 3.Are there () boats on the river? 4.---Do you have () brothers ?---Yes ,I have two brothers. 5.---Is there () tea in the cup? --- Yes,there is () tea in it ,but there isn’t milk. 6.I want to ask you() questions. 7.My little boy wants ()water to drink. 8.There are () tables in the room ,but there aren’t ( )chairs. 9.Would you like () milk? 10.Will you give me () paper? 复合不定代词的用法及练习 一.定义: 由 some，any，no，every 加上 -body,-one,-thing,-where构成的不定代词，叫做复合不定代词 . 二. 分类： 1.指人：含 -body 或 -one 的复合不定代词指人 . 2.含-thing 的复合不定代词指物。 3.含-where 的复合不定代词指地点。 三：复合不定代词： somebody =someone某人 something 某物，某事，某东西 somewhere在某处，到某处 anybody= anyone 任何人，无论谁 anything任何事物，无论何事，任何东西 anywhere 在任何地方 nobody=no one 无一人 nothing 无一物，没有任何东西 everybody =everyone每人，大家，人人 everything每一个事物，一切 everywhere 到处 , 处处 , 每一处

others的用法小结

others的用法小结 你们知道others的用法吗?我们一起来学习学习吧，下面就和大家分享，来欣赏一下吧。 others的用法 1、others相当于other+复数名词,如： 注意：所指的其他人并非其他所有的人,指其他部分的全部,请看“2、” ome people like walking.Some like running.Others like swimming. 有些人喜欢散步,有些人喜欢跑步,其他人喜欢游泳. He is always ready to help others. 他总是乐于助人. 2、the others 其余的，相当于the other的复数形式,特指某一范围内的其他全部,如： Two boys will go to the zoo,and the others will stay at home.

两个男孩将去动物园,其余的留在家里. Thirty of them are boy students and the others are girls. 他们当中三十人是男生,其他人是女生. other如果是指代词“另一个”时,必须要加“the” 而“one...the other...”用于单数,指“一个...另一个...” 如：I have two pensils.One is red,the other is blue. others 和some对比使用时,是“有些”的意思而不是做“其他”讲,如:Some cleaned the windows,others mopped the floor.有的擦窗户,有的擦地板. the others 是“其余的”意思,表示在一个范围内的其他全部,如:This dictionary is better than the others.这本字典比别[其余]的好. the other 是其中的“另一个”,如:Give me the other one; not this one.给我那一个,不是这一个.

some与any的用法区别

some与any的用法区别 一、一般说来，some用于肯定句，any用于否定句和疑问句。例如： She wants some chalk. She doesn’t want any chalk. Here are some beautiful flowers for you. Here aren’t any beautiful flowers. 二、any可与not以外其他有否定含义的词连用，表达否定概念。例如： He never had any regular schooling. In no case should any such idea be allowed to spread unchecked. The young accountant seldom (rarely, hardly, scarcely) makes any error in his books. I can answer your questions without any hesitation. 三、any可以用于表达疑问概念的条件句中。例如： If you are looking for any stamps, you can find them in my drawer. If there are any good apples in the shop, bring me two pounds of them. If you have any trouble, please let me know. 四、在下列场合，some也可用于疑问句。 1、说话人认为对方的答复将是肯定的。例如： Are you expecting some visitors this afternoon?（说话人认为下午有人要求，所以用some）Are you expecting any visitors this afternoon?(说话人不知道下午是否有人来，所以用any) Didn’t you give him some tickets?(说话人认为票已经给他了。问题的回答是：Sure I did) Did you give him any tickets?（说话人不知道是否给票了。回答可能是yes或no。） 2、款待用语或问句的实质等于提出一个要求并希望得到肯定的回答时。例如：

other the other和others的用法和区别

another与other的区别。 ①Some students like English and other students (others) like physics. 有些学生喜欢英语，有些学生喜欢物理。 【解析】other表示“别的”，“另外的”，只能与复数名词连用。但other前有冠词the 即可与单数名词连用。如： I have two pens. One is blue. The other (pen) is black. 我有两枝钢笔。一支是蓝色的，另一支是黑色的。 ②John did better than all the other players in the sport. 在那项运动中John比所有别的运动员都出色。 【解析】the other加复数名词指的是一定范围内“所有其余的人或事物”，是特指；而other 加复数名词却是没有明确范围的“另外的人或事物”。如： We must think more of other comrades. 我们必须多想想别的同志。 ③ This shirt is too large for me. Please show me another. 这件衬衫我穿太大。请另外拿一件我看看。（一般商店不会只有两件衬衫） 【解析】another, the other作代词的用法。The other表示“两个数量中的另一个”，表示特指，总数为俩；another表示“总数为三个以上中任意的另一个”，表示泛指。如： His parents both work in a hospital. One is a doctor and the other is a nurse. 他父母都在一家医院工作。一个是医生，一个是护士。（父母为两个人）

不定代词 other , another 的用法

不定代词other , another 的用法 1. Some people hope to be more successful while A simple want to feel more comfortable. A. the others B. others C. the other D. another 2. I have three pens. One is red, C two are black. A. another B. other C. the other D. the others 3.Mr Smith has two sons, D is a soldier, ______ is a doctor. A. One, another B. One, other C. This, the other D. One, the other 4. Children should be taught how to get along with C . A. another B. other C. others D. any other 5. There are many people in the park now. Some are boating, __D___ are walking along the lake. A. The other B. Other C. The others D.Others 6. The glass is broken. Go and get __B_____ . A. other B. another one C. others D. the other 三单元复习与提高 1.Liu Xiang ia an Olympic winner in the __B_____ hurdles (跨栏) . We are proud of him. A. 110-meters B. 110-meter C. 110 meter 2. –Dad, when will you be free? You agreed to go to the seaside with me four days ago. --I am sorry, Jean. But I think I will have a __B____ holiday soon. A. four day B. four—day C. four days D. four day 3. Enough sleep is good for health. If you _A___ for your favorite TV programs, you will feel sleepy. A. stay up B. set up(建立) C. pick up 4. She needs __A____ a rest, for she is working for over four hours. A. to have B. having C. have D. had 5.He is too young. He is not __A____ to join the army. A. old enough B. enough old C. too old D. old too 6. I must be back home _A_____ 8;00 pm/ A. by B. from C. for D. between 7.We were allowed __B___ here for a little longer time. A. staying B. to stay C. stayed D. to staying 8. The young man was often seen _A_____ by the lake. A. to draw B. to drawing C. draw D. drew ( 题中的主语与动词see为被动关系，而see用于被动语态时，结构为be seen to do …) 9．Our English teacher always tells us a verb must _B____ its subject in number and person. A. agree B. agree with C. agree to D. agree on Agree with ―同意某人说的话或建议‖ Agree to ―同意计划，建议，要求，条件等‖ Agree on ―就。。。达成一致‖ 10. Many students will be _B_____ if the class is boring. A. sleep B. sleepy C. sleeping D. slept 11. All the boys succeeded __C_____ the English examination. A. pass B. to pass C. passing D. on passing

关于some和any的用法和区别

关于some和any的用法和区别 some和any都是常见词汇，他们有共同点也有很多不一样的地方，你们知道他们饿区别在哪里吗?接下来小编在这里给大家带来some和any的用法和区别，我们一起来看看吧! some和any的用法和区别 一、some和any作为形容词或代词，可以用来说明或代替复数名词或不可数名词，表示不定量，意为一些，其区别是：对其所说明或代替的名词持肯定态度时，用some;持非肯定(否定或疑问)态度时，用any。 在以下句子中使用some： 1.肯定句(包括肯定的陈述句和祈使句以及反意疑问句中肯定的陈述部分)。如： There are some new books on the teachers desk. We have a lot of sugar. Take some with you, please. He bought some bread, didnt he? 2.持肯定态度的一般疑问句。如： Are there some stamps in that drawer? Didnt she give you some money? 3.表示请求或建议的一般疑问句，通常都希望得到对方肯定的答复，所以也用some。如： May I ask you some questions?

Would you like some tea? 4.特殊疑问句及选择疑问句。因为特殊疑问句和选择疑问句并不对some所说明或代替的名词表示疑问。如： Where can I get some buttons? Do you have some pens or pencils? 在以下句子中使用ANY： 1.否定句(包括否定的陈述句和祈使句以及反意疑问句中否定的陈述部分)。如： I cant give you any help now. Do not make any noise. There werent any trees here, were there? 2.含有除not以外的其他否定词或否定结构的句子。如： Jim hardly makes any mistakes in his homework. He went to London without any money in his pocket. She was too poor to buy any new clothes. 3.一般疑问句(持肯定态度的除外)。如： Did she buy any tomatoes yesterday? I want some paper. Do you have any? 4.条件状语从句。如： If you are looking for any ink, you can find it on my desk. If there are any good apples there, get me two kilos, please.

some,any,one ones those that的区别和用法

some和any的区别和用法 要表示"一些"的意思，可用some, any。 some 是肯定词，常用于肯定句；any是非肯定词，常用于否定句或疑问句。例如： There are some letters for me. There aren''t any letters for me. Are there any letters for me? I seldom get any sleep these days. any也常用于条件分句以及带有否定含义的句子中： If you have any trouble, please let me know. 如果你有任何麻烦，请让我知道。 I forgot to ask for any change. 我忘了要一些零钱。 当说话人期待肯定回答时，some也可用于疑问句, 比如当说话人期待来信时，他可以问道：Are there some letters for me? 当购物时向售货员提问或者主人向客人表示款待时，也可在疑问句中用some: Could I have some of these apples? some和any 既可以修饰可数名词又可以修饰不可数名词，some常用在肯定句中，而any则常用在否定和疑问句中。因此 some和any 的用法主要是考虑用在肯定句、疑问句还是否定句中，与名词的可数与否无关。 some意为“一些”，可作形容词和代词。它常修饰可数名词复数。如：some books一些书，some boys一些男孩，也可修饰不可数名词，如：some water一些水，some tea一些茶叶，some常用在肯定句中。any意为“任何一些”，它也可修饰可数名词复数或不可数名词，常用于疑问句和否定句。如： --I have some tea here. 我这儿有些茶叶。 --I can’t see any tea. 我没看见茶叶。 --Do you have any friends at school? 你在学校有些朋友吗? --I have some English books, they are my best friends. 我有英语书，它们是我最好的朋友。 但在表示建议，反问，请求的疑问句中，或期望得到肯定回答时，多用some而不用any。如：Would you like some coffee? 你要不要来点咖啡? What about some fruit juice? 来点水果汁如何? 当any表示“任何”的意义，起强调作用时，它可以用在肯定句中； Any student can answer this question.任何学生都可以回答这个问题。 选题角度： 辨析some和any的不同用法：some 常用在肯定句中，而any 则常用在否定和疑问句中。在表示建议，反问，请求的疑问句中，或期望得到肯定回答时，多用some而不用any。一般用于疑问句或否定句中，用于never, hardly, without等词之后，用于if / whether 之后。而some则用于肯定句中，用于建议或请求的疑问句中，用于预料会作肯定回答的疑问句中，用于表示反问的否定的疑句中。 如： 1. I’d been expecting ________ letters the whole morning, but there weren’t ________ for me. (全国卷)卷 A. some; any B. many; a few C. some; one D. a few; none