The modular height of an abelian variety and its finiteness property

吉林双辽市2017高考英语阅读理解一轮训练（二）【2016高考训练】阅读下列短文, 从给的四个选项 (A、B、C和D) 中, 选出最佳选项。

London's newest skyscraper (摩天大楼) is called the Shard and it cost about 430 million pounds to build.At a height of almost 310 metres，it is the tallest building in Europe.The Shard has completely changed the appearance of London.However，not everyone thinks that it is a change for the better.The Shard was designed by the famous Italian architect Renzo Piano.When he began designing the Shard for London，Piano wanted a very tall building that looked like a spire(尖顶)．He wanted the glass surfaces to reflect the sky and the city.The sides of the building aren't regular.So the building has an unusual shape.It looks like a very thin，sharp piece of broken glass.And that is how the building got the name：the Shard.Piano says that the spire shape of the Shard is part of a great London tradition.The shape reminds him of the spires of the churches of London or the tall masts(桅杆) of the ships that were once on the river Thames.The Shard has 87 floors.At the top，there is an observatory.At the moment the building is empty，but eventually there will be a five­star hotel.There will also be top quality restaurants，apartments and offices.Before building work began，a lot of people didn't want the Shard though the plans were approved.Now they are still unhappy about the Shard.Some critics say that such a tall skyscraper might be good in a city like New York，but not in London.They say that the best thing about the Shard is its spire shape.But that is the only thing.There is no decoration，only flat surfaces.The Egyptians did that 4,500 years ago.They also think the Shard is too big for London.It destroys the beauty of the city.Other critics don't like what the Shard seems to represent.They say that the Shard shows how London is becoming more unequal.Only very rich people can afford to buy the expensive private apartments and stay in the hotel.But the people who live near the Shard are among the poorest in London.So the Shard seems a symbol of the division in society between the very rich and the poor.The Shard now dominates the London skyline.It is not certain，however，that ordinary London citizens will ever accept it as a valuable addition to the city.1．London's newest skyscraper is called the Shard because of ________.A．its cost B．its sizeC．its shape D．its height2．When he designed the Shard，Piano wanted it to ________.A．change London's skylineB．inherit London's traditionC．imitate the Egyptian styleD．attract potential visitors3．The critics who refer to social division think the Shard ________.A．is only preferred by the richB．is intended for wealthy peopleC．is far away from the poor areaD．is popular only with Londoners4．Which would be the best title for the passage?A．The Shard：Cheers and ClapsB．The Shard：Work of a Great ArchitectC．The Shard：New Symbol of London?D．The Shard：A Change for the Better?答案：【语篇解读】本文是一篇说明文。

a rX iv:mat h /1111v1[mat h.DS]2Ja n2REDUNDANT PICARD–FUCHS SYSTEM FOR ABELIAN INTEGRALS D.NOVIKOV ∗,S.YAKOVENKO ∗∗Abstract.We derive an explicit system of Picard–Fuchs differential equa-tions satisfied by Abelian integrals of monomial forms and majorize its coef-ficients.A peculiar feature of this construction is that the system admitting such explicit majorants,appears only in dimension approximately two times greater than the standard Picard–Fuchs system.The result is used to obtain a partial solution to the tangential Hilbert 16th problem.We establish upper bounds for the number of zeros of arbitrary Abelian integrals on a positive distance from the critical locus.Under the additional assumption that the critical values of the Hamiltonian are distant from each other (after a proper normalization),we were able to majorize the number of all (real and complex)zeros.In the second part of the paper an equivariant formulation of the above problem is discussed and relationships between spread of critical values and non-homogeneity of uni-and bivariate complex polynomials are studied.Contents 1.Tangential Hilbert Sixteenth Problem,complete Abelian integrals and Picard–Fuchs equations 21.1.Abelian integrals and tangential Hilbert 16th problem 21.2.Abelian integrals and differential equations 31.3.Meandering of integral trajectories 31.4.Picard–Fuchs equations and systems of equations 41.5.Regularity at infinity and Gavrilov theorems 51.6.Redundant Picard–Fuchs system:the first main result 51.7.Corollaries:theorems on zeros 61.8.Equivariant formulation 61.9.Geometry of critical values of polynomials 7Acknowledgements72.Picard–Fuchs system in the hyperelliptic case72.1.Gelfand–Leray residue72.2.Division by polynomial ideals and 1-forms82.3.Derivation of the Picard–Fuchs system in the hyperelliptic case92.4.Bounds for the matrix norms102.5.Digression:doubly hyperelliptic Hamiltonians112 D.NOVIKOV,S.YAKOVENKO2.6.Discussion123.Derivation of the redundant Picard–Fuchs system123.1.Notations and conventions123.2.Normalizing conditions and quasimonic Hamiltonians133.3.Balanced Hamiltonians133.4.Lemma on bounded division143.5.Derivation of the redundant Picard–Fuchs system153.6.Abelian integrals of higher degrees173.7.Properties of the redundant Picard–Fuchs system174.Zeros of Abelian integrals away from the singular locus and relatedproblems on critical values of polynomials194.1.Heuristic considerations194.2.Meandering theorem and upper bounds for zeros of Abelian integrals204.3.Affine group action and equivariant problem on zeros of Abelianintegrals224.4.From Theorem3to Equivariant problem225.Critical values of polynomials235.1.Geometric consequences of quasimonicity235.2.Almost-homogeneity implies close critical values245.3.Dual formulation,limit and existential problems245.4.Parallel problems for univariate polynomials265.5.Spread of roots vs.spread of critical values for univariate moniccomplex polynomials265.6.Demonstration of Theorem5285.7.Existential bounds cannot be uniform295.8.Discussion:atypical values and singular perturbations29References301.Tangential Hilbert Sixteenth Problem,complete Abelianintegrals and Picard–Fuchs equations The main result of this paper is an explicit derivation of the Picard–Fuchs system of linear ordinary differential equations for integrals of polynomial1-forms over level curves of a polynomial in two variables,regular at infinity.The explicit character of the construction makes it possible to derive upper bounds for the coefficients of this system.In turn,application of the bounded meandering principle[17,15]to the system of differential equations with bounded coefficients allows to produce upper bounds for the number of complex isolated zeros of these integrals on a positive distance from the ramification locus.1.1.Abelian integrals and tangential Hilbert16th problem.If H(x,y)isa polynomial in two real variables,called the Hamiltonian,andω=P(x,y)dx+ Q(x,y)dy a real polynomial1-form,then the problem on limit cycles appearing in the perturbation of the Hamiltonian equation,dH+εω=0,ε∈(R,0)(1.1)REDUNDANT PICARD–FUCHS SYSTEM FOR ABELIAN INTEGRALS3 after linearization inε(whence the adjective“tangential”)reduces to the study of complete Abelian integralI(t)=I(t;H,ω)= H=tω,(1.2) where the integration is carried over a continuous family of(real)ovals lying on the level curves{H=t}.Problem1(Tangential Hilbert16th problem).Place an upper bound for the num-ber of real zeros of the Abelian integral I(t;H,ω)on the maximal natural domain of definition of this integral,in terms of deg H and degω=max(deg P,deg Q)+1.A more natural version appears after complexification.For an arbitrary complex polynomial H(x,y)having only isolated critical points,and an arbitrary complex polynomial1-formω,the integral(1.2)can be extended as a multivalued analytic function ramified over afinite set of points(typically consisting of critical values of H).The problem is to place an upper bound for the number of isolated complex roots of any branch of this function,in terms of deg H and degω.1.2.Abelian integrals and differential equations.Despite its apparently alge-braic character,the tangential Hilbert problem still resists all attempts to approach it using methods of algebraic geometry.Almost all progress towards its solution so far was based on using methods of analytic theory of differential equations.In particular,the(existential)generalfiniteness theorem by Khovanski˘ı–Var-chenko[12,24]claims that for anyfinite combination of d=degωand n=deg H the number of isolated zeros is indeed uniformly bounded over all forms and all Hamiltonians of the respective degree.One of the key ingredients of the proof is the so called Pfaffian elimination,an analog of the intersection theory for varieties defined by Pfaffian differential equations[13].Another important achievement,an explicit upper bound for the number of zeros in the elliptic case when H(x,y)=y2+p(x),deg p=3and forms of arbitrary degree,due to G.Petrov[21],uses the fact that the elliptic integrals I k(t)= x k−1y dx,k=1,2,in this case satisfy an explicit system of linearfirst order system of differential equations with rational coefficients.This method was later generalized for other classes of Hamiltonians whose level curves are elliptic (i.e.,of genus1),see[10,7,27]and references therein.The ultimate achievement in this direction is a theorem by Petrov and Khovanskii,placing an asymptotically linear in degωupper bound for the number of zeros of arbitrary Abelian integrals, with the constants being uniform over all Hamiltonians of degree n(unpub-lished).However,one of these constants is purely existential:its dependence on n is totally unknown.It is important to remark that all the approaches mentioned above,require a very basic and easily obtainable information concerning the differential equations(their mere existence,types of singularities,polynomial or rational form of coefficients,in some cases their degree).1.3.Meandering of integral trajectories.A different approach suggested in [14]consists in an attempt to apply a very general principle,according to which integral trajectories of a polynomial vectorfield(in R n or C n)have a controllable meandering(sinuosity),[17,15].More precisely,if a curve of known size is a part of an integral trajectory of a polynomial vectorfield whose degree and the magnitude4 D.NOVIKOV,S.YAKOVENKOof the coefficients are explicitly bounded from above,then the number of isolated intersections between this curve and any affine hyperplane in the ambient space can be explicitly majorized in terms of these data.The bound appears to be very excessive:it is polynomial in the size of the curve and the magnitude of the coefficients,but the exponent as the function of the degree and the dimension of the ambient space,grows as a tower(iterated exponent)of height4.In order to apply this principle to the tangential Hilbert problem,we consider the curve parameterized by the monomial integrals,t→(I1(t),...,I N(t)),I i(t)= H=tωi,whereωi,i=1,...,N are all monomial forms of degree d.Isolated zeros of the Abelian integral of an arbitrary polynomial1-formω= i c iωi correspond to isolated intersections of the above curve with the hyperplane c i I i=0.If this monomial curve is integral for a system of polynomial differential equations with explicitly bounded coefficients,then the bounded meandering principle would yield a(partial)answer for the tangential Hilbert16th problem.The system of polynomial(in fact,linear)differential equations can be written explicitly for the case of hyperelliptic integrals corresponding to the Hamiltonian H(x,y)=y2+p(x)with an arbitrary univariate potential p(x)∈C[x],see§2below and references therein.Application of the bounded meandering principle allowed us to prove in[14]that the number of zeros of hyperelliptic integrals is majorized by a certain tower function depending only on the degrees of n=deg H=deg p and d=degω.(Actually,it was done under an additional assumption that all critical values of p are real,but we believe that this restriction is technical and can be removed).1.4.Picard–Fuchs equations and systems of equations.In order to general-ize the construction from[14]for the case of arbitrary(not necessarily hyperelliptic) Hamiltonians it is necessary,among other things,to write a system of polynomial differential equations for Abelian integrals and estimate explicitly the magnitude of its coefficients.The mere existence of such a system is well known since times of Riemann if not Gauss.In today’s language,the monodromy group of any form depends only on the Hamiltonian.Denote byµthe rank of thefirst homology group of a typical affine level curve{H=t}⊂C2.Then for any collection of1-formsω1,...,ωµthe period matrix X(t)can be formed,whose entries are integrals ofωi over the cycles δ1(t),...,δµ(t)generating the homology.If the determinant of this matrix if not identically zero,then X(t)satisfies a linear ordinary differential equation of the form˙X(t)=A(t)X(t),A(·)∈Mat(C(t)),(1.3)µ×µwith a rational matrix function A(t).This system of equations is known under several names,from Gauss–Manin connection[20,especially p.18]to Picard–Fuchs system(of linear ordinary differential equations with rational coefficients,in full). We shall systematically use the last name.The rank of thefirst homology can be easily computed:for a generic Hamiltonian of degree n+1it is equal to n2.The degree deg A(t)can be relatively easily determined if the degrees of the formsωi are known.However,the choice of the formsωi may also be a difficult problem for some Hamiltonians.The matrix A(t)REDUNDANT PICARD–FUCHS SYSTEM FOR ABELIAN INTEGRALS5 apriori may have poles not only in the ramification points of the Abelian integrals, which leads to additional difficulties.But worst of all,this topological approach gives absolutely no control over the magnitude of the(matrix)coefficients of the rational(matrix)function A(t).1.5.Regularity at infinity and Gavrilov theorems.Part of these problems problems can be resolved.In particular,if the Hamiltonian is sufficiently regular at infinity,then all questions concerning the degrees,can be answered.Definition1.A polynomial H(x,y)∈C[x,y]of degree n+1is said to be regular at infinity,if one of the three equivalent conditions holds:1.its principal homogeneous part H,a homogeneous polynomial of degree n+1,is a product of n+1pairwise different linear forms;2. H has an isolated critical point(necessarily of multiplicityµ=n2)at theorigin(x,y)=(0,0);3.the level curve{ H=1}⊂C2is nonsingular.This condition means that after the natural projective compactification of the (x,y)-plane C2,all“interesting”things still happen only in thefinite part of the compactified plane.In particular,for a polynomial regular at infinity:1.all level curves{H=t}intersect the infinite line C P1∞⊂C P2transversally,2.all critical points{(x,y):dH(x,y)=0}are isolated and their number isexactlyµ=n2if counted with multiplicities,3.the rank of thefirst homology of any regular affine level curve{H=t}isµ=n2,4.the map H:C2→C1is a topological bundle over the set of the regular valuesof H,hence the Abelian integrals can be ramified only over the critical values of H.In[5]L.Gavrilov proved that for polynomials regular at infinity,the space of Abelian integrals isfinitely generated as a C[t]-module byµbasic integrals that can be chosen as integrals of anyνformsωi of degree 2n whose differentials form the basis of the quotient spaceΛ2/d H∧Λ1,whereΛk is the space of polynomial k-forms on C2.As a corollary,one can prove that the collection of these basic integrals satisfies a system of equations(1.3)of sizeµ×µwithµ=n2,and place an upper bound for the degree of the corresponding matrix function A(t).This system is minimal(irredundant):generically(for Morse Hamiltonians regular at infinity), all branches of full analytic continuation of an Abelian integral span exactlyµ-dimensional linear space.From this theorem one can also derive further information concerning the Picard–Fuchs ly,one can prove that if in addition to being regular at infinity,H is a Morse function on C2,then the matrix A(t)of the Picard–Fuchs system(1.3)has only simple poles(Fuchsian singularities)at the critical values of the Hamiltonian and only at them(the point t=∞is a regular though in general non-Fuchsian singularity).However,these results do not yet allow an explicit majoration of the coefficients(e.g.,the residue matrices)of the matrix function A(t)in(1.3).1.6.Redundant Picard–Fuchs system:thefirst main result.We suggest in this paper a procedure of explicit derivation of the Picard–Fuchs system of equations,based on the division by the gradient ideal H x,H y ⊂C[x,y]in the6 D.NOVIKOV,S.YAKOVENKOpolynomial ring.It turns out that if instead of choosingµ=n2forms of degree 2n constituting a basis modulo the gradient ideal,one takes allν=n(2n−1) cohomologically independent monomial forms of degree 2n,then the resulting Picard–Fuchs system can be written in the form(tE−A)˙X(t)=BX(t),A,B∈Matν×ν(C),(1.4) where E is the identity matrix,and X(t)is the rectangular periodν×µ-matrix.The procedure of deriving the system(1.4),being completely elementary,can be easily analyzed and upper bounds for the matrix norms A and B derived. These bounds depend on the magnitude of the all non-principal terms H− H of the Hamiltonian,relative to the principal part H.More precisely,we introduce a normalizing condition(quasimonicity)on the ho-mogeneous part:this condition plays the same role as the assumption that the leading term has coefficient1for univariate polynomials.The quasimonicity con-dition can be always achieved by an affine change of variables,provided that H is regular at infinity,hence it is not restrictive.Theorem2,ourfirst main result, allows to place an upper bound for the norms A + B in terms of the norm(sum of absolute values of all coefficients)of the non-leading part H− H,assuming thatH is quasimonic.1.7.Corollaries:theorems on zeros.The above information on coefficients of Picard–Fuchs system already suffices to apply the bounded meandering principle and obtain an explicit upper bound for the number of zeros of complete Abelian integrals away from the critical locus of the Hamiltonian(Theorem3),which seems to be thefirst known explicit result of that kind.In addition to this bound valid for some zeros and almost all Hamiltonians,one can apply results(or rather methods)from[22].If in addition to the quasimonicity and bounded lower terms,all critical values t1,...,tµof the Hamiltonian H are far away from each other(i.e.,a lower bound for|t i−t j|is known for i=j),then one can majorize the number of zeros on any branch of the Abelian integral by a function depending only on n,d and the minimal distance between critical values. The accurate formulation is given in Theorem4.1.8.Equivariant formulation.However,the description given by Theorem2, is not completely sufficient for further advance towards solution of the tangential Hilbert problem by studying zeros of Abelian integrals near the critical locus when the latter(or some part of it)shrinks to one point of high multiplicity.One reason is that in order to run an inductive scheme similar to that con-structed in[14],one has to make sure that the Hamiltonian H:C2→C1can be rescaled(using affine transformations in the preimage C2and the image C1)so that simultaneously:1.the critical values of H do not tend to each other(e.g.,their diameter isbounded from below by1),and2.the“non-homogeneous part”H− H is bounded by a constant explicitly de-pending on n(each of the two conditions can be obviously satisfied separately).Another,intrinsic reason is the equivariance(or,rather precisely,non-invariance of neither Theorem2nor Theorems3and4)by the above affine group action.InREDUNDANT PICARD–FUCHS SYSTEM FOR ABELIAN INTEGRALS7 order to be geometrically sound,all assertions should be related to a certain privi-leged affine chart on the t-plane.Since our future goal is to study a neighborhoodof the critical locus,it is natural to choose the privileged chart so that the critical locus will not shrink into one point.More detailed explanations and motivations are given in§4below,where we for-mulate several problems all in the following sense:for a polynomial whose principal homogeneous part is normalized(in a certain sense)and whose critical values are ex-plicitly bounded,it is required to place an upper bound for the“non-homogeneous”part,eventually after a suitable translation(which does not affect the principal part, naturally).1.9.Geometry of critical values of polynomials.The reason why several problems of the above type were formulated instead of just one,is very simple:we do not know a complete solution,so partial,existential or limit cases were con-sidered as intermediate steps towards the ultimate goal.In§5we prove that:•if a monic complex polynomial p(x)=x n+1+···∈C[x]has all critical values in the unit disk,then its roots form a point set of diameter<11(Theorem6)and hence by a suitable translation the norm of the non-principal part can be made 12n+1(this gives a complete solution in the univariate and hyperelliptic cases);•all critical values of a Hamiltonian regular at infinity,cannot simultaneously coincide unless the Hamiltonian is essentially homogeneous(Theorem5);•for any normalized principal part H there exists an upper bound for H− H (eventually after a suitable translation),provided that the critical values ofH are all in the unit disk(Corollary to Theorem5).All these are positive results towards solution of the problem on critical values.It still remains to compute the upper bound from the last assertion explicitly:the proof below does not provide sufficient information for that.However,it can be shown already in simple examples that this bound cannot be uniform over all homogeneous parts.As some of the linear factors from H approach too closely to each other,an explosion occurs and the non-principal part may be arbitrarily large without affecting the“moderate”critical values.The phenomenon can be seen as“almost occurrence”of atypical values,ramification points for Abelian integrals that are not critical values of H:such points are known to appear if the principal part H has a non-isolated singularity. Acknowledgements.We are grateful to J.-P.Fran¸c oise,L.Gavrilov,Yu.Ilya-shenko,A.Khovanskii,man,R.Moussu,R.Roussarie and Y.Yomdin for numerous discussions and many useful remarks.Bernard Teissier suggested an idea thatfinally developed into the proof of Lemma3below.Lucy Moser provided us with a counterexample(§5.6below).We are grateful to all our colleagues from Laboratoire de Topologie,Universit´e de Bourgogne(Dijon)and Departamento de Algebra,Geometr´ıa y Topolog´ıa,Uni-versidad de Valladolid,where a large part of this work has been done.They made our stays and visits very stimulating.2.Picard–Fuchs system in the hyperelliptic case 2.1.Gelfand–Leray residue.The derivative of an Abelian integral H=tωcan be computed as the integral over the same curve of another1-formθcalled the8 D.NOVIKOV,S.YAKOVENKOGelfand–Leray derivative(residue).More precisely,if a pair of polynomial1-forms ω,θsatisfies the identity dω=dH∧θ,then for any continuous family of cyclesδ(t) on the level curves{H=t}dREDUNDANT PICARD–FUCHS SYSTEM FOR ABELIAN INTEGRALS9 2.3.Derivation of the Picard–Fuchs system in the hyperelliptic case. Throughout this section we assume that H(x,y)=12y2+p(x) x i−1dx∧dy= 12x i−1y dx−b i(x)dy∧dH+a i(x)dx∧dy= 12I i+nj=1b ij I j(2.6)or,in the matrix form,(t−A)˙I=BI,I∈C n,A,B∈Mat n×n(C),(2.7) where,obviously,I j(t)= ωj are the Abelian integrals and I=(I1,...,I n)the column vector.10 D.NOVIKOV,S.YAKOVENKORemark1.The computation above does not depend on the choice of the cycle of integration,therefore the system of equations will remain valid if we replace the column vector I by the period matrix X(t)obtained by integrating all formsωi over all vanishing cyclesδj(t),j=1,...,n(see[1])on the hyperelliptic level curves.The matrices A,B can be completely described using the division process(2.3).Let x∗∈C be a critical point of p and t∗=p(x∗)the corresponding critical value.Then(2.3)imply that the column vector(1,x∗,x2∗,...,x n−1∗)∈C n is the eigenvector of A with theeigenvalue t∗,which gives a complete description(eigenbasis and eigenvalues)of A.Entries of the matrix B can be described similarly:b ij=0for j>i because of the assertion about degrees of b j(x),so B is triangular.The diagonal entries can be easily computed by looking at the leading terms:since p is monic,b i(x)=x in+1x i−1+···.Finally,the diagonal entries{−i2}n i=1form the spectrumof B.However,knowledge of the critical values of H is not yet sufficient to produce an upper bound for the norms A , B ,since the conjugacy by the Vandermonde matrix(whosecolumns are the above eigenvectors(1,x j,x2j,...,x n−1j )T,j=1,...,n)may increase ar-bitrarily the norm of the diagonal matrix diag(t1,...,t n),where t j are all critical values of H(or p,what is the same).On the contrary,a linear change in the space of1-forms that makes A diagonal,can increase in an uncontrollable way the norm of the matrix B, whose eigenbasis differs from the standard one by a triangular transformation.It is the explicit division procedure that allows to majorize the matrix norms.2.4.Bounds for the matrix norms.For a polynomial p∈C[x]let p be the sum of absolute values of its coefficients(sometimes it is called the length of p).It has the advantage of being multiplicative, pq p · q .Proposition1.If q=x n+···∈C[x]is a monic polynomial with q−x n =c, then any other polynomial f∈C[x]of degree d n can be divided with remainder,f(x)=b(x)q(x)+a(x),deg a n−1,(2.8)so thatb + a K f ,K=1+C+C2+···+C d−n,C=1+c= q .(2.9) Proof.The proof goes by direct inspection of the Euclid algorithm of univariate polynomial division.The assertion of the Proposition is trivial for q=x n:in this case the string of coefficients of r has to be split into two,and immediately we have the decomposition r=bx n+a with b + a = r .The general nonhomogeneous case is treated by induction.Suppose that the inequality(2.9)is valid for any polynomial f of degree d−1(for d=n−1 it is trivially satisfied by letting b=0and a= f).Take a polynomial f of degree d and write the identity f=bx n+a=bq+b(x n−q)+a=bq+ f, where the polynomial f=a+b(x n−q)is of degree d−1and has the norm explicitly bounded: f c b + a (1+c)( b + a ) C f .By the induction assumption, f can be divided, f= bq+ a,with the norms satisfying the inequality(2.9).Collecting everything together,we have f=(b+ b)q+ a and b+ b + a f +C f (1+C+···+C d−1−n) f (1+···+C d−n).As a corollary to this Proposition and the explicit procedure of the division, we obtain upper bounds for norms of the matrices A,B.Recall that we use theREDUNDANT PICARD–FUCHS SYSTEM FOR ABELIAN INTEGRALS11ℓ1-norm on the“space of columns”,so the norm of a matrix A=(a ij)ni,j=1isA =maxj=1,...,nni=1|a ij|(2.10)Theorem1.Suppose that p(x)=x n+1+ n−1i=0c i x i is a monic polynomial of degree n+1and the non-principal part of p is explicitly bounded: n−1i=0|c i| c.Then the entries of the matrices A,B determining the Picard–Fuchs system(2.7) are explicitly bounded:A +B n2(1+C+···+C n+1),C=1+c= p .(2.11) Proof.The derivative p′(x)is not monic,but the leading coefficient is explicitly known:p′(x)=(n+1)(x n+···) ,with the non-principal part denoted by the dots bounded by c in the sense of the norm.Applying Proposition1to q=p′/(n+1), we see that any polynomial can be divided by p′and the same inequalities(2.9) would hold(since n+1 1).Thus we have b i + a i K x i−1 p =KC,where K=1+C+···+C n,then obviously b′j n b j andfinally for the sum of matrix elements A,B occurring in the i th line,we produce an upper bound12 n(1+C+···+C n+1).Clearly,this means that every entry of thesematrices is majorized by the same expression and therefore for the matrixℓ1-norms on C n we have the required estimate.2.5.Digression:doubly hyperelliptic Hamiltonians.The algorithm suggest-ed above,works with only minor modifications for doubly hyperelliptic Hamiltonians having the form H(x,y)=p(x)+q(y)(the hyperelliptic case corresponds to q(y)= 1∂xy j−∂b∗j12 D.NOVIKOV,S.YAKOVENKObeing a2-form with coefficient of degrees n−1in x and m−1in y,is a linear combination of the forms dωij.Thus we have the equationsH dωij=dH∧ k,l=0B ij,klωkl+dF ij + k,l A ij,kl dωkl,i,k=0,...,n−1,j,l=0,...,m−1.Rescaling x and y by appropriate factors independently,we can assume that the polynomials p(x)and q(y)are both monic.Then all divisions will be bounded pro-vided that the norms p and q are explicitly bounded,and in a way completely similar to the arguments from§2.4,we can derive upper bounds for the matrix coefficients A ij,kl,B ij,kl.Thus the case of doubly hyperelliptic Hamiltonians does not differ much from the ordinary hyperelliptic case,at least as far as the Picard–Fuchs systems for Abelian integrals are concerned.2.6.Discussion.The Picard–Fuchs system written in the form(2.7)for a generic hyperelliptic Hamiltonian(with the potential p(x)being a Morse function on C), is a system remarkable for several instances:•it possesses only Fuchsian singularities(simple poles)both at allfinite singu-larities t=t j,j=1,...,n,and at infinity;•it has no apparent singularities:all points t j are ramification points for the fundamental system of solutions X(t)that is obtained by integrating all forms ωi over all vanishing cyclesδj(t)(the period matrix);•it is minimal in the sense that analytic continuations of any column of the period matrix X(t)along all closed loops span the entire space C n.•its coefficients can be explicitly bounded in terms of H .(All these observations equally apply to doubly hyperelliptic Hamiltonians.) In the next section we generalize this result for arbitrary bivariate Hamiltoni-ans.It will be impossible to preserve all properties,and we shall concentrate on the derivation of redundant system,eventually exhibiting apparent singularities, but all of them(including that at infinity)Fuchsian and with explicitly bounded coefficients.3.Derivation of the redundant Picard–Fuchs system3.1.Notations and conventions.Recall thatΛk denote the spaces of polynomial k-forms on C2for k=0,1,2.They will be always equipped with theℓ1-norms:the norm of a form is always equal to the sum of absolute values of all its coefficients. This norm behaves naturally with respect to the(wedge)product:for any two formsη∈Λk,θ∈Λl,0 k+l 2,we always have η∧θ η · θ .It is also convenient to grade the spaces of polynomial forms so that the degree of a k-form is the maximal degree of its(polynomial)coefficients plus k.Under this convention the exterior derivation is degree-preserving:deg dθ=degθ(unless dθ=0).An easy computation shows that dθ degθ· θ for any0-and1-form θ.On several occasions thefinite-dimensional linear space of k-forms of degree d will be denoted byΛk d.Ifω∈Λ1is a polynomial1-form and H∈Λ0,then by dω/dH is always de-noted the Gelfand–Leray derivative(2.1),while by dω。

七宝中学 2012 学年第二学期高一英语期末考试卷出卷： 施国华 审卷：尹知跃第I卷I. Listening ComprehensionSection A 1.20 分(每题 1 分 共 10 分)A. At a bakery. B. At a hotel. C. At a post office. D. At a birthday party. 2. A. Librarian and student. B. Operator and caller. C. Boss and secretary. D. Customer and repairman. 3. A. 30 minutes. B. 60 minutes. C. 90 minutes. D. 120 minutes. 4. A. A best salesman. B. A latest book. C. A good business. D. An exciting story. 5. A. By going on a diet. B. By eating fruit and vegetables. C. By having fewer meals. D. By doing physical exercise. 6. A. It was interesting and easy to follow. B. It was far beyond his understanding. C. It was as difficult as he had expected. D. It was long but easy to understand. 7. A. The traffic is always very heavy. B. The man needn’t go to the airport. C. The subway is crowded at rush hours. D. The man had better take the subway. 8. A. The woman prefers going out. B. The woman will make the decision. C. The man agrees with the woman. D. The man would rather stay inside. 9. A. The man can tell Kelly the news the next day. B. The man shouldn’t tell Kelly the news at school. C. The man should tell Kelly’s brother the news. D. The man had better tell Kelly the news by himself. 10. A. The match will be put off due to the rain. B. The rain isn’t likely to influence the man’s team. C. The result of the match is sure to be affected. D. The rain will help the man’s team win the match. Section B (每题 1 分共 6 分) Questions 11 through 13 are based on the following passage. 11. A. To get better jobs. B. To save more money. C. To move into a larger apartment. D. To receive further education. 12. A. It had family-friendly politics. B. It owned a child-care center on site. C. It offered its staff excellent benefits. D. It rewarded employees with apartments. 13. A. Satisfied. B. Surprised. C. Confused. D. Interested. Questions 14 through 16 are based on the following passage. 14. A. In 1185. B. In 1209. C. In 1386. 15. A. Because many early learning centers were set up. B. Because Oxford and Cambridge were established. C. Because the largest expansion of universities occurred. D. Because independent universities came into existence.1D. In 1836.16. A. Definition of universities. B. Development of universities. C. Finance for universities. D. Donations for universities. Section C （每题 1 分共 4 分） Blanks 17 through 20 are based on the following conversation. Complete the form. Write ONE WORD for each blank.Summer JobsWorking at KFC: Cutting grass: Working for a sports camp: Serving at a hotel: It’s __17__. One can earn $ __18__ each time. It’s a(n) __19__ job. One may get lots of __20__.Section A： （每题 1 分共 16 分）本题答案一律写到答题纸上！！ ！ II. Fill in the blanks with proper verb form or a suitable word:Social InsectsSome insects, such as ants, bees, and termites, live in organized communities and ____25____(consider)social insects. Ants are one of the ____26____ highly developed of the social insects. They live in colonies（集群）where their members are divided into a clearly defined fixed social system, which includes queens, workers and males. _____27_____ ant has its own purpose and duties within the colony. There are over 10,000 kinds of ants, with unique characteristics, such as the slave makers _____28_____ attack the nests of other ants for their young, or dairying ants which capture certain insects and keep them to be milked of liquids. Ants ___29___(go) to war and attack other ant colonies, even within their own type. ____30____(be) a social insect, they must communicate ____31____ others within their colony ____32____(send) warnings of danger or signal the location of a food supply. Sound Ants that make their nests in leaves or plants will tap the tail part of their bodies against the outside wall of the nest to signal for danger or let their colony mates know___33___ they've discovered a food supply. This tapping will send vibrations through the nest. Some ants will make a squeaking or buzzing sound by ____34____(rub )segments of their bodies together. This sends signals to warn of danger or requests assistance in transporting food. Scent Ants also communicate by scent. They have glands(腺) located on the different segments of their body. From those glands they can release a chemical ____35_____(call) a pheromone. The type of message they wish to convey – danger versus food--will determine which gland they will use to release the pheromone. They can also use pheromones to leave a trail, which will lead others in the colony to a newly ____36____ (discover) food supply. Greeting An ant colony may have just a few dozen members, or it can number in the millions. Since ants will attack even their own kind, it is important_____37______ members of a nest can identify others2in their immediate community. ____38____ two ants come face to face, odor（气味）____39____be a factor in communicating their identity. ____40____(use) their antennae, （触角） ants will smell each the other. After it is determined they are nest mates, they sometimes engage in a strange greeting ritual which involves the ants standing mouth to mouth. One ant then regurgitates（回吐） a drop of liquid into the other ant's mouth. Section B （每题 1 分共 9 分） A. screen F. previous B. similar G. fact C. fewer D. differently H. idea I. prevent E. strangely J. hang注意！ 从本题开始填涂答题卡一定要从 41 题起始按顺序涂黑答题卡！Most of us act ___41___ when we think we’re being watched. If we know someone has their eye on us, we act more properly and speak more carefully. But can this___42___ be used to ___43___crime ? Security staff at Newcastle University in England are testing that question, and so far they’ve had success, reports MSN News. Their___44___ was simple: ___45___ a picture of a pair of eyes above each bicycle rack (搁架) on campus to stop people from stealing bicycles. Since the university adopted this idea, the racks with the eyes above them have ―experienced 62 percent ___46___ thefts compared with the ___47___ year‖, say Newcastle officials. British Transport Police are trying a___48___ solution to stop crime on a train route between London and Essex. Just make sure your parents don’t hear about this story. If they do, they might hang giant eyes above the television ___49___ when you have exams coming up. III. Reading Comprehension Section A: （每题 1 分共 10 分） It happens every spring. Flowers ___50___ burst out everywhere on the same day, as if they were in tune with one another. ____51____how do plants ―know‖ when to flower? It is a question that has ___52___biologists for years. But a US research group may have finally answered it – the ___53___ lies in a protein called FKF1. This allows plants to ___54___ the differences in day lengths so that they can tell the seasons are changing. The study was published in the May 25 issue of the journal Science. Researchers spotted the FKF1 protein when they were studying a plant called Arabidopsis. They found that the FKF1 protein was a photoreceptor (感光器). This means it is___55___ to and can be activated (激活) by sunlight. Plants produce this protein every day in the late afternoon throughout the year. If there is no light at this time, ___56___ in winter when the sun goes down early, the protein won’t be activated. But when spring comes and the days get longer, the FKF1 protein can be activated by ___57___ and the plants ―know‖ it is time to flower. “The presence of light in the late afternoon is the___58___ for plants that the days are getting longer and that it is the optimal (最佳的) time for flowering,” explained Takato Imaizumi, assistant professor at the University of Washington and leader of the study. This daylight-sensing system also keeps plants from flowering when conditions are poor for growing, such as during autumn or winter when the weather is cold and days are short. ___59___ researchers have only studied how the FKF1 protein works in the Arabidopsis, they believe that the process is similar in many other more complex plants, including crops such as rice and wheat. This could be useful to the agricultural industry.350. A. automatically 51. A. But 52. A. exited 53. A. case 54. A. sense 55. A. important 56. A. for example 57. A. fertilizer 58. A. habit 59. A. MeanwhileB. suddenly B. So B. attracted B. fact B. measure B. sensitive B. by contrast B. protein B. style B. MoreoverC. slowly C. While C. fascinated C. secret C. evaluate C. likely C. in addition C. air C. signal C. BesidesD. angrily D. Or D. puzzled D. problem D. see D. necessary D. in short D. daylight D. order D. AlthoughSection B: （60—63 每题 1 分共 4 分 64—70 每题 2 分共 14 总分 18 分） Directions: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read. A Have you ever listened to young children talking in the playground? They are always boasting．They say things like, ―My Dad’s car is bigger than your Dad’s,‖ and ―My Mom is smarter than yours．‖ They particularly like to boast about their families． There were three little boys, Harry, Ted and Gavin, who were always boasting． Gavin was the worst．Everything about his family was always the best or the biggest or the most expensive． Whatever the others said, he could always go on better．One day when they were walking to school, Harry said, ―My father had a bath twice a week,‖ Ted spoke next．―That’s nothing,‖ he said．―Having a bath twice a week is dirty．My father has a bath every day, sometimes twice a day． Ted looked at Gavin． ‖ Now it was his turn． what could he say? ―This time,‖ Ted thought, But ―I’m going to win．‖ Gavin didn’t know what to say．He couldn’t say that his father had a bath three times a day． That was silly． walked on in silence． smiled at Harry, and Harry smiled He Ted back．They were sure that for once they had beaten Gavin．They reached the school gates．Still Gavin said nothing．―We’ve won,‖ Ted said to Harry, but he spoke too soon．On the way home, Gavin said, ―My Dad is so clean that he doesn’t have to bathe at all．‖ 60. According to the passage, the meaning of the word “boast” is __________． A. to sing highly of B. to talk big C. to go on better D. to compete 61. What did Harry boast about? A. The number of times his father had a bath every week． B. The size of the bathroom in his home． C. His father’s fancy car． D. His mother’s big family． 62. Who boasted the most? A. Ted. B. Harry and Ted. C. Harry and Gavin. D. Gavin. 63. What does “he spoke too soon” in the last paragraph mean? A. He spoke quickly． B. He should have spoken earlier． C. He was wrong． D. He couldn’t be understood．4B Here are four festivals in the UK. The Parklife weekender Date: 11—12 June 2011 Location: Platt Fields Park, Fallowfield, Manchester Cost: Weekend ticket￡58. 50. Day ticket ￡35.00 The Parklife Weekender Festival returns as a two—day event. Each Day will feature six areas in tents and a huge open—air main stage featuring the finest selection of local, national and international artists from across the globe．Inside you will find a large number of bars selling a wide range of alcoholic and non—alcoholic drinks. There will also be over 20 food traders offering food including organic and vegetarian dishes. Redfest Date : 22—23 July 2011 Location: Redhill, Surrey Cost: Adult: ￡49. 50. Under 14 ￡16. 50 Redfest in 2011 is a two—day music festival held every year featuring live performances on ―The Main Stage‖ and ―The Introducing Stage‖ and top DJ’S playing till 2 am in the ―Fresh Era Dance Tent‖．Now into its fifth year, Redfest has been known for supporting new band, solo artists and DJ’s. The Big Festival Date: 1—3 July 2011 Location: Clapham Common, London． Cost: Earlybird￡35, Adult￡50, ￡65 with 3 dishes, ￡75 with 5 dishes, Family ￡175, Kids free. The Big Festival 2011—The idea behind it is to take a village fete（户外游乐会） make it much bigger, stick it in the centre of London and have all income go towards charity. It will combine food, the family friendly entertainment and some top—quality live music including Soul H Soul, The Charlatans, Guillemots, The Mystery Jets and many more． Ben and Jerry’s Sundae in the Park Festival Date: 23—24 July 2011 Location: Heaton Park, Manchester Cost: to be confirmed Ben and Jerry’ s festival has now a second site in Manchester’s Heaton Park to spread the Peace, Love&Ice Cream up North. There will be fun—filled activities for everyone. including fairground attractions and social activities! Not to mention a traditional petting farm and a less traditional bare toe wrestling championship. There will be unlimited FREE Ben&Jerry’ s Ice Cream !64. If John and his girlfriend are going to enjoy arts at home and abroad on Friday, they should5pay_______ for the tickets. A. ￡70 B. ￡93．5 C. ￡100 D. ￡117 65. When was the first Redfest festival held? A. 2006. B. 2007. C. 2010. D. 2011. 66. The purpose of the passage is to___________. A. introduce four festivals in the UK B. tell us the history of festivals in the UK C. explain how festivals are celebrated in the UK D. show how important these festivals are for people in the UK C Why birds fly in the sky? Birds can fly – this is something that we take for granted. But have you ever thought about how they developed this enviable ability? It may be impossible for us to go back in time in person to witness how it happened, but scientists from Yale University, US, studied the fossils (化石) of ancient birds. These fossils are like recorded tapes of ancient times that took the scientists back 150 million years. Back then, there lived a kind of animal known as the Archaeopteryx. It is considered to be the earliest bird because it had feathers and wings — the key factors for a bird to fly. Meanwhile, it also had features of a reptile (爬行动物), such as teeth in its mouth, claws (爪子) on its wings, and a long, bony (有骨的) tail, which means the ancestors of birds were probably reptiles. In order to escape from dangerous animals, reptiles leaped from tree to tree. They gradually grew feathers on their arms so they could leap a little farther, then a little farther still. However, they were far from becoming true fliers because their wings and feathers were quite different from those of birds today. Looking closely at the fossils, scientists found that ancient feathers contained many layers of long feathers, like a penguin’s. This made the wings rather heavy and difficult for birds to move up and down. ―We don’t think these things could take off from the ground,‖ said Nicholas Longrich, of Yale University, lead author of the study. ―They can’t fly like a modern bird.‖ This is why scientists concluded that early feathers were only used by birds to glide (滑翔) from high trees, or keep warm, or maybe as camouflage (伪装), but not flying, according to National Geographic. Over the next millions of years, the evolution of feathers never stopped. Birds developed wings that included a layer of long feathers covered with short feathers on top. This structure allowed birds to separate their feathers more easily to create a wider span (翼展) and to change directions in midair. They have also grown special wrist bones (腕骨) so that they can spread their wings when flying and fold them into their sides when walking. “This new research is shedding light not just on how birds came to fly, but more specifically on how feathers came to be the way they are today – one of the most amazing and highly specialized structures in nature,‖ said Jakob Vinther, from the University of Bristol, UK. 67. According to the article, scientists conclude that early birds _____. A. could already fly high B. could only leap and glide C. only used feathers to keep warm D. could easily open their wings wide 68. What is vital in finding out how feathers came to be the way they are? A. Fossils of ancient birds.6B. Tapes of old times. C. Comparing features of reptiles and birds. D. Studying bones and structures of modern birds. 69. The word ―shedding light‖ in the last paragraph probably means________. A. falling off as part of a natural process in the light B. someone or something is seen or shown in a particular light C. providing new information that making sth easier to understand D. killing or injuring people, especially during a war or a fight 70. What is the main point of the article? A. To explore how birds looked 150 millions years ago. B. To look at the evolution of feathers that enabled birds to fly. C. To list all the fossils of ancient birds recently discovered. D. To explore how modern birds developed from ancient reptiles. Section C （本题每题 1 分共 4 分） Directions: Read the following text and choose the most suitable heading from A-E for each paragraph. There is one extra heading which you do not need.注意！本题填涂答题卡一律从 76 题起始号按顺序涂黑答题卡！！ ！A. B. C. D. E. Make a list of things to do Take care of yourself Try physical activity Make time for yourself Think positivelyWhile it is impossible to live completely free of stress, it is possible to prevent stress as well as reduce its effect when it can’t be avoided. The US Department of Health and Human Services offers the following suggestions for ways to deal with stress. 76._________________ When you are nervous, angry or upset, try releasing the pressure through exercise or physical activity. Running, walking, playing tennis, and working in your garden are just some of the activities you might try. 77. _________________ You should make every effort to eat well and get enough rest. If you easily get angry and can’t sleep well enough, or if you’re not eating properly, it will be more likely that you will fall into stressful situations. If stress repeatedly keeps you from sleeping, you should consult a doctor. 78. __________________ Schedule time for both work and entertainment. Don’t forget, play can be just as important to your overall well-being as work. You need a break from your dally routine to just relax and have fun. Go window-shopping or work on a hobby. Allow yourself at least a half hour each day to do something you enjoy. 79. __________________ Stress can result from disorganization and a feeling that ―there’s so much to do, and not enough time‖. Trying to take care of everything at once can be too much for you and as a result, you may not achieve anything. Instead, make a list of everything you have to do ,then do one thing7at a time, checking off each task as it is completed. Set out to do the most important task first. Section D 本题每题 2 分共 8 分 Directions: Read the passage carefully. Then answer the questions or complete the statements within10 words. Researchers have announced the result of two studies on the health effects of the drug aspirin. One study shows aspirin can sharply reduce the chance that a healthy, old man will suffer from a heart attack. The study offered two new results from earlier findings. It said taking one aspirin pill every other day helped only healthy men over the age of fifty. It also said aspirin gave the greatest protection against heart attacks to men with low blood cholesterol（胆固醇）levels. Earlier in the United States began a major aspirin study in the early 1980s. It included 22,000 healthy men doctors. All were between the ages of forty and eighty-four. More than 11,000 of the doctors took a harmless pill that contained no drug. The men did not know which kind of pill they were taking. The doctors who took aspirin suffered 44% fewer heart attacks than those taking the harmless pill. 139 men who took aspirin suffered from heart attacks． of them died. 239 men who did not Ten take aspirin suffered from heart attacks．Twenty-six of them died. The researchers said the doctors' study provides clear proof that taking aspirin can prevent a first heart attack in healthy, older men. They said, however, the result does not mean every man over the age of fifty should take aspirin. They said aspirin couldn't help men who do not eat healthy foods, who smoke cigarettes and who are fat. The researchers said men who think they would be helped by taking aspirin should talk with their doctors first. 80. We may learn that the new use of aspirin is to__________. 81. According to the researchers, aspirin can help those who __________. 82. What is the death percentage of people who suffered from heart attacks without taking aspirin according to the experiment data ? 83. What do the researchers advise us in the last paragraph? 第 II 卷 I. Translation （每题 3 分.共 15 分） Directions: Translate the following sentences into English, using the words given in the brackets.1. 这两个游戏在许多方面有明显的共同之处。

第一部分异域风光（五）21. Cambridge, a University Town剑桥——大学城早在两千年前，罗马人就曾在这地方安营扎寨，屯兵驻军。

在漫长的岁月里，剑桥只是个乡间集镇而已。

直到剑桥大学成立后，这个城镇的名字才渐为人所知，今天它是座10万人口的城市。

剑桥大学有近800年历史，孕育出60多位诺贝尔奖得主，共有31个学院。

When we say that Cambridge is a university town, we do not mean just that it is a town with a university in it. Manchester and Milan have universities. But we do not call them university towns. A university town is one where there is no clear separation between the university buildings and the rest of the city. The university is not just one part of the town, but it is all over the town. The heart of Cambridge has its shops, pubs, market places and so on, but most of it is university—colleges, faculties, libraries, clubs and other places for university staff and students. Students fill the shops, cafes, banks and churches, making these as well part of the university.The town was there first. Two Romans crossed there, and there were signs of building before Roman times (earlier than 43A.D.). In 1226, the authority of the university, the chancellor, was recognized by the king, and Cambridge became a centre of learning.At that time many of the students were too poor to afford lodgings. Colleges were opened so that students could live cheaply. This was the beginning of the college system which has continued at Cambridge up to the present day.The colleges were built with money from kings, queens, religious houses or other sources. One example is Clare College. It was first founded in 1326 as University Hall after the Black Death (a disease which killed half the population of England between 1349 and 1350). It was rebuilt with money from the countess of Clare. Today there are nearly thirty colleges. The newest are University College, founded in 1965, and Clare Hall, founded in 1966, both for graduates.With about 8,250 undergraduates and over 2,000 postgraduates, the city is a busy place in full term. Undergraduates are not allowed to keep cars in Cambridge, so nearly all of them use bicycles. Don't try to drive through Cambridge during the five minutes between lectures. If one has a lecture in Downing College at 9:55 and another in Trinity at 10, his bicycle must get him there through a boiling sea of other bicycles hurrying in all directions. If you are in Cambridge at five minutes to the hour any morning of full term, you will know that you are in a university town. Stop in some safe place, and wait.A. at least 1,960 years ago.B. some 780 years ago.C. about 43 A.D. D. after colleges were built there.3. Which of the following statements is true according to the passage? ______A. Clare College was rebuilt in 1966.B. University College refers to University Hall.C. The newest college is Clare Hall.D. There is a university and nearly thirty colleges in Cambridge.4. According to the last paragraph, what do you think is the feature of Cambridge? ______A. Many bicycles hurry in all directions during the five minutes between lectures.B. There are very few cars in the city.C. The city is too noisy and crowed.D. It is difficult to drive in the city.5. Which of the following statements is not considered correct according to your understanding? ______A. In the beginning many students were so poor that they couldn’t pay for accommodation in the university.B. Colleges were originally sep up for the purpose of offering students cheap lodgings.C. The passage mainly tells us about the history of Cambridge University.D. The passage mainly describes why Cambridge is called a university town. (CBCAD)22. Niagara Falls, Making You Stand in Awe and in Wonder令人敬畏又惊奇的尼亚加拉大瀑布被称为世界著名七大奇景之一的尼亚加拉大瀑布，以丰沛而浩瀚的水汽和磅礴的气势，使前来观赏的游人都为之震撼。