All Abelian Symmetries of Landau-Ginzburg Potentials

All Abelian Symmetries of Landau-Ginzburg Potentials
All Abelian Symmetries of Landau-Ginzburg Potentials

a r X i v :h e p -t h /9211047v 2 20 N o v 1992CERN-TH.6705/92

ITP–UH–10/92

hep-th/9211047

ALL ABELIAN SYMMETRIES OF LANDAU–GINZBURG POTENTIALS Maximilian KREUZER ?CERN,Theory Division CH–1211Geneva 23,SWITZERLAND and Harald SKARKE #Institut f¨u r Theoretische Physik,Universit¨a t Hannover Appelstra?e 2,D–3000Hannover 1,GERMANY ABSTRACT We present an algorithm for determining all inequivalent abelian symmetries of non-degenerate quasi-homogeneous polynomials and apply it to the recently constructed complete set of Landau–Ginzburg potentials for N =2superconformal ?eld theories with c =9.A complete calculation of the resulting orbifolds without torsion increases the number of known

spectra by about one third.The mirror symmetry of these spectra,however,remains at the same low level as for untwisted Landau–Ginzburg models.This happens in spite of the fact that the subclass of potentials for which the Berglund–H¨u bsch construction works features perfect mirror symmetry.We also make ?rst steps into the space of orbifolds with Z Z 2torsions by including extra trivial ?elds.

CERN-TH.6705/92

ITP–UH–10/92

October 1992

1Introduction

Landau–Ginzburg(LG)models[1,2]represent a fairly general framework for constructing N=2superconformal?eld theories,which are needed for supersymmetric string vacua[3]. They provide,for example,a link between exactly solvable models and Calabi–Yau compact-i?cations[4],and also contain large classes of such models as special cases.In general,LG theories cannot be solved exactly.Still,some basic information on the massless spectrum of the resulting string models can be extracted in a simple way from the superpotential owing to non-renormalization properties.As a bonus,on the other hand,a very e?cient algorithm for the calculation of the number of non-singlet representations of E6in(2,2)vacua can be derived from the formulae for charge degeneracies of LG orbifolds given in refs.[5,6].We therefore believe that it is worth while to invest some e?ort into the classi?cation of the string vacua that can be obtained in this way.

Recently the basis for this work has been laid by the enumeration of all(deformation classes of)non-degenerate potentials with central charge c=9[7,8].In the present paper, as a second step,we calculate all abelian orbifolds that can result from manifest symmetries at non-singular points in the moduli spaces of these potentials,disregarding however the possibility of discrete torsions[9,10](except for a modest probe into Z Z2torsions).To do so, we extend the results of Vafa and Intriligator[5,6]for the calculation of the chiral ring in orbifolds and prove an analogue of a well-known theorem for Calabi–Yau manifolds[11]in the LG context.

Theoretically,our results are interesting for the question of mirror symmetry[12,13,14] and for the classi?cation of N=2models.Unfortunately,for both of these(related)issues, our results are,in a sense,negative,as even many spectra with a large Euler number remain without mirror.From a phenomenological point of view,however,extensions of our calcula-tions towards including torsion and non-abelian orbifolds look very promising,because new spectra mainly arise in the realm of small particle content in the e?ective?eld theory.

In section2we show how the(non-singlet)massless spectrum of an orbifold can,in general, be obtained from the index and the dimension of the chiral ring in a very e?cient way.The fact that the computation of neither of these quantities requires explicit knowledge of a basis for the ring is vital for a complete investigation of all non-degenerate cases.Only if there are states with left-right charges(q L,q R)=(1,0)or(0,1)do we need extra information. We show how to extract the number of such states and that they can exist only if Witten’s index vanishes.Section3is quite technical and details how we construct,based upon the classi?cation of potentials[15],all possible(linear)abelian symmetries.The reader who is only interested in the results may wish to proceed to section4.Implications of our?ndings are then discussed in the?nal section.

2Hodge numbers

In this section we review the results of refs.[5,6]and use them to derive an e?cient algorithm for the calculation of the spectrum in case of vanishing torsions.We do,however,keep the discussion general as long as possible.

The basic information about the massless spectrum of(2,2)heterotic string models is contained in the chiral ring,i.e.the non-singular operator product algebra of chiral primary ?elds[16].These?elds have a linear relation between their conformal dimensions and their U(1)charges.Their charge degeneracies are conveniently summarized by the Poincar′e poly-nomial which,for left-and right-chiral?elds,is de?ned as

P(t,ˉt)=tr(c,c)t J0ˉtˉJ0.(1) Analogous generating functions can be de?ned for the charge degeneracies of Ramond ground states and anti-chiral?elds;they are related to one another by spectral?ow[2].

For N=2supersymmetric LG models,de?ned by a non-degenerate quasi-homogeneous superpotential

W(λn i X i)=λd W(X i),(2) the Poincar′e polynomial is given by[17]

P(t,ˉt)=P(tˉt)= 1?(tˉt)1?q i

27denote the numbers of

27and

27,h12=p11=n27,and the Euler number of the manifold

isχ=2(h11?h12)=2(p12?p11)=2(n

2

?q i±(θh i?1

1Namely those which would have h11=0,in contradiction with the requirement of the existence of a K¨a hler form.

and that the action of a group element g on that state is

g |h =(?1)K g K h ε(g,h )det g |h

|G | gh =hg (?1)N +K g K h +K gh ε(g,h ) θg i =θh i =0n i ?d

27

+2n 27+8p 01(for the second equality we have assumed p 01=p 02=p 10=p 20and the Poincar′e duality P (t,ˉt

)=(t ˉt )D P (1/t,1/ˉt )).As the chiral ?elds X i have left-right symmetric charges,all states in a given twisted sector contribute with the same sign to the index

(7).This sign can be computed from the ground state,for which,according to (5),J 0?ˉJ 0= θh i >0(2θh i ?1).As det h =(?1)K h ,this is nothing but K h +N ?N h mod 2,where N h is the number of untwisted

?elds in that sector.Correspondingly,(?1)K h+N?N h is the sign of the contribution with g=1 to the index(7).We thus obtain

P(1,1)=ˉχ=1

n i

(8)

for the dimension of the(c,c)-ring of the orbifold.

Given the analogue of the?rst Betti number,p1=p01+p10,we can use this information to compute n27and n

q i

,(9)

ˉχ=|G|?1 g∈G(?1)N g h∈G θg i=θh i=0q i?1

q i

,(11)χ=|G|?1 M?{X i} ?M?{X i}(?1)|M|n M n?M i:X i∈M∩?M q i?1

a natural bitwise representation for the sets M,namely setting the i th bit to1if M contains X i and to0otherwise,and of course this bit pattern can be identi?ed with an integer.The operator&provided by the programming language C(bitwise logical and)then represents the intersection of M and?M.

A fast algorithm using these tricks works the following way:

(1)Create an array of length2N containing the numbers n M.

(2)Create arrays with entries k?

M = M∩?M=?M n M n?M andˉk?M= M∩?M=?M(?1)|M|n M n?M.

(3)Calculate

χ=|G|?1 ?M?{X i}k?M i:X i∈?M q i?1

q i

,(14) evaluating the time-consuming product over rational numbers only once for each?M with

(k?

M ,ˉk?

M

)=(0,0).

2.2The?rst Betti number

The?nal ingredient we need for the calculation of the Hodge diamond is the number p01. In[7]we have shown that this number can be non-zero only if there is a subset M1of the set

of?elds{X i}and an element j1of the symmetry group such that j1acts like j on the?elds

X i∈M1and does not act at all on the remaining?elds.Furthermore,the contribution of

the?elds in M1to the central charge has to be3,i.e. X i∈M1(1?2q i)=1.The proof of this statement in[7]applies without modi?cation to the general case with arbitrary torsions and

K g.In addition,we have also shown there that,as for CY manifolds,p1>0impliesχ=0,

because any LG model with p1>0factorizes into the product of a torus times a conformal

?eld theory with c=6.Although the factorization property does not generalize to the case of an arbitrary orbifold,we will now show thatχ=0can still be concluded,so that we need to calculate p01only if the index vanishes.(Note that only twisted vacua can contribute to p01; thus the calculation of this number does not require explicit knowledge of the ring either.) In order to show this,we consider a state|j1 contributing to p01.Of course,it has to be invariant under the whole group,implying

(?1)K j1K gε(g,j1)=

det g

To prove the assertion,consider a symmetry g of W (X i )and let {X i }g denote the set of those X i that are themselves invariant under g .The restriction of W to {X i }g ,i.e.setting all other ?elds to 0,is also non-degenerate (in the language of [15],no ?elds in {X i }g ,and thus no links between these ?elds can point out of that set).Then,with g =j (j 1)?1,W restricted to M 1is a non-degenerate potential with D =1.Any twist h can be decomposed as h 1h 2,where h 1acts only on M 1and h 2acts only on the other ?elds.Consider the D =1torus twisted by j 1and h 1.The Poincar′e polynomial of the D =1torus is unique and all its entries come from states twisted by powers j p 1of j 1(unless there are at least two trivial ?elds leading to a doubling of the ground state;this would not a?ect our arguments,however).Therefore the h 1-twisted states cannot survive the j 1-projection unless h 1is itself a power of j 1(h 1does not project itself out).2According to eq.(6)the action of j 1on the h 1twisted sector for D =1is the same as the action of j 1on the h twisted sector for D =3.Thus the twists h that contribute to P (t,ˉt )belong to one of the following classes:Either p =±1,then the ?elds in M 1contribute a factor t or ˉt

,respectively,or p =0.In the latter case the ?elds in M 1contribute two states:q L =q R =0and q L =q R =1.If any of these four contributions is present,then the other three,with identical contributions from the remaining ?elds,will also occur.Therefore the complete Poincar′e polynomial has the form P (t,ˉt

)=(1+t )(1+ˉt )Q (t,ˉt ),and χ=?P (?1,?1)=0.

This factorization of the Poincar′e polynomial does not mean that the LG orbifold is a product of a c =3and a c =6theory.For the latter there seem to be only two possible spectra,namely those corresponding to the torus T 2or the K3surface,whereas we ?nd several di?erent c =9spectra with p 01=0.The reason is that not all states of a “would-be product”survive the group projections.Consider,for example,the 19with the symmetries j 1=Z Z 3[1,1,1,0,0,0,0,0,0],j 2=Z Z 3[0,0,0,1,1,1,1,1,1]and g =Z Z 3[0,1,2,0,1,2,0,0,0].Without g ,the Poincar′e polynomial would be

P (t,ˉt )=(1+t )(1+ˉt ) (1+t 2)(1+ˉt 2)+20t ˉt ,(16)

where the 20t ˉt stand for the (63)=20states X i X j X k |0 with 4≤i,j,k ≤9in the untwisted sector.The g projection reduces this number to 4+(43)=8,coming from states X 4X 5X i |0 and X i X j X k |0 ,6≤i,j,k ≤9.

3Finding the symmetries

This section is based on the criterion for non-degeneracy of a con?guration

2

These features cannot be generalized to orbifolds with torsion.It is straightforward to construct examples where twists with complex determinants on M 1contribute to the chiral ring.Still,in all the examples we know,the conclusion χ=0is true.

the skeletons without such double pointers invertible.They correspond to the polynomials that are non-degenerate with only N monomials and play an important role in mirror symmetry. Note that a skeleton already determines a con?guration and also?xes a maximal abelian phase symmetry,which of course contains any phase symmetry of the complete potential.

The crucial simpli?cation in considering abelian LG orbifolds is that we can assume any abelian symmetry to be diagonalized,i.e.to act as a phase symmetry.Thus we construct all possible inequivalent abelian symmetries by?rst constructing the inequivalent skeletons of a con?guration.Then we determine the maximal phase symmetry of such a skeleton.Finally we construct all subgroups of this symmetry that satisfy the relevant set of conditions.

3.1All skeletons of a con?guration

Constructing all skeletons of a con?guration is straightforward by choosing all possible combi-nations of pointers:X i can point at X j i?n i divides d?n j(here it is convenient to say a?eld points at itself i?it corresponds to a Fermat-type monomial Xαi;these“pointers”,however, do not count for the non-degeneracy criterion).In the case of permutation symmetries of the con?guration,i.e.if some n i are equal,this procedure,however,can generate many equivalent skeletons.It is convenient to represent each consistent skeleton for a given con?guration by an integer whose i th digit is the index of the target of the i th?eld.This implies a natural ordering.A simple concept for eliminating the redundancy is to compute for each skeleton the set of all permutations consistent with the quasi-homogeneity of the con?guration.Then we keep only those skeletons whose integer representation is minimal in the respective set of equivalent skeletons.This(admittedly crude)method requires only a few minutes to produce 106144inequivalent skeletons for the10838non-degenerate con?gurations with up to8?elds, which were constructed in refs.[7,8].For the unique con?guration with9?elds,however, which has the maximal permutation symmetry,this method wastes about a day of computing time.In that case we can alternatively use the algorithm of ref.[7]to compute all topologically inequivalent skeletons,which takes about a second.As a check for our programs we have used both algorithms to construct the2615di?erent skeletons with9?elds.

3.2The maximal abelian symmetry of a skeleton

We will see that for a given skeleton the maximal abelian phase symmetry is of order O= iαi j ( k≤l jαjk)?(?1)l j ,where the?rst product extends over all exponents of?elds

that are not members of a loop,while the second product extends over all loops j with l j respective?elds.This symmetry group can be represented by at most N generators g i,which we construct recursively.

We start with the?elds that are not members of a loop and always consider the origin of a pointer before its target.When arriving at the?eld Y with exponentβ,the typical situation is that the Y-dependence of the skeleton polynomial is given by

I

ZYβ+

Y Xαi i.(17)

i=1

As an induction hypothesis we assume that for each such?eld X i there is,so far,only one generator g i under which X i transforms,and that X i transforms with a phase1/αi,i.e.g i X i= exp(2πi/α)X i(for convenience we omit the obvious factor2πand thus have rational“phases”).

Let a i be the order of g i and U be the least common multiple of the a i.Choose a set of divisors t i of the a i such that U= t i and gcd(t i,t j)=gcd(t i,a i/t i)=1.Given the prime decomposition U= p n j j this means that each number p n j j divides a particular t i.The group generated by g i is therefore the direct product of the groups generated by g′i=(g i)t i and by (g i)a i/t i.

Allowing now Y to transform,but keeping Z?xed,we obviously enlarge the group order by the factorβ,because the Xαi i may haveβdi?erent phases under a group transformation. We now construct a generator g Y of order Uβ,which generates,together with the g′i and all generators g′k already constructed,the maximal phase symmetry that keeps Z?xed.Let the phases of X i and Y under g Y be?1/(αiβ)and1/β,respectively.All other?elds,pointing at some X i,transform with?1/βtimes their phase under g i.Note that we never use a relation involving the group order to change a phase to an equivalent one,so that all monomials are manifestly invariant,even if we enlarge the group order by a factor.Thus,if a path of length n points from some?eld X k to Y,then the inverse phase of X k under g Y is(?1)n times the product of all exponents along the path.The order of g Y isβtimes U,which is equal to the least common multiple of the products of the exponents along maximal paths pointing at Y. Furthermore,there are no non-trivial relations between g Y and the g′i,so that the group we have constructed has the correct order.

There is a slight complication if we eventually hit a?eld belonging to a loop.Within a loop,the transformation of all?elds is?xed by the phase of any particular one.Thus,for our purpose,a loop acts like a single?eld.Let Y j point at Y j+1for j

?b j

αi ,

?1

3.3All abelian symmetries of a skeleton

Obviously the result of the procedure described above is a direct product of cyclic groups

G=Z Z n

1×...×Z Z n

k

.Now we want to?nd all subgroups G of this group that ful?l the

following criteria:

(1)det g=1for all g∈G,

(2)Z Z d?G,

(3)There is a non-degenerate polynomial that is invariant under G.

Since Z Z a×Z Z b=Z Z a×b if gcd(a,b)=1our problem reduces to the construction of all sub-groups of Z Z p l1×...×Z Z p l

k

,for some prime number p,that ful?l these criteria.Then,for the second condition,Z Z d has to be replaced by its maximal subgroup whose order is a power of p.In addition,the existence of a non-degenerate polynomial with the symmetry group G has to be checked again after its subgroups corresponding to di?erent prime numbers have been combined.If we denote the generators of Z Z p l1×...×Z Z p l

k

by g1,...g k,they will have determinants det g i=exp(2πa i/p m i)with gcd(a i,p)=1and0≤m i≤l i.

We use the following algorithm for constructing the maximal subgroup with det=1:

(1)Find the maximal m i.

(2)If there are i,i′with m i=m i′and(say)l i≥l i′,replace g i by g i g?x i′,where x is chosen in such a way that xa i′=a i mod p m i.

(3)Repeat this until there is only one maximal m i,then replace g i by(g i)p.

(4)Repeat the whole procedure until det g i=1for all g i.

Denoting the result of this construction again by Z Z p l1×...×Z Z p l

k

,any subgroup will be

of the form Z Z

p?l1×...×Z Z

p?l?k

with generators?g i= k j=1gλij j,1≤i≤?k,0≤λij

course we severely overcount the subgroups in this way,for two reasons:A generator?g will be equivalent to?gλifλis not divisible by p,and the set of generators{?g a,?g b}is equivalent to {?g a?gλb,?g b}for anyλ.The?rst source of overcountings can be overcome in the following way: We note that the order of?g i is given by p?l i=max j(p l j/gcd(λij,p l j)).If we assume the j’s to be ordered,we can denote by?j(i)the?rst j for which p l j/gcd(λij,p l j)=p?l i.This allows us to choose the“normalization”λi?j(i)=p l j??l i.Overcountings of the second type can be avoided by demandingλij

This implies the following algorithm for constructing each subgroup exactly once:

(1)Fix the type(?l1,...,?l?

k )of the subgroup,with some ordering(e.g.?l i≥?l i+1).Of course

?l i ≤l i,if the same ordering for the l i’s is assumed.

(2)Choose?j(i)for each i,with some ordering if l i=l i+1(e.g.?j(i)

(3)Choose all otherλij’s subject toλij

At this point we check for the Z Z d by explicitly calculating all elements of Z Z

p?l1×...×Z Z

p?l?k

and comparing them with the p-projection of the generator of the Z Z d.Putting together the subgroups is straightforward.The check for non-degeneracy is based on the results of[15]and follows the route indicated in[7].

4Results

We have implemented these ideas in a C program,which we had running for about6days

(real time)on a workstation to produce the results presented below.Because of this rather

reasonable computing time we did not worry about the redundancy of calculating the same orbifolds several times for di?erent skeletons or even calculating equivalent moddings for a

particular skeleton in case of permutation symmetries of the potential.In general this redun-

dancy is not too bad,since there is an average of only10skeletons per con?guration and the

overlap occurs mainly for groups of low orders.It is particularly bad,however,for the19, i.e.the potential W= 9i=1X3i–one out of108759skeletons–which alone consumed more than80%of our computation time and required the calculation of about2million orbifolds,

producing eventually only23spectra,none of which was new.Obviously,for including all

torsions one will have to work harder on this part of the calculation.

In table I we give a detailed statistics of our results.3According to our organization of the computation we list,in the?rst6columns,the results for a?xed number of non-trivial?elds, and?nally combine the individual?gures.Our starting point was the list of non-degenerate con?gurations with c=9,obtained in refs.[7,8].Using the algorithms described in section3 we have then calculated for each con?guration all inequivalent skeletons,and for each skeleton all subgroups of the maximal phase symmetry that contain the canonical Z Z d and allow a non-degenerate invariant polynomial.As the Z Z d has negative determinant for even N,we have added a trivial?eld in that case before restricting to determinant1.The numbers of di?erent skeletons and symmetries that arise in this way are given in lines2and3.Then we list the maximal numbers of symmetries and di?erent orbifold spectra that can come from a single skeleton.

In the line denoted by“spectra”we list the total number of di?erent spectra obtained

from all con?gurations with the respective number of non-trivial?elds.The majority of these

spectra appear in pairs with n27and n

27=230.Note that the spectra with the largest numbers of particles have large positive

Euler number,in agreement with the expectation that orbifolding generically increases the Euler number.

Among the new models there is a number of spectra with p1=p01+p10>0,namely n27=n27=9for p1=6(we count,in this paper,spectra with di?erent p1as di?erent;our plots may thus in fact show up to5points fewer than is indicated in the?gure captions).In accordance with the results of section2,

?elds total

239051652567669471

767430575312162925774222615

17833532821396961116921876412324394

≤140≤140≤13506≤2664≤56632≤2052656

≤43≤28≤63≤39≤47≤47

227831822002101528985

25867519977171

invertible skeletons27299 spectra from ISK2730 Table I:Statistics of the calculation and results.(I)SK and SP denote

(invertible)skeletons and spectra.

all these spectra haveχ=0.In contrast with the canonical orbifolds of[7],however,they do not all correspond to tensor products of conformal?eld theories,as has been discussed in section2.This conclusion can also be drawn from the fact that the factors would have to be either the torus,with all coe?cients of the Poincar′e polynomial equal to1,or a model with D=2.By applying our program to the922inequivalent skeletons of the124non-degenerate con?gurations with D=2we have checked that,within our class of orbifolds,the Hodge diamond of the K3surface and the2-torus are the only possible spectra.This implies that only the last two of the above spectra can be products and is another check for the remarkably successful geometric interpretation of D=2models[18].Some of our spectra with p1>0have recently also been obtained from CP m coset models with non-diagonal modular invariants[20]. They can be compared with the80non-chiral spectra we have found with p1=0,which exist for n27=n

27≤66.To give a more detailed picture we have also included in that plot all other

spectra with n27+n

C(1,1,1,1,1,1,2,2,2)[4]with6non-trivial and3trivial?elds

and has the spectrumχ=b1=0,n27=n

C(1,1,1,1,1,1)[4],and twist by the group(Z Z4)2×Z Z2with the same action on the non-trivial?elds,but now with torsionsε(g1,g2)=ε(g2,g3)=?1andε(g1,g3)=1.Of course, we also have to use the appropriate signs K g and torsions with the canonical Z Z4as discussed in section2.These results give us a clear hint that realistic spectra with very low numbers of ?elds can be expected in the set of models with non-trivial torsions.

To check the construction of the mirror model by Berglund and H¨u bsch(BH)[14],which applies exactly to the invertible skeletons,we have veri?ed for a number of such skeletons that the inverted(or“transposed”)skeleton yields exactly the mirror spectra.In addition,we have examined the mirror symmetry of the complete set of spectra that come from invertible skeletons(let us recall that we use the term mirror symmetry in its na¨?ve sense of just rotating the Hodge diamond;we do not check the fusion rules).The BH construction does not imply that this space is exactly symmetric,since the mirror of a potential containing Xα+XY2 would contain a trivial?eld.Indeed,we found12spectra violating the mirror symmetry of the list of spectra of invertible skeletons we produced.An explicit check of these models shows that they are of the type described above,and that the inverted skeleton,which contains trivial ?elds,produces the correct spectrum.In fact,6of the missing mirror spectra are already contained in the set of spectra from non-invertible skeletons,whereas the other6are part of the above39spectra with Z Z2torsions.

The situation is drastically di?erent for non-invertible models.Of the1068spectra that cannot be obtained from invertible skeletons,810have no mirror spectrum.Figure3shows the258remaining spectra in this class,which do have mirrors.The ones without mirror are indicated by small dots in?g.3,and all of them are plotted separately in?g.4.It should be noted that the258spectra with mirrors all occur in the“dense”region of all spectra in?g.5. It is,therefore,not unlikely that their mirror pairings are purely accidental.A look at?g.2 shows that indeed in some of the low-lying regions of the plot of all spectra a high percentage of all possible combinations of Euler numbers divisible by4and even n27+n

27?3∈{13,15,16,17,20,26,27,29,31,34,37,42,

47,67,74}and n

27

.It belongs to the con?guration

n27n con?guration

1411?6C(2,3,3,4,4,5,5)[13]

(Z Z3)2 0010011

0100020 1815?6C(1,2,3,3,3,3,3)[9]

Z Z2[01010]

2522?6C(2,3,4,9,9)[27]

Z Z2[10001]

3128?6C(1,3,4,4,9)[21]

(Z Z2)2 00011

01100

3734?6C(2,3,5,15,20)[45]

Z Z2[10001]

4744?6C(2,3,7,21,30)[63]

Z Z3[02010]

5451?6C(1,2,4,13,19)[39]

Z Z2[01111]

7067?6C I(1,3,15,20,36)[75]

Z Z16[81261314101] 18216C I(2,3,5,8,9)[27]

Z Z2[0010010] 21246C I(1,1,2,2,5)[11]

Z Z3[12000]

32356C I(3,3,10,14,15)[45]

(Z Z2)2 11000

10001 40436C I(1,1,5,8,10)[25]

Z Z2[10001]

48516C I(1,5,6,18,25)[55]

n27n con?guration

12132C(5,6,7,8,9,11,12)[29]

Z Z2[0100010]

20212C(3,3,4,5,10)[25]

Z Z2[01001]

34352C(1,1,1,4,6)[13]

Z Z2[10001]

3332?2C(1,3,4,4,11)[23]

Z Z2[01010]

Table III:1-generation models

language of[15],each of the monomials X5Y and X4W in(19)can be interpreted as a pointer; then the other one belongs to the set of required links).

In table III we list the1-generation models.None of them has a mirror spectrum,and hence none of them comes from an invertible skeleton.With two exceptions,all values of the original inverse charge quantum d are prime.In fact,as for the untwisted case[7],all 1-and3-generation models have odd d,and hence an odd number of?elds.The number of 2-generation models in our list is33;26of them do not require a twist and18have negative Euler numbers.

It is signi?cant that the spectra without mirror that have the largest values of|χ|all come from untwisted LG models(compare?gs.1and4).The?rst12of these,with spectra(13,433), (17,341),(20,326),...,have large positive Euler numbers.As all our new spectra have much smaller particle content,it appears to be most unlikely that these spectra will eventually?nd their mirrors in the realm of(more general)LG orbifolds.

5Discussion and outlook

Considering the complete set of abelian symmetries of Landau–Ginzburg potentials,we have studied approximately250times as many models as previously,with canonically twisted theo-ries(there is,however,some redundancy in our constructions owing to the possible occurrence of the same symmetries for di?erent skeletons in a speci?c con?guration and to permutation symmetries of some skeletons).Doing so without pushing computer time to astronomical heights was only possible with an algorithm that was extremely e?cient,at least at its central part,i.e.at the calculation of the numbers of chiral generations and anti-generations from a given potential and symmetry.The numbers of spectra obtained,and the number of new features,however,did not rise in a comparable manner.Less than25%of the spectra we found were new compared with[7,8],and the overall impression of the plot of spectra in?g.5 is the same as in the pioneering work ref.[12].Although the new spectra arise primarily in

the range of low generation and anti-generation numbers,we have not found any models that look particularly promising from a phenomenological point of view.Yet,we believe that our results are quite interesting from the following points of view:

We have established now what we already found in[7],namely that mirror symmetry in the context of Landau–Ginzburg orbifolds occurs regularly for those and only for those models for which the Berglund–H¨u bsch construction works.In particular,among the spectra that remain without mirrors,there are a number of canonical LG models whose Euler numbers are much larger than what any orbifold contributed.This makes it appear very unlikely that a generalization of the BH contruction exists for LG orbifolds or for the related Calabi–Yau manifolds.In accordance with the results of[21],the lack of mirror symmetry also indicates that,for a complete classi?cation of rational N=2theories,we need to go beyond LG models. Still,in addition to their phenomenological use,the lessons we learn from them may be helpful also in that direction.

Obviously non-abelian symmetries and twists with non-trivial torsion[10]are good candi-dates for providing phenomenologically more realistic spectra.This is clear from the fact that our“smallest”3-generation model requires nearly twice as many fermions of opposite chirality as the well-known model of[22],which can be interpreted as a non-abelian Landau–Ginzburg orbifold.Furthermore,the few new spectra that we obtained with our excursion into the space of models with non-trivial Z Z2torsion are all in the area of very small particle numbers,and this set contains the model with the least value of n27+n

References

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C.Vafa and N.Warner,Phys.Lett.B218(1989)51

[2]W.Lerche,C.Vafa and N.P.Warner,Nucl.Phys.B324(1989)427

[3]W.Boucher,D.Friedan and A.Kent,Phys.Lett.B172(1986)316;

T.Banks,L.Dixon,D.Friedan and E.Martinec,Nucl.Phys.B299(1988)613;

C.Hull and E.Witten,Phys.Lett.B160(1985)398

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[5]C.Vafa,Mod.Phys.Lett.A4(1989)1169;Superstring Vacua,HUTP-89/A057preprint

[6]K.Intriligator and C.Vafa,Nucl.Phys.B339(1990)95

[7]M.Kreuzer and H.Skarke,hep-th/9205004,CERN-TH.6461/92and TUW-92-06preprint,

to appear in Nucl.Phys.B

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TUM-TP-142/92preprint

[9]C.Vafa,Nucl.Phys.B273(1986)592

[10]A.Font,L.E.Ib′a?n ez,F.Quevedo and A.Sierra,Nucl.Phys.B337(1990)119

[11]T.H¨u bsch,Calabi-Yau Manifolds(World Scienti?c,Singapore,1992);

P.Candelas,Lectures on Complex Manifolds,in Proceedings of the Trieste Spring School Strings1987

[12]P.Candelas,M.Lynker and R.Schimmrigk,Nucl.Phys.B341(1990)383

[13]B.R.Greene and M.R.Plesser,Nucl.Phys.B338(1990)15

[14]P.Berglund and T.H¨u bsch,hep-th/9201014,CERN-TH.6341/91and UTTG-32-91

preprint,to appear in Nucl.Phys.B

[15]M.Kreuzer and H.Skarke,hep-th/9202039,CERN-TH.6373/92and TUW-92-01preprint,

to appear in Commun.Math.Phys.

[16]D.Gepner,Nucl.Phys.B296(1988)757;N=2String Theory,in Proceedings of the

Trieste Spring School Strings1989,eds.M.Green et al.(World Scienti?c,Singapore, 1990)

[17]N.Bourbaki,Lie Groups and Lie Algebras(Hermann,Paris1971)chapter V,section5.1;

V.I.Arnold,S.M.Gusein-Zade and A.N.Varchenko,Singularities of Di?erentiable Maps (Birkh¨a user,Boston1985)Vol.I

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[19]M.Kreuzer,R.Schimmrigk and H.Skarke,Nucl.Phys.B372(1992)61

[20]G.Aldazabal,I.Allekotte,E.Andr′e s and C.N′u?n ez,hep-th/9209059,Bariloche preprint

[21]J.Fuchs and M.Kreuzer,hep-th/9210053,CERN-TH.6669/92preprint

[22]R.Schimmrigk,Phys.Lett.B193(1987)175

Figures 200400100300q

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q q q 27vs.Euler number χfor the 800new LGO spectra

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q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q

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vs.Euler number for the 39spectra with Z Z 2torsion (circles)and for the LG/LGO spectra with n 27+n -200-100-30010015050

Fig.3:n 27+n

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p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p

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27vs.Euler number χfor the 810spectra without mirror

of与for的用法以及区别

of与for的用法以及区别 for 表原因、目的 of 表从属关系 介词of的用法 (1)所有关系 this is a picture of a classroom (2)部分关系 a piece of paper a cup of tea a glass of water a bottle of milk what kind of football,American of soccer? (3)描写关系 a man of thirty 三十岁的人 a man of shanghai 上海人 (4)承受动作 the exploitation of man by man.人对人的剥削。 (5)同位关系 It was a cold spring morning in the city of London in England. (6)关于,对于 What do you think of Chinese food? 你觉得中国食品怎么样? 介词 for 的用法小结 1. 表示“当作、作为”。如: I like some bread and milk for breakfast. 我喜欢把面包和牛奶作为早餐。What will we have for supper? 我们晚餐吃什么?

2. 表示理由或原因,意为“因为、由于”。如: Thank you for helping me with my English. 谢谢你帮我学习英语。 Thank you for your last letter. 谢谢你上次的来信。 Thank you for teaching us so well. 感谢你如此尽心地教我们。 3. 表示动作的对象或接受者,意为“给……”、“对…… (而言)”。如: Let me pick it up for you. 让我为你捡起来。 Watching TV too much is bad for your health. 看电视太多有害于你的健康。 4. 表示时间、距离,意为“计、达”。如: I usually do the running for an hour in the morning. 我早晨通常跑步一小时。We will stay there for two days. 我们将在那里逗留两天。 5. 表示去向、目的,意为“向、往、取、买”等。如: let’s go for a walk. 我们出去散步吧。 I came here for my schoolbag.我来这儿取书包。 I paid twenty yuan for the dictionary. 我花了20元买这本词典。 6. 表示所属关系或用途,意为“为、适于……的”。如: It’s time for school. 到上学的时间了。 Here is a letter for you. 这儿有你的一封信。 7. 表示“支持、赞成”。如: Are you for this plan or against it? 你是支持还是反对这个计划? 8. 用于一些固定搭配中。如: Who are you waiting for? 你在等谁? For example, Mr Green is a kind teacher. 比如,格林先生是一位心地善良的老师。

The way常见用法

The way 的用法 Ⅰ常见用法: 1)the way+ that 2)the way + in which(最为正式的用法) 3)the way + 省略(最为自然的用法) 举例:I like the way in which he talks. I like the way that he talks. I like the way he talks. Ⅱ习惯用法: 在当代美国英语中,the way用作为副词的对格,“the way+ 从句”实际上相当于一个状语从句来修饰整个句子。 1)The way =as I am talking to you just the way I’d talk to my own child. He did not do it the way his friends did. Most fruits are naturally sweet and we can eat them just the way they are—all we have to do is to clean and peel them. 2)The way= according to the way/ judging from the way The way you answer the question, you are an excellent student. The way most people look at you, you’d think trash man is a monster. 3)The way =how/ how much No one can imagine the way he missed her. 4)The way =because

(完整版)介词for用法归纳

介词for用法归纳 用法1:(表目的)为了。如: They went out for a walk. 他们出去散步了。 What did you do that for? 你干吗这样做? That’s what we’re here for. 这正是我们来的目的。 What’s she gone for this time? 她这次去干什么去了? He was waiting for the bus. 他在等公共汽车。 【用法说明】在通常情况下,英语不用for doing sth 来表示目的。如: 他去那儿看他叔叔。 误:He went there for seeing his uncle. 正:He went there to see his uncle. 但是,若一个动名词已名词化,则可与for 连用表目的。如: He went there for swimming. 他去那儿游泳。(swimming 已名词化) 注意:若不是表目的,而是表原因、用途等,则其后可接动名词。(见下面的有关用法) 用法2:(表利益)为,为了。如: What can I do for you? 你想要我什么? We study hard for our motherland. 我们为祖国努力学习。 Would you please carry this for me? 请你替我提这个东西好吗? Do more exercise for the good of your health. 为了健康你要多运动。 【用法说明】(1) 有些后接双宾语的动词(如buy, choose, cook, fetch, find, get, order, prepare, sing, spare 等),当双宾语易位时,通常用for 来引出间接宾语,表示间接宾语为受益者。如: She made her daughter a dress. / She made a dress for her daughter. 她为她女儿做了件连衣裙。 He cooked us some potatoes. / He cooked some potatoes for us. 他为我们煮了些土豆。 注意,类似下面这样的句子必须用for: He bought a new chair for the office. 他为办公室买了张新办公椅。 (2) 注意不要按汉语字面意思,在一些及物动词后误加介词for: 他们决定在电视上为他们的新产品打广告。 误:They decided to advertise for their new product on TV. 正:They decided to advertise their new product on TV. 注:advertise 可用作及物或不及物动词,但含义不同:advertise sth=为卖出某物而打广告;advertise for sth=为寻找某物而打广告。如:advertise for a job=登广告求职。由于受汉语“为”的影响,而此处误加了介词for。类似地,汉语中的“为人民服务”,说成英语是serve the people,而不是serve for the people,“为某人的死报仇”,说成英语是avenge sb’s death,而不是avenge for sb’s death,等等。用法3:(表用途)用于,用来。如: Knives are used for cutting things. 小刀是用来切东西的。 This knife is for cutting bread. 这把小刀是用于切面包的。 It’s a machine for slicing bread. 这是切面包的机器。 The doctor gave her some medicine for her cold. 医生给了她一些感冒药。 用法4:为得到,为拿到,为取得。如: He went home for his book. 他回家拿书。 He went to his friend for advice. 他去向朋友请教。 She often asked her parents for money. 她经常向父母要钱。

常用介词用法(for to with of)

For的用法 1. 表示“当作、作为”。如: I like some bread and milk for breakfast. 我喜欢把面包和牛奶作为早餐。 What will we have for supper? 我们晚餐吃什么? 2. 表示理由或原因,意为“因为、由于”。如: Thank you for helping me with my English. 谢谢你帮我学习英语。 3. 表示动作的对象或接受者,意为“给……”、“对…… (而言)”。如: Let me pick it up for you. 让我为你捡起来。 Watching TV too much is bad for your health. 看电视太多有害于你的健康。 4. 表示时间、距离,意为“计、达”。如: I usually do the running for an hour in the morning. 我早晨通常跑步一小时。 We will stay there for two days. 我们将在那里逗留两天。 5. 表示去向、目的,意为“向、往、取、买”等。如: Let’s go for a walk. 我们出去散步吧。 I came here for my schoolbag.我来这儿取书包。 I paid twenty yuan for the dictionary. 我花了20元买这本词典。 6. 表示所属关系或用途,意为“为、适于……的”。如: It’s time for school. 到上学的时间了。 Here is a letter for you. 这儿有你的一封信。 7. 表示“支持、赞成”。如: Are you for this plan or against it? 你是支持还是反对这个计划? 8. 用于一些固定搭配中。如: Who are you waiting for? 你在等谁? For example, Mr Green is a kind teacher. 比如,格林先生是一位心地善良的老师。 尽管for 的用法较多,但记住常用的几个就可以了。 to的用法: 一:表示相对,针对 be strange (common, new, familiar, peculiar) to This injection will make you immune to infection. 二:表示对比,比较 1:以-ior结尾的形容词,后接介词to表示比较,如:superior ,inferior,prior,senior,junior 2: 一些本身就含有比较或比拟意思的形容词,如equal,similar,equivalent,analogous A is similar to B in many ways.

The way的用法及其含义(二)

The way的用法及其含义(二) 二、the way在句中的语法作用 the way在句中可以作主语、宾语或表语: 1.作主语 The way you are doing it is completely crazy.你这个干法简直发疯。 The way she puts on that accent really irritates me. 她故意操那种口音的样子实在令我恼火。The way she behaved towards him was utterly ruthless. 她对待他真是无情至极。 Words are important, but the way a person stands, folds his or her arms or moves his or her hands can also give us information about his or her feelings. 言语固然重要,但人的站姿,抱臂的方式和手势也回告诉我们他(她)的情感。 2.作宾语 I hate the way she stared at me.我讨厌她盯我看的样子。 We like the way that her hair hangs down.我们喜欢她的头发笔直地垂下来。 You could tell she was foreign by the way she was dressed. 从她的穿著就可以看出她是外国人。 She could not hide her amusement at the way he was dancing. 她见他跳舞的姿势,忍俊不禁。 3.作表语 This is the way the accident happened.这就是事故如何发生的。 Believe it or not, that's the way it is. 信不信由你, 反正事情就是这样。 That's the way I look at it, too. 我也是这么想。 That was the way minority nationalities were treated in old China. 那就是少数民族在旧中

of和for的用法

of 1....的,属于 One of the legs of the table is broken. 桌子的一条腿坏了。 Mr.Brown is a friend of mine. 布朗先生是我的朋友。 2.用...做成的;由...制成 The house is of stone. 这房子是石建的。 3.含有...的;装有...的 4....之中的;...的成员 Of all the students in this class,Tom is the best. 在这个班级中,汤姆是最优秀的。 5.(表示同位) He came to New York at the age of ten. 他在十岁时来到纽约。 6.(表示宾格关系) He gave a lecture on the use of solar energy. 他就太阳能的利用作了一场讲演。 7.(表示主格关系) We waited for the arrival of the next bus. 我们等待下一班汽车的到来。

I have the complete works of Shakespeare. 我有莎士比亚全集。 8.来自...的;出自 He was a graduate of the University of Hawaii. 他是夏威夷大学的毕业生。 9.因为 Her son died of hepatitis. 她儿子因患肝炎而死。 10.在...方面 My aunt is hard of hearing. 我姑妈耳朵有点聋。 11.【美】(时间)在...之前 12.(表示具有某种性质) It is a matter of importance. 这是一件重要的事。 For 1.为,为了 They fought for national independence. 他们为民族独立而战。 This letter is for you. 这是你的信。

(完整版)the的用法

定冠词the的用法: 定冠词the与指示代词this ,that同源,有“那(这)个”的意思,但较弱,可以和一个名词连用,来表示某个或某些特定的人或东西. (1)特指双方都明白的人或物 Take the medicine.把药吃了. (2)上文提到过的人或事 He bought a house.他买了幢房子. I've been to the house.我去过那幢房子. (3)指世界上独一无二的事物 the sun ,the sky ,the moon, the earth (4)单数名词连用表示一类事物 the dollar 美元 the fox 狐狸 或与形容词或分词连用,表示一类人 the rich 富人 the living 生者 (5)用在序数词和形容词最高级,及形容词等前面 Where do you live?你住在哪? I live on the second floor.我住在二楼. That's the very thing I've been looking for.那正是我要找的东西. (6)与复数名词连用,指整个群体 They are the teachers of this school.(指全体教师) They are teachers of this school.(指部分教师) (7)表示所有,相当于物主代词,用在表示身体部位的名词前 She caught me by the arm.她抓住了我的手臂. (8)用在某些有普通名词构成的国家名称,机关团体,阶级等专有名词前 the People's Republic of China 中华人民共和国 the United States 美国 (9)用在表示乐器的名词前 She plays the piano.她会弹钢琴. (10)用在姓氏的复数名词之前,表示一家人 the Greens 格林一家人(或格林夫妇) (11)用在惯用语中 in the day, in the morning... the day before yesterday, the next morning... in the sky... in the dark... in the end... on the whole, by the way...

蔡琴歌词大全

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出人头地 蔡琴 那个不想出人头地 说起来简单做起来不易 那个不愿登峰造极 首先要问一问你自己 哎哟朋友你有没有自私自利 哎哟朋友你有没有自暴自弃 哎哟朋友你有没有尽心尽力 不努力怎会出人头地 那个不想出人头地 那万丈高楼由平地造起 不要空想登峰造极 大海洋源流于小水滴 哎哟朋友你必须要先人后自 哎哟朋友你必须要自食其力 哎哟朋友你必须要自强不息 到头来终会出人头地 哎哟朋友你有没有自私自利 哎哟朋友你有没有自暴自弃 哎哟朋友你有没有尽心尽力 不努力怎会出人头地 出塞曲 LRC EDIT:洪枫sunhfone@https://www.360docs.net/doc/e88286493.html, 请为我唱一首出塞曲 用那遗忘了的古老言语 请用美丽的颤音轻轻呼唤 我心中的大好河山 那只有长城外才有的清香 谁说出塞歌的调子太悲凉 如果你不爱听 那是因为歌中没有你的渴望 而我们总是要一唱再唱 想着草原千里闪着金光 想着风沙呼啸过大漠 想着黄河岸啊阴山旁 英雄骑马壮 骑马荣归故乡初恋 演唱:蔡琴 夜已深沉月儿昏昏 寂寞地等侯你 等侯你诉一诉 别后岁月凄凉 我要你常记在你心里 我要你永远不忘记 我俩相依蔷薇花底 好想鸟儿双栖枝连李 千万种的情义 我要你快回到我身边 我要你相爱在一起 趁着今夜星月依稀 我俩重回蔷薇花底 大约在冬季 词、曲:齐秦 轻轻的我将离开你 请将眼角的泪拭去 漫漫长夜里未来日子里亲爱的你别为我哭泣 前方的路虽然太凄迷 请在笑容里为我祝福 虽然迎著风虽然下著雨我在风雨之中念著你 没有你的日子里 我会更加珍惜自己 没有我的岁月里 你要保重你自己 你问我何时归故里 我也轻声地问自己 不是在此时不知在何时我想大约会是在冬季 不是在此时不知在何时我想大约会是在冬季

“the way+从句”结构的意义及用法

“theway+从句”结构的意义及用法 首先让我们来看下面这个句子: Read the followingpassageand talkabout it wi th your classmates.Try totell whatyou think of Tom and ofthe way the childrentreated him. 在这个句子中,the way是先行词,后面是省略了关系副词that或in which的定语从句。 下面我们将叙述“the way+从句”结构的用法。 1.the way之后,引导定语从句的关系词是that而不是how,因此,<<现代英语惯用法词典>>中所给出的下面两个句子是错误的:This is thewayhowithappened. This is the way how he always treats me. 2.在正式语体中,that可被in which所代替;在非正式语体中,that则往往省略。由此我们得到theway后接定语从句时的三种模式:1) the way+that-从句2)the way +in which-从句3) the way +从句 例如:The way(in which ,that) thesecomrade slookatproblems is wrong.这些同志看问题的方法

不对。 Theway(that ,in which)you’re doingit is comple tely crazy.你这么个干法,简直发疯。 Weadmired him for theway inwhich he facesdifficulties. Wallace and Darwingreed on the way inwhi ch different forms of life had begun.华莱士和达尔文对不同类型的生物是如何起源的持相同的观点。 This is the way(that) hedid it. I likedthe way(that) sheorganized the meeting. 3.theway(that)有时可以与how(作“如何”解)通用。例如: That’s the way(that) shespoke. = That’s how shespoke.

for和of的用法

for的用法: 1. 表示“当作、作为”。如: I like some bread and milk for breakfast. 我喜欢把面包和牛奶作为早餐。 What will we have for supper? 我们晚餐吃什么? 2. 表示理由或原因,意为“因为、由于”。如: Thank you for helping me with my English. 谢谢你帮我学习英语。 Thank you for your last letter. 谢谢你上次的来信。 Thank you for teaching us so well. 感谢你如此尽心地教我们。 3. 表示动作的对象或接受者,意为“给……”、“对…… (而言)”。如: Let me pick it up for you. 让我为你捡起来。 Watching TV too much is bad for your health. 看电视太多有害于你的健康。 4. 表示时间、距离,意为“计、达”。如:

I usually do the running for an hour in the morning. 我早晨通常跑步一小时。 We will stay there for two days. 我们将在那里逗留两天。 5. 表示去向、目的,意为“向、往、取、买”等。如: Let’s go for a walk. 我们出去散步吧。 I came here for my schoolbag.我来这儿取书包。 I paid twenty yuan for the dictionary. 我花了20元买这本词典。 6. 表示所属关系或用途,意为“为、适于……的”。如: It’s time for school. 到上学的时间了。 Here is a letter for you. 这儿有你的一封信。 7. 表示“支持、赞成”。如: Are you for this plan or against it? 你是支持还是反对这个计划? 8. 用于一些固定搭配中。如:

way 用法

表示“方式”、“方法”,注意以下用法: 1.表示用某种方法或按某种方式,通常用介词in(此介词有时可省略)。如: Do it (in) your own way. 按你自己的方法做吧。 Please do not talk (in) that way. 请不要那样说。 2.表示做某事的方式或方法,其后可接不定式或of doing sth。 如: It’s the best way of studying [to study] English. 这是学习英语的最好方法。 There are different ways to do [of doing] it. 做这事有不同的办法。 3.其后通常可直接跟一个定语从句(不用任何引导词),也可跟由that 或in which 引导的定语从句,但是其后的从句不能由how 来引导。如: 我不喜欢他说话的态度。 正:I don’t like the way he spoke. 正:I don’t like the way that he spoke. 正:I don’t like the way in which he spoke. 误:I don’t like the way how he spoke. 4.注意以下各句the way 的用法: That’s the way (=how) he spoke. 那就是他说话的方式。 Nobody else loves you the way(=as) I do. 没有人像我这样爱你。 The way (=According as) you are studying now, you won’tmake much progress. 根据你现在学习情况来看,你不会有多大的进步。 2007年陕西省高考英语中有这样一道单项填空题: ——I think he is taking an active part insocial work. ——I agree with you_____. A、in a way B、on the way C、by the way D、in the way 此题答案选A。要想弄清为什么选A,而不选其他几项,则要弄清选项中含way的四个短语的不同意义和用法,下面我们就对此作一归纳和小结。 一、in a way的用法 表示:在一定程度上,从某方面说。如: In a way he was right.在某种程度上他是对的。注:in a way也可说成in one way。 二、on the way的用法 1、表示:即将来(去),就要来(去)。如: Spring is on the way.春天快到了。 I'd better be on my way soon.我最好还是快点儿走。 Radio forecasts said a sixth-grade wind was on the way.无线电预报说将有六级大风。 2、表示:在路上,在行进中。如: He stopped for breakfast on the way.他中途停下吃早点。 We had some good laughs on the way.我们在路上好好笑了一阵子。 3、表示:(婴儿)尚未出生。如: She has two children with another one on the way.她有两个孩子,现在还怀着一个。 She's got five children,and another one is on the way.她已经有5个孩子了,另一个又快生了。 三、by the way的用法

经典感恩歌曲推荐《念亲恩》

经典感恩歌曲推荐《念亲恩》《念亲恩》是由陈百强演唱的一首歌曲,由杨继兴作曲和作词,收录在同名专辑《念亲恩》中。 【歌词】 长夜空虚使我怀旧事 明月朗相对念母亲 父母亲爱心 柔善像碧月 怀念怎不悲莫禁 长夜空虚枕冷夜半泣 遥路远碧海示我心 父母亲爱心 柔善像碧月 常在心里问何日报 亲恩应该报 应该惜取孝道 惟独我离别 无法慰亲旁 轻弹曲韵梦中送 长夜空虚枕冷夜半泣 遥路远碧海示我心 父母亲爱心

柔善像碧月 常在心里问何日报长夜空虚使我怀旧事明月朗相对念母亲父母亲爱心 柔善像碧月 怀念怎不悲莫禁 长夜空虚枕冷夜半泣遥路远碧海示我心父母亲爱心 柔善像碧月 常在心里问何日报亲恩应该报 应该惜取孝道 惟独我离别 无法慰亲旁 轻弹曲韵梦中送 长夜空虚枕冷夜半泣遥路远碧海示我心父母亲爱心 柔善像碧月 常在心里问何日报

【歌手简介】 陈百强(Danny Chan)1958年9月7日生于香港,籍贯广东台山,中国香港歌手、演员。 1979年以专辑《First Love》在香港出道,同年凭借歌曲《眼泪为你流》获得香港十大中文金曲奖;1981年主演电影《失业生》。1983年凭借《今宵多珍重》获得香港十大劲歌金曲奖以及AGB观众抽签调查最受欢迎奖;同年他演唱的歌曲《偏偏喜欢你》入选香港十大中文金曲。1984年主演电影《圣诞快乐》。1985年推出个人专辑《深爱着你》。1986年主演电影《秋天的童话》。1987年由陈百强作曲的《我的故事》入选香港十大中文金曲。1988年、1989年获得两届叱咤乐坛流行榜“叱吒乐坛男歌手铜奖”。1989年凭借歌曲《一生何求》获得香港十大中文金曲以及十大劲歌金曲奖。 陈百强擅长演绎浪漫情歌,亦能自己创作歌曲。1993年10月25日,陈百强因为逐渐性脑衰竭而去世,终年35岁。XX年1月30日,在香港电台第三十二届十大中文金曲颁奖礼上,陈百强被追颁“金针奖”。 XX年8月31日,陈奕迅、梁咏琪、杨千、张敬轩、苏永康、梁汉文众星齐聚一堂以歌声纪念陈百强。

英语形容词和of for 的用法

加入收藏夹 主题: 介词试题It’s + 形容词 + of sb. to do sth.和It’s + 形容词 + for sb. to do sth.的用法区别。 内容: It's very nice___pictures for me. A.of you to draw B.for you to draw C.for you drawing C.of you drawing 提交人:杨天若时间:1/23/2008 20:5:54 主题:for 与of 的辨别 内容:It's very nice___pictures for me. A.of you to draw B.for you to draw C.for you drawing C.of you drawing 答:选A 解析:该题考查的句型It’s + 形容词+ of sb. to do sth.和It’s +形容词+ for sb. to do sth.的用法区别。 “It’s + 形容词+ to do sth.”中常用of或for引出不定式的行为者,究竟用of sb.还是用for sb.,取决于前面的形容词。 1) 若形容词是描述不定式行为者的性格、品质的,如kind,good,nice,right,wrong,clever,careless,polite,foolish等,用of sb. 例: It’s very kind of you to help me. 你能帮我,真好。 It’s clever of you to work out the maths problem. 你真聪明,解出了这道数学题。 2) 若形容词仅仅是描述事物,不是对不定式行为者的品格进行评价,用for sb.,这类形容词有difficult,easy,hard,important,dangerous,(im)possible等。例: It’s very dangerous for children to cross the busy street. 对孩子们来说,穿过繁忙的街道很危险。 It’s difficult for u s to finish the work. 对我们来说,完成这项工作很困难。 for 与of 的辨别方法: 用介词后面的代词作主语,用介词前边的形容词作表语,造个句子。如果道理上通顺用of,不通则用for. 如: You are nice.(通顺,所以应用of)。 He is hard.(人是困难的,不通,因此应用for.) 由此可知,该题的正确答案应该为A项。 提交人:f7_liyf 时间:1/24/2008 11:18:42

常用介词用法(for-to-with-of)

常用介词用法(for-to-with-of)

For的用法 1. 表示“当作、作为”。如: I like some bread and milk for breakfast. 我喜欢把面包和牛奶作为早餐。 What will we have for supper? 我们晚餐吃什么? 2. 表示理由或原因,意为“因为、由于”。如: Thank you for helping me with my English. 谢谢你帮我学习英语。 3. 表示动作的对象或接受者,意为“给……”、“对…… (而言)”。如: Let me pick it up for you. 让我为你捡起来。Watching TV too much is bad for your health. 看电视太多有害于你的健康。 4. 表示时间、距离,意为“计、达”。如: I usually do the running for an hour in the morning. 我早晨通常跑步一小时。 We will stay there for two days. 我们将在那里逗留两天。

5. 表示去向、目的,意为“向、往、取、买”等。如: Let’s go for a walk. 我们出去散步吧。 I came here for my schoolbag.我来这儿取书包。 I paid twenty yuan for the dictionary. 我花了20元买这本词典。 6. 表示所属关系或用途,意为“为、适于……的”。如: It’s time for school. 到上学的时间了。 Here is a letter for you. 这儿有你的一封信。 7. 表示“支持、赞成”。如: Are you for this plan or against it? 你是支持还是反对这个计划? 8. 用于一些固定搭配中。如: Who are you waiting for? 你在等谁? For example, Mr Green is a kind teacher. 比如,格林先生是一位心地善良的老师。

七八十年代歌曲

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The way的用法及其含义(一)

The way的用法及其含义(一) 有这样一个句子:In 1770 the room was completed the way she wanted. 1770年,这间琥珀屋按照她的要求完成了。 the way在句中的语法作用是什么?其意义如何?在阅读时,学生经常会碰到一些含有the way 的句子,如:No one knows the way he invented the machine. He did not do the experiment the way his teacher told him.等等。他们对the way 的用法和含义比较模糊。在这几个句子中,the way之后的部分都是定语从句。第一句的意思是,“没人知道他是怎样发明这台机器的。”the way的意思相当于how;第二句的意思是,“他没有按照老师说的那样做实验。”the way 的意思相当于as。在In 1770 the room was completed the way she wanted.这句话中,the way也是as的含义。随着现代英语的发展,the way的用法已越来越普遍了。下面,我们从the way的语法作用和意义等方面做一考查和分析: 一、the way作先行词,后接定语从句 以下3种表达都是正确的。例如:“我喜欢她笑的样子。” 1. the way+ in which +从句 I like the way in which she smiles. 2. the way+ that +从句 I like the way that she smiles. 3. the way + 从句(省略了in which或that) I like the way she smiles. 又如:“火灾如何发生的,有好几种说法。” 1. There were several theories about the way in which the fire started. 2. There were several theories about the way that the fire started.

keep的用法及of 、for sb.句型区别

keep的用法 1. 用作及物动词 ①意为"保存;保留;保持;保守"。如: Could you keep these letters for me, please? 你能替我保存这些信吗? ②意为"遵守;维护"。如: Everyone must keep the rules. 人人必须遵守规章制度。 The teacher is keeping order in class.老师正在课堂上维持秩序。 ③意为"使……保持某种(状态、位置或动作等)"。这时要在keep的宾语后接补足语,构 成复合宾语。其中宾语补足语通常由形容词、副词、介词短语、现在分词和过去分词等充当。如: 例:We should keep our classroom clean and tidy.(形容词) 我们应保持教室整洁干净。 You'd better keep the child away from the fire.(副词)你最好让孩子离火远一点。 The bad weather keeps us inside the house.(介词短语)坏天气使我们不能出门。 Don't keep me waiting for long.(现在分词)别让我等太久。 The other students in the class keep their eyes closed.(过去分词) 班上其他同学都闭着眼睛。 2. 用作连系动词 构成系表结构:keep+表语,意为"保持,继续(处于某种状态)"。其中表语可用形容词、副词、介词短语等充当。如: 例:You must look after yourself and keep healthy.(形容词) 你必须照顾好自己,保持身体健康。 Keep off the grass.(副词)请勿践踏草地。 Traffic in Britain keeps to the left.(介词短语)英国的交通是靠左边行驶的。 注意:一般情况下,keep后接形容词较为多见。再如: She knew she must keep calm.她知道她必须保持镇静。 Please keep silent in class.课堂上请保持安静。 3. ①keep doing sth. 意为"继续干某事",表示不间断地持续干某事,keep后不 能接不定式或表示静止状态的v-ing形式,而必须接延续性的动词。 例:He kept working all day, because he wanted to finish the work on time. 他整天都在不停地工作,因为他想准时完成工作。 Keep passing the ball to each other, and you'll be OK.坚持互相传球,你们就

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