a r
X i
v
:m
a
t h /
9
8
2
3
4
v
1
[
m
a
t h
.
F
A
]
6
F
e
b
1
9
9
8
DEFORMATION QUANTIZATION OF CERTAIN NON-LINEAR POISSON STRUCTURES BYUNG–JAY KAHNG Abstract.As a generalization of the linear Poisson bracket on the dual space of a Lie algebra,we introduce certain non-linear Poisson brackets which are “cocycle perturbations”of the linear Poisson bracket.We show that these special Poisson brackets are equivalent to Poisson brackets of central extension type,which re-semble the central extensions of an ordinary Lie bracket via Lie algebra cocycles.We are able to formulate (strict)deformation quantizations of these Poisson brackets by means of twisted group C ?–algebras.We also indicate that these deformation quantiza-tions can be used to construct some speci?c non-compact quantum groups.Introduction.Let M be a Poisson manifold.Consider C ∞(M ),the commutative algebra under pointwise multiplication of smooth functions on M .We attempt to deform the pointwise product of smooth functions into a noncommutative product,with respect to a parameter ,such that the direction of the deformation is given by the Poisson bracket on M .This problem of ?nding a deformation quanti-zation of M ([33],[1])is actually a problem dating back to the early days of quantum mechanics [34],[20].We are particularly interested in the settings where the deformed product of functions is again a function—in contrast to much of the literature on the subject involving formal power series,or the so-called “star products”.In this direction,Rie?el has been developing the no-tion of “strict”deformation quantization,in the C ?–algebra framework
[26,29].Here,in addition to the requirement that the deformed prod-uct of functions is again a function,the deformed algebra is required to have an involution and a C ?–norm.By using the C ?–algebra frame-work,one gains the advantage of being able to keep the topological and geometric aspects of the given manifold while we perform the quanti-zation.
Let h be a (?nite dimensional)Lie algebra.It is well-known [35]that the Lie algebra structure on h de?nes a natural Poisson bracket on the dual vector space h ?of h ,which is called a linear Poisson bracket .
2BYUNG–JAY KAHNG
This Poisson bracket is also called the“Lie–Poisson bracket”,to em-phasize the fact that it actually was already known to Lie.In[27] Rie?el showed that given the linear Poisson bracket on h?,a defor-mation quantization is provided by the convolution algebra structure on the simply connected Lie group H corresponding to h.In partic-ular,when h is a nilpotent Lie algebra,this is shown to be a strict deformation quantization.
In this paper,we wish to generalize the above situation and to include twisted group convolution algebras into the framework of deformation quantization.We?rst de?ne a class of Poisson brackets on the dual vector space of a Lie algebra,which contains the linear Poisson bracket as a special case.These Poisson brackets can actually be realized as “central extensions”of the linear Poisson bracket.We then show that twisted group convolution algebras provide deformation quantizations of these Poisson brackets.We obtain strict deformation quantizations when the Lie algebra is nilpotent.
In addition to its interest as a generalization of the deformation quantization problem into the non-linear situation,this result has a nice application to quantum groups.Quantum groups[13],[7]are usually obtained by suitably“deforming”ordinary Lie groups,and as suggested by Drinfeld[13],we expect to obtain quantum groups by deformation quantization of the so-called Poisson–Lie groups[19].In some cases, the compatible Poisson brackets on the Poisson–Lie groups are shown to be of our special type,in which case we can apply the result of this paper to obtain(strict)deformation quantizations of them.
This enables us to construct some speci?c non-compact quantum groups.Not only have we actually been able to show[16]that some of the earlier known examples of non-compact quantum groups[28],[31], [11],[18]are obtained in this way,but we also obtain a new class of non-compact quantum groups[15].Although the method of construction may seem rather naive,our new example is shown to satisfy some interesting properties,including the“quasitriangular”property.We will discuss our construction of quantum groups in a separate paper. This paper is organized as follows.In the?rst section,we review the de?nitions of Poisson brackets and the formulation of(strict)de-formation quantization.We also include a discussion on twisted group algebras,which are the main objects of our study.In the second section, we de?ne our special class of non-linear Poisson brackets,as motivated by the central extension of ordinary Lie brackets.We show in the third section that certain twisted group C?–algebras provide strict defor-mation quantizations of these special Poisson brackets.We use some non-trivial results on twisted group C?–algebras obtained by Packer
DEFORMATION QUANTIZATION OF POISSON STRUCTURES3 and Raeburn[21,22].We restrict our study to the strict deforma-tion quantization case,but some indications for generalization are also brie?y mentioned.
The essential part of this article is from the author’s Ph.D.thesis at U.C.Berkeley.I would like to express here my deep gratitude to Professor Marc Rie?el,without whose constant encouragement,show of interest and numerous suggestions,this work would not have been made possible.
1.Preliminaries
Let M be a C∞manifold,and let C∞(M)be the algebra of complex-valued C∞functions on M.It is a commutative algebra under point-wise multiplication,and is equipped with an involution given by com-plex conjugation.
De?nition1.1.By a Poisson bracket on M,we mean a skew,bilinear map{,}:C∞(M)×C∞(M)→C∞(M)such that the following holds:?{,}de?nes a Lie algebra structure on C∞(M).
(i.e.the bracket satis?es the Jacobi identity.)
?(Leibniz rule):{f,gh}={f,g}h+g{f,h},for f,g,h∈C∞(M). We also require that the Poisson bracket be real,in the sense that {f?,g?}={f,g}?.A manifold M equipped with such a bracket is called a Poisson manifold,and C∞(M)is a Poisson(*–)algebra. The deformation quantization will take place in C∞(M)(or to allow non-compact M,in C∞∞(M),which is the space of C∞functions van-ishing at in?nity).The Poisson bracket on M gives the direction of the deformation.Let us formulate the following de?nition for deformation quantization,which is the“strict”deformation quantization proposed by Rie?el[26].Depending on the situations,we may also be inter-ested in some other subalgebras of C∞(M),on which the deformation takes place.So the de?nition is formulated at the level of an arbitrary (dense)?–subalgebra A,for example the algebra of Schwartz functions. De?nition1.2.Let M be a Poisson manifold as above.Let A be a dense?–subalgebra(with respect to the C?–norm ∞)of C∞(M), on which{,}is de?ned with values in A.By a strict deformation quantization of A in the direction of{,},we mean an open interval I of real numbers containing0,together with a triple(× ,? , )for each ∈I,of an associative product,an involution,and a C?–norm (for× and? )on A,such that
4BYUNG–JAY KAHNG
1.For =0,the operations× ,? , are the original pointwise
product,involution(complex conjugation),and C?–norm(i.e.
sup-norm ∞)on A,respectively.
2.The completed C?-algebras A form a“continuous?eld”of C?–
algebras(In particular,the map → f is continuous for any f∈A.).
3.For any f,g∈A,
f× g?g× f
DEFORMATION QUANTIZATION OF POISSON STRUCTURES5 For convenience,let us denote byσr,r∈N the ordinary group cocycle on G de?ned byσr(x,y)=σ(x,y;r).Corresponding to the continuous?eld of cocyclesσ:r→σr,we can de?ne[25]the twisted convolution and involution on L1 G,C∞(N,A) .We de?ne,forφ,ψ∈L1 G,C∞(N,A) :
(φ?σψ)(x;r)= Gφ(y;r)αy ψ(y?1x;r) σr(y,y?1x)dy
and
φ?(x;r)=αx φ(x?1;r)? σr(x,x?1)??G(x?1).
We thus obtain a Banach?–algebra.Let us denote this algebra by
L1(G,N,A,σ),with the group actionαto be understood.We also de-
?ne C?(G,N,A,σ),the enveloping C?–algebra of L1(G,N,A,σ).There are also the notions of induced representations and regular representa-tions[5],[37],[25].So we may as well de?ne the reduced C?–algebra
C?r(G,N,A,σ).All these are more or less straightforward.
Compare this de?nition with the de?nition given in[5],where the twisted group convolution algebra has been formulated via a single cocycle.Nevertheless,the present de?nition is no di?erent from the usual one,since we can regardσalso as a single cocycle,taking values in UZM C∞(N,A) .The present formulation is useful when we study the continuity problem of the?elds of C?–algebras consisting of twisted group C?–algebras,by varying the cocycles.Recall that the continuous
?eld property is essential in the de?nition of the strict deformation quantization(De?nition1.2).
Using the universal property of full C?–algebras,and also taking ad-vantage of the property of the reduced C?–algebras that one is able to work with their speci?c representations,Rie?el in[25]gave an answer to the problem of the continuity of the?eld of C?–algebras C?(G,A,σr) r∈N, as follows.Here C?(G,A,σr)is the twisted group C?–algebra in the usual sense of[5].
Theorem1.4.Let G,A,αbe understood as above.Letσbe a contin-uous?eld over N ofα–cocycles on G.Then
?The?eld C?(G,A,σr) over N is upper semi-continuous.? C?r(G,A,σr) over N is lower semi-continuous.
?Thus,if each(G,A,α,σr)satis?es the“amenability condition”,
i.e.C?r(G,A,α,σr)=C?(G,A,α,σr),then it follows that the?eld of C?–algebras C?(G,A,σr) r∈N is continuous.
For the proof of the theorem and the related questions,we will refer the reader to[25],and the references therein.Note that by replacing
6BYUNG–JAY KAHNG
A with C∞(N,A)and by introducing a new base space,we may even consider a continuous?eld of the twisted group C?–algebras given by the cocycles of continuous?eld type(De?nition1.3).
The C?–algebra C?(G,N,A,σ)may be regarded as a C?–algebra of “cross sections”of the continuous?eld C?(G,A,σr) r∈N.It is called the C?–algebra of sections of a C?–bundle by Packer and Raeburn[22] (Compare this terminology with Fell’s notion of“C?–algebraic bun-dles”[14],which is considerably more general than is needed for our present purposes.).Actually in[22],the continuity problem of twisted group C?–algebras allowing both the cocycle and the action to vary continuously has been studied in terms of the aforementioned notion of section C?–algebra of a C?–bundle.Taking a related viewpoint, Blanchard in[3]has recently developed a framework for a general con-tinuous?eld of C?–algebras in terms of“C∞(X)–algebras”,where X in our case is the locally compact base space N.A C∞(X)–algebra is a certain C?–algebra having a C(X)module structure.See[3].
We conclude this section by quoting(without proof)a couple of deep theorems of Packer and Raeburn[21,22]on the structure of twisted group C?–algebras.We tried to keep Packer and Raeburn’s notation and terminology.Although some of them are di?erent from our no-tation,they are clear enough to understand.For example,A×α,u G denotes the twisted group C?–algebra(or“twisted crossed product”) C?(G,A,α,u).All this and more can be found in[21,22].These the-orems will be used later in the proof of our Theorem3.4,which is our main result.
Theorem1.5.([21])(Decomposition of twisted crossed products)Sup-pose that(A,G,α,u)is a separable twisted dynamical system and N is a closed normal subgroup of G.There exists a canonically determined twisted action(β,v)of G/N on A×α,u N such that:
A×α,u G～=(A×α,u N)×β,v G/N.
The next theorem is about the continuity of a?eld of twisted group C?–https://www.doczj.com/doc/5716786626.html,pare this with Theorem1.4,where we considered the continuity problem only when the twisting cocycle is varying.Mean-while,note in the theorem that G is assumed to be amenable(So by the“stabilization trick”of Packer and Raeburn[21],the amenability condition always holds for any quadruple(G,A,α,u).).
Theorem1.6.([22])Suppose A is the C?–algebra of sections of a sep-arable C?–bundle over a locally compact space X,and(α,u)is a twisted action of an amenable locally compact group G on A such that each ideal I x={a∈A:a(x)=0}is invariant.Then for each x∈X,there is a
DEFORMATION QUANTIZATION OF POISSON STRUCTURES 7natural twisted action α(x ),u (x )
on the quotient A/I x ,and A ×α,u G is the C ?–algebra of sections of a C ?–bundle over X with ?bers isomor-phic to (A/I x )×α(x ),u (x )G .
2.The non-linear Poisson bracket
Let us begin by trying to characterize the special Poisson brackets that will allow twisted group algebras to be deformation quantizations of them.Recall that the twisting of the convolution algebra structure in a twisted group algebra is given by group (2–)cocycles.Meanwhile any group cocycle for a locally compact group G having values in an abelian group N can be canonically associated with a central extension of G by N ,and actually all central extensions are essentially obtained in this way [14].
Since it is known [27]that ordinary group convolution algebras can be regarded as deformation quantizations of linear Poisson brackets,the above observations suggest that twisted group algebras will provide deformation quantizations of certain Poisson brackets which are,in a loose sense,“central extensions”of linear Poisson brackets.Although we have to make clear what we mean by this last statement,this is the main motivation behind the de?nition of our special type of Poisson bracket formulated below.
Let h be a (?nite–dimensional)Lie algebra and let us denote by g =h ?its dual vector space.As usual,we will denote the dual pairing between h and g by , .We will have to require later that h is a nilpotent or an exponential solvable Lie algebra because of some technical reasons to be discussed below,but for the time being we allow h to be a general Lie algebra.Recall [35]that we de?ne the linear Poisson bracket on the dual vector space g =h ?by
{φ,ψ}lin (μ)= [dφ(μ),dψ(μ)],μ
(2.1)where φ,ψ∈C ∞(g )and μ∈g .Here dφ(μ)and dψ(μ)has been naturally realized as elements in h .
We wish to de?ne a generalization of this Poisson bracket by allowing a suitable “perturbation”of the right-hand side of equation (2.1).This will be done via a certain Lie algebra 2–cocycle on h ,denoted by ?,having values in C ∞(g ).That is,we will consider the Poisson brackets of the form:
{φ,ψ}(μ)= [dφ(μ),dψ(μ)],μ +? dφ(μ),dψ(μ);μ .
(2.2)As above,we regard dφ(μ)and dψ(μ)as elements in h .
8BYUNG–JAY KAHNG
Compare equation(2.2)with the de?nition of the linear Poisson bracket.In the linear Poisson bracket case,the Lie bracket takes val-ues in h,the elements of which can be regarded as(linear)functions contained in C∞(g),via the dual pairing.That is,the right-hand side of equation(2.1)can be viewed as the evaluation atμ∈g of a C∞–function,[X,Y]∈h?C∞(g),where X=dφ(μ)and Y=dψ(μ). In the“perturbed”case,the right-hand side of equation(2.2)may be viewed as the evaluation atμ∈g of a C∞–function,[X,Y]+?(X,Y)∈C∞(g),where X=dφ(μ)and Y=dψ(μ).So to make sense of the Poisson brackets of the type given by equation(2.2),we will?rst study the“perturbation”of the Lie bracket on h by a https://www.doczj.com/doc/5716786626.html,ter,we will?nd some additional conditions for the cocycle?such that the equation(2.2)indeed gives a well-de?ned Poisson bracket on g.
Let V be a U(h)–module,possibly in?nite dimensional.Consider a 2–cocycle?for h having values in V.It is a skew-symmetric,bilinear map from h×h into V such that d?=0(For more discussion on cohomology of Lie algebras,see the standard textbooks on the subject [6,§5],[17].).When V is further viewed as an abelian Lie algebra,the space h⊕V can be given a Lie algebra structure[4],[17]which becomes a central extension Lie algebra of h by V:
(X,v),(Y,w) h⊕V= [X,Y],X·w?Y·v+?(X,Y)
for X,Y∈h and v,w∈V.Here the dot denotes the module action.In particular,when V is assumed to be a trivial U(h)–module,we have: (X,v),(Y,w) h⊕V= [X,Y],?(X,Y) .
(2.3)
Let us slightly modify this“central extension”picture as follows,so that we are able to consider a Lie bracket on h+V,where we now allow h∩V=0in general.Clearly,h∩V is a subspace of h.However,since h is already equipped with its given Lie bracket and since V will be assumed to be an abelian Lie algebra,it is only reasonable to consider the case in which h∩V is an abelian subalgebra of h.For simplicity,we will further assume that h∩V is a central subalgebra of h,which means that V is a trivial U(h)–module.Let us denote this central subalgebra by z.Without loss of generality,we may assume that z is the center of h.In this case,we just replace V by an extended abelian Lie algebra, still denoted by V,satisfying h∩V=z.
Let us look for a trivial U(h)–module V,which we will view as an abelian Lie algebra,such that h∩V=z is the center of h.Since we eventually want to de?ne a V–valued cocycle for h,from which we construct a bracket operation on C∞(g),we also require that V is contained in C∞(g).So let us consider the subspace q=z⊥of g,and
DEFORMATION QUANTIZATION OF POISSON STRUCTURES 9choose as our V the following:
V =C ∞(g /q )?C ∞(g ).
Here the functions in V =C ∞(g /q )have been realized as functions in C ∞(g ),by the “pull-back”using the natural projection p :g →g /q .Since any X ∈h can be regarded as a linear function contained in C ∞(g )via the dual pairing,we can see easily that X ∈h ∩V ?C ∞(g )if and only if X,μ+ν = X,μ for all μ∈g ,ν∈q .It follows imme-diately that h ∩V =z .On the other hand,consider the representation ad ?.For any X ∈h and any μ∈g ,we have ad ??X (μ)=ν∈q ,since for any Y ∈z ,we have Y,ν = Y,ad ??X (μ) = [X,Y ],μ =0.It follows that ad ?X (f )(μ)=f ad ??X (μ) =f (ν)=0,
for any f ∈V =C ∞(g /q ).By the natural extension of ad ?to U (h ),we can give V the trivial U (h )–module structure.
Remark.When h has a trivial center,the space V will be just {0}.To avoid this problem,we could have considered C ∞(g )H ,the space of Ad ?H –invariant C ∞functions on g .It is always nonempty (It contains the so-called Casimir elements [32].).It also satis?es h ∩C ∞(g )H =z and can be given the trivial U (h )–module structure.However,it does not satisfy the following property (Lemma 2.1),which we need later when we de?ne our Poisson bracket.For this reason,we choose our V as it is de?ned above.At least for nilpotent h ,which is the case we are going to study most of the time,this is less of a problem since h has a non-trivial center.
Lemma 2.1.Let V ?C ∞(g )be de?ned as above.Then for any func-tion χ∈V and for any μ∈g ,we have:dχ(μ)∈z .
Proof.Since V ?C ∞(g ),it follows that dχ(μ)∈h .Recall that dχ(μ)de?nes a linear functional on g =h ?by dχ(μ),ν =
d dt t =0
χ(μ)=0.Since ν∈q is arbitrary,we thus have:dχ(μ)∈z .
10BYUNG–JAY KAHNG
Let us now turn to the discussion of de?ning a(perturbed)bracket operation on h+V,which will enable us to formulate our special Poisson bracket on C∞(g).Since h and V are subspaces of h+V,there exists a (linear)surjective map,h→(h+V)/V,whose kernel is h∩V=z.We thus obtain a vector space isomorphism,in a canonical way,between (h+V)/V and h/z.The map from h+V onto h/z is a canonical one, which extends the canonical projection of h onto h/z.Therefore,it is reasonable to consider a cocycle for h/z having values in V(viewed as a trivial U(h/z)–module)and use it to de?ne a bracket operation on h+V.Let?be such a cocycle for h/z.
Remark.Note that in this setting,the cocycle?can naturally be iden-ti?ed with a cocycle??for h having values in V(considered as a trivial U(h)–module),satisfying the following“centrality condition”:
??(Z,Y)=??(Y,Z)=0,
(2.4)
for any Z∈z and any Y∈h.In fact,we may de?ne??as??(X,Y)=?(˙X,˙Y),where˙X denotes the image in h/z of X under the canonical projection.For this reason,we will from time to time use the same notation,?,to denote both?and??.
By viewing?as a cocycle for h,we can de?ne,as in equation(2.3), a Lie bracket on h⊕V:
(X,v),(Y,w) h⊕V= [X,Y],?(X,Y) .
To de?ne a bracket operation on h+V,consider the natural surjective map from h⊕V onto h+V,whose kernel is:
δ={(Z,?Z):Z∈z}?h⊕V.
Sinceδis clearly central with respect to the Lie bracket[,]h⊕V given above,it is an ideal.Therefore,h+V=(h⊕V)/δis a Lie algebra. The Lie bracket on it is given by:
(2.5)
[X+v,Y+w]h+V=[X,Y]+?(X,Y),X,Y∈h,v,w∈V which is the given Lie bracket on h plus a cocycle term.In this sense, equation(2.5)may be considered as a“perturbed Lie bracket”of the given Lie bracket on https://www.doczj.com/doc/5716786626.html,pare this with equation(2.2),where the linear Poisson bracket(given by the Lie bracket on h)is“perturbed”by a certain cocycle?.
Using the observation given above as motivation,let us de?ne more rigorously our Poisson bracket on g=h?.This is,in fact,a“cocycle perturbation”of{,}lin on g.
DEFORMATION QUANTIZATION OF POISSON STRUCTURES11 Theorem2.2.Let h be a Lie algebra with center z and let g=h?be the dual vector space of h.Consider the vector space V=C∞(g/q)?C∞(g)as above,where q=z⊥.Let us give V the trivial U(h)–module structure.Let?be a Lie algebra2–cocycle for h having values in V, satisfying the centrality condition.That is,?is a skew-symmetric, bilinear map from h×h into V such that:
? X,[Y,Z] +? Y,[Z,X] +? Z,[X,Y] =0,X,Y,Z∈h satisfying:?(Z,Y)=?(Y,Z)=0for Z∈z and any Y∈h.Then the bracket operation{,}?:C∞(g)×C∞(g)→C∞(g)de?ned by {φ,ψ}?(μ)= [dφ(μ),dψ(μ)],μ +? dφ(μ),dψ(μ);μ is a Poisson bracket on g.
Remark.If we denote dφ(μ)and dψ(μ)by X and Y,as elements in h,the right-hand side of the de?nition of the Poisson bracket may be viewed as the evaluation atμ∈g of a C∞–function,[X,Y]+?(X,Y)∈h+V∈C∞(g).Note that this expression is just the Lie bracket on h+V de?ned earlier by equation(2.5).
Proof.Since dφ(μ)and dψ(μ)can be naturally viewed as elements in h,it is easy to see that{,}?is indeed a map from C∞(g)×C∞(g) into C∞(g).The skew-symmetry and bilinearity are clear.
To verify the Jacobi identity,consider the functionsφ1,φ2,φ3in C∞(g).We may write{φ2,φ3}?as:
{φ2,φ3}?(μ)={φ2,φ3}lin(μ)+χ(μ),
whereχis a function in V.We therefore have:
d {φ2,φ3}? (μ)= dφ2(μ),dφ3(μ) +dχ(μ).
The?rst term in the right hand side is the di?erential of the linear Poisson bracket,which is rather well known[35].Moreover,sinceχ∈V,it follows from Lemma2.1that dχ(μ)∈z,which is“central”with respect to both[,]and?.We thus have:
φ1,{φ2,φ3}? ?(μ)
= dφ1(μ),d({φ2,φ3}?)(μ) ,μ +? dφ1(μ),d({φ2,φ3}?)(μ);μ = dφ1(μ),[dφ2(μ),dφ3(μ)] ,μ +? dφ1(μ),[dφ2(μ),dφ3(μ)];μ , and similarly for φ2,{φ3,φ1}? ?and φ3,{φ1,φ2}? ?.So the Jacobi identity for{,}?follows from that of the Lie bracket[,]and the cocycle identity for?.That is,
φ1,{φ2,φ3}? ?(μ)+ φ2,{φ3,φ1}? ?(μ)+ φ3,{φ1,φ2}? ?(μ)=0.
12BYUNG–JAY KAHNG
Finally,since d(φψ)=(dφ)ψ+φ(dψ)for anyφ,ψ∈C∞(g),the Leibniz rule for the bracket is also clear.
Remark.This is the special type of Poisson bracket we will work with from now on.The linear Poisson bracket{,}lin on g=h?is clearly of this type,since it corresponds to the case when the cocycle?is trivial.Meanwhile when?is a scalar-valued cocycle,we obtain the so-called a?ne Poisson bracket[2],[30].A?ne Poisson structures oc-cur naturally in the study of symplectic actions of Lie groups with general(not necessarily equivariant for the coadjoint action)moment mappings.The notion of a?ne Poisson brackets has generalization also to the groupoid level.See[12]or[36].
Suppose we are given a Poisson bracket of our special type,{,}?, on g=h?.To discuss its(strict)deformation quantization,it is useful to observe that{,}?can be viewed as a“central extension”of the linear Poisson bracket on the dual vector space of the Lie algebra h/z. This actually follows from the fact that the Lie bracket[,]h+V given by equation(2.5)can be transferred to a Lie bracket on h/z⊕V,which turns out to be a central extension of the Lie bracket on h/z.Let us make this observation more precise.
Consider the exact sequence of Lie algebras,
0→zι→hρ→h/z→0
such thatιandρare the injection and the quotient map,respectively. Let us?x a linear mapτ:h/z→h such thatρτ=id.In this case,the exactness implies that the map
ω0:(x,y)→ι?1 [τ(x),τ(y)]?τ([x,y]h/z)
(2.6)
is well-de?ned from h/z×h/z into z,and it is actually a Lie algebra cocycle for h/z having values in z.See[4],[17].Then the Lie bracket on h can be written as follows:
[X,Y]=τ [ρ(X),ρ(Y)]h/z +ι ω0(ρ(X),ρ(Y)) ,X,Y∈h.
(2.7)
Since we have been regarding the center z as a subalgebra of h such that h∩V=z,we may ignore the mapιand view z and its image in h or V as the same.Thenτis actually the map that determines the vector space isomorphism between h/z⊕V and h+V.Under this isomorphism and by using equation(2.7),the Lie bracket[,]h+V of equation(2.5)is transferred to a Lie bracket on h/z⊕V de?ned by:
(2.8)
(x,v),(y,w) h/z⊕V=[x,y]h/z+ω0(x,y)+?(x,y)=[x,y]h/z+ω(x,y).
DEFORMATION QUANTIZATION OF POISSON STRUCTURES13 Hereω0is the cocycle de?ned in equation(2.6)and we regarded?as a cocycle for h/z,as assured by an earlier remark.For convenience,we introduced a new(V–valued)cocycleωfor h/z,as a sum of the two cocyclesω0and?.Then it is clear that equation(2.8)de?nes a central extension of the Lie bracket[,]h/z,where the extension is given by the cocycleω.We can now de?ne a Poisson bracket on g modeled after this central extension type Lie bracket.
Theorem2.3.Let h be a Lie algebra with center z and let us?x the mapsρandτgiven above.Let g=h?.Consider the vector space V?C∞(g)de?ned above and let us give V the trivial U(h/z)–module structure.Supposeωis a Lie algebra cocycle for h/z having values in V.Then the bracket operation{,}ω:C∞(g)×C∞(g)→C∞(g) de?ned by
{φ,ψ}ω(μ)= τ([˙dφ(μ),˙dψ(μ)]h/z),μ +ω ˙dφ(μ),˙dψ(μ);μ is a Poisson bracket on g.Here˙X denotes the image of X under the canonical projectionρof h onto h/z.
Proof.De?ne?:h/z×h/z→V by
(2.9)
?(x,y)=ω(x,y)?ω0(x,y),
whereω0is the cocycle for h/z de?ned in equation(2.6).Then the discussion in the previous paragraph implies that the bracket{,}ωis equivalent to the Poisson bracket{,}?given in Theorem2.2.There-fore,it is clearly a Poisson bracket on g.
Although the present formulation depends on the choice of the map τand hence is not canonical,this Poisson bracket is,by construction, equivalent to the canonical Poisson bracket given in Theorem2.2.The relationship between them is given by equation(2.9).In particular,if we consider the cocycleω0of equation(2.6)in place ofω,so that?is trivial,we obtain the linear Poisson bracket{,}lin on g.There-fore,to?nd a(strict)deformation quantization of our Poisson bracket {,}?of Theorem2.2,we may as well try to?nd a(strict)deforma-tion quantization of the central extension type Poisson bracket{,}ω. This change in our point of view is useful when we work with speci?c examples,where the choice of coordinates are usually apparent.
3.Twisted group C?–algebras as deformation
quantizations
As we mentioned earlier,we expect that twisted group C?–algebras will be deformation quantizations of the Poisson brackets of“central extension”type.These are in fact the special type of Poisson brackets
14BYUNG–JAY KAHNG
we de?ned in the previous section.A more canonical description has been given in Theorem2.2,while an equivalent,“central extension”type description has been given in Theorem2.3.
Let us from now on consider the Poisson bracket{,}ωon g=h?, as de?ned in Theorem2.3.For convenience,we will?x the map τ:h/z→h and identify h/z with its imageτ(h/z)?h underτ. To?nd a deformation quantization of{,}ω,we will look for a group (2–)cocycle,σ,for the Lie group H/Z of h/z,corresponding to the Lie algebra cocycleω.Then we will form a twisted group C?–algebra of H/Z withσ,which we will show below will give us a strict deformation quantization of C∞(g)in the direction of{,}ω.By the equivalence of the Poisson brackets{,}ωand{,}?,this may also be interpreted as giving a strict deformation quantization of C∞(g)in the direction of{,}?.This result will be a generalization of the result by Rie?el [27]saying that an ordinary group C?–algebra C?(H)provides a defor-mation quantization of C∞(h?)in the direction of the linear Poisson bracket on h?.
Recall that the cocycleωprovides a Lie bracket on the space h/z⊕V. If we restrict this Lie bracket to h/z,we obtain the map[,]ω:h/z×h/z→h/z⊕V de?ned by:
[x,y]ω=[x,y]h/z+ω(x,y).
(3.1)
For the time being,to make our book keeping simpler,let us denote by k and K the Lie algebra h/z and its Lie group H/Z.We now try to construct a group-like structure corresponding to[,]ω.From equation(3.1),we expect to obtain a cocycle extension of the Lie group K=H/Z via a certain group cocycle corresponding toω.As a?rst step,let us consider the following Baker–Campbell–Hausdor?series for k⊕V,ignoring the convergence problem for the moment.De?ne
S(X,Y)=X+Y+1
12
[X,[X,Y]k⊕V]k⊕V+
1
S( X, Y)for
=0in R.For =0,we let S
(X,Y)=X+Y.
Lemma3.1.Let ∈R be?xed and let S be as above.Then we have, at least formally(ignoring the convergence problem),
S X,S (Y,Z) =S S (X,Y),Z
S (X,?X)=0,S (X,0)=S (0,X)=X
for X,Y,Z∈k⊕V.
DEFORMATION QUANTIZATION OF POISSON STRUCTURES15 Proof.Since(X,Y)→1
16BYUNG–JAY KAHNG
Also from the lemma,we have:
R (x,?x)=0,R (x,0)=R (0,x)=0.
As we mentioned above,this proposition does not make much sense unless we clear up the convergence problem of the series S .Note that the“multiplication”,? ,on k is only locally de?ned.The de?nition of R (,)is even more di?cult because of the fact that it takes values in an in?nite dimensional vector space V.Fortunately,in some special cases,these convergence problems do become simpler.Let us mention a few here.
When the cocycle is known to take values in a?nite dimensional subspace W of V,the space k+W becomes a?nite dimensional Lie algebra such that the convergence of S is obtained at least locally in a neighborhood of0.But this case is rather uninteresting,because the cocycle extension becomes just another(?nite dimensional)Lie algebra.The corresponding Poisson bracket obtained as in Theorem 2.3is just the linear Poisson bracket on the dual space of this extended Lie algebra.In particular,whenω=ω0,the series S becomes just the Baker–Campbell–Hausdor?series for the Lie algebra h/z+z=h and the corresponding Poisson bracket is the linear Poisson bracket on h?. Meanwhile,when the Lie algebra k is nilpotent,the series S becomes a?nite series and hence always converges,whether or not the cocycle ωtakes values in an in?nite dimensional vector space.In particular, x? y and R (x,y)can be de?ned for any x,y∈k.
Despite this attention to detail which we have to make,there are still possibilities for generalization.There are some special cases of ex-ponential solvable Lie algebras that do not fall into one of these cases but whose convergence problem(at least locally)can still be managed. Meanwhile in a formal power series setting,since the convergence prob-lem becomes less crucial,the result of the above proposition is still valid for any Lie algebra.But in these general settings,we no longer expect to obtain“strict deformation quantizations”as we do below.For this, some generalized notion of a group cocycle,in a“local”sense,needs to be developed.This search for a correct,weaker notion of deformation quantizations,will be postponed as a future project.
Since we wish to establish a“strict”deformation quantization(in the sense of De?nition1.2)of our special type of a Poisson bracket,we will consider the case when the Lie algebra k is nilpotent.So from now on,let us assume that the Lie algebra h is nilpotent.Then k=h/z also becomes nilpotent.Let us choose and?x a basis for h(for example,we can take the“Malcev basis”[10])such that elements of z can be written
DEFORMATION QUANTIZATION OF POISSON STRUCTURES 17as z =(0,z )and elements of k =h /z can be written as x =(x,0).Recall that we have de?ned our V as the space C ∞(g /q ),where q is the subspace q =z ⊥of g .
By Proposition 3.2,we obtain a group cocyle R for the nilpotent Lie group K = h /z ,? .Here and from now on,if g is a Lie algebra with the corresponding (simply connected)Lie group G ,we will denote by G for the (simply connected)Lie group corresponding to g whose Lie bracket is given by [,].Since R is a continuous function–valued cocycle having values in V =C ∞(g /q ),it is more convenient to introduce instead the following (continuous)family of ordinary T –valued cocycles,r →σr .Below and through out the rest of the paper,e (t )denotes the function exp (2πi )t .Also ˉe (t )=exp (?2πi )t .The proof of the following proposition is immediate from Proposition 3.2.
Proposition 3.3.For a ?xed r ∈g /q ,de?ne the map σr :h /z ×h /z →
T by σr (x,y )=ˉe R (x,y ;r ) =exp (?2πi )R (x,y ;r )
where R (x,y ;r )is the evaluation at r of R (x,y )∈C ∞(g /q ).Then σr is a smooth normalized group cocycle for K having values in T .That is,
σμ (y,z )σμ (x,y ? z )=σμ (x,y )σμ (x ? y,z )
σμ (x,0)=σμ (0,x )=1
for x,y,z ∈K = h /z ,? .Moreover,if we ?x x,y ∈K ,then μ→σμ (x,y )is a C ∞–function from g /q into T .
Remark.Since the functions in V =C ∞(g /q )are invariant under the coadjoint action of H ,by an observation made earlier,they are also invariant under the coadjoint action of K .Thus the cocycle condition of the proposition can also be interpreted as the condition for a normal-ized α–cocycle in De?nition 1.3,where αin this case is the coadjoint action of K .Although this interpretation is not directly needed in our discussion below,this still suggests a possibility of future generaliza-tion.
Since σ :r →σr is a continuous ?eld of normalized T –cocycles
(De?nition 1.3)for the Lie group K ,it follows that we can,as in section 1,de?ne a twisted convolution algebra L 1 K ,C ∞(g /q ) .For f,g in L 1 K ,C ∞(g /q ) ,we have:(f ?σ g )(y ;r )= h /z
f (x ;r )
g (x ?1? y ;r )σr (x,x ?1? y )dx.
(3.3)
18BYUNG–JAY KAHNG
Here dx denotes the left Haar measure for the(nilpotent)Lie group
K =(h/z,? ),which is just a?xed Lebesgue measure for the under-lying vector space k=h/z.We will show below that as approaches 0,the family of twisted convolution algebras L1 K ,C∞(g/q) ∈R provides a deformation quantization of our Poisson bracket{,}ωon
g.Since we prefer to?nd a deformation at the level of continuous functions on g,we need to develop suitable machinary.
Choose a Lebesgue measure,dX,on h and Lebesgue measures,dx and dz,on h/z and z respectively,such that we have:dX=dxdz. These Lebesgue measures will be Haar measures for the(nilpotent)Lie groups corresponding to the Lie algebras h,h/z,and z.In particular, Haar measure for K =(h/z,? )is a Lebesgue measure dx for h/z. Meanwhile,we may write the dual vector space g=h?as g=q⊕(g/q), a direct product of subspaces,where q=z⊥.By elementary linear algebra,we can realize q and g/q as dual vector spaces of k=h/z and z,respectively.Therefore,we are able to choose dual(Plancherel) measures,dq and dr,for q and g/q such that dμ=dqdr becomes a Plancherel measure for g=h?.All these measures are essentially Lebesgue measures.
We then de?ne the Fourier transform,F,between the spaces of
Schwartz functions S(h)and S(g)by
(F f)(μ)= h f(X)ˉe X,μ dX,f∈S(h)
and the inverse Fourier transform,F?1,from S(g)to S(h)by
(F?1φ)(X)= gφ(μ)e X,μ dμ,φ∈S(g).
Here again,e(t)=exp (2πi)t andˉe(t)=exp (?2πi)t .Our choice of the Plancherel measure means that we have F?1(F f)=f for all f∈S(h)and F(F?1φ)=φfor allφ∈S(g).This is the Fourier inversion theorem.Let us also de?ne the partial Fourier transform,∧, from S(h/z×g/q)to S(g)=S(q×g/q)by
f∧(q;r)= h/z f(x;r)ˉe x,q dx.
Its inverse Fourier transform∨is similarly de?ned from S(g)to S(h/z×g/q)by replacingˉe with e in the de?nition.Again,we have the Fourier inversion theorem:(f∧)∨=f for all f∈S(h/z×g/q)and(φ∨)∧=φfor allφ∈S(g).
DEFORMATION QUANTIZATION OF POISSON STRUCTURES19 We are now ready to state and prove our main theorem.We show that given our Poisson bracket{,}ωon g,its strict deformation quan-tization is essentially given by a family of twisted group C?–algebras.
Theorem3.4.Let h be a nilpotent Lie algebra.Let the notation be as above and letωbe a Lie algebra cocycle for h/z having values in V=C∞(g/q).Suppose that g=h?is equipped with a Poisson bracket of our special type,{,}ω,as in Theorem2.3.Then there exists a dense(with respect to the usual ∞norm)subspace,A?S(g)of C∞(g)such that for a?xed ∈R,the following operation,× ,de?ned by:
(φ× ψ)=(φ∨?σ
ψ∨)∧,φ,ψ∈A
(3.4)
is a well-de?ned multiplication on A.Here?σ
is the twisted convolu-tion de?ned in equation(3.3).Moreover,the following properties hold:?We can de?ne a suitable involution,? ,and a C?–norm, ,on
A such that the C?–completion of(A,× ,? )with respect to
de?nes a C?–algebra A .
?For ∈R,the C?–algebras A form a continuous?eld of C?–algebras.
? A,× ,? , ∈R is a strict deformation quantization of A?C∞(g)in the direction of the Poisson bracket(1/2π){,}ωon g.
In particular,we have:
φ× ψ?ψ× φ2π{φ,ψ}ω →0
(3.5)
as →0.
Proof.(Step1).The twisted convolution,equation(3.3),has been de?ned between functions on h/z×g/q.To de?ne a multiplication between functions on g,we use the partial Fourier transform to transfer the twisted convolution to S(g). Although S(h/z×g/q)?L1 K ,C∞(g/q) ,it is in general not true that S(h/z×g/q)is an algebra under the twisted convolution?σ
, unless the cocycle is trivial.Still,at least on C∞c(h/z×g/q),the C∞–functions on h/z×g/q with compact support,the twisted convolution is closed.This result actually corresponds to a similar result in the trivial cocycle case(i.e.the crossed products[24]),and the proof is also done by straightforward calculation.So for the purpose of proving the theorem,we may take C∞c(h/z×g/q)as the subspace on which the twisted convolution is closed.
We will let A be the image of C∞c(h/z×g/q)in S(g)under the partial Fourier transform,∧.By the inverse partial Fourier transform,∨,the
20BYUNG–JAY KAHNG
subspace A is carried back onto C∞c(h/z×g/q).Therefore,it follows immediately that equation(3.4)de?nes a closed multiplication on A?S(g).Since C∞c(h/z×g/q)is dense in S(h/z×g/q)?L1 K ,C∞(g/q) with respect to the L1–norm,it is clear that A is dense in S(g)?
C∞(g/q)with respect to the ∞norm.
Since we have de?ned our deformed multiplication on A to be isomor-phic to the twisted convolution on C∞c(h/z×g/q)?L1 h/z,C∞(g/q) , we may also transfer other structures on the twisted convolution al-gebra to A via partial Fourier transform.On the twisted convolution algebra,the involution is given by the following formula:
f?(x;r)=
be sure的用法有多少 对be sure的用法常感困惑不解，现将其主要用法归纳如下。 1．be sure＋of／about+动名词或名词，意为“确信……”；“对……有把握”。但在接名词时，be sure of侧重指主语对某抽象事物的确信无疑；而besure about则侧重指主语对某具体事物的确信无疑。例如： He is sure of success．他自信会成功的。 I'm sure of his honesty．我肯定他是诚实的。 However，he went on to explain that he was not sure about two things—thegrammar and some of the idioms．不过，他接着就解释说，他在语法和某些习惯用语方面都没多大把握。 If you are not sure about the situation in the world，you can read the newspaper every day．如果你对世界形势不大了解，你可以每天看看报纸。 若后面要接反身代词，则只能用be sure of，即be sure of oneself，意为“有自信心”。例如： Joan will sit for an important examination next week，but she is not sure of herself．琼下周要参加一个重要考试，但她对自己没有十分把握。 2．be sure+不定式，表示说话人对句子主语作出的判断，认为句子主语“必定”、“必然会”、“准会”如何如何。例如： It is sure to rain．天一定会下雨。 He is sure to have known about that．他准会知道那件事。 He told me I was sure to get a warm welcome．他告诉我，说我准会受到热情接待的。 然而be sure to do用在祈使句中，不是表示判断，而是表示对对方的要求，意为“务必要”，“一定要”。例如： Be sure not to forget it！千万别忘记呀！ Be sure to send my regards to your mother．务必代我问候你母亲。 Be sure not to do that again．一定不要再干那种事了。 3．be sure+宾语从句，表示主句主语对宾语从句中涉及的事物所作出的判断，意为“确信某事一定会……”。例如： I'm not sure whether I've met him before．我不能确定以前是否见到过他。 We are sure he will make great progress this term．我们确信他这学期一定会取得巨大进步。 sure一词在中学英语中是一个常见的词，但你对它的用法掌握了多少呢？本期我们就来谈谈sure 的一些用法。 1. 用作形容词
disappointed的用法小结 今天给大家整理了disappointed的用法，快来一起学习吧，下面就和大家分享，来欣赏一下吧。 disappointed用法 disappointed的解释 a. 失望的 disappointed的例句 The condition or feeling of being disappointed. 失望，扫兴失望的状态或沮丧的感觉 If you were expecting our life in the country to be the New Jerusalem, youre going to be very disappointed. 如果你期望我们在乡下的生活会像天堂那样美丽，那你是会非常失望的。 失望别再只说disappointed，这些才是老外最常用的表达
我们竭尽全力，希望都与人友好相处。不幸的是，大家并非总能如此，因此我们会有表达失望之情的需要，对自己或对别人失望。 对于这些情况，记住表达失望情绪的正确方式就很重要了。 换句话说，我们谈话的对象是谁?彼此是什么关系?我们会根据自己是在交友或工作的场合来使用不同的短语。 对自己感到失望： I wish I (我希望自己)+ Past Simple (一般过去时)= Present Disappointments 表达现在的失望情绪 将I wish(我希望)与一般过去时连用，以表达你对现在发生的某件事感到失望。这与使用非真实条件句来表达假想中的事相类似。 I wish I had a better job. 我希望能找到更好的工作。 I wish I had more time for my family. 我希望自己能有更多的时间陪伴家人。 I wish I spoke Italian. 我希望自己会说意大利语。
英语中的sure和certain是近义形容词，均有“肯定的；确信的；有把握的”的意思，它们的用法也大体相似，许多情况下可以通用，但它们之间也有一些差别，现归纳如下： 一、相同点 1. 两者都能用于“be sure/certain + about /of短语”句型中，表示“对（某事）有把握”。主语必须是人，about/of之后多跟名词、代词。如： I am sure/certain of his returning. 我确信他会回来。 He is quite sure/certain of /about it. 他对这事很有把握。 2. 两者都能用于be sure/certain to do sth句型中，表示“一定会做某事”，主语可以是人，也可以是物。如： Spring is sure/certain to follow winter. 冬天过后一定是春天。 They are sure/certain to come. 他们一定会来的。 3. 两者都能用于“be sure/certain +从句”句型中，表示“确信……”，主语只能是人，而不能是物。如： Tom is sure/certain that I put the key on the table. 汤姆确信我把钥匙放在桌子上了。 We are sure/certain that the book will be of great help to us.我们相信这本书对我们会有很大帮助。 4. 两者都能用于“make sure/certain +从句”结构中，表示“确定；弄清楚”，主语只能是人。如： You must make sure/certain when the bus will leave. 你必须弄清汽车何时出发。 5. “I ' m not sure…”与“I ' m not certain…”意思相同。如： I ' m not sure/certain where he is. 我不能确定他在哪儿。 I ' m not sure/certain how to do it. 我不能确定如何做这件事。
remind vt.使想起；提醒 remind sb．to do sth．提醒某人做某事 remind sb．从句提醒某人…… remind sb．of／about sth．提醒某人某事 reminder n．使人忆起某事的人或物；提示 The old photos remind me of the old days． 使想起 If my father forgets it, I hope you would remind him. 如果我父亲忘了的话, 希望你能提醒他一下。 When I reminded her, she nodded her head. 我提醒她时, 她点了点头。 Please remind me in case I should forget. 如果我忘了, 请你提醒我一下。 The lady didn't like to be reminded. 那位太太不喜欢人们提醒她。 Be sure to remind her to come back early. 一定要提醒她早点回来。 He reminded himself to thank her for the present. 他提醒自己要感谢她送的礼物。 He reminded me to lock up the house when I go out. 他提醒我外出时把屋子锁起来。 These notes will remind you what to say.
这些笔记可以帮你记起要说哪些话。 These notes will remind you how to present your ideas. 这些笔记可以帮你记起怎样讲清楚你的意思。 He reminded me when to go. 他提醒我是不是我该走了。 She reminded me that I hadn't written to Mother. 她提醒我还没有给母亲写信。 That suddenly reminded her that she had promised to ring them up. 这突然使她想起她说过要给他们打电话的。 Please remind him that we are leaving at six tomorrow morning. 请提醒他我们明晨6时出发。 I must remind him that time is money. 我必须提醒他时间就是金钱。 That reminds me that I have a meeting to attend this evening. 这使我想起今晚我还有一个会要参加。 May I remind you that time will soon be up? 我可以提醒你时间快到了吗? The sight of the clock remind-ed me that I was late. 看到时钟使我想起我迟到了。
Be sure 和make sure 怎样区别 be sure是一种潜意识,意为确定 I'm sure that no one will love me more than he does. 我确定没有人会比他更爱我了. make sure则是主观地想要证实某事,即确认 I just want to make sure that everything is ok. 我只想确认一切安好. 翻译地有些牵强,但一时也想不起什么更贴切的词汇了. Be sure意思是已经sure了： I am sure you will do well on the exam if you study hard. Are you sure he's the guy you were talking about? Make sure强调过程，从不sure到sure或从sure到更sure的过程: Make sure you will bring your umbrellas tomorrow in case it rains. I will make sure that everyone gets a book. make sure... make sure表示“务必”，“确信”，“弄明白”，后面常接of/about sth.或that引导的宾语从句。例如: Make sure(that) you will arrive there on time.你务必准时到这。 I know there s a train this afternoon，but I must make sure of the time.我知道今天下午有列火车，但我必须弄清楚(发车)时间。 注意:make sure后通常不接不定式。 误:Make sure to shut the windows. 正:Make sure that you shut the windows. 正:Make sure that the windows are shut. be sure的用法有多少 对be sure的用法常感困惑不解，现将其主要用法归纳如下。 1．be sure＋of／about+动名词或名词，意为“确信……”；“对……有把握”。但在接名词时，be sure of侧重指主语对某抽象事物的确信无疑；而besure about则侧重指主语对某具体事物的确信无疑。例如：
高中英语知识点大全（18）：be certain/ be sure的用法 1、be certain/ be sure be uncertain about意思是“对……不确定（没把握）” uncertain的词根是certain,意思是“确信的，有把握的”，常用于以下结构： (1)be certain(sure) to do sth.“肯定会做……”（表示某事将要发生）。如： He is certain(sure) to come next Sunday. (2)be certain(sure) of/ a bout sth.“确信、有把握”（表示某个人的思想状态）。如： We are certain/ sure of victory. (3)名词从句作主语时，一般多用certain. It is certain that he will come. 2、be different from 与……不同 Your idea is different from mine. 你的想法和我的不同。 对比：make sb./ sth. different from使某人/某物不同于…… Her special accent makes her different from others. 她特殊的口音使她与众不同。 3、be familiar with/be familiar to be familiar with的主语是有生命的事，意为“某人对人、事熟悉”；be familiar to 的主语是无生命的事物，意为“某人/事为某人所熟悉”，对比:He is very familiar with the names of plants in English.他很熟悉植物的英语名称。Suzhou and Hangzhou are familiar to many foreigners.苏州和杭州为许多外国人所熟悉。 I’m not familiar with European history./European history is not familiar to me.我对欧洲历史不太熟悉。
There are many computers in the office. Make sure ___the door before you leave. A you will lock B you lock C for locking D locking 答案为B。 make sure 1）宾语从句2)of/about... 3)what/how/when/...to do sth 本题考察的是make sure后面接宾语从句的用法，当从句中表示的是将来的事情时，通常采用一般现在时态来表示。 例句： 1.Make sure that you floss your teeth every night. 记得每天晚上都要用牙线来清洁牙齿。 2.You must believe you are the best and then make sure you are. 你必须要相信自己是最佳的，然后努力使自己成为最佳。 3.This is important to make sure you are losing body fat and not muscle. 这一点很重要的确保您正在失去的是身体脂肪，而不是肌肉。 4.Please put the cup in a safe place. Make sure both of us could see it. 请将样品杯放在一个安全的地方。确保我们双方都能看得到。 5.It can also be a great help to make sure you type a password correctly. 它也可以成为一个伟大的帮助，以确保您键入一个密码正确。 6.How to make money with Shareware? How can I make sure my investment is worthwhile? 怎样才能赚钱？怎样才能确保我在共享软件上面的投资是值得的。 7.They make sure, too, that the children get enough rest and play, along with their education. 他们还要弄清楚孩子们在受教育的同时，能够得到足够的休息和娱乐。 8.Teenagers and adults can write lists and establish ways to make sure they do important tasks. 青少年和承认可以写清单和确立方法来确保他们做重要的任务。 9.After that, the technicians test samples of all the milk to make sure it is safe and healthy做完那些，技术员 会测试所有母乳的样品以确保母乳是安全和健康的。 10.Coordinate with the Maintenance Technician to make sure all the abnormal were communicated efficiently. 保持与设备维护技术员有效沟通，关注所有设备异常情况。 11.Our assets walk out of the door each evening. We have to make sure that they come back the next morning. 我们的资产每个晚上都要走出我们的大门，我们必须确保他们第二天早上还能回来。
there be 用法小结 1. 基本结构 There be + 主语+ 地点/ 时间状语。如： There is a computer in the room. 房间里有一台电脑。 There are two TV plays every evening. 每晚有两场电视剧。 2. 主谓一致 要采取就近一致原则，和靠近be 的主语一致。如： There is a pen, two rulers in the box. 盒子里有一只钢笔，两把尺子。 There are two boys and a teacher at the school gate. 门口有两个男孩，一个老师。3. 主语后的动词形式 在there be 句型中，主语与动词是主动关系时用现在分词；是被动关系时用过去分词。如： There is a purse lying on the ground. 地上有一个钱包。 There are five minutes left now. 现在还有5分钟。
4. 反意疑问句。 反意疑问句应与there be 对应，而不是依据主语。如： There is a radio on the table, isn't there? 桌子上有一台收音机，是吧？ There are more than fifty classes in your school, aren't there? 你们班有50多名学生，是吧？ 5. there be 与have 的替换 there be 表示所属时可与have 替换。 There is nothing but a book in my bag. = I have nothing but a book in my bag. 包里只有一本书。 6. there be 后接不定式时常用主动形式表示被动意义。如： There is a lot of work to do. 有许多工作要做。 注意：当该句型主语是something, anything, nothing 等不定代词时，后面的不定式用主动形式或被动形式，意义各不同。 There is nothing to do. 没有事可做。 There is nothing to be done. 没有办法(束手无策)。
bear 1 1.忍受，忍耐，经受住 ?She was afraid she wouldn’t be able to bear the pain. 她恐怕自己忍受不了这种痛苦。 ?Overcrowding makes prison life even harder to bear. 过度拥挤使监狱生活更加难以忍受。 ?Make the water as hot as you can bear . 把水温调到你能忍受的最高温度。 ?The humiliation was more than he could bear . 这样的羞辱是他万万不能忍受的。 ?Black people continue to bear the brunt of most racial violence (= have to deal with the most difficult or damaging part ) . 在大多数种族暴力事件中，黑人仍然是首当其冲的受害群体。?Passengers could be insulting, and stewardesses just had to grin and bear it (= accept it without complaining ) . 乘客有时会破口大骂，而乘务员只能微笑着默默忍受。 ?Experts were worried the financial system would not be able to bear the strain . 专家担心现行金融体系承受不了这样的压力。2.can’t bear sth接受不了某事 ?Please don’t leave me. I couldn’t bear it. 请你不要离开我，我受不了。
there be 用法小结 默认分类 2008-12-24 15:48 阅读626 评论1 字号：大中小 form:https://www.doczj.com/doc/5716786626.html,/mb/1/ReadNews.asp?NewsID=420744 1. 基本结构 There be + 主语 + 地点/ 时间状语。如： There is a computer in the room. 房间里有一台电脑。 There are two TV plays every evening. 每晚有两场电视剧。 2. 主谓一致 要采取就近一致原则，和靠近 be 的主语一致。如： There is a pen, two rulers in the box. 盒子里有一只钢笔，两把尺子。 There are two boys and a teacher at the school gate. 门口有两个男孩，一个老师。 3. 主语后的动词形式 在there be 句型中，主语与动词是主动关系时用现在分词；是被动关系时用过去分词。如： There is a purse lying on the ground. 地上有一个钱包。 There are five minutes left now. 现在还有5分钟。 4. 反意疑问句。 反意疑问句应与 there be 对应，而不是依据主语。如： There is a radio on the table, isn't there? 桌子上有一台收音机，是吧？ There are more than fifty classes in your school, aren't there? 你们班有50多名学生，是吧？ 5. there be 与 have 的替换 there be 表示所属时可与 have 替换。
八年级(下)固定搭配和用法（Unit 1-10） 1、fall& be fall in love with...爱着某人/某物(瞬间动作) be in love with...爱上某人/某物(持续动作) 2、keep keep (on) doing sth.一直/不断地做某事 keep sb. Doing sth.被某人一直做某事 3、use be used by sb 被某人使用(被动语态) 4、help help sb (to) do sth.帮助某人做某事. help sb with sth. 帮助某人某事 with the help of...= with one's help of在......的帮助下 help sb. off with sth.帮某人脱去...... help sb. on with sth.帮某人穿上...... help sb. over...帮助某人渡过/越过 5、such& so such+a/an+（adj)+单数名词 so+adj+a/an/the+单数名词 6、agree agree to do sth 同意做某事 agree with sb 赞成某人 agree on/upon 就某事达成一致意见 agree to+计划/建议/安排 agree with+从句/名词 agree about+讨论内容 agree in 表示观点或原则上相同，达成一致 disagree 也有同样的用法 7、try try one's best to do sth 尽某人最大能力做某事 =do one's best to do sth try to do sth 尽力做某事 try doing sth 试着做某事 try...on 试穿 try out sth...尝试 8、there be : there is/are+sb/sth+doing sth有某人/某物在做某事 9、seem seem like +n 好像似乎是...... seem+adj/n 似乎是 seem to do sth 好像 it seems (that)... 10、be yourself 保持个性/做自己help yourself to do sth.请自便/随意做某事unit2 1、play sth for sb=play sb sth 为某人播放某物 同样用法有:make、pour、buy、cook、 2、enough ...enough to...足够......以至于...... (否定形式）：not...enough to...不够......以至于 3、argue argue with sb 争吵；争辩 argue on/about sth 争辩某事 argue against 反驳 argue sb out of sth/doing sth. 说服某人不做某事 4、give sb sth = give sth to sb 给某人某物
英语中各种词的用法 介词是一种用来表示词与词, 词与句之间的关系的词。在句中不能单独作句字成分。介词后面一般有名词代词或相当于名词的其他词类，短语或从句作它的宾语。介词和它的宾语构成介词词组，在句中作主语, 状语，表语，补语或介词宾语。例如: Most of the students went to the classroom. 大部分学生去了教室。 We play basketball on the sports ground. 我们在操场上打蓝球。 介词常与动词,形容词,名词一起构成固定搭配。 belong to 属于rely on 依靠 talk to 同...谈话be afraid of 害怕 be strict with对...严格 介词一般放在名词之前。但它后面的介词宾语是疑问代词,疑问副词或者关系代词时，这些词提到了前面而只剩下介词在后了。 Where do you come from? 你是哪儿人? Who are you talking to? 你在跟谁谈话呢? What do you study for? 你为了什么而学习? 介词在英语词汇中所占比例很小，但它们的用法却非常灵活,复杂。以下是一个例子: about 关于,附近,大约,周围,随身. I have bought a book about Shakespearean. 我买了一本有关莎士比亚的书。 1. WITH (1)v+with (a) v+with begin, mix, agree, deal, fight, meet, play, quarrel, do, fool, reason, correspond, comply, settle. (b) v + sth (sb) + with + sth (sb) compare, provide, supply, feed, replace, combine, equip, furnish,
1.最好的 best ?He won the best actor award. 他获得最佳男演员奖。 ?What’s the best way to cook this fish? 这鱼怎么烧最好？ ?The best thing to do is to stop worrying. 最好的办法就是不要再担心。?Our pilots are given the best possible training. 我们的飞行员接受最好的训练。 ?We use only the very best ingredients. 我们只用最好的原料。 it’s best to do sth?It’s best to go later in the season. 最好是当季晚些时候去。 easily the best/by far the best (=much better than anything else)绝对是最好的?John’s idea is by far the best option. 约翰的主意绝对是最好的选择。 3.best dress/shoes/clothes etc最好的礼服/鞋子/衣服等〔指留待特殊场合穿戴的衣物等〕?I put on my best suit for the wedding. 我穿上最好的一套衣服参加婚礼。 best 2 1.最好地?It works best if you let it warm up first. 如果先把它预热一下，使用效果最佳。 2.最，极SYN MOST ?You know him best – you should ask him. 你最熟悉他，应该你去问他。 3.as best you can尽最大努力，竭力?I’ll try and fix it as best I can. 我会尽力把它修好。 4.had best应该，最好?We’d best be getting back. 我们最好回去。 best 3 1.the best最佳；至上；至善?We all want the best for our children. 我们都希望给孩子最好的东西。 the person or thing that is better than any other最好之人；最佳之物?She’s the best of the new young writers. 她是新一代年轻作家中最优秀的。 2.do your best尽力而为；尽最大努力?As long as you do your best, we’ll be happy. 只要你尽力而为，我们就满意了。 do your best to do sth?She did her best to make him comfortable. 她尽力让他感到舒服。 3.at best充其量；至多?The campaign was at best only partially successful. 这场宣传活动充其量只是部分成功。?The technique is at best ineffective and at worst dangerous. 这个方法说轻一点是无效，说重一点是危险。 4.to the best of your knowledge/belief/ability etc就某人所知/在某人看来/尽某人所能等?I’m sure he’ll do the work to the best of his ability. 我相信他会尽己所能做好这项工作。 5.the best of sth最好的某事物?We wish him the best of luck with this venture. 我们祝愿他在这个风险项目上顺顺利利。 6.with the best of intentions/for the best of reasons出于好心，出于好意?I’m sure he went there with the best of intentions. 我相信他去那里是出于好心。 7.the best of both worlds两全其美；各取其长
cheers 说到cheers这个词，你一定不会感到陌生。它在众多社交场合被经常使用，只要我们端起酒杯，互相致意，那句熟悉的声音便会在耳旁，异口同声地想起——“CHEERS”。 可是，你知道吗？当歪果仁对你说“Cheers”，可不一定是“干杯”哦！如果会错会对方的好意，那个时候，你收到的也许会是“尴尬而不是礼貌的微笑”了！ 来看几个释义，感受一下它的不同使用语境。 1、欢呼雀跃，踊跃参与。 当你开心的蹦跶起来时，你就会为某事欢呼（Cheer），积极踊跃地参与到某项活动中去。来看看例句： The crowd cheered as she went up to the steps to the bandstand. 当她走上演奏台的台阶时，人群欢呼起来。 The audience greets his speech with a loud cheer. 听众对他的演讲报以高声欢呼。 The audience gave a great cheer when he scored. 当他得分的时候，观众们爆发出热烈的欢呼。 2、感到振奋 当你因某件事而摆脱失落情绪，说明你已经开始振奋自己了。这个时候你就可以用cheer来形容自己的心情。 Stephen noticed that the people around him looked cheered by
his presence. 斯蒂芬注意到周围的人因为他的到场而振奋起来。 Cheer up! I'm sure you'll feel better tomorrow. 振作起来！我肯定你明天会好些的。 Our teachers often cheer us up in class every day. 老师常常使我们每天在课堂上都很振奋。 看过了以上两个释义及其使用语境，我们再来看看cheers在英美语中用法的不同。 英式英语中的Cheers 在英式英语中，cheers有三种释义：干杯、谢谢、再见。 来看看牛津词典对它的解释：
英语祈使句的分类及用法详解 根据祈使句的语用意义和语气的强弱，祈使句可以分为四类：表示 命令的祈使句都带有强制性，要求对方必须服从，言辞肯定，态度严肃。与 表示命令的祈使句相比，表示请求的祈使句的语气要舒缓一些，可以使用语 气词“吧、啊”，主语可以出现，也可以不出现。表示禁止的祈使句明确表示 禁止对方做什幺事情，言辞强硬，态度坚决，不用语气词。表示劝阻的祈使 句语调比较平缓，常用语气词“吧、啊”。 1英语祈使句的分类祈使句表请求、命令、叮嘱、邀请、劝告等。祈 使句分为第二人称祈使句及第—、三人称祈使句两大类。1.第二人称祈使句 通常用来向听话者发出命令，提出要求或建议。这种祈使句的主语you通常 不表示出来，而是以动词原形开头。如：Standup!Don’tworryabout!但如果说 话时有多人在场，就得把主语表示出来，或加呼语，以便指明是向谁提出请 求或发出命令。如：Parentswithchildrengothefront!带孩子的家长到前面去! Marycleanthewindows，andyouboyswashthefloor!玛丽擦窗户，你们男孩子洗地板!Comein，everybody!每个人都进来!有时将主语表示出来是为了加强语气或 表示“不高兴”、“厌烦”、“鄙视”等情绪。如：Don’tyoubelateagain!你可别再迟 到了!Yougetoutofhere!你给我滚出去!Mindyourownbusiness，you!你少管闲事!2.第一、三人称祈使句是以第一人称和第三人称代词或者名词等作为祈使的对象， 这类祈使句通常以let为引导词表建议。如：Let’ sgo!Letusgohome!Lethimbehereby10o’clock.祈使句除用谓语动词表示外，还 可用名词、副词、动词短语等表示。如：Help!Patience!Quickly!Handsup!
精品文档 英语词类 英语中的词可以根据词义、语法功能和形式特征分为十大类，即名词(noun)、代词(pronoun)、形容词(adjective)、副词(adverb)、动词(verb)、数词(numeral)、冠词(article)、介词(preposition)、连词(conjunctions)和感叹词(int erjection)。 英语代词的用法全归纳 一、定义与分类 代词是代替名词及起名词作用的短语或句子的词。代词根据其意思和用法可分为人称代词、物主代词、反身代词、指示代词、相互代词、疑问代词、连接代词、关系代词、不定代词九类。综观历年高考情况，在这九类代词中，不定代词一直是高考英语的重点。 二：人称代词，物主代词，反身代词用法概述 这三种代词都有人称（第一、二、三人称）的变化、数（单、复数）的变化，性（阴性、阳性、中性）以及格（主格、宾格）的变化。 物主代词又包括两种形式：形容词性物主代词和名词性物主代词。特别注意：形容词性物主代词its没有相应的名词性物主代词。 这三种代词的形式变化表如下：
精品文档．精品文档
三、人称代词的用法：人称代词是用来指代人、动物或事物的代词。它必须在人1) 定义（阴以及性数（单数、复数）、称(第一人称、第二人称、及第三人称) 性、阳性、中性）三方面与被指代的名词一致。Tom is a boy, and he is a student. 如：I am a student. Mary is very pretty, and she likes singing. The boys are students, and they are in the room. The doy is small. It is Tom's. 人称代词的句法功能2) A) 人称代词有主格和宾语之分：主格用作主语，宾格用作宾语。 B) 人称代词的主格形式在在句中作主语和表语。 ) 主语). : I like music(如主语She is a teacher.(精品文档．精品文档 She and I are good friends(主语). Neither she nor I am student. I saw the boys this morning. ——)? Are you sure it was they(表语——C) 人称代词的宾格在句子作动词的宾语，或者介词的宾语。). I saw him at the party(宾语如：) I haven't seen them recently.(宾语) 作介词宾语I bought a book for them.() 间接宾语=I bought them a book.(☆注意：在口语中，人称代词的主格形式作表语时，常常可以用宾格替换。a.It
beg的用法(总结) 第1篇 beg的意思 vi。vt。恳求（原谅）；乞讨；回避（问题）； beg的用法 1。口语中说的I beg your pardon能够有多种用法 (1) 用于没听清或没听懂对方的话，请对方再说一遍（可只说Pardon，说时用升调）。如 I beg your pardon―I didn’t hear what you said。请再说一遍――我没听见你刚才说的话。 (2) 用于道歉，比用I’m sorry更正式。如 I beg your pardon。I suppose I should have knocked。对不起，我想我本来应当敲门的。 (3) 用于表示生气或气愤。如
Pardon me，but this is my coat。对不起，这是我的外套。 I beg your pardon but the woman you’re insulting happens to be my wife。请你尊重些，你侮辱的这个女人正是我的妻子。 (4) 用于对别人的冒犯或无礼表示威胁。如 AYou cunt! 你这笨蛋! BI beg your pardon。你再说一遍。 (5) 用于引起别人的注意。如 I beg your pardon; can you tell me the way to the station? 对不起，你能告诉我去车站怎样走吗? 2。有时可后接that从句，从句谓语通常用虚拟语气。如 He begged that he (should) be allowed to leave。他恳求让他离开。 3。用于go begging，现代英语中主要表示“（东西等）没有人要”“（职位
There be结构的一些特例 There be结构在英语中是一种常见的倒装结构，它的最一般的形式为：there 后接动词be的各种时态，然后是主语，再后常有附加短语表达时间或地点，其中there是引导词，没有实在意义，常弱读为[ ] 1.含有情态动词的there be结构 多数情态动词，如：can, may, must, should, ought to, used to等都可以用于there be结构，并且在此结构中分别保留其原有的情态意义。例如： (1).There should have been someone on duty all the time. 本来，每时每刻都应有人值班的。 (2). There used to be a small pine wood near where I live. 在我的住所附近曾经有一片小松林。 (3).There must be something wrong with the machine. 这机器一定是有故障了。 2.含有半助动词的there be结构 半助动词本身有一定的含义，并且用法较为固定，常用在引导词there之后，be之前。这些半助动词有：appear to, happen to, chance to, seem to, tend to, prove to, turn out to, be about to, be sure to, be certain to, be likely to等。例如： (4). There dosen't seem to have been any difficulty over the money question. 似乎不存在经济上的问题。 (5).There's sure to be a well somewhere nearby. 附近肯定有一口井。 (6). There's likely to be a large audience in the theatre. 剧院里可能有大量的观众。