a r X i v :m a t h -p h /9903021v 2 28 O c t 1999Commutative Geometries are Spin Manifolds
A.Rennie email:arennie@https://www.doczj.com/doc/ad16215509.html,.au February 7,2008Abstract In [1],Connes presented axioms governing noncommutative geometry.He went on to claim that when specialised to the commutative case,these axioms recover spin or spin c geometry depending on whether the geometry is “real”or not.We attempt to ?esh out the details of Connes’ideas.As an illustration we present a proof of his claim,partly extending the validity of the result to pseudo-Riemannian spin manifolds.Throughout we are as explicit and elementary as possible.1Introduction The usual description of noncommutative geometry takes as its basic data unbounded Fred-holm modules,known as K -cycles or spectral triples.These are triples (A ,H ,D )where A is an involutive algebra represented on the Hilbert space H .The operator D is a closed,unbounded operator on H with compact resolvent such that the commutator [D ,π(a )]is a bounded opera-tor for every a ∈A .Here πis the representation of A in H .We also suppose that we are given an integer p called the degree of summability which governs the dimension of the geometry.If p is even,the Hilbert space is Z 2-graded in such a way that the operator D is odd.In [1],axioms were set down for noncommutative geometry.It is in this framework that Connes states his theorem recovering spin manifolds from commutative geometries.Perhaps the most
important aspect of this theorem is that it provides su?cient conditions for the spectrum of a C ?-algebra to be a (spin c )manifold.It also gives credence to the idea that spectral triples obeying the axioms should be regarded as noncommutative manifolds.
Let us brie?y describe the central portion of the proof.Showing that the spectrum of A is actu-ally a manifold relies on the interplay of several abstract structures and the axioms controlling their representation.At the abstract level we can de?ne the universal di?erential algebra of A ,denoted ??(A ).The underlying linear space of ??(A )is isomorphic to the chain complex from which we construct the Hochschild homology of A .In the commutative case,there is also another de?nition of the di?erential forms over A .This algebra, ?
?(A ),is skew-commutative and when the algebra A is ‘smooth’,[2],it coincides with the Hochschild homology of A .For a commutative algebra it is always the case that Hochschild homology contains ??(A )as a direct summand.
The axioms,among other things,ensure that we end up with a faithful representation of ?
?(A ).The process begins by constructing a representation of ??(A )from a representation πof A ,which we may assume is faithful.This is done using the operator D introduced above by setting
π(δa )=[D ,π(a )]?a ∈A ,(1)
1
2BACKGROUND2 where??(A)is generated by the symbolsδa,a∈A.There are three axioms/assumptions
controlling this representation.The?rst is Connes’?rst order condition.This demands that
[[D,π(a)],π(b)]=0for all a,b∈A,at least in the commutative case.It turns out that
the kernel of a representation of??(A)obeying this condition is precisely the image of the
Hochschild boundary.Thus our representation descends to a representation of Hochschild
homology HH?(A)???(A),and is moreover faithful.
The algebra??D(A):=π(??(A))is no longer a di?erential algebra.To remedy this,one
quotients out the‘junk’forms,and these turn out to be the submodule generated over A by
graded commutators and the image of the Hochschild boundary.Thus the algebra,Λ?D(A),
that we arrive at after removing the junk is skew-commutative,and we will show that it is
isomorphic to ??(A).We will then prove that the representation of Hochschild homology with values inΛ?D(A)is still faithful,showing thatΛ?D(A)～= ??(A)～=HH?(A).This is a necessary, though not su?cient,condition for the algebra A to be smooth.Note that by virtue of the
?rst order condition,both??D(A)andΛ?D(A)are symmetric A bimodules,and so may be
considered to be(left or right)modules over A?A.
Returning to the axioms,the critical assumption to show that the spectrum is indeed a manifold
is the existence of a Hochschild p-cycle which is represented by1or the Z2-grading depending
on whether p is odd or even respectively.The main consequence of this is that this cycle is
nowhere vanishing as a section ofπ(?p(A)).As it is a cycle,we know that it lies in the skew-
commutative part of the algebra,and this will allow us to?nd generators of the di?erential
algebra over A and construct coordinate charts on the spectrum.Indeed,the non-vanishing
of this cycle is the most stringent axiom,as it enforces the underlying p-dimensionality of the
spectrum.
Thus we have a two step reduction process
??(A)→??D(A)→Λ?D(A),(2) and the third axiom referred to above is needed to control the behaviour of the intermediate algebra??D(A),as well as to ensure that our algebra is indeed smooth.Recalling that D is required to be closed and self-adjoint,we demand that for all a∈A
δn(π(a)),δn([D,π(a)])are bounded for all n,δ(x)=[|D|,x].(3) It is easy to imagine that this could be used to formulate smoothness conditions,but it also forces??D(A)to be a(possibly twisted)representation of the complexi?ed Cli?ord algebra of the cotangent bundle of the spectrum.As the representation is assumed to be irreducible,we will have shown that the spectrum is a spin c manifold.Inclusion of the reality axiom will then show that the spectrum is actually spin.The detailed description of these matters requires a great deal of work.
We begin in Section2with some more or less standard background results.These will be
required in Section3,where we present our basic de?nitions and the axioms,as well as in
Section4where we state and prove Connes’result.Section5addresses the issue of abstract
characterisation of algebras having“geometric representations.”
2Background
Although there are now several good introductory accounts of noncommutative geometry,eg
[3],to make this paper as self-contained as possible,we will quote a number of results necessary
for the proof of Connes’result and the analysis of the axioms.
2BACKGROUND3 2.1Pointset Topology
The point set topology of a compact Hausdor?space X is completely encoded by the C?-algebra
of continuous functions on X,C(X).This is captured in the Gel’fand-Naimark Theorem. Theorem1For every commutative C?-algebra A,there exists a Hausdor?space X such that
A～=C0(X).If A is unital,then X is compact.
In the above,C0(X)means the continuous functions on X which tend to zero at in?nity.In the
compact case this reduces to C(X).We can describe X explicitly as X=Spec(A)={maximal
ideals of A}={pure states of A}={unitary equivalence classes of irreducible representations}.
The weak?topology on the pure state space is what gives us compactness,and translates into
the topology of pointwise convergence for the states.While this theorem provides much of the
motivation for the use of C?-algebras in the context of“noncommutative topology,”much of
their utility comes from the other Gel’fand-Naimark theorem.
Theorem2Every C?-algebra admits a faithful and isometric representation as a norm closed
self-adjoint?-subalgebra of B(H)for some Hilbert space H.
In the classical(commutative)case there are a number of results showing that we really can
recover all information about the space X from the algebra of continuous functions C(X).
For instance,closed sets correspond to norm closed ideals,and so single points to maximal
ideals.The latter statement is proved using the correspondence pure state?kernel of pure
state;this is the way that the Gel’fand-Naimark theorem is proved,and is not valid in the
noncommutative case,[3].There are many other such correspondences,see[4],not all of which still make sense in the noncommutative case,for the simple reason that the three descriptions
of the spectrum no longer coincide for a general C?-algebra,[3].On this cautionary note,let
us turn to the most important correspondence;the Serre-Swan theorem,[5].
Theorem3Let X be a compact Hausdor?space.Then a C(X)-module V is isomorphic(as
a module)to a moduleΓ(X,E)of continuous sections of a complex vector bundle E→X if
and only if V is?nitely generated and projective.
We abbreviate?nitely generated and projective to the now common phrase?nite projective.
This is equivalent to the following.If V is a?nite projective A-module,then there is an
idempotent e2=e∈M N(A),the N×N matrix algebra over A,for some N such that
V～=eA N.Thus V is a direct summand of a free module.We would like then to treat?nite projective C?-modules as noncommutative generalisations of vector bundles.Ideally we would
like the idempotent e to be a projection,i.e.self-adjoint.Since every complex vector bundle
admits an Hermitian structure,this is easy to formulate in the commutative case.In the
general case we de?ne an Hermitian structure on a right A-module V to be a sesquilinear map
·,· :V×V→A such that?a,b∈A,v,w∈V
1) av,bw =a? v,w b,
2) v,w = w,v ?,
3) v,v ≥0, v,v =0?v=0.
A?nite projective module always admits Hermitian structures(and connections,for those
looking ahead).Such an Hermitian structure is said to be nondegenerate if it gives an isomor-
phism onto the dual module.This corresponds to the usual notion of nondegeneracy in the
classical case,[3].We also have the following.
2BACKGROUND4 Theorem4Let A be a C?-algebra.If V is a?nite projective A-module with a nondegenerate Hermitian structure,then V～=eA N for some idempotent e∈M N(A)and furthermore e=e?, where?is the composition of matrix transposition and the?in A.
We shall regard?nite projective modules over a C?-algebra as the noncommutative version of (the sections of)vector bundles over a noncommutative space.As a last point before moving on,we note that for every bundle on a smooth manifold,there is an essentially unique smooth bundle.For more information on all these results,see[3,6].
2.2Algebraic Topology
Bundle theory leads us quite naturally to K-theory in the commutative case,and to further demonstrate the utility of de?ning noncommutative bundles to be?nite projective modules,we ?nd that this de?nition allows us to extend K-theory to the noncommutative domain as well. Moreover,in the commutative case(at least for?nite simplicial complexes),K-theory recovers the ordinary cohomology via the Chern character,and the usual cohomology theories make no sense whatsoever for a noncommutative space.Thus K-theory becomes the cohomological tool of choice in the noncommutative setting.
This is not the whole story,however,for we know that in the smooth case we can formulate ordinary cohomology in geometric terms via di?erential forms.Noncommutatively speaking, di?erential forms correspond to the Hochschild homology of the algebra of smooth functions on the space,whereas the de Rham cohomology is obtained by considering the closely related theory,cyclic homology,[6].The latter,or more properly the periodic version of the theory, see[2,6],is the proper receptacle for the Chern character(in both homology and cohomology).
We can de?ne K-theory for any C?-algebra and K-homology for a class of C?-algebras and certain of their dense subalgebras.For any of these algebras A,elements of K?(A)may be regarded as equivalence classes of Fredholm modules[(H,F,Γ)].These consist of a represen-tationπ:A→B(H),an operator F:H→H such that F=F?,F2=1,and[F,π(a)]is compact for all a∈A.IfΓ=1,the class de?ned by the module is said to be odd,and it resides in K1(A).IfΓ=Γ?,Γ2=1,[Γ,π(a)]=0for all a∈A andΓF+FΓ=0,then we call the class even,and it resides in K0(A).For complex algebras,Bott periodicity says that these are essentially the only K-homology groups of A.For the case of a commutative C?-algebra, C0(X),this coincides with the(analytic)K-homology of X.A relative group can be de?ned as well,along with a reduced group for dealing with locally compact spaces(non-unital algebras), [7,8,9,10].
The K-theory of the algebra A may be described as equivalence classes of idempotents,K0(A), and unitaries,K1(A),in M∞(A)and GL∞(A)respectively.Again,relative and reduced groups can be de?ned.We will?nd it useful when discussing the cap product to denote elements corresponding to actual idempotents or unitaries by[(e,N)],or[(u,N)]respectively,with e,u∈M N(A).Much of what follows could be translated into KK-theory or E-theory,but we shall be content with the simple presentation below.
The duality pairing between K-theory and K-homology can be broken into two steps.First one uses the cap product,described below,and then acts on the resulting K-homology class with the natural index map.This map,Index:K0(A)→K0(C)=Z is given by
Index([(H,F,Γ)])=Index(1?Γ
2
).(4)
2BACKGROUND5 On the right we mean the usual index of Fredholm operators,and the beauty of the Fredholm module formulation is that the Index map is well-de?ned.The Index map can also be de?ned on K1(A),but as the operators in here are all self-adjoint,it always gives zero.For this reason we will avoid mention of the“odd”product rules for the cap product.
The cap product,∩,in K-theory will allow us to formulate Poincar′e duality in due course.It is a map
∩:K?(A)×K?(A)→K?(A)(5) with the even valued terms given on simple elements by the following rules:
K0(A)×K0(A)→K0(A) [(e,N)]∩[(H,F,Γ)]=[(e H N+⊕e H N?, 0e?F??1N e
e?F?1N e0 e00?e )],(6)
where H±=1±Γ
2H)2N, 0u u?0 , 1+F
2?1N
)].(7)
The cap product turns K?(A)into a module over K?(A).Usually,one is given a Fredholm module,μ=[(H,F,Γ)],whereΓ=1if the module is odd,and then considers the Index as a map on K?(A)via Index(k):=Index(k∩μ),for any k∈K?(A).
Let us for a minute suppose that A is commutative.If X=Spec(A)is a compact?nite simplicial complex,there are isomorphisms
K?(X)?Q～=K?(A)?Q ch?→H?(X,Q)(8) and
K?(X)?Q～=K?(A)?Q ch?→H?(X,Q)(9) given by the Chern characters.Here H?and H?are the ordinary(co)homology of X.Note that a(co)homology theory for spaces is a homology(co)theory for algebras.
Important for us is that these isomorphisms also preserve the cap product,both in K-theory and ordinary(co)homology,so that the following diagram
K?(X)?Q∩μ→K?(X)?Q
ch?↓ch?↓
H?(X,Q)∩ch?(μ)
→H?(X,Q)
commutes for anyμ∈K?(X).So if X is a?nite simplicial complex satisfying Poincar′e Duality in K-theory,that is there existsμ∈K?(C(X))such that∩μ:K?(C(X))→K?(C(X))is an isomorphism,then there exists[X]=ch?(μ)∈H?(X,Q)such that
∩[X]:H?(X,Q)→H?(X,Q)(10)
2BACKGROUND 6is an isomorphism.If ch ?(μ)∈H p (X,Q )for some p ,then we would know that X satis?ed Poincar′e duality in ordinary (co)homology,which is certainly a necessary condition for X to be a manifold.
We note in passing that K -theory has only even and odd components,whereas the usual (co)homology is graded by Z .This is not such a problem if we replace (co)homology with periodic cyclic homology(co).In the case of a classical manifold it gives the same results as the usual theory,but it is naturally Z 2-graded.Though we do not want to discuss cyclic (co)homology in this paper,we note that the appropriate replacement for the commuting square above in the noncommutative case is the following.
K ?(A )?Q
∩μ→K ?(A )?Q ch ?↓ch ?↓H per ?(A )?Q ∩ch ?(μ)→H ?per
(A )?Q Moreover,the periodic theory is the natural receptacle for the Chern character in the not necessarily commutative case,provided that A is an algebra over a ?eld containing Q .For more information on Poincar′e duality in noncommutative geometry,including details of the induced maps on the various homology groups,see [6,11].2.3Measure On the analytical front we have to relate the noncommutative integral given by the Dixmier trace to the usual measure theoretic tools.This is achieved using two results of Connes;one building on the work of Wodzicki,[12],and the other on the work of Voiculescu,[13].For more detailed information on these results,see [6]and [12,13].
To de?ne the Dixmier trace and relate it to Lebesgue measure,we require the de?nitions of several normed ideals of compact operators on Hilbert space.The ?rst of these is
L
(1,∞)(H )={T ∈K (H ):N n =0μn (T )=O (log N )}(11)
with norm
T 1,∞=sup N ≥21
T T ?arranged in decreasing order and
repeated according to multiplicity so that μ0(T )≥μ1(T )≥....This ideal will be the domain of de?nition of the Dixmier trace.Related to this ideal are the ideals L (p,∞)(H )for 1
L (p,∞)(H )={T ∈K (H ):
N n =0μn (T )=O (N 1?1
N 1?1
p i =1,
then the
product T 1···T n ∈L (1,∞)(H ).In particular,if T ∈L (n,∞)(H )then T n ∈L (1,∞)(H ).
2BACKGROUND7 We want to de?ne the Dixmier trace so that it returns the coe?cient of the logarithmically divergent part of the trace of an operator.Unfortunately,since(1/log N) Nμn(T)is in general only a bounded sequence,we can not take the limit in a well-de?ned way.The Dixmier trace is usually de?ned in terms of linear functionals on bounded sequences satisfying certain properties.One of these properties is that if the above sequence is convergent,the linear functional returns the limit.In this case,the result is independent of which linear functional is used.So,for T∈L(1,∞)(H)with T≥0,we say that T is measurable if
? T:=lim N→∞1
p(2π)p S?M trace Eσ?p(P)(x,ξ)√
2BACKGROUND8 Since the Dixmier trace acts on operators on Hilbert space we might expect it to be related to measure theory via the spectral theorem.Indeed this is true,but we must backtrack a little into perturbation theory.
The Kato-Rosenblum theorem,[16],states that for a self-adjoint operator T on Hilbert space, the absolutely continuous part of T is(up to unitary equivalence)invariant under trace class perturbation.This result does not extend to the joint absolutely continuous spectrum of more than one operator.Voiculescu shows that for a p-tuple of commuting self-adjoint operators (T1,...,T p),the absolutely continuous part of their joint spectrum is(up to unitary equiva-lence)invariant under perturbation by a p-tuple of operators(A1,...,A p)with A i∈L(p,1)(H). This ideal is given by
L(p,1)(H)={T∈K(H):
∞ n=0n1
2BACKGROUND 9Theorem 8Let p and D be as above,with D ?1∈L (p,∞)(H )and suppose that A is an invo-lutive subalgebra of B (H )such that [D,a ]is bounded for all a ∈A .Then
1)Setting τ(a )=? a |D |?p de?nes a trace on A .This trace is nonzero if k L (p,1)(A )=0.
2)Let p be an integer and a 1,...,a p ∈A commuting self-adjoint elements.Then the absolutely continuous part of their spectral measure
μac (f )= E ac f (x )m (x )d p x (21)
is absolutely continuous with respect to the measure
τ(f )=τ(f (a 1,...,a p ))=? f |D |?p ,?f ∈C ∞c (R p ).(22)
Combining the results on the Wodzicki residue and these last results of Voiculescu and Connes,we will be able to show that the measure on a commutative geometry is a constant multiple of the measure de?ned in the usual way.
The hypothesis of invertibility used in the above theorems for the operator D can be removed provided ker D is ?nite dimensional.Then we can add to D a ?nite rank operator in order to obtain an invertible operator,and the Dixmier trace will be unchanged.For these purely measure theoretic purposes,simply taking D ?1=0on ker D is ?ne.More care must be taken with ker D in the de?nition of the associated Fredholm module;see [6].
2.4Geometry
We will avoid discussion of cyclic (co)homology in this paper,as we do not absolutely require it.More important for this paper,is the universal di?erential algebra construction,and its relation to Hochschild homology.
The (reduced)Hochschild homology of an algebra A with coe?cients in a bimodule M is de?ned in terms of the chain complex C n (M )=M ??A
?n with boundary map b :C n (M )→C n ?1(M )b (m ?a 1?···?a n )=ma 1?a 2?···?a n
+n ?1 i =1(?1)i m ?a 1?···?a i a i +1?···?a n
+(?1)n a n m ?a 1?···?a n ?1
(23)Here ?A =A /C .In the event that M =A ,we denote C n (M ):=C n (A )and the resulting homology by HH ?(A ),otherwise by HH ?(A ,M ).Though we will be concerned with the commutative case,the general setting uses M =A?A op with bimodule structure a (b ?c op )d =abd ?c op ,with A op the opposite algebra of A .With this structure it is clear that
HH ?(A ,A ?A op )～=A ?A op ?HH ?(A )～=A ?HH ?(A )?A op .(24)
We will also require topological Hochschild homology.Suppose we have an algebra A which is endowed with a locally convex and Hausdor?topology such that A is complete.This is equivalent to requiring that for any continuous semi-norm p on A there are continuous semi-norms p ′,q ′such that p (ab )≤p ′(a )q ′(b )for all a,b ∈A .In particular,the product is (separately)continuous.Any algebra with a topology given by an in?nite family of semi-norms
2BACKGROUND10 in such a way that the underlying linear space is a Frechet space satis?es this property,and moreover we may take p′=q′so that multiplication is jointly continuous.In constructing the topological Hochschild homology,we use the projective tensor product??instead of the usual
tensor product,in order to take account of the topology of A.This is de?ned by placing on the algebraic tensor product the strongest locally convex topology such that the bilinear map (a,b)→a?b,A×A→A?A is continuous,[6,18].If A is complete with respect to this Frechet topology,then the topological tensor product is also Frechet and complete for this topology. The resulting Hochschild homology groups are still denoted by HH?(A).Provided that these groups are Hausdor?,all the important properties of Hochschild homology,including the long exact sequence,carry over to the topological setting.
The universal di?erential algebra over an algebra A is de?ned as follows.As an A bimodule,?1(A)is generated by the symbols{δa}a∈A subject only to the relationsδ(ab)=aδ(b)+δ(a)b. Note that this implies thatδ(C)={0}.This serves to de?ne both left and right module structures and relates them so that,for instance,
aδ(b)=δ(ab)?δ(a)b.(25)
We then de?ne
?n(A)=
n
i=1?1(A)(26)
and
??(A)= n=0?n(A),(27)
with?0(A)=A.Thus??(A)is a graded algebra,and we make it a di?erential algebra by setting
δ(aδ(b1)···δ(b k))=δ(a)δ(b1)···δ(b k),(28) and
δ(ωρ)=δ(ω)ρ+(?1)|ω|ωδ(ρ)(29) whereωis homogenous of degree|ω|.If A is an involutive algebra,we make??(A)an involutive algebra by setting
(δ(a))?=?δ(a?),(ωρ)?=(ρ)?(ω)?a∈A,ρ,ω∈??(A).(30) With these conventions,??(A)is a graded di?erential algebra,with graded di?erentialδ.
It turns out,[2],that the chain complex used to de?ne Hochschild homology HH?(A)is the same linear space as??(A).So
C n(A)～=?n(A)(31)
(a0,a1,...,a n)→a0δa1···δa n.(32) The algebraic and di?erential structures of these spaces are di?erent,however.The relation between b andδis known,and is given by
b(ωδa)=(?1)|ω|[ω,a](33) forω∈?|ω|(A)and a∈A.We will make much use of this relation in our proof.
For a commutative algebra,there is another de?nition of di?erential forms.We retain the
2BACKGROUND11 de?nition of?1(A),now regarded as a symmetric bimodule,but de?ne ?n(A)to beΛn A?1(A), the antisymmetric tensor product over A.This has the familiar product
(a0δa1∧···∧δa k)∧(b0δb1···∧δb m)=a0b0δa1∧···∧δa k∧δb1∧···∧δb m(34) and is a di?erential graded algebra forδas above.For smooth(smooth algebras are automat-ically unital and commutative)algebras,see[2]for this technical de?nition, ??(A)～=HH?(A) as graded algebras,[2],though they have di?erent di?erential structures.It can also be shown that the Hochschild homology of any commutative and unital algebra A contains ??(A)as a direct summand,[2].For the smooth functions on a manifold,Connes’exploited the locally convex topology of the algebra C∞(M)and the topological tensor product to prove the analo-gous theorem for continuous Hochschild cohomology and de Rham currents on the manifold,[6]. We also use the universal di?erential algebra to de?ne connections in the algebraic setting.So suppose that E is a?nite projective A module.Then it can be shown,[6],that connections, in the sense of the following de?nition,always exist.
De?nition1A(universal)connection on the?nite projective A module E is a linear map ?:E→?1(A)?E such that
?(aξ)=δ(a)?ξ+a?(ξ),?a∈A,ξ∈E.(35) Note that this de?nition corresponds to what is usually called a universal connection,a con-nection being given by the same de?nition,but with?1(A)replaced with a representation of ?1(A)obeying the?rst order condition.The distinction will not bother us,but see[3,6].
A connection can be extended to a map??(A)?E→??+1(A)?E by demanding that ?(φ?ξ)=δ(φ)?ξ+(?1)|φ|(φ?1)?(ξ)whereφis homogenous of degree|φ|,and extending by linearity to nonhomogenous terms.
If there is an Hermitian structure on E,and we can always suppose that there is,then we may ask what it means for a connection to be compatible with this structure.It turns out that the appropriate condition is
δ(ξ,η)E=(ξ,?η)E?(?ξ,η)E.(36) We must explain what we mean here.If we write?ξas ωi?ξi,then the expression(?ξ,η)E means (ωi)?(ξi,η)E,and similarly for the other term.As real forms,that is di?erentials of self-adjoint elements of the algebra,are anti-self adjoint,we see the need for the extra minus sign in the de?nition.Note that some authors have the minus sign on the other term,and this corresponds to their choice of the Hermitian structure being conjugate linear in the second variable.
As a last point while on this subject,if?is a connection on a?nite projective??(A)module E,then[?,·]is a connection on??(A)“with values in E”.Here the commutator is the graded commutator,and our meaning above is made clear by
[?,ω]?ξ=δω?ξ+(?1)|ω|ω?ξ?(?1)|ω|ω?ξ
=δω?ξ.(37) This will be important later on when discussing the nature of D.
3DEFINITIONS AND AXIOMS12 3De?nitions and Axioms
We shall only deal with normed,involutive,unital algebras over C satisfying the C?-condition: aa? = a 2.To denote the algebra or ideal generated by a1,...,a n,possibly subject to some relations,we shall write a1,...,a n .We write,for H a Hilbert space,B(H),K(H) respectively for the bounded and compact operators on H.We write Cliff r+s,Cliff r,s,for the
+···+s
?···?). Cli?ord algebra over R r+s with Euclidean signature,respectively signature(r
The complex version is denoted C liff r+s.It is important to note that while our algebras A are imbued with a C?-norm,we do not suppose that it is complete with respect to the topology determined by this norm,nor that this norm determines the topology of A.
Typically in noncommutative geometry,we consider a representation of an involutive algebra
π:A→B(H)(38) together with a closed unbounded self-adjoint operator D on H,chosen so that the represen-tation of A extends to(bounded)representations of??(A).This is done by requiring
π(δa)=[D,π(a)].(39) So in particular,commutators of D with A must be bounded.We then set
π(δ(aδ(b1)···δ(b k)))=[D,π(a)][D,π(b1)]···[D,π(b k)].(40) We say that the representation of??(A)is induced from the representation of A by D.With the?-structure on??(A)described in the last section,πwill be a?-morphism of??(A).We shall frequently write[D,·]:??(A)→??+1(A),where we mean that the commutator acts only on elements of A,notδA,as de?ned above.
Asπis only a?-morphism of??(A)induced by a?-morphism of A,and not a map of di?erential algebras,we should not expect from the outset that it would encode the di?erential structure of??(A).In fact,it is well known that we may have the situation
π( i a iδb i1···δb i k)=0(41)
while
π( iδa iδb i1···δb i k)=0.(42) These nonzero forms are known as junk,[6].To obtain a di?erential algebra,we must look at
π(??(A))/π(δkerπ).(43) The pejorative term junk is unfortunate,as the example of the canonical spectral triple on a spin c manifold shows(and as we shall show later with one extra assumption on|D|and the representation).This is given by the algebra of smooth functions A=C∞(M)acting as multiplication operators on H=L2(M,S),where S is the bundle of spinors and D is taken to be the Dirac operator.In this case,[6],the induced representation of the universal di?erential algebra is(up to a possible twisting by a complex line bundle)
π(??(A))～=Cliff(T?M)?C=C liff(T?M)(44)
3DEFINITIONS AND AXIOMS13 and
π(??(A))/π(δkerπ)～=Λ?(T?M)?C.(45) Clearly it is the irreducible representation of the former algebra which encodes the hypothesis ‘spin c’.We will come back to this throughout the paper,and examine it more closely in Section 5.With the above discussion as some kind of motivation,let us now make some de?nitions. De?nition2A smooth spectral triple(A,H,D)is given by a representation
π:??(A)?A op→B(H)(46) induced from a representation of A by D:H→H such that
1)[π(φ),π(b op)]=0,?φ∈??(A),b op∈A op
2)[D,π(a)]∈B(H),?a∈A
3)π(a),[D,π(a)]∈∩∞m=1Domδm whereδ(x)=[|D|,x].
Further,we require that D be a closed self-adjoint operator,such that the resolvent(D?λ)?1 is compact for allλ∈C\R.
The?rst condition here is Connes’?rst order condition,and it plays an important r?o le in all that follows.It is usually stated as[[D,a],b op]=0,for all a,b∈A.A unitary change of representation on H given by U:H→H sends D to U D U?=D+U[D,U?].When it is important to distinguish between the various operators so obtained,we will write Dπ.Note that the smoothness condition can be encoded by demanding that the map t→e it|D|be?it|D| is C∞for all b∈π(??(A)).This condition also restricts the possible form of|D|and so D. This will in turn limit the possibilities for the product struture in??D(A),eventually showing us that it must be the Cli?ord algebra(up to a possible twisting).It is of some importance that,as mentioned,the Hochschild boundary on di?erential forms is given by
b(ωδa)=(?1)|ω|[ω,a].(47) A representation of??(A)induced by D and the?rst order condition satis?es
π?b=0.(48) That is,Hochschild boundaries are sent to zero byπ.First this makes sense,as HH?(A) is a quotient of C?(A),which has the same linear structure as??(A).Second,it gives us a representation of Hochschild homology
π:HH?(A)→π(??(A)).(49) Ifπ:A→π(A)is faithful,then the resulting map on Hochschild homology will also be injective.The reason for this is that the kernel ofπwill be generated solely by elements satisfying the?rst order condition;i.e.Hochschild boundaries.We will discuss this matter and its rami?cations further in the body of the proof.
To control the dimension we have two more assumptions.
De?nition3For p=0,1,2,...,a(p,∞)-summable spectral triple is a smooth spectral triple with
1)|D|?1∈L(p,∞)(H)
2)a Hochschild cycle c∈Z p(A,A?A op)with
π(c)=Γ(50) where if p is oddΓ=1and if p is even,Γ=Γ?,Γ2=1,Γπ(a)?π(a)Γ=0for all a∈A and ΓD+DΓ=0.
3DEFINITIONS AND AXIOMS14 Note that condition1)is invariant under unitary change of representation.Condition2)is a very strict restraint on potential geometries.We will writeΓorπ(c)in all dimensions unless we need to distinguish them.As a last de?nition for now,we de?ne a real spectral triple.
De?nition4A real(p,∞)-summable spectral triple is a(p,∞)-summable spectral triple to-gether with an anti-linear involution J:H→H such that
1)Jπ(a)?J?=π(a)op
2)J2=?,J D=?′D J,JΓ=?′′ΓJ,
where?,?′,?′′∈{?1,1}depend only on p mod8as follows:
p0*******
?11?1?1?1?111
(51)
?′1?1111?111
?′′1×?1×1×?1×
We will learn much about the involution from the proof,but let us say a few words.First, the mapπ(a)→π(a)op in part1),De?nition4,is C linear.Though we won’t deal with the noncommutative case in any great detail in this paper,let us just point out an interesting feature.The requirement Jb?J?=b op and the?rst order condition allow us to say that
[D,π(a?b op)]=π(b op)[D,π(a)]+π(a)[D,b op]
=π(b op)[D,π(a)]+?′π(a)J[D,π(b)]J?.(52)
This allows us to de?ne a representationπ(??(A op))=?′π(??(A))op.Then,just as A op commutes withπ(??(A)),A commutes withπ(??(A))op.In the body of the paper we introduce a slight generalisation of the operator J which allows us to introduce an inde?nite metric on our manifold.We leave the details until the body of the proof.As noted in the introduction, in the commutative case we?nd thatπ(??(A))is automatically a symmetric A-bimodule,so when A is commutative,we may replace A?A op by A.
With this formulation(i.e.hiding all the technicalities in the de?nitions)the axioms for noncommutative geometry are easy to state.A real noncommutative geometry is a real(p,∞)-summable spectral triple satisfying the following two axioms:
i)Axiom of Finiteness and Absolute Continuity
Settingμ=[(H,D,π(c))]∈KR p(A?A op),we require that the cap product byμis an isomorphism;
K?(A)∩μ→K?(A).(54) Note thatμdepends only on the homotopy class ofπ,and in particular is invariant under unitary change of representation.If we want to discuss submanifolds(i.e.unfaithful represen-tations of A that satisfy the de?nitions/axioms)we would have to considerμ∈K?(π(A)),and the isomorphism would be between the K-theory and the K-homology ofπ(A).Equivalently,
3DEFINITIONS AND AXIOMS15 we could considerμ∈K?(A,kerπ),the relative homology group,[9].We will not require this degree of generality.To see thatμde?nes a KR class,note that J·J?provides us with an involution onπ(??(A?A op)).It may or may not be trivial,but always allows us to regard the K-cycle obtained fromμas Real,in the sense of[11].
Axiom i)controls a great deal of the topological and measure theoretic structure of our ge-ometry.Suppose A is a commutative algebra.Since
π(A)is isomorphic to a bundle of continuous sectionsΓ(X,E) for some complex bundle E→X.Here X=Spec(
√
(U D U?)?(U D U?)=
3DEFINITIONS AND AXIOMS16 However,(U|D|U?)2=U|D|2U?=U D2U?so
|U D U?|=U|D|U?.(59) From this(U|D|U?)?1=U|D|?1U?,and
|U D U?|?p=U|D|?p U?.(60) It is a general property of the Dixmier trace,[6],that conjugation by bounded invertible operators does not alter the integral(this is just the trace property).So
? |U D U?|?p=? U|D|?p U?= |D|?p.(61) Furthermore,for any a∈A,
? Uπ(a)U?|U D U?|?p=? π(a)|D|?p,(62)
showing that integration is well-de?ned given
1)the unitary equivalence class[π]ofπ
2)the choice of c∈Z p(A,A?A op).
For this reason we do not need to distinguish between? |Dπ|?p and? |D UπU?|?p and we will simply write D in this context.Anticipating our later interest,we note that? |D|2?p is not invariant.Sending D to U D U?sends? |D|2?p to
? (U D U?)2|U D U?|?p=? (D+A)2U|D|?p U?
=? (D2+{D,A}+A2)U|D|?p U?
=? (D2U|D|?p U?)+? ({D,A}+A2)|D|?p
=? (D2+{D,A}+A2)|D|?p?? U[D2,U?]|D|?p,(63)
where A=U[D,U?]and{D,A}=D A+A D.For this to make sense we must have[D,U?] bounded of course.It is important for us that we can evaluate this using the Wodzicki residue when D is an operator of order1on a manifold.Note that when this is the case,U[D2,U?]is a?rst order operator,and the contribution from this term will be from the zero-th order part of a?rst order operator.
4STATEMENT AND PROOF OF THE THEOREM17 4Statement and Proof of the Theorem
4.1Statement
Theorem9(Connes,1996)Let(A,H,D,c)be a real,(p,∞)-summable noncommutative geometry with p≥1such that
i)A is commutative and unital;
ii)πis irreducible(i.e.only scalars commute withπ(A)and D).
Then
1)The space X=Spec(
12 X R√
.(65)
(4π)p/2Γ(p/2+1)
Remark:Since,as is well known,every spin manifold gives rise to such data,the above theo-rem demonstrates a one-to-one correspondence(up to unitary equivalence and spin structure preserving di?eomorphisms)between spin structures on spin manifolds and real commutative geometries.
4.2Proof of1)and2)
Without loss of generality,we will make the simplifying assumption thatπis faithful on A. This allows us to identify A withπ(A)?B(H),and we will simply write A.Asπis a?-homomorphism,the norm closure,
A)is a compact,Hausdor?space.Since A is dense in its norm closure,each state on A(de?ned with respect to the C?norm of A)extends to a state on the closure,by continuity.Recall that we are assuming that A is imbued with a norm such that the C?-condition is satis?ed for elements of A;thus here we mean continuity in the norm. Hence Spec(A)=Spec(
A～=C(X).So let p∈
4STATEMENT AND PROOF OF THE THEOREM18 So(1?2p)[D,p]=0implying that[D,p]=0.By the irreducibility ofπ,we must have p=1 or p=0.Hence A contains no non-trivial projections and X is connected.Note that the irreducibility also implies that[D,a]=0unless a is a scalar.Also,as there are no projections, any self-adjoint element of A has only continuous spectrum.
The reader will easily show that equation(64)does de?ne a metric on X,[20].The topology de?ned by this metric is?ner than the weak?topology,so functions continuous for the weak?topology are automatically continuous for the metric.Furthermore,elements of A are also Lipschitz,since for any a∈A, [D,a] ≤1,we have|a(x)?a(y)|≤d(x,y).Thus for any a∈A we have|a(x)?a(y)|≤ [D,a] d(x,y).Later we will show that the metric and weak?topologies actually agree.This will follow from the fact that A,and so
A,which is equivalent to the metrizability of X. This will complete the proof of1)and2),but it will have to wait until we have learned some more about A.The last point of2)is that the metric is invariant under unitary transformations. That is if U:H→H is unitary
[U D U?,UaU?]=U[D,a]U?? [U D U?,UaU?] = [D,a] .(67) So while D will be changed by a unitary change of representation
D:=Dπ→U D U?:=D UπU?=Dπ+U[Dπ,U?],(68) commutators with D change simply.For this reason,when we only need the unitary equivalence class ofπ,we drop theπ,and write??D(A)forπ(??(A)),where the D is there to remind us that this is the representation of??(A)induced by D and the?rst order condition.
4.3Proof of3)and remainder of1)and2)
Before beginning the proof of3),which will also complete the proof of1)and2),let us outline our approach,as this is the longest,and most important,portion of the proof.
4.3.1Generalities.This section deals with the various bundles involved,their Hermi-tian structures and their relationships.We also analyse the structure of??D(A)andΛ?D(A), particularly in relation to Hochschild homology.
4.3.2X is a p-dimensional topological manifold.We show that the elements of A involved in the Hochschild cycle
π(c)=Γ= i a i0[D,a i1]···[D,a i p](69)
provided by the axioms generate A.This is done in two steps.Results from4.3.1show that Γis antisymmetric in the[D,a i j],and this is used to show that?1D(A)is?nitely generated by the[D,a i j]appearing inΓ.The second step then uses the long exact sequence in Hochschild homology to show that A is?nitely generated by the a i j and1∈A.In the process we will also show that X is a topological manifold.
4.3.3X is a smooth manifold.We show here that A is C∞(X),and in particular that X is a smooth manifold.After proving that the weak?and metric topologies on X agree,we show that A is closed under the holomorphic functional calculus,so that the K-theories of A and C(X)agree.At this point we will have completed the proof of1)and2).
4.3.4X is a spin c manifold.The form of the operators D,|D|and D2is investigated. The main result is that D2is a generalised Laplacian while D is a generalised Dirac operator,in the sense of[27].This allows us to show that??D(A)is(at least locally)the Cli?ord algebra of
4STATEMENT AND PROOF OF THE THEOREM19 the complexi?ed cotangent bundle.This is su?cient to show that the metric given by equation (64)agrees with the geodesic distance on X.As the representation of??D(A)is irreducible, we will have completed the proof that X is a spin c manifold.
4.3.5X is spin.It is at this point that we utilise the real structure.Furthermore,we reformulate Connes’result to allow a representation of the Cli?ord algebra of an inde?nite metric.This will necessarily involve a change in the underlying topology,which we do not investigate here.
4.3.1Generalities.
The axiom of?niteness and absolute continuity tells us that H∞=∩m≥1Dom D m is a?nite projective A module.This tells us that H∞～=e A N,as an A module,for some N and some e= e2∈M N(A).Furthermore,from what we know about A and Spec(A),H∞is also isomorphic to a bundle of sections of a vector bundle over X,say H∞～=Γ(X,S).These sections will be of some degree of regularity which is at least continuous as A?C(X).This bundle is also imbued with an Hermitian structure(·,·)E:H∞×H∞→A such that(aψ,bη)S=a?(ψ,η)S b etc,which provides us with an interpretation of the Hilbert space as H=L2(X,S,? (·,·)|D|?p).We will return to the important consequences of the Hermitian structure and the measure theoretic niceties of the above interpretation later.
As mentioned earlier,A is a Frechet space for the locally convex topology coming from the family of seminorms
a , δ(a) , δ2(a) ,...?a∈A,δ(a)=[|D|,a].(70) Note that the?rst semi-norm in this family is the C?-norm of A,and thatδn(a)makes sense for all a∈A by hypothesis.As the?rst semi-norm is in fact a norm,this topology is Hausdor?. In fact our hypotheses allow us to extend these seminorms to all of??D(A),and it too will be complete for this topology.
Now let us turn to the di?erential structure.The?rst things to note are that DH∞?H∞, AH∞?H∞and [D,a] <∞?a∈A.The associative algebra??D(A)is generated by A and [D,A],so??D(A)H∞?H∞.In other words
??D(A)?End(Γ(X,S))～=End(e A N)～={B∈M N(A):Be=eB}.(71) The most important conclusion of these observations is that??D(A)and so?1D(A)are?nite projective over A,and so are both(sections of)vector bundles over X,the former being(the sections of)a bundle of algebras as well.Moreover,the irreducibility hypothesis tells us that Γ(X,S)is an irreducible module(of sections)for the algebra(of sections)??D(A).
The?nite generation of??D(A)can be used to show that the algebra A is?nitely generated. We will then be dealing with a?nitely generated algebra and this can be used to show that the weak?and metric topologies agree.When we have completed the proofs of these matters, 1)will have been proved completely.
So what is??D(A)?The central idea for studying this algebra is the?rst order condition.When we construct this representation of??(A)fromπand A using D,the?rst order condition forces us to identify the left and right actions of A on??D(A),at least in the commutative case.Assuming as we are that the representation is faithful on A,we see that the ideal kerπis generated by the?rst order condition,
kerπ= ωa?aω a∈A,ω∈??(A)= ?rst order condition .(72)
4STATEMENT AND PROOF OF THE THEOREM20 So forω=δf of degree1and a∈A,aδf?δfa∈kerπand
(δf)(δa)+(δa)(δf)∈δkerπ.(73) Equation(72)ensures thatπ?b=0,as Image(b)=kerπ,so that we have a well-de?ned faithful representation of Hochschild homology
π:HH?(A)→??D(A).(74) If we write d=[D,·],we see that the existence of junk is due to the fact thatπ?δ=d?π, and that this may be traced directly to the?rst order condition.Let us continue to write ??D(A):=π(??(A))and also writeΛ?D(A):=π(??(A))/π(δkerπ),and note that the second algebra is skew-commutative,and a graded di?erential algebra for the di?erential d=[D,·]. Note that this notation di?ers somewhat from the usual,[6].Note that?1D(A)andΛ1D(A)are the same?nite projective A module,and we denote them both byΓ(X,E)for some bundle E→X,where as before we do not specify the regularity of the sections,only that they are at least continuous for the weak?topology and Lipschitz for the metric topology.
The next point to examine isδkerπ=Image(δ?b).This is easily seen to be generated by kerπ=Image(b)and graded commutators(δa)ω?(?1)|ω|ωδa,forω∈??(A)and a∈A. Thus the image ofπ?δ?b in??D(A)is junk,and this is graded commutators.As elements of the form appearing in equation(73)generateπ(δkerπ),it is useful to think of equation(73) as a kind of‘pre-Cli?ord’relation.In particular,controlling the representation of elements of δkerπwill give rise to a representation of the Cli?ord algebra as well as the components of the metric tensor.More on that later.
To help our analysis,de?neσ:??(A)→??(A)byσ(a)=a for all a∈A andσ(ωδa)= (?1)|ω|(δa)ω,for|ω|≥0.Then,[2],we have b?δ+δ?b=1?σ,on??(A).Asπ?b=0,and Image(π?δ?b)=Junk,we have Image(π?(1?σ))=Junk.
Thus while
π(??(A))=??D(A)～=??(A)/ Image(b) ,(75) passing to the junk-free situation gives
??(A)/ Image(b),Image(1?σ) ～=Λ?D(A)～= ??(A).(76) It is easy to see that b(1?σ)=(1?σ)b,so that ker b is preserved by1?σ.In fact,1?σsends Hochschild cycles to Hochschild boundaries.For if bc=0for some element c∈C n(A), then
(1?σ)c=(bδ+δb)c=bδc(77) which is a boundary.So ker b is mapped into Image(b)under1?σand so when we quotient by Image(1?σ)we do not lose any Hochschild cycles.
So,πdescends to a faithful representation of Hochschild homology with values inΛ?D(A).In general,the Hochschild homology groups of a commutative and unital algebra contain ??(A) as a direct summand,[2],but we have shown that in fact
HH?(A)～=Λ?D(A)～= ??(A).(78) This is certainly a necessary condition for the algebra A to be smooth,but more important for us at this point is that all Hochschild cycles are antisymmetric in elements of?1D(A).In particular,π(c)=Γ=0inΛp D(A)and is totally antisymmetric.
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第一章 公司简介 主营大码男童装，年龄阶段为8-15岁 第二章 店铺及平台简介 天猫- XXX旗舰店天猫- XXX旗舰店 第三章 部门人事架构 客服主管售前客服售后客服CRM专员
第四章 部门规章制度 一、目的 更好地规范在职人员的工作行为及高压线，使日常工作流畅化 二、范围 客服部所有在职人员 三、内容 １．上班时间【售前】 早班客服：０８：３０－１６：３０ 晚班客服：１5：００－23 ：００【在家值班】【周日在家值班】 上班时间【售后】 统一班次：０9：30 －１8：３０【中午轮流用餐】 ＰＳ：每月休４天，法定假期轮休，按排班表进行轮班，若有突发情况，可与当日其它同事换班； ２．薪资说明 售前客服薪资组成架构：底薪＋提成【绩效考核分数决定提成点】＋奖金 售后客服薪资组成架构：底薪＋绩效【绩效考核分数判定绩效百分比】＋奖金３．行为规范 １）．不得穿着暴露、不得着拖鞋上班，违者第一次警告，第二次扣绩效分２分
营销类书籍排行榜 【篇一：营销类书籍排行榜】 no.10 《从谷底到山巅》top销售之路 大学刚毕业的农村女孩张华、拥有2年销售经验的冯刚、工作4年 的工业原料销售员赵凯、从事销售7年的前销售冠军秦超、全球知 名企业的销售总监杰克——一5位资历、追求各不相同的销售人员 因不满现状而进入传说中神秘的“金牌销售大学”，遭遇到传奇教练 陶浦(top)的魔鬼训练……但都在一年后缔造了属于自己的销售奇迹！你将“偷学”到支付高昂学费才学得到的顶级销售秘籍，火速晋级为 受人敬仰的top销售！不但给新销售员迅速成为王牌销售员指出速 通之路，更给老销售员提供了突破工作瓶颈的必胜法宝！ no.9 《99%的人都用错了销售技巧》 日本销售大王让你业绩翻五倍连续8年成为“销售大王”的河濑和幸 亲身传授销售诀窍，助你业绩翻5倍。作者42岁时从公司的一名员 工调职为一名销售员，从最初的毫无业绩到现在的与各大公司签订 合作协议的自由销售员，他闯下了一片属于自己的天地。现在作者 能够在2个小时内向顾客成功推销300瓶价值4000日元的美容液，在一天内卖掉50台8000日元的自行车，再加上200瓶价值2300 日元的橄榄油，他的销售手段令店内员工瞠目结舌。销售大王的独 家销售技巧，让你轻松成为一名优秀的销售员。 no.8 《成交高于一切》 大客户销售十八招没有成交，谈何销售？成交是营销的终极目的， 也是企业生存的命脉。在销售活动中，永远都只有两个硬道理：第一，卖出去；第二，卖上价。本书围绕“成交”这一概念，运用四维 成交法，展开大客户销售十八招，招招紧扣成交，招招落到实处。 其目的就在于帮助销售人员切实练好基本功，拒绝失败的借口，真 正做到用业绩说话，同时也有力地解决企业中普遍存在的“中场盘带 过多，欠缺临门一脚”的问题。本书方法重于理论，易教、易学、易 复制。实战、有效、会做——成交才是硬道理！ no.7 《顶尖导购的秘密》 （60位一线导购，从未外传的销售绝招，你来，只能告诉你）顶尖 导购不仅懂得顾客的购买心理，更懂得根据顾客的反应适度地应对 进退，适时地激发调动起顾客的购买欲望。顶尖导购的销售，绝不
五大销售技巧和话术 要想成为一名出色的推销员，掌握一些推销技巧是必不可少的，那么，有效实用的推销技巧有哪些?下文就介绍了推销高手们总结的五大推销技巧，可供参考! 步骤/方法 1.推销技巧一：厉兵秣马 兵法说，不打无准备之仗。做为销售来讲，道理也是一样的。很多刚出道的促销员通常都有一个误区，以为销售就是要能说会道，其实根本就不是那么一回事。记得那时候我们培训了将近一个月，从产品知识到故障分析，从企业历史到销售技巧，每一个环节都反复练习，直至倒背如流。那时候我们同事之间经常互相打趣说咱都成了机器人了。我记得当时为了调试出一个最佳音乐效果，一没有顾客在场，我就专心致志地一个键一个键的反复试验，持续了将近一个星期，终于得到了自己满意的效果。 每次轮到自己休息，我总喜欢到各个卖场去转转：一来调查一下市场，做到心中有数。现在的顾客总喜欢讹促销员，哪里哪里有多么便宜，哪里哪里又打多少折了，如果你不能清楚了解这些情况，面对顾客时将会非常被动。二来可以学习一下别的促销员的技巧，只有博采各家之长，你才能炼就不败金身! 2.推销技巧二：关注细节 现在有很多介绍促销技巧的书，里面基本都会讲到促销员待客要主动热情。但在现实中，很多促销员不能领会到其中的精髓，以为热情就是要满面笑容，要言语主动。 其实这也是错误的，什么事情都要有个度，过分的热情反而会产生消极的影响。 热情不是简单地通过外部表情就能表达出来的，关键还是要用心去做。所谓精诚所至，金石为开!随风潜入夜，润物细无声，真正的诚就是想顾客所想，用企业的产品满足他们的需求，使他们得到利益。 3.推销技巧三：借力打力 销售就是一个整合资源的过程，如何合理利用各种资源，对销售业绩的帮助不可小视。作为站在销售第一线的促销员，这点同样重要。 我们经常在街头碰到骗子实施诈骗，其中一般都有一个角色—就是俗称的托，他的重要作用就是烘托气氛。当然，我们不能做违法的事，但是，我们是不是可以从中得到些启发呢?我在做促销员的时候，经常使用一个方法，非常有效，那就是和同事一起演双簧。特别是对一些非常有意向购买的顾客，当我们在价格或者其他什么问题上卡住的时候，我常常会请出店长来帮忙。一来表明我们确实很重视他，领导都出面了，二来谈判起来比较方便，只要领导再给他一点小实惠，顾客一般都会买单，屡试不爽!当然，如果领导不在，随便一个人也可以临时客串一下领导。关键是要满足顾客的虚荣心和爱贪小便宜的坏毛病。
SPIN销售四大模式-方法和技巧 SPIN技法由四种类型的提问构成，每一种提问都有不同的目的。以销售一个自动仓储设备系统为例（能解决买方可能存在的仓储能力不足、出错机率高、服务不及时等问题），卖方可用以下的问题组合来发掘和引导潜在客户（如一个物流中心）的需求和购买决定： 一、有关现状的提问(Situation Questions)。了解有关客户组织与现状的背景信息，如：现在货物仓储采用什么作业方式？共存储多少不同种类的货物？高峰期最多有多少产品需要仓储？ 二、有关问题的提问(Problem Questions)。发现和理解客户的问题、困难和不满，如：目前的仓储能力您是否满意？货物存储品种太多，差错率高吗?高峰期的仓储服务跟得上吗？ 三、有关影响之提问(Implication Questions)。发掘问题不解决将给客户带来的不利后果，如：仓储能力有限对成本控制和业务增长有何影响？仓储差错会不会影响到营运效率和客户满意度？高峰时货 物不能及时处置会有什么不利？ 四、有关需求与回报之提问(Need-Payoff Questions)。取得客户对于解决问题后的回报与效益的看法，将讨论推进到行动和承诺阶段，如：如果仓储能力得以充分利用，可增加多少收入？您考虑过用更先进的自动仓储系统来消除差错吗？高峰时的及时服务能为您带来什 么正面影响？
我们还可以从下面这样一段销售对话案例中进一步了解SPIN技法： 卖方：你们现在的复印机使用得如何？有什么不满意的地方？ 买方：没什么不满意，用得挺好。 卖方：影印效果是不是令人满意呢？ 买方：就是有时复印图像时黑黑的。 卖方：你们经常复印有图像的文件吗？ 买方：是的，尤其在投标中，70%的文件都有图像。 卖方：用这些黑黑的图像会对你们的投标有什么影响吗？ 买方：当然，这种复印质量很影响我们中标的。 卖方：如果单单因为图像质量差而失了标，你觉得这意味着什么？ 买方：我们从来不敢去这样想。 卖方：你们现在有什么解决办法来解决这个问题呢？ 买方：关键的投标我们都拿出去印。 卖方：那这样做在时间上来得及吗？ 买方：一般还可以。
怎样提高销售成交?反客为主,用提问引导你的客户。用提问引导你的客户 所谓用提问引导你的客户，就是先陈述一个事实，然后再针对这个事实发问，让对方给出相应信息。确实，以提问的方式引导客户可以起到让对方易于接受的作用。 那么，销售人员如何用提问的方式引导客户呢？下面我们介绍几种方法： 一、肯定性诱导提问 肯定性诱导提问法是对肯定性说法、诱导性说法以及提问的说话方法三种方式的同时运用。首先是肯定性说法，即使用正面性用语——“很受人欢迎的”。其次是诱导性说法——“这种产品有大小两种，不知您愿选择哪一种，不过我想是不是大的比较好呢？”最后是提问的方法——“这位先生，您要如何使用呢？” 二、与类似问题相比较 简单地说，就是利用客户的随身物品作为一个实际的例子来说服客户。 比如，小陈是学习软件的推销员。有一次，一位客户在看了产品简介之后，还想要看看所要购买软件的内容：“我应根据所要买的产品内容是 否适合我来确定买不买，对不对？” 小陈：“您说得没错，可是出版这本书的出版社非常有名，我希望您能相信一流的出版社。先生，可以问一下您的笔记本电脑是什么品牌吗？” 客户：“是国产产品。” 小陈：“哦！您买这台电脑的时候是否先把它拆开看一下里面的部件呢？” 客户：“没有。” 小陈：“我想你在看过电脑后，即便认为电脑质量没问题，也是因为相信这家公司的信誉和服务才买下它的。同样，买汽车的时候你也不能把 车子拆开看一下引擎吧？还有买药品的时候你无法从100元一盒的药品中，挑选其中一颗拿起来品尝，试试其功效后，才决定购买与否。虽然不同品牌的产品，也有可能有许多的价
格差异，但若是你分不出品质的好坏，我认为你应该依据厂商的信誉来购买。买这部学习软件也是一样，您应信任出版商的声誉。” 三、拆分问题引导 在推销价格昂贵的产品时，这个方法十分有效。一位销售人员经常在推销一套价格不菲的家具时，多次利用拆分问题来说服客户客户：“这件家具太贵了。” 销售人员：“您认为贵了多少？” 客户：“贵了1000多元。” 销售人员：“那么现在就假设贵了1000元整。”这时销售人员拿出了随身带的笔记本，在上面写下1000元给目标客户看。 销售人员：“先生，你想这套家具你肯定至少打算能够用10年再换吧？” 客户：“是的。” 销售人员：“那么，依照你所想的也就是每年多花了100元，您指的是不是就是这样？” 客户：“对，我就是这样认为的。” 销售人员：“1年100元，每个月该是多少钱？” 客户：“哦！每个月大概就是8块多点吧！” 销售人员：“好，就算是8.5元吧。你每天至少要用两次吧，早上和晚上。” 客户：“有时更多。” 销售人员：“我们保守估计为l天2次，那也就是说1个月你将用60次。所以，假如这套家具每月多花了8.5元，那每次就多花了不到0. 15元。” 客户：“是的。” 销售人员：“那么每天不到1毛5分，却能让你的家变得利落和整洁，让你不再为东西没合适地方放而苦恼、发愁，而且还起到装饰作用，你不觉得很划算吗？”
销售技巧和话术培训 销售员是企业销售的主力军之一，为企业的发展进步做出巨大贡献，好的销售员能成功树立商品的形象，并且能销售出商品，形成二次顾客。所以，良好的销售员技巧培训是企业明智的选择。 徐清祥老师的销售技巧和话术培训能深入的讲解销售的技巧，并与实际相结合，让促销员受益匪浅，为企业形成良好销售回路。 主讲老师：徐清祥 课程时间：2天 课程背景： 在竞争激烈的销售市场上，精确掌握销售循环前、中、后的销售技巧与全方位的客户服务，是赢得最佳商机的关键。从销售的基本理论方法、实战技巧到销售的十大步骤、销售人员的自我管理，为销售人员专业技能训练提供了一整套的解决方案。如果你要掌握销售人员应具备的正确与基础的技巧，并吸收专家和成功者的经验与智慧，以资借鉴，减少自己摸索的时间、精力和犯错的机会，那就快走近我们的课程，唯有行动才能解决问题。 课程预定: 0371- 课程提纲： 第一部分：自我管理与规划能力 一、准备: 1、精神状态的准备： 静坐; 良好的睡眠.
2、体力的准备: 饮食; 运动. 3、专业知识的准备: 商品名称; 商品内容; 使用方法; 商品特征; 售后服务; 交货期与交货方式; 价格与付款方式; 研究同行业竞争中的商品; 材料来源与生产过程; 相关商品. 4、广泛知识的准备: 心理学; 管理学; 人际关系学; 行销学; 国学; 公众演说. 5、对顾客的准备.
使自己达到巅峰状态: 改变自己的动作: 1)自我确认; 2)自我放松. 二、改变自己的心态: 学习的心态; 积极的心态; 老板的心态; 感恩的心态; 付出的心态; 持之以恒的心态; 平衡的心态; 自律的心态; 宽容的心态; 拒绝找借口的心态. 三、问自己一些好的问题: 发生这些事对自己有什么好处? 自己学到了什么? 自己现在可以做什么? 第二部分：塑造良好的职业形象能力。 一、拥有良好的形象:
实用文档 . 《SPIN销售巨人》读后感 这次读的书是《SPIN销售巨人——大订单销售训练手册》。感觉理论篇和 实践篇内容差不多。 这本书是针对大订单销售，也就是说我们应该先分清什么是大订单，什么是小订单。书中说通常一个电话就可以敲定或者交易金额很少的销售就是小生意，或者称之为小订单。根据我的理解，现在除了零售业，别的订单生意差不多都可以算成是大订单销售了。这样更突显出对针对大订单销售技巧的需求了。 销售分为四步：初步接触、需求调查、能力证实和晋级承诺。其中需求调查是最关键的，由这个步骤可以明确知道客户的购买欲望和需求，以便达成生意。SPIN包括四个环节：背景问题（Situation Question）、难点问题（Problem Question）、暗示问题（Implication Question）和需求—效益问题（Need-Payoff Implication）。书中说SPIN可以帮助大订单销售，是所有大客户销售经理必须掌握的专业技能。现在许多国际化大公司都在使用这套方法。 书中说，最成功的销售人员就是那些提问最多的人，但不是任何问题。成功的销售人员会提出很精明的问题，并且趋向于以一个特殊的序列提这些问题。记得之前罗钧总开的销售会议上也说过，要抓住客户的痛点，痛点找准了，离订单成功也就不远了。这样看来，会提问成为一名优秀的销售人员的必备武器，虽然现在还不具备，但平日多加练习，善于总结，这项技能应该也是可以获得的。 读完书的感觉是这本书一直在批判之前的理论或者说结论。确实从这本书中能学到好多知识，但对任何问题都要采取一分为二的态度，任何理论之所以能传播开来，都有一定的道理。书中的理论也应该经过不断的练习才能成为自己的财富，才能为自己所用。不经过练习，读过的知识也就只是读过了，怎样获得晋级承诺，什么情况是暂时中断，我们都无从知晓。 读书的时候有种很强烈的感觉，这本书是给有一定销售知识基础的人准备的，其中说的开场白技巧、收场白技巧等等，我还不知道具体都是指什么。有时读起来有点费力或者说有些地方还不能理解得很透彻。突然觉得自己的知识好少，需要学的地方还有很多很多。 陈杭（Sara）
SPIN?销售技巧概览
在当今的情势下，我们不得不开始学会倾听客户的声音，而不只是不停地说。这说起来容易，但做起来难！然而，藉由SPIN?来提升销售人员的顾问式销售技巧，使销售流程更为系统化，为RSA在竞争激烈的保险市场上建立了独特的竞争优势。 RSA 销售人员能做的最有说服力的事就是向客户展示他们能提供客户想要的。 您或许会问，这是理所当然的吧？是。但为什么典型的销售人员却花95％ 的时间在做别的事情？我们独一无二涉及约 40,000 个企业对企业（B to B）销售拜访的观察研究，揭露了成功销售拜访中销售人员使用的行为，但这些行为实际被运用的比例却非常地少。 好消息是荷士卫（Huthwaite）确切地知道你的销售人员应该做些什么以及如何做。SPIN?销售技巧是世界上最被广泛验证成功的销售模式，它提供了一个框架让销售人员去构建销售的对话，去发掘和发展客户需求，并且为产品独特之处创造价值。它给予销售人员最大化展现解决方案效能的技能。 许多的销售技巧来来去去，偶尔有些宣称以研究为基础。但在世界上最杰出的公司中，SPIN?仍是最被广泛应用并且能够产生最大成效的销售技巧。SPIN?建立在坚实且历久弥新的基础上，这是为什么我们每年用SPIN?帮助12,000名销售代表发展他们销售技巧的原因。 我们帮您解决以下这些问题： ?“客户就是不明白为什么我们价格比较高。”?“我们像一般商品一样被对待，而不是高价 值的供应商或者合作伙伴。” ?“我们不擅长与专业的买家打交道。” ?“我们的销售人员出身技术背景，所以他们 通常喜欢谈论产品的技术特点，而不是挖 掘客户的业务问题。” ?“我们要确保我们的销售人员为我们产品的 独特区分点创造价值。” ?“我们的人知道他们在做什么，但我们需要 再加强他们的技巧，并让他们从个人到团 队以至于整个公司保持一致。” ?“我们需要能够贯穿公司经营的共同语言以 及衡量成功方法的技巧。” ?“我们需要的不光是培训。” ?“我们要提高客户体验的质量。” 通过提供： 一系列重要的口语技巧让销售人员能够有效挖掘和发展客户的需求，随后展现产品价值并做出高冲击力的利益陈述。曾经有一位客户这么说过：“SPIN?是企业对企业销售（B to B）唯一的方法，这些行为要求销售人员远离以产品为导向的推销行为和不灵活的销售话术，专注在以客户为导向、以价值为驱动的顾问式销售对话。 带给您： y创造价值一致可靠的方法。 y更多利润率较高的销售。 y更高水平的客户满意度，忠诚度和保留率。 y缩短销售周期。 y更少的反对意见。 y用共同的语言分析、计画以及报告销售活动。
酒店销售技巧与话术 酒店在日常经营过程中，是最需要客流量的一种行业之一，对于酒店来说，如果客流量断了，那么将会给整个企业带来巨大的损失，而酒店销售技巧和话术是维持客户不销售，形成二次甚至多次消费的重要条件。 一、酒店客房销售技巧和话术 1. 把握客人的特点 不同的客人有不同的特点，对酒店也有不同的要求。总台人员要注意客人的衣着打扮、言谈举止以及随行人数等方面把握客人的特点(年龄、性别、职业、国籍、旅游动机等)，进而根据其需求特点和心理，做好有针对性的销售。 (1)商务客人通常是因公出差，对房价不太计较，但要求客房安静，光线明亮(有床头灯)，办公桌宽大，服务周到、效率高，酒店及房内办公设备齐全[如安装有宽带网和直拨电话以及电脑、传真机等现代化设备]，有娱乐项目。 (2)旅游客人要求房间外景色优美，房间内干净卫生，但经济承受能力有限，比较在乎房间价格。 (3)度蜜月的客人喜欢安静、不受干扰且配有一张大床的双人房。 (4)知名人士、高薪阶层及带小孩的父母喜欢住套房。 (5)年老的和有残疾的客人喜欢住在靠近电梯和餐厅的房间。 (6)朋友多的喜欢住多床房 2. 熟悉客房 总台员工在销售客房时应明确自己是在销售客房，而非销售价格。因此总台员工应熟悉客房，在销售时，应多做客房的具体描述，而尽量少提及价格。具体如下： (1)总台人员在销售客房时，必须对客房做适当的描述，以减弱客房价格的分量，突出客房能够满足客人需要的特点。 (2)比如，不能只说：“一间198的客房，您要不要?”而应说，“一间刚装修过的、宽敞的房间”“一间舒适、安静、能看到临街景色的客房”“一间具有有民族特色的、装修豪华的07客房”，等等。这类形容词是无穷无尽的，只有这样容易为客人所接受。 (3)要准确地描述客房，必须首先了解客房的特点。这是对总台员工的最基本要求之一，比如带他们参观客房，并由专人讲解客房的特点，以加深印象。 3. 从高到低报价 在总台销售服务过程中，不可避免要谈到房价，向客人报房价也需要一定的技巧，即要从高到底报价。 (1)总台人员可根据客人的特点，向他推荐两中乃至三种不同价格的、可供比较的客房，让客人自己选择。 (2)如果是推荐一种客房，就会使客人失去比较的机会，推出价格范围，应考虑到客人的特点。 (3)一般来说，报房价由高价格退到较底价格比较适宜。例如：可以说“配备先进麻将机和电脑设备，装修宽敞的是208元”“靠近绿色菜圃，麻将房隔音效果极佳的客房是188元”“进出
销售技巧和话术经典语 1. 对销售人员来说，销售学知识是必须掌握的，没有学问作为根基的销售，只能视为投机，无法真正体会销售的妙趣。 2. 一次成功的推销不是一个偶然发生的故事，它是学习计划以及一个销售人员的知识和技巧运用的结果。 3. 推销完全是常识的运用，但只有将这些为实践所证实的观念运用在积极者身上，才能产生效果。 4. 在取得一鸣惊人的成绩之前，必须做好枯燥的准备工作。 5. 推销前的准备、计划工作，决不可疏忽轻视，有备而来才能胜券在握。准备好推销工具、开场白，该问的问题、该说的话，以及可能的回答。 6. 事前的充分准备和现场的灵感所综合出来的力量，往往很容易瓦解坚强对手而获得成功。 7. 最优秀的销售人员是那些态度最好、商品知识最丰富、服务最周到的销售人员。 8.对与公司有关的资料、说明书、广告等，均必须努力研讨、熟记。同时要收集竞争对手的广告、宣传资料、说明书等加以研讨、分析，以便做到知己知彼，采取相应对策。 9.销售人员必须多读些有关经济、销售方面的书籍、杂志，尤其必须每天阅读报纸，了解国家、社会消息、新闻大事，拜访客户这往往是最好的话题，且不致于孤陋寡闻，见识浅薄。 10. 获取订单的道路是从寻找客户开始的，培养客户比眼前的销售量更重要。如果停止补充新顾客、销售人员就不再有成功之源。 11. 对客户无易的交易也必然对销售人员有害，这是最重要的一条商业道德准则。 12.在拜访客户时，销售人员应当奉行的准则是即使跌倒也要抓一把沙，意思是销售人员不能空手而归，即使推销没有成交，也要让客户为你介绍一位新客户。 13. 选择客户、衡量客户的购买意愿与能力，不要将时间浪费在犹豫不决的人身上。 14. 强烈的第一印象的重要规则，是帮助别人感到自己的重要。 15. 准时赴约，迟到意味着：我不尊重你的时间，迟到是没有任何借口的。 假使无法避免迟到的发生，你必须在约定时间之前打通电话过去道歉，再继续未完成的推销工作。 16. 向可以做出购买决策的人推销，如果你的销售对象没有权力说买的话，你是不可能卖出什么东西的。 17. 每个销售人员都应当认识到，只有目不转睛地注视着你的可户，销售才能成功。 18、有计划且自然的接近客户，并使客户觉得有益处，而能顺利进行商洽，是销售人员必须事前努力准备的工作与策略 19、销售人员不可能与他拜访的每一位客户达成交易，他应当努力去拜访更多的客户来提高成交百分比。 20、要了解你的客户，因为他们决定着你的业绩。 21、在成为一个优秀的销售人员之前，你要成为一个优秀的调查员，你必须
SPIN销售法 目录 什么是SPIN销售法？ SPIN销售模式的4个步骤 特点、优点与购买利益 什么是SPIN销售法？ SPIN销售模式的4个步骤 特点、优点与购买利益 编辑本段什么是SPIN销售法？ SPIN销售法是尼尔·雷克汉姆（Neil Rackham）先生创立的。尼尔·雷克汉姆先生的SPIN销售法是在IBM和Xerox等公司的赞助下通过对众多高新技术营销高手的跟踪调查提炼完成的。 营销活动一般要经历4个周期阶段：1、开场启动阶段；2、调研交流阶段；3、能力展示阶段；4、买卖承诺阶段。只有上一个阶段完成了才能进入到下一个阶段，但是第二阶段即调研交流阶段是最关键的，在这一阶段的表现将在很大程度上决定营销成功与否，很多营销失败就是营销人员将重点放在了其他阶段而在第二阶段浅尝辄止。 SPIN销售法提供了一种巧干的高效系统方法。 SPIN销售法其实就是情景性(Situation)、探究性(Problem)、暗示性(Implication)、解决性(Need-Payoff)问题四个英语词组的首位字母合成词，因此SPIN销售法就是指在营销过程中职业地运用实情探询、问题诊断、启发引导和需求认同四大类提问技巧来发掘、明确和引导客户需求与期望，从而不断地推进营销过程，为营销成功创造基础的方法。 SPIN销售法教人如何找到客户现有背景的事实，引发客户说出隐藏的需求，放大客户需求的迫切程度，同时揭示自己策的价值或意义。使用 SPIN 策略，销售人员还能够全程掌控长时间销售过程中客户细微的心理变化。 SPIN销售法从谈话提问技巧和谈话条理性角度另外提供了一种全新的营销理念和方法，并为不少欧美高新技术公司所倚重，财富100 强中的半数以上公司也利用它来训练营销人员．
1.销售技巧和话术经典语句一： 2.推销前的准备、计划工作，决不可疏忽轻视，有备而来才能胜券在握。准备好推 销工具、开场白，该问的问题、该说的话，以及可能的回答。 3.对与公司有关的资料、说明书、广告等，均必须努力研讨、熟记。同时要收集竞 争对手的广告、宣传资料、说明书等加以研讨、分析，以便做到知己知彼，采取相应对策。 4.强烈的第一印象的重要规则，是帮助别人感到自己的重要。准时赴约，迟到意味 着：我不尊重你的时间，迟到是没有任何借口的。假使无法避免迟到的发生，你必须在约定时间之前打通电话过去道歉，再继续未完成的推销工作。 5.每个销售人员都应当认识到，只有目不转睛地注视着你的客户，销售才能成功。 6.有计划且自然的接近客户，并使客户觉得有益处，而能顺利进行商洽，是销售人 员必须事前努力准备的工作与策略 7.销售人员不可能与他拜访的每一位客户达成交易，他应当努力去拜访更多的客户 来提高成交百分比。 8.要了解你的客户，因为他们决定着你的业绩。 9.相信你的产品是销售人员的必要条件，这份信心会传给你的客户，如果你对自己 的商品没有信心，你的客户对他自然也没有信心，客户与其说是因为你说话的逻辑水平高而被说服，倒不如说他是被你的深刻信心所说服的。业绩好的销售人员经得起失败，部分原因是他们对于自己和所推销的产品有不折不扣地信心。
10.对于销售人员而言，最有价值的东西莫过于时间，了解和选择客户，是让销售人 员把时间和力量放在最有可能购买的人身上，而不是浪费在不能购买你的产品的人身上。 11.客户没有高低之分，却有等级之分，依客户等级确定拜访的次数、时间，可以使 销售人员的时间发挥出最大的效能。 12.接近客户一定不可千篇一律公式化，必须事先有充分准备，针对各类型的客户， 采取最合适的方式及开场白。 13.推销的机会往往是稍纵即逝，必须迅速、准确地判断，细心留意，以免错失良 机，更应努力创造机会。 14.把精力集中在正确的目标，正确的使用时间及正确的客户，你将拥有推销的老虎 之眼。 15.推销的黄金准则是你喜欢别人怎样对你，你就怎样对待别人，推销的白金准则是 按人们喜欢的方式待人。 16.让客户谈论自己，让一个人谈论自己，可以给你大好的良机去挖掘共同点，建立 好感并增加完成推销的机会 17.推销必须有耐心，不断的拜访，以免操之过急，亦不可掉以轻心，必须从容不 迫，察言观色，并在适当的时机促成交易。. 18.客户拒绝推销，切勿泄气，要进一步说服顾客并设法找出顾客拒绝的原因。再对 症下药。 19.为帮助顾客而销售，而不是为了提成而销售
SPIN销售工具 一、是什么： 针对大额产品销售具有金额相对较大、顾客心理变化大、做决定周期比较长等特点，美国Huthwaite公司的销售咨询专家尼尔·雷克汗姆与其研究小组分析了35000多个销售实例，与10000多名销售人员一起到各地进行工作，观察他们在销售会谈中的实际行为，研究了116个可以对销售行为产生影响的因素和27个销售效率很高的国家，耗资100万美元，历时12年，于1988年正式对外公布了SPIN模式——这项销售技能领域中最大的研究项目成果，她引起了人们对销售技巧认识的一次新的革命，推动了销售技巧的进一步完善。 二、为什么要使用这个工具： 实践表明，被培训过的人在销售额上比同一公司的参照组（没有被培训过）的销售员平均提高了17%。 三、SPIN四类问题的顺序如何使用: 1．背景问题。 即有关买方现在的业务和状况的事实或背景。如“你们用这些设备有多久啦”或“您能和我谈谈公司的发展计划吗”尽管背景问题对于收集信息大有益处，但如果过多，则会令买方厌倦和恼怒。因此，询问的时候要把握两个原则：（1）数量不可太多；（2）目的明确，问那些可以开发成为明确的需求，并是你的产品或服务可以解决的难题方面的问题。 2.难点问题。 即发现客户的问题、难点和不满，而卖方的商品和服务正是可以帮助客户解决他们的这些问题、难点和不满的问题。如“这项操作是否很难执行”或“你担心那些老机器的质量吗”难点问题为订单的开展提供了许多原始资料。因为难点问题有一定的风险性，所以许多缺乏经验的销售人员很难把握提问的时机。 3.暗示问题。 即扩大客户的问题、难点和不满，使之变得清晰严重，并能够揭示出所潜伏的严重后果的问题。如“这个问题对你们的远期利益有什么影响吗”或“这对客户的满意程度有什么影响吗”暗示问题就是通过积聚潜在顾客难题的严重性，使它大到足以付诸行动的程度。询问暗示问题的困难在于措辞是否恰当和问题的数量是否适中，因为它往往使潜在顾客心情沮丧、情绪低落。如果还没有提问过背景问题和难点问题，过早引入暗示问题往往使潜在顾客产生不信任甚至拒绝。 4．需求－效益问题。 销售人员通过问这类问题，描述可以解决顾客难题的对策，让顾客主动告诉你，你提供的这些对策让他获利多少。如“如果把它的运行速度提高10%对您是否有利呢”或“如果我们可以将其运行质量提高，那会给你怎样的帮助呢”这些都是典型的需求－效益问题。需求－效益问题对组织购买行为中的那些影响者最有效，这些影响者会在你缺席的情况下担当起你的角色，将你的对策提议交给决策者，并通过自身的努力给决策者施加一定的影响。销售人员最易犯的错误就是在积聚起问题的严重性之前过早地介绍对策，在潜在顾客没有认识到问题的严重性之前会为你的需求开发设置障碍。因此，问需求－效益问题的最佳时机是：在你通过暗示问题建立起买方难题的严重性后，而又在你描述对策之前。在每笔生意中，出色的销售人员较之一般销售人员所问的需求-效益问题要多10倍。 四、SPIN销售工具如何设计问题: 例子--背景问题(S) 了解买家的背景情况（为你能提供产品服务）可能的信息来源： 现在使用的是什么类型的存储补给系统？共存储多少不同种类的货物？在比较典型的时期中有多少产品要补给？仓库管理者分销服务的小册子——仓库管理者 例子--难点问题（P） 我们的产品或服务能提供更好对策的难题有可能存在这种难题的客户： 目前，正在使用的X型好的存储补给系统有那些不足的地方？例如：内存不足使用不方便员工产生抱怨……。 例子--内含型问题（I） 买方难题(我们可以解决的) 内含(使内含问题更加严重)： 员工产生抱怨对你生产会有什么影响呢？例如：质量受到影响压力可以导致员工抗拒竞争竞争中失去技术领先的地
概论 本章要点：跟单员的定义，工作界定，工作定位，工作特点，跟单员的素质，工作内容及知识，技能要求及跟单员工作的重要性。 一。跟单员的定义：(什么是跟单员？) 跟单员是指在企业运作过程中，以客户订单为依据，跟踪产品，跟踪服务运作流向的专职人员。(不能兼职，替代) 所有围绕着订单去工作，对出货交期负责的人，都是跟单员。 二。跟单员的工作界定： 跟单员广泛存在于订单型生产企业和进出口外贸企业中，跟单员的工作性质与特点随企业的规模与性质而有所区别，但跟单员总的来说是作为业务跟单与生产跟单而存在的。 1. 业务跟单：对客户进行跟进。尤其是已对本公司的产品已有了兴趣，有购买意向的人 进行跟进。以缔结业务，签定合同为目标的一系列活动。对外叫业务员or业务助理。 三。跟单员的工作定位：(工作性质) 跟单员是业务员。他的工作不仅仅是被动的接受订单，而是要主动的进行业务开拓，对准客户实施推销跟进，以达成订单为目标，既进行业务跟单。因此，跟单员要： (1) 寻找客户：通过各种途径寻找新客户，跟踪老客户。 (2) 设定目标：主要客户和待开发的客户。我们的工作着重点及分配的工作时间。 (3) 传播信息：将企业产品的信息传播出去。 (4) 推销产品：主动与客户接洽，展示产品，为获取订单为目的。 (5) 提供服务：产品的售后服务，及对客户的服务。 (6) 收集信息：收集市场信息，进行市场考察。 (7) 分配产品：产品短缺时先分配给主要客户。 跟单员是业务助理；跟单员在许多时候扮演业务经理助理的角色，他们协助业务经理接待， 管理，跟进客户，因此跟单员要： (1) 函电的回复： (2) 计算报价单 (3) 验签订单 (4) 填对帐表 (5) 目录，样品的寄送与登记 (6) 客户档案的管理 (7) 客户来访接待 (8) 主管交办事项的处理 (9) 与相关部门的业务联系跟单员是协调员：跟单员对客户所订产品的交货进行跟踪，即进行生产跟踪。跟踪的要点是生产进度，货物报关，装运等。因此，在小企业中，跟单员身兼数职，既是内勤员，又是生产计划员，物控员，还可能是采购员。在大企业，则代表企业的业务部门向生产制造部门催单要货，跟踪出货。 四。跟单员的工作特点： 跟单员的工作几乎涉及的企业的每一个环节，从销售，生产，物料，财务，人事到总务都会有跟单员
直入人心的八大销售技巧和话术 推销话术其实就是说服技巧，推销员通过信息传递，让对方改变观点的过程。怎样才能让对方改变态度呢？最有效的方法是抓住对方的心理需求，利用心理需求来制定说服策略，从而改变他的态度。关于人的需求和需求层次，可以参看《马斯洛需求层次论》。 销售话术的任务实际上是推销一种象征性满足人的心理的方式，这种需求往往是隐含的潜意识。高明的销售话术就是瞄准说服对象的潜意识，将潜意识转化为一种动力。这样就知道了卖啤酒其实卖的是文化，卖可口可乐其实卖的是活力和正宗，卖彩券、保险是卖的未来期望。消费者行为学认为，消费者行为在很大程度上取决于隐藏在它们内心的心理需要，而将销售话术与这些心理需要结合起来，就会打动他们，让他们改变态度。 第一大销售话术：安全感 人总是趋利避害的，内心的安全感是最基本的心理需求，用安全感来说服客户是最常用的销售话术。这种说服随处可见，比如保险销售话术中基本都是从安全保障为出发点来说服的。汽车销售话术中，说这种汽车的安全系统对于保证出行中的家庭很有效，对于买车的人肯定是一个有力的论点。比如卖房子，对客户说物价上涨、房价上涨，资金缩水，不如投资房屋来得安全。再比如卖设备说，购买这台设备，可以让客户的体验更好，吸引更多的客户，而如果不买，你的竞争对手就会买，会把你的客户抢走。 安全感的反面是恐惧感，如果安全感打动不了客户，那你不妨用恐惧感吓唬他一下。卖儿童智力玩具的说，不要让孩子输在起跑线上，就是一种吓唬；让客户观察皮肤里面的螨虫来推销化妆品，也是一种吓唬；日本一个保险公司推销员用录音机模拟死人到阴间和阎王的对话，讲述由于没有给家人购买保险，而受到惩罚的事情，更是吓唬。吓唬可能是最有效的推销话术。 第二大销售话术：价值感 每个人都希望自己的个人价值得到认可。汶川大地震中，有乞丐主动为灾区捐款，除了是善心之外，恐怕也有一份希望得到社会认可的潜意识。抓住价值感，也是的一个重点。劝说买保险，你可以说：“给家人买保险就是买平安，这是作为父亲和丈夫的职责。”、“这台设备用上以后，公司的工作效率会大大提高，这说明你这个设备部主任的慧眼识货。”。推销烤肉机“当丈夫拖着疲惫的身子回来，他多么渴望吃一顿美味可口的饭菜，当妻子将美味的烤肉端上来的时候，丈夫的心该有多幸福？”呵呵，如果你小嘴这样会说，那妻子如果不买，我都想鄙视她一下。 第三大销售话术：自我满足感
今麦郎公司物流部 管理手册 1
今麦郎公司制面事业部 物 流 部 管 理 手 册
目录 前言 第一章组织定位与职能 第二章部门管理细则……………………………………………… 一、物流部日常管理规定(重新拟定)………………… 二、物流部办公室5S管理制度………………………………… 第三章组织结构、岗位描述、考核方案…………………………… 一、物流部组织结构……………………………………………… 1、物流部组织结构图……………………………………… 2、外埠营运科结构图……………………………………… 二、物流部岗位职责描述………………………………………… 1、物流部部长岗位职 责…………………………………… 2、综合管理部经理岗位职 责…………………………………… 1
3、运输公司经理岗位职责………………………………… 4、营运部经理岗位职责……………………………… 5、外埠营运科长岗位职责 6、总部营运组长岗位职责………………………………… 7、信息管理组长岗位职责…………………………………… 8、信息员岗位职责…………………………………… 9、订单员岗位职责 10、办票员岗位职责…………………………… 11、订单调度员岗位职责…………………………………… 12、货物调度员岗位职责……………………………………… 三、物流部考核方案 1、物流部部长考核方案……………………………………… 2、综合管理部经理考核方案 3、运输公司经理考核方案 4、营运部经理考核方案 5、外埠营运科长考核方案 2
6、总部营运组长考核方案 7、信息管理组长考核方案 8、物流商考核方案 第四章物流业务管理规范…………………………………………… 一、订单管理规定………………………… 二、自提车管理规定………………………………………… 三、促销产品配发管理规定 第五章物流业务标准流程………………………………………………… 一、订发货流 程………………………………………………… 二、订单处理流 程……………………………………………… 三、司机装车流程 四、增值税票办理流 程…………………………………………… 五、运费报批流程 六、运价申请流程 七、物流商选择标准 八、招投标业务流程 第六章物流业务各类表单 3
销售技巧：五大销售技巧和话术 推销技巧一：厉兵秣马 兵法说，不打无准备之仗。做为销售来讲，道理也是一样的。很多刚出道的促销员通常都有一个误区，以为销售就是要能说会道，其实根本就不是那么一回事。记得那时候我们培训了将近一个月，从产品知识到故障分析，从企业历史到销售技巧，每一个环节都反复练习，直至倒背如流。那时候我们同事之间经常互相打趣说咱都成了机器人了。我记得当时为了调试出一个最佳音乐效果，一没有顾客在场，我就专心致志地一个键一个键的反复试验，持续了将近一个星期，终于得到了自己满意的效果。 每次轮到自己休息，我总喜欢到各个卖场去转转：一来调查一下市场，做到心中有数。现在的顾客总喜欢讹促销员，哪里哪里有多么便宜，哪里哪里又打多少折了，如果你不能清楚了解这些情况，面对顾客时将会非常被动。二来可以学习一下别的促销员的技巧，只有博采各家之长，你才能炼就不败金身! 推销技巧二：关注细节 现在有很多介绍促销技巧的书，里面基本都会讲到促销员待客要主动热情。但在现实中，很多促销员不能领会到其中的精髓，以为热情就是要满面笑容，要言语主动。其实这也是错误的，什么事情都要有个度，过分的热情反而会产生消极的影响。 热情不是简单地通过外部表情就能表达出来的，关键还是要用心去做。所谓精诚所至，金石为开!随风潜入夜，润物细无声，真正的诚就是想顾客所想，用企业的产品满足他们的需求，使他们得到利益。 推销技巧三：借力打力
销售就是一个整合资源的过程，如何合理利用各种资源，对销售业绩的帮助不可小视。作为站在销售第一线的促销员，这点同样重要。 我们经常在街头碰到骗子实施诈骗，其中一般都有一个角色—就是俗称的托，他的重要作用就是烘托气氛。当然，我们不能做违法 的事，但是，我们是不是可以从中得到些启发呢?我在做促销员的时候，经常使用一个方法，非常有效，那就是和同事一起演双簧。特 别是对一些非常有意向购买的顾客，当我们在价格或者其他什么问 题上卡住的时候，我常常会请出店长来帮忙。一来表明我们确实很 重视他，领导都出面了，二来谈判起来比较方便，只要领导再给他 一点小实惠，顾客一般都会买单，屡试不爽!当然，如果领导不在， 随便一个人也可以临时客串一下领导。关键是要满足顾客的虚荣心 和爱贪小便宜的坏毛病。 推销技巧四：见好就收 销售最惧的就是拖泥带水，不当机立断。根据我的经验，在销售现场，顾客逗留的时间在5-7分钟为最佳!有些促销员不善于察言观色，在顾客已有购买意愿时不能抓住机会促成销售，仍然在喋喋不 休地介绍产品，结果导致了销售的失败。所以，一定要牢记我们的 使命，就是促成销售!不管你是介绍产品也好，还是做别的什么努力，最终都为了销售产品。所以，只要到了销售的边缘，一定要马上调 整思路，紧急刹车，尝试缔约。一旦错失良机，要再度钩起顾客的 欲望就比较困难了，这也是刚入门的促销员最容易犯的错误。 推销技巧五：送君一程 销售上有一个说法，开发一个新客户的成本是保持一个老客户成本的27倍!要知道，老客户带来的生意远比你想象中的要多的多。 我在做促销员的时候，非常注意和已成交的顾客维持良好关系，这 也给我我带来了丰厚的回报。其实做起来也很简单，只要认真地帮 他打好包，再带上一声真诚的告别，如果不是很忙的话，甚至可以 把他送到电梯口。有时候，一些微不足道的举动，会使顾客感动万 分!
大客户营销技巧之 大客户及大客户营销概述 讲师：闫治民 课程大纲 第一章大客户及大客户营销概述 第二章质量型大客户的开发和沟通 第三章如何实现高效的项目谈判策略 第四章招投标过程管理与项目成功实施 第五章高绩效的大客户关系管理实效策略 第一章大客户与大客户营销概述 -什么是大客户? 作为大客户至少包含以下元素之一： (1)与本公司事实上存在大订单并至少有1-2年或更长期连续合约的，能带来相当大的销售额或具有较大的销售潜力； (2)有大订单且是具有战略性意义的项目客户； (3)对于公司的生意或公司形象，在目前或将来有着重要影响的客户； (4)有较强的技术吸收和创新能力； (5)有较强的市场发展实力。 -大客户与消费品的客户差异
1、时间长 2、干扰因素多 3、客户理性化 4、决策结果影响大 5、竞争激烈 -大客户营销的特点 1、竞争性日趋激烈 2、大客户自身日益成熟 3、增值销售机会较多（案例：利乐） 4、更重视与客户建立长期的合作关系 5、对销售代表的能力要求越来越高 -大客户营销对营销人员的新要求 1、知识面要宽，知识层次要深 2、高水平人际沟通技巧 3、正确的态度 4、良好的个人素质 营销人员要成为问不倒，不仅有极强的专业能力，既是营销专业也是技术专家，更要有广博的知识素养。
(1) 保证大客户能够成为销售订单的稳定来源 20%客户带来公司80%的业务。从企业的角度来看，80%的项目和收益来自于只占其客户总数20%的大客户，而数量众多的中小客户所带来的零散项目却只占其营业收益的20%。当然，这数字随企业的具体经营范围和特点，在具体的比例上有所差异，但大客户对企业而言具有重要意义则是毋庸置疑的。 两个典型的统计数据 一个美国商业银行的统计数据: 客户组占银行耗费的银行对银行 客户总数比例客户管理时间的利润贡献 10 18 93 30 28 17 60 54 -10 一个美国零售商的统计数据: 客户组占零售商对零售商的 客户总数比例利润贡献 4 37 11 27 14 16 ----- ----- 29 80