Supertwistor formulation for higher dimensional superstrings

超级转基因工程的英语作文Title: The Enigma of Transgenic Superorganisms。

In the realm of scientific innovation, the topic of "Super Transgenic Engineering" is a captivating puzzle, shrouded in intrigue and potential. This isn't your typical lab experiment; it's a futuristic leap into the realm of genetic engineering, where the boundaries blur between nature and nurture.Imagine a lab where scientists wield the power of DNA like a master sculptor, shaping organisms to surpass their natural limits. This "Super Transgenic" creation is notjust a creature, but a fusion of technology and biology, a living testament to human ingenuity. It's not about adding genes, but about rewriting them, crafting a new genetic tapestry.The first question that arises is, "Why?" The answer lies in the quest for efficiency, productivity, and evensurvival. These organisms, designed to outperform their wild counterparts, could revolutionize agriculture, providing higher yields with less resources. They could also serve as bioremediators, cleaning up pollutants in a heartbeat.However, the ethical conundrum looms large. The line between 'man-made' and 'natural' becomes increasingly blurred. Do we have the right to play God with life? The debate rages on, with some advocating for responsible use, while others fear the potential misuse.Then, there's the question of sustainability. Can these superorganisms coexist with their environment? Will they create a genetic imbalance, disrupting the delicate balance of ecosystems? The answer is not yet clear, as we're still navigating the uncharted waters of their long-term effects.In the end, Super Transgenic Engineering is a double-edged sword. It promises to shape the future, but at what cost? The scientific community, armed with curiosity and responsibility, must navigate this complex terrain withcaution, ensuring that our pursuit of knowledge does not surpass our responsibility to nature.So, when you ask, "Who am I?", in this context, I am not just a name, but a symbol of the power and responsibility that comes with the creation of these extraordinary beings. I am the embodiment of the ever-evolving dance between science and the natural world.。

小学上册英语第1单元测验卷英语试题一、综合题(本题有100小题，每小题1分，共100分.每小题不选、错误，均不给分)1.What is the largest breed of dog?A. ChihuahuaB. BeagleC. Great DaneD. PoodleC2.What do we call a person who travels to different countries for pleasure?A. TouristB. TravelerC. ExplorerD. NomadA3.The process of making biodiesel involves _______ oils.4.What is the sound a cow makes?A. BarkB. MeowC. MooD. RoarC5.The girl has a kind ________.6.The giraffe is known for its long ________________ (脖子).7.I thin k it’s fun to go ________ (参加舞会).8.What do you call the action of making something less dirty?A. CleaningB. TidyingC. OrganizingD. DustingA9.My friend is a great __________ (听众) and always supports me.10.The _____ (sun/cloud) is shining.11.The horse gallops across the ________________ (田野).12.The ________ (香味) of flowers can be pleasant.13.The __________ (铁路) connects different cities.14.What do you drink from?A. PlateB. CupC. TableD. ChairB15.He is _____ (tall/short) like his father.16.The chemical symbol for xenon is ______.17.What is the name of the planet known as the "Red Planet"?A. VenusB. MarsC. JupiterD. SaturnB18.I want to be a ___ (scientist/artist).19.The __________ is often unpredictable in spring. (天气)20.What is the capital city of Burkina Faso?A. OuagadougouB. Bobo DioulassoC. BanforaD. Koudougou21.I like to draw _____ in my notebook.22.What is the primary color of a lemon?A. GreenB. YellowC. RedD. BlueB23.What is the primary color of a lemon?A. GreenB. YellowC. RedD. Orange24.We can grow __________ (蔬菜) in our backyard.25.I see a ___ in the garden. (flower)26.The _____ (土壤改良) can enhance plant growth.27.The baby is _______ (笑着).28. A __________ is a characteristic of a substance that can be observed without changing it.29.My sister enjoys __________ (参加) local festivals.30.I can ________ my bike.31.What is the term for a baby cow?A. CalfB. FoalC. KidD. Lamb32.We had a _________ (玩具交换) at school, and I got a new _________ (玩具).33. A __________ is a natural elevation of the Earth's surface.34.The chemical formula for cobalt(II) nitrate is _____.35.He is _______ at playing soccer.36. A _______ can help illustrate how energy is transferred in a circuit.37.The _______ (鲸鱼) breaches the surface.38.I enjoy taking my camera to capture beautiful ______ (瞬间).39.I enjoy making ______ (手工艺品) from recycled materials. It’s a fun way to be creative and eco-friendly.40.What do you call the time before noon?A. AfternoonB. EveningC. MorningD. Midnight答案:C41.I like to bake ______ (美味) treats for my friends.42.I think being a __________ (志愿者) is very important.43.The ________ (植物资源管理) is crucial.44.The hamster runs in its _______ (仓鼠在它的_______里跑).45.What is the capital of Malta?A. VallettaB. MdinaC. RabatD. BirkirkaraA46.The dog is _____. (barking/sleeping/jumping)47.What is the capital city of Slovakia?A. BratislavaB. KošiceC. PrešovD. Nitra48.My favorite _____ is a friendly little puppy.49.I love to ______ (与家人一起) explore different cuisines.50. A _____ is a large body of gas and dust in space.51. A ________ (蝎子) has a stinger and can be dangerous.52.We use ______ (草) to make lawns green.53.What is the name of the first person to walk on the moon?A. Yuri GagarinB. Neil ArmstrongC. Buzz AldrinD. John Glenn54.What is the name of the famous scientist who formulated the laws of gravity?A. Albert EinsteinB. Isaac NewtonC. Galileo GalileiD. James Clerk Maxwell55.Rabbits are known for their strong ______ (后腿).56.Which season is the hottest?A. WinterB. SpringC. SummerD. AutumnC57.My brother collects ____ (stamps) from different countries.58.This boy, ______ (这个男孩), enjoys fishing.59.What do you call the distance between two points?A. LengthB. WidthC. HeightD. Measurement60.What do we call the process of a liquid turning into a gas?A. FreezingB. MeltingC. EvaporationD. CondensationC61.I enjoy _______ (listening) to classical music.62.The ______ helps with the communication between cells.63.My ________ (姐姐) loves to bake cookies and cakes.64. A ______ (刺猬) curls up when it feels threatened.65.The chemical formula for calcium chloride is ______.66.I love camping in the mountains. My favorite part is __________.67.The _______ (The Age of Enlightenment) emphasized reason and scientific thought in society.68.Many plants are ______ (适应性强) to their surroundings.69.I can see a ___ (bird) in the tree.70.The main component of cell membranes is ______.71.My favorite game is ______ (国际象棋).72.Plant cells have ______ that capture sunlight.73.I love to watch the __________ dance in the wind. (树叶)74.What do we call the act of traveling to different countries?A. ExploringB. AdventuringC. TouringD. VacationingC75.ration of Independence was signed in ________ (1776). The Decl76.What is the capital of Nepal?A. KathmanduB. PokharaC. BhaktapurD. LalitpurA77.The parrot repeats everything it _________. (听到)78.Which food is made from milk?A. BreadB. CheeseC. RiceD. PastaB79.My teacher is very __________ (耐心).80.The __________ is a famous city known for its beaches and nightlife. (迈阿密)81.The ant carries food back to its ______ (巢).82.My uncle is a skilled ____ (blacksmith).83.小龙虾) scuttles across the riverbed. The ____84.What is the capital of Tunisia?A. TunisB. SfaxC. KairouanD. Bizerte85.His favorite food is ________.86.What do you call a place where animals are kept for public viewing?A. FarmB. ZooC. AquariumD. ParkB87.The process of breaking down food is a type of _____ reaction.munity gardens promote ______ (邻里关系).89.What do you call a collection of stories or articles published together?A. AnthologyB. NovelC. MagazineD. JournalA90.The capital city of Vietnam is __________.91. f Enlightenment emphasized reason and __________. (科学) The Age92.island) is surrounded by water on all sides. The ____93.My friend’s _________ (玩具飞机) can fly high in the sky!94.The Stone Age is known for the use of ________ tools.95.What is the main ingredient in pizza?A. RiceB. DoughC. MeatD. SaladB96.The ice cream is ________ cold.97.What is the fastest land animal?A. ElephantB. CheetahC. HorseD. LionB98.Solar systems can contain a variety of celestial _______.99.My favorite thing about school is ________ (友谊).100.An ecosystem is a community of living organisms interacting with their ______ environment.。

A Super-Twisting Algorithm for Systems of Relative Degree MoreThan OnePost ASCC2013paperAbstract—This paper presents a data-driven homogeneous continuous super-twisting algorithm for systems with relative de-gree more than one.The conditions offinite-time convergence to an equilibrium are obtained demonstrating that the equilibrium can be moved as close to the origin as necessary,increasing a value of the control gain.The paper concludes with numerical simulations illustrating performance of the designed algorithms.Index Terms—Sliding mode control,super-twisting,systems of relative degree more than oneI.I NTRODUCTIONIt is well known that the classical discontinuous sliding mode control providesfinite-time convergence for a system of relative degree one[1].Afinite-time stabilizing control for a system of relative degree two is realized using the twisting algorithm[2],where the second order sliding mode control is also discontinuous.Both algorithms are robust with respect to bounded disturbances.On the other hand,using a continuous second-order sliding mode super-twisting algorithm[3],a state of a relative degree one system can be stabilized along with itsfirst derivative.The super-twisting algorithm is robust with respect to unbounded disturbances satisfying a Lipschitz con-dition.Thefinite-time convergence of the designed algorithms is conventionally established using geometrical techniques[2], [3],direct Lyapunov method[4],[5],[6],or homogeneity approach[7],[8].The explicit Lyapunov functions for their second-order super-twisting algorithms can be found in[6]. The homogeneity approach,mentioned even in the classical book[9],was consistently developed in the mentioned papers and applied to the observer design in[10].Various modifications of the sliding mode technique have always been actively used in industrial applications([11], [12],[13],[14],[15],[16],[17]),including data-driven ones ([18],[19],[20],[21]).This paper presents a data-driven homogeneous continuous super-twisting algorithm for systems with relative degree more than one.First,the case of relative degree two is addressed.The conditions offinite-time conver-gence to an equilibrium are obtained in demonstrating that the equilibrium can be moved as close to the origin as necessary, increasing a value of the control gain.Then,a modification of the designed algorithm enablingfinite-time convergence to the origin is proposed,based on the knowledge of the equilibrium from the previous run.Finally,the robustness of the designed data-driven algorithm is discussed.Similar results are then obtained for systems of relative degree more then two. The paper concludes with numerical simulations illustrating performance of the designed algorithms.The simulation results are discussed and demonstrated in a number offigures.The paper is organized as follows.The problem statement is given in Section2.A super-twisting-like control algorithm for systems of relative degree two is designed in Section3.The corresponding examples are provided in Section4.A super-twisting-like control algorithm for systems of relative degree more than two is presented in Section5and illustrated by examples in Section6.The proofs of all theorems are given in Appendix.A brief conference version of this paper was presented in[22].II.C ONTROL P ROBLEM S TATEMENT Consider a conventional dynamic system of relative degree two˙x1(t)=x2(t),x1(t0)=x10,(1)˙x2(t)=u(t),x1(t0)=x20,where x(t)=[x1(t),x2(t)]∈R2is the system state and u(t)∈R is the control input.In the classical second-order sliding mode control theory, afinite-time stabilizing control for the system(1)is designed using the twisting algorithm[2]in the formu(t)=−k1sign(x1(t))−k2sign(x2(t)),(2) where k1,k2>0are certain positive constants,and the signum function of a scalar x is defined as sign(x)=1,if x>0, sign(x)=0,if x=0,and sign(x)=−1,if x<0([9]).On the other hand,if a scalar dynamic system is of relative degree one˙x(t)=u(t),x(t0)=x0,(3) a continuousfinite-time stabilizing control for the system(3) can be designed using the super-twisting algorithm[3]as followsu(t)=−λ|x(t)|1/2sign(x(t))−α∫tt0sign(x(s))ds,(4)whereλ>0,α>0are certain positive constants.Note that applying the continuous control(4)to the system(3) results in a second-order sliding mode,i.e.,both x(t)and ˙x(t)converge to zero for afinite time.In other words,the continuous control(4)yieldsfinite-time convergence similar to that produced by a classical discontinuous sliding mode control u(t)=−Ksign(x(t)),where K>0is sufficiently large, for the system(3).In this paper,we propose a homogeneous super-twisting-like continuous modification of the twisting control algorithm (2)as followsu(t)=−λ1|x1(t)|1/3sign(x1(t))−−λ2|x2(t)|1/2sign(x2(t))−α∫tt0sign(x2(s))ds,(5)where λ1,λ2>0,α>0are certain positive constants.It would be demonstrated that the designed continuous control (5)works similarly to the twisting control (2),i.e.,results in finite-time convergence of both states x 1(t )and x 2(t )of the system (1).The announced result is formalized in the next section and then proved in Appendix.III.S UPER -T WISTING A LGORITHM FOR R ELATIVED EGREE T WO S YSTEMS The result for the control law (5)is given as follows.Theorem 1.Consider a dynamic system (1)of relative degree two.Then,the modified super-twisting control law (5)yields finite-time convergence of both states x 1(t )and x 2(t )to a point [x 1f ,0].Proofs of all the theorems are given in Appendix.Remark 1.Theorem 1ensures finite-time stability of the system (1)with the control law (5)with respect to an equilibrium point [x 1f ,0]located in the manifold x 2=0,which is however different from the origin.Nonetheless,the equilibrium point [x 1f ,0]could be moved as close to the origin as necessary,increasing a value of the control gain λ1>0,inview of the inequality |x 1f |1/3≤λ−11αT 1,where T 1is the finite convergence time.Although Remark 1underlines that the control law (5)cannot lead both states of the system (1)to the origin,this problem can be solved using the control law proposed in the following theorem.Theorem 2.Consider a dynamic system (1)of relative degree two.If upon applying the control law (5),the system (1)was finite-time stabilized at a point [x 1f ,0],it can be stabilized at the origin from the same initial condition [x 10,x 20]using the control lawu (t )=−λ1|x 1(t )+x 1f |1/3sign (x 1(t )+x 1f )−−λ2|x 2(t )|1/2sign (x 2(t ))−α∫tt 0sign (x 2(s ))ds .(6)Consider now a system (1)in presence of a disturbance:˙x 1(t )=x 2(t ),x 1(t 0)=x 10,(7)˙x 2(t )=u (t )+ξ(t ),x 1(t 0)=x 20,where ξ(t )satisfies the Lipschitz condition with a constant L .The system (7)can still be stabilized at a point [x 1f ,0]in view of the following theorem.Theorem 3.Consider a dynamic system (7)of relative degree two in presence of a disturbance ξ(t )satisfying the Lipschitz condition with a constant L .Then,the modified super-twisting control law (5)yields finite-time convergence of both states x 1(t )and x 2(t )to a point [x 1f ,0],provided that the following conditions hold for control gains:α>L ,λ22>2(α+L )2/(α−L ).IV.E XAMPLES :I.R ELATIVE D EGREE T WOThis section presents examples of designing a finite-time stabilizing regulator for a dynamic system (1)of relative degree two,based on the modified super-twisting regulator (5)in Theorems 1–3.1.Consider a linear system (1).The modified super-twisting regulator (4)is applied with the control gains selected as λ1=20,λ2=10,α=1.The initial conditions are assigned as x 10=1000,x 20=1000.The obtained results are shown in Fig.1.The final value of x 1(t )is equal to x 1f =0.0254.The initial conditions x 10=1000,x 20=−1000yield x 1f =−0.0004.The obtained results are shown in Fig.2.2.Consider the linear system from the previous example,assigning the same initial conditions,x 10=1000,x 20=1000and x 10=1000,x 20=−1000,and applying the control law (6),with λ1=20,λ2=10,α=1to both the considered cases,assuming x 1f =0.0254and x 1f =−0.0004.It can be observed from Figs.3and 4and their amplifications around the final time,Figs.5and 6,respectively,that the system is stabilized at the origin in each case.3.Consider a linear system (7)with disturbance ξ(t )=sin (1000t ).Again,the modified super-twisting regulator is applied with the control gains selected as λ1=20,λ2=10,α=1.The initial conditions are assigned as x 10=1000,x 20=1000.The obtained results are shown in Fig.7.The final value of x 1(t )is equal to x 1f =0.0244.The initial conditions x 10=1000,x 20=−1000yield x 1f =−0.0007.The obtained results are shown in Fig.8.This example clearly demonstrates that the sufficient condi-tions for the control gains in Theorem 3are too conservative,and the finite-time convergence takes place with much relaxed values.In particular,the value of constant L in this example is equal to 1000,due to high-frequency sinusoidal oscillations sin (1000t ).V.S UPER -T WISTING A LGORITHM FOR R ELATIVE D EGREEM ORE T HAN T WO S YSTEMS The main result can be generalized as follows.Consider a dynamic system of relative degree n >2˙x 1(t )=x 2(t ),x 1(t 0)=x 10,(8)˙x 2(t )=x 3(t ),x 2(t 0)=x 20,···˙x n (t )=u (t ),x n (t 0)=x n 0,using the notation for the system (1).We propose a gener-alization the super-twisting-like continuous control algorithm (5)as followsu (t )=−v 1(t )−v 2(t )−...−v n (t )+v n +1(t ),(9)where v i (t )=λi |x i (t )|γi sign (x i (t )),i =1,...,n ,v n +1(t )=|(s (t ))|γ/(1−γ)sign (s (t )),s (t )=−α∫tt 0sign (x n (s ))ds ,and certain positive constants λ1,...,λn >0,α>0and exponents γi ,i =1,...,n ,are assigned according to [7]to yield homogeneous finite-time convergence of all the states of the closed-loop system (8),(9).Namely,γi ∈(0,1),i =1,...,n satisfy the recurrent relations γi −1=γi γi +1/(2γi +1−γi ),i =2,...,n ,γn +1=1,and γn =γ,where γbelongs to an interval(1−ε,1),ε>0.Theorem8.1in[7]establishes that there exists such anε>0that the homogeneous(in view of definition ofγi)closed-loop system(8),(9)is globallyfinite-time stable at the origin.It would be demonstrated that the designed continuous con-trol(7)works similarly to the twisting control(5),i.e.,results infinite-time convergence of the states x1(t),x2(t),...,x n(t)of the system(6).The announced result is formalized in the next theorem and then proved in Appendix.Theorem 4.Consider a dynamic system(8)of rela-tive degree n>2.Then,the modified super-twisting con-trol law(9)yieldsfinite-time convergence of the states x1(t),...,x n−1(t),x n(t)to a point[x1f,...,x(n−1)f,0]. Remark2.Theorem4ensuresfinite-time stability of the system(8)with the control law(9)with respect to an equilib-rium point[x1f,...,x(n−1)f,0]located in the manifold x n=0, which is however different from the origin.Then,it follows from the equations(8)that x2f=x3f=...=x(n−1)f=0. Thus,the equilibrium point is given by[x1f,0...,0,0]and located in the manifold x2=x3=...=x n=0.Nonetheless, the equilibrium point[x1f,0...,0,0]could be moved as close to the origin as necessary,increasing a value of the control gainλ1>0in such a way thatλ1,...,λn still correspond to a Hurwitz polynomial s n+λn s n−1+...+λ2s+λ1(see[7]), in view of the inequality|x1f|1/3≤λ−11αT1,where T1is the finite convergence time.Although Remark2underlines that the control law(9) cannot lead all states of the system(1)to the origin,this problem can be solved using the control law proposed in the following theorem.Theorem 5.Consider a dynamic system(8)of relative degree n>2.If upon applying the control law(9),the system (8)wasfinite-time stabilized at a point[x1f,...,x(n−1)f,0],it can be stabilized at the origin from the same initial condition [x10,...,x(n−1)0,x n0]using the control lawu(t)=−w1(t)−w2(t)−...−w n−1(t)−v n(t)+v n+1(t), where w i(t)=λi|x i(t)+x i f|γi sign(x i(t)+x i f),i=1,...,n−1.Consider now a system(8)in presence of a disturbance:˙x1(t)=x2(t),x1(t0)=x10,(10)˙x2(t)=x3(t),x2(t0)=x20,···˙x n(t)=u(t)+ξ(t),x n(t0)=x n0,whereξ(t)satisfies the Lipschitz condition with a con-stant L.The system(10)can still be stabilized at a point [x1f,...,x(n−1)f,0]in view of the following theorem. Theorem6.Consider a dynamic system(10)of relative degree n>2in presence of a disturbanceξ(t)satisfying the Lipschitz condition with a constant L.Then,the modified super-twisting control law(9)yieldsfinite-time convergence of the states x1(t),...,x n−1(t),x n(t)to a point[x1f,...,x(n−1)f,0], provided that the following conditions hold for control gains:α>L,λ2n>2(α+L)2/(α−L).VI.E XAMPLES:II.R ELATIVE D EGREE M ORE T HAN T WO This section presents examples of designing afinite-time stabilizing regulator for a dynamic system(8)of a relative degree more than two,based on the modified super-twisting regulator(9)in Theorems4–6.4.Consider a linear3D system˙x1(t)=x2(t),x1(t0)=x10,(11)˙x2(t)=x3(t),x2(t0)=x20,˙x3(t)=u(t),x3(t0)=x30,The modified super-twisting regulator(9)u(t)=−λ1|x1(t)|1/4sign(x1(t))−λ2|x2(t)|1/3sign(x2(t))−λ3|x3(t)|1/2sign(x3(t))−α∫tt0sign(x3(s))ds,(12)is applied with the control gains selected asλ1=λ2=20,λ3=10,α=1.The initial conditions are assigned as x10= x20=x30=1000.The obtained results are shown in Fig.9. Thefinal value of x1(t)is equal to x1f=−6.03×10−9.The initial conditions x10=x30=1000,x20=−1000yield x1f=−0.00017.The obtained results are shown in Fig.10.5.Consider a linear system(11)with disturbanceξ(t)= sin(1000t).Again,the modified super-twisting regulator is applied with the control gains selected asλ1=λ2=20,λ3=10,α=1.The initial conditions are assigned as x10= x20=x30=1000.The obtained results are shown in Fig.11.Thefinal value of x1(t)is equal to x1f=−1.38×10−7. The initial conditions x10=x30=1000,x20=−1000yield x1f=−1.13×10−5.The obtained results are shown in Fig.12.This example again demonstrates that the sufficient condi-tions for the control gains in Theorem6are too conservative, and thefinite-time convergence takes place with much relaxed values.In particular,the value of constant L in this example is equal to1000,due to high-frequency sinusoidal oscillations sin(1000t).VII.A PPENDIXA.Proof of Theorem1.The system(1),(5)can be recast in the time-invariant form˙x1(t)=x2(t),x1(t0)=x10,(13)˙x2(t)=−λ1|x1(t)|1/3sign(x1(t))−−λ2|x2(t)|1/2sign(x2(t))+x3(t),x2(t0)=x20,˙x3(t)=−αsign(x2(t)),x3(t0)=0.The vectorfield f in the right-hand side of(13)can be represented as the sum of two homogeneous vectorfields, f=g1+g2,where g1=[x2,−λ1|x1(t)|1/3sign(x1(t))−ρλ2|x2(t)|1/2sign(x2(t)),0],ρ∈(0,1),and g2=[0,−(1−ρ)λ2|x2(t)|1/2sign(x2(t))+x3(t),−αsign(x2(t))]of homo-geneity degree m2=−1.The homogeneity degree m1for g1 can be selected as m1=−2<m2.Thefield g1provides thefinite-time stability at a point[0,0,x3(t0)]in view of its homo-geneity and Lyapunov function V(x1,x2)=λ1(3/4)|x1(t)|4/3 +(1/2)|x2(t)|2.Thefield g2corresponds to a super-twisting algorithm[3],which converges to a point[x1f,0,0]for afinite time.The theorem assertion now follows from Theorem7.4in [7],taking into account that the Lyapunov function for super-twisting has a continuous total derivative in time along the trajectory,so the results of Theorem6.2,Lemma4.2and the inequalities(34)-(36)from[7]hold.B.Proof of Theorem2.The theorem assertion follows from the fact that the equilib-rium[x1f,0,x3f]of the time-invariant representation(13)for the system(1),(5)can be moved to a point[0,0,x3f]using the coordinate change x1−x1f,x2,x3.C.Proof of Theorem3.The system(7)can be recast in the time-invariant form˙x1(t)=x2(t),x1(t0)=x10,(14)˙x2(t)=−λ1|x1(t)|1/3sign(x1(t))−−λ2|x2(t)|1/2sign(x2(t))+x3(t),x2(t0)=x20,˙x3(t)=−αsign(x2(t))+˙ξ(t),x3(t0)=0. where˙ξ(t)exists and is bounded for almost all t≥t0.The the-orem assertion follows from Theorem1and the convergence conditions for a super-twisting algorithm[3].D.Proof of Theorem4.The system(8),(9)can be recast in the time-invariant form˙x1(t)=x2(t),x1(t0)=x10,(15)˙x2(t)=x3(t),x2(t0)=x20,···˙x n−1(t)=x n(t),x n−1(t0)=x(n−1)0,˙x n(t)=−v1(t)−v2(t)−...−v n(t)++|x n+1(t)|γ/(1−γ)sign(x n+1(t)),x n(t0)=x n0,˙x n+1(t)=−αsign(x n(t)),x n+1(t0)=0. Similarly to(13),the vectorfield f in the right-hand side of(13)can be represented as the sum of two homogeneous vectorfields,f=g1+g2,where g1=[x2,x3,...,x n,−v1(t)−v2(t)−...v n−1(t)−ρv n(t),0],ρ∈(0,1),of homogeneity de-gree m1=(γ−1)/γ<0,and g2=[0,0,...,0,−(1−ρ)v n(t)+ v n+1(t),−αsign(x n(t))].The homogeneity degree m2for g2 can always be selected greater than m1,m2>m1=(γ−1)/γ, by virtue of exponentγ/(1−γ)for v n+1,making the entire system(15)homogeneous.Thefield g1provides thefinite-time stability at a point[0,...,0,x n+1(t0)]in view of Theorem8.1in [7].Thefield g2introduces a modification of a super-twisting algorithm[3],which converges to a point[x1f,...,x(n−1)f,0,0] for afinite time,in view of Lyapunov function V(x n,x n+1)=α|x n(t)|+(1−γ)|x n+1(t)|1/(1−γ).The theorem assertion now follows from Theorem7.4in[7],taking into account that V(x n,x n+1)has a continuous total derivative in time along the trajectory,so the results of Theorem6.2,Lemma4.2and the inequalities(34)-(36)from[7]hold. E.Proof of Theorem5.The theorem assertion follows from the fact that the equi-librium[x1f,...,x(n−1)f,0,x(n+1)f]of the time-invariant rep-resentation(15)for the system(8),(9)can be moved to a point[0,...,0,0,x(n+1)f]origin using the coordinate change x1−x1f,...,x n−1−x(n−1)f,x n,x n+1.F.Proof of Theorem6.The system(10)can be recast in the time-invariant form˙x1(t)=x2(t),x1(t0)=x10,˙x2(t)=x3(t),x2(t0)=x20,···˙x n−1(t)=x n(t),x n−1(t0)=x(n−1)0,˙x n(t)=−v1(t)−v2(t)−...−v n(t)++|x n+1(t)|γ/(1−γ)sign(x n+1(t)),x n(t0)=x n0,˙x n+1(t)=−αsign(x n(t))+˙ξ(t),x n+1(t0)=0. where˙ξ(t)exists and is bounded for almost all t≥t0.The the-orem assertion follows from Theorem4and the convergence conditions for a super-twisting algorithm[3].Fig.1.Graphs of the linear system(1)upon applying the control law(4) with initial conditions x10=1000,x20=1000.Thefinal value of x1(t)is equal to x1f=0.0254.R EFERENCES[1]V.I.Utkin,Sliding Modes in Control and Optimization,Springer,1992.[2] A.Levant,“Sliding mode and sliding accuracy in sliding mode control,”International Journal of Control,V ol.58,1993,pp.1247–1263. [3] A.Levant,“Robust exact differentiation via sliding mode technique,”Automatica,V ol.34,1998,pp.379–384.[4] A.Polyakov and A.Poznyak,“Lyapunov function design forfinite-timeconvergence analysis:Twisting controller for second-order sliding mode realization,”Automatica,V ol.45,2009,pp.444–448.Fig.2.Graphs of the linear system(1)upon applying the control law(4) with initial conditions x10=1000,x20=−1000.Thefinal value of x1(t)is equal to x1f=−0.0004.Fig.3.Graphs of the linear system(1)upon applying the control law(6) with initial conditions x10=1000,x20=1000and x1f=0.0254.[5] A.Polyakov and A.Poznyak,“Reaching time estimation for super-twisting second order sliding mode controller via Lyapunov function design,”IEEE Trans.on Automatic Control,V ol.54,2009,pp.1951–1955.[6]J.A.Moreno and M.Osorio,“Strict Lyapunov functions for the super-twisting algorithm,”IEEE Trans.on Automatic Control,V ol.57,2012, pp.1035–1040.[7]S.P.Bhat and D.S.Bernstein,“Geometric homogeneity with appli-cations tofinite-time stability,”Mathematics of Control,Signals,and Systems,V ol.17,2005,pp.101–127.[8] A.Levant,“Homogeneity approach to high-order sliding mode design,”Automatica,V ol.41,2005,pp.823–830.[9] A.F.Filippov,Differential Equations with Discontinuous RighthandSides,Kluwer,1988.[10]W.Perruquetti,T.Floquet,and E.Moulay,“Finite-time observers:appli-cations to secure communication,”IEEE Trans.on Automatic Control,Fig.4.Graphs of the linear system(1)upon applying the control law(6) with initial conditions x10=1000,x20=−1000and x1f=−0.0004.Fig.5.Graphs of the linear system(1)upon applying the control law(6) with initial conditions x10=1000,x20=1000and x1f=0.0254.(Amplified) V ol.53,2008,pp.356–360.[11] E.Kayacan,Y.Oniz,and O.Kaynak,“A grey system modeling approachfor sliding-mode control of antilock braking system,”IEEE Trans.on Industrial Electronics,V ol.56,2009,pp.3244–3252.[12]Y.Xia,Z.Zhu,M.Fu,and S.Wang,“Attitude tracking of rigid spacecraftwith bounded disturbances,”IEEE Trans.on Industrial Electronics,V ol.57,2010,pp.647–659.[13]Y.Xia,Z.Zhu,and M.Fu,“Back-stepping sliding mode control formissile systems based on extended state observer,IET Control Theory and Applications,V ol.5,2011,pp.93–102.[14] E.Kayacan,O.Cigdem,and O.Kaynak,“Sliding mode control approachfor online learning as applied to type-2fuzzy neural networks and its experimental evaluation,”IEEE Trans.on Industrial Electronics,V ol.59,2012,pp.3510–3520.[15]H.Gao,W.Zhan,H.R.Karimi,X.Yang,and S.Yin,“Allocation of ac-tuators and sensors for coupled-adjacent-building vibration attenuation,”IEEE Trans.on Industrial Electronics,V ol.60,2013,pp.5792–5801.Fig.6.Graphs of the linear system(1)upon applying the control law(6)with initial conditions x10=1000,x20=−1000and x1f=−0.0004.(Amplified)Fig.7.Graphs of the linear system(7)with disturbanceξ(t)=sin(1000t) upon applying the control law(4)with initial conditions x10=1000,x20= 1000.Thefinal value of x1(t)is equal to x1f=−0.0244.[16]Z.Zhu,D.Xu,J.Liu,and Y.Xia,“Missile guidance law based onextended state observer,”IEEE Trans.on Industrial Electronics,V ol.60,2013,pp.5882–5891.[17]H.Li,X.Jing,H.R.Karimi,“Output-feedback-based control for vehiclesuspension systems with control delay,”IEEE Trans.on Industrial Electronics,V ol.61,2014,pp.436–446.[18]L.Wu,P.Shi,and H.Gao,“State estimation and sliding-mode control ofMarkovian jump singular systems,IEEE Trans.on Automatic Control, V ol.55,2010,pp.1213–1219.[19]L.Ma,Z.Wang,Y.Bo,and Z.Guo,“Robust H∞sliding mode control fornonlinear stochastic systems with multiple data packet losses,”Intern.Journal of Robust and Nonlinear Control,V ol.22,2012,pp.473–491.[20]L.Wu,X.Su,and P.Shi,“Sliding mode control with bounded L2gain performance of Markovian jump singular time-delay systems,”Automatica,V ol.48,2012,pp.1929–1933.[21]J.Hu,Z.Wang,H.Gao,and L.K.Stergioulas,“Robust sliding modeFig.8.Graphs of the linear system(7)with disturbanceξ(t)=sin(1000t) upon applying the control law(4)with initial conditions x10=1000,x20=−1000.Thefinal value of x1(t)is equal to x1f=−0.0007.Fig.9.Graphs of the linear system(11)upon applying the control law(9)with initial conditions x10=x20=x30=1000.Thefinal value of x1(t)is equalto x1f=−6.03×10−9.control for discrete stochastic systems with mixed time delays,randomlyoccurring uncertainties,and randomly occurring nonlinearities,”IEEETrans.on Industrial Electronics,V ol.59,2012,pp.3008–3015. [22]M.V.Basin and P.Rodriguez-Ramirez,“A super-twisting algorithm forsystems of relative degree more than one,”Proc.9th Asian ControlConference,Istanbul,Turkey,2013.Fig.10.Graphs of the linear system(11)upon applying the control law (9)with initial conditions x10=x30=1000,x20=−1000.Thefinal value of x1(t)is equal to x1f=−0.00017.Fig.11.Graphs of the linear system(11)with disturbanceξ(t)=sin(1000t) upon applying the control law(9)with initial conditions x10=x20=x30= 1000.Thefinal value of x1(t)is equal to x1f=−1.38×10−7.Fig.12.Graphs of the linear system(11)with disturbanceξ(t)=sin(1000t) upon applying the control law(9)with initial conditions x10=x30=1000, x20=−1000.Thefinal value of x1(t)is equal to x1f=−1.13×10−5.。

a rXiv:h ep-th/067243v311A pr28hep-th/0607243DISTA-UPO-06DFTT-18/2006July,2006Abstract As been recently pointed out,physically relevant models derived from string theory re-quire the presence of non-vanishing form fluxes besides the usual geometrical constraints.In the case of NS-NS fluxes,the Generalized Complex Geometry encodes these informa-tions in a beautiful geometrical structure.On the other hand,the R-R fluxes call for supergeometry as the underlying mathematical framework.In this context,we analyze the possibility of constructing interesting supermanifolds recasting the geometrical data and RR fluxes.To characterize these supermanifolds we have been guided by the fact topological strings on supermanifolds require the super-Ricci flatness of the target space.This can be achieved by adding to a given bosonic manifold enough anticommuting co-ordinates and new constraints on the bosonic sub-manifold.We study these constraints at the linear and non-linear level for a pure geometrical setting and in the presence of p-form field strengths.We find that certain spaces admit several super-extensions and we give a parameterization in a simple case of d bosonic coordinates and two fermionic coordinates.In addition,we comment on the role of the RR field in the construction of the super-metric.We give several examples based on supergroup manifolds and coset supermanifolds.1IntroductionRecently the pure spinor formulation of superstrings[1,2]and the twistor string theory [3]have promoted supermanifolds as a fundamental tool to formulate the string theory and supersymmetric models.The main point is the existence of backgroundfields of extended supergravity theory(and therefore of10dimensional superstrings)which are the p-forms of the fermionic sector of superstrings.Thosefields can be coupled to worldsheet sigma modelsfields by considering the target space as a supermanifold.Indeed,as shown in[4,5]by adding to the bosonic space some anticommuting coordinatesθ’s,one can easily encode the informations regarding the geometry and the p-formfluxes into the supermetric of a supermanifold.In the present paper,we provide a preliminary analysis of the construction of supervarieties on a given bosonic space with or without p-forms (these are usually described by a bispinor Fµνsince they couple to target space spinors). We have to mention the interesting results found in[6,7,8,9]stimulated by the work [3,10].There the case of super-CY is analyzed and it is found that by limiting the number of fermions,the constraints on the bosonic submanifold of the supermanifold become very stringent.The super-CY spaces were introduced in string theory in paper[11].There the sigma models with supertarget spaces,their conformal invariance and the analysis of some topological rings were studied.In addition,we would like also to make a bridge with the recent successes of the Generalized Complex Geometry[12,13]to construct N=1 supersymmetric vacua of type II superstrings[14].In that context,theflux of the NS-NS background contributes to the generalized geometry and space is no longer CY.In the same way for a super-CY its bosonic subspace does not need to be a CY.The differential geometry of supermanifolds is a generalization of the usual geometry bosonic manifolds extended to anticommuting coordinates.There one can define super-vectorfields,superforms and the Cartan calculus.In addition,one can extend the usual Levi-Civita calculus to the fermionic counterparts defining super-Riemann,super-Ricci and super-Weyl tensorfields.A tensor has fermionic components and each component is a superfield.Due to the anticommuting coordinates,one defines also a metric on a supermanifold which can be generically given byds2=G(mn)dx m⊗dx n+G mµdx m⊗dθµ+G[µν]dθµ⊗dθν(1.1) where G(mn)(x,θ),G mµ(x,θ),G[µν](x,θ)are superfields and the tensor product⊗respects the parity of the differentials dx m,dθµ.The superfields are polynomials ofθ’s and the coef-ficients of their expansions are ordinary bosonic and fermionicfields.For example the ex-(x) pansion of G mn(x,θ)=G(0)mn(x)+...has a direct physical interpretation:G(0)mn=g(body)mn where the latter is the metric of the bosonic submanifold and the higher components are fixed by requiring the super-Ricciflatness.The physical interpretation of Gµm and Gµνis more interesting.Indeed,as suggested by the pure spinor string theory[1,15,16]the first components of Gµνcan be identified with a combination of RRfield strengths(in the text,this point will be clarified further).Thefirst component of Gµm arefixed by consistency and,in a suitable gauge,it coincides with a Dirac matrix.This is not the only way in which the RRfields determine the geometrical structure of the supermanifold.We learnt from recent analysis[17,18,19,20]in superstrings and, consequently in supersymmetricfield theory,that the superspace can be deformed by the existence of RRfield strength Fµν(x,θ){θµ,θν}=Fµν.(1.2) This has been only verified for constant RRfield strengths.Notice that a given bispinor Fµνis decomposed into a symmetric F(µν)and antisym-metric part F[µν].Therefore,it is natural to identify the antisymmetric part with the fermion-fermion(FF)components G[µν](x)of the supermetric(1.1)and the symmetric one with the non-vanishing r.h.s.of(1.2).This is very satisfactory since it provides a complete mapping between the superstring backgroundfields and the deformations of the superspace:the non-commutativity and the non-flatness.Indeed,the presence of a superfield Gµν(x,θ)in the supermetric modifies the supercurvature of the superspace. So,finally we can summarize the situation in the following way:given the backgrounds g mn and the NSNS b mn they are responsible for the deformation of the bosonic metric and of the commutation relations between bosonic coordinates[x m,x n]=θ[mn](where θmn=(b−1)mn).On the other side,the RRfields deforms the supermetric and the anti-commutation relations.There is another important aspect to be consider.The anticommuting deformation of superspace is treatable whether thefield strength Fµνof the RRfield is constant and,in four dimensions[19],whether is self-dual.In that case,there is no back-reaction of the metric and it is a solution of the supergravity equations.In the case of antisymmetric RR fields,which deform the supermetric(1.1),we have to impose a new condition in order that the worldsheet sigma model is conformal and therefore treatable.We require that the supermanifold is super-RicciflatR MN=0.(1.3) In this way,we can still use some of the conventional technique of conformalfield theory and of topological strings to study such models.The main point is that even if there is back-reaction,this is under control since the total superspace is super-Ricciflat[4,5,21, 22,23].We would like to remind the reader that the super-Ricciflatness is the condition which guarantees the absence of anomalies in the case of B-model topological strings on supermanifolds[24].This is equivalent to the well-known condition for conventional B-model on bosonic target spaces.The latter have to be Calabi-Yau which are Ricci-flat K¨a hler spaces.A main difference is that the CY must be a3-fold in order to compensate the ghost anomaly.In the case of supermanifold this is given by a virtual dimension which is essentially the difference between the bosonic and fermionic dimensions[25].In the A-model,the CY conditions is required for open string in presence of D-branes(see[24]for a complete discussion).1We start with a metric on a given space,for example the round sphere metric for S n, or Fubini-Study metric for CP n.Then we construct the corresponding supermanifold by adding anticommuting coordinates and additional components of the metric(in some cases,the extension of the metric to a super-metric is not enough and some non-vanishing torsion components are necessary)and we require the super-Ricciflatness.As a conse-quence,we notice that thefirst coefficient of Gµνmust be different from ly,we need a non-vanishing component of the fermion-fermion part of the metric.However,in general for a given space,no invariant spinorial tensor exists to provide such component. Therefore,we argue that this should be identified with p-formfluxes needed to sustain the solution of the supergravity.As a matter of fact,super-Ricciflatness does not seems to relate the value of theflux with the geometry of the bosonic submanifold.In fact,we will see in the forthcoming sections,the value of theflux isfixed by a suitable normalization which is implemented by a gauge condition on the supermetric.So,we conclude that super Ricciflatness turns out to be a necessary ad useful condition,but it does not seem to be sufficient to characterize the vacuum of the theory.We have to divide the analysis in two parts:thefirst one is the analysis of linearized equations and the second one concerns the analysis of the non-linear theory.It turns out that for the linear equation one gets very strong constraints on the bosonic metric such as the scalar curvature must be zero(see also[6,7,9]).This is compatible with theflat or Ricciflat space,but certainly is not true for a sphere or any Einstein manifold.Therefore, it is unavoidable to tackle the non-linear analysis to see whether these constraints can be weakened.Let us explain briefly the lines of analysis.The Ricci-flatness condition(1.3) can be analyzed by expanding the super-metric in components.Since we are interested only in the vacuum solutions,we decompose the superfields of the super-metric into bosonic components by setting to zero all fermionic ones.This implies a set of equations for each single component.Among them,a subset of these equations can be easily solved; indeed,the equations for the components to the order n are solved in terms of some components at the n+1order.However,the equations for highest-components cannot be solved algebraically and in fact,by plugging the solutions of the lower orders in these equations,onefinds high-derivative equations for the lowest components.To show that our analysis is not purely academic,we give some examples of super-manifolds modeled on some bosonic submanifold.Of course,the most famous example is P SU(2,2|4)(from AdS/CFT correspondence)and SL(4|4)(appearing in twistor string theory).Moreover,we give examples based on S n,V(p,q)and CP n.As a by-product we construct a super-Hopffibration connecting the different types of supercoset manifolds. One interesting example is the construction of a supermanifold with T(1,1)as a bosonic submanifold.In addition,we propose a construction which can be applied to any Einstein spaces,and in particular to Sasaki-Einstein space and K¨a hler-Einstein.At the momentwe have not explored the role of Killing spinors,Killing vectors and the role of super-symmetry in the construction.We only argued that if the supermanifold is obtained as a single coset space and not a direct product of coset space,it has more chances to have an enhanced supersymmetry.The relation between supersymmetry and the construction of super-Ricciflat supermanifold will be explored in forthcoming papers.More important,it is not yet established the relation between the solution of supergravity and the super-Ricci flatness.The paper is organized as follows:in sec.2we recall some basic fact about super-geometry..In sec.3,we present a general strategy,we study the linearized version of the constraints and,in the case of d bosonic coordinates and2fermionic directions, we analyze completely the non-linear system.In sec.4,we study the construction of supercosets modeled on bosonic cosets.In sec.5,we add thefluxes(RRfield strengths). Appendices A and B contain some related material.2Elements of supergeometryWe adopt the definition of supermanifold given in[30]based on a superalgebra with the coarse topology.However,this is not the only way and there is an equivalent formulation based on sheaves andflag manifolds[31,32].The equivalence turns out to be useful for studying topological strings on supermanifolds[33].2.1SupermanifoldsA supermanifold M(m|n)is called Riemannian if it is endowed with a metric supertensor which is a real non-singular commuting tensorfield G MN satisfying the following graded-symmetry condition:G MN=(−1)MN G NM.(2.1) The meaning of(2.1)becomes more evident if one distinguishes between bosonic and fermionic components and rewrites the metric supertensor as a block supermatrix GG MN= g mn jµν (2.2)being g an m×m real symmetric commuting matrix,j an n×n imaginary antisymmetric commuting matrix and h an m×n imaginary anticommuting one.Obviously thefirst component of the superfield of the even-even block g is an ordinary metric for the body M(m)of the supermanifold M(m|n).Body means the bosonic submanifold.Note that the supermetric G is preserved by the orthosymplectic supergroup OSp(m|n) whose bosonic submanifold is the direct product SO(m)×Sp(n)of an orthogonal group encompassing the Lorentz transformations of the metric g with a symplectic group mixing the odd directions.Thus,we immediately see that the number n of fermionic dimensions(in the paper this is also denoted by d F)over a Riemannian supermanifold must be even, otherwise Sp(n)would not be defined.The vielbein formalism can be constructed in the same way:supervielbeins will carry a couple of superindices whose different parity combinations induce the matrix block formE A M= E a m Eαµ (2.3)with commuting diagonal blocks and anticommuting off-diagonal blocks.The correct formula to pass from supervielbeins to supermetrics isG MN=(−1)MA E A MηAB E B N(2.4)whereηdenotes theflat superspace metric diag(m1,...,1,−1,...,−1,nσ2,...,σ2).Butwhat isηµν?A question that has been raised several times in the literature(see for example [34]for a review and some comments),and it turns out that theflat super-metric in the fermionic directionsηµνcannot be always defined for any models.For example,in the case of10d supergravity IIB there is no constant tensor which can be used to raise and lower the spinorial indices and to contact to Weyl indices.However,it is also known that a typical bispinor emerges in the quantization of superstrings in the fermionic sector of the theory and these states are known as the RRfields.Indeed,in the case of constant RRfields,one can identify the fermionic components of theflat supermetric with constant RRfield strength.This will be discuss in sec.5.The inverse of a supermetric G MN is computed by inverting the supermatrix G MN. In order that the inverse exists it must be that det(G mn)and det(Gµν)do not vanish.2.2K¨a hler supermetricsA very important subclass of supermetrics consists of so-called K¨a hler supermetrics,which are a straightforward generalization of the usual K¨a hler metrics living on ordinary man-ifolds.As one can expect,the components G M¯N of such a supermetric are obtained applying superderivatives-denoted by“,”-to a superpotential K:G M¯N=K,M¯N.(2.5) As a consequence the body of a K¨a hler supermanifold with a supermetric G M¯N originated by a superpotential K is a K¨a hler manifold endowed with the metric g m¯n obtained by applying two derivatives to the bosonic part of K according to g(body)m¯n=∂m∂¯n K|θ=0. 2.3Superconnections and covariant superderivativeThe idea of connection and covariant derivative have their relative supergeometric coun-terparts.Indicating superconnections withΓ,one writes the defining rules for covariantsuperderivatives-denoted by“;”-as follows:Φ;M=Φ,M,(2.6a)X M;N=X M,N+(−1)P(M+1)X PΓM P N,(2.6b)ΦM;N=ΦM,N−ΦPΓP MN.(2.6c) A superconnection over a supermanifold M(m|n)can be chosen in many ually this freedom is restricted by demanding that the superconnection be metric-compatible, i.e.that the covariant superderivative of the supermetric vanishes,and that the coeffi-cientsΓM NP be graded-symmetric in their lower indices.These two requirements uniquely determine the superconnection to be given by the super Christoffel symbolsM N P =1andR MN=(−1)MN R NM,(2.13) while Bianchi identities readR MNP Q;R+(−1)P(Q+R)R MNQR;P+(−1)R(P+Q)R MNRP;Q=0,(2.14)1R NMN;−2 T M NP−(−1)Q(N+1)G MQ G NR T R QP−(−1)Q(P+1)+NP G MQ G P R T R QN .(2.20) The corresponding Ricci supercurvature can be obtained adding a few terms to the ex-pression valid with vanishing torsion:R MN=(2.10)+(−1)P(M+1) −K P MP;N+(−1)P(M+Q)K P QP K Q MN+(−1)NP K P MN;P−(−1)N(M+P+Q)K P QN K Q MP .(2.21) In terms of the super-vielbeins,the torsion tensor is given byT A BC E C∧E B=T A≡dE A+ΩA B∧E B.(2.22)3Supermanifolds on given bosonic manifoldsIn the present section we are interested in studying examples of supermanifolds which have a given bosonic submanifold.We construct a corresponding supermanifold by adding the anticommuting coordinates and by requiring that the supermanifold is super-Ricciflat.Before proceeding we would like to give an example of a simple model where the procedure for constructing a supermanifold is illustrated.Let us consider the superfield Φ(x,θ)and we set to zero all fermionic components.Then,Φ= nΦn(θ2)n.We can consider the naive generalization of the Klein-Gordon(KG)equation∂m∂mφ+13!Φ30=(∂m∂m+∂µ∂µ)Φ+13!Φ30+2Φ2=0,∂m∂mΦ2+13!∂2Φ30+112Φ50=0.(3.3)This is a weaker condition onΦ0than the usual KG equation and it must be solved in order that the superfieldΦ0can be thefirst component of the superfieldΦwhich solves the new KG equation.The present example has no physical relevance,but it illustrates the problem appearing in the extension of a given bosonic metric to a supermetric and the extension of the bosonic equations to super-equations.Even in this case,we assume that the background has no fermions and we add only superfields with bosonic components.We have to mention that equations similar to(3.1)where already proposed in several papers in thefirst years of supersymmetry(see for example[36,37,38]).However,this construction was abandoned since the resulting theories possess ghosts and higher-spin fields(see for example[34]and the references therein).Here,we do not pretend to interpret the super-Ricciflatness asfield equations with dynamical content,but we consider these equations a way of recasting the informations on the manifold together the RRfluxes in a single mathematical structure.This is very similar to the case of NS-NSfluxes and the generalized geometry of[12,13].3.1General FrameworkIn the present section,we give the general equations of the super-Ricci tensor R MN by expanding the supermetric in components.These are very useful in order to single outthe structure of the equations.Indeed,it can be observed that at each order there is free component of the supermetric that is used to solve the corresponding equation.First, we show that every equation at the level lower than the maximumθ-expansion can be easily algebraically solved,then we show that the remaining equations are high deriva-tives and they provide consistency conditions for the construction of the corresponding supermanifold.As in the above example,using the iterative equations,one derives the highest derivative equation.This is rather cumbersome since the iterative procedure tends to explode soon in long unmanageable expressions.Wefirst tame the linearized system and then we tackle the non-linear one.For the former we are able to provide a complete analysis;for the latter only the case(d|2)will be discussed.In order to study the structure of the equation for super Ricci-flat manifold it is convenient to consider the bosonic and fermionic components separately.For that we display the few terms coming from the superfield expansion of the supermetric:1G mn=g mn+w mνρστθτθσθρ+O(θ5),61Gµν=hµν+hρσ(hσρ;(mn)−t m[σρ];n−t n[σρ];m+f mnσρ)(3.7)2−1Rµν=−12hρσ(lµνσρ+lσρµν+lµρσν+lσνµρ)−14hρσg mn(2t m(µρ)−hµρ,m)(2t n(νσ)−hνσ,n)+O(θ2)=0.(3.8)Eq.(3.7)can be easily solved in terms of f mnρσ.In the same way,(3.8)is solved by a suitable combination of lµνρσ.At the next order,a free bosonicfield will appear from the expansion of the superfields G MN and a new equationfixes it.This iterative procedure allows us to solve the complete set of equations,starting from a given bosonic metric g mn (and with aflat supermetric in the fermionic sector g[µν]),in terms of the components of G MN.At the end of the iterative procedure,wefind the consistency conditions on the bosonic manifold.Wefirst analyze the linearized equations and we expand around the flat space and around a Ricciflat(R(body)mn)bosonic space.It turns out that variation of the bosonic Ricci tensor must satisfy a set of differential equations(with a high-derivative differential operator)and some algebraic conditions(the space must have vanishing scalar curvature).However,some of the constraints are weaker in the non-linear theory.Furthermore,one can also impose additional structure in the superspace such as a su-percomplex structure.Some simplifications are in order in the case of a K¨a hler manifolds.On a K¨a hler supermanifold the hermitian metric supertensor comes from a K¨a hler superpotential,K,by the following formula:G M¯N=K,M¯N.(3.9) As a consequence the Ricci super-tensor components in holomorphic coordinates reduce to the formR M¯N=−(−1)P+Q G¯P Q G Q¯P,M¯N+(−1)P+Q+S+M(P+R)G P¯Q G¯QR,M G R¯S G¯SP,¯N =−(ln sdet G),M¯N(3.10)We can expand again the supermetric in the odd coordinates(θµ,θ¯µ)G m¯n=g m¯n+hρ¯σ,m¯nθ¯σθρ+12f m¯n¯ρ¯σθ¯σθ¯ρ+O(θ4),G m¯ν=h¯νρ,mθρ+t m¯ν¯ρθ¯ρ+O(θ3), Gµ¯n=tµ¯nρθρ+hµ¯ρ,¯nθ¯ρ+O(θ3),Gµ¯ν=hµ¯ν+lµ¯νρ¯σθ¯σθρ+12lµ¯ν¯ρ¯σθ¯σθ¯ρ+O(θ4),f m¯n¯ρ¯σ=f∗n¯mσρ,tρ¯mσ=t∗m¯ρ¯σ,lµ¯ν¯ρ¯σ=l∗ν¯µσρ,(3.11)and obtain the equivalent of the super Ricci-flatness equation,R MN=0,for g,h,l and t.The last line expresses the constraints coming from the hermiticity of the K¨a hlerpotential.At the lowest order in the odd variables one getsR m¯n=R(body)m¯n−h¯ρσhσ¯ρ,m¯n−hµ¯νh¯νρ,m hρ¯σh¯σµ,¯n+O(θ2)==R(body)m¯n+(ln(det h)),m¯n+O(θ2)=0,Rµ¯ν=−g¯p q hµ¯ν,q¯p−h¯ρσlσ¯ρµ¯ν−hρ¯σh¯σµ,p g p¯q hρ¯ν,¯q+hρ¯σt p¯σ¯νg p¯q tρ¯qµ+O(θ2)=0,where R(body)mn is the Ricci tensor on the body of the Ricci-flat K¨a hler supermanifold(SCY)and h=(hµ¯ν),from whichR(body)2 (−1)MN H QN;MP−(−1)Q(M+N)H MN;QP+H QM;NP−(−1)P(M+N)H QP;MN ,(3.14) which,making parity of indices explicit,splits intoδR mn=12(−1)Q G P Q H Qm;µP+H Qµ;mP−(−1)Q H mµ;QP−(−1)P H QP;mµ ,(3.15b)δR µν=12g pqH qm ;np +H qn ;mp −H mn ;qp −H qp ;mn+12g pqH qm ;µp +H qµ;mp −H mµ;qp −H qp ;mµ−12g pqH qµ;νp −H qν;µp −H µν;qp −H qp ;µν+12g pqH (2k )qm [τ1...τ2k ];np +H (2k )qn [τ1...τ2k ];mp −H (2k )mn [τ1...τ2k ];qp−H (2k )qp [τ1...τ2k ];mn+12gpq(2k +2)H (2k +2)qm [µτ1...τ2k +1];p +H (2k +1)qµ[τ1...τ2k +1];mp −H (2k +1)mµ[τ1...τ2k +1];qp−(2k +2)H (2k +2)qp [µτ1...τ2k +1];m−(k +1)jρσ(2k +3)H (2k +3)σm [µρτ1...τ2k +1]+H (2k +2)σµ[ρτ1...τ2k +1];m+(2k +3)H (2k +3)mµ[σρτ1...τ2k +1]+H (2k +2)σρ[µτ1...τ2k +1];m ,(3.18b)δR (2k )µν[τ1...τ2k ]=1Analizing the recursive structure of(3.18)’s,one realizes that the only independentfields which effectively play a role in constraining g mn are obtained by contracting odd indices with j as follows:S(2k) mn =H(2k)mn[τ1...τ2k]jτ1τ2···jτ2k−1τ2k,(3.19a)T(2k+1) m =H(2k+1)mτ1[τ2...τ2k+2]jτ1τ2···jτ2k+1τ2k+2,(3.19b)U(2k)=H(2k)τ1τ2[τ3...τ2k+2]jτ1τ2jτ3τ4···jτ2k+1τ2k+2,(3.19c)V(2k)=H(2k)τ1τ2[τ3...τ2k+2]jτ1τ3jτ2τ4···jτ2k+1τ2k+2.(3.19d) From these definitions one has a few identities:V(0)=0,U(d F)=d F V(d F),S(0)mn=δg mn.(3.20) The calculation is simplified by a gauge-fixing of the supermetric(3.6):a convenient choice isG mµθµ=0,Gµνθνθµ=jµνθνθµ,(3.21) which,in terms of variations,becomesH mµθµ=0,Hµνθνθµ=0.(3.22) We notice that some of thefields defined in(3.19)are vanishing because of the gauge-fixing:T(1)m=0,U(0)=0.(3.23) Introducing S,T,U and V into(3.18)’s,one obtains in principle fourfield equations but one of them follows from the other ones when Ricci curvature of the body vanishes.The remaining threefield equations corresponding to super Ricci-flatness are(2k+2)(2k+1)S(2k+2)mn=(2k+1) T(2k+1)m;n+T(2k+1)n;m −U(2k);mn−g pq S(2k)qm;np+S(2k)qn;mp−S(2k)mn;qp−S(2k)qp;mn ,(3.24a) (2k+2)(2k+1) U(2k+2)+V(2k+2) =2 U(2k)+V(2k) ,(3.24b)2S(2k)mm−S(2k)mnmn;=2 U(2k)+V(2k) .(3.24c) Recalling the identity(3.20)and the gauge-fixing(3.23),from here wefindU(2k)+V(2k)=0,U(d F)=V(d F)=0,(3.25) so that(3.24b)is automatically solved and(3.24c)with k=0shows that,at linear order, the curvature scalar on the body has to be zero:R(body)=0.(3.26)Besides this algebraic constraint on the Ricci curvature,also some differential consis-tency condition has to be taken into account that comes from the remaining constraints. IntroducingˆS(2k+2) mn def=(2k+2)!S(2k+2)mn−(2k+1)! T(2k+1)m;n+T(2k+1)n;m +(2k)!U(2k);mn(3.27)into(3.24a)and(3.24c),one can rewrite them asˆS(2k+2) mn =2ˆS(2k)mn+2ˆS pq(2k)R(body)pmqn(k>0)(3.28)withˆS(2) mn =−2δR(body)mn,ˆS(d F+2)mn=0.(3.29)In other words,at linear order,the only condition to require in addition to(3.26)isL d F/2R R(body)mn=0(3.30) where LRis the Lichnerowicz operator defined on Ricci-flat spaces byL R X mn=2X mn+2X pq R(body)pmqn.(3.31) In conclusion,at linear order,a Ricci-flat manifold can be deformed in the body of a Ricci-flat supermanifold only if conditions(3.26)and(3.30)are satisfied,and,in that case,an overlying supergeometrical structure is constrained by(3.25),(3.28)and(3.29). As a consequence,if a consistent deformation is possible,an infinite number of Ricci-flat supermanifolds with gauge-fixed supermetric can be constructed over the same bosonic body.It would be interesting to compare the present result with the results of paper[9].We plan to explore deeply the relation between the super-Ricciflatness equations and the moduli space of the supervarieties.3.3Non-linear equationsSince a complete general analysis is rather cumbersome,we give the complete expressions only in the case of(d|2)-dimensional supermanifolds.We show that equations at the lowest order can be indeed solved by some free components of the superfields and we derive the equations for the highest components.Following the scheme given in section 3.2,by substituting the lower-order results one gets thefinal constraints on the metric of the bosonic submanifold.As in the present case there are just two fermionic directions,theθ-expansion of the supermetric with the gauge-fixing(3.21)is given byG mn=g mn+12f(2)θ2(3.32)。

a r X iv:mat h /2818v1[mat h.AG ]23Aug202TORELLI’S THEOREM FOR HIGH DEGREE SYMMETRIC PRODUCTS OF CUR VES NAJMUDDIN F AKHRUDDIN Abstract.We show that two smooth projective curves C 1and C 2of genus g which have isomorphic symmetric products are isomorphic unless g =2.This extends a theorem of Martens.Let k be an algebraically closed field and C 1and C 2two smooth projective curves of genus g >1over k .It is a consequence of Torelli’s theorem that if Sym g −1C 1∼=Sym g −1C 2,then C 1∼=C 2.The same holds for the d ’th symmetric products,for 1≤d <g −1as a consequence of a theorem of Martens [2].In this note we shall show that with one exception the same result continues to hold for all d ≥1,i.e.we have the following Theorem 1.Let C 1and C 2be smooth projective curves of genus g ≥2over an algebraically closed field k .If Sym d C 1∼=Sym d C 2for some d ≥1,then C 1∼=C 2unless g =d =2.It is well known that there exist non-isomorphic curves of genus 2over C with isomorphic Jacobians.Since the second symmetric power of a genus 2curve is isomorphic to the blow up of the Jacobian in a point,it follows that our result is the best possible.Proof of Theorem.Let C i ,i =1,2be two curves of genus g >1with Sym d C 1∼=Sym d C 2for some d ≥1.Since the Albanese variety of Sym d C i ,d ≥1,is iso-morphic to the Jacobian J (C i ),it follows that J (C 1)∼=J (C 2).If d ≤g −1,the theorem follows immediately from [2],since the image of Sym d C i in J (C i )(after choosing a basepoint)is W d (C i ).Note that in this case it sufices to have a birational isomorphism from Sym d C 1to Sym d C 2.Suppose g ≤d ≤2g −3.Then the Albanese map from Sym d C i to J (C i )is surjective with general fibre of dimension d −g .Interpreting the fibres as complete linear systems of degree d on C i ,it follows by Serre duality that the subvariety ofJ (C i )over which the fibres are of dimension >d −g is isomorphic to W 2g −2−d (C i ).Therefore if Sym d C 1∼=Sym d C 2,then W 2g −2−d (C 1)∼=W 2g −2−d (C 2),so Martens’theorem implies that C 1∼=C 2.Now suppose that d >2g −2and g >2.By choosing some isomorphism we identify J (C 1)and J (C 2)with a fixed abellian variety A .If φ:Sym d C 1→Sym d C 2is our given isomorphism,from the universal property of the Albanese morphism we obtain a commutative diagramSym d C 1φSym d C 2π2A12NAJMUDDIN FAKHRUDDINwhere theπi’s are the Albanese morphisms corresponding to some basepoints and f is an automorphism of A(not necessarily preserving the origin).By replacing C2 with f−1(C2)we may then assume that f is the identity.Since d>2g−2,the mapsπi,i=1,2make Sym d C i into projective bundles over A.By a theorem of Schwarzenberger[5],Sym d C i∼=P roj(E i),where E i is a vector bundle on A of rank d−g+1with c j(E i)=[W g−j(C i)],i=1,2,0≤j≤g−1,in the group of cycles on A modulo numerical equivalence.Sinceφis an isomorphism of projective bundles,it follows that there exists a line bundle L on A such that E1∼=E2⊗L.Letθi=[W g−1(C i)],so by Poincar´e’s formula[W g−j(C i)]=θj i/j!,i=1,2,i≤j≤g−1.The lemma below implies thatθg−1i =θg−12in the group of cyclesmodulo numerical equivalence on A.Sinceθg i=g!,this implies thatθ1·[C2]=g. By Matsusaka’s criterion[3],it follows that W g−1(C1)is a theta divisor for C2, which by Torelli’s theorem implies that C1∼=C2.If d=2g−2and g>2,then we can still apply the previous argument.In this case we also have that Sym d(C i)∼=P roj(E i),i=1,2but E i is a coherent sheaf which is not locally free.However on the complement of some point of A it does become locally free and the previous formula for the Chern classes remains valid.The above argument clearly does not suffice if g=2.To handle this case we shall use some properties of Picard bundles for which we refer the reader to[4]. Suppose that d>2and C i,i=1,2are two non-isomorphic curves of genus2with Sym d C1∼=Sym d ing the same argument(and notation)as the g>2case, it follows that there exist embeddings of C i,i=1,2,in A and a line bundle L on A such that E1∼=E2⊗L and L⊗d−1∼=O(C1−C2)(we identify C i,i=1,2with their images).For i≥1,let G i denote the i-th Picard sheaf associated to C2,so that P roj(G i)∼= Sym i(C2).(G i is the sheaf denoted by F2−i in[4]and G d∼=E2).There is an exact sequence([4,p.172]):0→O A→G i→G i−1→0(1)for all i>1.We will use this exact sequence and induction on i to compute the cohomology of sheaves of the form E1⊗P∼=E2⊗L⊗P,where P∈P ic0(A).Considerfirst the cohomology of G1,which is the pushforward of a line bundle of degree1on a translate of C2.Since we have assumed that C1≇C2,it followsthat C1·C2>2.Since C21=C22=2,deg(L|C2)=(C1−C2)·C2/(d−1)>0.By Riemann-Roch it follows that h j(A,G1⊗L⊗P),j=1,2is independent of P,except possibly for one P if deg(L|C2)=1,and h2(A,G1⊗L⊗P)=0since G1issupported on a curve.Now C1·C2>2also implies that c1(L)2<0.By the index theorem,it follows that h0(A,L⊗P)=h2(A,L⊗P)=0and h1(A,L⊗P)is independent of P. Therefore by tensoring the exact sequence(1)with L⊗P and considering the long exact sequence of cohomology,we obtain an exact sequence(2)0→H0(A,G i⊗L⊗P)→H0(A,G i−1⊗L⊗P)→H1(A,L⊗P)→H1(A,G i⊗L⊗P)→H1(A,G i−1⊗L⊗P)→0 and isomorphisms H2(A,G i⊗L⊗P)→H2(A,G i−1⊗L⊗P)for all i>1.By induction,it follows that H2(A,G i⊗L⊗P)=0for all i>0.Since the Euler characteristic of G i⊗L⊗P is independent of P,the above exact sequence(2)TORELLI’S THEOREM FOR HIGH DEGREE SYMMETRIC PRODUCTS OF CURVES3 along with induction shows that for all i>0and j=0,1,2,h j(G i⊗L⊗P)is independent of P,except for possibly one P.In particular,this holds for i=d hence h j(A,E1⊗P)is independent of P except again for possibly one P.We obtain a contradiction by using the computation of the cohomology of Picard sheaves in Proposition4.4of[4]:This implies that h1(A,E1⊗P)is one or zero depending on whether the point in A corresponding to P does or does not lie on a certain curve (which is a translate of C1).Lemma1.Let X be an algebraic variety of dimension g≥3and let E i,i=1,2 be vector bundles on X of rank r.Suppose c1(E i)=θi,c j(E i)=θj i/j!for i=1,2 and j=2,3(j=2if g=3),and E1∼=E2⊗L for some line bundle L on X.Then θj1=θj2for all j>1(j=2if g=3).Proof.Since E1∼=E2⊗L,c1(E1)=c1(E2)+rc1(L),hence c1(L)=(θ1−θ2)/r.For a vector bundle E of rank r and a line bundle L on any variety,we have the following formula for the Chern polynomial([1],page55):c t(E⊗L)=rk=0t k c t(L)r−k c i(E).Letting E=E1,E⊗L=E2,and expanding out the terms of degree2and3,one easily sees thatθj1=θj2for j=2and also for j=3if g>3.(Note that this only requires knowledge of c j(E i)for j=1,2,3).Since any integer n>1can be written as n=2a+3b with a,b∈N,the lemma follows.We do not know whether the theorem holds if k is not algebraically closed.Remark.S.Ramanan has suggested it may also be possible to prove the theorem when g>2by computing the Chern classes of the push forward of the tangent bundle of Sym d C to J(C).Acknowledgements.We thank A.Collino for some helpful correspondence,in par-ticular for informing us about the paper[2],and S.Ramanan for the above remark.References[1]W.Fulton,Intersection theory,Springer-Verlag,Berlin,second ed.,1998.[2]H.H.Martens,An extended Torelli theorem,Amer.J.Math.,87(1965),pp.257–261.[3]T.Matsusaka,On a characterization of a Jacobian variety,Memo.Coll.Sci.Univ.Kyoto.Ser.A.Math.,32(1959),pp.1–19.[4]S.Mukai,Duality between D(X)and D(ˆX)with its application to Picard sheaves,NagoyaMath.J.,81(1981),pp.153–175.[5]R.L.E.Schwarzenberger,Jacobians and symmetric products,Illinois J.Math.,7(1963),pp.257–268.School of Mathematics,Tata Institute of Fundamental Research,Homi Bhabha Road,Mumbai400005,IndiaE-mail address:naf@math.tifr.res.in。

小学下册英语第6单元真题[含答案]考试时间：90分钟（总分:110）A卷一、综合题(共计100题共100分)1. 填空题：The ______ (小鸟) flies south for the winter.2. 选择题：What is the capital of France?A. LondonB. ParisC. BerlinD. Madrid答案: B3. 填空题：I share my secrets with my ____.4. 选择题：What is the name of the first artificial satellite launched into space?A. SputnikB. ApolloC. VoyagerD. Gemini答案: A5. 选择题：What is the main source of energy for all living things?A. WaterB. FoodC. SunlightD. Air答案:C. Sunlight6. 听力题：The chemical formula for potassium phosphate is _______.The main purpose of a catalyst is to speed up a ______.8. 听力题：A __________ is a visual representation of the periodic table.9. 选择题：What is the name of the first artificial satellite?A. Vanguard 1B. Explorer 1C. Sputnik 1D. Luna 110. 选择题：What is the opposite of "hot"?A. ColdB. WarmC. BoilingD. Spicy11. 听力题：I have _____ (两) pets at home.12. 听力题：I can ___ (throw) a ball far.13. 填空题：The ________ was a key treaty that marked the end of hostilities.14. 选择题：What is the capital of Kyrgyzstan?A. BishkekB. OshC. Jalal-AbadD. Talas15. 填空题：The __________ (历史的理解) requires critical thinking.16. 选择题：What do we call a person who studies animals?A. BotanistB. ZoologistC. GeologistD. ArchaeologistThe _____ (自然环境) is important for biodiversity.18. 选择题：What is the capital of France?A. BerlinB. MadridC. ParisD. Rome答案:C19. conservation) of resources is essential for sustainability. 填空题：The ____20. 填空题：_____ (阳光) is essential for photosynthesis.21. 选择题：Which food is made from milk?A. BreadB. CheeseC. RiceD. Pasta22. 填空题：The first successful vaccine for measles was developed in ________.23. 选择题：What do we call the movement of people from one place to another?A. MigrationB. TransportationC. TravelD. Relocation答案: A24. 听力题：The _____ (草坪) is freshly mowed.25. 听力题：The chemical symbol for carbon is ______.26. 填空题：The lotus flower grows in ______ (水) and is very beautiful.27. 听力题：I have a _____ (journal) for my thoughts.My sister has a __________ (乐观的) view on life.29. 填空题：My sister loves her _________ (玩具马) that she brushes every day.30. 选择题：What is the primary color of a honeydew melon?A. GreenB. YellowC. OrangeD. Red31. 选择题：Which of these colors is a secondary color?A. YellowB. RedC. OrangeD. Blue答案: C32. 填空题：A ______ (植物的种植) can be fulfilling.33. 选择题：What do we call a person who studies languages?A. LinguistB. PhilologistC. SociologistD. Anthropologist答案: A34. 选择题：What is the capital of Italy?A. RomeB. ParisC. BerlinD. Madrid答案:A35. 听力题：The main gas that causes acid rain is __________.36. 听力题：We will _______ (have) a barbecue this weekend.My brother likes to learn about ____ (space).38. 听力题：The first woman to fly in space was _______ Ride.39. 听力题：An alloy is a mixture of two or more _____, usually metals.40. 填空题：The ancient Chinese built the Great Wall to prevent _____.41. 听力题：The chemical symbol for calcium is _____ (Ca).42. 填空题：A bumblebee is important for _______ (授粉).43. 听力题：The chemical name for common sugar is _______.44. 听力题：She is _____ (running) a marathon.45. 填空题：The _______ (小长颈鹿) reaches for leaves in tall trees.46. 选择题：How many colors are in a rainbow?A. FiveB. SixC. SevenD. Eight答案:C47. 听力题：The rabbit is ______ in the garden. (hiding)48. Depression started in the United States in ______ (1929年). 填空题：The Grea49. 选择题：What shape has three sides?A. SquareB. RectangleC. TriangleD. Circle答案: C50. 听力题：I can ________ (jump) very high.51. 听力题：The ________ (stars) are bright in the sky.52. 听力题：Galaxies are huge systems of stars, gas, and ______.53. 选择题：What is the capital of Japan?a. Beijingb. Tokyoc. Seould. Bangkok答案:b54. 填空题：We made a ________ house for birds.55. 听力题：Chemical reactions can be classified based on whether they are _____ or spontaneous.56. 填空题：The _______ (猪) is known for its intelligence.57. 听力题：A solar eclipse happens when the moon blocks the _____.58. 选择题：What do you call the main character in a story?A. VillainB. ProtagonistC. NarratorD. Supporting character答案:B59. 填空题:My favorite vegetable is ________.60. 填空题：The __________ (历史的研究方法) shapes conclusions.A _______ can help to test the effects of temperature changes on materials.62. 选择题：What is the value of 5 × 5 15?A. 10B. 15C. 20D. 25答案:A63. 听力题：The teacher is very ________.64. 选择题：How many continents are there in the world?A. FiveB. SixC. SevenD. Eight65. 填空题：My dad likes to _______ (动词) on weekends. 他觉得这个活动很 _______ (形容词).66. 填空题：A _______ (金鱼) can live for many years with proper care.67. 填空题：The __________ (历史的启发性思维) encourages exploration.68. 填空题：My grandma is a wonderful __________ (谈话者) who shares stories.69. 选择题：What do we call the chemical process that occurs when plants take in carbon dioxide and release oxygen?A. RespirationB. PhotosynthesisC. TranspirationD. Fermentation答案: B. Photosynthesis70. 听力题：The ______ is known for her leadership skills.The first successful pancreas transplant was performed in ________.72. 填空题：In winter, I like to play with my ________ (玩具名) inside. It keeps me ________ (形容词) and entertained when it’s cold outside.73. 听力题：We are going ________ a trip.74. 填空题：My friend is very __________ (友好).75. 填空题：The __________ will help us know if we need to prepare for rain. (天气预报)76. 听力题：The cat is on the _____ (桌子).77. 填空题：I enjoy _______ (参加) music classes after school.78. 听力题：Chemicals are often measured in ________.79. 填空题：The sun gives us ______ (光和热).80. 选择题：What do you call a person who creates art?A. ArtistB. PainterC. SculptorD. All of the above答案:D81. 填空题：The _____ (园艺技巧) can be learned through practice.82. 选择题：What do you call the process of growing plants?A. GardeningB. FarmingC. PlantingD. Sowing答案:A83. 填空题：The frog jumps from ______ to ______.84. sustainable transit) options encourage eco-friendly travel. 填空题：The ____85. 听力题：Parrots can ______ human speech.86. 填空题：A ________ (植物文化) enriches society.87. 选择题：What is the main language spoken in the UK?A. SpanishB. FrenchC. EnglishD. German答案:C88. 填空题：Planting trees can help combat ______ change. (种树可以帮助抵御气候变化。

a r X i v :m a t h /0411523v 1 [m a t h .Q A ] 23 N o v 2004Twisted representations of vertex operatorsuperalgebrasChongying Dong 1and Zhongping ZhaoDepartment of Mathematics,University of California,Santa Cruz,CA 95064AbstractThis paper gives an analogue of A g (V )theory for a vertex operator superalgebra V and an automorphism g of finite order.The relation between the g -twisted V -modules and A g (V )-modules is established.It is proved that if V is g -rational,then A g (V )is finite dimensional semisimple associative algebra and there are only finitely many irreducible g -twisted V -modules.1IntroductionThe twisted sectors or twisted modules are basic ingredients in orbifold conformal field theory (cf.[FLM1],[FLM2],[FLM3],[Le1],[Le2],[DHVW],[DVVV],[DL2],[DLM2]).The notion of twisted module [FFR],[D]is derived from the properties of twisted vertex operators for finite automorphisms of even lattice vertex operator algebras constructed in [Le1],[Le2]and [FLM2],also see [DL2].In this paper we study the twisted modules for an arbitrary vertex operator superalgebra following [Z],[KW]and [DLM2].An associative algebra A (V )was introduced in [Z]for every vertex operator algebra V to study the representation theory for vertex operator algebra.The main idea is to reduce the study of representation theory for a vertex operator algebra to the study of represen-tation theory for an associative algebra.This approach has been very successful and the irreducible modules for many well-known vertex operator algebras have been classified by using the associative algebras.This theory has been extended to the vertex operator superalgebras in [KW]and has been further generalized to the twisted representations for a vertex operator algebra in [DLM2].This paper is a “super analogue”of [DLM2].We construct an associative algebra A g (V )for any vertex operator superalgebra V together with an automorphism g of finite order.Then the vacuum space of any admissible g -twisted V -module becomes a module for A g (V ).On the other hand one can construct a ‘universal’admissible g -twisted V -module from any A g (V )-module.This leads to a one to one correspondence between the set of inequivalent admissible g -twisted V -modules and the set of simple A g (V )-modules.As in the case of vertex operator algebra,if V is g -rational then A g (V )is a finite dimensional semisimple associative algebra.The ideas of this paper and other related papers are very natural and go back to the theory of highest weight modules for Kac-Moody Lie algebras and other Lie algebras with triangular decompositions.In the classical highest weight module theory,the highestweight or highest weight vector determines the highest weight module structure to some extend(different highest weight modules can have the same highest weight).The role of the vacuum space for an admissible twisted module is similar to the role of the highest weight space in a highest weight module.So from this point of view,the A g(V)theory is a natural extension of highest weight module theory in the representation theory of vertex operator superalgebras.A vertex operator superalgebra has a canonical automorphismσof order2arising from the structure of superspace.Theσ-twisted modules which are called the Ramond sector in the literature play very important roles in the study of geometry.Important topological invariants such as elliptic genus and certain Witten genus can be understood as graded trace functions on the Ramond sectors constructed from the manifolds.It is expected that the theory developed in this paper will have applications in geometry and physics.Since the setting and most results in this paper are similar to those in[DLM2]we only provide the arguments which are either new or need a lot of modifications.We refer the reader to[DLM2]for details.The organization of this paper is similar to that of[DLM2].We review the definition of vertex operator superalgebra and define various notions of g-twisted V-modules in section2.In section3,we introduce the algebra A g(V)for VOSA V.Section4is devoted to the study of Lie superalgebra V[g]which is kind of twisted affinization of V.A weak g-twisted V-module is naturally a V[g]-module.In section5,we construct the functorΩwhich sends a weak g-twisted V-module to an A g(V)-module.We construct another functor L from the category of A g(V)-modules to the category of admissible g-twisted V-modules in Section6.That is,for any A g(V)-module U we can construct a kind of“generalized Verma module”¯M(U)which is the universal admissible g-twisted V-module generated by U.It is proved that there is a1-1correspondence between the irreducible objects in these two categories.Moreover if V is g-rational,then A g(V)is a finite dimensional semisimple associative algebra.We discuss some examples of vertex operator superalgebras constructed from the free fermions and their twisted modules in Section7.2Vertex Operator superalgebra and twisted mod-ulesWe review the definition of vertex operator superalgebra(cf.[B],[FLM3],[DL1])and various notions of twisted modules in this section(cf.[D],[DLM2],[FFR],[FLM3],[Z]).Recall that a super vector space is a Z2-graded vector space V=V¯0⊕V¯1.The elements in V¯0(resp.V¯1)are called even(resp.odd).Let˜v be0if v∈V¯0,and1if v∈V¯1.Definition2.1.A vertex operator superalgebra is a12Z+V n=V¯0⊕V¯1.(2.1)with V¯0= n∈Z V n and V¯1= n∈112(m3−m)δm+n,0c;(2.6)dz0 Y(u,z1)Y(v,z2)−(−1)˜u˜v z−10δ z2−z1z2 Y(Y(u,z0)v,z2).(2.9) whereδ(z)= n∈Z z n and(z i−z j)n is expanded as a formal power series in z j.Throughout the paper,z0,z1,z2,etc.are independent commuting formal variables.Such a vertex operator superalgebra may be denoted by V=(V,Y,1,ω).In the case V¯1=0,this is exactly the definition of vertex operator algebra given in[FLM3].Definition2.2.Let V be a vertex operator superalgebra.An automorphism g of V is a linear automorphism of V preservingωsuch that the actions of g and Y(v,z)on V are compatible in the sense thatgY(v,z)g−1=Y(gv,z)for v∈V.Note that any automorphism of V commutes with L(0)and preserves each homoge-neous space V n.As a result,any automorphism preserves V¯0and V¯1.Let Aut(V)be the group of automorphisms of V.There is a special automorphism σ∈Aut(V)such thatσ|V¯0=1andσ|V¯1=−1.It is clear thatσis a central element of Aut(V).Fix g∈Aut(V)of order T0.Let o(gσ)=T.Denote the decompositions of V into eigenspaces with respect to the actions of gσand g as followsV=⊕r∈Z/T Z V r∗(2.10)V=⊕r∈Z/T0ZV r(2.11) where V r∗={v∈V|gσv=e2πir/T v}and V r={v∈V|gv=e2πir/T0v}Definition2.3.A weak g-twisted V-module M is a vector space equipped with a linear mapV→(End M)[[z1/T0,z−1/T0]v→Y M(v,z)= n∈1T0+Zu n z−n−1;(2.12)u l w=0for l>>0;(2.13)Y M(1,z)=Id M;(2.14) z−10δ z1−z2−z0 Y M(v,z2)Y M(u,z1)=z−12 z1−z0z2 Y M(Y(u,z0)v,z2).(2.15)Following the arguments in[DL1]one can prove that the twisted Jacobi identity is equivalent to the following associativity formula(z0+z2)k+r T0Y M(Y(u,z0)v,z2)w.(2.16) where w∈M and k∈Z+s.t z k+rz2 −r/T0δz1−z0Lemma2.4.The associativity formula(2.16)is equivalent to the following: (z0+z2)m+s T Y M(Y(u,z0)v,z2)wfor u∈V s∗and some m∈1T Y M(u,z)w involves only nonnegative integral powers of z.Proof:Let u∈V r.It is enough to prove that wt u+sT0are congruent modulo Z.It is easy to see that s≡T2˜u+2r modulo Z if T0is odd.Thus wt u+s2˜u+r2˜u and wt u are congruentmodulo Z,the result follows immediately.Equating the coefficients of z−m−11z−n−12in(2.17)yields[u m,v n]=∞i=0 m i (u i v)m+n−i.(2.18)We may also deduce from(2.12)-(2.15)the usual Virasoro algebra axioms,namely that if Y M(ω,z)= n∈Z L(n)z−n−2then[L(m),L(n)]=(m−n)L(m+n)+1dzY M(v,z)=Y M(L(−1)v,z)(2.20) (cf.[DLM1]).The homomorphism and isomorphism of weak twisted modules are defined in an ob-vious way.Definition2.5.An admissible g-twisted V-module is a weak g-twisted V-module M which carries a1T Z+M(n)(2.21)satisfyingv m M(n)⊆M(n+wt v−m−1)(2.22) for homogeneous v∈V.Definition2.6.An ordinary g-twisted V-module is a weak g-twisted V-moduleM= λ∈C Mλ(2.23) such that dim Mλisfinite and forfixedλ,M nThe admissible g-twisted V-modules form a subcategory of the weak g-twisted V-modules.It is easy to prove that an ordinary g-twisted V-module is admissible.Shifting the grading of an admissible g-twisted module gives an isomorphic admissible g-twisted V-module.A simple object in this category is an admissible g-twisted V-module M such that0and M are the only graded submodules.We say that V is g-rational if every admissible g-twisted V-module is completely reducible,i.e.,a direct sum of simple admissible g-twisted modules.V is called rational if V is1-rational.V is called holomorphic if V is rational and V is the only irreducible V-module up to isomorphism.If M=⊕n∈1T Z+M(n)∗(2.24)where M(n)∗=Hom C(M(n),C).The vertex operator Y M′(a,z)is defined for a∈V via Y M′(a,z)f,u = f,Y M(e zL(1)(−z−2)L(0)a,z−1)u (2.25) where · denotes the natural paring between M′and M.Then we have the following [FHL]:Lemma2.7.(M′,Y M′)is an admissible g−1-twisted V-module.Lemma2.7is needed in the proof of several results in Section6although we do not intend to give these proofs(cf.[DLM2]).3The associative algebra A g(V)Let r be an integer between0and T−1(or T0−1).We will also use r to denote its residue class modulo T or T0.For homogeneous u∈V r∗,we setδr=1if r=0andδr=0 if r=0.Let v∈V we defineu◦g v=Res z (1+z)wt u−1+δr+rz1+δrY(u,z)v(3.1)where(1+z)αforα∈C is to be expanded in nonnegative integer powers of z.Let O g(V) be the linear span of all u◦g v and define the linear space A g(V)to be the quotient V/O g(V).We will use A(V),O(V),u◦v,when g=1.The A(V)was constructed in[KW] and if V is a vertex operator,A g(V)was constructed in[DLM2].Lemma3.1.If r=0then V r∗⊆O g(V).Proof:The proof is the same as that of Lemma2.1in[DLM2].Let I=O g(V)∩V0∗.Then A g(V)≃V0∗/I(as linear spaces).Since O(V0∗)⊂I, A g(V)is a quotient of A(V0∗).We now define a product ∗g on V which will induce an associative product in A g (V ).Let r,u and v be as above and setu ∗gv =Res z (Y (u,z )(1+z )wt uT+nzY (v,z )u ∈O (V 0∗)and(iii)u ∗v −(−1)˜u ˜v v ∗u −Res z (1+z )wt u −1Y (u,z )v ∈O (V 0∗).Proof:See the proofs of Lemmas 2.1.2and 2.1.3of[Z]bynoting thatY (u,z )v ≡(−1)˜u ˜v (1+z )−wtu −wtv Y (v,−zz Y (c,z )u(3.4)andu∗c≡Res z(1+z)wt c−1z0 Y(c,z1)Y(a,z2)b−(−1)˜c˜a z−10δ z2−z1z2 Y(Y(c,z0)a,z2)b.(3.6) Forε=0or1,(3.6)implies:xε=Res z1(1+z1)wt c−εTz1Y(c,z1)(1+z2)wt a−1+δr+rz1+δr2Y(a,z2)b=(−1)˜a˜c Res z1Res z2(1+z1)wt c−εTz1(1+z2)wt a−1+δr+rz1+δr2z−12δ z1−z0z1(1+z2)wt a−1+δr+rz1+δr2Y(a,z2)Y(c,z1)b+Res z2Res z(1+z2+z0)wt c−εTTz1Y(c,z1)b+∞i,j=0(−1)j wt c−εi Res z2(1+z2)wt a−1+δr+r z j+2+δr2Y(c i+j a,z2)b=(−1)˜c˜a Res z2(1+z2)wt a−1+δr+rz1+δr2Y(a,z2)Res z1(1+z1)wt c−εT+j+1−εNext we prove that∗g is associative.We need to verify that(a∗b)∗c−a∗(b∗c)∈O g(V0∗)for a,b,c∈V0∗.A straightforward computation using the twisted Jacobi identity gives(a∗b)∗c=wt a i=0(a i−1b)∗c=wt ai=0 wt a i Res w(Y(a i−1b,w)(1+w)wt(a i−1b)wc)=Res w Res z−w(Y(Y(a,z−w)b,w)(1+z)wt a(1+w)wt bw(z−w)c)−(−1)˜a˜b Res w Res z(Y(b,w)Y(a,z)(1+z)wt a(1+w)wt bwc)−(−1)˜a˜b∞i=0Res w Res z(Y(b,w)Y(a,z)(−1)i+1z i w−i−1(1+z)wt a(1+w)wt bzwc)mod O g(V0∗)≡a∗(b∗c)mod O g(V0∗)Thus A g(V)≃V0∗T0,t−1dt f(t) g(t).(4.1) (see[B]).Then the tensor productL(V)=C[t1T0]⊗V.(4.2)is a vertex superalgebra with vertex operatorY (f (t )⊗v,z )(g (t )⊗u )=f (t +z )g (t )⊗Y (v,z )u.(4.3)The L (−1)operator of L (V )is given by D =dT 0)(t m ⊗ga ).(4.4)Let L (V,g )be the g -invariants which is a vertex sub-superalgebra of L (V ).Clearly,L (V,g )=⊕T 0−1r =0tr/T 0C [t,t −1]⊗V r .(4.5)Following [B],we know thatV [g ]=L (V,g )/D L (V,g )(4.6)is a Lie superalgebra with bracket[u +D L (V,g ),v +D L (V,g )]=u 0v +D L (V,g ).(4.7)For short let a (q )be the image of t q ⊗a ∈L (V,g )in V [g ].Then we have Lemma 4.1.Let a ∈V r ,v ∈V s and m,n ∈Z .Then(i)[ω(0),a (m +r T 0a (m −1+rT 0),b (n +s T 0ia ib (m +n +r +sTZ -graded.Since D increases degree by 1,D L (V,g )is a graded subspace ofL (V,g )and V [g ]is naturally 1TZV [g ]n .By Lemma 4.1,V [g ]is a1TZV [g ]±n .Lemma 4.2.V [g ]0is spanned by elements of the form a (wt a −1)for homogeneous a ∈V 0∗.Proof:Let a∈V.Then the degree wt a−n−1of a(n)is0if and only if a∈V0¯andn=wt a−1or a∈V T0/2¯1and n=wt a−1.The bracket of V[g]0is given by[a(wt a−1),b(wt b−1)]=∞j=0 wt a−1j a j b(wt(a j b)−1).(4.10)Set o(a)=a(wt a−1)for homogeneous a∈V0∗and extend linearly to all a∈V0∗. This gives a linear mapV0∗→V[g]0,a→o(a).(4.11) As the kernel of the map is(L(−1)+L(0))V0∗,we obtain an isomorphism of Lie super-algebras V0∗/(L(−1)+L(0))V0∗∼=V[g]0.The bracket on the quotient of V0∗is given by[a,b]= j≥0 wt a−1j a j b.Lemma4.3.Let A g(V)Lie be the Lie superalgebra of the associative algebra A g(V)intro-duced in section3such that[u,v]=u∗g v−(−1)˜u˜v v∗g u.Then the map o(a)→a+O g(V) is an onto Lie superalgebra homomorphism from V[g]0to A g(V)Lie.Proof:Recall that I=O g(V)∩V0∗.So we have a surjective linear mapV[g]0∼=V0∗/(L(−1)+L(0))V0∗→V0∗/I≃A g(V),o(a)→a+(L(−1)+L(0))V0∗→a+I.(4.12) The Lie homomorphism follows from[o(a),o(b)]=∞j=0 wt a−1j o(a j b).and[a+O g(V),b+O g(V)]≡a∗g b−(−1)˜a˜b b∗g a≡∞j=0 wt a−1j a j b≡Res z(1+z)wt a−1Y(a,z)b mod O g(V0∗)≡∞i=0 wt a−1i a i b mod O g(V0∗).5The functorΩThe main purpose in this section is to construct a covariant functorΩfrom the category of weak g-twisted V-modules to the category of A g(V)-modules(cf.Theorem5.1).Let M be a weak g-twisted V-module.We define the space of“lowest weight vectors”to beΩ(M)={w∈M|u wt u+n w=0,u∈V,n≥0}.The main result in this section says thatΩ(M)is an A g(V)-module.Moreover if f:M→N is a morphism in weak g-twisted V-modules,the restrictionΩ(f)of f toΩ(M)is an A g(V)-module morphism.Note that if M is a weak g-twisted V-module then M becomes a V[g]-module such that a(m)acts as a m.Moreover,M is an admissible g-twisted V-module if and only if M is a1TY(u,z)v.z2The argument in the Proof of Theorem2.1.2in[Z]with suitable modification giveso(u∗v)=o(u)o(v).Note that o(L(−1)u+L(0)u)=0and(L(−1)u+L(0)u)∗v=u◦v.We immediately have o(u◦v)=0onΩ(M).Ifa=Res z (1+z)wt c−1+rzY(u,z)v,we can use Lemma2.4.Since z wt u−1+rT Y M(u,z0+z2)Y M(v,z2)w=(z2+z0)wt u−1+rT2to(5.1)yields0=Res z0Res z2z−10zwt v−rT Y M(Y(u,z0)v,z2)w=∞i=0 wt u−1+rTi o(u i−1v)w=o Res z(1+z)wt u−1+r z Y M(u,z)v w=o(a)w(5.2) as required.If M is a nonzero admissible g-twisted V-modules we may and do assume that M(0) is nonzero with suitable degree shift.With these conventions we haveProposition5.2.Let M be a simple admissible g-twisted V-module.Then the following hold(i)Ω(M)=M(0).(ii)Ω(M)is a simple A g(V)-module.Proof:The proof is the same as in[DLM2].6Generalized Verma modules and the functor LIn this section we focus on how to construct admissible g-twisted V-modules from a given A g(V)-module U.We use the same trick which was used in[DLM2]to do this.We will define two g-twisted admissible V-modules¯M(U)and L(U).The¯M(U)is the universal admissible g-twisted V-module such that¯M(U)(0)=U and L(U)is smallest admissible g-twisted V-module whose L(U)(0)=U.Just as in the classical highest weight module theory,L(U)is the unique irreducible quotient of¯M(U)if U is simple.We start with an A g(V)-module U.Then U is automatically a module for A g(V)Lie.By Lemma4.3U is lifted to a module for the Lie superalgebra V[g]0.Let V[g]−act trivially on U and extend U to a P=V[g]−⊕V[g]0-module.Consider the induced moduleM(U)=Ind V[g]P(U)=U(V[g])⊗U(P)U(6.1)T0Zv(m)z−m−1(6.2)Then Y M(U)(v,z)satisfies condition(2.12)-(2.14).By Lemma4.1(ii),the identity(2.18) holds.But this is not good enough to establish the twisted Jacobi identity for the action (6.2)on M(U).Let W be the subspace of M(U)spanned linearly by the coefficients of(z0+z2)wt a−1+δr+r T Y(Y(a,z0)b,z2)u(6.3) for any homogeneous a∈V r∗,b∈V,u∈U.We set¯M(U)=M(U)/U(V[g])W.(6.4) Proposition6.1.Let M be a V[g]-module such that there is a subspace U of M satisfying the following conditions:(i)M=U(V[g])U;(ii)For any a∈V r∗and u∈U there is k∈wt a+Z+such that(z0+z2)k+r T Y(Y(a,z0)b,z2)u(6.5) for any b∈V.Then M is a weak V-module.Proof:We only need to prove the twisted Jacobi identity,which is equivalent to com-mutator relation(2.17)and the associativity(2.16).But the commutator formula is built in already as M is a V[g]-module.By Lemma2.4,the assumption(ii)can be reformulated as follows:(ii’)For any a∈V r and u∈U there is k∈Z+such that(z0+z2)k+r T0Y(Y(a,z0)b,z2)u(6.6) Since M is a V[g]-module generated by U it is enough to prove that if u satisfies(ii’) then c n u also satisfies(ii’)for c∈V and n∈1T0Y(c i a,z0+z2)Y(b,z2)u=(z2+z0)k2+r+sT0Y(a,z0+z2)Y(c i b,z2)u=(z2+z0)k2+r+sT0+n−k1>k2+r+s(z 0+z 2)k +rT 0c n Y (a,z 0+z 2)Y (b,z 2)u −(−1)˜a ˜c (−1)˜b ˜c∞ i =0n i (z 0+z 2)k +r T 0Y (a,z 0+z 2)Y (c i b,z 2)u=(−1)˜a ˜c (−1)˜b ˜c (z 0+z 2)k +rT 0+n −iY (Y (c i a,z 0)b,z 2)u−(−1)˜b ˜c∞ i =0n iz n −i2(z 2+z 0)k +rT 0c n Y (Y (a,z 0)b,z 2)u−(−1)˜a ˜c (−1)˜b ˜c ∞ i =0n i (z 2+z 0)k +r T 0Y (c i Y (a,z 0)b,z 2)u+(−1)˜a ˜c (−1)˜b ˜c ∞ i =0∞ j =0n j j iz n −i2(z 2+z 0)k +rT 0c n Y (Y (a,z 0)b,z 2)u−(−1)˜a ˜c (−1)˜b ˜c ∞ i =0n i (z 2+z 0)k +r T 0Y (c i Y (a,z 0)b,z 2)u+(−1)˜a ˜c (−1)˜b ˜c ∞ j =0∞ i =jn j n −ji −jz n −i2(z 2+z 0)k +rT 0c n Y (Y (a,z 0)b,z 2)u−(−1)˜a ˜c (−1)˜b ˜c∞ i =0n i z n −i 2(z 2+z 0)k +rT 0c n Y (Y (a,z 0)b,z 2)u−(−1)˜a ˜c (−1)˜b ˜c (z 2+z 0)k +rT 0Y (Y (a,z 0)b,z 2)c n u,=(z 2+z 0)k +rThe proof is complete.Applying Proposition6.1to¯M(U)gives the following main result of this section. Theorem6.2.¯M(U)is an admissible g-twisted V-module with¯M(U)(0)=U and with the following universal property:for any weak g-twisted V-module M and any A g(V)-morphismφ:U→Ω(M),there is a unique morphism¯φ:¯M(U)→M of weak g-twisted V-modules which extendsφ.As in[DLM2]we also haveTheorem6.3.M(U)has a unique maximal graded V[g]-submodule J with the property that J∩U=0.Then L(U)=M(U)/J is an admissible g-twisted V-module satisfying Ω(L(U))∼=U.L defines a functor from the category of A g(V)-modules to the category of admissible g-twisted V-modules such thatΩ◦L is naturally equivalent to the identity.We have a pair of functorsΩ,L between the A g(V)-module category and admissible g-twisted V-module category.AlthoughΩ◦L is equivalent to the identity,L◦Ωis not equivalent to the identity in general.The following result is an immediate consequence of Theorem6.3.Lemma6.4.Suppose that U is a simple A g(V)-module.Then L(U)is a simple admissible g-twisted V-module.Using Lemma6.4,Proposition5.2(ii),Theorems6.2and6.3gives:Theorem6.5.L andΩare equivalent when restricted to the full subcategories of com-pletely reducible A g(V)-modules and completely reducible admissible g-twisted V-modules respectively.In particular,L andΩinduces mutually inverse bijections on the isomor-phism classes of simple objects in the category of A g(V)-modules and admissible g-twisted V-modules respectively.We now apply the obtained results to g-rational vertex operator superalgebras to obtain:Theorem6.6.Suppose that V is a g-rational vertex operator superalgebra.Then the following hold:(a)A g(V)is afinite-dimensional,semi-simple associative algebra(possibly0).(b)V has onlyfinitely many isomorphism classes of simple admissible g-twisted mod-ules.(c)Every simple admissible g-twisted V-module is an ordinary g-twisted V-module.(d)V is g−1-rational.(e)The functors L,Ωare mutually inverse categorical equivalences between the cate-gory of A g(V)-modules and the category of admissible g-twisted V-modules.(f)The functors L,Ωinduce mutually inverse categorical equivalences between the category offinite-dimensional A g(V)-modules and the category of ordinary g-twisted V-modules.The proof is the same as that of Theorem8.1in[DLM2].7ExamplesIn this section we discuss the well known vertex operator superalgebras constructed from the free fermions and their twisted modules.In particular we compute the algebra A g (V )and classify the irreducible twisted modules using A g (V ).The classification results have been obtained previously in [Li2]with a different approach.Let H = li =1C a i be a complex vector space equipped with a nondegenerate symmet-ric bilinear form (,)such that {a i |i =1,2,...l }form an orthonormal basis.Let A (H,Z +12}subject to the relation [a (n ),b (m )]+=(a,b )δm +n,0.Let A +(H,Z +12,n >0},andmake C a 1-dimensional A +(H,Z +12)=A (H,Z +12)C∼=Λ[a i(−n )|n >0,n ∈Z +1∂a i (−n )if n is positive and by multiplication by a i (n )if nis negative.The V (H,Z +12Zso thatV (H,Z +12Zwe define a normal ordering:b 1(n 1)···b k (n k ):=(−1)|σ|b i 1(n i 1)···b i k (n i k )such that n i 1≤···≤n i k where σis the permutation of {1,...,k }by sending j to i j .For a ∈H set Y (a (−1/2),z )=n ∈12)···b k (−n k −12)where n i arenonnegative integers.We setY (v,z )=:(∂n 1b 1(z ))···(∂n k b k (z )):where ∂n =1dz)n .Then we have a linear map:V (H,Z +12))[[z,z −1]]v→Y (v,z )=n ∈Z v n z −n −1(v n ∈End V (H,Z +12li =1a i (−32).The following result is well known (cf.[FFR],[KW]and [Li1]).Theorem7.1.(V(H,Z+12)for i=1,...,l.We have already mentioned in Section2that any vertex operator superalgebra has a canonical automorphismσsuch thatσ=1on V¯0andσ=−1on V¯1.Note thatV(H,Z+12)¯1.We next discuss theσ-twisted V(H,Z+1∂b i(−n)∗if n is nonnegative and multiplication by b i(n)if n is nega-tive.Similarly,b i(n)∗acts as∂2)-module such thatY V(H,Z)(u(−12)is isomorphic to thematrix algebra M2k×2k(C)and V(H,Z)is the unique irreducibleσ-twisted V(H,Z+12)is isomorphic to thematrix algebra M2k×2k(C).Since g=σ,the decomposition(2.10)becomes V=V0∗.By lemma3.2(i),Res z (1+z)1z2+ma i(z)v= s≥0c s a i(−m+s−3lies in O σ(V (H,Z +12s.This implies that a i (−m −32+s )vmod O σ(V (H,Z +12))is spanned by b 1(−1/2)s 1···b k (−1/2)s k b ∗1(−1/2)t 1···b ∗k (−1/2)t kwith s i ,t i =0,1.As a result,dim A σ(V (H,Z +12)-module.By Theorem 5.1,Ω(V (H,Z ))is a simple A σ(V (H,Z +12)≥dim Ω(V (H,Z ))=22k .This forces dim A σ(V (H,Z +12))∼=M 2k ×2k (C ).We now deal with the case dim H =2k +1for some nonnegative integer k.Then H can be decomposed into:H =k i =1C b i +k i =1C b ∗i +C ewith (b i ,b j )=(b ∗i ,b ∗j )=0,(b i ,b ∗j )=δi,j ,(e,b i )=(e,b ∗i )=0,(e,e )=2.Let A (H,Z )be the associative algebra generated same as above,and A (H,Z )+be the subalgebra generated by {b i (n ),b ∗i (m ),e (n )|m,n ∈Z ,m >0,n ≥0,i =1,···,k }and make C a 1-dimensional A (H,Z )+-module so that b i (n )1=0for n ≥0and b ∗i (m )1=e (m )1=0for m >0,i =1,···,k.SetV (H,Z )=A (H,Z )⊗A (H,Z )+C∼=Λ[b i (−n ),b ∗i (−m ),e (−m )|n,m ∈Z ,n >0,m ≥0]and letW (H,Z )=Λ[b i (−n ),b ∗i (−m ),e (−n )|n,m ∈Z ,n >0,m ≥0]=W (H,Z )even⊕W (H,Z )odd be the decomposition into the even and old parity subspaces.Also defineV ±(H,Z )=(1±e (0))W (H,Z )even ⊕(1∓e (0))W (H,Z )odd .ThenV (H,Z )=V +(H,Z )⊕V −(H,Z )and V ±(H,Z )are irreducible A (H,Z )-modules.The actions of b i (n ),b ∗i (n )are the same as before.The e (n )acts as 2∂2),z )=u (z )=n ∈Zu (n )z −n −1/2for u ∈H.Proposition7.3.If dim H=2k+1is odd,then Aσ(V(H,Z+12)has exactly two irreducibleσ-twisted modulesV±(H,Z)up to isomorphism.Proof:The proof is similar to that of Proposition7.2.Note that the automorphismσof V(H,Z+12)as follows:For any a1(−n1)···a s(−n s)∈V(H,Z+1/2),τ(a1(−n1)a2(−n2)···a m(−n m))=(τa1)(−n1)(τa2)(−n2)···(τa m)(−n m).Let o(τσ)=N.We decompose H into eigenspaces with respect to theτσandτas follows:H=⊕r∈Z/N Z H r∗(7.4)H=⊕r∈Z/N0ZH r(7.5) where H r∗={v∈H|τσv=e2πir/N v},and H r={v∈H|τv=e2πir/N0v}.Let l0=dim H0∗.As before we need to consider two separate cases:l0is even or odd. If l0=2k0for some nonnegative integer k0,we haveH0∗=k0i=1C h i+k0 i=1C h∗iwith(h i,h j)=(h∗i,h∗j)=0,(h i,h∗j)=δi,j.Let l r=dim H r∗with r=0.If r=N−r,wefix bases b r,1,b r,2,···b r,lr∈H r∗and b∗r,1,b∗r,2,···b∗r,lr∈H(N−r)∗such that(b r,i,b∗r,j)=(b∗r,j,b r,i)=δi,j.If r=N−r,let{c1,c2,···c lN 2∗.Then M= N−1r=1Λ[b(−n)|n∈r2)-module so that for u∈H r∗, Y M(u(−1N +Zu(n)z−n−1/2(see[Li2]).Note that b r,i(n)acts as∂∂c i(−n)if n is positive andacts as multiplication by c i(n)if n is negative.Also,h i(n)acts as∂∂h i(−n)if n is positive,and acts as multiplication by h∗i(n)if nis nonnegative.One can easily calculate thatΩ(M)=Λ[h∗i(0)|h∗i∈H0∗,i=1,2,···k0]. So dimΩ(M)=2k0.Proposition7.4.If dim H0∗=l0=2k0then M= N−1r=1Λ[b(−n)|n∈r2)-module.Proof:As in the proof of Proposition7.2,it is sufficient to show that dim Aτ(V(H,Z+ 1N−1 z1+m a(z)b=∞l=0 r2l a(−m−12)).So using the same calculation done in Proposition7.2,we conclude that Aτ(V)is spanned byh1(−1/2)s1···h k0(−1/2)s k0h∗1(−1/2)t1···h∗k(−1/2)t k0with s i,t i=0,1.Hence dim Aτ(V(H,Z+1N+Z,1≤r≤N−1,n>0] are irreducibleτ-twisted V(H,Z+1∂e(−n)if n>0andas multiplication by e(n)if n≤0.The proof of Proposition7.4gives Proposition7.5.If dim H0∗=2k0+1is odd,V(H,Z+1[DVVV]R.Dijkgraaf,C.Vafa,E.Verlinde and H.Verlinde,The operator algebra of orbifold models,Comm.Math.Phys.123(1989),485-526.[DHVW]L.Dixon,J.Harvey,C.Vafa and E.Witten,Strings on orbifolds,Nucl.Phys.B261(1985),651;II,Nucl.Phys.B274(1986),285.[D] C.Dong,Twisted modules for vertex algebras associated with even lattice,J.of Algebra165(1994),91-112.[DL1] C.Dong and J.Lepowsky,Generalized Vertex Algebras and Relative Vertex Operators,Progress in Math.,Vol.112,Birkh¨a user Boston,1993.[DL2] C.Dong and J.Lepowsky,The algebraic structure of relative twisted vertex operators,J.Pure and Applied Algebra110(1996),259-295.[DLM1] C.Dong,H.Li and G.Mason,Regularity of rational vertex operator algebras, Adv.Math.132(1997),148–166.[DLM2] C.Dong,H.Li and G.Mason,Twisted representations of vertex operator alge-bras,Math.Ann.310(1998),571–600.[FFR]Alex J.Feingold,Igor B.Frenkel and John F.X.Ries,Spinor Construction of Vertex Operator Algebras,Triality,and E(1)8,Contemporary Math.121,1991.[FHL]I.Frenkel,Y.Huang and J.Lepowsky,On axiomatic approaches to vertex oper-ator algebras and modules,Mem.Amer.Math.Soc.1041993.[FLM1]I.B.Frenkel,J.Lepowsky and A.Meurman,A natural representation of the Fischer-Griess Monster with the modular function J as character,Proc.Natl.A81(1984),3256-3260.[FLM2]I.B.Frenkel,J.Lepowsky and A.Meurman,Vertex operator calculus,in:Math-ematical Aspects of String Theory,Proc.1986Conference,San Diego.ed.byS.-T.Yau,World Scientific,Singapore,1987,150-188.[FLM3]I.B.Frenkel,J.Lepowsky and A.Meurman,Vertex Operator Algebras and the Monster,Pure and Applied Math.,Vol.134,Academic Press,1988.[FZ]I.Frenkel and Y.Zhu,Vertex operator algebras associated to representations of affine and Virasoro algebras,Duke Math.J.66(1992),123-168.[KW]V.Kac and W.Wang,Vertex operator superalgebras and representations,Con-tem.Math.,AMS Vol.175(1994),161-191.[Le1]J.Lepowsky,Calculus of twisted vertex operators,Proc.Natl.Acad A 82(1985),8295-8299.。

super twisting sliding mode control讲解Super twisting sliding mode control is a robust control technique that has gained significant attention in thefield of control engineering. It is designed to address the problem of uncertainties and disturbances in a dynamic system, allowing for precise and stable control even in the presence of external factors that can affect system dynamics.超滑模控制是一种鲁棒控制技术，在控制工程领域得到了广泛关注。

它旨在解决动态系统中的不确定性和干扰问题，即使在存在可能影响系统动态的外部因素的情况下，也能实现精确稳定的控制。

The basic principle behind super twisting sliding mode control is to create a sliding surface that guarantees convergence towards the desired trajectory while ensuring fast error reduction. This sliding surface is created by combining two terms: a reaching term and a switching term. The reaching term adjusts the rate at which the error converges towards zero, while the switching term ensures rapid error reduction.超滑模控制的基本原理是创建一个滑动面，确保收敛到期望轨迹并实现快速误差减小。

小学上册英语第二单元真题(含答案)英语试题一、综合题(本题有50小题，每小题1分，共100分.每小题不选、错误，均不给分)1 The ______ helps maintain the balance of nature.2 The fruit is _____ and ripe. (fresh)3 A __________ is formed through the erosion of softer rock.4 The _______ can help improve your mood.5 Which part of the plant absorbs water?A. LeavesB. StemC. RootsD. Flower答案:C6 A _____ (草丛) is home to many small creatures.7 What is the name of the famous painting of a girl holding a pearl earring?A. The Girl with a Pearl EarringB. The Mona LisaC. The Birth of VenusD. The Starry Night8 The Earth’s ______ is responsible for its magnetic field.9 My dad loves to watch __________ (体育比赛) on TV.10 The _______ (猪) is known for its intelligence.11 A ______ (城市花园) can beautify neighborhoods.12 I have a _____ (pen/pencil) in my backpack.13 Which of these animals is a reptile?A. FrogB. SnakeC. RabbitD. Fish14 The cat is __________ the tree.15 The ______ (小鱼) swims gracefully in the tank.16 The ________ is the line of latitude at degrees.17 A _______ is a change that produces one or more new substances.18 How many seasons are there in a year?A. TwoB. ThreeC. FourD. Five19 A __________ is an area where two tectonic plates meet.20 A _____ (植物活动) can raise awareness about conservation.21 The Earth's atmosphere protects us from harmful _______ rays.22 My dog loves to chase after ______ (飞碟).23 Which insect makes honey?A. AntB. ButterflyC. Bee答案:C24 The ______ (小鸽子) coos softly as it perches on a ______ (电线).25 I enjoy reading ________ stories before bed.26 We eat _____ (dinner/breakfast) in the evening.27 The _______ (狗) loves to bark at mailmen.28 The ______ helps us understand mathematics.29 Did you see that _____ (小狗) chasing its tail?30 What do we call the outer layer of the Earth?A. CoreB. MantleC. CrustD. Atmosphere答案: C31 What is the term for a baby horse?A. CalfB. FoalC. KidD. Lamb答案:B32 A __________ is a mixture of liquids that can be separated.33 Kittens are very _________. (顽皮)34 The chemical formula for iron(III) oxide is ______.35 What is the base of a pyramid?A. TopC. BottomD. Edge答案:C36 I want to learn ________ (新技能).37 The ______ enjoys reading mystery novels.38 The ________ (社区援助) supports those in need.39 The capital of Brunei is _______.40 I can create a __________ (名词) with my __________ (玩具名).41 A dolphin's playful nature attracts many ________________ (游客).42 What is the capital of Thailand?A. BangkokB. HanoiC. ManilaD. Kuala Lumpur答案:A43 The chemical symbol for cerium is ______.44 A reaction that occurs rapidly is called a ______ reaction.45 What do you use to see in the dark?A. CandleB. FlashlightC. LampD. Mirror46 I like to collaborate with friends using my ________ (玩具).47 What do you call a person who studies plants?A. BiologistB. GardenerC. BotanistD. Zoologist答案:C48 What is the name of the famous American singer known for her hit song "Someone Like You"?A. AdeleB. Taylor SwiftC. Katy PerryD. Billie Eilish答案: A. Adele49 The city of Nicosia is the capital of _______.50 What instrument is used to measure temperature?A. BarometerB. ThermometerC. SpeedometerD. Altimeter51 What is the name of the sport played with a ball and a net?A. SoccerB. TennisC. BaseballD. Golf答案: B52 biome) refers to a large ecological area. The ____53 The rabbit hops ______. (兔子_______跳。