On the wonderful compactification

合集下载

天津市河东区2022届高三英语下学期第二次模拟考试试题含解析

天津市河东区2022届高三英语下学期第二次模拟考试试题含解析

天津市河东区2022届高三英语下学期第二次模拟考试试题本试卷分为第Ⅰ卷(选择题)、第Ⅱ卷(非选择题)两部分,共130分,考试时间100分钟。

第Ⅰ卷1至8页,第Ⅱ卷9至10页。

答题时,将第Ⅱ卷的答案填涂在答题卡上,将第Ⅱ卷答案填写在答题卡上。

祝各位考生考试顺利!第Ⅰ卷(选择题)第一部分:英语知识运用第一节:单项填空从A、B、C、D四个选项中,选出可以填入空白处的最佳选项,并在答题卡上将该项涂黑。

例:Stand over there ______ you’ll be able to see it better.A. orB. andC. butD. while答案是B.1. —I’m afraid I can’t finish the work that I have never done before.—______. It’s said that it will not be difficult.A. Take your timeB. Take it easyC. I can’t believe itD. Never mind【答案】B【解析】【详解】考查情景交际和习惯表达。

句意:—我担心我不能完成这份我以前从未做过的工作。

—别紧张。

据说这并不难。

A. Take your time慢慢来;B. Take it easy放轻松,别紧张;C. I can’t believe it 真不敢相信;D. Never mind没关系。

根据句意可知,空处上句描述的是我担心完不成任务,下句说这项任务并不难,可以看出,空处应该是在告诉对方不要紧张,放轻松(Take it easy)。

故选B项。

2. The newly-built school was much bigger than she ______ at first.A. was thinkingB. thoughtC. had thoughtD. has thought【答案】C【解析】【详解】考查时态。

COMAC航空科技英语等级考试B样题

COMAC航空科技英语等级考试B样题

COMAC航空科技英语等级考试B样题COMAC航空科技英语等级考试B1(技术类)样题Test Time:120 minutes部门_______________ 姓名____________ ⼯号____________Part I Listening (30%)Section 1 Conversation and Lecture(10%)Directions: In this section, you will hear a long conversation and a lecture. At the end of the conversation or the lecture, you will hear five questions. The conversation, the lecture and the questions will be spoken ONLY ONCE. After you hear a question, you must choose the best answer from the four choices.Conversation1. A) They get you directly to holiday destination.B) Their tickets can be bought on the internet.C) They offer excellent services to customers.D) They’re much cheaper than famous airlines.2. A) They have sprung up recently and become successful.B) They change prices on the basis of customers’ demand.C) They always offer travelers the extremely cheap flight.D) They do much advertising but few people ever watch it.3. A) By travelling before public holidays.B) By buying tickets a day in advance.C) By booking at the very last minute.D) By flying at peak time like Fridays.4. A) They try every possible means to reduce expenses.B) They charge different prices depending on demand.C) They don’t serve any food on any of their flights.D) They have increased the speed of their aero planes.5. A) They only offer cheap tickets online.B) They fail to offer satisfactory service.C) They spend little time on the ground.D) They fly to and from smaller airports.Lecture6. A) The 845m2 wing area is large enough to park 70 cars.B) The plane has the potential to carry 550 passengers.C) The tail is about as long as the Great Sphinx in Egypt.D) The two deck fuselage is as high as a 7-storey building.7. A) It is as economical to run as a common jet.B) It burns more fuel than other jumbo jets.C) It can fly an amazing 15,000 km non-stop.D) It can carry more fuel than other planes.8. A) Toulouse in France.B) England and Wales.C) All over the Europe.D) Spain and Germany.9. A) It is remarkably expensive.B) It is impressively efficient.C) It is a nation-wide project.D) It is extremely complicated.10.A) The expenses.B) The designing.C) The electronics.D) The cooperation.Section 2 Compound Dictation (10%)Directions: In this section, you will hear a passage TWICE. You have its script in the following, but with eleven blanks in it. You are required to fill in the first eight blanks with the exact words you have just heard. For last three blanks, you can either use the exact words you have just heard or write down the main points in your own words. Remember, there will be a pause for the last three blanks.Laurence Barron, President of Airbus China, defended the A380 superjumbo jet as its safety performance has been called into question.“The Qantas A380 suffered an (1) ______________ engine failure, a fairly rare event, which also damaged the aircraft itself. The aircraft performed as expected and (2) ______________ safely, so no, there is nothing wrong with the A380. It’s a (3)______________ aircraft.”Barron also says the engine issue will not (4) ______________ next summer’s scheduled delivery of the A380 to China Southern Airlines, the only (5) ______________ carrier to purchase the plane.Meanwhile, Barron explains that the lack of orders for its A350 aircraft, which is under development, from Chinese carriers is due to the country’s (6) ______________ planning structure.“The Chinese government, as you are well aware, works on a 5-year-plan basis, and they are about to (7) ______________ the 12th 5-year-plan which runs from 2011 to 2015. The A350 deliveries that we can offer are now in the what will become the 13th 5-year-plan period.”Eric Chen, Airbus China‘s Vice President, adds that the Chinese carriers’ timid (8) _____________ to the A350 is due to its competing product, Boeing’s 787.“Several years ago, Chinese airlines ordered more than 60 Boeing 787’s and for various reasons, airlines lack this kind of courage and determination to be a launching customer for a new program again. In other words, we are buying the bill for our rivals’ dilemma and consequences.”(9) __________________________________________________________________________________________________________________________________“I don’t really understand the world ‘challenge’. Our industry is challenging. There are lots of challenges but this is not a challenge, this is a competitor. (10) __________________________________________________________________________ Beverly Wyse, Vice President of Boeing’s 737 program, says Boeing is open to work with C919’s manufacturer.“I think (11) __________________________________________________________________________________________________________________________”Four Chinese airlines and two aircraft leasing companies have signed agreements to purchase 100 C919’s as launching costumers.Section 3 Listening and translating (10%)Directions: In this section you are going to hear five short passages. You will hear them ONLY ONCE. In each of these passages some of the sentences are already printed. You are required to translate the missing parts into Chinese. After each of the passages there will be a pause lasting one and a half minutes. The pause is intended for you to do the translation.1)The ARJ21-700 jetliner, China’s first self-designed aircraft, will undertake itsmaiden flight before the end of the year. COMAC chairman, Zhang Qingwei says this first homegrown regional jet has aroused great interest from aviation companies at home and abroad.”I just came back from the United States and Canada. ___________________________________________________________________________________________________________________________________________________________________________________________.”2)Nine top tier US manufacturing companies won competitive contracts to buildand supply the aviation system for China‘s new aircraft program, the C919.Airport infrastructure needs are filling opportunities in the US companies as well.___________________________________________________________________________________________________________________________________________________________________________________________________ 3)The Deputy Chief of the China’s Civic Aviation Administration, Xia Xinghua,says more cooperation is crucial for the Chinese side. “Firstly, we need to strengthen our cooperation on sustainable security development, expanding the relationship in a pragmatic way.________________________________________________________________________________________________________________________________________________________________________”4)The Transportation Secretary Ray LaHood said that the review would becomprehensive covering design, manufacturing and assembly of the Dreamliner.Michael Huerta of the Federal Aviation Administration said emphasis would be put on electrical systems and how these and the plane’s sophisticated mechanical systems interact._____________________________________________________________________________________________________________________________________________________________________________________.5)Though the tricycle arrangement may be most popular today, that was not alwaysthe case. The tail wheel undercarriage dominated aircraft design for the first four decades of flight and is still widely used on many small piston-engine planes.What makes this form of landing gear most attractive is its simplicity. Another potential advantage results from the fact that_____________________________ __________________________________________________________________________________________________________________________________ Part II Reading Comprehension(30%)Section 1 Skimming and Scanning (10%)Directions: In this section there are 10 incomplete statements. Based on the following passage, please complete the statements with the information given in the passage.Commercial aviation is an essential component of the global economy. The cost of aviation fuel is directly determined by the prevailing world price of oil, and it accounts for a major proportion of airplane operating costs. Several airline companies now add a fuel surcharge to the ticket cost of a commercial flight to compensate for the recent rapid rise in fuel costs. World oil prices are expected to remain high for several years. The prospect of sustained high aviation fuel prices could propel airline companies to seek alternative aviation fuels. Seeking alternative fuel could become paramount(最⾼的)for the airlineindustry should the peak-oil phenomenon actually occur.Breakthroughs and ResearchIt may become possible for super-cooled liquid hydrogen(氢)to eventually be used as an alternative fuel for some types of commercial airline service. Other alternative fuels may include high-density energy-storage technologies that result from breakthroughs in research in the areas of nanotechnology(纳⽶技术) and in high-temperature superconductivity(超导性). High-temperature superconductivity holds great promise for use in high-density energy-storage technology. Advances in nanotechnology could enable superconductive materials to eventually be manufactured at a cost that could justify their application in airliner propulsion. Electrical Storage and PropulsionEnergy stored in a superconductive storage technology could power electric motors that drive the identical propulsion fans that are found at the front-end of modern, “high-bypass” turbo-fan aircraft engines. Such fans provide up to 90% of the propulsive thrust of the turbo-fan engine. Each electrically powered propulsion fan may be driven by multiple (induction) lightweight electric motors during take-off. Some electric motors would “cut-out”under reduced power demand at cruising altitude so that the remaining motors will operate at higher efficiency (electric motors have poor part-load efficiency). Coanda fans may propel subsonic commercial aircraft that use high-density electrical storage technology. Such units were originally developed by physicist Henri Coanda and can operate at comparable efficiency and at comparable flight speeds as turbine-driven propulsion fans. Electrically powered aircraft that use either turbine propulsion fans or Coanda fans could be flown in thinner air at higher altitude (up to 65,000-feet) to reduce energy consumption (less drag on aircraft) on extended flights.The cooler air found at such altitudes could assist in keeping the superconductive energy storage systems functioning properly.Superconductive energy storage systems used in future commercial aircraft would likely be cooled by liquid nitrogen(氮). Both systems would need to be frequently recharged, which would likely be both energy-intensive as well as time consuming.It may be possible to design the energy storage systems along with their cooling systems to be removed and replaced during shorts layovers—such technology could help reduce the turn-around time of the aircraft. The introduction of superconductive energy storage systems in commercial aircraft in the long-term future would require that future airport terminals be equipped with power generation technology at or near the premises.Power GenerationThe number of electrically powered and hydrogen powered road and railway vehicles would likely increase during a post peak-oil period. Commuter aircraft that operate short-haul service could be powered by ethanol(⼄醇) or by hydrogen while future supersonic aircraft could use liquid hydrogen as fuel. The commercial aviation industry of the future (post peak oil) could likely require vast amounts of electric power to recharge superconductive energy storage systems, recharge liquid nitrogen cooling systems as well as to generate, compress and supercool large amounts of hydrogen.Modern commercial aircraft are energy intensive during take-off. Airports that serve metropolitan areas presently process continual processions of large long-distance aircraft during peak periods. Such aircraft could require between 300-Mw-hr and 1000-Mw-hr of power to undertake trans-oceanic flights at subsonic speed. The power requirements of a future electrically based commercial aviation industry could likely overwhelm the power generation industry of most developed nations. Major international airports may eventually need to generate electric power on-site to meet the energy needs of future fleets of electrically powered and hydrogen-fueled commercial aircraft. Airport power stations may be nuclear; use hydrogen fusion or be based on some other unconventional power generation technology that is still subject to research.Energy StorageThe ability to store large amounts of energy at or near major airports could gain importance during a post peak-oil period. Electric power could be purchased from the grid during their off-peak periods and put into short-term storage. Airport power stations that encounter off-peak periods could replenish(装满) airport energy storage systems that may include superconductive storage, flow batteries, hydraulic storage in hydroelectric dams in nearby mountains (coastal airports) or off-site pneumatic storage (subterranean salt domes that were emptied). Air that is exhausted from pneumatic storage systems may be sufficiently cold to assist in “replenishing” liquid nitrogen super-cooling systems.Power Regulation (Airports)Power stations that provide energy for air transportation use may have to be excluded from the regulatory framework. Most of the electrically powered airliners that will be recharged would be “foreign”owned, that is, the owners would be domiciled in adifferent jurisdiction(司法权) to where the aircraft would be recharged. The idea of regulators in one jurisdiction looking after the interests of parties who live, do business and pay taxes in another jurisdiction is quite ludicrous. Power stations that supply a future airline industry with electric power would need to be regulatory-free despite the “foreign”airline owners being “captive”customers. It would be possible for power to be supplied to a single airport by several small providers who compete against each other. Power providers and airline companies could negotiate deals, perhaps even on a daily basis. ConclusionFuture scientific breakthroughs are likely to occur in both nanotechnology and in superconductivity. High-density energy storage technologies could be the likely result and appear in the distant future. Electrically powered commercial aircraft that fly at subsonic speeds could appear in the future irrespective of whether or not peak-oil actually occurs. Alternative liquid fuels that are cost-competitive to fossil oil are also likely to appear and find applications in aviation. Large ground-effect aircraft (地效飞⾏器)that fly above water and that carry either passengers or freight between coastal cities are also likely appear in the future.1.The prospect of sustained high aviation fuel prices could propel airline companiesto seek _______________________.2.Breakthroughs in nanotechnology could enable _______________________ to beavailable in their application in airliner propulsion.3.Coanda fans were first developed by _______________________ .4._______________________ could be used to cool superconductive energystorage system used in future commercial aircraft.5._______________________, which operates short-haul service, could be poweredby ethanol(⼄醇) or by hydrogen.6.Future airport power stations may be_______________________; use hydrogenfusion or be based on some other unconventional power generation technology. 7.During a post peak-oil period, the ability to_______________________ at or nearmajor airports could gain importance.8.Power stations that provide energy for air transportation use are likely to be_______________________ from the regulatory framework.9.Electrically powered commercial aircraft that fly at _______________________speeds could appear in the future.10.Aircrafts flying above water and carrying either passengers or freight betweencoastal cities are called _______________________.Section 2 Reading Comprehension (10%)Directions: The following passage is followed by some questions. For each of them there are four choices marked A), B), C) and D). You should choose the best answer from the four choices.Living standards have soared during the twentieth century, and economists expect them to continue rising in the decades ahead. Does that mean that we humans can look forward to increasing Happiness?Not necessarily, warns Richard A. Easterlin, an economist at the University of Southern California, in his new book, Growth Triumphant: The Twenty-first Century in Historical Perspective. Easterlin concedes that richer people are more likely to report themselves as being happy than poorer people are. But steady improvements in the American economy have not been accompanied by steady increases in people’s self-assessments of their own Happiness.The explanation for this paradox(悖论) may be that people become less satisfied over time with a given level of income. In Easterlin’s word: “As incomes rise, the aspiration level does too, and the effect of this increase in aspirations is to vitiate (破坏) the expected growth in Happiness due to higher income.”Money can buy Happiness, Easterlin seems to be saying, but only if one’s amounts get bigger and other people aren’t getting more. His analysis helps to explain sociologist Lee Rainwater’s finding that Americans’perception of the income “necessary to get along” rose between 1950 and 1986 in the same proportion as actual per capita income. We feel rich if we have more than our neighbors, poor if we have less, and feeling relatively well-off is equated with being happy.Easterlin’s findings, challenge psychologist Abraham Maslow’s “hierarchy(等级) of wants” as a reliable guide to future human motivation. Maslow suggested that as people’s basic material wants are satisfied they seek to achieve nonmaterial or spiritual goals. But Easterlin’s evidence points to the persistence of materialism.“Despite a general level of affluence never before realized in the history of the world.” Easterlin observes, “Material concerns in the wealthiest nations today are as pressing as ever and the pursuit of material need as intense.” The evidence suggests there is no evolution toward higher order goals. Rather, each step upward on the ladder of economic development merely stimulates new economic desires that lead the chase ever onward.Needs are limited, but not greeds. Science has developed no cure for envy, so our wealth boosts our Happiness only briefly while shrinking that of our neighbors. Thus the outlook for the future is gloomy in Easterlin’s view. “The triumph of economic growth is not a triumph of humanity over material wants; rather, it is the triumph of material wants over humanity.”1.What does Easterlin warn in his new book?A)Humans can look forward to increasing happiness with soaring livingstandards.B)Humans might not be able to enjoy increasing happiness with soaring livingstandards.C)Richer people tend to report themselves as being happy more than poorerpeople do.D)Richer people tend to report themselves as being happy less than poorerpeople do.2.Which of the following statements may account for the paradox(悖论) mentionedin paragraph 3?A)People become less satisfied though the income rises over time.B)A general level of affluence never before realized in the history of the world.C)Though the American economy improved steadily, there isn’t a steadyincrease in people’s self-assessments of their own happiness.D)As incomes rise, there will be an increase in the aspiration level, which willhamper the expected growth in Happiness due to higher income.3.Whose finding is against the theory of “Hierarchy of wants”?A)Easterlin’s B) Maslow’s C) Rainwater’s D) Lee’s4.According to Easterlin, the outlook of the future of happiness is ________.A)bright B) sad C) unclear D) thrilling5.From the quotation in the end of the passage (paragraph 7), we can infer that___________?A)The triumph of economic growth results in more humanity.B)The triumph of economic growth results in more material wants.C)Humanity contributes more to the triumph of economic growth.D)Material wants contributes more to the triumph of economic growth.Section 3 Short Answer Questions (10%)Directions: Read the following passage and then answer the questions. The answer should not be more than 25 words.The maximum allowable weight for an aircraft is determined by design considerations. However, the maximum operational weight may be less than the maximum allowable weight due to such considerations as high-density altitude or high-drag field conditions caused by wet grass or water on the runway. The maximum operational weight may also be limited by the departure or arrival airport’s runway length.One important preflight consideration is the distribution of the load in the aircraft. Loading the aircraft so the gross weight is less than the maximum allowable is not enough. This weight must be distributed to keep the center of gravity (CG) within the limits specified in the POH or AFM.If the CG is too far forward, a heavy passenger can be moved to one of the rear seats or baggage can be shifted to a rear compartment. If the CG is too far aft, passenger weight or baggage can be shifted forward. The fuel load should be balanced laterally: the pilot should pay special attention to the POH or AFM regarding the operation of the fuel system, in order to keep the aircraft balanced in flight. Weight and balance of a helicopter is far more critical than for an airplane. With some helicopters, they may be properly loaded for takeoff, but near the end of a long flight when the fuel tanks are almost empty, the CG may have shifted enough for the helicopter to be out of balance laterally or longitudinally. Before making any long flight, the CG with the fuel available for landing must be checked to ensure it will be within the allowable range.Changes of fixed equipment may have a major effect upon the weight of the aircraft. The replacement of older, heavy electronic equipment with newer, lightertypes results in a weight reduction, which will probably cause the CG to shift and must be computed and annotated in the weight and balance record.Repairs and alteration are the major sources of weight changes. The A&P mechanic must compute the CG and record the new empty weight and EWCG in the aircraft weight and balance record.The A&P mechanic or repairman conducting an annual or condition inspection must ensure the weight and balance data in the aircraft records is current and accurate. It is the responsibility of the pilot in command to use the most current weight and balance data when operating the aircraft.Questions:1.What conditions might cause the operational weight of a plane to be less than themaximum allowable weight?2.What should be done if the CG is too far aft in an aircraft?3.Why is the weight and balance for a helicopter far more critical than for anairplane?4.According to the passage, what might lead to weight changes and cause the CG toshift in an aircraft?5.Who are responsible for recording and using the most current and accurate data ofthe weight and balance?Part III Translation (15%)Section 1 English-Chinese Translation (10%)Direction: In this section there are two passages in English. Please read these passages and translate the underlined parts into Chinese.Passage 1The airplane propeller consists of two or more blades and a central hub to which the blades are attached. 1) Each blade of an airplane propeller is essentially a rotating wing. As a result of their construction, the propeller blades are like airfoils and produce forces that create the thrust to pull, or push, the airplane through the air.The power needed to rotate the propeller blades is furnished by the engine. The engine rotates the airfoils of the blades through the air at high speeds, and the propeller transforms the rotary power of the engine into forward thrust.2) An airplane moving through the air creates a drag force opposing its forward motion. Consequently, if an airplane is to fly, there must be a force applied to it that is equal to the drag, but acting forward. This force is called “thrust.”Passage 2Aircraft flight control systems are classified as primary and secondary. 3) The primary control systems consist of those that are required to safely control an airplane during flight. Secondary control systems improve the performance characteristics of the airplane, or relieve the pilot of excessive control forces. Those included in the primary control systems are the ailerons, elevator (or stabilator), and rudder. Examples of secondary control systems are wing flaps and trim systems.Airplane control systems are carefully designed to provide a natural feel, and at the same time, allow adequate responsiveness to control inputs. 4) At low airspeeds, the controls usually feel soft and sluggish, and the airplane responds slowly to controlapplications. At high speeds, the controls feel firm and the response is more rapid.Movement of any of the three primary flight control surfaces changes the airflow and pressure distribution over and around the airfoil. These changes affect the lift and drag produced by the airfoil/control surface combination, and allow a pilot to control the airplane about its three axes of rotation.Design features limit the amount of deflection of flight control surfaces. For example, control-stop mechanisms may be incorporated into the flight controls, or movement of the control column and/or rudder pedals may be limited. The purpose of these design limits is to prevent the pilot from inadvertently overcontrolling and overstressing the aircraft during normal maneuvers.5) A properly designed airplane should be stable and easily controlled during maneuvering. Control surface inputs cause movement about the three axes of rotation. The types of stability an airplane exhibits also relate to the three axes of rotation. Section 2 Chinese-English Translation (5%)Direction: In this section there are five sentences in Chinese. Please translate them into English.1.太阳能动⼒飞机的平均飞⾏时速为70公⾥,暂时不会对商⽤飞机构成威胁。

广州2024年11版小学三年级第5次英语上册试卷(含答案)

广州2024年11版小学三年级第5次英语上册试卷(含答案)

广州2024年11版小学三年级英语上册试卷(含答案)考试时间:90分钟(总分:110)B卷一、综合题(共计100题共100分)1. 选择题:Which fruit is red and often associated with teachers?A. BananaB. AppleC. OrangeD. Grape2. 听力题:An island formed from volcanic activity is called a ______ island.3. 听力题:The chemical formula for ethyl alcohol is __________.4. 选择题:What is the main ingredient in butter?A. MilkB. EggC. SugarD. Flour5. 填空题:In spring, the ________ (花朵) bloom and the ________ (树叶) grow green.6. 选择题:What is the name of the famous giant panda in the zoo?A. Bao BaoB. Ling LingC. Tien TienD. Shin Shin7. 填空题:__________ (化学反应) can release or absorb energy.8. 填空题:The __________ (历史的价值意义) guide actions.9. 填空题:I have a toy _____ that can transform.10. 听力题:Countries are divided into smaller regions called ________.11. 听力题:A mammal is an animal that has hair or ______.12. 填空题:The _____ (植物图谱) can serve as an educational tool.13. 选择题:What do we call the person who fixes cars?A. TeacherB. MechanicC. ChefD. Artist14. 听力题:The chemical formula for nitric acid is ______.15. 填空题:A _______ (小蝴蝶) lands gently on a flower.16. 选择题:What type of animal is a dolphin?A. FishB. AmphibianC. MammalD. Reptile答案:C17. 填空题:My teacher gives us _______ (名词) for homework. 我觉得 _______ (形容词).18. 选择题:What do you call a person who writes stories?A. AuthorB. EditorC. JournalistD. Poet答案:A19. 填空题:In ancient Mesopotamia, people created one of the first systems of __________. (书写)20. 填空题:The __________ is known for its stunning natural landscapes. (新西兰)21. 填空题:Many plants change color in ______ (秋天).22. 填空题:The ________ (民族) in our region celebrate festivals.23. 填空题:Certain plants can ______ (增强) the local economy.24. 听力题:A flashlight uses a _______ to produce light.25. 填空题:Planting native species can help support local ______ (生态).26. 填空题:The dolphin loves to play in the _______ (海浪).27. 选择题:How many players are on a volleyball team?A. 6B. 7C. 8D. 928. 听力题:The park is _____ (near/far) from here.29. 填空题:My ________ (玩具名称) makes me feel special.30. 选择题:What do you call a person who participates in a sport?A. AthleteB. PlayerC. CompetitorD. All of the above答案:D31. 听力题:A chemical reaction can release ______.32. 填空题:The kitten loves to chase a ______.33. 填空题:We will have a ________ (旅行) next week.34. 选择题:What is the main source of energy for the Earth?A. The MoonB. The SunC. The StarsD. The Wind35. 选择题:What do you call a room used for cooking?A. KitchenB. Dining roomC. Living roomD. Bedroom答案:A36. 选择题:Which animal is known for its ability to fly?A. DogB. FishC. BirdD. Cat答案:C37. 听力题:Chemical reactions can be influenced by temperature and _____.38. 听力题:The chemical symbol for platinum is _______.39. 填空题:The ________ was a key event in the history of civil rights in the United States.The process of separating components based on their density is called ______.41. 填空题:A ____(collaborative project team) works towards a common goal.42. 听力题:We are going to ______ (host) a party soon.43. 填空题:My dad is a __________ (律师).44. 选择题:What is the largest organ inside the human body?A. LiverB. HeartC. BrainD. Lung答案: A45. 听力题:The chemical symbol for carbon is __________.46. 填空题:My cousin is very . (我的表兄弟/表姐妹很。

小学上册第十四次英语第4单元期中试卷

小学上册第十四次英语第4单元期中试卷

小学上册英语第4单元期中试卷英语试题一、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 What is the main ingredient in a bagel?A. BreadB. DoughC. WaterD. Flour答案:D2 I can ______ (克服) challenges with determination.3 An alkali metal is found in group _____ of the periodic table.4 I want to be a ________ (演员) when I grow up.5 My mom loves to _______ (动词) new recipes. 她的厨艺非常 _______ (形容词).6 What is the name of the famous building in India?A. Taj MahalB. ColosseumC. Eiffel TowerD. Great Wall7 I enjoy drawing ______ (漫画) characters and creating my own ______ (故事).8 Energy is the ability to do ______.9 She is a great ________.10 What is the capital of Mexico?A. CancunB. GuadalajaraC. Mexico CityD. Monterrey答案: C. Mexico City11 Which planet is known as the Red Planet?A. JupiterB. SaturnC. MarsD. Neptune答案:C12 The sloth's diet consists mainly of ________________ (叶子).13 The _____ (海豹) basks on the rocks in the sun.14 What do we call the study of weather?A. GeographyB. MeteorologyC. ClimatologyD. Ecology答案:B15 I see a _____ (朋友) at the coffee shop.16 I planted _____ (草) last week.17 The ______ is very supportive.18 I like to go ___. (hiking)19 The ________ (生态网络) connects habitats.20 My favorite dish is ______ (沙拉).21 What is the name of the fairy tale character who had a sleeping curse?A. Snow WhiteB. Sleeping BeautyC. CinderellaD. Rapunzel答案:B22 The chemical formula for chromium trioxide is _______.23 The flowers are _______ (很美丽)。

我最喜欢的中国动画片英语作文120字

我最喜欢的中国动画片英语作文120字

我最喜欢的中国动画片英语作文120字全文共6篇示例,供读者参考篇1My Favorite Chinese CartoonDo you know what my favorite cartoon is? It's a really cool and exciting Chinese cartoon called "Legend of Deification"! I just love everything about it – the characters, the storyline, the animation, and the awesome action scenes. Let me tell you all about it!The story takes place in a world filled with cultivators who can use powerful spiritual energy and martial arts skills. The main character is Yang Ling, a young boy who gets caught up in an epic adventure after meeting a mysterious old master. Yang Ling has to go on a journey to unlock his true potential and become a legendary deity.Along the way, he meets so many amazing characters that become his friends and allies. There's Ling'er, a beautiful young cultivator who is super strong and fights with her spirit sword. Then there's Zhen Chan, a hilarious talking monkey who uses trickery and martial arts. My favorite character is probably HuaChu, the kind-hearted princess who can control plants and nature energies.The animation in Legend of Deification is just breathtaking! The fight scenes are incredibly well choreographed, with the characters doing all sorts of flips, kicks, and using magical abilities. The backgrounds are lush and vibrant, filled with sprawling temples, misty mountains, and mystical realms. The music score is epic too, adding to the sense of adventure.What I love most though is the deeper meaning behind the story. It teaches values like courage, perseverance, friendship, and fighting for what's right against powerful evil forces. Yang Ling and his friends have to overcome so many challenges and villains bent on destroying the world. With willpower and working together, they're able to grow stronger and prevail.I get so excited whenever a new episode comes out! I've watched the whole series like five times already. My friends and I evenact out scenes from it during recess, pretending to use magic and martial arts. We have so much fun!I really hope the creators make more seasons because the story is far from over. Yang Ling still has much more to learn and achieve on his journey to become a true deity. I can't wait to see what other awesome realms, characters, and adventures he'llencounter next. The Legend of Deification series is a masterpiece of Chinese animation!I'm definitely going to keep watching and re-watching it over and over. For any kid who loves action, magic, martial arts, and an epic mystical fantasy tale, you have to check out Legend of Deification! It's my hands-down favorite cartoon ever. What's yours?篇2My Favorite Chinese CartoonHi there! My name is Lily and I'm 10 years old. Today I want to tell you all about my absolute favorite cartoon ever - Ne Zha! It's a Chinese animated movie that came out a few years ago, but I've probably watched it like a hundred times since then. I love everything about it!The story is so cool and exciting. It's based on an ancient Chinese myth about this young boy named Ne Zha who is born with amazing superpowers. His parents were so scared of his powers that they tried to kill him when he was just a baby! But Ne Zha was too powerful and couldn't be killed. As he grew up, he learned to use his fire powers, wind powers, and cool magicweapons to fight evil monsters and demons that were terrorizing the human world.My favorite part is when Ne Zha takes on the Huge Skull Monster, which is this massive skeleton beast thing. The fight scenes are so epic! Ne Zha uses his fire skills and his magic wind fire wheels to battle the monster. There's flames and destruction everywhere as they smash through buildings and mountains. I was on the edge of my seat the whole time! In the end, after an insane struggle, Ne Zha is able to burn the Huge Skull Monster down to ashes with his fire powers. Phew!But Ne Zha isn't just about the awesome action scenes. The movie also has a really cool message about being true to yourself and not letting others tell you what you should be. See, even though Ne Zha has all these incredible abilities, his own parents were terrified of him and thought he was a monster. A lot of people judged Ne Zha for being different. But he didn't listen to their hate and just stayed focused on using his powers to protect the good people. I think that's such an inspiring message for kids like me - to have confidence in who you are and not care what the haters say!I also really love the vibrant, colorful animation style in this movie. All the magic powers and fight scenes just pop off thescreen with dazzling reds, oranges, and blues. The backgrounds showing ancient Chinese cities and landscapes are rendered in incredible detail too. The entire movie is just a feast for the eyes! Whenever I watch it, I feel like I'm being transported to this awesome fantasy world.The music in Ne Zha is a total banger as well. It's got these soaring, epic orchestral scores during the action sequences that get your heart pumping. But it also has quieter, more emotional themes for the tender family moments between Ne Zha and his parents. I've probably listened to the soundtrack like a zillion times!What really makes Ne Zha special though is that crazy cliffhanger ending. I won't spoil it for those who haven't seen it yet, but let's just say it's pretty insane! My mind was blown the first time I watched it. That ending has set things up perfectly for Ne Zha 2 to be released soon (I'm counting down the days!). I can't wait to see what other adventures Ne Zha will go on using his epic powers. Rumor has it the action sequences in the sequel will be even crazier than the first movie. Eeeek I'm so excited!Do yourselves a favor and go watch Ne Zha right now if you haven't already seen it. Whether you're a kid like me or a grown-up, I'm telling you you're going to be blown away. Withits dazzling visuals, thrilling action, relateable characters, and inspirational themes, Ne Zha is a true masterpiece of Chinese animation. It deserves to be appreciated all over the world! For me, it will always be my #1 all-time favorite cartoon. I could rewatch this movie every day and never get bored. Ne Zha, you really are the ultimate heroic bada**!篇3My Favorite Chinese CartoonDo you like cartoons? I love cartoons! My favorite cartoon ever is a Chinese cartoon called Pleasant Goat and Big Big Wolf. It's about these cool goats that live on a farm, and there's a big bad wolf who is always trying to eat them. But the goats are too smart, and they trick the wolf every time! It's so funny to watch.The main characters are the Pleasant Goats - there's Westuardo, Sparky, Linny, Merri, and Sluggy. Westuardo is the leader and he's really brave. Sparky is smart and likes to invent things. Linny is kind of a bully but he's still one of the good guys. Merri is a girl goat who likes to dance and sing. And Sluggy just sleeps all the time! Then there's the wolf, whose name is Wolf-y. He's big and dumb and always falls for the goats' tricks.In every episode, Wolf-y tries to catch the goats so he can eat them for dinner. But the goats are too clever for him! They use their smarts and teamwork to outsmart Wolf-y. Sometimes they dress up in disguises. Sometimes they build elaborate traps. Once they even made a giant robot to fight Wolf-y! No matter what plan Wolf-y comes up with, the goats can always foil it. The goats win every time, and Wolf-y ends up failing miserably. It's hilarious to watch him get all mad and stomp around when his plans don't work.My favorite character is definitely Sparky. He's just so smart and creative! He's always inventing cool new gadgets to help the goats escape Wolf-y. Like in one episode, he made these cool jet packs so the goats could fly away. In another episode, he invented boxing glove arrows to punch Wolf-y in the face! Sparky is always using his brain to get them out of trouble. I want to be an inventor just like him when I grow up.The animation in Pleasant Goat is so bright and vibrant too. All the colors just pop right off the screen! The character designs are really fun and exaggerated. Like Wolf-y has this huge snout and big teeth that make him look extra goofy and not-too-scary for kids. And the goats' eyes are these big bubbly circles that are just so cute and expressive.But my favorite part of the show is the humor. The jokes are just so clever and silly at the same time. There's lots of visual humor with sight gags and funny background animations happening. But there are also great wordplay jokes and hysterical dialogue too. Like when Wolf-y gets so mad at the goats, he'll start ranting and raving with his silly wolf accent. It never fails to make me laugh out loud!I've watched Pleasant Goat and Big Big Wolf so many times, but it never gets old. Every episode is a new hilarious adventure with more gags, more gadgets from Sparky, and more foiled plans from that dumb old Wolf-y. Watching it just makes me feel so happy and giggly.If you've never seen Pleasant Goat before, you are seriously missing out! It's the funniest, most clever, most creatively animated cartoon ever made. Once you start watching it, you'll be hooked too. The jokes will have you rolling on the floor laughing, I promise! You'll fall in love with those cute goats, and you'll root for them every time as they outwit that big bad wolf again. Pleasant Goat and Big Big Wolf is an absolute must-watch for cartoon fans of any age! It's my favorite cartoon ever, and I hope you'll love it just as much as I do.篇4My Favorite Chinese CartoonHi there! My name is Xiaoming, and I'm 9 years old. Today, I want to tell you all about my favorite Chinese cartoon – it's called "Pleasant Goat and Big Big Wolf"! I just love this show so much. Whenever I'm feeling down or bored, watching an episode always cheers me up and makes me laugh.The main characters are a group of adorable goats who live on Green Pasture Mountain. There's Westin, Sparky, Merrybell, Lazy, and more. They're all really good friends and love going on adventures together. But they have to watch out for the Big Big Wolf and his pesky little wolf sons who are always trying to catch and eat them!The Big Big Wolf is so funny. He's big and loud and loves to make up silly plans to trap the goats. But no matter what he tries, the goats are always too smart for him and find a way to escape.I laugh so hard whenever the wolf gets outsmarted and ends up falling into his own traps! His two little sons are super naughty too and are always getting into trouble.Even though the wolves cause a lot of chaos, they don't really mean any harm. Deep down, they have good hearts. Sometimes the goats even help the wolves out of sticky situations. Like in one episode, the Big Big Wolf gets his headstuck in a honeypot, and the goats have to figure out a way to free him. It's moments like these that show how the goats and wolves actually care about each other, despite all the silliness.My favorite character is probably Sparky. He's the bravest of all the goats and is always coming up with clever ideas to escape the wolves' plots. Sparky is also really athletic and good at sports like soccer. I wish I could be as fast and fearless as him! Lazy is hilarious too – he's always napping and complaining about having to do work. I can definitely relate to being lazy sometimes haha!The show has the perfect mix of humor, adventure, and heartwarming lessons about friendship, perseverance, and being a good person. The songs are super catchy too – I'm always walking around singing "Wonderful Goat Wonderland" to myself. Whenever I'm feeling sad or angry, this show reminds me to stay positive and look on the bright side of life.What I really love about Pleasant Goat though, is how creative and imaginative it is. The writers are always coming up with fun new worlds and adventures for the characters to explore. One week they might be stranded on a deserted island, and the next they're in outer space fighting evil aliens! No matter whatcrazy situation they're in, the goats always stick together as a team.The show reminds me of the importance of using my imagination and playing outside with my friends. Wheneverwe're bored, my buddies and I like to pretend we're the goats going on epic journeys and outwitting the Big Big Wolf. Sometimes we even build goat and wolf ears out of paper to really get into character!I also like how the cartoon teaches good morals without being too preachy. Like there's an episode where the wolves finally catch the goats, but then realize that eating them would go against their values of kindness. So instead of eating them, they let the goats go. It shows that even your greatest enemies deserve to be treated with compassion.The animation is so vibrant and colorful too. All the characters are drawn in a totally unique style that's unlike any other cartoon I've seen. The landscapes are breathtaking, from the rolling green hills of the pasture to the vast deserts and forests the goats explore. WheneverI watch, I feel like I'm being transported to a magical wonderland.That's why Pleasant Goat and Big Big Wolf will always be my favorite Chinese cartoon. It has everything – action, laughter,valuable life lessons, and characters you can't help but fall in love with. I've watched every single episode like a million times, but it never gets old! I'm so glad my parents introduced me to this classic show from their childhood. It's truly a masterpiece that spans generations.Honestly, the only downside is that there isn't enough merchandise! I have all the DVDs, books, and toys...but I want more. A Pleasant Goat theme park would be my absolute dream come true. I'd get to walk through the rolling green hills and Maybe even meet the Big Big Wolf himself! Imagine getting a big hug from that big silly guy. A Goat fan can dream, right?Well, thanks for reading about my favorite show! I could ramble about Pleasant Goat all day, but I better stop here. If you've never seen it before, I super duper recommend checking it out. Just be warned – once you start watching, you won't be able to stop! It's that Goat-tastic!篇5My Favorite Chinese CartoonHi there! My name is Li Ming and I'm 10 years old. I love watching cartoons and my all-time favorite is an amazingChinese cartoon called Pleasant Goat and Big Big Wolf. It's so funny and exciting! Let me tell you all about it.Pleasant Goat and Big Big Wolf is about a group of goats who live on Green Grass Mountain. The main characters are Weslie the goat, Sparky the elephant, and Gungun the wolf cub. But the star of the show is definitely Wolnie - he's a big grey wolf who is constantly trying to catch and eat the goats, but he never succeeds because they always outsmart him using their clever tricks.Wolnie is so silly and clumsy. In one episode, he dresses up like a mail carrier to try and sneak into the goats' village. But when Sparky asks him for the password, Wolnie can't remember it so he just shouts "Open Sesame!" Then the next thing you know, the ground swallows him up and he gets stuck in a pit! It's so hilarious when things like that happen to Wolnie. My friends and I laugh so hard our stomachs hurt.My favorite character is definitely Sparky the pink elephant. He's super strong and brave, but also really kind and loyal to his goat friends. Whenever Wolnie is causing trouble, Sparky rushes in to chase him away or rescue the goats from his traps. Like in one episode, Wolnie ties up all the goats and is about to eatthem when suddenly Sparky appears and knocks Wolnie away with his trunk! Sparky is the coolest.The animation in Pleasant Goat is so vibrant and colorful. All the characters are drawn in a really fun, exaggerated cartoon style. And the scenery is gorgeous, with lush green fields, sparkling streams, and majestic mountains. Sometimes the characters break into song and there are big dancing numbers too! The songs are so catchy and I find myself humming them all the time at school.But what I really love most about Pleasant Goat are all the jokes and slapstick humor. The writers are just brilliant at coming up with hilarious visual gags and witty dialogue that makes me crack up every time. Like when Wolnie gets hit on the head and sees cartoon birdies tweeting around his noggin. Or when one of the characters makes a really silly pun and groans "Oh no, another bomb!" Because their jokes are so clever and unexpected, you never know what's going to happen next.I've been watching Pleasant Goat since I was a tiny kid and I'll never get tired of it. Whenever I'm feeling sad or grumpy, I just pop in a Pleasant Goat DVD and within five minutes I'm smiling and laughing again. It's that good at cheering me up! Thejokes never get old, no matter how many times I've seen an episode before.All my friends are obsessed with Pleasant Goat too. At school, we're always talking about our favorite episodes and repeating the funniest lines from the show. Sometimes we even act out scenes for each other at recess! I'll be Wolnie trying (and failing) to catch Weslie, while my friend plays the part of Sparky swooping in to save the day. We have so much fun recreating the hijinks from Pleasant Goat.My parents love it when I watch Pleasant Goat because they know it's not just mindless silliness - it actually has really positive messages about friendship, perseverance, and using your brains instead of brawn. The goats embody virtues like loyalty, teamwork, and quick thinking. And even bumbling Wolnie learns lessons about the importance of hard work and never giving up on your goals. So while the show is super funny and entertaining, it's also篇6My Favorite Chinese CartoonI absolutely love watching Chinese cartoons! There are so many amazing ones to choose from, but my absolute favorite is"Qian Nuren You Jian." It's an incredible martial arts show about a young girl named Qian Nu who learns kung fu from her teacher Zhang Sanfeng.The animation in Qian Nuren You Jian is just breathtaking. The fight scenes are choreographed so beautifully, with Qian Nu doing all sorts of flips, kicks, and punches against different opponents. Her kung fu moves look so cool and powerful! The colors are also really vibrant and pretty, especially Qian Nu's bright red outfit. Whenever there's a big fight, the animators make the backgrounds look all blurry so you can focus just on the awesome martial arts action.But it's not just the visuals that make me love this show. The story is fantastic too! Qian Nu is such an inspiring main character. Even though she's just a young girl, she never gives up on her dream of becoming a kung fu master. She works incredibly hard, training every single day with her wise teacher Zhang Sanfeng. I love how determined and persistent she is.Qian Nu also has to overcome lots of challenges beyond just learning kung fu. Her parents were killed when she was very young, so she has to deal with that tragic loss. But she never lets it weigh her down or make her bitter. She stays positive andkindhearted throughout everything. Her strength and optimism despite her difficult circumstances make me really admire her.The other characters are awesome too! Zhang Sanfeng is the perfect wise old mentor. He's stern but caring, always pushing Qian Nu to improve while also looking out for her wellbeing. Then there's the villains like Xuan Zhen and Wu Dang who use their kung fu skills for evil purposes. They make for really exciting adversaries that Qian Nu has to face. My favorite dynamic is probably between Qian Nu and her fellow student Wei Zhongxian though. He starts off as her rival, jealous of her skills, but they eventually become good friends who support each other.I've watched every single episode multiple times, and I'm still not sick of Qian Nuren You Jian! The beautiful visuals, engaging story, and wonderful characters make it, in my opinion, the greatest Chinese cartoon ever created. Whenever I'm feeling down, I just throw on a few episodes and instantly my mood brightens. This show has inspired me to never give up on my dreams, just like the determined Qian Nu. It's a true masterpiece that I'll cherish forever.I really want to learn kung fu someday, just like Qian Nu! Of course, I'd never use it for violence like the villains in the show.Only for self-defense and exercise. Mostly I'm just in awe of all the incredible martial arts skills on display. The way the fighters can leap through the air, perform all those crazy kicks and flips, and take down multiple foes at once is just mesmerizing to watch. I try my best to mimic some of the simpler moves at home, but I'm sure I look pretty silly compared to Qian Nu's gracefulness.My parents just think it's a "weird phase" I'm going through, but my love for Qian Nuren You Jian is definitely not some temporary thing. This show has honestly helped shape who I am as a person. Qian Nu's perseverance and compassion are qualities I really admire and hope to embody. Whenever I'm struggling with something difficult, I think "What would Qian Nu do in this situation?" And the answer is: never quit and always be kind. Such simple advice, but it's stuck with me.I really want to meet the creators of the show someday so I can thank them for making such an amazing piece of art that has had such a profound impact on me. The visionaries who conceived of Qian Nu's world, crafted her journey, designed the captivating animation and choreographed those mind-blowing fight sequences - they deserve all the praise in the world as far as I'm concerned. Cartoons are so often disregarded as simple"kid's shows," but masterworks like Qian Nuren You Jian demonstrate just how powerful and meaningful the animation art form can be.。

八年级英语形容词和副词比较级单选题80题

八年级英语形容词和副词比较级单选题80题

八年级英语形容词和副词比较级单选题80题1. Tom is taller than Jerry, but Jerry is ____ than Tom.A. strongB. strongerC. strongestD. the strongest答案:B。

本题考查形容词比较级的用法。

A 选项“strong”是原级;C 选项“strongest”是最高级;D 选项“the strongest”是最高级,且最高级前需要有定冠词“the”。

根据句意,此处是两者比较,要用比较级,“strong”的比较级是“stronger”,所以选B。

2. The dress is more beautiful than that one, but that one is ____ than this one.A. cheapB. cheaperC. cheapestD. the cheapest答案:B。

本题考查形容词比较级。

A 选项“cheap”是原级;C 选项“cheapest”是最高级;D 选项“the cheapest”是最高级且需定冠词。

这里是两者比较,“cheap”的比较级是“cheaper”,故选B。

3. Lucy's hair is longer than Lily's, but Lily's hair is ____ than Lucy's.A. blackB. blackerC. blackestD. the blackest答案:B。

此题考查形容词比较级。

A 选项“black”是原级;C 选项“blackest”是最高级;D 选项“the blackest”是最高级且需定冠词。

因为是两者比较头发的颜色,“black”的比较级是“blacker”,所以选B。

4. This bag is heavier than that one, but that one is ____ than this one.A. bigB. biggerC. biggestD. the biggest答案:B。

何为好的艺术作品英语作文

何为好的艺术作品英语作文Art has always been a significant part of human culture, reflecting our emotions, ideas, and the world around us. But what constitutes a good piece of art? This question has intrigued both artists and art enthusiasts for centuries. While the answer may vary from person to person, there are several elements that are commonly agreed upon ascontributing to the quality of an artwork.1. Emotional Impact: A good artwork has the power to evoke emotions. Whether it's joy, sorrow, surprise, or even discomfort, the ability to make the viewer feel something is a mark of a successful piece.2. Technical Skill: The artist's mastery of their medium is often a point of admiration. This could be the precision of a draftsman's line, the vibrant use of color in a painting, or the intricate detail in a sculpture.3. Originality: Originality is key in art. A good artwork brings something new to the table, offering a fresh perspective or a novel approach to an old subject.4. Cultural Relevance: Art that speaks to the zeitgeist, reflecting the social, political, or cultural climate, often resonates more deeply with the audience.5. Aesthetic Appeal: While beauty is subjective, there is aninherent aesthetic quality that many good artworks possess. This could be in the form of balance, proportion, or harmony within the piece.6. Conceptual Depth: A good artwork often has layers of meaning. It invites the viewer to look beyond the surface and engage with deeper themes or narratives.7. Craftsmanship: The care and attention to detail that goes into creating an artwork can greatly enhance its quality. This is evident in the quality of the materials used and the execution of the artist's vision.8. Timelessness: Great art often stands the test of time. It remains relevant and continues to engage viewers across generations.9. Innovation: A good artwork can push the boundaries of what is considered possible within its medium, inspiring others to think differently about art.10. Connection to the Viewer: Ultimately, a good artwork creates a connection with the viewer. It speaks to them on a personal level, making them feel seen, understood, or challenged.In conclusion, while the criteria for what makes a good artwork can be subjective and vary widely, these elements are frequently cited as indicators of artistic excellence. Art that manages to incorporate several of these qualities is often celebrated and remembered as truly great.。

八年级英语电影评论语言表达技巧单选题50题

八年级英语电影评论语言表达技巧单选题50题1. In the movie "Avengers", the action scenes were _______ amazing.A. reallyB. veryC. quiteD. too答案:A。

“really”强调程度深,“非常”;“very”表示“很”;“quite”意为“相当”;“too”常用于“too...to...”结构,表示“太……而不能……”。

在这个语境中,“really”更能突出动作场景令人惊叹的程度。

2. The acting in the film "Titanic" was _______ perfect.A. nearlyB. almostC. mostlyD. nearly always答案:B。

“nearly”表示“几乎,差不多”,常与否定词连用;“almost”意思是“几乎,差不多”,语气比“nearly”更接近;“mostly”表示“主要地,大部分”;“nearly always”是错误表达。

这里“almost”最能准确表达表演接近完美的程度。

3. The special effects in "Avatar" were _______ spectacular.A. extremelyB. highlyC. greatlyD. very much答案:A。

“extremely”表示“极其,非常”,程度最强;“highly”常用于“be highly + 形容词”结构;“greatly”常修饰动词;“very much”一般修饰动词。

此句用“extremely”强调特效极其壮观。

4. The story of "The Lion King" is _______ interesting.A. muchB. moreC. mostD. very答案:D。

非常了得英语作文

非常了得英语作文1. Wow, that performance was absolutely incredible! I was blown away by the talent and passion on stage.2. I can't believe how quickly she finished that project. She's really on top of her game.3. The food at that new restaurant was out of this world. I've never tasted anything like it before.4. Did you see the way he handled that difficult situation? He's truly a master at problem-solving.5. The way she speaks multiple languages fluently is just mind-blowing. I wish I had that skill.6. I was so impressed by her ability to stay calm under pressure. She handled the whole situation with grace and poise.7. The way he can play the guitar is just unreal. I could listen to him for hours.8. Her dedication to her craft is truly inspiring. She puts in so much hard work and it really shows in her performances.9. I can't get over how talented she is. She'sdefinitely going places in her career.10. The way he can connect with people on such a deep level is truly a gift. He has a way of making everyone feel heard and understood.。

小学下册第十二次英语第一单元自测题[含答案]

10. 填空题: The _______ (The Great Depression) brought economic hardship to millions.
11. 填空题: A _______ (小海豚) is known for its intelligence.
12. 听力题: A ____ is known for its ability to jump great distances.
49. 填空题: A pig's sense of smell is incredibly ______ (敏锐).
50. 听力题: An element's atomic number tells you the number of ______ in its nucleus.
51. 听力题: My mom enjoys making ____ (smoothies) for breakfast.
5. 听力题: She has a _____ (new/old) dress.
6. 选择题: What do you call the outer layer of the Earth? A. Core B. Mantle C. Crust D. Layer 答案:C
7. 填空题: The ________ (环境适应策略) is developed over time.
25. 听力题: The chemical symbol for iron is _______.
26. 填空题: The seal can dive deep into the ______ (海洋).
27. 选择题: What is the capital of Greece? A. Athens B. Rome C. Istanbul D. Cairo 答案:A
  1. 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
  2. 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
  3. 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。

a r X i v :0801.0456v 1 [m a t h .A G ] 3 J a n 2008ON THE WONDERFUL COMPACTIFICATIONSAM EVENS AND BENJAMIN F JONESAbstract.These lecture notes explain the construction and basic properties of the wonderful compactification of a complex semisimple group of adjoint type.An appendix discusses the more general case of a semisimple symmetric space.Contents 1.Introduction 11.1.Notation 22.Construction and basic properties of the wonderful compactification 22.1.Definition of the compactification 22.2.Geometry of the open affine piece 42.3.Global Geometry 62.4.Description of the G ×G -orbits 82.5.Geometry of orbits and their closures parison of different compactifications 133.1.Independence of highest weight 133.2.Lie algebra realization of the compactification 174.Cohomology of the compactification 204.1.T ×T -fixed points on Xputation of cohomology225.Appendix on compactifications of general symmetric spaces24References 271.IntroductionThe purpose of these notes is to explain the construction of the wonderful compactifi-cation of a complex semisimple group of adjoint type.The wonderful compactification of a symmetric space was introduced by DeConcini and Procesi [5],and has been extensively12S.EVENS AND B.JONESstudied in algebraic geometry.Of particular interest are the recent proofs of the Manin conjecture for the compactification in[16]and[12].Intuitively,the wonderful compact-ification gives information about the group at infinity.In this connection,we mention the paper by He and Thomsen[14]showing that closures of different regular conjugacy classes coincide at infinity.The wonderful compactification has been used by Ginzburg in the study of character sheaves[11],and by Lusztig in his study of generalized character sheaves[15].It is closely related to the Satake compactification and to various analytic compactifications[3].It also plays an important role in Poisson geometry,as it is crucial for undersanding the geometry of a moduli space of Poisson homogeneous spaces[8,9]. This list is not meant to be exhaustive,but to give some idea of how the wonderful com-pactification is related to the rest of mathematics.The reader may consult[18]and[19] for further references.The appendix contains a construction of the wonderful compactification of a complex symmetric space due to DeConcini and Springer[7].This construction does not specialize to the construction in Section2in the case when the symmetric space is a group,but it is equivalent,and the two constructions are conceptually similar.These notes are based on lectures given by thefirst author at Hong Kong University of Science and Technology and by both authors at Notre Dame.They are largely based on work of DeConcini,Procesi,and Springer which is explained in the papers[5],[7], and the book[4].The notes add little in content,and their purpose is to make the simple construction of the wonderful compactification more accessible to students and to mathematicians without an extensive background in algebraic groups and algebraic geometry.We would like to thank Hong Kong University of Science and Technology for hospitality during preparation of afirst draft of these notes,and would like to thank Jiang-Hua, Dragan Milicic,Allen Moy,Xuhua He,Francois Ledrappier,and Dennis Snow for useful comments and discussions.1.1.Notation.In these notes,an algebraic variety is a complex quasi-projective variety, not necessarily irreducible.A subvariety is a locally closed subset of a variety.2.Construction and basic properties of the wonderful compactification We explain how to prove that the wondeful compactification of a semisimple complex group G of adjoint type is smooth,and describe its G×G-orbit structure.2.1.Definition of the compactification.Let G be a complex connected semisimple group with trivial center.Let˜G be the simply connected cover of G,and choose a maximal torus˜T contained in a Borel subgroup˜B of˜G.We denote their images in G by T⊂B. We denote Lie algebras of algebraic groups by the corresponding gothic letter,so the Lie algebras of T⊂B⊂G,and of˜T⊂˜B⊂˜G,are t⊂b⊂g.Let X∗(˜T)be the group of characters of˜T.We write this group additively,so ifλ,µ∈X∗(˜T)we have by definition3 (λ+µ)(t)=λ(t)µ(t)for all t∈˜T.LetΦ⊂X∗(˜T)be the roots of˜T in˜G,and take the positive rootsΦ+to be the roots of˜T in˜B.Let{α1,...,αl}be the corresponding set of simple roots,where l=dim(˜T).With these choices,forλ,µ∈X∗(˜T),we sayλ≥µifλ−µ= α∈Φ+nααwhere the nαare nonnegative integers.There is an embedding of the character group X∗(T)֒→X∗(˜T)as the characters that are trivial on the center ofG.Ifλis in the image of this embedding,we may computeλ(t)for t∈T.Let˜B−be the opposite Borel of˜G containing˜T and let B−be its image in G.Let U and U−be the unipotent radicals of˜B and˜B−.The group homomorphism˜G→G restricts to an isomorphism on any unipotent subgroup,and we identify U and U−with their images in G.If W is a representation of˜G,set Wµfor theµ-weight space for the ˜T-action.If v∈Wµ,then U−·v⊂v+ φ<µWφ.Fix an irreducible representation V=V(λ)of G with regular highest weightλandchoose a basis v0,...,v n of T-weight vectors of V with the following properties:(1)v0has weightλ;(2)For i∈{1,...,l},v i has weightλ−αi;(3)Letλi be the weight of v i.Then ifλi<λj,then i>j.Remark2.1.For i=0,...,l,dim(Vλi)=1.Remark2.2.U−·v i−v i is in the span of the v j for j>i.The induced action of˜G on the projective space P(V)factors to give an action of G.Define P0(V)={[ n i=0b i v i]:b0=0}⊂P(V).The affine open set P0(V)∼={v0+ n i=1b i v i}∼=C n.Further,P0(V)is U−-stable by Remark2.2applied to v0.Also,the morphism U−→U−·[v0]is an isomorphism of algebraic varieties.Indeed,the stabilizer of[v0]in G is B,so the stabilizer in U−of[v0]is trivial,and the result follows.Lemma2.3.U−→U−·[v0]is an isomorphism between U−and the closed subvarietyU−·[v0]of P0(V).We have proved everything but the last statement,which follows by the followingstandard fact.Lemma2.4.Let A be a unipotent algebraic group and let X be an affine A-variety.Then every A-orbit in X is closed.˜G acts on V∗by the dual action,g·f=f◦g−1.Choose a dual basis v∗0,...,v∗n ofV∗.Then each v∗i is a weight vector for˜T of weight−λi,and v∗0is a highest weight vector with respect to the negative of the above choice of positive roots.Note that P0(V)={[v]:v∗0(v)=0}.Define P0(V∗)={[ n i=0b i v∗i]:b0=0}.Then U acts on P0(V∗),and the morphism U→U·[v∗0]is an isomorphism of varieties.We state the analogue of Lemma2.3for this action,which follows by reversing the choice of positive roots.4S.EVENS AND B.JONESLemma2.5.U→U·[v∗0]is an isomorphism between U and the closed subvariety U·[v∗0] of P0(V∗).˜GטG acts on End(V)by the formula(g1,g2)·A=g1Ag−12and on V⊗V∗by linearextension of the formula(g1,g2)·v⊗f=g1·v⊗g2·f.Then the canonical identification V⊗V∗∼=End(V)is˜GטG-equivariant,and we treat this identification as an equality. Then{v i⊗v∗j:i,j=0,...n}is a basis of End(V).As before,the˜GטG-action on P(End(V))descends to give a G×G-action.We consider the open set defined byP0={[ a ij v i⊗v∗j]:a00=0}={[A]∈P(End(V)):v∗0(A·v0)=0}.It follows from the above that U−T×U preserves P0.Indeed,we use that fact that U−T preserves P0(V)and U preserves P0(V∗).Define an embeddingψ:G→P(End(V))byψ(g)=[g],and note thatψis G×G-equivariant,where G×G acts on G by(g1,g2)·x=g1xg−12.Let X=ψ(U−T U)in P0.Define Z=λ(˜t)v k⊗v∗k].5 Butλk(˜t)l i=1αi(t)n ikwhereλk=λ− l i=1n ikαi.Thus,ψ(t)=[v0⊗v∗0+li=11l i=1αi(t)n ik v k⊗v∗k].Define F:C l→P0byF(z1,...,z l)=[v0⊗v∗0+li=1z i v i⊗v∗i+ k>l(l i=1z n ik i)v k⊗v∗k].Then F is a closed embedding,so to prove the Lemma,it suffices to identify the image with6S.EVENS AND B.JONESNote also that there is an isomorphismσ:G×G→G×G given byσ(x,y)=(y,x). Consider the isomorphismγ:V⊗V∗→V∗⊗V such thatγ(v⊗f)=f⊗v.Then γ◦a=σ(a)γ,for a∈G×G.Proof of Theorem2.8.The last Lemma applied to Lemma2.9implies that(2.1)X0∼=(U−×U)×β−1(e,e).Butβ−1(e,e)is closed in X0and by Lemma2.9,ψ(T)⊂β−1(e,e).Thus,Z=7(1)If H is a subgroup of K,then K/H is projective if and only if H is parabolic.(2)A nonzero vector v in V is a highest weight vector if and only if its stabilizer K[v] in K is parabolic.If v1,v2are highest weight vectors in V,then there is g∈K such that g·v1=v2.By(1)and(2),it is clear that if v is a highest weight vector,K·[v]∼=K/K[v]is projective, and thus closed in P(V).Let[u]∈P(V)be such that K·[u]∼=K/K[u]is closed and therefore projective.Then K[u]is parabolic by(1),so u is a highest weight vector.Thus, by(2),K·[u]=K·[v].Q.E.D.Lemma2.12.Let A be an algebraic group and let W be a A-variety with a unique closed orbit Y.Let V⊂W be open and suppose V∩Y is nonempty.Then W=∪g∈A g·V. Proof.The union∪g∈A g·V is clearly open,so W−∪g∈A g·V is closed,and is A-stable. We assume it is nonempty and argue by contradiction.Recall that if an algebraic group acts on a variety,then it has a closed orbit.Thus,there exists a closed orbit Z for A on W−∪g∈A g·V.Since Z is a closed orbit for A in W,then Z=Y.But Y⊂W−∪g∈A g·V by assumption.It follows that W−∪g∈A g·V is empty.Q.E.D.Lemma2.13.Let W⊂X=Q=∪a∈G×G a·(Q∩X0=Q.This gives(2),and in the case Q=ψ(G),we obtain8S.EVENS AND B.JONESNow(1)follows by Theorem2.8since X is a union of smooth open sets.To prove(3), note that by(2),Q′.Since Q and Q′are both open in their closures,they coincide.Q.E.D.2.4.Description of the G×G-orbits.We classify the G×G-orbits in X and show they have smooth closure.For I⊂{1,...,l},letZ I:={(z1,...,z l)∈Z:z i=0∀i∈I}and letZ0I:={(z1,...,z l)∈Z I:z j=0∀j∈I}.Then Z I is the closure of Z0I,Z I∼=C l−|I|and Z0I∼=(C∗)l−|I|.Moreover,by the proof of Lemma2.7,T∼=(C∗)l acts on Z∼=C l by(a1,...,a l)·(z1,...,z l)=(a1z1,...,a l z l)in appropriate coordinates,so Z is(T×{e})-stable,and the Z0I are exactly the T×{e}-orbits in Z.The next lemma follows from the above discussion.Lemma2.15.Set z I=(z1,...,z l),where z i=1if i∈I,and z i=0if i∈I.Then Z0I=(T×{e})·z I.For i=1,...,l,Z i:=Z{i}is a hypersurface,and Z I=∩i∈I Z i.LetΣI=χ(U−×U×Z I)and letΣ0I=χ(U−×U×Z0I)The following result follows easily from Theorem2.8.are precisely the U−T×U orbits in X0,and the closureΣI Proposition2.16.TheΣ0IofΣ0I in X0is isomorphic to C dim(G)−|I|.In particular,U−T×U has precisely2l orbits on X0,and all these orbits have smooth closure.Remark2.17.Let Q be a G×G-orbit in X.Then9(2)S i=Σi is contained in X−ψ(G).Thus,Σi=Sαfor someα,and Sα∩X0=Σi. The result follows.Q.E.D. Lemma2.20.The following hold:(1)Let S I=∩i∈I S i.Then S J⊂S I if I⊂J and S I∩X0=ΣI.(2)Let S0I=S I−∪I J S J.Then S0I=(G×G)·z I is a single G×G-orbit and S0I=S0Jimplies I=J.(3)S I=∪a∈G×G a·ΣI,and in particular S I is smooth.Proof.Thefirst claim of(1)is obvious from the definition and the second claim follows from Lemma2.19(3).For(2),it is clear that S0I∩X0=Σ0I.Since the S i are G×G-stable, S I is G×G-stable,so S0I is G×G-stable.Let x,y∈S0I.By Proposition2.14(1),there are a and b in G×G such that a·x and b·y are in X0.Since a·x and b·y are in S0I by G×G-stability,they are inΣ0I.Thus,by Proposition2.16,there is c∈U−T×U such that a·x=cb·y.Part(2)follows since clearly z I∈S0I.Part(3)follows by the above assertions and Proposition2.14(2).Q.E.D. Remark2.21.Recall the following definition.Let X be a smooth variety with hyper-surface Z.We say Z is a divisor with normal crossings at x∈Z if there is an open neighborhood U of x such that Z∩U=D1∪···∪D k is a union of hypersurfaces andD i1∩···∩D ijis smooth of codimension j for each distinct j-tuple{i1,...,i j}in{1,...,k}.A standard example is to take X=C n and Z to be the variety defined by the vanishing of z1...z k.Then Z is the union of the hyperplanes given by vanishing of z i,and in the complex analytic setting,every divisor with normal crossings is locally of this nature. Theorem2.22.G×G has2l orbits in X,given by S0I=(G×G)·z I where I is a subset of{1,...,l}.In particular,all orbits have smooth closure,and the pair(X,X−ψ(G))is a divisor with normal crossings.Proof.Let Q⊂X be a G×G-orbit.Then10S.EVENS AND B.JONES2.20(2).Again by Lemma2.20(2),there are exactly2l orbits.SinceΣI is the closure ofΣ0I in X0,it follows that S I is the closure of S0I in X using Lemma2.12.It is clear from Propositon2.16that(X0,X0−ψ(U−T U))is a divisor with normal crossings.The last assertion follows from Proposition2.14(1)and group action invariance of the divisor with normal crossings property.Q.E.D.2.5.Geometry of orbits and their closures.We now want to understand the geom-etry of S0I and S I.We show that the orbit S0Ifibers over a product of generalizedflag varieties,withfiber a semisimple group.In this picture,the closure S Ifibers over the same product of generalizedflag varieties,and thefiber is the wonderful compactification of the semisimple group.For a subset I⊂{1,...,l},let∆I={αi:i∈I}.LetΦI be the roots that are in the linear span of the simple roots in∆I,and letl I=t+ α∈ΦI gαLet p I=l I+u and let p−I =l I+u−.Let u I and u−Ibe the nilradicals of p I andp−I ,respectively.Note that p∅=g and p{1,...,l}=b.Let P I,P−I,U I,U−I,and L I bethe corresponding connected subgroups of G.Let Z(L I)be the center of L I and let G I=L I/Z(L I)be the adjoint group of L I and denote its Lie algebra by g I.For a Lie algebra a,let U(a)be its enveloping algebra.Let V I=U(l I)·v0.Then V I is a l I-stable submodule of V and V I may be identified with the irreducible representation of g I of highest weightλ.Lemma2.23.The following hold:(1)V I is p I-stable,and u I annihilates V I.(2)Let Q={g∈G:g·V I=V I}.Then Q=P I.Proof.For(1),by the Poincare-Birkhoff-Witt Theorem,U(p I)=U(l I)U(u I),and p I-stability of V I follows by the theorem of the highest weight.Since u I is an ideal of p I, the space of u I-invariants V u I is l I-stable and meets V I.Since it is nonzero and V I is an irreducible l I-module,V I=V u I.For(2),note that it follows from(1)that P I⊂Q.Hence Q is parabolic so it is connected.To show P I=Q,it suffices to show p I=q,where q is the Lie algebra of Q. There is t∈Z(L I)such thatαi(t)=1for every simple rootαi∈∆i.Then if˜t∈˜T is a preimage of t,Ad˜t acts byλ(˜t)on V I,and if x∈g−αiforαi simple,Ad˜t has weightλ(˜t)Lemma2.24.The following hold:(1)LetJ={k∈{0,...,n}:λk=λ− i∈∆I n iαi,n i≥0}Then the{v j:j∈J}form a basis of V I.(2)Recall z I={(ǫ1,...,ǫl)∈Z I},whereǫi=1if i∈I andǫi=0if i∈I.Thenz I=[pr VI ],where pr VIis the projection on V I such that pr VI(v k)=0if k∈J.Proof.(1)is routine and is left to the reader.For(2),recall that for{z1,...,z l}∈Z, the corresponding class in P(End(V))is[v0⊗v∗0+li=1z i v i⊗v∗i+ k>l l i=1z n ik i v k⊗v∗k]whereλk=λ− l i=1n ikαi.It follows thatz I=[ k∈J v k⊗v∗k].But k∈J v k⊗v∗k is easily identified with pr V I.Q.E.D.We now compute the stabilizer of(G×G)at the point z I∈S0I=(G×G)·z I.Proposition2.25.The stabilizer(G×G)zIof z I in G×G is{(xu,yv):u∈U I,v∈U−I,x∈L I,y∈L I,and xy−1∈Z(L I)}In particular,S0Ifibers over G/P I×G/P−Iwithfiber G I.Proof.Suppose(r,s)·z I=z I for(r,s)∈G×G.Then[r pr VI s−1]=[pr VI]so r preservesthe image V I of pr VI .Thus,by Lemma2.23,r∈P I.Further,if r∈U I,then r·[pr VI]=pr VIsince U I acts trivially on V I.Recall the identificationγ:End(V)→End(V∗)from the proof of2.9.Let V∗I= U(p−)v∗0⊂V∗.V∗I is an irreducible representation of L I with lowest weight−λ.One caneasily check thatγ(pr VI )=pr V∗I,andγ((x,y)·A)=(y,x)·γ(A),for(x,y)∈G×G andA∈End(V).Hence(s,r)·[pr V∗I ]=γ((r,s)·[pr VI])=γ([pr VI])=[pr V∗I].As above,it follows that s preserves V∗I.Thus,by Lemma2.23applied to the oppositeparabolic,s∈P−I .Further,if s∈U−I,then[pr VI◦s]=[pr VI].Now let r=xu and let s=yv with x,y∈L I,u∈U I and v∈U−I.Thenr·[pr VI ]·s−1=x·[pr VI]·y−1.In particular,xy−1acts trivially on P(V I).ButZ(L I)={g∈L I:g·[v]=[v],∀[v]∈P(V I)},so xy−1∈Z(L I).Now we consider the projection(G×G)/(G×G)zI →(G×G)/(P I×P−I)withfiber(P I×P−I )/(G×G)zI∼=(LI×L I)/{(x,y):x∈L I,y∈L I,and xy−1∈Z(L I)}The morphism(L I×L I)/{(x,y):x∈L I,y∈L I,and xy−1∈Z(L I)}→L I/Z(L I)given by(a,b)→ab−1is easily seen to be an isomorphism.Q.E.D. To understand the closure S I of S0I,we embed the compactification of a smaller group into X.First,note that we may embed End(V I)into End(V)by using the mapi,j∈J a ij v i⊗v∗j→i,j∈{0,...,n}b ij v i⊗v∗jwhere b ij=a ij if i,j∈J,and b ij=0otherwise.This map induces an embedding P(End(V I))→P(End(V)),and we will regard P(End(V I)) as a closed subvariety of P(End(V)).It follows from definitions that z K∈P(End(V I))⇐⇒I⊂K.Define a morphism L I→P(End(V))byψI(g)=[g pr VI ].ThenψI descends to amorphismψI:L I/Z(L I)=G I→P(End(V)).Note that the image ofψI is in P(End(V I)). Recall that the center of G I is trivial,and note that V I is an irreducible representation of a cover of G I with regular highest weight.We set X I=(*)Let A be a group with subgroup B.Let Y be a B-set.Then B-orbits in Y correspond to A-orbits in A×B Y via B·y→A·(e,y).Moreover,the stabilizer B y coincides with the stabilizer A(e,y).It follows that the stabilizer(G×G)yK =(P I×P−I)zK.But U I Z(L I)×U−IZ(L I)actstrivially on X I since itfixes each z K with I⊂K,so(P I×P−I )zKis the preimage of(G I×G I)zK in P I×P−I.By Proposition2.25applied to G I and X I,it follows that(P I×P−I )zK={(xu,yv):u∈U K,v∈U−K,x∈L K,y∈L K,and xy−1∈Z(L K)},which coincides with the stabilizer(G×G)zK .Thus,χis injective when restricted to aG×G orbit.To complete the proof thatχis injective,it suffices to show that every G×G-orbit contains some y K,I⊂K.This is an easy consequence of(*)and the classification of G I orbits in its wonderful compactification X I,which follows from Theorem2.22.We leave the routine details to the reader.Q.E.D.parison of different compactifications3.1.Independence of highest weight.For completeness,we show that X is indepen-dent of the choice of the regular,dominant highest weight.This result is not especially surprising sinceλdoes not appear in the statements describing the G×G-orbit structure. The proof mostly follows that of DeConcini and Springer([7],Proposition3.10).We may carry out the wonderful compactification construction using regular,dominant weightsλ1andλ2to get varieties X1⊂P(End V(λ1))and X2⊂P(End V(λ2))respectively. We denote the class of the identity by[id1]∈X1and[id2]∈X2.We define X∆=(T×{e})·([id1],[id2])where closure is taken in the open set X10×X20⊂X1×X2.Lemma3.2.Z∆∼=C l and p i:Z∆→Z i is a(T×{e})-equivariant isomorphism. Proof.This is a straightforward calculation using coordinates as in Lemma2.7.Choose a basis of weight vectors{v i}i=0..n for V(λ1)and a basis of weight vectors{w i}i=0..m forV (λ2)satisfying properties (1)-(3)preceding Remark 2.1.Then,for t ∈T ,(t ×{e })·([id 1],[id 2])=(t ×{e })·([ n i =0v ∗i ⊗v i ],[ m i =0w ∗i ⊗w i ])=([v ∗0⊗v 0+ li =11αi (t )w ∗i⊗w i +···])where the terms indicated by ···in the two factors have coefficients which are polynomial in the 1αi (t )replaced by z i identifies Z ∆with C l and the claim follows.Q.E.D.Recall the embedding from Theorem 2.8,χi :U −×U ×Z i →X iThis is an isomorphism to X i 0.Let X ∆0:=p −1i (X i 0)⊂X ∆,and consider the subset V = a ∈G ×G a ·X ∆0of X ∆.Define χ∆:U −×U ×Z ∆→X ∆in the same manner as χfrom Theorem 2.8.We have a commutative diagram:(3.1)U −×U ×Z ∆id ×id ×p i ∼=X ∆p iX i .We will use the following two lemmas,which we prove below.Lemma 3.3.χ∆is an embedding onto the open set X ∆0.Lemma 3.4.p i |V is injective.We assume these for now and prove that p i is an isomorphism.Proof of Proposition 3.1.Consider the commutative diagram:(3.2)U −×U ×Z ∆id ×id ×p i ∼=X ∆0p iX i 0.Since χi :U −×U ×Z i →X i 0is surjective,p i :X ∆0→X i 0is surjective using the abovecommutative diagram.Then p i :V →X i is surjective by Proposition 2.14(1).Further,p i :V →X i is injective by Lemma 3.4,and hence an isomorphism since X i is smooth.Thus V is complete,so V=X∆,since X∆is irreducible and dim(V)=dim(X∆).Q.E.D. We now prove the two lemmas.Proof of Lemma3.3.First,χ∆is an embedding sinceχi◦(id×id×p i|Z∆)is an embedding and the diagram(3.1)commutes.Let Y denote the image ofχ∆.Commutativity of(3.1)and Theorem2.8imply that Y⊂X∆0.Consider the maps=χ∆◦(id×id×p i)−1◦(χi)−1:X i0→X∆0.Note that s is a section for the map p i over X i0,i.e.,p i◦s=id.Consider the compositionf=s◦p i|X∆0.It is routine to check that f|Y=id|Y.Y and X∆0have the same dimensionand X∆0is irreducible since it is an open subset of the irreducible variety X∆.Hence, there is an open,dense subset of X∆0contained in Y.Then f is the identity on an open, dense subset of X∆0,so f is the identity on X∆0.Hence,Y=X∆0.Q.E.D.Proof of Lemma3.4.For I⊂{1,...,l},let z∆I =(z1I,z2I),so p i(z∆I)=z i I.Then thestabilizer(G×G)z∆I =(G×G)z1I∩(G×G)z2I,which coincides with(G×G)z iIfor i=1,2,since(G×G)z1I =(G×G)z2Iby Proposition2.25.Thus,p i:V→X i is injective whenrestricted to the orbits through some z∆I.By Lemma3.3,each U−T×U-orbit on X∆0 meets Z∆,so it meets some z∆I.The Lemma follows.Q.E.D. We now prove a result related to Proposition3.1which will be used in the sequel.Let E be a representation of˜GטG.Then the˜GטG action on P(E)descends to an action of G×G on P(E).Suppose there exists a point[x]∈P(E)such that(G×G)[x]=G∆. We may embed G into P(E)by the mappingψ:G→P(E)given byψ(g)=(g,e)·[x]. Let X(E,[x])be the closure ofψ(G)in P(E).Let Xλ=X(End(V),[id V])when V is irreducible of highest weightλ.Ifλis regular,then Xλis of course smooth and projective with known G×G-orbit structure by Theorem2.22.Let W1,...,W k be a collection of irreducible representations of G of highest weights µ1,...,µk.Let W=W1⊕···⊕W k.Let F be a representation of˜GטG.Let V have highest weightλas before.When useful,we will denote the irreducible representation of highest weightλby V(λ).Proposition3.5.Suppose eachµj is of the formµj=λ− n iαi with all n i nonnegative integers.Then,X(End(V)⊕End(W)⊕F,[id V⊕id W⊕0])∼=Xλ.Proof.It is immediate from definitions thatX(End(V)⊕End(W)⊕F,[id V⊕id W⊕0])∼=X(End(V)⊕End(W),[id V⊕id W]),i.e.,nothing is lost by setting F=0.Indeed,the G×G-orbit through[id V⊕id W⊕0] lies inside the closed subvariety P(End(V)⊕End(W))of P(End(V)⊕End(W)⊕F),and this implies the claim.Define X′=X(End(V)⊕End(W),[id V⊕id W])for notational simplicity.Consider the open subset˜P(End(V)⊕End(W))={[A⊕B]:A=0}of P(End(V)⊕End(W)).The projectionπ:˜P(End(V)⊕End(W))→P(End(V))given by[A⊕B]→[A] is a morphism of varieties.We claim that˜P(End(V)⊕End(W))is(G×G)-stable.Indeed,if A=0and(x,y)∈˜GטG,then xAy−1is nonzero since x and y are invertible.It follows that if[A⊕B]∈˜P(End(V)⊕End(W)),then(x,y)·[A⊕B]∈˜P(End(V)⊕End(W)),which establishes the claim.Let w0(j)be a highest weight vector for W j,and let w0(j)∗be a nonzero vector of weight−µj of W∗j,normalized so w0(j)∗(w0(j))=1.Let w0= k j=1w0(j)and let w∗0= k j=1w0(j)∗.Then w∗0(w0)=k.Define an open subset P0(End(V)⊕End(W))as the set of[A⊕B]such that v∗0(A·v0)=0and w∗0(B·w0)=0.It is easy to see that P0(End(V)⊕End(W))is T×T-stable.Let Z′be the closure ofψ(T)in P0(End(V)⊕End(W)).Claim:Z′∼=C l.The proof of this claim is essentially the same as the proof of Lemma2.7.Indeed,we may computeψ(t)=(t,e)·[id V⊕id W]in the same manner as in Lemma2.7.In2.7, the coordinates z i are essentially given by1(G×G)zI acts by scalars on U(l I)·w0(j).This implies injectivity ofπ,and completesthe proof of the Proposition.Q.E.D. Remark3.6.In the statement of Proposition3.5,we can replace End(W)with End(W1)⊕···⊕End(W k),and we can replace id W with c1id W1⊕···⊕c k id Wk,where c1,...,c k arescalars.Indeed,when all c j=1,this follows because End(W1)⊕···⊕End(W k)is a (G×G)-submodule of End(W),so we may compute the closure in the smaller space.Moreover,in this embedding,id W=id W1⊕···⊕id Wk.To see that we can put scalars infront of each summand follows from an easy analysis of the argument.3.2.Lie algebra realization of the compactification.We give another realization of X,in which no choice of highest weight is used.This realization is used in[8,9]to give a Poisson structure on the wonderful compactification.Let n=dim(G).G×G acts on Gr(n,g⊕g)through the adjoint action.Let g∆={(x,x):x∈g},the diagonal subalgebra.Then the stabilizer in(G×G)of g∆is G∆={(g,g):g∈G},so(G×G)·g∆∼= (G×G)/G∆∼=G.Let(G×G)·g∆,where the closure is computed in Gr(n,g⊕g).Since Gr(n,g⊕g) is projective,G∼=X.In particular,G,and then apply the machinery developed above.Embed i:Gr(n,g⊕g)֒→P(∧n(g⊕g))via the Plucker embedding.That is,if U∈Gr(n,g⊕g)has basis u1,...,u n,we map U→i(U)=[u1∧···∧u n].It is well-known that the Plucker embedding is a closed embedding.Denote[U]for i(U)∈P(∧n(g⊕g)). Note that∧n(g⊕g)is naturally a representation of G×G by taking the exterior power of the adjoint representation.Clearly,i:(G×G)·g∆→(G×G)·[g∆]is an isomorphism.Since the Plucker embedding is a closed embedding,it follows that i:(G×G)·[g∆]is an isomorphism. To prove Proposition3.7,we apply Proposition3.5to prove X∼=(3){(0,Eα):α∈Φ},and note that(0,Eα)generates the unique weight space ofweight(0,α).We obtain a basis of weight vectors of∧n(g⊕g)as follows.Let A and B be subsets ofΦsuch that|A|+|B|≤n and let R be a subset of the vectors in(1)of cardinality n−|A|−|B|.For each triple(A,B,R)as above,we define a weight vectorx A,B,R=∧α∈A(Eα,0) ∧β∈B(0,Eβ) ∧i=1,...,n−|A|−|B|K i,where the K i are the vectors in the subset R.Then the collection{x A,B,R}is a basis of weight vectors of∧n(g⊕g),and it is routine to check that the weight of x A,B,R is ( α∈Aα, β∈Bβ).where A and B are subsets ofΦ.Take A=Φ+and B=−Φ+and let R0be the subset of basis vectors from(1)with K1=(H1,H1),...,K l=(H l,H l).Then we let v0=xΦ+,−Φ+,R0,so v0is a weightvector with weight(2ρ,−2ρ).The weight of any x A,B,R is(λ,µ)whereλ≤2ρand µ≥−2ρ.It follows immediately that v0is a highest weight vector of∧n(g⊕g)of highest weight(2ρ,−2ρ)relative to the Borel subgroup B×B−of G×G.Further,note that [v0]=[t∆+u⊕u−],where t∆={(x,x):x∈t},and u⊕u−is the Lie algebra of the unipotent radical of B×B−.Now we show that v0∈E.Choose H∈t such thatαi(H)=1for every simple root αi.Defineφ:C∗→T byφ(eζ)=exp(ζH),ζ∈C.We claim that:(3.3)limz→∞(φ(z),e)·[g∆]=[v0],where z∈C∗.To check this claim,takev∆=∧α∈Φ+(Eα,Eα) ∧i=1,...,l(H i,H i) ∧β∈Φ+(E−β,E−β),and note that[v∆]=[g∆].We decompose(φ(z),e)·v∆into a sum of22r terms(2r=|Φ|)in the span of weightvectors x A,B,R0,with R0={(H1,H1),...,(H l,H l)}.To get the22r terms,for each rootγ∈Φ,decompose(Eγ,Eγ)=(Eγ,0)+(0,Eγ),and choose one of the two summands in each term in the product.Each term corresponds to a choice of the subset A⊂Φ,and then B=Φ−A.For a rootα,let ht(α)= k i,whereα=k1α1+···+k lαl.Then we compute(φ(z),e)·x A,B,R=∧α∈A(z ht(α)Eα,0) ∧i=1,...,l(H i,H i) ∧β∈B(0,Eβ).Thus(φ(z),e)·x A,B,R0=z n A x A,B,R,where n A= α∈A ht(α).Let n0= α∈Φ+ht(α).It follows easily from properties of roots that n0≥n A for any(A,B,R0),and n0=n A if and only if A=Φ+and B=−Φ+.。

相关文档
最新文档