Chern class formulas for quiver varieties
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马懿,喻康杰,肖雄峻,等. 二氢杨梅素对梨酒抗氧化活性及风味物质的影响[J]. 食品工业科技,2023,44(10):107−115. doi:10.13386/j.issn1002-0306.2022100292MA Yi, YU Kangjie, XIAO Xiongjun, et al. Effects of Dihydromyricetin on Antioxidant Activity and Flavor Substances of Pear Wine[J]. Science and Technology of Food Industry, 2023, 44(10): 107−115. (in Chinese with English abstract). doi:10.13386/j.issn1002-0306.2022100292· 研究与探讨 ·二氢杨梅素对梨酒抗氧化活性及风味物质的影响马 懿1,2, *,喻康杰1,2,肖雄峻1,2,谢李明1,2,魏紫云1,2,熊 蓉1,2,禹 潇1,2,黄慧玲1,2(1.四川轻化工大学生物工程学院,四川宜宾 644000;2.四川省酿酒专用粮工程技术研究中心,四川宜宾 644000)摘 要:为推进梨酒减硫增质进程,促进天然抗氧化剂在梨酒中的应用。
本文通过单独使用二氢杨梅素(Dihydromyricetin ,DMY )或与SO 2联合使用酿造库尔勒香梨酒,分别测定各组梨酒自由基清除率、总酚含量、总黄酮含量、酒体色度及风味物质,研究了DMY 对梨酒抗氧化活性以及风味物质的影响。
结果表明,各组梨酒基础理化指标无显著性差异(P >0.05),发酵均能正常完成;H4组(SO 2 30 mg/L 、DMY 100 mg/L )抗氧化活性最佳,DPPH 自由基清除率为70.08%,ABTS +自由基清除率达到95.89%;单独使用DMY (浓度为150或200 mg/L )或与SO 2联合使用时,均能促进酒体总酚及总黄酮的生成(P <0.05)。
用英语介绍法国的一道菜作文French cuisine is renowned worldwide for its exquisite flavors, intricate techniques, and rich cultural heritage. One dish that exemplifies the essence of French culinary artistry is the classic Coq au Vin, a beloved poultry dish that has captivated the hearts and palates of food enthusiasts across the globe.The origins of Coq au Vin can be traced back to the rural regions of France, where it was initially a humble peasant dish. The name "Coq au Vin" translates to "rooster in wine," reflecting the dish's humble beginnings when farmers would use tough, older roosters in the stew to tenderize the meat. Over time, as French cuisine evolved and refined, Coq au Vin became a culinary masterpiece, showcasing the country's exceptional ingredients and the ingenuity of its chefs.At the heart of this dish is the interplay between the succulent chicken and the rich, velvety sauce. The chicken, typically a whole bird or a combination of legs and thighs, is carefully selected for its quality and flavor. The meat is first seared to lock in its natural juices and develop a beautiful caramelized crust. This step is crucial, as itlays the foundation for the depth of flavor that will permeate the entire dish.The next step in the preparation of Coq au Vin is the creation of the sauce, which is the true star of the show. The recipe calls for the use of a full-bodied red wine, often a Burgundy or a Pinot Noir, which lends its robust and complex flavors to the dish. The wine is combined with a flavorful broth, typically made from chicken or beef, along with a variety of aromatic vegetables such as onions, carrots, and mushrooms.As the dish simmers, the flavors of the wine, broth, and vegetables meld together, creating a harmonious and intoxicating aroma that fills the kitchen. The long, slow cooking process allows the tough connective tissues in the chicken to break down, resulting in a tender and succulent meat that practically falls off the bone.But Coq au Vin is more than just a simple stew – it is a culinary masterpiece that showcases the attention to detail and the mastery of technique that is so integral to French cuisine. The dish is often garnished with crispy lardons (small pieces of bacon), pearl onions, and fresh parsley, adding a delightful contrast of textures and flavors.One of the most fascinating aspects of Coq au Vin is the way it has evolved over time, reflecting the diverse regional influences and thecreativity of French chefs. While the basic recipe remains largely unchanged, there are numerous variations that have emerged, each with its own unique twist.For example, in the Burgundy region of France, where the dish is believed to have originated, the Coq au Vin is often made with the region's renowned Pinot Noir wine, lending a distinct and elegant flavor profile. In other parts of the country, chefs may incorporate different types of mushrooms, such as the prized porcini, or experiment with the addition of pearl onions or even small potatoes.Regardless of the specific variations, one thing remains constant –the unwavering commitment to quality ingredients and the meticulous attention to detail that is the hallmark of French cuisine. Coq au Vin is a dish that truly embodies the essence of French culinary culture, where the pursuit of perfection is not just a goal, but a way of life.As you savor each bite of this iconic French dish, you can't help but be transported to the charming countryside, where the aroma of simmering chicken and the rich, velvety sauce wafts through the air, beckoning you to indulge in the simple pleasures of life. Coq au Vin is not just a meal – it is a celebration of the French culinary heritage, a testament to the power of patience and the rewards of mastering the art of slow cooking.Whether you're a seasoned chef or a passionate home cook, the opportunity to recreate this classic French dish in your own kitchen is a true privilege. It is a culinary journey that allows you to connect with the rich history and traditions of France, while also exploring your own creativity and culinary skills. So, the next time you're in the mood for a truly exceptional dining experience, consider the timeless and beloved Coq au Vin – a dish that will transport you to the heart of French cuisine and leave you with a newfound appreciation for the art of gastronomy.。
红茶果醋发酵工艺优化魏建敏,陈燕,陈莉,卢红梅(贵州省发酵工程与生物制药重点实验室,贵州大学酿酒与食品工程学院,贵州贵阳 550025) 摘要:为确定红茶果醋的最佳工艺参数,本试验以红茶、石榴为原料,接种酵母菌、醋酸菌进行发酵。
在单因素实验的基础上,以茶果汁比、酵母菌接种量、加糖量为因素进行酒精发酵阶段的正交实验,以初始酒精度、醋酸菌接种量、发酵温度为因素进行醋酸发酵阶段的正交实验。
结果表明:优化后的最佳工艺条件为:酒精发酵阶段,微波浸提4 min,茶水比5:100,茶果汁比1:5,酵母菌接种量0.2%,加糖量16 g/100 mL;醋酸发酵阶段,装瓶量30%,初始酒精度7% vol,醋酸菌接种量12%,30 ℃发酵。
在此条件下,发酵的红茶果醋总酸含量为6.48 g/100 mL、茶多酚含量为397.85 mg/L,富含16种氨基酸,总含量为5.93 mg/mL,其中必需氨基酸占总含量的38.45%,其色泽明亮、香气协和、风味独特,理化及卫生指标符合国家相关标准。
本研究结果可为红茶果醋的开发和工业化生产提供理论依据。
关键词:茶果醋;红茶;石榴;酒精发酵;醋酸发酵文章篇号:1673-9078(2021)06-89-97 DOI: 10.13982/j.mfst.1673-9078.2021.6.1030 Optimization of the Fermentation Technology of Black Tea VinegarWEI Jian-min, CHEN Y an, CHEN Li, LU Hong-mei(Guizhou Key Laboratory of Fermentation Engineering and Biopharmacy, School of Liquor and Food Engineering,Guizhou University, Guiyang 550025, China)Abstract: In order to determine the best technological parameters of black tea fruit vinegar, black tea and pomegranate were used as raw materials in this study, yeast and acetic acid bacteria were inoculated for fermentation. Based on the single factor experiments, the orthogonal experiment of alcohol fermentation stage was carried out with tea juice ratio, yeast inoculation amount and sugar content as factors, and the orthogonal experiment of acetic acid fermentation stage was carried out with the initial alcohol precision, acetic acid bacteria inoculation amount and fermentation temperature as factors. The results showed that the optimal fermentation conditions were as follows: in alcohol fermentation stage, microwave extraction was 4 minutes, the ratio of tea to water was 5:100, the ratio of tea to juice was 1:5, yeast inoculation amount was 0.2%, sugar addition amount was 16 g/100 mL; in acetic acid fermentation stage, bottling amount was 30%, initial ethanol content was 7% vol, acetic acid bacteria inoculation amount was 12%, fermentation at 30 ℃. Under the optimized conditions, the total acid ofblack tea fruit vinegar was 6.48 g/100 mL, and the content of tea polyphenols was 397.85 mg/L. It was rich in 16 kinds of amino acids, and the total content was 5.93 mg/mL, among which the essential amino acids accounted for 38.45% of the total content. Its color was bright, aroma was harmonious, flavor was unique, the physical and chemical and health indicators met the relevant national standards. It provided a theoretical basis for the development and industrial production of black tea vinegar.Key words: tea fruit vinegar; black tea; pomegranate; alcohol fermentation; acetic acid fermentation引文格式:魏建敏,陈燕,陈莉,等.红茶果醋发酵工艺优化[J].现代食品科技,2021,37(6):89-97WEI Jian-min, CHEN Y an, CHEN Li, et al. Optimization of the fermentation technology of black tea vinegar [J]. Modern Food Science and Technology, 2021, 37(6): 89-97红茶属于全发酵茶,通过典型的萎凋、揉捻、发酵、干燥等工艺流程精制而成,因冲泡后的茶汤和叶底色呈红色而得名,是目前世界上生产和贸易量最大收稿日期:2020-11-08基金项目:贵州省科技支撑计划项目(黔科合支撑[2019]2371号)作者简介:魏建敏(1996-),女,硕士研究生,研究方向:食品生物工程 通讯作者:陈莉(1975-),女,副教授,研究方向:食品生物技术 的茶类[1-3]。
制作法式巧克力蛋糕配覆盆子英语作文Title: Making a French Chocolate Cake with Raspberry Topping French Chocolate Cake with Raspberry Topping is a delectable dessert that combines the rich flavors of chocolate with the refreshing sweetness of raspberries. This exquisite dessert is perfect for special occasions or as a treat for yourself. Here is a detailed guide on how to make this delightful French delicacy.Ingredients:1 cup allpurpose flour1 cup granulated sugar1/2 cup unsweetened cocoa powder1/2 teaspoon baking powder1/2 teaspoon baking soda1/2 teaspoon salt1/2 cup vegetable oil2 large eggs1 teaspoon vanilla extract1/2 cup hot water1/2 cup fresh raspberriesInstructions:1. Preheat the Oven: Preheat your oven to 350°F (175°C) and grease a 9inch round cake pan.2. Prepare the Dry Ingredients: In a mixing bowl, sift together the flour, sugar, cocoa powder, baking powder, baking soda, and salt. Mix well to combine all the dry ingredients.3. Mix Wet Ingredients: In a separate bowl, whisk together the vegetable oil, eggs, and vanilla extract until well combined.4. Combine Wet and Dry Ingredients: Slowly pour the wet ingredients into the dry ingredients, stirring constantly to avoid lumps. Mix until the batter is smooth.5. Add Hot Water: Gradually pour in the hot water while stirring the batter. The batter will be thin, but that is normal.6. Bake the Cake: Pour the batter into the prepared cake pan and bake in the preheated oven for about 3035 minutes, or untila toothpick inserted into the center comes out clean.7. Cool the Cake: Allow the cake to cool in the pan for 10 minutes before transferring it to a wire rack to cool completely.8. Prepare the Raspberry Topping: In a small saucepan, heat the fresh raspberries over low heat, stirring occasionally until they break down and form a saucelike consistency.9. Assemble the Cake: Once the cake has cooled, spread the raspberry topping over the top of the cake, ensuring even coverage.10. Serve and Enjoy: Slice the French Chocolate Cake with Raspberry Topping into pieces and serve. The combination of rich chocolate and sweet raspberries is sure to delight your taste buds.This French Chocolate Cake with Raspberry Topping is a delicious and elegant dessert that is bound to impress your family and friends. Enjoy the delightful flavors and textures of this decadent treat, and savor every bite of this exquisite creation.。
高一英语新人教选择性必修一知识点盘点
Unit 6
ingredient: 成分
recipe: 食谱
nal: 传统的
cuisine: 烹饪风格
worldwide: 全世界的
culinary: 烹饪的
masterpiece: 杰作
___: 异国情调的
flavor: 味道
___: 呈现
appetite: 食欲
___: 素食主义者
preference: 偏好
中国传统烹饪以多样的口味和使用新鲜食材而闻名。
A masterpiece of French cuisine。
___.
这道法国烹饪的杰作将精致的口味与精美的呈现方式结合在一起。
我喜欢辣食,所以四川菜是我最喜欢的。
She has a strong ___.
她对素食菜肴有很强的偏好。
我们为什么不试试这个新食谱呢?它制作简单而且味道美味。
___ cuisine.
我建议我们去那家餐厅。
他们供应正宗的意大利菜。
填空题目:根据短文内容填写空缺处的适当单词。
叙述经历:描述自己的一次特别的用餐经历,包括所去的餐厅、点的菜和感受等。
本文对《高一英语新人教选择性必修一》教材中的 Unit 6 进行
了知识点盘点,涵盖了词汇、重点句型和语法、阅读理解以及写作
任务等方面的内容。
希望本文能为学生们学习和复习 Unit 6 提供一
些参考和帮助。
2021年第4期252020美国数学竞赛(八年级)中图分类号:G 424.79文献标识码:A文章编号:1005 - 6416(2021 )04 - 0025 - 061. 卢卡在一次学校集资活动中卖自己制作的柠檬水.制作的过程中,他加入的水是糖 浆的4倍,糖浆是柠檬汁的2倍.他使用了 3 杯柠檬汁,则他能制作( )杯柠檬水.(A )6 (B )8 (C )12 (D )18 (E )242. 四位好友在周末为邻居整理庭院,分别获得15美元、20美元、25美元和40美元 的报酬.他们决定平分他们的酬劳,则酬劳为 40美元的人一共需要给其他人()美元•(A )5 (B )10 (C )15(D )20(E )253.凯瑞在6 x 8平方英尺的花园中全部 种植草莓.凯瑞能够在一平方英尺上种植4 株草莓,并且每株草莓平均结10颗草莓.则 预计她能收获()颗草莓•(A )560 (B )960 (C )1 120(D)l 920(E )3 8404.如图1,有三个边长依次增大的正六 边形.在原有正六边形点阵的基础上增多一 层,形成下一层正六边 _形内的点阵.则第四层 正六边形内有( )个点.(A ) 35(B ) 37(C ) 39(D ) 43(E ) 49图l5. —个容器内装有$的菠萝汁,将这些的果汁量是容器容量的百分之().(A )5 (B )10 (C )15(D )20(E )256J 、fi、C 、£>、£坐在一辆共有5节车厢 的小火车上,每节车厢只能坐一个人.已知£» 坐在最后一节车厢,4紧挨着£但在£后面, S 坐在4的前面车厢,在C 和fi之间至少间 隔一个人.则()坐在正中间的位置.(A)/l (B)fi (C )C (D )Z) (E )E 7.在2 020到2 400之间有( )个整数,其四位数字顺序是严格递增的,且各不相同(如2 357是满足要求的一个整数).(A )9 (B )10 (C )15(D )20(E )288•里卡尔多有2 020枚硬币,其中,一些是1分的,剩下的是5分的,且至少有一 枚1分的,也至少有一枚5分的.则里卡尔 多可能拥有的最大金额与最小金额的差是 ()分.(A )8 062 (B )8 068 (C )8 072(D )8 076(E )8 0829.阿卡什的生日蛋糕 是一个4 x 4 x 4的正方 体,如图2.蛋糕的顶部和 四个侧面都涂有糖霜,底 部没有糖霜.若将蛋糕切 成64个1 x 1 x 1的小正方体,则恰有两个面都有 糖霜的小正方体蛋糕有((A )12 (B )16 (C )18菠萝汁平均倒人5个空杯子.则每个杯子中(D )20 (E)24j . ! = 1 x 2x ,..x n •若 12x ;V !,则iV的值为(C )12(A )6 (D )2012.对于正整数i正整数Af满足5!x 9!:()•(A )10 (B)ll (D )13(E )1413. 贾马尔的一个抽屉里有6只绿色的袜子,18只紫色的袜子和12只橘色的袜子. 现又添加了一些紫色的袜子后,发现此时从 抽屉里随机抽取一只袜子是紫色的概率为 60% .则贾马尔一共添加了()只袜子.(A )6 (B )9 (C )12 (D )18 (E )2414.牛顿郡有20座城市,这些城市的人口数如图4.所有城市的人口数的平均数是 图4中的水平虚线.则这20座城市的总人数10. 扎拉搜集了 4颗弹珠,分别是玛瑙石的、大黄蜂宝石的、钢的和虎皮石的.她要在 一个架子上将这4颗弹珠排成一排展示,但 是不让钢弹珠和虎皮石弹珠挨在一起.则有 ()种排法.(A )6 (B )8 (C )12 (D )18 (E )2411. 放学后,玛雅和内奥米前往6英里处的一个海滩.玛雅骑自行车去,内奥米坐公交 车去.图3显示了他们行程中的时间与路程 关系.则内奥米和玛雅两人的平均速度的差 值为()英里/小时•最接近选项(Q-<8 0006 0004 0002 000城市图4(A )65 000 (B )75 000 (C )85 000(D )95 000 (E ) 105 00015. 若;c 的15%等于y 的20%,则;k 是工的百分之().(A )5(B )35(C )75(D )133y (E )30016. 如图5,4,5,…,F 六个点代表1,2,…,6这六个不同的数字.五条直线中的每一 条都经过其中的一些点.将每条直线上的点 对应的数相加,可以得到五个和数,且这五个 数之和为47•则点S 对应的数为().(A)l (B )2 (C )3 (D )4 (E )517•在2 020的所有约数中,有()个数的约数多于3个(如12有六个约数,分别为 1、2、3、4、6 和 12).(A )6 (B )7 (C )8 (D )9 (E )1018.如图6,矩形/IfiCD内接于半圆,为半圆的直径,似=16,/^=狀=9.则矩形的面积为()•)英里/小时.内奥米117111/1」1/11/111 ./5 10 15 20 25时间(分钟)图3(B )12 (C )18(E )2426中等数学(W 梂)被圏2021年第4期27^16FDA E 图6(A )240 (B )248 (C )256(D )264(E )27219. 若一个数是由两个不同数字交替组成,则称此数为“交替数”(如2 020、37 373 都是交替数,而3 883、123 123不是交替数)• 则有()个五位交替数能被15整除.(A )3 (B )4 (C )5 (D )6 (E )820. —位科学家走在一片森林中,她将一 排的5棵树的高度用整数记录下来.她注意 到每一棵树的高度要么是右边树高度的2 倍,要么是右边树高度的一半.不幸的是由于 下雨,她的记录 本上的一些数据 丢失了,其中,横 线处为丢失的数 据(如表1).基 于她的观察,她 能够重新确定这 些数.则这些树 的平均高度为 ()米.(A )22.2 (B )24.2 (C )33.2(D )35.2 (E )37.221.如图7,一个棋盘上有64个黑白相间的方格,格分别位于最下面一行、最上面一行.一枚棋 子从P 处开始移动,每一次移动都是移动 到所在格的上面一行 相邻的白色格内.则 图7一共有()种不同的路径,使得恰用七步,棋子从P 移动到<?(图中是其中一种移表1第一颗树米第二棵树11米第三颗树米第四棵树米第五棵树米平均高度• 2米动的方式).(A )28(B )30 (C )32(D )33(E )3522.将正整数/V 输人到图8中的一台机 器中,机器按照程序输出一个正整数(如八= 7,则机器输出3x 7+1 =22).N若yv 为偶数/V"2若/V 为奇数3/V +1图8现将输出的结果继续输人到机器中,一 直重复五次,则输出为26,即7 — 22 — 11 — 34 -»■ 1752 — 26.若输人/V 值,进行六次操作,最终得到1,N —> _ —► _ —> _ —> _ —y _ —► 1则所有这些正整数iv的和为()_(A )73 (B )74 (C )75(D )82(E )8323. 将五个不同的奖品分给三名学生,每名学生至少得到一个奖品•则共有( )种不同的分配方法.(A )120 (B )150 (C )180(D )210(E )24024.在一个大正方形的区域上铺有r a 2块灰色方砖,每一块方砖的边长均为s,每块方 砖周围均有宽度为d的边.图9是n = 3的 情况•当n = 24时,576块方砖覆盖了整 个大正方形区域的64%.则此时4的值S为().(A )蠢(B )16(C )笞(D )I (E 4□ □□□ □□ □ □□图928中等数学25.矩形、/?2和正方形S,、S2、S3按照图10方式摆放成一个长为3 322、宽为2 020的矩形•则S2的边长为()•S,r2s2S,(A)651 (D)662m i〇(B)655(E)666(C)656参考答案1.E.由已知条件得7jC:糖楽:柠檬汁=4x2: 2:1.故卢卡一共制作了 3 x8 =24杯柠檬水.2. C.四位朋友一共获得报酬15 +20 +25 +40 = 100(美元),平分后每人获得25美元.因此,酬劳为40美元的人需要拿出15 美元.3. D.由题意,知凯瑞的花园为48平方英尺,能种植48 x4 = 192株草莓•因此,预计收获 192 x10 = 1 920 颗草莓.4. B.设第〃个六边形内有、个点.则"1=1,hn=hn-i +6(n - l) (n€Z+,n^2).故 /i2 = 1 + 6 = 7,/i3 = 7+12= 19,/i4 =19 + 18 =37.5. C.每个杯子里的果汁占容器容量的比例为3_ 1_ 3 _ 15=20=100'6. A.将车厢的顺序记为□□□□□,其中,最左边表示第一节车厢,最右边表示最后一节车厢.由第一个条件,知□□□□/).由第二个条件,知必为如下三种情况之一:□DEAD, \JEAUD,EAUBD.(1)由fi在4前,知排列为和,均不满足最后一个条件;(2) 得排列满足所有条件;(3) 显然不满足fi在4前.因此,唯一满足条件的顺序是B£4C£>,即坐在中间位置的是儿7. C.由于各位数字是严格递增的,于是,第二位数字不能为1和2.又四位数不能大于2 400,则第二位数字不能为4.从而,第二位数字只能为3.故后两位只能选自集合14,5,6,7,8,9丨.6x5因此,共有G1x215个.8. C.显然,里卡尔多拥有的5分最多,则他拥有的金额就最大,此时,有2 019x5+1分.类似地,他拥有的1分最多,则他拥有的金额就最小,此时,有2 019 x 1 +5分.因此,他能拥有的最大金额和最小金额的差为(2 019 x5 +1)-(2 019 xl +5)=2 019 x4 -4=2018 x4 =8 072.9. D.注意到,对于不含底面的小正方体,每一条棱的中间两块均恰有两个面有糖霜,一共有12 -4 =8条这样的棱•于是,有8 x2 = 16个满足要求的小正方体.当含有底面时,4个角的小正方体满足要求.2021年第4期29因此,共有16 +4 =20个小正方体蛋糕恰有两个面有糖霜.10. C.记s、:r分别表示钢的和虎皮石的,先排由题意,知这五个和数分别为4+ fi+ c、A+E + f\C + D + E、B + D、B+F.将其相加得2A +3B+2C+2D+2E+2F = 47s和r.因为s、;r不能相邻,所以,只能是snrn,snnr,nsnr,以及交换S、r的情况,共有3 x 2种方式.另外两个位置只有2种方式.因此,一共有2 x6 = 12种不同的排法.11.E.由图3,知内奥米在这6英里中用了 10分钟,于是,她的速度为+=36英里/小时.6"类似地,玛雅的速度为12英里/小时.故平均速度差为36 -12 =24英里/小时.12. A.注意到,5! =120.则 120 x9! =12 x/V!=> /V! =10 x9! =10!4 /V = 10.=?>2(A+B + C + D+E + F) + B =47.又4到F是1到6这六个不同的数,则A +B +C +D +E +F = 1+2 + •■• +6 =21.故 5=47 -21 x2 =5.17. B.由 2 020 = 22 x 5 x 101,知 2 020 有 12 个 约数,分别为 1、2、4、5、10、20、101、202、404、505、1010、2 020,其中,1、2、4、5、101的约数个数不大于3.故满足要求的约数有12 -5 =7个.18.A.记半圆圆心为〇,其直径为9 + 16+9 =34.于是,OC= 17.由对称性,得〇为A4的中点.贝I J 0/) = 〇4 =8.在Rt A〇况中,由勾股定理得13. B.当贾马尔添加了 x只紫色袜子之后,抽 屉里共有18 + x只紫色袜子,于是,袜子总数 为6+18 +12 + x= 36 + a:.这表明,他任意选取一只袜子是紫色的概率为.36 +x14. D.由图4,知虚线位于4 500和5 000正 中,即4 750处.于是,20座城市的总人数约 为 4 750 x 20 =95 000.15.C.由已知有〇. 15尤15 75:20 =1〇0 ':0. 20y.则LX 16. E.CD =V l72 -82= 15.故矩形的面积为16 x15 =240.19. B.一个数若能被5和3同时整除,则这个 数能被15整除.于是,这个数的个位数为5 或〇,且各位数字之和为3的倍数.若这个数的个位数字为〇,则由于此数为五位数,故首位也为0,显然不可能.从而,个位数必然为5,且这个数为5口5口5.记未知的数字为1则5 +1+5 + i+ 5=0(mod 3)=> 2^=0(mod 3).因为2与3互素,所以,$为3的倍数,即*只能取〇、3、6、9.故只有 50 505、53 535、56 565、59 595 这 四个数满足题意.30中等数学20. B.注意到,每棵树的高度均为整数.由第二棵树的高度为11米,知第一棵与 第三棵树的高度之和必为22米.下面考虑第四棵树的高度.它不可能为11米,否则,第五棵树只能 为22米,于是,五棵树的高度之和的平均值 的末尾不为0.2.从而,第四棵树的高度为44米.此时,前 四棵树高度之和为99,因此,第五棵树的高度不能为88,而只能为22米.故五棵树的平均高度为22^99 = 1|1=24_2(^)_21. A.除了位置P,棋子的每一个位置都是从它下面一行与之相邻的两个位置中的一个移 动过来的.这表明,从P移动到这个位置的方法数,就是移动到它下面一行与之相邻两格的方法数之和.如图11的白格中数值即表示移动到这个格的方法数.图11从而,移动到位置<?共有28种不同方法(即不同路径).22. E.从1开始逆向推导.考虑所有可能的情况.则|1(->|2}-^|4}->|1,8|^|2,16}-> |4,5,32|-^| 1,8,10,64}.例如,对于2,一定是由4得到(因为没有正整数使得3n +1 = 2),但16可以由32 或5得到(因为^=3 x5 +1 =16).通过构造,最后一个集合就是通过六次操作得到1的所有数组成的集合.因此,所求结果为1 +8 + 10 +64 =83.23. B.若没有“每名学生至少得到一个奖品”的限制,一共有35 =243种不同分配方法.下面考虑其中不满足要求的情况,即至 少有一个人没有奖品的情况.若某名学生没有得到奖品,则有25 =32 种方式.因为是三名学生,所以,共有3 x32 =96 种,其中,恰有两名学生没有得到奖品的情况 被重复计数,而某两名学生没有得到奖品的方法只有1种,三名学生中选定两名学生有3种方法.从而,有1 x3 =3种方法•因此,满足题目要求的方法种数为243 -96 + 3 =150.24. E.由题意,知阴影部分的面积为(2如)2.注意到,没有被覆盖的部分有23 +2 = 25行和25列.于是,整个大正方形的边长为2知+251则(245)2{2As+25d)264100=>245 _ 8 _ 424s+25d= \0=~5=>120s =965+ 100^d67 =25'25. A.记正方形足的边长为~则5, + 52 + s3 =3 322.类似地 h_ s2 + s3= 2 020.两式相减得2s2=1 302 s2 =651.(吴建平提供张广民翻译)。
食品研究与开发F ood Research And Development圆园20年8月第41卷第15期DOI :10.12161/j.issn.1005-6521.2020.15.015作者简介:王兵琦(1994—),女(汉),在读硕士,研究方向:农产品加工及贮藏。
*通信作者:郎秀杰(1989—),女,助教,硕士,研究方向:农产品加工及贮藏;郭成宇(1959—),女,教授,本科,研究方向:农产品加工及贮藏。
速冻沙果营养成分分析及其饮料配方的优化王兵琦,郎秀杰*,郭成宇*,孔凡品,马吉瑶(齐齐哈尔大学食品与生物工程学院,黑龙江省果蔬杂粮饮品工程技术研究中心,黑龙江齐齐哈尔161006)摘要:对速冻沙果和土右旗沙果、123苹果、沈农2号苹果、嘎啦、红勋5号和新疆2号的营养成分对比分析。
结果表明,速冻沙果含有较为丰富的维生素C 、蛋白质、锌、灰分和烟酸;以速冻沙果为原料,采用单因素试验和响应面分析方法对速冻沙果饮料的最佳生产配方进行研究,以感官评价和沉淀率为判断依据确定最佳的产品配方:果汁添加量为30%、柠檬酸添加量为0.2%、白砂糖添加量为7%、黄原胶添加量为0.4%,此时产品口感饱满,风味醇厚,有良好的稳定体系。
最终对产品的流变性、粒径分布和营养成分进行检测分析,最终获得的产品中含有丰富的硒元素(1.79mg/kg )和烟酸(250mg/L )。
关键词:速冻沙果;配方;营养成分;饮料;响应面Analysis of the Influencing Factors and Nutritional Components of the Basic Formula ofQuick-frozen Sand Fruit BeverageWANG Bing-qi ,LANG Xiu-jie *,GUO Cheng-yu *,KONG Fan-pin ,MA Ji-yao(College of Food and Biological Engineering ,Qiqihar University ,Beverage Engineering Technological Research Center of Fruit-vegetables and Coarse Cereals of Heilongjiang Province ,Qiqihar 161006,Heilongjiang ,China )Abstract :The quality of quick -frozen sand fruit and the nutritional composition of sand fruit in Tuyouqi County ,123apple ,shennong 2apple ,gala ,hongxun 5and Xinjiang 2were compared.The results showed that the quick -frozen fruit was rich in vitamin C ,protein ,zinc ,ash and niacin.With frozen sand fruit as raw material ,single factor test and response surface analysis were used to study the best production process ,parameters and optimal production formula of quick-frozen fruit beverage ,the optimum process parameters andproduct formulation were determined based on sensory evaluation and precipitation rate ,the juice content was 30%,the amount of citric acid added was 0.2%,the amount of sugar added was 7%,and the amount ofxanthan gum was 0.4%,at this time ,the product tastes full ,the flavor was mellow ,and there was a good stability system.Finally ,the rheological properties ,particle size distribution and nutritional composition of the product were analyzed.The final product was rich in selenium (1.79mg/kg )and niacin (250mg/L ).Key words :quick-frozen sand fruit ;formula ;nutritional composition ;drinks ;response surface引文格式:王兵琦,郎秀杰,郭成宇,等.速冻沙果营养成分分析及其饮料配方的优化[J].食品研究与开发,2020,41(15):80-87WANG Bingqi ,LANG Xiujie ,GUO Chengyu ,et al.Analysis of the Influencing Factors and Nutritional Components ofthe Basic Formula of Quick-frozen Sand Fruit Beverage[J].Food Research and Development ,2020,41(15):80-87应用技术80沙果(Malus asiatica Nakai),呈现扁圆形,直径4cm~5cm,黄色或红色,味道有些酸涩,含有丰富的维生素、有机酸[1],具有除烦消食、积滞和化、止渴生津、明目泄火、驱虫杀虫、解毒的作用。
巧克力西葫芦快手面包丨健康烘焙
Quick Bread系列。
双重巧克力,软糯湿润,和以前的快手面包一样惊喜,全麦也能
做出蛋糕质感。
西葫芦起湿润作用,味道上完全没有存在感。
夏天西葫芦大丰收,是美国人吃西葫芦快手面包的季节。
美国人爱吃西葫芦面包,因为此面包非常符合他们对甜点的要求:moist.
地道的美式做法基本都是糖油各200克起步。
我婆婆做的西葫芦面包,看起来不起眼,棕乎乎的马芬样,第一
次吃觉得太好吃了,连吃了3个,后来自己做饭后看见配料,发现难
怪以为在吃蛋糕。
这个改良版本同样好吃,只靠香蕉和黑巧的甜度。
酸奶、香蕉、西葫芦都让面包湿润得不像全麦做的。
【甜度】类似65%左右的黑巧,喜欢甜口的人,可以在干性材料里增加20-50克红糖。
或吃时抹一些蜂蜜、果酱。
————————————【计量单位】标准烘焙量杯、量勺单
位
1杯 = 240ml(毫升)
1大勺 =1 tablespoon = 15ml(毫升)
半大勺=0.5 tablespoon=7.5ml(毫升)。
The New American PlateComfort FoodsWhat Are Comfort Foods? The New American PlateProportionPortion Size Adjusting Comfort Foodsfor Health Sensational SubstitutionsRecipes3 4 5 6 10 12 14Comfort Foods Table of ContentsMore than ever, Americans who choose food for both taste and health are turning to AICR’s New American Plate. They’re filling their plates with two-thirds (or more) vegetables, fruits, whole grains and beans and one-third (or less) fish, poultry or red meat. They’ve heard that experts recommend a mostly plant-based diet to help reduce the risk of chronic diseases like cancer and to maintain a healthy weight.Traditionally, many comfort foods are high in calories and fat, laden with but-ter, cream, whole milk and cheese, and lacking in the nutrients and protective phytochemicals (unique plant substances) that vegetables and fruits have to offer. The good news is you don’t have tostop eating your favorite comfort foods to reap the health benefits of the New American Plate. These dishes just needa little remodeling to help you reach the “2⁄3 to 1⁄3” ratio.Start by making a few healthy adjust-ments to traditional recipes. This brochure contains 10 modified recipesfor some of your favorite comfort foods. Serve them in smaller portions, accompanied by an extra helping of vegetables. Try the suggested menus that precede each recipe to make these foods part of a health-protective meal. By mak-ing simple adjustments, eating for a healthy weight and a healthy life can be comforting, too.What Are Comfort Foods? Comfort foods can be defined as feel-good, hearty foods that are both nourishing and nurturing. They are frequently craved in moments of unhappiness, and, interestingly enough, during times of celebration. Com-fort foods are what we ate at grandma’s house, after a long day at school or what mom served when we were sick. As adults, we relish flavors from the comforting past. These foods take us back to a time when life was easier and someone else made the hard decisions.Besides the nostalgic feelings they evoke, it’s the textures and mouthfeel that make comfort foods so appealing. They are generally characterized by moist, creamy, soft, mashed, rich or still-warm textures, and are known for having a relatively high fat content.Age, regional origin and ethnic background all have a bearing on which items people consider comfort foods. Many Americans include foods like macaroni and cheese, beef stew, chicken soup, chili, meatloaf, mashed potatoes, pizza, spaghetti, choc-olate chip cookies and rice pudding. Since many of these foods come from a time when the relationship between diet and disease was not well known, these foods are often less than healthful. But rather than pass up the foods we crave, we can make simple adjustments to increase their nutritional value. The result: “com-forting” foods that fit well on the New American Plate.The New American Plate AICR and its affiliate, the W orld Cancer Research Fund in the U.K., assembledan expert panel of scientists to reviewthe existing research on the connec-tion between diet and cancer. The panel reviewed more than 4,500 studies con-ducted around the world and issued a landmark report entitled Food, Nutrition and the Prevention of Cancer: a global per-spective. This report clearly shows the link between a predominantly plant-baseddiet and reduced cancer risk. Eating more vegetables and fruits, exercising regularly and maintaining a healthy weight could cut cancer rates by 30 to 40 percent.The New American Plate is based on the recommendations from this expert report. It isn’t a short-term “diet” to use for quick weight loss, but a new approach to eating for better health. It emphasizes the kinds of foods that can significantly reduce our risk for disease. It also shows how to enjoy all foods in sensible portions. That is, it pro-motes a healthy weight as just one part of an overall healthy lifestyle.At the center of the New American Plate are a variety of vegetables, fruits, whole grains and beans. These plant-based foods are rich in protective substances that can help keep us in good health and reduce the risk of many types of cancer. They are also naturally low in calories. When plant-based foods fill our plates, we’re able to eat more filling and satisfying meals — all for fewer calories than the typical American diet.AICR Diet and Health Guidelines for Cancer Prevention1.Choose a diet rich in a variety ofplant-based foods.2. Eat plenty of vegetables and fruits.3. Maintain a healthy weight and bephysically active.4. Drink alcohol only in moderation,if at all.5. Select foods low in fat and salt.6. Prepare and store food safely.And always remember…Do not use tobacco in any form. ProportionThe traditional American plate contains a large piece of meat, a small serving of veg-etables and some form of potatoes or rice.This plate provides too many calories and too few nutrients to decrease disease risk or help us maintain a healthy weight. It certainly won’t help us reach the 5 to 10with reduced cancer risk.T o accomplish that, you have to change the proportion of foods on your plate. That is, you have to increase the variety of plant-based foods and decrease the amount of animal protein. The New American Plate aims for two-thirds (or more) vegetables, fruits, whole grains and beans and one-third (or less) fish, poultry or red meat. Portion SizeWhen it comes to reaching a healthy weight, it’s not just what you eat, but also how much you eat that matters. Choosing appropriate portion sizes is essential.Look at the chart on page 7 for a listof standard serving sizes from the U.S. Department of Agriculture (USDA). In contrast with what we usually eat, these serving sizes may seem remarkably small. For example, many people eat three cups of pasta at a sitting. Some restaurants are known to serve six or eight cups on a plate. Y et USDA defines a standard serving size of pasta as one-half cup.Try an experiment at your next meal. Mea-sure out your usual portion size onto a plate or bowl. Make a mental note of how much of your plate or bowl is covered by this portion.After checking the chart, measure out a standard serving size of the same food onto another plate or bowl. Compare the two plates. Ask yourself how many standard servings go into the portion you normally eat. If your weight is satisfactory, you areprobably eating the right number ofare overweight, the first thing you should consider is reducing the number of stan-dard servings in your regular portions. Decrease your portion size gradually so that you will be less likely to notice the change. Even small reductions add up to substantial health benefits. Remember, maintaining the right pro-portion of plant foods to animal foods is important to your long-term health. So reduce the portion sizes on your plate, but maintain the “2⁄3 to 1⁄3” proportion.A fad diet that has not stood up to rigorous scientific testing is no way to lose weight. Obesity became an epidemic in this country at the same time portion sizes grew enor-mous. It is likely you can reach a healthy weight on your own by simply reducing the size of the portions you eat and exercising more. If you still do not see your weight gradually moving in a healthy direction, contact your doctor or a registered dietitian for a more individualized plan.The bottom line is this: A diet that lowers cancer risk is also a diet that helps maintain a healthy weight.Researchers are finding growing evidence that overweight and obesity help increase risk for developing certain cancers. By fol-lowing the New American Plate advice to eat a mostly plant-based diet, get regular exercise and eat smaller portions, you can help prevent disease and keep your weightwithin a healthy range.3. Serve Smaller PortionsComfort foods are often served as “meals in themselves” – a bowl full of beef stew or a plate piled high with macaroni ’n cheese. T o help your favorite dishes fit the “2⁄3 to 1⁄3” ratio, serve a smaller portion and have a side salad and a serving of vegetables with your meal.4. Save Some Foods for Special OccasionsSome recipes may be impossible to modify while maintaining the same level of flavor and texture. For example, if you are looking for the taste and mouthfeel of a cheeseburger and fries, a turkey burger with fat-free cheese and oven-baked fries may not satisfy you. So, if necessary, enjoy traditional comfort foods less frequently and in smaller portions. You don’t need to completely give up eating the foods you love.1. Make SubstitutionsPrepare lowfat comfort foods by substituting one ingredient for another. Instead of using cream in a recipe, try evaporated skim milk or lowfat (1 percent) or reduced fat (2 per-cent) milk. If your recipe calls for butter, alternatives such as olive oil or light tub margarine may do the trick. Check AICR’s “Sensational Substitutions” chart on page 12 for more ideas on how to increase the nutri-tion and lower the fat content in your favorite recipes.2. Add Healthful IngredientsTry adding a few healthful ingredients such as vegetables, fruits, whole grains and beans to enhance the nutritional quality of the recipe. Add diced bell peppers and zucchini to your spaghetti sauce or a variety of beans to your chili recipe. The trick to preparing healthful comfort foods is to carefully adjust your recipe to preserve the same mouthfeel and flavor of the original while sneaking in extra health benefits.Adjusting Comfort Foods for HealthComfort foods don’t have to be unhealthy. Most can be made more nutritious without losing their appeal. Try these suggestions for adjusting your favorite recipes to fit theMacaroni ’n CheeseBoth nourishing and nurturing, macaroni and cheese tops the list of comfort foods. This creamy, fuss-free recipe is full of healthy ingredients.2 cups uncooked whole wheat elbow macaroni1 Tbsp. butter or margarine 1 onion, finely chopped 1 garlic clove, minced1 small red bell pepper, finely sliced 1 small green bell pepper, finely sliced 1 1⁄2 cups lowfat milk1⁄4cup grated Parmesan cheese 1 cup shredded reduced fat, sharp or extra-sharp cheddar cheese 1⁄2cup fat-free sour cream Salt and freshly ground black pepper, to taste1⁄2tsp. paprika In large saucepan, cook macaroni according to package directions. Drain and return to pan. Set aside. In large skillet, heat butter or margarine over medium heat; sauté onion and garlic until onion is translucent. Add bell peppers and sauté 2 more minutes, stir-ring constantly. Add to macaroni. In small bowl, combine milk, Parmesan, cheddar and sour cream. Add to macaroni and cook 10 minutes over low or medium heat, stir-ring constantly, until cheese is completely melted and macaroni is piping hot. Add salt and pepper to taste. Sprinkle with paprika to garnish.Makes 8 servings. Per serving: 198 calories, 6 g total fat (4 g saturated fat), 27 g carbohy-drates, 11 g protein, 3 g dietary fiber, 93 mg sodium.RecipesAICR has taken some of the recipes that have been treasured for generations and made them lower in fat, calories and sodium. W e’ve also added health-protec-tive plant-based ingredients, while still retaining the great taste that made these foods family favorites.pot, sauté onion, carrots, leeks and garlic, 5 minutes. Return beef to pot. Add toma-toes, tomato paste, broth and oregano. Add water; bring to boil. Reduce heat to low; simmer until beef is nearly tender, about 50 minutes. Add potatoes. Cover partially; simmer until beef and potatoes are tender. Add green beans and kale. Cook another 6-8 minutes. Season with salt and pepper, to taste.Makes 6 servings. Per serving: 200 calories, 5 g total fat (1 g saturated fat), 29 g carbohy-drates, 13 g protein, 6 g dietary fiber, 303 mg sodium.Chicken SoupWhat better way to chase away the blues than to indulge in this warm, homey treat? Y ou’ll be reminded of the days when mom or grandma could comfort your colds with homemade soups. This simple, healthier recipe will take off the chill on a rainy day.Beef StewHere is an easy and healthful recipe that can smooth out a bad day.1 Tbsp. olive oil1⁄2lb. beef stew meat, cut into 1-inch cubes1 large onion, diced2 medium carrots, sliced 1 cup diced leeks3 garlic cloves, chopped1 can (14 oz.) diced tomatoes 1 can (6 oz.) tomato paste1 can (14 oz.) fat-free, reduced sodium beef broth1-2 Tbsp. dried oregano 1 cup water2 medium potatoes, cubed1 package (10 oz.) frozen green beans 1 cup chopped kaleSalt and freshly ground black pepper, to taste In large nonstick pot, heat oil over medium-high heat. Add beef, sauté until brown, about 5 min-utes. Remove from pot and set aside. In sameRecipe continues on p. 20Start reshaping your diet by looking at your plate. Is the greater proportion of your meal plant–based? Are your portion sizes appropriate to your activity level? The reci-pes beginning on page 14 modify traditional comfort foods by adding health without sacrificing taste. Comfort foods can now have a place in a meal that is two-thirds vegetables, fruits, whole grains and beansComfort Foods2 skinless, boneless chicken breasts (about 8 oz.), cut into 1-inch pieces 1 bay leaf8 cups fat-free, reduced sodium chicken broth2 cups sliced celery1 1⁄2 cups chopped green onions2 cups sliced carrots2 garlic cloves, chopped1 cup sliced zucchini2 cups peeled, diced potatoes or cooked noodles1 tsp. minced fresh parsley1 tsp. snipped fresh chives1 1⁄2 tsp. ground coriander (optional)Freshly ground black pepper,to tasteIn soup pot or deep pan, combine chicken, bay leaf, broth, celery, green onions, carrots, garlic, zucchini and potatoes, if using. Bring to boil. Reduce heat and let simmer about 20 minutes or until chicken and vegetables are tender. Just before serving, remove bay leaf and add parsley, chives, coriander and noodles, if using. Season with pepper, to taste.Makes 8 servings. Per serving: 94 calories, <1 g total fat (<1 g saturated fat), 13 g carbohy-drates, 10 g protein, 3 g dietary fiber, 636 mg sodium.Note: Y ou can also make chicken and vegetable soup using a whole chicken. In this case, boil whole chicken and vegetables in 8 cups of water. When chicken is tender, remove from pot and cool. Discard chicken bones and skin. Chop meat into chunks and return to pot. Refrigerate for 24 hours. Before serving, carefully remove fat, season with fresh herbs and heat through.This simple and rewarding dish will allow you to enjoy your favorite fare with the New American Plate in mind.1 Tbsp. olive oil1 medium yellow onion, chopped1 medium green bell pepper, cut in1⁄2-inch pieces1 Tbsp. finely chopped garlic1-3 jalapeño peppers, seededand minced1 Tbsp. ground cumin2 tsp. ground ancho chile, or1 Tbsp. chili powder1 tsp. dried oregano2 cans (15 oz. each) pinto beans,rinsed and drained1 cup canned diced tomatoes2 cups vegetable broth, divided2 Tbsp. masa or cornmealSalt and freshly ground black pepper,to taste1⁄4 cup chopped cilantro (optional)MeatloafOften called “the mother of all comfort foods,” meatloaf is a hearty dish that has been cherished by American families for many generations. T o transform it, we substituted 1⁄2 pound of ground turkey and 1⁄2pound of ground turkey breast for the usual 1 pound of ground chuck, and added lots of chopped veggies. The adaptedrecipe turned out beautifully, with a savory yet sweet taste. The same mixture could be used to make a juicy burger – another comfort food favorite. Serve it on a whole wheat bun.1⁄2 pound ground turkey breast 1⁄2 pound ground turkey 1⁄3 cup ketchup1 cup unseasoned breadcrumbs, preferably whole wheat 3⁄4cup finely chopped onion 1 tsp. dried basil 2 tsp. dried oregano 2 garlic cloves, minced 1 large egg1⁄2cup shredded carrots 1⁄4cup chopped fresh parsley 1 1⁄4 cups green bell pepper, minced 1⁄4cup red bell pepper, minced Salt and freshly ground black pepper, to taste3 Tbsp. ketchup (optional topping)Preheat oven to 350 degrees. In large bowl, combine all ingredients, except for extra ketchup. Place mixture in 9×5-inch non-stick loaf pan. Bake 1 hour, uncovered. Let stand 10 minutes before serving. Spread extra ketchup on top, if desired. Cut into slices and serve.In Dutch oven, heat oil over medium-high heat. Sauté onion, bell pepper and garlic until onion is translucent, about 4 minutes. Add jalapeño pepper, cumin, ancho chile or chili powder and oregano. Stir until spices are fragrant, about 1 minute. T ake care not to let them burn. Add beans, tomatoes and all but 3 tablespoons of vegetable broth. Bring chili to boil. Reduce heat and sim-mer, uncovered, 10 minutes. Meanwhile, in small bowl, combine masa or cornmeal and remaining broth, stirring to make a smooth mixture. Add to chili, blending well. Stir frequently to prevent sticking. Add salt and pepper, to taste. Simmer 10 more minutes. For best flavor, refrigerate 1-2 hours. Before serving, reheat and sprinkle cilantro on top, if desired.Makes 5 servings. Per serving: 227 calories, 5 g total fat (<1 g saturated fat), 35 g carbohy-drates, 12 g protein, 12 g dietary fiber, 574 mg sodium.Makes 5 servings. Per serving: 276 calories, 9 g total fat (2 g saturated fat), 25 g carbo-hydrates, 23 g protein, 2 g dietary fiber, 463 mg sodium.The changes here are minimal but signifi-cant. Instead of using whole milk we used nonfat milk, which offers the same fluffy mashed potato texture, yet fewer calories and less fat per serving. Soft tub margarine is lower in unhealthful saturated fat and trans-fat than stick varieties. Just watch that portion size.4 medium russet potatoes, scrubbed and quartered1⁄2cup hot skim milk 1 1⁄2 Tbsp. soft tub margarine or whipped butterSalt and freshly ground black pepper, to tasteIn large saucepan, place potatoes with enough water to cover. Bring to boil. Reduce heat and simmer until potatoes are tender, about 25 minutes. Drain well. Transfer potatoes to bowl. Using an electric mixer or potato masher, mash potatoes, gradually adding hot milk and margarine orbutter until smooth and fluffy. Add salt and pepper, to taste. Serve immediately.Makes 6 servings. Per serving: 121 calories, 3 g total fat (<1 g saturated fat), 22 g car-bohydrates, 3 g protein, 2 g dietary fiber, 55 mg sodiumBubbling cheese, golden crust and tender vegetables: pizza is one of the foods that best satisfies cravings. If you have the time to make a fiber-rich whole wheat dough from scratch, here is a trouble-free recipe. Or use whole wheat dough mix, which is available in many specialty food stores.Crust:1 1⁄2 tsp. dry yeast3⁄4cup warm water (105-115 degrees)1 cup whole wheat flour 1 cup all-purpose flour 1⁄4tsp. salt 1 Tbsp. olive oilcooked crust; top with vegetable mixture. Sprinkle with Parmesan and sage. Bake 20 minutes more or until crust is golden brown.Makes 8 servings. Per serving: 178 calories, 4 g total fat (1 g saturated fat), 29 g carbohy-drates, 8 g protein, 4 g dietary fiber, 339 mg sodium.This meatless sauce bears a resemblance to old-fashioned Italian spaghetti sauce due to the sweetness of mixed vegetables. Canned tomatoes offer convenience and the garlic and onions create a zesty aroma. 1 Tbsp. olive oil1 medium onion, chopped1⁄4cup chopped green bell pepper 1 medium zucchini, chopped 1 medium carrot, grated 1 Tbsp. dried oregano 1 Tbsp. dried basilT opping:Nonstick cooking spray 1⁄2cup water 1⁄4cup chopped onion 1⁄2Tbsp. dried thyme 1⁄2Tbsp. dried basil 4 cups sliced mushrooms 2 cups chopped zucchini 1 red bell pepper, choppedSalt and freshly ground black pepper, to taste1 cup ready-made pizza sauce or pasta sauce1⁄2cup grated Parmesan cheese 1⁄2Tbsp. chopped fresh sage In small bowl, dissolve yeast in warm water. In large bowl, combine flours and salt. Add yeast mixture and stir until it forms a ball (adding a bit more water if necessary). Place dough on lightly floured surface and knead, adding more flour if necessary, 3-5 minutes until smooth and elastic. In large bowl coated with oil, place dough and turn it to oil the top. Cover and let rise in warm place, free from drafts, until doubled in bulk, about 1 1⁄2 hours. Meanwhile, spray large nonstick skillet and heat on medium-high heat. Add water, onion, thyme, basil and mushrooms and bring to boil. Add zucchini and bell pepper; reduce heat, and simmer 2 minutes. Season with salt and pepper, to taste. Discard liquid and set vegetables aside. Preheat oven to 350 degrees. Punch dough down; roll into 12-inch circle. Place dough on large sprayed baking sheet and bake 7-10 minutes (make sure baking sheet is not too close to bottom of oven). Remove from oven and raise temperature to 400 degrees. Spread pizza sauce on pre-Our chocolate chip cookies have delighted everyone who has tried them. Try these treats with a tall glass of lowfat or nonfat milk. And remember that although these are probably lower in fat and calories than your usual recipe, they still aren’t low in calories and fat – so portion size matters!Nonstick cooking spray 1⁄4cup packed brown sugar 3 Tbsp. granulated sugar 3 Tbsp. butter or margarine 1⁄2tsp. vanilla extract 1 large egg 3 Tbsp. water1 Tbsp. dried thyme1⁄2cup fat-free, reduced sodium vegetable or chicken broth2 cans (28 oz. each) whole peeled tomatoes, drained and crushed8 oz. spaghetti, preferably whole wheat 7 garlic cloves, chopped 1 cup chopped mushrooms 2 Tbsp. tomato pasteSalt and freshly ground black pepper, to taste2 Tbsp. freshly grated Parmesan cheese (optional)In heavy saucepan, heat oil over medium heat. Add onion, green pepper, zucchini and carrot and sauté 2 minutes. Add orega-no, basil, thyme and broth. Stir in tomatoes. Reduce heat to low and simmer, uncovered, 40 minutes, stirring occasionally. Mean-while, cook spaghetti according to package directions. T o vegetable mixture, add garlic, mushrooms, tomato paste, salt and black pepper, to taste. Raise heat to medium, cover and cook 10 minutes. When spa-ghetti is done, drain and transfer to warm serving bowl. Add sauce and toss. Garnish with cheese, if desired.Makes 6 servings. Per serving: 254 calories, 3 g total fat (<1 g saturated fat), 48 g car-bohydrates, 9 g protein, 5 g dietary fiber, 575 mg sodium.Rice PuddingRemember this heartening, creamy dessert from your childhood? Cinnamon, apples and pears add a distinctive fruity flavor. It might take some time to prepare, but it is worth the wait.4 cups lowfat milk 1 cup brown rice 2-3 cinnamon sticks Pinch of salt 1⁄4cup sugar 1 tsp. vanilla extract Pinch of nutmeg1⁄4cup nonfat vanilla yogurt 2 apples, peeled and diced 1 pear, peeled and dicedGround cinnamon (optional)3⁄4 cup all-purpose flour1⁄2 cup “white” whole-wheat flour(see note)1⁄3cup toasted wheat germ 3⁄4tsp. baking soda 1⁄4tsp. salt 1⁄2cup semisweet mini chocolate chips Preheat oven to 350 degrees. Lightlyspray baking sheet with nonstick spray. In medium bowl, cream sugars with butter or margarine. Stir in vanilla, egg and water. Sift together flours, toasted wheat germ, bak-ing soda and salt; stir into creamed mixture. Stir in mini chocolate chips. Drop dough by heaping teaspoonfuls onto baking sheet and flatten slightly with fork. Bake 10-12 minutes. Allow cookies to cool for a few minutes on baking sheet before removing to cool completely on wire racks.Makes 24 cookies. Per cookie: 73 calories, 3 g total fat (1 g saturated fat), 12 g carbohy-drates, 2 g protein, 1 g dietary fiber, 68 mg sodium.Note: “White” whole wheat flour has a milder flavor than regular whole wheat flour. It is available in most supermarkets and whole food markets.cinnamon sticks and salt to simmer. Reduce heat to low. Cover and gently simmeruntil rice is very tender and milk is almost absorbed, stirring occasionally (about 1 hour). Add sugar, vanilla and nutmeg and stir to blend over low heat until mixtureis very thick (about 15 minutes). Remove cinnamon sticks. Stir yogurt and 3⁄4 of fruit into rice pudding. Transfer to large bowl.T op with remaining fruit and sprinkle with ground cinnamon, if desired. Serve warm. Makes 8 servings. Per serving: 202 calories,2 g total fat (1 g saturated fat), 40 g carbo-hydrates, 7 g protein,3 g dietary fiber,106 mg sodium.Note: For softer consistency, add more milk andcook longer.Request additional brochures: (single copies free)• Simple Steps to Prevent Cancer• Moving T oward a Plant-Based Diet • A Healthy W eight for LifeCall the toll-free Nutrition Hotline:Dial 1-800-843-8114 to leave a message for a registered dietitian, who will return your call. Monday-Friday, 9:00 a.m.–5:00 p.m. ET.For more delicious, healthy recipes, look for The New American Plate Cookbook (University of California Press) in your bookstore, avail-able March 2005. Another great source of AICR recipes is .Editorial Review CommitteeRitva Butrum, Ph.D.AICR Senior Science Advisor Karen Collins, M.S., R.D. Nutrition Consultant Elaine Feldman, M.D.Medical College of Georgia David Heber, M.D., Ph.D.UCLA Center for Human Nutrition Jan Kasofsky, Ph.D., R.D.Capital Area Human Services District, LouisianaLaurence Kolonel, M.D., Ph.D. University of HawaiiMelanie Polk, M.M.Sc., R.D., F.A.D.A. AICR Director of Nutrition Education AICR Executive StaffAbout AICRThe American Institute for Cancer Research is the third largest cancer charity in the U.S. and focuses exclusively on the link between diet and cancer. The Institute provides a wide range of education programs that help mil-lions of Americans learn to make changes for lower cancer risk. AICR also supports innovative research in cancer prevention and treatment at universities, hospitals and research centers across the U.S. The Institute has provided more than $70 million in fund-ing for research in diet, nutrition and cancer. AICR is a member of the W orld Cancer Research Fund International.Need More Help?For free publications or to make a memorial donation, please contact us.American Institute for Cancer Research1759 R Street NW, P.O. Box 97167 W ashington, DC 20090-7167 1-800-843-8114 or 202-328-7744 AICR’s message about proportion and portion size comes to you in a variety of vehicles:• Brochures: The New American Plate, One-Pot Meals, V eggies, Breakfast• NAP Serving Size Finder: single copy free • Small NAP Poster (81⁄2×11"): single copy free • Large NAP Poster (17×23"): $2.00 each • NAP Place mat (11×17"): $12.00 (set of four)All these materials make great teaching tools or healthy reminders for your home. T o order, call AICR toll-free at 1-800-843-8114.Bulk order discounts are available for health professionals.Y ou can help provide for future cancer research and education through a simple bequest in your will to the American Institute for Cancer Research. Consult with your attorney when first writing your will, or to add a simple paragraph to your existing will. Y our bequest to help in the war against cancer can be a cash amount, a gift of the remainder of your estate or a portion of the remainder, after obligations to your family and loved ones are met.Y our attorney will need to know:AICR’s official name:American Institute for Cancer Research AICR’s mailing address:1759R Street NW, W ashington, DC 20009 AICR’s telephone number:202-328-7744AICR’s identification:A not-for-profit organization under Section 501(c)(3) of the Internal Revenue Code AICR’s tax-exempt IRS number:52-1238026For further information, contact AICR’sGift Planning Department at 1-800-843-8114.Prepared by the American Institute for Cancer Research.Latest revision, November 2004.Copyright © 2002。
a rX iv:mat h /98441v1[mat h.AG ]7Apr1998CHERN CLASS FORMULAS FOR QUIVER V ARIETIES ANDERS SKOVSTED BUCH AND WILLIAM FULTON 1.Introduction Our goal in this paper is to prove a formula for the general degeneracy locus Ωr associated to an oriented quiver of type A n .If we are given a sequence of vector bundles and vector bundle maps E 0φ1−→E 1φ2−→E 2→···→E n −1φn −→E n on an algebraic variety X ,and a collection r =(r ij )0≤i<j ≤n of non-negative inte-gers,there is a degeneracy locus Ωr =Ωr (E •)=Ωr (E •,φ•)defined by Ωr ={x ∈X |rank E i (x )→E j (x ) ≤r ij ∀i <j }.(1.1)This is a closed subscheme of X ;locally,where the bundles are trivial,this is defined by vanishing of the minors of size r ij +1in the product of matrices giving the map φj ◦···◦φi +1from E i to E j ,for all i <j .Not all rank conditions give reasonable loci.Those that do—and the only ones we will consider—are characterized by the conditions r ij ≤r i,j −1and r ij ≤r i +1,j for all i <j ,and r i +1,j −1−r i,j −1−r i +1,j +r ij ≥0for all i <j −1,(1.2)where we set r ii =rank(E i ).In fact,rank conditions satisfying (1.2)are the only conditions that can actually occur ,i.e.for which one can have equality in (1.1).When the maps are sufficiently generic,each such Ωr is irreducible,of codimension d (r )= i<j (r i,j −1−r ij )(r i +1,j −r ij ).(1.3)When n =1,the formula for Ωr is the well-known Giambelli-Thom-Porteous formula,which we recall in order to introduce some notation.For a map φ:E →F of vector bundles of ranks e and f ,and a non-negative integer r ≤min(e,f ),Ωr is the locus where φhas rank at most r .The formula for Ωr is the Schur polynomials (e −r )f −r (F −E ),which is defined as follows.Define cohomology classes hi by the formulah i =c (E ∨)/c (F ∨),where c (E ∨)=1−c 1(E )+c 2(E )−···is the total Chern class,andthe division is carried out formally;in particular,h 0=1and h i =0for i <0.For any sequence λ=(λ1,...,λp )of non-negative integers,sets λ(F −E )=det(h λi +j −i )1≤i,j ≤p .In the Giambelli-Thom-Porteous formula,λ=(e −r )f −r denotes the sequence e −r repeated f −r times.In a Schur determinant s λ(F −E ),λwill usually be a2ANDERS SKOVSTED BUCH AND WILLIAM FULTONpartition,i.e.a weakly decreasing sequence,but later we will also need this notation whenλis not a partition.Our general formula for the locusΩr,when r is any set of rank conditions satisfying(1.2),has the formλcλ(r)sλ(E•),where the sum is over sequencesλ=(λ(1),λ(2),...,λ(n)),with eachλ(i)a parti-tion.The class sλ(E•)is defined to besλ(E•)=sλ(1)(E1−E0)·sλ(2)(E2−E1)·...·sλ(n)(E n−E n−1).The coefficients cλ(r)are certain integers for which we give an inductive formula.A second purpose of this paper is to introduce these integers cλ(r),which we regard as generalized Littlewood-Richardson coefficients.We have a conjectured formula for cλ(r)as the number of sequences(T1,T2,...,T n)of Young tableaux, with T i of shapeλ(i),satisfying certain conditions.This formula has been proved when the number of bundles is at most four,but it appears to be a difficult com-binatorial problem to prove it in general.In[8]a special case of this situation was studied,where the rank conditions are given by a permutation w.For mapsG1→G2→···→G m→F m→F m−1→···→F1with rank(G i)=rank(F i)=i,and w∈S m+1,letΩw={x∈X|rank(G q(x)→F p(x))≤r w(p,q)∀p,q≤m},where r w(p,q)=#{i≤p|w(i)≤q}.These loci are special cases of the lociΩr described in this paper.The formulas given here therefore specialize to the universal double Schubert polynomials S w(c•(F•);c•(G•))for these loci.Since these universal Schubert polynomials specialize to quantum and double Schubert polynomials([5], [13],[7]),we derive formulas for these important polynomials.These formulas appear to be new even for the single Schubert polynomials S w(x).Among the loci considered here are the varieties of complexes,which are the loci Ωr with r ij=0for j−i≥2.In this case the formula for the coefficients cλ(r)is particularly simple,and it agrees with our general conjectured formula.In Section2we discuss the lociΩr in more detail,state the main theorem,and derive the main applications.This includes a precise statement of what it means for a polynomial cλ(r)sλ(E•)to give a formula for a locusΩr.This statement implies the assertion that if X is non-singular andΩr has the expected codimension d(r),then(1.4)[Ωr]= λcλ(r)sλ(E•)in the Chow group A d(r)(X).However,weaker assertions can be made when X is singular or the mapsφi are less generic.At the end of Section3we sketch a generalization,which is based on explicit resolutions of singularities of these loci.The coefficients cλ(r)are determined by the geometry,if this assertion is inter-preted correctly.We will see in Section2that cλ(r)depends only on the differences r i,j−1−r ij and r i+1,j−r ij.This allows the ranks of the bundles E i to be taken large compared to the expected codimension d(r);if the Chern classes of the bundles are independent,the coefficients cλ(r)are then uniquely determined by(1.4).FORMULAS FOR QUIVER V ARIETIES3 Much of the work in a project of this kind—discovering the shape of the formula—is invisible in thefinal product,which has a short proof(given in Section3).Inparticular,it came as a pleasant surprise to us that the polynomials for all the loci Ωr can be written as a linear combination of the polynomials sλ(E•).We know of no reason for this other than the proof of the explicit formula.That the coefficients cλ(r)appear to be non-negative is even more surprising.The conjectured formula for the coefficients is discussed in more detail in the final Section4;proofs of the combinatorial assertions made there can be found in [4].We are particularly grateful to S.Fomin,who provided an involution on pairs of tableaux which gave us the strongest evidence for the conjectured formula,and who has collaborated with us on the combinatorial aspects of this problem.Thanks also to M.Haiman and M.Shimozono for their responses to combinatorial questions. The Schubert package[12]was useful for calculations.2.Quiver varieties;the theorem and applications2.1.The Main Theorem.Given vector bundles E0,...,E n on a variety X,letH be the direct sum of the bundles Hom(E i−1,E i),i.e.H=Hom(E0,E1)×X Hom(E1,E2)×X···×X Hom(E n−1,E n). Writing E i for the pullback of E i to H,we have a universal or tautological sequence of bundle mapsE0Φ1−→ E1Φ2−→ E2→···→ E n−1Φn−−→ E n(2.1)on H.For this universal case,it is a theorem of Lakshmibai and Magyar[14]that for r satisfying(1.2),the scheme Ωr=Ωr( E•)for(2.1)is reduced and irreducible, of codimension d(r).Moreover, Ωr is a Cohen-Macaulay variety if X is Cohen-Macaulay.(Earlier Abeasis,del Fra,and Kraft[2]had shown,in characteristic zero,that the reduced scheme( Ωr)red is Cohen-Macaulay.)Note that,when the bundles are trivial,H is a Cartesian product of X and a product M of spaces of matrices,and Ωr is a product of X with the corresponding locus in M;it is this locus in M that is studied in[1],[2],and[14].The statement that“the polynomial P= cλ(r)sλ(E•)is a formula for the locusΩr”has the usual meaning in intersection theory(cf.[6,§14],[10,App.A]). It implies that when X is non-singular and codim(Ωr,X)=d(r),then[Ωr]= λcλ(r)sλ(E•)in the Chow group A d(r)X,where[Ωr]is the cycle defined by the schemeΩr. For arbitrary X and mapsφi,there is a well defined cycle classΩr in the Chowgroup A m−d(r)(Ωr),where m=dim(X),whose image in A m−d(r)(X)is the class cλ(r)sλ(E•)∩[X].WheneverΩr has codimension d(r)in X,Ωr is a positive cycle supported onΩr;if X is Cohen-Macaulay,or more generally if depth(Ωr,X)=d(r), this cycle is[Ωr],but if X is not Cohen-Macaulay the coefficient of a component of Ωr inΩr may be smaller than the length ofΩr at its generic point.These classes Ωr are compatible with the basic constructions of intersection theory,exactly as in [6,Thm.14.3].4ANDERS SKOVSTED BUCH AND WILLIAM FULTON In fact,to give mapsφi:E i−1→E i for all i is the same as giving a section s:X→H of the bundle H,andΩr=s−1( Ωr).The general classΩr is constructed by intersecting Ωr⊂H with the(regular)embedding s:X→H,i.e.Ωr=s![ Ωr],where s!:A∗( Ωr)→A∗(Ωr)is the refined intersection[6,§6].As in[6,§14],the general properties of these classes follow from this construction.It therefore suffices to prove the corresponding formula on H,i.e.that[ Ωr]= λcλ(r)sλ( E•)∩[H]in A N−d(r)(H),where N=dim(H)=dim(X)+ n i=1e i−1e i,e i=rank(E i).It is natural to arrange the rank conditions in a triangular array:r00r11r22r33···r nnr01r12r23···r n−1,nr02r13···r n−2,nr03···r n−3,n...r0nIt is useful to replace each small trianglee froccurring in this array by the rectangle of width e−r and height f−r.FORMULAS FOR QUIVER V ARIETIES5 correspond to the rectangular array:6ANDERS SKOVSTED BUCH AND WILLIAM FULTONHere the sum is over all sequences(σ(1),...,σ(n−1))and(τ(1),...,τ(n−1)) of partitions,with|σ(i)|+|τ(i)|=|µ(i)|,such that the length ofσ(i)is at most the height of R i,i.e.ℓ(σ(i))≤e i−r i.Define I(i)to be(R i+σ(i),τ(i−1))for 1≤i≤n,whereτ(0)andσ(n)are taken to be the empty partition.One uses the rules just given to write each S(I(1),...,I(n))as0or±S(λ)for a unique λ=(λ(1),...,λ(n)),thus arriving at a polynomial cλ(r)S(λ). Main Theorem.The formula forΩr is cλ(r)sλ(E•).This theorem will be proved in the next section.Wefirst interpret it in the case where the rectangular array has only two non-empty rows,i.e.R ij=∅if j−i> 2.In this case the inductive polynomial dµS(µ)is just S(µ),for µ=(R02,R13,...,R n−2,n).For a rectangle shape R,the Littlewood-Richardson coefficient c Rσ,τvanishes unlessσand the180◦rotation ofτfit together to make R, in which case c Rσ,τ=1.FORMULAS FOR QUIVER V ARIETIES7 so the array of rectangles is8ANDERS SKOVSTED BUCH AND WILLIAM FULTONThe number of connections between the left column with q dots and the right column with p dots is the numberr w(p,q)=#{i≤p|w(i)≤q}.There are the maximal number of connections between two columns on the left or between two on the right.This means that for a sequence E•of bundle mapsG1→G2→···→G m→F m→F m−1→···→F1with rank(G i)=rank(F i)=i,the locusΩr(E•)defined in the introduction is exactly the locusΩw defined in[8],with the same scheme structure.In[8]“universal Schubert polynomials”S w(c•(F•);c•(G•))were constructed,which represent the lociΩw.From the fact that the formula for a locus is unique,we deduce the following corollary.Corollary2.With r determined by w as above,S w(c•(F•);c•(G•))= λcλ(r)sλ(E•).When these bundle maps are specialized so that each G i−1→G i is an inclusion of bundles,and each F i→F i−1is a surjection,then S w(c•(F•);c•(G•))becomes the double Schubert polynomialS w(x1,...,x m;y1,...,y m)of Lascoux and Sch¨u tzenberger;here we set x i=c1(ker(F i→F i−1))and y i= c1(G i/G i−1).The right side of the formula in this corollary also simplifies in this case.It follows from the definition that for a partitionτ,we havesτ(G i−G i−1)= (y i)q ifτ=(q),q≥00otherwiseandsτ(F i−1−F i)= (−x i)p ifτ=(1)p,p≥00otherwise.Thus sλ(E•)=0unlessλ=((q2),(q3),...,(q m),τ,(1)p m,...,(1)p2),in which case sλ(E•)=(−1)p2+···+p m x p22···x p m m y q22···y q m m sτ(x/y),where sτ(x/y)=det(hτi +j−i), h k= (1−y i)/ (1−x j).Our formula thereforewrites S w(x;y)as a signed sum of monomials in the x2,...,x m,y2,...,y m times Schur polynomials sτ(x/y).When all variables y i are set equal to zero,we have only the terms with q2=···=q m=0,and this writes the ordinary Schubert polynomial S w(x)=S w(x;0)as a signed sum of monomials in x2,...,x m times (symmetric)Schur polynomials sτ(x).Unlike the inductive construction of Schubert polynomials from high degree to low degree,our formulas are simplest for those of low degree.FORMULAS FOR QUIVER V ARIETIES9 For example,for w=3142∈S4,the corresponding array of rectangles is10ANDERS SKOVSTED BUCH AND WILLIAM FULTONLet G =G 0×X H ,with projection π:G →H .Let 0→A i →E i →Q i →0be the universal exact sequences on G 0,and hence also on G .Let Z ⊂G be the intersection of the zero-schemes of the canonical maps E i −1→E i →Q i ,i.e.Z =Z (E 0→Q 1)∩Z (E 1→Q 2)∩···∩Z (E n −1→Q n ).On Z we have maps E i −1→A i for 1≤i ≤n .Composing these with the inclusions A i −1⊂E i −1we get a sequence A •of bundles and bundle maps on Z :A 1→A 2→···→A n .Let ¯r denote the rank conditions obtained by omitting the top row of the trian-gular array for r ,and let Ω¯r (A •)⊂Z be the locus given by these maps and rankconditions.It is easy to see that Ω¯r (A •)is mapped into Ωr (E •)by π.Now Z is isomorphic to the bundle n i =1Hom(E i −1,A i )over G 0,and we have a canonical projectionρ:Z =n i =1Hom(E i −1,A i )−→n i =2Hom(A i −1,A i )=H ′.Denote by Ω′the universal locus Ω¯r (A •)of H ′.Then Ω¯r (A •)in Z is the inverseimage of Ω′by ρ.Since the maps on H ′are universal,it follows that Ω′is irreducible,and therefore Ω¯r (A •)is an irreducible subscheme of Z .Ω′Ωr (E •)G 0FORMULAS FOR QUIVER V ARIETIES 11Furthermore [Z ]= ni =1c top (Hom(E i −1,Q i ))∩[G ],so [Z ]=n i =1s R ′i (Q i −E i −1)∩[G ],where R ′i =(e i −1)e i −r i .Since π∗[Ω¯r (A •)]=[Ωr (E •)]by Lemma 1,we are therefore reduced to proving the identity π∗ µc µ(¯r )s µ(A •)·n i =1s R ′i (Q i −E i −1)∩[G ] = λc λ(r )s λ(E •)∩[H ].(3.1)For this we need the following basic Gysin formula of Pragacz [18,Prop.2.2],whose proof comes from [11],cf.[10,App.F].Lemma 2.Let E and F be vector bundles of ranks e and f on a variety X .Let 0≤d ≤min(e,f ).Let G =Gr(d,F )be the Grassmann bundle,with projection π:G →X and universal exact sequence 0→A →F →Q →0.Let q =f −d ,R =(e −d )q ,and R ′=(e )q .For any partitions λand µ,with λof length at most q ,and any α∈A ∗(X ),π∗(s R ′+λ(Q −E )s µ(A −E )∩π∗α)=s R +λ,µ(F −E )∩α.We also need the following special case of the factorization formula of Lascoux and Sch¨u tzenberger [16]and Berele and Reger [3],cf.[18].Lemma 3.Let E and F be vector bundles of ranks e and f .Let R =(e )f .Let λbe a partition of length at most f .Thens λ(F )s R (F −E )=s R +λ(F −E ).Note that this identity follows from Lemma 2.Finally we need the basic identity[17,§1.5]:Lemma 4.For bundles E 1,E 2,and E 3,and a partition µ,s µ(E 3−E 1)= c µστs σ(E 2−E 1)s τ(E 3−E 2),the sum over partitions σand τwith |σ|+|τ|=|µ|,with c µστthe Littlewood-Richardson coefficient.Now we can prove (3.1).First use Lemma 4to replace each factor s µ(i )(A i +1−A i )that occurs in each s µ(A •)on the left side of (3.1)by the sumc µ(i )σ(i ),τ(i )s σ(i )(E i −A i )s τ(i )(A i +1−E i )= c µ(i )σ(i ),τ(i )s σ(i )(Q i )s τ(i )(A i +1−E i ).Note that s σ(i )(Q i )=0if ℓ(σ(i ))>rank(Q i )=e i −r i .Next use Lemma 3toreplace each s σ(i )(Q i )·s R ′i (Q i −E i −1)in the result by s R ′i +σ(i )(Q i −E i −1).The left side of (3.1)becomesµc µ(¯r )σ(i ),τ(i ) n −1 i =1c µ(i )σ(i ),τ(i ) ·π∗n i =1s R ′i +σ(i )(Q i −E i −1)s τ(i −1)(A i −E i −1)∩[G ] .12ANDERS SKOVSTED BUCH AND WILLIAM FULTONFinally,n applications of Lemma 2yields π∗ n i =1s R ′i +σ(i )(Q i −E i −1)s τ(i −1)(A i −E i −1)∩[G ] =n i =1s R i +σ(i ),τ(i −1)(E i −E i −1)∩[H ],and this gives the required formulaµc µ(¯r )σ(i ),τ(i ) n −1 i =1c µ(i )σ(i ),τ(i ) n i =1s R i +σ(i ),τ(i −1)(E i −E i −1)∩[H ]=λc λ(r )s λ(E •)∩[H ].Although we have stated it for varieties over a field,the theorem (and its proof)extend readily to schemes of finite type over a regular base,as in [6,§20].3.3.A generalization.There is a generalization of the theorem which may be useful in its own right,and which gives some insight into the proof.(It is not needed in this paper.)Fix E 0,...,E n on X ,and r =(r ij )satisfying (1.2).Let H = Hom(E i −1,E i ),on which the tautological bundle maps are universal,and one has the universal locus Ωr ⊂H .Let π:F →H be the partial flag bundle parameterizing flags in each E j of ranks r 0j ,r 1j ,...,r j −1,j .Let E 0j ⊂E 1j ⊂···⊂E j −1,j ⊂E j denote the tautological flags of vector bundles on F ,1≤j ≤n .Let Z ⊂F be the locus on which the image of E i,j −1by the map E j −1→E j is contained in the subbundle E ij ,i.e.Z is the subscheme defined by the vanishing of all maps E i,j −1→E j /E ij for i <j .One sees as in Lemma 1that πmaps Z birationally onto Ωr .In fact,if X is non-singular,this construction gives a canonical resolution of singularities of the universal locus Ωr .It is easy to see that the class of Z is given by [Z ]=i<j z ij ,where z ij =c top (Hom(E i,j −1,E i +1,j /E ij )).Consider a path γthrough the triangular array for r ,going from r 00to r nn .The path must be a union of line segments between neighboring rank conditions,and it must intersect any vertical line at mostonce.FORMULAS FOR QUIVER V ARIETIES13 F(γ)corresponding to the rank conditions passed through by the path.(In the illus-tration,m=7,and the bundle sequence is E00,E01,E02,E12,E22,E23,E34,E44.) LetΩr(γ)⊂F(γ)be the subscheme defined by the conditions that each mapE i,j−1→E j/E ij vanishes for r ij on or above the path,and rank(E ip→E j)≤r ijfor r ip on or above the path and p≤j.The canonical maps F→F(γ)→X map Z birationally ontoΩr(γ)which in turn is mapped birationally ontoΩr.Ourgoal is to give a formula for the class ofΩr(γ).To do this,we define a formallinear combinationΦ(γ)of symbols S(λ),whereλis a sequence of partitions,one for each line segment inγ.The formula forΩr(γ)is obtained by replacing eachS(λ(1),...,λ(m))by m i=1sλ(i)(A i−A i−1),and multiplying the result by z ij, the product over all i,j such that r ij is on or aboveγ.We defineΦ(γ)inductively.Ifγis the lowest possible path,going from r00tor0n to r nn,thenΦ(γ)=S(∅,...,∅),where the empty partition∅is repeated2n times.Otherwise,we canfind a pathγ′that is equal toγexcept it goes lower at one place,in one of the following ways:γ′γCase1:14ANDERS SKOVSTED BUCH AND WILLIAM FULTONthat for r in a vertical line,and replacing each rectangle by its ing the basic identity that s λ(F −E )=s λ′(E ∨−F ∨),where λ′is the transpose of λ,we find thatc λ∨(r ∨)=c λ(r ),(4.1)where,if λ=(λ(1),...,λ(n )),we put λ∨=(λ(n )′,...,λ(1)′).It can happen that for some k ,all of the rank conditions rank(E i →E k )≤r ik and rank(E k →E j )≤r kj follow from other rank conditions.This happens when,in the rectangle diagram,all the rectangles on the two 45◦lines descending from position k are empty.FortheexampleG1→G 2→G 3→F 3→F 2→F 1considered at the end of Section 2,with rank conditions r coming from w =3142∈S 4,the bundles G 1and F 2are inessential in this way.If an inessential bundle E k is omitted,one has a shorter sequence E ′•:E 0→···→E k −1→E k +1→···→E n ,with the map E k −1→E k +1being φk +1◦φk ,and corresponding rank conditions r ′;the array of rectangles for r ′is obtained by omitting the 45◦lines of empty rectangles and moving all rectangles below up a row.For example,if G 1and F 2are omitted from the example,one has G 2→G 3→F 3→F 1,with rectangular array124311411123411214311214311123411123444444FORMULAS FOR QUIVER V ARIETIES15 entry of T ij must be strictly smaller than any entry of T kl if R kl lies in the wedge cut out by45◦lines below R ij,i.e.if k≤i and l≥j with(k,l)=(i,j).From this array of rectangular tableaux we will construct a set of n-tuples of tableaux(T1,...,T n)that we call factor sequences.Our conjecture is that cλ(r) is the number of factor sequences(T1,...,T n)such that T i has shapeλ(i)for 1≤i≤n.Wefirst explain this for n=3,where we start with an array of rectangular tableaux:A B CD EFFactor F into a product F=F1·F2of tableaux.Pass F1up to the left,and multiply it to D from the right.Pass F2up to the right and multiply it to E from the left.Then factor the results:D·F1=D1·D2and F2·E=E1·E2.Pass the results up to the left and right,arriving at tableaux A·D1,D2·B·E1, and E2·C.This gives a factor sequence(T1,T2,T3)=(A·D1,D2·B·E1,E2·C).In general one proceeds by induction.A factor sequence for the given array of rectangular tableaux is obtained by forming a factor sequence(S1,...,S n−1)for the array of the bottom n−1rows.Factor each S i arbitrarily into S i=P i·Q i. Then(T1,...,T n)=(T01·P1,...,Q i−1·T i−1,i·P i,...,Q n−1·T n−1,n)is a factor sequence for the given array.Conjecture.cλ(r)is the number of factor sequences(T1,...,T n)of shapeλ= (λ(1),...,λ(n))that can be made from a given array of rectangular tableaux.The conjecture has a number of consequences:C1.Each cλ(r)is a non-negative integer.C2.The coefficients cλ(r)depend only on the rectangles R ij,not on their sides.This means that if one of the sides of a rectangle R ij is0,the length of the other side is irrelevant.(The algorithm of the main theorem shows this when the height of a rectangle is0,but not when the width is0.)Implicit in the conjecture is the assertionC3.The number of factor sequences of shapeλis independent of choice offixed tableaux T ij.Granting C3,it is not hard to see that the conjectured formula for the cλ(r) satisfies the duality(4.1).For this one chooses the T ij so that no entry appears more than once,and uses the fact that factoring a tableau T with distinct entries into P·Q is equivalent to factoring its conjugate T′into Q′·P′.It is also not difficult to verify that the conjectured formula satisfies the property(4.2)for omitting inessential bundles.The conjecture is true for the case where R ij is empty for j−i>2.This follows from the description in Corollary1of Section2,together with the fact that for a tableau T of rectangular shape R,for eachσandτthatfit together to form R, there is a unique factorization T=P·Q with P of shapeσand Q of shapeτ; conversely,any factorization of T has factors of shapes thatfit together to form R.16ANDERS SKOVSTED BUCH AND WILLIAM FULTON The conjecture has been proved when n≤3.More generally,it is proved when R ij is empty for j−i>3and no two non-empty rectangles in the third row are adjacent.The proof depends on a wonderful involution on pairs of tableaux produced for us by S.Fomin.This proof is given in[4].One reason that the combinatorial formula is hard to work with is that a given factor sequence can arise in many ways by the algorithm that produces them.At first glance it would appear that to tell if some(T1,...,T n)is a factor sequence, one would have to test all possible ways of carrying out the sequence of factorings. However,there is a direct test.For this,define P i to be the part of T i lying to the right of the rectangle R i,and define Q i−1to be everything lying below T i−1,i and P i:FORMULAS FOR QUIVER V ARIETIES17 There are analogous conjectures for the coefficients of the polynomials for the more general loci described in§3.3.In particular,all these coefficients should also be positive.Details will be given in[4].References[1]S.Abeasis and A.Del Fra,Degenerations for the representations of an equioriented quiverof type A m,Boll.Un.Mat.Ital.Suppl.1980,no.2,157–171.[2]S.Abeasis,A.Del Fra,and H.Kraft,The geometry of representations of A m,Math.Ann.256(1981),401–418.[3]A.Berele and A.Regev,Hook Young diagrams,combinatorics and representations of Liesuperalgebras,Bull.Amer.Math.Soc.8(1983),337–339.[4]A.S.Buch,Combinatorics of degeneracy loci,to appear.[5]S.Fomin,S.Gelfand,and A.Postnikov,Quantum Schubert polynomials,J.Amer.Math.Soc.10(1997),565–596.[6]W.Fulton,Intersection Theory,Ergebnisse der Mathematik und ihrer Grenzgebiete(3),vol.2,Springer-Verlag,Berlin,1984,1998.[7],Universal Schubert polynomials,To appear in Duke Math.J.,1998.[9],Polynˆo mes de Schubert,C.R.Acad.Sci.Paris S´e r.I Math.294(1982),447–450.[17]I.G.Macdonald,Symmetric functions and Hall polynomials,Oxford University Press,1979,1995.[18]P.Pragacz,Enumerative geometry of degeneracy loci,Ann.Sci.´Ecole Norm.Sup.(4)21(1988),413–454.Department of Mathematics,University of Chicago,Chicago,IL60637E-mail address:abuch@Department of Mathematics,University of Chicago,Chicago,IL60637E-mail address:fulton@。