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https://www.hackmath.net/en/calculator/divisors?n=13
Divisors Calculator n=13 Divisors Calculator Enter number Divisors of the number 13:1, 13 Number of divisors:2 The sum of its divisors:14 The given number 13 is prime. Input a positive integer and this calculator will calculate: • the complete list of divisors of the given number • the sum of its divisors, • the number of divisors What is a divisor? If we have a natural number p, then the number q is a divisor of the number p if it is true that it divides it without a remainder. So there is a natural number k such that kq=p. The number p is also called the divisor, q is the divisor, and k is the quotient. Every natural number has at least two divisors. The number 1 (one) and itself, because 1p=p1=p. They are called trivial divisors. The number 0 (zero) is not a natural number but has infinitely many divisors - all numbers except zero. The set of divisors can be calculated for all natural numbers. The usual procedure is to prime factorization the number p and obtain the divisors as a product of permutations of the prime factors. The list of divisors can also be obtained for small numbers by dividing the number p by numbers from 1 to the square root of p. Divisors have many uses in mathematics. All maths calculators Math Practice Problems 18530 Worksheets © 2025 HackMath.net | contact | en | cz | sk
10401
https://secure-media.collegeboard.org/apc/ap08_calculus_ab_q6.pdf
AP® CALCULUS AB 2008 SCORING GUIDELINES Question 6 © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. Let f be the function given by ( ) ln x f x x = for all 0. x > The derivative of f is given by ( ) 2 1 ln . x f x x − ′ = (a) Write an equation for the line tangent to the graph of f at 2. x e = (b) Find the x-coordinate of the critical point of f. Determine whether this point is a relative minimum, a relative maximum, or neither for the function f. Justify your answer. (c) The graph of the function f has exactly one point of inflection. Find the x-coordinate of this point. (d) Find ( ) 0 lim . x f x + → (a) ( ) 2 2 2 2 ln 2 , e f e e e = = ( ) ( ) 2 2 2 4 2 1 ln 1 e f e e e − ′ = = − An equation for the tangent line is ( ) 2 2 4 2 1 . y x e e e = − − 2 : ( ) ( ) 2 2 1 : and 1 : answer f e f e ⎧ ′ ⎪ ⎨ ⎪ ⎩ (b) ( ) 0 f x ′ = when . x e = The function f has a relative maximum at x e = because ( ) f x ′ changes from positive to negative at . x e = 3 : 1 : 1 : relative maximum 1 : justification x e = ⎧ ⎪ ⎨ ⎪ ⎩ (c) ( ) ( ) 2 4 3 1 1 ln 2 3 2ln x x x x x f x x x − − − −+ ′′ = = for all 0 x > ( ) 0 f x ′′ = when 3 2ln 0 x −+ = 3 2 x e = The graph of f has a point of inflection at 3 2 x e = because ( ) f x ′′ changes sign at 3 2. x e = 3 : ( ) 2 : 1 : answer f x ′′ ⎧ ⎨ ⎩ (d) 0 ln lim x x x + → = −∞ or Does Not Exist 1 : answer ©2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. ©2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. ©2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. ©2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. ©2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. ©2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® CALCULUS AB 2008 SCORING COMMENTARY Question 6 Overview This problem presented students with a function f defined by ( ) ln x f x x = for together with a formula for 0, x > ( ). f x ′ Part (a) asked for an equation of the line tangent to the graph of f at 2. x e = In part (b) students needed to solve ( ) 0 f x ′ = and determine the character of this critical point from the supplied ( ). f x ′ In part (c) students had to demonstrate skill with the quotient rule to obtain a formula for ( ) f x ′′ and solve ( ) 0 f x ′′ = to find the x-coordinate of what was promised to be the only point of inflection for the graph of f. Part (d) tested students’ knowledge of properties of ln x to determine the limit of ( ) f x as 0 . x + → Sample: 6A Score: 9 The student earned all 9 points. Sample: 6B Score: 6 The student earned 6 points: 2 points in part (a), 1 point in part (b), 3 points in part (c), and no points in part (d). In part (a) the student identifies the value of the function and the derivative at 2 x e = and correctly writes a tangent line equation. In part (b) the student correctly identifies x e = but classifies it as neither a minimum nor a maximum so only earned the first point. In part (c) the student gives the correct second derivative and correctly solves the equation. In part (d) the student’s answer is not correct. Sample: 6C Score: 4 The student earned 4 points: no points in part (a), 2 points in part (b), 2 points in part (c), and no points in part (d). In part (a) the student does not identify the value of the derivative at 2 x e = and does not write an equation of the tangent line. In part (b) the student correctly identifies x e = and classifies x e = as a maximum but does not give a justification. The student gives only the definition of a maximum. In part (c) the student earned 2 points by correctly applying the product rule to find the second derivative. The student makes an arithmetic error in solving the equation and did not earn the third point. In part (d) the student has an acceptable answer but did not earn the point because of the reference to L’Hospital’s Rule. This limit is not a candidate for L’Hospital’s Rule. © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.
10402
https://study.com/academy/lesson/what-is-a-power-function-equations-and-graphs.html
Power Function | Definition, Formula & Examples - Lesson | Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Math Courses Math 105: Precalculus Algebra Power Function | Definition, Formula & Examples Contributors: Gauri Girme, Cathryn Jackson Author Author: Gauri Girme Show more Instructor Instructor: Cathryn Jackson Show more Understand what Power Function is. Learn about Power Function Equation and how to find Power Function. Get practical insights through examples of Power Function. Updated: 11/21/2023 Table of Contents What Is a Power Function? Examples of Power Function Lesson Summary Show FAQ What are power function examples? Power function includes linear functions, quadratic functions, cubic functions, and square root functions, provided that they are single term. Some examples of the power functions are : 3x^2, 56P^3 , x^(1/2), x^(-3). What is a power function? A one-term function with a variable as the base and a constant as the exponent. The exponent in the case of a power function is always a constant and a real number. The general form of a power function is, f(x)=k. x raised to n. How do you find a power function? The single term function in the form f(x)=k.x^(n) is a power function. A power function can be found if any 2 points are given or a graph of the function is given. Create an account LessonTranscript VideoQuizCourse An error occurred trying to load this video. Try refreshing the page, or contact customer support. You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it now Already registered? Log in here for access Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. Video lessons Quizzes and worksheets Classroom integration Lesson plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. Teacher Try it now Go back Coming up next: Radical Functions | Graph, Equation & Examples You're on a roll. Keep up the good work! Take QuizWatch Next Lesson Replay Just checking in. Are you still watching? Yes! Keep playing. Your next lesson will play in 10 seconds 0:01 Power Functions 2:36 Graphs of Power Functions 5:17 Fractional Power Functions 7:45 Lesson Summary QuizCourseView Video OnlySaveTimeline 303K views Recommended lessons and courses for you Related LessonsRelated Courses ##### Integer Exponents | Multiplication, Division & Examples 6:06 ##### Adding and Subtracting Scientific Notation | Steps & Examples 7:40 ##### Power Rule For Exponents | Overview & Examples 3:32 ##### Exponents with Fractional Bases 5:00 ##### Cube Root Definition, Examples & Charts 3:36 ##### Using Properties of Exponents to Create Equivalent Expressions 7:19 ##### Exponential Form | Definition & Examples 4:24 ##### Radical Expressions | Parts & Examples 4:00 ##### Irrational Numbers | Definition, Types & Examples 6:36 ##### Power of a Quotient Rule | Equation & Examples 4:33 ##### Cube Root | Symbol, List & Properties 5:19 ##### How to Simplify an Expression with Parentheses & Exponents 8:07 ##### Zero Exponent Rule | Overview & Examples 4:32 ##### Proportions | Study.com ACT® Math Test Prep 7:20 ##### Square & Cube Roots of Monomials 4:59 ##### Quotient of Powers Definition, Properties & Examples 4:58 ##### Product of Powers | Definition, Property & Examples 5:29 ##### Quotient Rule for Exponents | Overview & Examples 4:10 ##### Product of Square Roots Rule: Definition & Example 4:57 ##### Negative Exponents Definition, Rules & Examples 4:29 ##### Math 101: College Algebra ##### Math 103: Precalculus ##### CLEP College Algebra Study Guide and Exam Prep ##### CLEP Precalculus Study Guide and Exam Prep ##### CUNY Assessment Test in Math: Practice & Study Guide ##### Intermediate Algebra for College Students ##### TECEP College Algebra Study Guide and Exam Prep ##### Algebra I: High School ##### Common Core Math - Functions: High School Standards ##### CAHSEE Math Exam: Test Prep & Study Guide ##### Algebra II: High School ##### Precalculus: High School ##### UExcel Precalculus Algebra: Study Guide & Test Prep ##### Holt McDougal Algebra 2: Online Textbook Help ##### Prentice Hall Algebra 2: Online Textbook Help ##### McDougal Littell Algebra 2: Online Textbook Help ##### Math 101: College Algebra ##### Math 103: Precalculus ##### CLEP College Algebra Study Guide and Exam Prep ##### CLEP Precalculus Study Guide and Exam Prep What Is a Power Function? ------------------------- Ever worked with a function that has just one term? Most likely, you have been using a power function. Because this function type is so broad, while studying mathematics, one may already come across a type of power function without even realizing it. A single-term function made up of a real number, a coefficient, and a variable raised to a fixed real value is called a power function. A coefficient is a constant. This constant is multiplied by the variable whose exponent is any real number. Power function definition: A one-term function with a variable as the base and a constant as the exponent. The exponent in the case of a power function is always a constant and a real number. The general form of a power function is, f(x)=k x n Some common power functions are as follows: Area of a square =a 2 Area of a circle =π r 2 Volume of a sphere =4 3 π r 3. The purpose of a power function is to return a number raised to a power. That it calculates the power of the variable. Power Function Formula The power function f(x) is represented by the below formula, f(x)=k x n This formula means f(x)=k×(x raised to a power a) where x is any variable, k is a constant which is not equal to 0, and n is a real number. When n is positive, the value of a function is proportional to the n th power of x. When n is negative, the value of a function is inversely proportional to the n th power of x.Here are some power functions: −7 x 2,3 n,5 p 3,78 z 12. For any single-term function, if the exponent is a variable and the base is a constant, then it is an exponential function and not a power function. For example, 4 n,2 x. Derivation of a Power Function For the power function f(x)=k x n, its derivative is obtained by placing the variable's exponent in front of it. Multiply it by the coefficient to get the result. Reduce the exponent by one. ∴d d x[f(x)]=k n x n−1 How to Find Power Function It is possible to find the equation of a power function from its graph or from any two points on the graph. Steps for that are as follows: Substitute the given point values in the general form of the power function k x n. Try to retain either k or n in one of the equations. Determine the value of k and n and substitute them into the general form of power function. How to find the power function that passes through points (2,16) and (3,54). So, for (2,16)→16=k(2)n∴k=16 2 n..(1) for (3,54)→16=k(3)n∴k=54 3 n..(2) From (1) and (2), 16 2 n=54 3 n 2 3 2 n=3 3 3 n 2 3−n=3 3−n This equation can be true if 3−n=0.∴n=3 Putting the value of n into equation (1), k=16 2 3∴k=2 Putting the value of n and k in the general form of a power function, the required power function can be obtained f(x)=2 x 3. Power Function Rules: Below are the rules to find a power function: It must be a single-term function. It should be in the form of a coefficient multiplied by the base and its exponent. The exponent is always a real number. A coefficient is a constant that is not equal to 0. Power Function Graphs: The graph of a power function will depend on the values of k and n. The symmetry and end-behavior properties of power functions need to be considered while plotting their graphs. If the graph of a function f is symmetric with respect to the y-axis, the function is even. If the graph of a function f is symmetric with respect to the origin, the function is odd. Graph of even functions where the coefficient is positive: For the negative value of x, the function is decreasing, and for the positive value of x, the function is increasing. So, the graph is opening upward. The graph for a function f(x)=x 2 is shown in the figure. The graph of Even Function Graph of even functions where the coefficient is negative: Here, for the negative values of x, the function is increasing, and for the positive values of x, it is decreasing. So, the graph is opening downward. Now, the graph of f(x)=−1 x 2 is shown in the figure : The graph of minus x raised to 2 Graph of odd functions where the coefficient is positive: Here, when x is negative, function is increasing, and when x is positive, function is increasing as well. So, on the left side, it is going down, and on the right side, it is going up. The graph of odd function with positive coefficient Graph of odd functions where the coefficient is negative: For both positive and negative values of x, function is decreasing. So, on the left side, it is going up, and on the right side, it is going down. The graph of a odd function with negative coefficient Below is the graph of a power function with negative power and with positive and negative coefficients, respectively. Here, the graph of f(x)=x−2 shows the empty space near origin. The graph of x raised to negative coefficient Now, the graph of f(x)=−x 2 is again has empty space near the origin. Graph of a function with a positive fractional power is as shown in the figure. Graph of a function with a negative fractional power will be like this: The graph of x raised to a negative fraction Power function equations: There are other parent functions that are also power functions if they are single term. Constant function: y=a Linear Function: y=x Quadratic function: y=x 2 Cubic functions: y=x 3 Square root functions: y=x Reciprocal function: y=1 x To unlock this lesson you must be a Study.com memberCreate an account Examples of Power Function -------------------------- Below are some examples of power function. Example 1: Determine whether the function is power function or not. (a)f(x)=−7 x 2.5 x f(x)=−35 x 3 Now, it is a single-term function with its coefficient and exponents being real numbers. Hence,f(x) is a power function. (b)g(x)=−(x+2)2 g(x)=−x 2−2 x−4 Even in the simplified form, the function g(x) is not a single-term function. Hence, g(x) is not a power function. Example 2: Determine the end behavior of the following function: m(x)=−6 x 4 To determine the end behavior of a power function, check the sign of the coefficient and exponent values. The above function has a negative coefficient and a positive exponent, so the graph is expected to open downward. Example 3: Find the power function m(x) that passes through the points (4, -6) and (9, -9). The general form of a power function is, f(x)=k x n Substituting each point value in the general form of power function, For (4,−6)→−6=k(4)n k=−6 4 n..(1) For (9,−9)→9=k(−9)n k=−9 9 n..(2) From (1) and (2), −6 4 n=−9 9 n−2 2 2 n=−3 3 2 n This will be true when both sides are equal to 1, which makes exponents equal to 0. 1−2 n=0∴n=1 2 Putting the value of n in equation (1) to get the value of k, k=−6 4 1 2∴k=−3 Putting the values of k and n in the general form of a power function, m(x)=−3 x 1 2∴m(x)=−3 x To unlock this lesson you must be a Study.com memberCreate an account Lesson Summary -------------- According to the power function definition, a single-term function made up of a real number, a coefficient, and a variable raised to a fixed real value is called a power function. The power function formula is given by, f(x)=k x n, where x is any variable, k is a constant which is not equal to 0, and n is a real number. The functions 3 x 2,67 t 5,x 4 5,g are some examples of power functions. The purpose of the power function is to return a number raised to a power. In the case of any single-term function, if the exponent is a variable and the base is a constant, then it is an exponential function and not a power function. The derivative of the power function is obtained by k n x n−1. It is possible to find a power function when any two points on a function or a graph of a function are given. The symmetry, end-behavior of power functions are considered while drawing the power function graph. The lesson focused on the appearance of graphs of even, odd, and fractional functions. To specify the structure of power functions, power function rules can be utilized. To unlock this lesson you must be a Study.com memberCreate an account Video Transcript Power Functions Savanna is studying the paths of asteroids, comets, and other bodies that fly through space. She notices that as a certain comet gets closer to the earth, the path of the comet curves and moves away. Savanna wants to create mathematical functions of the comet and asteroid paths as they come closer to the earth. This way, she can predict the paths of certain comets and asteroids. Savanna can use her knowledge of power functions to create equations based on the paths of the comets. A power function is in the form of f(x) = kx^n, where k = all real numbers and n = all real numbers. You can change the way the graph of a power function looks by changing the values of k and n. If n is greater than zero, then the function is proportional to the n th power of x. This basically means that the two graphs would look the same. Here is a graph showing x^4: So in this graph, n is greater than zero. Here is the graph of f(x) = x^4. There is no difference between the two graphs. If n is less than zero, then the function is inversely proportional to the n th power of x. That means you will see the graph sort of flipped. Let's look at our graph of x^4 again. Now let's look at the graph of x^-4. Notice that this graph has an empty space near the origin. It almost creates a cut-out section. In a power function, k represents the constant of proportionality. This means that the shape of the line on the graph will not change depending on the value of k, but the placement of the line on the graph will change. Take a look at this graph to see what I mean. The blue line on this graph is the equation f(x) = x^3, and the green line is the equation f(x) = 5 x^3. Notice that when we add the 5 in front of the x, the shape of the graph stays the same, but the line moves closer to the origin. These concepts will become more important as you study calculus, but you do need to keep them in mind as you explore power functions. Graphs of Power Functions This is the graph of f(x) = x^2. You've probably seen this type of function a lot; the shape it creates is a parabola. In this graph, k = 1 and n = 2. This is the graph of f(x) = -x^2. Here k = -1 and n = 2. Notice that this graph is the opposite of the first graph you saw. The only thing that changed in the equation is the negative sign on the k value. Often, the negative sign will indicate the opposite or reverse. These types of functions, functions that contain the x^2 value, are called quadratic functions. Here is a graph of the function f(x) = x^4. Notice that the bottom of this function sort of widens, but it never crosses over the x-axis. This means that all of the y-coordinates of this function are positive. Let's look what happens when we make n negative in this function. This time the lines on the graph split into two sections, but the line still does not cross the x-axis. This is the graph of f(x) = x^-4. Savanna is studying comets that go straight towards the earth, veer off to the side, and then keep going in a straight line past the earth. She is probably looking at comets that make the path f(x) = x^3. Take a look at this graph and see if it matches Savanna's description. Looks pretty close, huh? What about the function f(x) = x^-3? What would that graph look like? Wow! Notice that, like the other graphs that had negative exponents, the lines on the graph sort of separate into two different directions. However, unlike the graph f(x) = x^-4, shown with blue lines, the graph of f(x) = x^-3, shown with green lines, has negative y-coordinates. This is because the power in this function is odd, which will give you a negative result. You'll notice that functions with an even power are symmetrical across the y-axis and functions with an odd power are symmetrical about the origin. You can learn more about symmetry in the Graph Symmetry chapter of this course. Fractional Power Functions Savanna is now studying the path of asteroids. This time, she needs to find the difference between the asteroids that go past the earth and the asteroids that collide with the earth. She has two different asteroids she is studying. The first has the function of f(x) = x^1/3, and the second asteroid has the function of f(x) = x^1/4. This is the graph of f(x) = x^1/3. You'll see that the line on this graph goes from the positive side, curves, and then to the negative side of the graph. We can assume that the asteroid that creates this path will not collide with the earth. This is the graph of f(x) = x^1/4. You'll see that the line on this graph goes from the positive side and stops at the origin. We can assume that the asteroid that creates this path will collide with the earth. Now that we've looked at positive fractional power functions, let's look at some negative fractional power functions. The first graph, represented with green lines, is the function f(x) = x^-3/5. The second graph, represented with a blue line, is the function f(x) = x^-1/4. Notice that the function with the even denominator is located only on the positive side of the x- and y-axis. Notice that these two functions, f(x) = x^-4 and f(x) = x^-1/4, look very similar. Notice that the only differences in these graphs are the positions of the curves of the lines. You'll see that all of the numbers in the powers of the two functions are odd numbers. Lastly, we must consider the functions that have powers with improper fractions, such as this graph. Notice that this graph does not contain any negative x- or y-coordinates. This graph represents the function f(x) = x^5/2. Now that you've seen all of these graphs, how can you remember which functions go with which? Let's summarize. Lesson Summary A power function appears in the form f(x) = kx^n, where k = all real numbers and n = all real numbers. The three main types of power functions are even, odd, and fractional functions. You will see these in both positive and negative forms. Here are the graphs of the functions f(x) = x^2 and f(x) = x^4. These graphs are shaped similarly because they both have positive and even powers. These graphs are similar because they both contain positive, odd numbers in their powers. The first graph represents the function f(x) = x^3, and the second graph represents f(x) = x^5/2. Notice these graphs have a nice sloping curve close to the origin. These graphs are similar because they all have positive, fractional powers. Here are the graphs of the functions f(x) = x^-4, f(x) = x^-3, f(x) = x^-3/5, and f(x) = x^-1/4. These functions are similar because they have negative powers. Also, don't forget when you are graphing power functions to graph with a curvy line. Learning Outcomes Watch this video lesson as you pursue these goals: Define and use power functions Recall the equation form for a power function and the write down three main types Accurately identify whether a graph is an even or odd powered function Graph a power function Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock your education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a member Already a member? Log in Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. Video lessons Quizzes and worksheets Classroom integration Lesson plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. Teacher Try it now Go back Create an account to start this course today Start today. Try it now Math 105: Precalculus Algebra 13 chapters 121 lessons 12 flashcard sets Chapter 1 Mathematical Modeling Writing & Evaluating Algebraic Expressions for Two-Dimensional Geometric Figures 4:10 min Linear Model | Equation & Examples 11:00 min Applying Systems of Linear Equations to Breakeven Point: Steps & Example 5:44 min Applying Systems of Linear Equations to Market Equilibrium: Steps & Example 10:09 min Maximum & Minimum Values of a Parabola | Overview & Formula 9:54 min Using Quadratic Functions to Model a Given Data Set or Situation 5:53 min Chapter 2 Linear Equations & Inequalities Linear Equations | Definition, Formula & Solution 7:28 min Forms of a Linear Equation | Overview, Graphs & Conversion 6:38 min Abstract Algebraic Examples and Going from a Graph to a Rule 10:37 min Undefined & Zero Slope Graph | Definition & Examples 4:23 min Linear Equation | Parts, Writing & Examples 8:58 min Equation of a Line Using Point-Slope Formula 9:27 min Inequality Signs in Math | Symbols, Examples & Variation 7:09 min Graphing Inequalities | Definition, Rules & Examples 7:59 min Graphing Inequalities | Overview & Examples 12:06 min System of Equations in Algebra | Overview, Methods & Examples 8:39 min How Do I Use a System of Equations? 9:47 min How to Solve System of Equations Word Problems | Method & Example 6:23 min Chapter 3 Quadratic Functions Conic Sections | Overview, Equations & Types 6:22 min Factoring Quadratic Equations Using Reverse Foil Method 8:50 min Factoring Quadratic Equations | Solution & Examples 7:35 min How to Complete the Square | Method & Examples 8:43 min Completing the Square Practice Problems 7:31 min How to Solve a Quadratic Equation by Factoring 7:53 min Quadratic Function | Formula, Equations & Examples 9:20 min How to Solve & Graph Quadratic Inequalities 6:14 min Graphing a System of Quadratic Inequalities: Examples & Process 8:52 min Chapter 4 Rational Expressions and Functions How to Multiply and Divide Rational Expressions 8:07 min Adding & Subtracting Rational Expressions | Overview & Examples 8:02 min Rational Equations | Definition, Formula & Examples 7:58 min Horizontal and Vertical Asymptotes 7:47 min Graphing Rational Functions That Have Linear Polynomials: Steps & Examples 7:55 min Graphing Rational Functions That Have Polynomials of Various Degrees: Steps & Examples 8:59 min Domain & Range of Rational Functions | Definition & Graph 5:50 min Chapter 5 Polynomial Functions of a Higher Degree Adding, Subtracting & Multiplying Polynomials | Steps & Examples 6:53 min Factoring Polynomials Using Quadratic Form: Steps, Rules & Examples 8:38 min Polynomial Long Division | Overview & Examples 8:05 min Synthetic Division of Polynomials | Method & Examples 6:51 min Factor & Remainder Theorem | Definition, Formula & Examples 7:00 min Dividing Polynomials with Long and Synthetic Division: Practice Problems 10:11 min Using the Rational Zeros Theorem to Find Rational Roots 8:45 min Fundamental Theorem of Algebra | Definition, Examples & Proof 7:39 min Writing a Polynomial Function With Given Zeros | Steps & Examples 8:59 min Chapter 6 Geometry Basics Finding the Area & Circumference of a Circle 7:24 min How to Graph a Circle | Equation & Examples 8:32 min Completing the Square | Formula, Examples & Importance 6:21 min Equation of a Circle | Formula, Forms & Examples 9:41 min Distance Formula | Overview & Examples 5:27 min Midpoint | Formula & Examples 3:33 min Chapter 7 Functions Overview Domain & Range of a Function | Definition, Equation & Examples 8:32 min Viewing now Power Function | Definition, Formula & Examples 9:26 min Up next Radical Functions | Graph, Equation & Examples 6:47 min Watch next lesson Graphing Square Root & Cube Root Functions 6:51 min Points of Discontinuity | Overview, Types & Examples 6:26 min Piecewise Function | Definition, Evaluation & Examples 7:22 min Piecewise Functions | Graph & Examples 5:00 min Domain of Piecewise Functions | Notation & Method 4:34 min Transformations: How to Shift Graphs on a Plane 7:12 min Greatest Integer Function | Definition, Graph & Equation 4:53 min Chapter 8 Function Operations How to Add, Subtract, Multiply and Divide Functions 6:43 min How to Compose Functions 6:52 min Applying Function Operations Practice Problems 5:17 min Compounding Functions and Graphing Functions of Functions 7:47 min Domain & Range of Composite Functions | Steps & Examples 5:58 min Inverse Functions | Definition, Methods & Calculation 6:05 min Inverse Function | Graph & Examples 7:31 min One to One Function | Definition, Graph & Examples 4:11 min Limit of a Function | Definition, Rules & Examples 5:15 min One-Sided Limits and Continuity 4:33 min Functions in Real Life | Applications & Examples 5:36 min Quadratic Equations in Real Life | Overview, Uses & Examples 5:22 min Chapter 9 Graph Symmetry Symmetric Graphs | X-Axis, Y-Axis & Algebraic Symmetry 11:19 min Recognizing Symmetry Graphically, Algebraically & Numerically About the Origin 6:17 min Line of Symmetry | Definition, Graph & Equation 8:07 min Even & Odd Functions | Formulas, Graphs & Examples 10:02 min Real-World Applications of Function Symmetry Symmetry in Piecewise and Discontinuous Functions Chapter 10 Exponential and Logarithmic Functions Exponential Function | Definition, Equation & Examples 7:24 min Exponential Growth & Decay | Formula, Function & Graphs 8:41 min Exponential Functions | Transformation, Graphs & Examples 5:51 min Natural Base e | Overview & Importance 4:47 min Logarithms | Overview, Process & Examples 5:23 min Evaluating Logarithms | Properties & Examples 6:45 min Inverse of Log Functions | Definition & Examples 7:09 min Exponentials, Logarithms & the Natural Log 8:36 min Graphing Logarithmic Functions | Overview & Examples 8:08 min Logarithmic Properties | Product, Power & Quotient Properties 5:11 min Practice Problems for Logarithmic Properties 6:44 min Solving Logarithmic Equations | Properties & Examples 6:50 min Change-of-Base Formula for Logarithms | Rules & Examples 4:56 min Exponential Equations | Definition, Solutions & Examples 6:18 min Applications of Exponential Growth and Decay Chapter 11 Essentials of Trigonometry Trigonometric Functions | Definition, Formula & Examples 6:40 min Inverse Trigonometric Functions | Definition, Problems & Examples 5:24 min Trigonometric Identities Definition, Formulas & Examples 4:38 min Alternate Forms of Trigonometric Identities 5:13 min Solve Trigonometric Equations with Identities & Inverses 5:44 min How to Solve Trigonometric Equations for X 4:57 min Solving Trigonometric Equations | Steps & Examples 7:18 min Solving Trigonometric Equations with Infinite Solutions 6:32 min Law of Sines Formula & Examples 6:04 min Law of Cosines | Definition & Equation 8:16 min Chapter 12 Introduction to the Derivative Velocity and the Rate of Change 2:54 min Slopes and Rate of Change 3:11 min Derivatives: The Formal Definition 4:02 min Graphing the Derivative from Any Function 15:26 min Tangent Lines and Instantaneous Rate of Change Calculating Derivatives of Polynomial Equations 10:25 min Chapter 13 Studying for Math 105 Mathematical Modeling Flashcards Linear Equation & Inequality Flashcards Quadratic Function Flashcards Rational Expressions & Functions Flashcards Higher Degree Polynomial Function Flashcards Absolute Value Equations & Inequalities Flashcards Complex Number Flashcards Geometry Overview Flashcards Function Operations Flashcards Graph Symmetry Flashcards Exponential & Logarithmic Function Flashcards Derivatives Flashcards Math 105: Precalculus Algebra Formulas & Properties Related Study Materials Power Function | Definition, Formula & Examples LessonsCoursesTopics ##### Integer Exponents | Multiplication, Division & Examples 6:06 ##### Adding and Subtracting Scientific Notation | Steps & Examples 7:40 ##### Power Rule For Exponents | Overview & Examples 3:32 ##### Exponents with Fractional Bases 5:00 ##### Cube Root Definition, Examples & Charts 3:36 ##### Using Properties of Exponents to Create Equivalent Expressions 7:19 ##### Exponential Form | Definition & Examples 4:24 ##### Radical Expressions | Parts & Examples 4:00 ##### Irrational Numbers | Definition, Types & Examples 6:36 ##### Power of a Quotient Rule | Equation & Examples 4:33 ##### Cube Root | Symbol, List & Properties 5:19 ##### How to Simplify an Expression with Parentheses & Exponents 8:07 ##### Zero Exponent Rule | Overview & Examples 4:32 ##### Proportions | Study.com ACT® Math Test Prep 7:20 ##### Square & Cube Roots of Monomials 4:59 ##### Quotient of Powers Definition, Properties & Examples 4:58 ##### Product of Powers | Definition, Property & Examples 5:29 ##### Quotient Rule for Exponents | Overview & Examples 4:10 ##### Product of Square Roots Rule: Definition & Example 4:57 ##### Negative Exponents Definition, Rules & Examples 4:29 ##### Math 101: College Algebra ##### Math 103: Precalculus ##### CLEP College Algebra Study Guide and 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Fractions How to Define a Zero and Negative Exponent Exponent Rules: Review & Practice Simplifying Algebraic Expressions with Rational Exponent The Distributive Property & Negative Exponents Factorials | Study.com ACT® Math Test Prep Simplifying Expressions with Exponents | Overview & Examples How to Simplify Roots of Roots Create an account to start this course today Used by over 30 million students worldwide Create an account Explore our library of over 88,000 lessons Search Browse Browse by subject College Courses Business English Foreign Language History Humanities Math Science Social Science See All College Courses High School Courses AP Common Core GED High School See All High School Courses Other Courses College & Career Guidance Courses College Placement Exams Entrance Exams General Test Prep K-8 Courses Skills Courses Teacher Certification Exams See All Other Courses Study.com is an online platform offering affordable courses and study materials for K-12, college, and 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Isosceles and Equilateral Triangles (practice) | Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. We're a nonprofit that relies on support from people like you. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. Please help keep Khan Academy free, for anyone, anywhere forever. Select gift frequency One time Recurring Monthly Yearly Select amount $10 $20 $30 $40 Other Give now By donating, you agree to our terms of service and privacy policy. Skip to lesson content Scholarship exam Class 8 MH Course: Scholarship exam Class 8 MH>Unit 3 Lesson 4: Triangle - Properties of a triangle Angle relationships in a Triangle Side length inequality of Triangles Isosceles and Equilateral Triangles Centroid of a Triangle Incenter, Circumcenter, Orthocenter Congruence of Triangles Math> Scholarship exam Class 8 MH> Geometry> Triangle - Properties of a triangle © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Isosceles and Equilateral Triangles Google Classroom Microsoft Teams Problem In isosceles triangle △P Q R‍, P Q=P R‍ and P S‍ is an altitude to side Q R‍. If m∠Q P S=28‍ degrees. What is the measure of base angle ∠P R Q‍? Choose 1 answer: Choose 1 answer: (Choice A) 28‍ degrees A 28‍ degrees (Choice B) 56‍ degrees B 56‍ degrees (Choice C) 32‍ degrees C 32‍ degrees (Choice D) 62‍ degrees D 62‍ degrees Report a problem Do 4 problems Skip Check Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Accept All Cookies Strictly Necessary Only Cookies Settings Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. 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http://math0.bnu.edu.cn/~liujun/courseware/calculus_A_I/Chapter3/3-3.pdf
第三章微分中值定理与导数应 用 微分中值定理与导数应用(第三讲) 1 第三讲：导数的应用 微分中值定理与导数应用(第三讲) 2 x y o ) (x f y = x y o ) (x f y = a b A B 0 ) ( ³ ¢ x f 0 ) ( £ ¢ x f a b B A 3.1 函数单调性的判别法 微分中值定理与导数应用(第三讲) 3 1、 0 ) ( > ¢ x f ，则 ( ) f x 在 ] , [ b a 上单调增加； 设 ( ) y f x = 在[ , ] a b 上连续，在( , ) a b 内可导. 引理 若在( , ) a b 内 2、 0 ) ( < ¢ x f ，则 ( ) f x 在 ] , [ b a 上单调减少. 证 ), , ( , 2 1 b a x x Î " , 2 1 x x < 且 用拉氏定理得， ), )( ( ) ( ) ( 1 2 1 2 x x f x f x f -¢ = -x . 2 1 x x < ¢ x f , 0 ) ( > ¢ x f 则 ). ( ) ( 1 2 x f x f > \ . ] , [ ) ( 上单调增加 在 b a x f y = \ , 0 ) ( ) 2 ( < ¢ x f , 0 ) ( < ¢ x f 则 ). ( ) ( 1 2 x f x f < \ . ] , [ ) ( 上单调减少 在 b a x f y = \ , 0 1 2 > - x x ! 内， 所以，在 ) , ( b a 微分中值定理与导数应用(第三讲) 5 例1 解 e 1, x y¢ = -! , ) 0 , ( 内 在-¥ , 0 < ¢ y 函数单调减少； \ , ) , 0 ( 内 在 +¥ , 0 > ¢ y . 函数单调增加 \ ( , ). D = -¥ +¥ 且 讨论函数 e 1 x y x = --的单调性. 微分中值定理与导数应用(第三讲) 6 注意:函数的单调性是一个区间上的性质，要用 单调区间求法： 问题: 函数在定义区间上不是单调的，但在各个 部分区间上单调． 导数在这一区间上的符号来判定，而不能用一 点处的导数符号来判别一个区间上的单调性． 若函数在其定义域的某个区间内是单调的，则该 区间称为函数的单调区间. 微分中值定理与导数应用(第三讲) 7 导数等于零的点和不可导点是单调区间的可能分 界点． 方法: . , ) ( ) ( 0 ) ( 数的符号 然后判断区间内导 的定义区间 来划分函数 不存在的点 的根及 用方程 x f x f x f ¢ = ¢ 微分中值定理与导数应用(第三讲) 8 例2 解 . 3 12 9 2 ) ( 2 3 的单调区间 确定函数 -+ -= x x x x f ). , ( : +¥ -¥ D ! 12 18 6 ) ( 2 + -= ¢ x x x f ) 2 )( 1 ( 6 --= x x 得， 解方程 0 ) ( = ¢ x f . 2 , 1 2 1 = = x x 时， 当 1 < < ¥ -x , 0 ) ( > ¢ x f 上单调增加； 在 ] 1 , (-¥ \ 时， 当 2 1 < < x , 0 ) ( < ¢ x f 上单调减少； 在 ] 2 , 1 [ \ 时， 当 +¥ < < x 2 , 0 ) ( > ¢ x f 上单调增加； 在 ) , 2 [ +¥ \ 微分中值定理与导数应用(第三讲) 9 单调区间为 ， ] 1 , (-¥ ， ] 2 , 1 [ ). , 2 [ +¥ 例3 . ) ( 3 2 的单调区间 确定函数 x x f = 解 ). , ( : +¥ -¥ D ! ) 0 ( , 3 2 ) ( 3 ¹ = ¢ x x x f 微分中值定理与导数应用(第三讲) 10 . , 0 导数不存在 时 当= x 时， 当 0 < < ¥ -x , 0 ) ( > ¢ x f 上单调增加； 在 ) , 0 [ +¥ \ 时， 当 +¥ < < x 0 , 0 ) ( < ¢ x f 上单调减少； 在 ] 0 , (-¥ \ 单调区间为 ， ] 0 , (-¥ ). , 0 [ +¥ 3 2 x y = 微分中值定理与导数应用(第三讲) 11 例4 证 . ) 1 ln( , 0 成立 试证 时 当 x x x + > > ), 1 ln( ) ( x x x f + -= 设 ( ) 0. 1 x f x x ¢ = > + 上单调增加； 在 ) , 0 [ +¥ \ , 0 ) 0 ( = f ! 时， 当 0 > \ x , 0 ) 0 ( ) ( ) 1 ln( = > = + -f x f x x ). 1 ln( x x + > 即 注意: 区间内个别点导数为零不影响区间的单调性. 如, , 3 x y = , 0 0 = ¢ = x y . ) , ( 上单调增加 但在 +¥ -¥ (0, ) +¥ 上连续，在 内可导，且 则函数 在 ( ) f x [0, ) +¥ 微分中值定理与导数应用(第三讲) 12 o x y a b ) (x f y = 1 x 2 x 3 x 4 x 5 x 6 x o x y o x y 0 x 0 x 3.2 函数的极值及求法 则称 ) ( 0 x f 为 ) (x f 的一个极小值. 则称 ) ( 0 x f 为 ) (x f 的一个极大值. 微分中值定理与导数应用(第三讲) 13 极值的定义： 设函数 ) (x f 在区间 ) , ( b a 内有定义，且 ) , ( 0 b a x Î 若存在 0 x 的某邻域,对其中异于 0 x 的任意点x 均有 若存在 0 x 的某邻域,对其中异于 0 x 的任意点x 均有 ) ( ) ( 0 x f x f < ) ( ) ( 0 x f x f > 函数的极大值与极小值统称为极值,使函数取得 极值的点称为极值点. 微分中值定理与导数应用(第三讲) 14 . ) ( ) 0 ) ( ( ) ( 的驻点 叫做函数 的实根 即方程 的导数为零的点 使函数 x f x f x f = ¢ ( ) ( ) . f x f x 函数 的驻点和导数不存在的点叫做函数 的临界点 微分中值定理与导数应用(第三讲) 15 函数极值的求法： 设 ) (x f 在点 0 x 处具有导数,且 在 0 x 处取得极值,那末必定 0 ( ) 0 f x ¢ = . 定理(必要条件) 注意: . , ) ( 是极值点 但函数的驻点却不一定 点 的极值点必定是它的驻 可导函数 x f 如, , 3 x y = , 0 0 = ¢ = x y . 0不是极值点 但= x (2)当 0 x x < 时， 0 ) ( < ¢ x f ；当 0 x x > 时， 0 ) ( > ¢ x f ， 则 ) ( 0 x f 为极小值. (3)当 ) (x f ¢ 在点 0 x 的两侧同号时, ) (x f 在 0 x 无极值. (1) 当 0 x x < 时， 0 ) ( > ¢ x f ；当 0 x x > 时， 0 ) ( < ¢ x f ， 则 ) ( 0 x f 为极大值. 微分中值定理与导数应用(第三讲) 16 定理(第一充分条件) 设函数 ) (x f 在临界点 0 x 的某去心邻域内可导， 且 0 ) ( ¹ ¢ x f ,若 微分中值定理与导数应用(第三讲) 17 x y o x y o 0 x 0 x + --+ (是极值点情形) x y o x y o 0 x 0 x + --+ (非极值点情形) 如图所示： 微分中值定理与导数应用(第三讲) 18 求函数极值的步骤: ); ( ) 1 ( x f ¢ 求导函数 ; ) 2 ( 求全部临界点 . ) 4 ( 小值 求出所有的极大值和极 ; ) ( ) 3 ( 的符号，求极值点 检查临界点附近 x f ¢ 微分中值定理与导数应用(第三讲) 19 例5 解 . 5 9 3 ) ( 2 3 的极值 求出函数 + --= x x x x f 9 6 3 ) ( 2 --= ¢ x x x f ， 令 0 ) ( = ¢ x f . 3 , 1 2 1 = -= x x 得驻点 列表讨论 x ) 1 , ( --¥ ) , 3 ( +¥ ) 3 , 1 (-1 -3 ) (x f ¢ ) (x f + -+ ­ 0 0 ¯ ­ 极 大 值 极 小 值 ) 3 ( f 极小值 . 22 -= ) 1 (-f 极大值 , 10 = ) 3 )( 1 ( 3 -+ = x x 微分中值定理与导数应用(第三讲) 20 5 9 3 ) ( 2 3 + --= x x x x f M m 图形如下 微分中值定理与导数应用(第三讲) 21 例6 解 . ) 2 ( 1 ) ( 3 2 的极值 求出函数 --= x x f ) 2 ( ) 2 ( 3 2 ) ( 3 1 ¹ --= ¢ -x x x f . ) ( , 2 不存在 时 当 x f x ¢ = 时， 当 2 < x ; 0 ) ( > ¢ x f 时， 当 2 > x . 0 ) ( < ¢ x f . ) ( 1 ) 2 ( 的极大值 为 x f f = \ . ) ( 在该点连续 但函数 x f 注意: 函数的不可导点也可能是函数的极值点. M 微分中值定理与导数应用(第三讲) 22 设 ) (x f 在 0 x 处二阶可导,且 0 ( ) 0 f x ¢ = ,则 定理(第二充分条件) (1) 当 0 ( ) 0 f x ¢¢ < 时, ) (x f 在 0 x 处取得极大值; (2) 当 0 ( ) 0 f x ¢¢ > 时, ) (x f 在 0 x 处取得极小值； (3) 当 0 ) ( 0 = ¢ ¢ x f 时, 不确定. 微分中值定理与导数应用(第三讲) 23 时， 当 0 < Dx ) ( ) ( 0 0 x f x x f ¢ > D + ¢ 有 , 0 = 时， 当 0 > Dx ) ( ) ( 0 0 x f x x f ¢ < D + ¢ 有 , 0 = 所以,函数 ) (x f 在 0 x 处取得极大值. 类似证（2）. 证 ) 1 ( x x f x x f x f x D ¢ -D + ¢ = ¢ ¢ ® D ) ( ) ( lim ) ( 0 0 0 0 ! , 0 < 异号， 与 故 x x f x x f D ¢ -D + ¢ ) ( ) ( 0 0 微分中值定理与导数应用(第三讲) 24 例7 解 . 20 24 3 ) ( 2 3 的极值 求出函数 --+ = x x x x f 24 6 3 ) ( 2 -+ = ¢ x x x f ， 令 0 ) ( = ¢ x f . 2 , 4 2 1 = -= x x 得驻点 ) 2 )( 4 ( 3 -+ = x x , 6 6 ) ( + = ¢ ¢ x x f ! = -¢ ¢ ) 4 ( f ! , 0 18 < -) 4 (-f 故极大值 , 60 = = ¢ ¢ ) 2 ( f , 0 18 > ) 2 ( f 故极小值 . 48 -= 20 24 3 ) ( 2 3 --+ = x x x x f 图形如下 微分中值定理与导数应用(第三讲) 25 M m 注意: 当 0 ( ) 0 f x ¢¢ = 时， ( ) f x 在点 0 x 处不一定取得 极值，此时，仍需用第一充分条件判别. 微分中值定理与导数应用(第三讲) 26 单调性的判别是拉格朗日中值定理的重要应用. 定理中的区间换成其它有限或无限区间，结论 仍然成立. 利用函数的单调性可以确定某些方程实根的个 数和证明不等式. 3.3 小结与思考题 微分中值定理与导数应用(第三讲) 27 极值是函数的局部性概念:极大值可能小于极小 值,极小值可能大于极大值. 函数的极值必在临界点取得. 判别法 第一充分条件; 第二充分条件. (注意使用条件) 微分中值定理与导数应用(第三讲) 28 思考题 若 0 ) 0 ( > ¢ f ，是否能断定 ) (x f 在原点 的充分小的邻域内单调递增？ 微分中值定理与导数应用(第三讲) 29 思考题解答 不能断定. 例 ï î ï í ì = ¹ + = 0 , 0 0 , 1 sin 2 ) ( 2 x x x x x x f = ¢ ) 0 ( f ) 1 sin 2 1 ( lim 0 x x x D × D × + ® D 0 1 > = 但 0 , 1 cos 2 1 sin 4 1 ) ( ¹ -+ = ¢ x x x x x f 微分中值定理与导数应用(第三讲) 30 π ) 2 1 2 ( 1 + = k xk 当 时， 0 π ) 2 1 2 ( 4 1 ) ( > + + = ¢ k x f k π 2 1 k xk = 当 时， 0 1 ) ( < -= ¢ k x f 注意可以任意大，故在 点的任何邻 域内， 都不单调递增． k 0 0 = x ) (x f 微分中值定理与导数应用(第三讲) 31 思考题 下命题正确吗？ 如果 0 x 为连续函数 ) (x f 的极小值点，那 么必存在 0 x 的某个邻域，在此邻域内 ) (x f 在 0 x 的左侧下降，而在 0 x 的右侧上升. 微分中值定理与导数应用(第三讲) 32 思考题解答 不正确． 例 ï î ï í ì = ¹ + + = 0 , 2 0 ), 1 sin 2 ( 2 ) ( 2 x x x x x f 当 0 ¹ x 时， = -) 0 ( ) ( f x f ) 1 sin 2 ( 2 x x + 0 > 于是 0 = x 为 ) (x f 的极小值点. 微分中值定理与导数应用(第三讲) 33 当 0 ¹ x 时， 当 0 ® x 时， , 0 ) 1 sin 2 ( 2 ® + x x x 1 cos 在–1和1之间振荡 因而 ) (x f 在 0 = x 的两侧都不单调. 故命题不成立． x x x x f 1 cos ) 1 sin 2 ( 2 ) ( -+ = ¢ 微分中值定理与导数应用(第三讲) 34 一、 填空题： 1、 函数 1 12 3 2 2 3 + --= x x x y 单调区间为_. 2、 函数 2 1 2 x x y + = 在区间[-1,1]上单调_， 在_上单调减. 3、函数 2 2 ln x x y -= 的单增区间为_， 单减区间为___. 课堂练习题 微分中值定理与导数应用(第三讲) 35 三、 证明：当 0 > x 时， ) 1 ln( ) 1 ( 1 x x x + + > -e . 四、 设 ) (x f 在[ b a, ]上连续，在( b a, )内 0 ) ( > ¢ ¢ x f , 试证明对于[ b a, ]上任意两点 1 x ， 2 x 有 2 ) ( ) ( ) 2 ( 2 1 2 1 x f x f x x f + < + 二、 设函数 ) (x f 在 ] 1 , 0 [ 上可导，且 1 ) ( 0 < < x f .又 对于 ) 1 , 0 ( 内的一切, ( ) 1 x f x ¢ ¹ .证明： x x f = ) ( 在 ) 1 , 0 ( 内有惟一实根. 微分中值定理与导数应用(第三讲) 36 一、1、 ) , 2 [ ], 1 , ( +¥ --¥ 单调增加, ] 2 , 1 [-单调减少； 2、增加, ) , 1 [ ], 1 , ( +¥ --¥ ； 3、 ] 1 , 0 ( ], 1 , ( ); , 1 [ ], 1 , 0 ( ), 0 , 1 [ ], 1 , ( --¥ +¥ ---¥ . 二、提示：1、利用零点定理证明根的存在性； 2、利用罗尔定理证明根的惟一性. 三、提示：两次利用单调性. 课堂练习题答案 微分中值定理与导数应用(第三讲) 37 四、提示、设 ). ( 2 1 2 1 0 x x x + = （1）利用拉氏中值定理： 0 1 2 0 1 2 0 1 2 0 ( ) ( ) ( ) ( ) ( ) ( ) f x f x f x f x f f x x x x x x --¢ ¢ = < = -- （2）利用泰劳公式： ) 2 , 1 ( = i 2 0 0 0 0 1 ( ) ( ) ( )( ) ( )( ) 2 i i i i f x f x f x x x f x x x ¢ ¢¢ = + -+ - 微分中值定理与导数应用(第三讲) 38 一、 填空题： 1、 极值反映的是函数的___性质. 2、 若 ) (x f y = 在 0 x x = 可导，则它在 0 x 处有极值 的必要条件为__. 3、 3 2 ) 1 ( 2 3 --= x y 的极值点为___； 3 1 ) 1 ( 2 3 + -= x y 的极值为 __. 4、 设 î í ì £ + > = 0 , 1 0 , ) ( 3 x x x x x f x ,则 _ x = 时， y = 为 极小值；当 _ x = 时， y = 为极大值. 课堂练习题 微分中值定理与导数应用(第三讲) 39 二、求下列函数的极值： 1、 方程 2 e 0 x y y + = 所确定的函数 ) (x f y = ； 2、 2 e , 0, 0, 0, x x y x -ì ¹ ï = í = ï î . 三、 设 ) (x f 是具有二阶导函数的偶函数， 且 0 ) ( ¹ ¢ ¢ x f ， 证明： 0 = x 为函数 ) (x f 的极值点. 微分中值定理与导数应用(第三讲) 40 一、1、局部； 2、 0 ) ( 0 = ¢ x f ； 3、1, 无； 4、 1 , 0 , e , e 1 e 3 1 ---. 二、1、极小值 1 ) 0 ( -= y ； 2、极小值 0 ) 0 ( = y . 课堂练习题答案 微分中值定理与导数应用(第三讲) 41 问题: 如何研究曲线的弯曲方向? x y o A B C 3.4 曲线的凹凸性及其判别法 微分中值定理与导数应用(第三讲) 42 x y o 1 x 2 x ) (x f y = 图形上任意弧段位 于所张弦的上方 x y o ) (x f y = 1 x 2 x 图形上任意弧段位 于所张弦的下方 曲线凹凸的特点： 则称 ) (x f 在 ) , ( b a 内的图形为凸弧. 则称 ) (x f 在 ) , ( b a 内的图形为凹弧； 微分中值定理与导数应用(第三讲) 43 凹凸弧的定义： 设 ) (x f 在 ) , ( b a 内连续，若对 ) , ( b a 内的任意两点 , , 2 1 x x 恒有 , 2 ) ( ) ( ) 2 ( 2 1 2 1 x f x f x x f + < + 若对 ) , ( b a 内的任意两点 , , 2 1 x x 恒有 , 2 ) ( ) ( ) 2 ( 2 1 2 1 x f x f x x f + > + (凸)的，则称 ) (x f 在 ] , [ b a 上的图形是凹(凸)弧. 若 ) (x f 在 ] , [ b a 上连续，且在 ) , ( b a 内的图形是凹 微分中值定理与导数应用(第三讲) 44 曲线凹凸的判定： x y o ) (x f y = x y o ) (x f y = a b A B 递增 ) (x f ¢ a b B A 0 > ¢ ¢ y 递减 ) (x f ¢ 0 < ¢ ¢ y 微分中值定理与导数应用(第三讲) 45 判别法： 设 ) (x f 在 ] , [ b a 内连续，在 ) , ( b a 内二阶可导, 若在 ) , ( b a 内 （1） , 0 ) ( > ¢ ¢ x f 则 ) (x f 为 ] , [ b a 上的凹弧； （2） , 0 ) ( < ¢ ¢ x f 则 ) (x f 为 ] , [ b a 上的凸弧. 微分中值定理与导数应用(第三讲) 46 例8 . 3 的凹凸性 判断曲线 x y = 解 , 3 2 x y = ¢ ! , 6x y = ¢ ¢ 时， 当 0 < x , 0 < ¢ ¢ y 为凸的； 在 曲线 ] 0 , (-¥ \ 时， 当 0 > x , 0 > ¢ ¢ y 为凹的； 在 曲线 ) , 0 [ +¥ \ . ) 0 , 0 ( 点 是曲线由凸变凹的分界 点 注意： 微分中值定理与导数应用(第三讲) 47 曲线的拐点及其求法 连续曲线上凹凸弧的分界点称为曲线的拐点. 若 ) (x f 在 ) , ( 0 0 d d + -x x 内存在二阶导数,则 注意: 拐点处的切线必在拐点处穿过曲线. 拐点的求法： 1、必要条件 ( ) ) ( , 0 0 x f x 为拐点的必要条件是 0 ( ) 0 f x ¢¢ = . 微分中值定理与导数应用(第三讲) 48 , 0 ) ( ) ( 0 0 = ¢ ¢ x f x x f 的邻域内二阶可导且 在 设 ; )) ( , ( , ) ( ) 1 ( 0 0 0 为拐点 则点 异号 两侧 若在 x f x x f x ¢ ¢ 2、第一充分条件 3、第二充分条件 , 0 ) ( ) ( 0 0 = ¢ ¢ x f x x f 的邻域内三阶可导且 在 设 . )) ( , ( , ) ( ) 2 ( 0 0 0 非拐点 则点 同号 两侧 若在 x f x x f x ¢ ¢ . ) ( )) ( , ( , 0 ) ( 0 0 0 的拐点 是曲线 则 x f y x f x x f = ¹ ¢ ¢ ¢ 微分中值定理与导数应用(第三讲) 49 例9 . 1 4 3 3 4 凸的区间 的拐点及凹 求曲线 + -= x x y 解 ) , ( : +¥ -¥ D ! , 12 12 2 3 x x y -= ¢ ). 3 2 ( 36 -= ¢ ¢ x x y , 0 = ¢ ¢ y 令 . 3 2 , 0 2 1 = = x x 得 x ) 0 , (-¥ ) , 3 2 ( +¥ ) 3 2 , 0 ( 0 3 2 ) (x f ¢ ¢ ) (x f + -+ 0 0 凹 凸 凹 拐点 拐点 ) 1 , 0 ( ) 27 11 , 3 2 ( 微分中值定理与导数应用(第三讲) 50 ). , 3 2 [ ], 3 2 , 0 [ ], 0 , ( +¥ -¥ 凹凸区间为 微分中值定理与导数应用(第三讲) 51 例10 . ) π 2 0 ( cos sin 的拐点 求曲线 £ £ + = x x x y 解 , sin cos x x y -= ¢ , cos sin x x y --= ¢ ¢ . sin cos x x y + -= ¢ ¢ ¢ , 0 = ¢ ¢ y 令 . 4 7 , 4 3 2 1 p = p = x x 得 2 ) 4 3 ( = p ¢ ¢ ¢ f , 0 ¹ 2 ) 4 7 ( -= p ¢ ¢ ¢ f , 0 ¹ 内曲线有拐点为 在 ] 2 , 0 [ p \ ). 0 , 4 7 ( ), 0 , 4 3 ( p p 微分中值定理与导数应用(第三讲) 52 . ) ( )) ( , ( , ) ( 0 0 0 的拐点 是连续曲线 也可能 点 不存在 若 x f y x f x x f = ¢ ¢ 注意: 微分中值定理与导数应用(第三讲) 53 例11 . 3 的拐点 求曲线 x y = 解 , 0时 当 ¹ x , 3 1 3 2 -= ¢ x y , 9 4 3 5 --= ¢ ¢ x y . , , 0 均不存在 是不可导点 y y x ¢ ¢ ¢ = , 0 , ) 0 , ( > ¢ ¢ -¥ y 内 但在 ; ] 0 , ( 上是凹的 曲线在-¥ , 0 , ) , 0 ( < ¢ ¢ +¥ y 内 在 . ) , 0 [ 上是凸的 曲线在 +¥ . ) 0 , 0 ( 3 的拐点 是曲线 点 x y = \ 微分中值定理与导数应用(第三讲) 54 曲线的弯曲方向——凹凸性; 改变弯曲方向的点——拐点; 凹凸性的判定. 拐点的求法1, 2. 3.5 小结与思考题 微分中值定理与导数应用(第三讲) 55 思考题 设 ) (x f 在 ) , ( b a 内二阶可导，且 0 ) ( 0 = ¢ ¢ x f ， 其中 ) , ( 0 b a x Î ， 则 , ( 0 x )) ( 0 x f 是否一定为曲 线 ) (x f 的拐点？举例说明. 微分中值定理与导数应用(第三讲) 56 思考题解答 因为 0 ) ( 0 = ¢ ¢ x f 只是 , ( 0 x )) ( 0 x f 为拐点 的必要条件， 故 , ( 0 x )) ( 0 x f 不一定是拐点. 例 4 ) ( x x f = ) , ( +¥ -¥ Î x 0 ) 0 ( = ¢ ¢ f 但 ) 0 , 0 ( 并不是曲线 ) (x f 的拐点. 微分中值定理与导数应用(第三讲) 57 一、 填空题： 1、 若函数 ) (x f y = 在（ b a, ）可导，则曲线 ) (x f y = 在( b a, )内为凹弧的充要条件是__. 2、 曲线上_的点称为曲线的拐点 . 3、 曲线 ) 1 ln( 2 x y + = 的拐点为_. 4、 曲线 ) 1 ln( x y + = 拐点为_. 课堂练习题 微分中值定理与导数应用(第三讲) 58 二、 利用函数图形的凹凸性，证明不等式： 2 2 2 ) ( ) ( 2 y x y x + > + ， ) ( y x ¹ . 三、 试决定 2 2 ) 3 ( -= x k y 中k 的值,使曲线在拐点处 的法线通过原点 . 微分中值定理与导数应用(第三讲) 59 一、1、 内 在 ) , ( b a ， ) (x f ¢ 递增或 0 ) ( > ¢ ¢ x f ； 2、凹凸部分的分界点；3、 ) 2 ln , 1 ( ), 2 ln , 1 (-； 4、不存在. 二、 . 2 ) ( ) ( ) 2 ( , ) ( 2 y f x f y x f x x f + < + \ = 且为凹弧 ! 三、 8 2 ± = k . 课堂练习题答案 微分中值定理与导数应用(第三讲) 60 1、渐近线 3.6 函数的图形的描绘 当曲线 ) (x f y = 上的一个动点P 沿此曲线移 向无穷远时， 如果点P 到某定直线L 的距离趋向 于零，那么直线L 称为曲线 ) (x f y = 的渐近线. 微分中值定理与导数应用(第三讲) 61 水平渐近线 ) ( 轴的渐近线 平行于x . ) ( ) ( ) ( lim ) ( lim 的一条水平渐近线 就是 那么 为常数 或 如果 x f y b y b b x f b x f x x = = = = -¥ ® +¥ ® 如, , arctan x y = 有水平渐近线两条: . 2 , 2 p -= p = y y 微分中值定理与导数应用(第三讲) 62 铅直渐近线 ) ( 轴的渐近线 垂直于x . ) ( ) ( lim ) ( lim 0 0 0 的一条铅直渐近线 就是 那么 或 如果 x f y x x x f x f x x x x = = ¥ = ¥ = -+ ® ® 如, , ) 3 )( 2 ( 1 -+ = x x y 有铅直渐近线两条: . 3 , 2 = -= x x 微分中值定理与导数应用(第三讲) 63 斜渐近线 . ) ( ) , ( 0 )] ( ) ( [ lim 0 )] ( ) ( [ lim 的一条斜渐近线 就是 那么 为常数 或 如果 x f y b ax y b a b ax x f b ax x f x x = + = = + -= + --¥ ® +¥ ® 斜渐近线求法: , ) ( lim a x x f x = ¥ ® . ] ) ( [ lim b ax x f x = -¥ ® . ) ( 的一条斜渐近线 就是曲线 那么 x f y b ax y = + = 微分中值定理与导数应用(第三讲) 64 注意: ; ) ( lim ) 1 ( 不存在 如果 x x f x ¥ ® , ] ) ( [ lim , ) ( lim ) 2 ( 不存在 但 存在 ax x f a x x f x x -= ¥ ® ¥ ® . ) ( 不存在斜渐近线 可以断定 x f y = 例12 . 1 ) 3 )( 2 ( 2 ) ( 的渐近线 求 -+ -= x x x x f 解 ). , 1 ( ) 1 , ( : +¥ -¥ ! D 微分中值定理与导数应用(第三讲) 65 = + ® ) ( lim 1 x f x ! , ¥ -= -® ) ( lim 1 x f x , ¥ + . 1是曲线的铅直渐近线 = \ x = ¥ ® x x f x ) ( lim ! 又 ) 1 ( ) 3 )( 2 ( 2 lim -+ -¥ ® x x x x x , 2 = ] 2 1 ) 3 )( 2 ( 2 [ lim x x x x x --+ -¥ ® 1 ) 1 ( 2 ) 3 )( 2 ( 2 lim ---+ -= ¥ ® x x x x x x , 4 = . 4 2 是曲线的一条斜渐近线 + = \ x y 微分中值定理与导数应用(第三讲) 66 的两条渐近线如图 1 ) 3 )( 2 ( 2 ) ( -+ -= x x x x f 微分中值定理与导数应用(第三讲) 67 利用函数特性描绘函数图形，步骤如下： 第一步 第二步 确定函数 ) (x f y = 的定义域,对函数进行奇 偶性、周期性、曲线与坐标轴交点等性态的讨论, 求出函数的一阶导数 ( ) f x ¢ 和二阶导数 ( ) f x ¢¢ ; 求出方程 ( ) 0 f x ¢ = 和 ( ) 0 f x ¢¢ = 在函数定义 域内的全部实根， 用这些根同函数的间断点或导数 不存在的点把函数的定义域划分成几个部分区间. 2、函数图形的描绘 微分中值定理与导数应用(第三讲) 68 第三步确定在这些部分区间内 ( ) f x ¢ 和 ( ) f x ¢¢ 的符 号，并由此确定函数的增减性与极值及曲线的凹凸 性与拐点（可列表进行讨论） ； 第四步确定函数图形的水平、铅直渐近线、斜渐 近线以及其他变化趋势; 第五步 描出与方程 ( ) 0 f x ¢ = 和 ( ) 0 f x ¢¢ = 的根对应 的曲线上的点，有时还需要补充一些点，再综合 前四步讨论的结果画出函数的图形. 微分中值定理与导数应用(第三讲) 69 作图举例： 例13 . 2 ) 1 ( 4 ) ( 2 的图形 作函数 -+ = x x x f 解 , 0 : ¹ x D 非奇非偶函数,且无对称性. , ) 2 ( 4 ) ( 3 x x x f + -= ¢ . ) 3 ( 8 ) ( 4 x x x f + = ¢ ¢ , 0 ) ( = ¢ x f 令 , 2 -= x 得驻点 , 0 ) ( = ¢ ¢ x f 令 . 3 -= x 得特殊点 ] 2 ) 1 ( 4 [ lim ) ( lim 2 -+ = ¥ ® ¥ ® x x x f x x , 2 -= ; 2 -= y 得水平渐近线 微分中值定理与导数应用(第三讲) 70 ] 2 ) 1 ( 4 [ lim ) ( lim 2 0 0 -+ = ® ® x x x f x x , +¥ = . 0 = x 得铅直渐近线 列表确定函数升降区间,凹凸区间及极值点和拐点: x ) 3 , ( --¥ ) , 0 ( +¥ ) 2 , 3 ( --3 -) 0 , 2 (-) (x f ¢ ) (x f + -+ 0 0 ) (x f ¢ ¢ 2 -0 --+ + -不存在 拐点 极小值 间 断 点 3 -) 9 26 , 3 ( ---+ 微分中值定理与导数应用(第三讲) 71 : 补充点 ); 0 , 3 1 ( ), 0 , 3 1 ( + -), 2 , 1 ( --A ), 6 , 1 ( B ). 1 , 2 ( C 作图 x y o 2 -3 -2 1 1 1 -2 -3 -6 A B C 微分中值定理与导数应用(第三讲) 72 2 ) 1 ( 4 ) ( 2 -+ = x x x f 微分中值定理与导数应用(第三讲) 73 例14 解 ), , ( : +¥ -¥ D 偶函数, 图形关于y轴对称. 2 2 ( ) e , 2π x x x j -¢ = -( ) 0, x j¢ = 令 , 0 = x 得驻点 ( ) 0, x j¢¢ = 令 . 1 , 1 = -= x x 得特殊点 1 :0 ( ) 0.4. 2π W x j < £ » 2 2 ( 1)( 1) ( ) e . 2π x x x x j -+ -¢¢ = -2 2 1 lim ( ) lim e 2 x x x x j p -®¥ ®¥ = ! , 0 = . 0 = y 得水平渐近线 作函数 2 2 1 ( ) e 2π x x j -= 的图形. 微分中值定理与导数应用(第三讲) 74 x ) 1 , ( --¥ ) , 1 ( +¥ ) 0 , 1 (-1 -) 1 , 0 ( ) (x j¢ ) (x j + -+ 0 0 ) (x j¢ ¢ 0 1 --+ + -拐点 极大值 p 2 1 ) 2 1 , 1 ( e p -列表确定函数升降区间,凹凸区间及极值点与拐点: 0 拐点 ) 2 1 , 1 ( e p x y o 1 1 -1 2π + --微分中值定理与导数应用(第三讲) 75 2 2 1 ( ) e 2π x x j -= 微分中值定理与导数应用(第三讲) 76 例15 . 1 ) ( 2 3 的图形 作函数 + --= x x x x f 解 ), , ( : +¥ -¥ D 无奇偶性及周期性. ), 1 )( 1 3 ( ) ( -+ = ¢ x x x f ). 1 3 ( 2 ) ( -= ¢ ¢ x x f , 0 ) ( = ¢ x f 令 . 1 , 3 1 = -= x x 得驻点 , 0 ) ( = ¢ ¢ x f 令 . 3 1 = x 得特殊点 : 补充点 ), 0 , 1 (-A ), 1 , 0 ( B ). 8 5 , 2 3 ( C 列表确定函数升降、凹凸区间及极值点与拐点: 微分中值定理与导数应用(第三讲) 77 x ) 3 1 , ( --¥ ) , 1 ( +¥ ) 3 1 , 3 1 (-3 1 -) 1 , 3 1 ( + -+ 0 3 1 1 --+ + -拐点 极大值 27 32 ) 27 16 , 3 1 ( 0 ) (x f ¢ ) (x f ) (x f ¢ ¢ 极小值 0 x y o ) 0 , 1 (-A ) 1 , 0 ( B ) 8 5 , 2 3 ( C 1 1 -3 1 3 1 -微分中值定理与导数应用(第三讲) 78 1 2 3 + --= x x x y 微分中值定理与导数应用(第三讲) 79 函数图形的描绘综合运用函数性态的研究,是导 数应用的综合考察. x y o a b 最 大 值 最 小 值 极 大 值 极 小 值 拐 点 凹的 凸的 单增 单减 ) (x f y = 3.7 小结与思考题 微分中值定理与导数应用(第三讲) 80 思考题 两坐标轴 0 = x ， 0 = y 是否都是 函数 x x x f sin ) ( = 的渐近线？ 微分中值定理与导数应用(第三讲) 81 思考题解答 0 sin lim = ¥ ® x x x ! 0 = \ y 是其图象的水平渐近线. 0 = \ x 不是其图象的渐近线. ¥ ¹ = ® 1 sin lim 0 x x x ! x x y sin = 微分中值定理与导数应用(第三讲) 82 一、 填空题： 1、 曲线 1 e x y = 的水平渐近线为__； 2、 曲线 1 1 -= x y 的水平渐近线为___， 铅直渐近线为__. 二、 描出下列函数的图形： 1、 2 2 ) 1 ( -= x x y ； 2、 x y sin ln = ； 3、 x x y 1 + = . 课堂练习题 微分中值定理与导数应用(第三讲) 83 一、1、 1 = y ； 2、 1 , 0 = = x y . 二、1、 课堂练习题答案 微分中值定理与导数应用(第三讲) 84 2、 3、 微分中值定理与导数应用(第三讲) 85 最值的求法： o x y o x y b a o x y a b a b . ] , [ ) ( ] , [ ) ( 方法求得 用本节的 上的最大值与最小值可 在 点，则 限个临界 上连续，并且至多有有 在 若函数 b a x f b a x f 3.8 导数的应用 微分中值定理与导数应用(第三讲) 86 求最值的步骤： 1. 求函数的临界点; 2. 求区间端点及临界点的函数值,比较大小,最 大者即最大值,最小者即最小值. 注意:如果区间内只有一个极(大或小)值,则这 个极(大或小)值就是最(大或小)值。 微分中值定理与导数应用(第三讲) 87 应用举例： 例16 解 ) 1 )( 2 ( 6 ) ( -+ = ¢ x x x f ! . ] 4 , 3 [ 14 12 3 2 2 3 上的最大值与最小值 的在 求函数 -+ -+ = x x x y 得 解方程 , 0 ) ( = ¢ x f . 1 , 2 2 1 = -= x x 计算 = - ) 3 ( f ; 23 = - ) 2 ( f ; 34 = ) 1 ( f ; 7 ; 142 = ) 4 ( f 微分中值定理与导数应用(第三讲) 88 ， 最大值 142 ) 4 ( = f 比较得 . 7 ) 1 ( = f 最小值 14 12 3 2 2 3 + -+ = x x x y 微分中值定理与导数应用(第三讲) 89 点击图片任意处播放\暂停 例17 一汽车从河的北岸A处以1千米/分钟的 速度向正北行驶，同时一摩托车从河的南岸B 处向正东追赶，速 度为2千米/分钟． 问摩托车何时与汽 车相距最近？ 微分中值定理与导数应用(第三讲) 90 解 0.5千米 (1)建立两车相距函数关系： ( ), t B 设为摩托车从处出发起 追赶的时间分 两车相距 2 2 ) 2 4 ( ) 5 . 0 ( ) ( t t t s -+ + = 4千米 B× A × ) (t s . ) ( ) 2 ( 的最小值点 求 t s s = = ¢ ) (t s . ) 2 4 ( ) 5 . 0 ( 5 . 7 5 2 2 t t t -+ + -, 0 ) ( = ¢ t s 令 得唯一驻点 . 5 . 1 = t . 5 . 1 分钟距离最近 处起追赶后 故摩托车从B 微分中值定理与导数应用(第三讲) 91 实际问题求最值应注意: (1) 建立目标函数; (2) 求最值; 最小值． 值即为所求的最大值或 点，则该点的函数 若目标函数只有唯一驻 1800 50 100 x æ ö -é ù -ç ÷ ê ú ë û è ø 微分中值定理与导数应用(第三讲) 92 例18 某房地产公司有50套公寓要出租，当租 金定为每月1800元时，公寓会全部租出去，当 租金每月增加100元时，就有一套公寓租不出 去，而租出去的房子每月需花费200元的整修 维护费．试问房租定为多少可获得最大收入？ 解设房租为每月元，租出去的房子有 x （元）. 1800 50 100 x -é ù - ê ú ë û 套，每月总收入为： 0( ) R x ( 200) x = -微分中值定理与导数应用(第三讲) 93 ( ) ( 200) 68 100 x R x x æ ö = --ç ÷ è ø 令 1 ( ) 68 ( 200) 100 100 x R x x æ ö æ ö ¢ = -+ --ç ÷ ç ÷ è ø è ø 70 . 50 x = -0 ) ( = ¢ x R 3500 x Þ = （唯一驻点） 故每月每套租金为3500元时收入最高。 最大收入为： 3500 ( ) (3500 200) 68 100 R x æ ö = --ç ÷ è ø 108 900( ). = 元 微分中值定理与导数应用(第三讲) 94 点击图片任意处播放\暂停 例19 形面积最大． 所围成的三角 及 线 处的切线与直 使曲线在该点 上求一点， 曲边 成一个曲边三角形，在 围 及抛物线 ， 由直线 8 0 8 0 2 2 = = = = = = x y x y x y x y 微分中值定理与导数应用(第三讲) 95 解 如图, ), , ( 0 0 y x P 设所求切点为 为 则切线PT ), ( 2 0 0 0 x x x y y -= -, 2 0 0 x y = ! ), 0 , 2 1 ( 0 x A \ ) 16 , 8 ( 2 0 0 x x B -), 0 , 8 ( C T x y o P A B C ) 16 )( 2 1 8 ( 2 1 2 0 0 0 x x x S ABC --= \ D ) 8 0 ( 0 £ £ x 微分中值定理与导数应用(第三讲) 96 , 0 ) 16 16 64 3 ( 4 1 0 2 0 = ´ + -= ¢ x x S 令 解得 ). ( 16 , 3 16 0 0 舍去 = = x x 8 ) 3 16 ( -= ¢ ¢ s ! . 0 < . 27 4096 ) 3 16 ( 为极大值 = \s . 27 4096 ) 3 16 ( 最大者 为所有三角形中面积的 故 = s 微分中值定理与导数应用(第三讲) 97 注意最值与极值的区别. 最值是整体概念而极值是局部概念. 实际问题求最值的步骤. 3.8 小结与思考题 微分中值定理与导数应用(第三讲) 98 思考题 若 ) (a f 是 ) (x f 在 ] , [ b a 上的最值， 且 ) (a f ¢ 存 在，那么是否一定有 0 ) ( = ¢ a f ？ 微分中值定理与导数应用(第三讲) 99 思考题解答 不成立. 因为最值点不一定是内点. 例 x x f y = = ) ( ] 1 , 0 [ Î x 在 有最小值，但 0 = x + (0) (0) 1 0. f f ¢ ¢ = = ¹ 微分中值定理与导数应用(第三讲) 100 一、 填空题： 1.最值可在处取得. 2.函数 2 3 3 2 x x y -= ( 5 0 £ £ x )的最大值为；最 小值为_. 3.函数 2 100 x y -= 在[0,8]上的最大值为；最 小值为__. 二、 求函数 x x y 16 2 -= ( 0 < x )的最小值. 三、 求数列 þ ý ü î í ì n n 2 2 的最大项 . 课堂练习题 微分中值定理与导数应用(第三讲) 101 一、1、区间端点及临界点； 2、最大值 175 ) 5 ( = y , 最小值 1 ) 1 ( -= y ； 3、10, 6. 二、当 2 -= x 时，函数有最小值12. 课堂练习题答案 三、 数列 þ ý ü î í ì n n 2 2 的第3 项： 2 3 3 9 2 8 = .
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https://www.youtube.com/watch?v=IVBIgPRfyA0
What Are Real-World Examples Of Entropy? - Physics Frontier Physics Frontier 2150 subscribers 1 likes Description 67 views Posted: 1 Apr 2025 What Are Real-World Examples Of Entropy? In this engaging video, we take a closer look at the fascinating concept of entropy and its impact on our everyday lives. We will break down what entropy means and how it manifests in various scenarios, from the natural world to everyday experiences. By examining real-world examples, we aim to illustrate how disorder and randomness influence the systems around us. We will explore how the melting of ice demonstrates the transition from order to disorder, and how aging in living organisms reflects the gradual increase in entropy over time. Additionally, we will discuss the thermodynamic principles at play when a hot cup of coffee cools down and the mixing of substances like cream in coffee. Finally, we will touch on the broader implications of entropy in the universe, highlighting its role in the evolution of cosmic systems. This video is designed to provide a clear understanding of how entropy operates in different contexts. Join us for this informative journey into the world of physics, and don't forget to subscribe for more engaging content on scientific concepts! ⬇️ Subscribe to our channel for more valuable insights. 🔗Subscribe: Entropy #Physics #Thermodynamics #Science #MeltingIce #AgingProcess #HeatTransfer #Coffee #MixingSubstances #Universe #CosmicEntropy #NaturalWorld #EverydayPhysics #ScientificConcepts #LearningPhysics #PhysicsEducation About Us: Welcome to Physics Frontier, your gateway to the captivating world of physics and astronomy! Dive deep with us as we explore the mysteries of the universe, from the birth of stars in nebulae to the enigmatic nature of black holes. Our channel covers a rich tapestry of topics, including celestial mechanics, cosmology, and the latest discoveries from NASA and the Leibniz-Institute. Transcript: what are real world examples of entropy imagine walking into a room where everything is perfectly organized the books are lined up by size the papers are stacked neatly and the furniture is arranged symmetrically now picture that same room after a party the books are scattered papers are crumpled and chairs are out of place this shift from order to disorder is a simple illustration of a concept known as entropy entropy is a measure of disorder or Randomness in a system it plays a significant role in various real world scenarios and understanding it can help us grasp how systems evolve over time one common example of entropy can be observed in the melting of ice when ice is in its solid state the molecules are closely packed and organized as the ice melts into water the molecules move more freely resulting in a higher level of disorder this transition from solid to liquid demonstrates how entropy increases as a system changes another example can be found in the process of Aging living organisms start with a certain level of organization and complexity over time as cells divide and repair mechanisms slow down the organization diminishes leading to a state of increased disorder this natural progression illustrates how entropy affects biological systems in the realm of thermodynamics consider a hot cup of coffee left on a table initially the coffee is hot and the surrounding air is cooler over time heat will transfer from the coffee to the air until both reach a similar temperature this process represents an increase in entropy as the energy becomes more evenly distributed between the coffee and its environment another instance of entropy can be seen in the mixing of different substances when you add cream to Coffee the initially separate layers blend together creating a uniform mixture this mixing process increases disorder showcasing how entropy operates in everyday life lastly think about the universe itself the second law of thermodynamics states that the total entropy of an isolated system can never decrease over time as the universe expands it moves towards a state of Maximum entropy where energy is evenly distributed and no usable energy remains this Cosmic perspective highlights the fundamental nature of entropy in shaping the fate of the universe these examples illustrate how entropy manifests in various aspects of Our Lives from the melting of ice to the aging process and even the behavior of the universe understanding these realworld applications of entropy can provide a clearer picture of how disorder and Randomness influence the world around us
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https://www.ahajournals.org/doi/10.1161/strokeaha.107.490672
Diagnostic Criteria of Vascular Dementia in CADASIL | Stroke Skip to main content Advertisement Become a member Volunteer Donate Journals BrowseCollectionsSubjectsAHA Journal PodcastsTrend Watch ResourcesCMEAHA Journals @ MeetingsJournal MetricsEarly Career Resources InformationFor AuthorsFor ReviewersFor SubscribersFor International Users Alerts 0 Cart Search Sign inREGISTER Quick Search in Journals Enter search term Quick Search anywhere Enter search term Quick search in Citations Journal Year Volume Issue Page Searching: This Journal This JournalAnywhereCitation Advanced SearchSearch navigate the sidebar menu Sign inREGISTER Quick Search anywhere Enter search term Publications Arteriosclerosis, Thrombosis, and Vascular Biology Circulation Circulation Research Hypertension Stroke Current Issue Archive Journal Information About Stroke Author Instructions Editorial Board Information for Advertisers Features Stroke Alert Podcast Blogging Stroke Early Career Program Journal Awards Journal of the American Heart Association Circulation: Arrhythmia and Electrophysiology Circulation: Cardiovascular Imaging Circulation: Cardiovascular Interventions Circulation: Cardiovascular Quality & Outcomes Circulation: Genomic and Precision Medicine Circulation: Heart Failure Stroke: Vascular and Interventional Neurology Annals of Internal Medicine: Clinical Cases Information For Authors For Reviewers For Subscribers For International Users Submit & Publish Submit to the AHA Editorial Policies Open Science Value of Many Voices Publishing with the AHA Open Access Information Resources AHA Journals CME AHA Journals @ Meetings Metrics AHA Journals Podcast Network Early Career Resources Trend Watch Professional Heart Daily AHA Newsroom Current Issue Archive Journal Information About Stroke Author Instructions Editorial Board Information for Advertisers Features Stroke Alert Podcast Blogging Stroke Early Career Program Journal Awards Submit Reference #1 Research Article Originally Published 7 February 2008 Free Access Diagnostic Criteria of Vascular Dementia in CADASIL Sarah Benisty, MD, Karen Hernandez, MSc, Anand Viswanathan, MD, PhD, Sonia Reyes, MSc, Annie Kurtz, MSc, Michael O’Sullivan, PhD, Marie-Germaine Bousser, MD, Martin Dichgans, MD, PhD, and Hugues Chabriat, MD, PhDAuthor Info & Affiliations Stroke Volume 39, Number 3 19,191 30 Metrics Total Downloads 19,191 Last 12 Months 1,064 Total Citations 30 Last 12 Months 0 View all metrics Track CitationsAdd to favorites PDF/EPUB Contents Abstract Methods Results Discussion Acknowledgments References eLetters Information & Authors Metrics & Citations View Options References Figures Tables Media Share Abstract Background and Purpose— Subcortical ischemic vascular dementia (SIVD) is a major subtype of vascular dementia (VaD). Recently, the diagnostic criteria of VaD have been modified to encompass this entity. Application of these criteria in CADASIL, a genetic model of SIVD, may help to better assess their significance. The aim of this study was to compare different sets of diagnostic criteria of VaD in a population of CADASIL patients. Methods— Different sets of diagnostic criteria of VaD (DSMIV, ICD10, standard NINDS-AIREN, modified NINDS-AIREN for SIVD) were applied to 115 CADASIL patients. Diagnosis of VaD was made through 2 steps: (1) diagnosis of dementia and (2) association of dementia to lesions of vascular origin. The percentage of patients satisfying the different sets and the concordance between these criteria was analyzed. Results— At least 1 set of criteria was satisfied for diagnosis in 29 subjects with dementia. In this group of patients, the sensitivity of the DSM IV, ICD 10, and standard NINDS-AIREN criteria for VaD was, respectively, 79%, 72%, and 45%. In contrast, the sensitivity of the NINDS-AIREN criteria for SIVD was 90%. The incomplete sensitivity of these last criteria was related to the absence of focal signs in some patients. The neuroimaging criteria were satisfied in all patients with dementia. Conclusions— The modified NINDS-AIREN criteria of SIVD are the most sensitive VaD criteria in CADASIL. Among these criteria, the neuroimaging criteria, although poorly specific to dementia, have a complete sensitivity. In contrast, focal signs were inconstant in CADASIL patients with dementia. Vascular dementia (VaD) is the second most common cause of dementia after Alzheimer disease.1 In clinical research, various criteria are proposed for diagnosis of VaD, particularly in therapeutic trials. They are based on 2 major requirements: (1) clinical diagnosis of dementia and (2) determination of its vascular origin. The latter requirement is most problematic because of the frequent overlap between cerebrovascular and degenerative disorders, particularly in the elderly. Four sets of criteria have been essentially used for diagnosis of VaD: Diagnostic and Statistical Manual of Mental Disorders, fourth edition (DSM IV),2 Alzheimer’s Disease Diagnostic and Treatment Centers (ADDTC),3 International Statistical Classification of Diseases, 10th revision (ICD 10),4 and National Institute of Neurological Disorders and Stroke Association Internationale Pour la Recherche et l’Enseignement en Neurosciences (NINDS-AIREN) criteria.5 Previous analyses of these criteria showed that they were not all equivalent6–9 In addition, because these criteria were proposed to cover the wide spectrum of VaD, none of them is able to discriminate the main subtypes of vascular dementia such as multi-infarct dementia, strategic infarct dementia, or subcortical ischemic vascular dementia (SIVD).1 SIVD may represent one of the most common forms of VaD. It can be caused by various types of small vessel disease that lead to lacunar infarcts and white matter lesions restricted to subcortical areas.10 This entity was considered homogenous enough to require specific diagnostic criteria. Recently, a set of research criteria derived from the NINDS-AIREN criteria of vascular dementia was proposed.11 These criteria rely on both clinical and radiological data. Further evaluation of their sensitivity and specificity is now needed. In contrast to sporadic small vessel diseases frequently detected during aging in association with degenerative disorders, CADASIL (cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy) is a rare disease with onset occurring between age 40 and 50 years.12 CADASIL is caused by mutations of the Notch 3 gene on chromosome 19.13 It is recognized as the most frequent hereditary cause of SIVD and as a model of “pure” VaD. In CADASIL, brain MRI shows widespread white matter lesions associated with lacunar infarcts of variable extent or number and developing from the third decade.14 The main clinical manifestations include attacks of migraine with aura, recurrent subcortical stroke, mood disturbances, and a progressive cognitive decline leading to dementia.12 The cognitive profile in CADASIL is characteristic of VaD and similar to that reported in sporadic SIVD.15 Because symptoms in CADASIL develop several decades before the onset of common degenerative diseases, confounding by concomitant medical conditions is considerably reduced. Therefore, CADASIL patients with dementia can be used as a reference population to understand the limitations of diagnostic criteria currently used for VaD or recently proposed for SIVD. In the present study, we applied different sets of diagnostic criteria for VaD in a large group of CADASIL patients to determine in this population: (1) the sensitivity of the these diagnostic criteria with respect to all demented patients detected in the cohort; (2) the sensitivity of subcriteria used for attribution of the vascular origin; and (3) the sensitivity of the modified NINDS-AIREN criteria for the diagnosis of SIVD. Methods Subjects One hundred fifteen CADASIL patients followed-up in Lariboisiere Hospital (Paris, France) were included in the study. Their mean age was 53.7±11.7 years, the sex ratio (male/female) was 1.01, and education level was from 3 to 7 (maximal). Diagnosis was confirmed by identification of a typical mutation in the Notch 3 gene. Informed consent was obtained from each subject or from a close relative if the subject was too severely disabled to give written consent. The study was approved by an independent ethics committee. Clinical Data Clinical and demographic data were collected at the time of inclusion with detailed baseline neurological examination including the assessment of presence or absence of gait abnormalities, Parkinsonian syndrome, pyramidal signs, sensory deficit, visual field defect, dysphagia, dysarthria, or ocular palsy. Neuropsychological Evaluation Neuropsychological evaluation was performed in patients without severe depression after evaluation by the psychologist. Global cognitive function was assessed by the Mini-Mental State Examination16 and Mattis Dementia Rating Scale (from 0 to 144 [best score]).17 Memory was assessed by an explicit verbal memory test adapted from the Grober and Buschke procedure.18,19 Executive functions were assessed by subtest of the Wechsler Adult Intelligence Scale-Revised (Similarities and Block Design),20 the Trail Making Test B,21 the Wisconsin Card Sorting test (revised version by Nelson),22 and both semantic and letter verbal fluency. Except for the Wisconsin test, performances in each test were converted into z scores using normative data reported from the literature and data from healthy populations.19 In addition to these neuropsychological tests, activities of daily living and disability attributable to cognitive impairment were assessed based on a French translation of Lawton’s scale of Instrumental Activities of Daily Living (IADL).23 The scoring of this scale was based on the interview of the patient or a close relative. Each of 8 IADL (ability to handle finances, to use the telephone, mode of transport, responsibility for own medication, etc) was assessed according to a 3- to 5-grade scale. Then the patients were given a score of 1 if they were considered nonrestricted or were given a score of 0 if they were restricted (score 0). A global score ranging from 0 to 8 was obtained. Magnetic Resonance Imaging MRI scans were obtained by the use of a 1.5-T system (Signa General Electric Medical Systems) as already reported.24 The extent of white matter lesions was visually assessed on FLAIR images based on the recommendations specified in the criteria and described further. The number and location of lacunar infarcts, defined as hypointense foci of diameter >3 mm on T1 scans or on FLAIR images, were also recorded when they were not located in areas with high prevalence of widened perivascular spaces.25,26 Diagnosis of Vascular Dementia In the present cohort, 4 different sets of diagnostic criteria (DSMIV, ICD-10, ADDTC, NINDS-AIREN) were considered for the diagnosis of VaD. The diagnosis was performed using a 2-step procedure: (1) diagnosis of dementia and (2) association of the dementia to lesions of vascular origin. Diagnosis of Dementia For the DSM IV, ICD-10, and standard NINDS-AIREN criteria for VaD, although slight differences in its formulation, the diagnosis of dementia in each patient was made by 1 of 2 expert neurologists, after discussion and agreement with the neuro-psychologist who performed the cognitive testing, according to a unique definition based on the DSM IV criteria for dementia.2 For the modified NINDS-AIREN research criteria for SIVD,11 the diagnosis of dementia was made separately based on available neuropsychological data and using a criteria-based algorithm. This definition of dementia includes a memory deficit that can be mild or severe and characterized by impaired recall, relative intact recognition, less severe forgetting, and benefit from cues, and a dysexecutive syndrome. This cognitive deficit should interfere with complex (executive) activities of daily living and not be attributable to physical effects of the cerebrovascular disease alone. Thus, in the present study, a patient was considered to be demented when presenting with a cognitive deficit characterized by reduced performances (z scores ≤1.65) for the total free recall (memory impairment) of the Grober and Buschke procedure and for at least 2 of the following tests of executive function: Similarities, Cubes, Trail Making Test B, verbal fluency, or <6 criteria found with the Wisconsin test. Additionally, these cognitive alterations had to be associated with at least 2 impaired instrumental activities of daily living as reflected by IADL score ≤6 (IADL items were preferred to ADL items because they are altered earlier in the course of dementia27). Attribution of Dementia to Lesions of Vascular Origin According to DSMIV, ICD 10, ADDTC, NINDS-AIREN, and Modified NINDS-AIREN Diagnostic Criteria of VaD The different set of criteria considered in this study require various neuroimaging evidence of cerebrovascular disease, focal neurological signs, or history of stroke with a temporal relationship with dementia. DSM IV criteria for VaD require only neuroimaging evidence of cerebrovascular disease judged to be etiologically related to cognitive alterations.2 ICD10 criteria for VaD require neuroimaging evidence of cerebrovascular disease “reasonably judged to be etiologically related to dementia” and the presence of focal neurological findings such as unilateral spastic weakness of the limb, unilaterally increased tendon reflexes, extensor plantar response, or pseudobulbar palsy.4 The ADDTC criteria for possible VaD include clinical and neuroimaging evidence of Binswanger disease, but without any specification for this “evidence.”28 To avoid redundancy with other sets of criteria in the absence of clear definition of “Binswanger’s disease,” we chose not to include these criteria in the present analysis. According to the standard NINDS-AIREN criteria,5 the diagnosis of probable VaD in patients with dementia requires the following conditions: (1) presence of focal signs at neurological examination such as hemiparesis, lower facial weakness, Babinski sign, sensory deficit, hemianopia, or dysarthria; (2) evidence of a temporal relationship between the onset of dementia and stroke or an abrupt deterioration or fluctuating/stepwise course; and (3) presence on brain imaging of extensive periventricular white matter lesions involving at least 25% of the total white matter or multiple basal ganglia and white matter lacunes. Finally, the NINDS-AIREN modified research criteria for subcortical ischemic vascular dementia11 require clinical evidence of cerebrovascular disease, ie, focal signs such as hemiparesis, lower facial weakness, Babinski sign, sensory deficit, dysarthria, gait disorders, extrapyramidal signs (others signs such as early urinary symptoms, behavioral symptoms, dysphagia were considered to “support” the diagnosis; and brain imaging evidence of cerebrovascular disease. Two types of cases were distinguished by visual assessment on MRI: (1) predominantly white matter cases characterized by extending periventricular and deep white matter lesions: extending caps or irregular halo (>10 mm broad) and diffuse confluent hyperintensities (>25 mm) and at least 1 lacunar infarct in the deep gray matter; and (2) predominant “lacunar cases” in which multiple lacunes ≥5 in the deep gray matter were associated with at least moderate white matter lesions. Statistical Procedure The different percentages of patients satisfying the various sets of diagnostic criteria for dementia as well as for vascular dementia were estimated in the whole cohort. The sample of patients satisfying at least one set of criteria for dementia was considered as the reference population for analysis of sensitivity. This population, the largest sample with demented subjects (to reduce bias related to the definition of dementia), was chosen to compare different sets of diagnostic criteria. The results obtained after using the diagnostic criteria for dementia are presented first with the main clinical characteristics of the demented population. Thereafter, the MRI and clinical features of demented patients are detailed according to the different sets of diagnostic criteria. The distribution of vascular dementia and concordance between these sets of criteria (for each pair of sets, percentage of patients satisfying both criteria) are then reported. Results Diagnosis of Dementia Among the 115 patients included in the study, 23 (20%) subjects fulfilled the DSM IV criteria for dementia and 28 (24%) patients satisfied the conditions of the modified NINDS-AIREN criteria of dementia for SIVD (Table 1). Note that 34 patients who had significant impairment in executive functions did not have dementia (29% of the whole cohort). There were 22 overlapping cases between these sets of criteria and a total of 29 patients who satisfied either the first set of criteria or the second (concordance between the 2 definitions on the whole cohort of 94%). This group of 29 patients served as the reference group for sensitivity analysis. Their mean age was 62±8.8 years (50.8±11.3 years in the group of patients without dementia), and sex ratio (M/F) was 2.2 (0.8 in the group of patients without dementia). Only a single patient was considered to be demented according to DSM IV criteria but did not fulfill the algorithm derived from the modified NINDS-AIREN criteria of dementia for SIVD (this case did not show any significant impairment of total free recall). Six patients without dementia according to DSM IV criteria fulfilled the conditions required in the modified NINDS-AIREN criteria for SIVD (“nonoverlapping cases”). Average scores of these patients for global cognitive scales and IADL are presented in Table 2. Among the 22 overlapping cases, 15 (70%) had a severe disability (IADL ≤1), whereas the IADL was ≥4 in only 2 patients (8%). In contrast, for 4 out of the 6 “nonoverlapping cases,” the IADL was ≥4 and none had IADL ≤1. Open in Viewer Table 1. Diagnosis of Dementia According to the DSMIV Criteria and to the Definition of the Modified NINDS-AIREN Criteria for SIVD in 115 CADASIL Patients | | n (%) | | The % was obtained on the whole population. | | DSM IV dementia criteria | 23 (20) | | Cognitive syndrome described in the modified NINDS-AIREN criteria for SIVD | 28 (24) | | z score ≤1.65 for total free recall of the Grober procedure | 38 (33) | | z score ≤1.65 in ≥2 tests for dysexecutive syndrome | 62 (54) | | IADL ≤6 | 34 (29) | | Concordance (% of overlapping cases with and without dementia) | 108/115 (94) | Expand Table Open in Viewer Table 2. Average Scores of MMSE, MDRS, and IADL (mean value±SD, range) in Patients Satisfying Both DSMIV and Modified NINDS-AIREN Criteria of Dementia and in Patients Only Satisfying the Modified NINDS-AIREN Criteria of Dementia | | DSMIV and Modified NINDS-AIREN Criteria of Dementia (n=22) | Only Modified NINDS-AIREN Criteria of Dementia (n=6) | --- | | MMSE scores and MDRS were obtained in only 15 of the patients with dementia. Eight subjects were too severely disabled to complete the tasks (27% of all patients with dementia). | | MMSE | 16.4±5.7 | 26.6±3.4 | | MDRS | 84.9±27.3 | 119±9.5 | | IADL | 0.87±1.63 | 4.3±1.2 | Expand Table Attribution of Dementia to Lesions of Vascular Origin According to DSMIV, ICD 10, NINDS-AIREN, and Modified NINDS-AIREN Diagnostic Criteria of VaD In our cohort, all 29 patients who satisfied at least 1 set of diagnostic criteria for dementia presented on MRI with extensive white matter lesions of vascular origin. Therefore, all CADASIL patients fulfilling the DSM IV criteria for dementia (n=23) were considered to satisfy the DSMIV criteria of VaD. Among them, 21 patients also had a pyramidal syndrome or pseudobulbar palsy and thus satisfied the ICD10 criteria. Finally, only 13 of them had a clinical history of stroke associated with focal signs and consequently satisfied the standard NINDS-AIREN criteria of probable VaD (Table 3). Thus, in this group of 23 patients with dementia, the sensitivity of the DSM IV, ICD 10, and standard NINDS-AIREN criteria for VaD was, respectively, 100%, 91%, and 56%. Open in Viewer Table 3. Diagnosis of VaD According to Different Sets of Criteria in 115 CADASIL Patients | VaD Criteria | DSM IV Criteria | ICD-10 Criteria | Standard NINDS-AIREN Criteria | Modified SIVD NINDS-AIREN Criteria | --- --- | CVD indicates cerebrovascular disease; NS, not specified; WML, white matter lesions. | | z score ≤1.65 for total free recall of the Grober procedure. | | z score ≤1.65 in ≥2 tests for dysexecutive syndrome. | | IADL ≤6. | | †Focal signs among the following: hemiparesis, pyramidal signs, sensory deficit, dysarthria, gait disorders, and extrapyramidal signs. | | Definition of dementia | Cognitive deterioration with memory impairment, IADL impaired | Memory deficit, dysexecutive syndrome, IADL impaired | | Attribution of dementia to lesions of vascular origin | Neuroimaging evidence of CVD (NS) | Neuroimaging evidence of CVD such as extensive WML or multiple basal ganglia and WM lacunar infarcts and history of stroke and temporal relationship | Extensive WML and ≥l lacune in deep gray matter or multiple deep gray matter lacunar infarcts | | | ±Focal Signs (NS) | Focal Signs (pyramidal signs) | Focal Signs† | | N of patients | 23 | 21 | 13 | 26 | | Total with dementia, % | 79 | 72 | 45 | 90 | | Total patients, % | 20 | 18 | 11 | 23 | Expand Table All 28 patients satisfying the modified NINDS-AIREN criteria of dementia for SIVD presented with extending caps and confluent hyperintensities in the white matter and at least 1 lacunar infarct in the deep gray matter (range of total number of lacunes from 2 to 39). Thus, all fulfilled the imaging criteria of the modified NINDS-AIREN definition of SIVD and could be considered as “predominantly white matter” cases. However, the complete modified NINDS-AIREN criteria for SIVD were fulfilled by only 26 out of the 28 patients because 2 patients did not show any focal signs at neurological examination. Thus, in this group of 28 patients with dementia, the sensitivity of the modified NINDS-AIREN criteria for SIVD was 93%. In addition, 29 out of 86 (33%) CADASIL patients without dementia also fulfilled the imaging diagnostic criteria of SIVD in the present cohort. If the whole group of patients with dementia was considered as the reference group (n=29), the overall sensitivity of the DSM IV, ICD 10, standard NINDS-AIREN criteria for VaD, and for modified NINDS-AIREN criteria for SIVD was, respectively, 79%, 72%, 45%, and 90%. The detailed results at the individual level after the application of the different sets of diagnostic criteria are presented in Figure. The concordance between the different sets of criteria was incomplete and varied between 86% and 98.2% as detailed in Table 4. This incomplete concordance is illustrated at the individual level by Figure. Finally, the associated clinical features in demented patients according to the modified NINDS-AIREN criteria are detailed in Table 5. Gait disturbances and pyramidal signs were detected in 92% of this population, whereas focal deficits such as sensory or visual field defects were detected in <12%. Open in Viewer Figure.Distribution of vascular dementia according to the different set of criteria. Note that 1 patient satisfying only the modified NINDS-AIREN criteria for dementia but not the complete criteria for SIVD is not included here. Open in Viewer Table 4. Concordance Between Criteria for the Diagnosis of Vascular Dementia in 115 CADASIL Patients | Applied Criteria | Concordance % | --- | | DSM IV/ICD 10 | 98.2 | | DSM IV/standard NINDS-AIREN for probable VaD | 91.3 | | DSM IV/modified NINDS-AIREN for SIVD | 95.6 | | ICD 10/standard NINDS-AIREN for probable VaD | 93 | | ICD 10/modified NINDS-AIREN for SIVD | 93 | | Standard NINDS-AIREN for probable VaD/ modified NINDS-AIREN for SIVD | 86 | Expand Table Open in Viewer Table 5. Clinical Features of Patients Satisfying the Modified NINDS-AIREN Criteria for SIVD (n=26) | | Modified NINDS-AIREN Criteria for SIVD, n=26 (%) | | Pyramidal signs | 24 (92) | | Gait disorders | 24 (92) | | Urinary disorders | 21 (80) | | Clinical history of ≥1 stroke | 15 (57) | | Dysarthria | 14 (54) | | Swallowing problems | 10 (38) | | Extrapyramidal symptoms | 5 (19) | | Sensory deficit | 3 (11) | | Visual field defect | 2 (7) | Expand Table Discussion In this study, we applied different sets of diagnostic criteria of VaD to a cohort of 115 CADASIL patients who were at various stages of the disease. Our study shows that, in a large population of patients with subcortical brain lesions of vascular origin, dementia as defined by the modified criteria for SIVD is observed in a larger population than dementia as defined by the DSMIV criteria, mainly developed for the diagnosis of Alzheimer disease. In addition, in the demented patients satisfying both sets of criteria, ≈70% of subjects had a severe disability. In contrast, none of those exclusively diagnosed by the modified criteria for SIVD (“nonoverlapping cases”) had a similar status and two-thirds had IADL ≥4. Such results suggest that the subcriteria for dementia derived from the modified NINDS-AIREN criteria for SIVD are more sensitive than the DSMIV criteria to early stages of dementia occurring in a small vessel disease of the brain. These data also emphasize the importance of executive dysfunction in the clinical expression of SIVD, and further confirm that the traditional definition of dementia focused on memory impairment is actually too “Alzheimerized”30 for dementia caused by small vessel diseases.31 In addition, in the present study, the use of both sets of criteria of dementia circumscribed less than one-third of the cohort patients. Most of them were at an advanced stage of the disease and 27% patients were even unable to complete the cognitive evaluation because they were too severely impaired. Of note, about one third of subjects in the cohort were outside the demented group but had significant executive alterations in the absence of severe disability. This group of patients may fall within the recently individualized category designated as “vascular mild cognitive impairment.”31,32 The use of a cut-off value for global cognitive scales is a frequent alternative approach to the diagnosis of dementia, particularly in large-scale studies. Herein, the data showed that the mean Mini-Mental State Examination score of demented patients at the early stage of dementia (nonoverlapping cases satisfying only the modified NINDS-AIREN criteria) is 26.6, beyond the usually accepted cut-off score of 24 used for dementia in degenerative disorders.33 Conversely, the mean Mattis Dementia Rating Scale score of these subjects was below the usual cut-off score of 127 found in the literature.34.The underestimation of executive dysfunction in the Mini-Mental State Examination compared with the Mattis Dementia Rating Scale may explain this discrepancy. These results support that the Mattis Dementia Rating Scale, used as a global scale, should be preferred to delineate the group of subjects with dementia in CADASIL and, probably, in other small vessel disorders. The application of different sets of VaD diagnostic criteria in the CADASIL cohort resulted in incompletely overlapping groups as previously reported in different populations.6–9 The DSM IV and ICD-10 criteria for VaD, as well as the standard NINDS-AIREN criteria, were found to be of variable sensitivity in demented patients with CADASIL. The comparison of the Standard NINDS-AIREN criteria for VaD and the modified research ones for SIVD showed that the latter criteria significantly increase the diagnostic sensitivity, which remains, however, <100%. The positive “history of stroke” when used as a criterion was responsible for the poor sensitivity of the NINDS-AIREN definition of probable VaD in our population. In contrast to some subtypes of VaD such as multi-infarct VaD or poststroke dementia,1 for which this criterion is a prerequisite, its use in CADASIL patients appears less valid because the cognitive decline may be progressive and independent of stroke events.35 In addition, our data also show that although the presence of focal signs of stroke was present in most demented subjects, 2 cases did not fulfill the modified NINDS-AIREN criteria for SIVD because of the lack of any focal deficit. These results are in line with those of Pohjaswaara et al36 who previously reported that patients with sporadic SIVD can present without any focal signs of stroke. This suggests that this criterion should be considered as a supportive symptom rather than as a prerequisite in the definition of SIVD. Interestingly, we also observed that the presence of gait disorders and pyramidal symptoms was much more frequent than sensory deficits or visual field defects in CADASIL patients, which suggests that all focal symptoms of stroke are not of equivalent sensitivity for defining SIVD. Imaging criteria are also considered for diagnosis of VaD. In our cohort, all patients with dementia showed extensive white matter lesions and at least 1 lacunar infarct in the deep gray matter as required in the “Binswanger type” definition of SIVD. Extensive white matter lesions on T2 and FLAIR are significantly correlated with clinical severity or executive performances in small vessel diseases.37 However, they may be insufficiently discriminating for the diagnosis of dementia. In a population of poststroke elderly patients, Ballard et al29 observed that the amount of white matter lesions, and the cut-off built into the standard NINDS-AIREN imaging criteria, did not discriminate between patients with and without dementia. In the present cohort, both extensive white matter lesions and the presence of a lacunar infarct were detected in one-third of patients without dementia, which further supports that the NINDS-AIREN imaging criteria of SIVD have poor specificity in this subtype of dementia. The total amount of lacunar lesions and also the degree of atrophy were recently found to be the most potent imaging predictors of cognitive impairment in CADASIL.24,25,38 Further studies may help to determine whether a detailed analysis of location of small deep infarcts, the exact measurement of their load, or the estimation of cerebral atrophy can further improve the diagnosis of SIVD.39 The main limitation of this study concerns the small number of patients with dementia recruited in the cohort. However, this is compensated by the strong homogeneity of the group of patients having an identical vascular disease. In the absence of consensus in the definition of dementia, the calculation of sensitivity required the choice of a reference population for patients with dementia (the largest sample of demented subjects in the present study). Another limitation of this study is that specificity of SIVD criteria could not be determined herein because we evaluated diagnostic criteria in a population with all patients having a vascular disorder. Other populations with dementia but without vascular disorders are needed for this purpose. Conclusion In conclusion, our results demonstrate that the modified NINDS-AIREN criteria proposed for SIVD actually improve the diagnosis of VaD related to small vessel diseases and suggest that some refinements may further improve its sensitivity. The present data suggest that some clinical features such as a positive history of stroke or presence of focal signs should not be mandatory for diagnosis of SIVD. In contrast, whether new imaging features such as the load of lacunar infarctions, the location of such lesions, or estimation of the degree of atrophy may improve the diagnosis of SIVD will need further investigation. Acknowledgments Sources of Funding This work was supported by a PHRC grant AOR 02-001 (DRC/APHP) and performed with the help of ARNEVA (Association de Recherche en Neurologie Vasculaire), Hôpital Lariboisiere, France. Disclosures None. References 1. O’Brien JT, Erkinjuntti T, Reisberg B, Roman G, Sawada T, Pantoni L, Bowler JV, Ballard C, DeCarli C, Gorelick PB, Rockwood K, Burns A, Gauthier S, DeKosky ST. Vascular cognitive impairment. Lancet Neurol . 2003; 2: 89–98. Crossref PubMed Google Scholar a [...] cause of dementia after Alzheimer disease. b [...] ischemic vascular dementia (SIVD). c [...] multi-infarct VaD or poststroke dementia, 2. American Psychiatric Association. Diagnostic and statistical manual of mental disorders. Washington, DC: American Psychiatric Press; 1994. Google Scholar a [...] Mental Disorders, fourth edition (DSM IV), b [...] based on the DSM IV criteria for dementia. c [...] related to cognitive alterations. 3. 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Sign In to Submit a Response to This Article Information & Authors Information Authors Information Published In Stroke Volume 39 • Number 3 • 1 March 2008 Pages: 838 - 844 PubMed: 18258841 Copyright © 2008. Versions You are viewing the most recent version of this article. 7 February 2008: Previous PDF (Version 1) History Received: 13 April 2007 Revision received: 26 July 2007 Accepted: 14 August 2007 Published online: 7 February 2008 Published in print: 1 March 2008 Permissions Request permissions for this article. Request permissions Keywords CADASIL diagnostic criteria leukoariosis vascular dementia Authors Affiliations Expand All Sarah Benisty, MD From Department of Geriatric Medicine (S.B.) and Department of Neurology (K.H., S.R., A.K., M.-G.B., H.C.), University Paris VII, France; Department of Neurology and Clinical Trials Unit (A.V.), Massachusetts General Hospital and Harvard Medical School, Boston, Mass; Department of Neurology (M.O., M.D.), Klinikum Grosshadern, Ludwig-Maximilians-University, Munich, Germany. View all articles by this author Karen Hernandez, MSc From Department of Geriatric Medicine (S.B.) and Department of Neurology (K.H., S.R., A.K., M.-G.B., H.C.), University Paris VII, France; Department of Neurology and Clinical Trials Unit (A.V.), Massachusetts General Hospital and Harvard Medical School, Boston, Mass; Department of Neurology (M.O., M.D.), Klinikum Grosshadern, Ludwig-Maximilians-University, Munich, Germany. View all articles by this author Anand Viswanathan, MD, PhD From Department of Geriatric Medicine (S.B.) and Department of Neurology (K.H., S.R., A.K., M.-G.B., H.C.), University Paris VII, France; Department of Neurology and Clinical Trials Unit (A.V.), Massachusetts General Hospital and Harvard Medical School, Boston, Mass; Department of Neurology (M.O., M.D.), Klinikum Grosshadern, Ludwig-Maximilians-University, Munich, Germany. View all articles by this author Sonia Reyes, MSc From Department of Geriatric Medicine (S.B.) and Department of Neurology (K.H., S.R., A.K., M.-G.B., H.C.), University Paris VII, France; Department of Neurology and Clinical Trials Unit (A.V.), Massachusetts General Hospital and Harvard Medical School, Boston, Mass; Department of Neurology (M.O., M.D.), Klinikum Grosshadern, Ludwig-Maximilians-University, Munich, Germany. View all articles by this author Annie Kurtz, MSc From Department of Geriatric Medicine (S.B.) and Department of Neurology (K.H., S.R., A.K., M.-G.B., H.C.), University Paris VII, France; Department of Neurology and Clinical Trials Unit (A.V.), Massachusetts General Hospital and Harvard Medical School, Boston, Mass; Department of Neurology (M.O., M.D.), Klinikum Grosshadern, Ludwig-Maximilians-University, Munich, Germany. View all articles by this author Michael O’Sullivan, PhD From Department of Geriatric Medicine (S.B.) and Department of Neurology (K.H., S.R., A.K., M.-G.B., H.C.), University Paris VII, France; Department of Neurology and Clinical Trials Unit (A.V.), Massachusetts General Hospital and Harvard Medical School, Boston, Mass; Department of Neurology (M.O., M.D.), Klinikum Grosshadern, Ludwig-Maximilians-University, Munich, Germany. View all articles by this author Marie-Germaine Bousser, MD From Department of Geriatric Medicine (S.B.) and Department of Neurology (K.H., S.R., A.K., M.-G.B., H.C.), University Paris VII, France; Department of Neurology and Clinical Trials Unit (A.V.), Massachusetts General Hospital and Harvard Medical School, Boston, Mass; Department of Neurology (M.O., M.D.), Klinikum Grosshadern, Ludwig-Maximilians-University, Munich, Germany. View all articles by this author Martin Dichgans, MD, PhD From Department of Geriatric Medicine (S.B.) and Department of Neurology (K.H., S.R., A.K., M.-G.B., H.C.), University Paris VII, France; Department of Neurology and Clinical Trials Unit (A.V.), Massachusetts General Hospital and Harvard Medical School, Boston, Mass; Department of Neurology (M.O., M.D.), Klinikum Grosshadern, Ludwig-Maximilians-University, Munich, Germany. View all articles by this author Hugues Chabriat, MD, PhD From Department of Geriatric Medicine (S.B.) and Department of Neurology (K.H., S.R., A.K., M.-G.B., H.C.), University Paris VII, France; Department of Neurology and Clinical Trials Unit (A.V.), Massachusetts General Hospital and Harvard Medical School, Boston, Mass; Department of Neurology (M.O., M.D.), Klinikum Grosshadern, Ludwig-Maximilians-University, Munich, Germany. View all articles by this author Notes Correspondence to Prof Hugues Chabriat, Department of Neurology, University Paris 7, 2 rue Ambroise Paré, 75010 Paris, France. E-mail hugues.chabriat@lrb.aphp.fr Metrics & Citations Metrics Citations 30 Metrics Article Metrics View all metrics Downloads Citations No data available. 19,191 30 Total 6 Months 12 Months Total number of downloads and citations Citations Download Citations If you have the appropriate software installed, you can download article citation data to the citation manager of your choice. Select your manager software from the list below and click Download. Please select your download format: [x] Direct Import Hugues Chabriat, Cerebral Autosomal Dominant Arteriopathy with Subcortical Infarcts and Leukoencephalopathy (CADASIL), Stroke Genetics, (117-137), (2024). Crossref Hugues Chabriat, Saskia Lesnik Oberstein, Cognition, mood and behavior in CADASIL, Cerebral Circulation - Cognition and Behavior, 3, (100043), (2022). Crossref Firoz Akhter, Alicia Persaud, Younis Zaokari, Zhen Zhao, Donghui Zhu, Vascular Dementia and Underlying Sex Differences, Frontiers in Aging Neuroscience, 13, (2021). Crossref E. Ouin, E. Jouvent, Spectre clinico-radiologique des maladies des petits vaisseaux cérébraux, La Revue de Médecine Interne, 41, 7, (459-468), (2020). Crossref Shuhui Cui, Ning Chen, Mi Yang, Jian Guo, Muke Zhou, Cairong Zhu, Li He, Cerebrolysin for vascular dementia, Cochrane Database of Systematic Reviews, (2019). Crossref H. Chabriat, M.-G. Bousser, Cadasil, EMC - Neurologia, 17, 3, (1-14), (2017). Crossref Sara N. Rushia, Ernst Garcon, Joel R. Sneed, The Implications of CADASIL as a Genetic Model of Vascular Depression, The American Journal of Geriatric Psychiatry, 25, 7, (728-729), (2017). Crossref Gretchen E. Tietjen, CADASIL ☆, Reference Module in Neuroscience and Biobehavioral Psychology, (2017). Crossref Nenad Đukić, Mihailo Nešković, Milorad Ševković, Đorđe Radak, Vascular dementia revised, Medicinska istrazivanja, 50, 2, (23-29), (2016). Crossref Fatemeh Mohammadian, Maryam Noroozian, Shahriar Nafissi, Farzad Fatehi, Blink Reflex May Help Discriminate Alzheimer Disease From Vascular Dementia, Journal of Clinical Neurophysiology, 32, 6, (505-511), (2015). Crossref See more Loading... View Options View options PDF and All Supplements Download PDF and All Supplements Download is in progress PDF/EPUB View PDF/EPUB Figures Open all in viewer Figure.Distribution of vascular dementia according to the different set of criteria. Note that 1 patient satisfying only the modified NINDS-AIREN criteria for dementia but not the complete criteria for SIVD is not included here. Go to FigureOpen in Viewer Tables Open all in viewer Table 1. Diagnosis of Dementia According to the DSMIV Criteria and to the Definition of the Modified NINDS-AIREN Criteria for SIVD in 115 CADASIL Patients Go to TableOpen in Viewer Table 2. Average Scores of MMSE, MDRS, and IADL (mean value±SD, range) in Patients Satisfying Both DSMIV and Modified NINDS-AIREN Criteria of Dementia and in Patients Only Satisfying the Modified NINDS-AIREN Criteria of Dementia Go to TableOpen in Viewer Table 3. Diagnosis of VaD According to Different Sets of Criteria in 115 CADASIL Patients Go to TableOpen in Viewer Table 4. Concordance Between Criteria for the Diagnosis of Vascular Dementia in 115 CADASIL Patients Go to TableOpen in Viewer Table 5. Clinical Features of Patients Satisfying the Modified NINDS-AIREN Criteria for SIVD (n=26) Go to TableOpen in Viewer Media Share Share Share article link Copy Link Copied! Copying failed. Share FacebookLinkedInX (formerly Twitter)emailWeChatBluesky References References 1. O’Brien JT, Erkinjuntti T, Reisberg B, Roman G, Sawada T, Pantoni L, Bowler JV, Ballard C, DeCarli C, Gorelick PB, Rockwood K, Burns A, Gauthier S, DeKosky ST. Vascular cognitive impairment. Lancet Neurol . 2003; 2: 89–98. Crossref PubMed Google Scholar a [...] cause of dementia after Alzheimer disease. b [...] ischemic vascular dementia (SIVD). c [...] multi-infarct VaD or poststroke dementia, 2. American Psychiatric Association. Diagnostic and statistical manual of mental disorders. Washington, DC: American Psychiatric Press; 1994. Google Scholar a [...] Mental Disorders, fourth edition (DSM IV), b [...] based on the DSM IV criteria for dementia. c [...] related to cognitive alterations. 3. Chui HC, Victoroff JI, Margolin D, Jagust W, Shankle R, Katzman R. Criteria for the diagnosis of ischemic vascular dementia proposed by the state of California Alzheimer’s disease diagnostic and treatment centers. Neurology . 1992; 42: 473–480. Go to Citation Crossref PubMed Google Scholar 4. World Health Organization. The ICD-10 classification of mental and behavioral disorders: Diagnostic criteria for research. Geneva: WHO; 1993. Google Scholar a [...] of Diseases, 10th revision (ICD 10), b [...] plantar response, or pseudobulbar palsy. 5. Roman GC, Tatemichi TK, Erkinjuntti T, Cummings JL, Masdeu JC, Garcia JH, Amaducci L, Orgogozo JM, Brun A, Hofman A, et al. Vascular dementia: Diagnostic criteria for research studies. Report of the ninds-airen international workshop. Neurology . 1993; 43: 250–260. Crossref PubMed Google Scholar a [...] en Neurosciences (NINDS-AIREN) criteria. b [...] to the standard NINDS-AIREN criteria, 6. Wetterling T, Kanitz RD, Borgis KJ. Comparison of different diagnostic criteria for vascular dementia (ADDTC, DSM-IV, ICD-10, NINDS-AIREN). Stroke . 1996; 27: 30–36. Crossref PubMed Google Scholar a [...] showed that they were not all equivalent b [...] reported in different populations. 7. Pohjasvaara T, Mantyla R, Ylikoski R, Kaste M, Erkinjuntti T. Comparison of different clinical criteria (DSM-III, ADDTC, ICD-10, NINDS-AIREN, DSM-IV) for the diagnosis of vascular dementia. National institute of neurological disorders and stroke-association internationale pour la recherche et l’enseignement en neurosciences. Stroke . 2000; 31: 2952–2957. Crossref PubMed Google Scholar 8. Gold G, Bouras C, Canuto A, Bergallo MF, Herrmann FR, Hof PR, Mayor PA, Michel JP, Giannakopoulos P. Clinicopathological validation study of four sets of clinical criteria for vascular dementia. Am J Psych . 2002; 159: 82–87. Crossref PubMed Google Scholar 9. Lopez OL, Kuller LH, Becker JT, Jagust WJ, DeKosky ST, Fitzpatrick A, Breitner J, Lyketsos C, Kawas C, Carlson M. Classification of vascular dementia in the cardiovascular health study cognition study. Neurology . 2005; 64: 1539–1547. Crossref PubMed Google Scholar 10. Roman GC, Erkinjuntti T, Wallin A, Pantoni L, Chui HC. Subcortical ischaemic vascular dementia. Lancet Neurol . 2002; 1: 426–436. Go to Citation Crossref PubMed Google Scholar 11. Erkinjuntti T, Inzitari D, Pantoni L, Wallin A, Scheltens P, Rockwood K, Roman GC, Chui H, Desmond DW. Research criteria for subcortical vascular dementia in clinical trials. J Neural Transm Suppl . 2000; 59: 23–30. PubMed Google Scholar a [...] criteria of vascular dementia was proposed. b [...] NINDS-AIREN research criteria for SIVD, c [...] for subcortical ischemic vascular dementia 12. Chabriat H, Vahedi K, Iba-Zizen MT, Joutel A, Nibbio A, Nagy TG, Krebs MO, Julien J, Dubois B, Ducrocq X, et al. Clinical spectrum of CADASIL: A study of 7 families. Cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy. Lancet . 1995; 346: 934–939. Crossref PubMed Google Scholar a [...] occurring between age 40 and 50 years. b [...] cognitive decline leading to dementia. 13. Joutel A, Corpechot C, Ducros A, Vahedi K, Chabriat H, Mouton P, Alamowitch S, Domenga V, Cecillion M, Marechal E, Maciazek J, Vayssiere C, Cruaud C, Cabanis EA, Ruchoux MM, Weissenbach J, Bach JF, Bousser MG, Tournier-Lasserve E. Notch3 mutations in CADASIL, a hereditary adult-onset condition causing stroke and dementia. Nature . 1996; 383: 707–710. Go to Citation Crossref PubMed Google Scholar 14. Chabriat H, Levy C, Taillia H, Iba-Zizen MT, Vahedi K, Joutel A, Tournier-Lasserve E, Bousser MG. Patterns of MRI lesions in CADASIL. Neurology . 1998; 51: 452–457. Go to Citation Crossref PubMed Google Scholar 15. Charlton RA, Morris RG, Nitkunan A, Markus HS. The cognitive profiles of CADASIL and sporadic small vessel disease. Neurology . 2006; 66: 1523–1526. Go to Citation Crossref PubMed Google Scholar 16. Folstein MF, Folstein SE, McHugh PR. “Mini-mental state.” A practical method for grading the cognitive state of patients for the clinician. J Psychiatr Res . 1975; 12: 189–198. Go to Citation Crossref PubMed Google Scholar 17. Mattis S. Mental status examination for organic mental syndrome in elderly patients. In: Bellak LKT, ed. Geriatric psychiatry: A handbook for psychiatrists and primary care physicians. New York: Grune&Stratton 1976: 77–121. Go to Citation Google Scholar 18. Van der Linden MWC, Bruyer R, Ansay C, Seron X. Educational level and cued recall performance in older and young adults. Psychologica Belgica . 1993; 33: 37–47. Go to Citation Google Scholar 19. Buffon F, Porcher R, Hernandez K, Kurtz A, Pointeau S, Vahedi K, Bousser MG, Chabriat H. Cognitive profile in CADASIL. J Neurol Neurosurg Psych . 2006; 77: 175–180. Go to Citation Crossref PubMed Google Scholar 20. Wechsler D. Wechsler adult intelligence scale revised (WAIS-R). 1981. les Editions du Centre de Psychologie Appliqué, 1991. Go to Citation Google Scholar 21. Reitan RM. The relation of the trail making test to organic brain damage. J Consult Psychol . 1955; 19: 393–394. Go to Citation Crossref PubMed Google Scholar 22. Nelson HE. A modified card sorting test sensitive to frontal lobe defects. Cortex . 1976; 12: 313–324. Go to Citation Crossref PubMed Google Scholar 23. Lawton MP, Brody EM. Assessment of older people: Self-maintaining and instrumental activities of daily living. Gerontologist . 1969; 9: 179–186. Go to Citation Crossref PubMed Google Scholar 24. Jouvent E, Viswanathan A, Mangin JF, O’Sullivan M, Guichard JP, Gschwendtner A, Cumurciuc R, Buffon F, Peters N, Pachai C, Bousser MG, Dichgans M, Chabriat H. Brain atrophy is related to lacunar lesions and tissue microstructural changes in CADASIL. Stroke . 2007; 38: 1786–1790. Crossref PubMed Google Scholar a [...] Medical Systems) as already reported. b [...] of cognitive impairment in CADASIL. 25. Viswanathan A, Gschwendtner A, Guichard JP, Buffon F, Cumurciuc R, O’Sullivan M, Holtmannspotter M, Pachai C, Bousser MG, Dichgans M, Chabriat H. Lacunar lesions are independently associated with disability and cognitive impairment in CADASIL. Neurology . 2007; 69: 172–179. Go to Citation Crossref PubMed Google Scholar 26. Viswanathan A, Guichard JP, Gschwendtner A, Buffon F, Cumurcuic R, Boutron C, Vicaut E, Holtmannspotter M, Pachai C, Bousser MG, Dichgans M, Chabriat H. Blood pressure and haemoglobin a1c are associated with microhaemorrhage in CADASIL: A two-centre cohort study. Brain . 2006; 129: 2375–2383. Crossref PubMed Google Scholar 27. Barberger-Gateau P, Commenges D, Gagnon M, Letenneur L, Sauvel C, Dartigues JF. Instrumental activities of daily living as a screening tool for cognitive impairment and dementia in elderly community dwellers. J Am Geriatr Soc . 1992; 40: 1129–1134. Go to Citation Crossref PubMed Google Scholar 28. Chui HC, Mack W, Jackson JE, Mungas D, Reed BR, Tinklenberg J, Chang FL, Skinner K, Tasaki C, Jagust WJ. Clinical criteria for the diagnosis of vascular dementia: A multicenter study of comparability and interrater reliability. Arch Neurol . 2000; 57: 191–196. Go to Citation Crossref PubMed Google Scholar 29. Ballard CG, Burton EJ, Barber R, Stephens S, Kenny RA, Kalaria RN, O’Brien JT. NINDS AIREN neuroimaging criteria do not distinguish stroke patients with and without dementia. Neurology . 2004; 63: 983–988. Go to Citation Crossref PubMed Google Scholar 30. Sachdev P. Vascular cognitive disorder. Int J Geriatr Psych . 1999; 14: 402–403. Go to Citation Crossref PubMed Google Scholar 31. Roman GC, Sachdev P, Royall DR, Bullock RA, Orgogozo JM, Lopez-Pousa S, Arizaga R, Wallin A. Vascular cognitive disorder: A new diagnostic category updating vascular cognitive impairment and vascular dementia. J Neurol Sci . 2004; 226: 81–87. Crossref PubMed Google Scholar a [...] dementia caused by small vessel diseases. b [...] as “vascular mild cognitive impairment.” 32. Bowler JV. Vascular cognitive impairment. J Neurol Neurosurg Psych . 2005; 76 (Suppl 5): v35–v44. Crossref PubMed Google Scholar 33. Madureira S, Verdelho A, Ferro J, Basile AM, Chabriat H, Erkinjuntti T, Fazekas F, Hennerici M, O’Brien J, Pantoni L, Salvadori E, Scheltens P, Visser MC, Wahlund LO, Waldemar G, Wallin A, Inzitari D. Development of a neuropsychological battery for the leukoaraiosis and disability in the elderly study (LADIS): Experience and baseline data. Neuroepidemiology . 2006; 27: 101–116. Go to Citation Crossref PubMed Google Scholar 34. Lucas JA, Ivnik RJ, Smith GE, Bohac DL, Tangalos EG, Kokmen E, Graff-Radford NR, Petersen RC. Normative data for the Mattis dementia rating scale. J Clin Exp Neuropsychol . 1998; 20: 536–547. Go to Citation Crossref PubMed Google Scholar 35. Desmond DW, Moroney JT, Lynch T, Chan S, Chin SS, Mohr JP. The natural history of CADASIL: A pooled analysis of previously published cases. Stroke . 1999; 30: 1230–1233. Go to Citation Crossref PubMed Google Scholar 36. Pohjasvaara T, Mantyla R, Ylikoski R, Kaste M, Erkinjuntti T. Clinical features of MRI-defined subcortical vascular disease. Alzheimer Dis Assoc Disord . 2003; 17: 236–242. Go to Citation Crossref PubMed Google Scholar 37. Sachdev PS, Brodaty H, Valenzuela MJ, Lorentz L, Looi JC, Wen W, Zagami AS. The neuropsychological profile of vascular cognitive impairment in stroke and TIA patients. Neurology . 2004; 62: 912–919. Go to Citation Crossref PubMed Google Scholar 38. Peters N, Holtmannspotter M, Opherk C, Gschwendtner A, Herzog J, Samann P, Dichgans M. Brain volume changes in CADASIL: A serial MRI study in pure subcortical ischemic vascular disease. Neurology . 2006; 66: 1517–1522. Crossref PubMed Google Scholar 39. Vermeer SE, Prins ND, den Heijer T, Hofman A, Koudstaal PJ, Breteler MM. Silent brain infarcts and the risk of dementia and cognitive decline. N Engl J Med . 2003; 348: 1215–1222. Go to Citation Crossref PubMed Google Scholar Advertisement Recommended October 2007 The Contribution of Medial Temporal Lobe Atrophy and Vascular Pathology to Cognitive Impairment in Vascular Dementia António J. Bastos-Leite, Wiesje M. van der Flier, Elisabeth C.W. van Straaten, Salka S. Staekenborg, Philip Scheltens, and [...] Frederik Barkhof +2 authors April 2008 Vascular Subcortical Hyperintensities Predict Conversion to Vascular and Mixed Dementia in MCI Patients Stéphanie Bombois, Stéphanie Debette, Amélie Bruandet, Xavier Delbeuck, Christine Delmaire, Didier Leys, and [...] Florence Pasquier +3 authors December 2005 Infratentorial Abnormalities in Vascular Dementia António J. Bastos Leite, Wiesje M. van der Flier, Elisabeth C. W. van Straaten, Philip Scheltens, and [...] Frederik Barkhof +1 authors Advertisement Submit a Response to This Article Close Compose eLetter Title: Comment text: Contributors (all fields are required) Remove Contributor First Name: Last Name: Email: Affiliation: Add Another Contributor Statement of Competing Interests Competing Interests? 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https://www.youcubed.org/tasks/fraction-puzzles/
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https://artofproblemsolving.com/community/c2899h1240082_concurrent_lines?srsltid=AfmBOooYrgWp6gS9gk0yb55b2kA7RZ5kXhvTXvr5MS2n4uODKnUp3vGo
Geometry : Concurrent lines Community » Blogs » Geometry » Concurrent lines Sign In • Join AoPS • Blog Info Geometry ======== Concurrent lines by Puzzled417, May 9, 2016, 4:53 PM 1995 IMO #1: Let be four distinct points on a line, in that order. The circles with diameters and intersect at and . The line meets at . Let be a point on the line other than . The line intersects the circle with diameter at and , and the line intersects the circle with diameter at and . Prove that the lines are concurrent. Solution Let be the points in which respectively meet . We then see that and thus . Similarly from we have . Thus since and must be on the same side of , we conclude that . Generalization Let be four points on a line. The circles with diameters and intersect at and . The line meets at . Let be a point on the line other than . The line intersects the circle with diameter at and , and the line intersects the circle with diameter at and . Then the lines are concurrent if and only if the order of points on the line is one of . This post has been edited 4 times. Last edited by Puzzled417, May 10, 2016, 3:59 PM EGMOEGMO 2.31IMOIMO 1995-1 No comments Post comment Comment 0 Comments Puzzled417 Archives May 2016 Concentric circles Concurrent lines Convex quadrilaterals External isosceles triangles Figure seen from every point at a right angle April 2016 Equal angles from a point Lucas Circles BAMO 2012 #6 The Pitot Theorem Excircle/incircle result Proof of excenter Proof of incenter Proving the orthocenter exists using radical axes Radical axis and a tangent to circles Power of a Point Intersecting Chords Theorem Simson Line Right Angles on Incircle Chord Lemma March 2016 Thales' theorem iff Three tangents lemma Thomson cubic An algebraic system Rational sided and angled triangle January 2016 Proof of incircle property December 2014 Anyone know how to solve? Shouts Submit thanks . It helped me by student1212, Oct 13, 2024, 7:03 AM Nice Blog by makethan, Oct 7, 2024, 1:18 AM 3 year bump by JackLi123, Mar 17, 2024, 6:23 PM nice blog ：D by Penguinlna, Jan 14, 2024, 2:07 AM two year bump bro by huashiliao2020, Jun 12, 2023, 4:33 AM One year bump moment they left aops lmao by the_mathmagician, Nov 27, 2021, 6:12 PM nice blog by fukano_2, Jul 8, 2020, 7:03 AM hello bye by LLL2019, Jun 13, 2020, 6:33 AM Nice asymptote figures by freeman66, Jun 5, 2020, 11:50 PM hello EGMO radical Axis theorem by JustKeepRunning, Jul 28, 2019, 10:35 PM Anyone??? by hellomath010118, Jun 15, 2019, 8:28 AM hello this blog might be ded by mamis511, Feb 6, 2018, 3:50 AM wait wow this blog is amazing. I've done a lot of these problems before though by checkmatetang, Sep 11, 2016, 6:26 PM EGMO PROBLEM 2.12 SECTION 2.3 RADICAL AXES MAYBE by Awesomekid05, Sep 11, 2016, 1:59 AM and in EGMO.... by algebra_star1234, Aug 31, 2016, 4:30 PM I agree by Eugenis, Aug 27, 2016, 2:13 AM Haha your blog posts look suspiciously alike the topics found in 106 Geo... by shiningsunnyday, Apr 11, 2016, 2:12 AM could i have contrib? plz keep this blog alive by oceanair, Dec 13, 2015, 5:11 PM hiiiiiiiii by fxiong2002, Nov 1, 2015, 4:47 PM 19 shouts Tags EGMOIMOUSAMO1.451982-14 IMO SL1982-8 IMO SL1982-9 IMO SL2.12BAMOBAMO 2012-6EGMO 1.44EGMO 1.48EGMO 1.6EGMO 2.11EGMO 2.18EGMO 2.20EGMO 2.25EGMO 2.27EGMO 2.30EGMO 2.31EGMO 2.32EGMO 2.5IMO 1995-1USAMO 1997-2USAMO 1998-2 About Owner Posts: 593 Joined: Jun 5, 2013 Blog Stats Blog created: Jun 11, 2013 Total entries: 25 Total visits: 23655 Total comments: 18 Search Blog Something appears to not have loaded correctly. Click to refresh. a
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https://math.stackexchange.com/questions/61963/clique-and-independent-set-proof
Skip to main content Clique and Independent Set Proof Ask Question Asked Modified 13 years, 11 months ago Viewed 6k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I am currently working on an exercise that is described like so: Prove that a graph G has a clique of size k if and only if G¯¯¯¯ has an independent set of size k, where G¯¯¯¯ is the complement of G. (Note for if and only if proofs: if you wish to prove a statement of the form "A if and only if B", then you must prove "if A then B" and "if B then A"). Proofs are not my strong point and the class notes on this section is very vague. I'm not sure how to go about beginning this proof. I can't visually imagine in my head how proving one graph with a clique size equal to its complement's independent set would provide proof for all future graphs. Can someone please break this down in layman's terms for the thoroughly confused? graph-theory proof-writing Share CC BY-SA 3.0 Follow this question to receive notifications edited Sep 5, 2011 at 6:15 Srivatsan 26.8k77 gold badges9595 silver badges148148 bronze badges asked Sep 5, 2011 at 5:54 raphnguyenraphnguyen 31911 gold badge55 silver badges1111 bronze badges 1 I added a [proof-writing] tag since it seemed appropriate. I also fixed the TeX for you. Hope it's ok. – Srivatsan Commented Sep 5, 2011 at 6:01 Add a comment | 1 Answer 1 Reset to default This answer is useful 2 Save this answer. Show activity on this post. By definition, e is an edge of G if and only if e is not an edge in G¯¯¯¯ (this is a more standard notation for complement). If you have a clique of size k in G, then you have k vertices where every possible edge between them is included. Thus, in G¯¯¯¯, none of these edges are present, and therefore these k vertices form an independent set of size k. Similarly, if you start with an independent set in G¯¯¯¯, there is a corresponding clique in G. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Sep 5, 2011 at 6:09 Austin MohrAustin Mohr 26.3k44 gold badges7575 silver badges128128 bronze badges 5 I couldn't figure out how to type out the complement G. You are all over this board with helpful information! So for my proof, I would write i) e is an edge of G (if and only if arrow) e is not an edge of the complement G. ii. prove this for an instance of G with clique size k? I'm not sure how graph theory proofs are expected to be written. – raphnguyen Commented Sep 5, 2011 at 6:17 Your part i) is just the definition of graph complement. The proof is really as straightforward as it sounds - don't complicate it. A clique is a cluster of vertices with all possible edges. An independent set is a cluster of vertices with no edges. So, if you have one, taking the complement gives you the other. – Austin Mohr Commented Sep 5, 2011 at 6:20 Your explanations are always crystal clear. As soon as you explain it, it clicks for me. I guess I am just more confused as to how I am expected to format the proof. Are graph theory proofs usually done in the format of discrete mathematics proofs? P(n) = ?, take an instance of P(0) to find that this statement is true, then prove P(n+1)? Or is it generally accepted to provide a proof by explanation and not equations. – raphnguyen Commented Sep 5, 2011 at 6:28 You're describing proof by induction. You'll probably find plenty of chances to use it in graph theory, but it isn't necessary here. All "proof" means is "an explanation why something is true". If you can explain it clearly without any fancy techniques like induction, then don't use them. – Austin Mohr Commented Sep 5, 2011 at 6:50 Awesome. Discrete mathematics really pounded induction proofs into my head, so I kept trying to apply the same method for this proof. Thanks again and have a great Labor Day! – raphnguyen Commented Sep 5, 2011 at 7:06 Add a comment | You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions graph-theory proof-writing See similar questions with these tags. Featured on Meta Will you help build our new visual identity? Upcoming initiatives on Stack Overflow and across the Stack Exchange network... 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https://www.seas.upenn.edu/~jean/diffgeom-spr-II.pdf
Differential Geometry and Lie Groups A Second Course Jean Gallier and Jocelyn Quaintance Department of Computer and Information Science University of Pennsylvania Philadelphia, PA 19104, USA e-mail: jean@cis.upenn.edu © Jean Gallier Please, do not reproduce without permission of the authors March 22, 2025 2 To my daughter Mia, my wife Anne, my son Philippe, and my daughter Sylvie. To my parents Howard and Jane. Preface This book is written for a wide audience ranging from upper undergraduate to advanced graduate students in mathematics, physics, and more broadly engineering students, especially in computer science. Basically, it covers topics which belong to a second course in differential geometry. The reader is expected to be familiar with the theory of manifolds and with some elements of Riemannian geometry, including connections, geodesics, and curvature. Some familiarity with the material presented in the following books is more than sufficient: Tu (the first three chapters), Warner (the first chapter and parts of Chapters 2, 3, 4), Do Carmo (the first four chapters), Gallot, Hulin, Lafontaine (the first two chapters and parts of Chapter 3), O’Neill (Chapters 1 and 3), and Gallier and Quaintance , which contains all the preliminaries needed to read this book. The goal of differential geometry is to study the geometry and the topology of manifolds using techniques involving differentiation in one way or another. The pilars of differential geometry are: (1) Riemannian metrics. (2) Connections. (3) Geodesics. (4) Curvature. There are many good books covering the above topics, and we also provided our own account (Gallier and Quaintance ). One of the goals of differential geometry is also to be able to generalize “calculus on Rn” to spaces more general than Rn, namely manifolds. We would like to differentiate functions f : M →R defined on a manifold, optimize functions (find their minima or maxima), and also to integrate such functions, as well as compute areas and volumes of subspaces of our manifold. The generalization of the notion of derivative of a function defined on a manifold is the notion of tangent map, and the notions of gradient and Hessian are easily generalized to manifolds equipped with a connection (or a Riemannian metric, which yields the Levi-Civita connection). However, the problem of defining the integral of a function whose domain is a manifold remains. 3 4 One of the main discoveries made at the beginning of the twentieth century by Poincar´ e and ´ Elie Cartan, is that the “right” approach to integration is to integrate differential forms, and not functions. To integrate a function f, we integrate the form fω, where ω is a volume form on the manifold M. The formalism of differential forms takes care of the process of the change of variables quite automatically and allows for a very clean statement of Stokes’ theorem. The theory of differential forms is one of the main tools in geometry and topology. This theory has a surprisingly large range of applications, and it also provides a relatively easy access to more advanced theories such as cohomology. For all these reasons, it is really an indispensable theory, and anyone with more than a passable interest in geometry should be familiar with it. In this book, we discuss the following topics. (1) Differential forms, including vector-valued differential forms and differential forms on Lie groups. (2) An introduction to de Rham cohomology. (3) Distributions and the Frobenius theorem. (4) Integration on manifolds, starting with orientability, volume forms, and ending with Stokes’ theorem on regular domains. (5) Integration on Lie groups. (6) Spherical harmonics and an introduction to the representations of compact Lie groups. (7) Operators on Riemannian manifolds: Hodge Laplacian, Laplace–Beltrami Laplacian, and Bochner Laplacian. (8) Fibre bundles, vector bundles, principal bundles, and metrics on bundles. (9) Connections and curvature in vector bundles, culminating with an introduction to Pontrjagin classes, Chern classes, and the Euler class. (10) Clifford algebras, Clifford groups, and the groups Pin(n), Spin(n), Pin(p, q) and Spin(p, q). Topics (3)-(7) have more of an analytic than a geometric flavor. Topics (8) and (9) belong to the core of a second course on differential geometry. Clifford algebras and Clifford groups constitute a more algebraic topic. These can be viewed as a generalization of the quaternions. The groups Spin(n) are important because they are the universal covers of the groups SO(n). Since this book is already quite long, we resolved ourselves, not without regrets, to omit many proofs. We feel that it is more important to motivate, demystify, and explain 5 the reasons for introducing various concepts and to clarify the relationship between these notions rather than spelling out every proof in full detail. Whenever we omit a proof, we provide precise pointers to the literature. We must acknowledge our debt to our main sources of inspiration: Bott and Tu , Br¨ ocker and tom Dieck , Cartan , Chern , Chevalley , Dieudonn´ e [32, 33, 34], do Carmo , Gallot, Hulin, Lafontaine , Hirzebruch , Knapp , Madsen and Tornehave , Milnor and Stasheff, Morimoto , Morita , Petersen , and Warner . The chapters or sections marked with the symbol ⊛contain material that is typically more specialized or more advanced, and they can be omitted upon first (or second) reading. Acknowledgement: We would like to thank Eugenio Calabi, Ching-Li Chai, Ted Chinburg, Chris Croke, Ron Donagi, Harry Gingold, H.W. Gould, Herman Gluck, David Harbater, Julia Hartmann, Jerry Kazdan, Alexander Kirillov, Florian Pop, Steve Shatz, Jim Stasheff, George Sparling, Doran Zeilberger, and Wolfgand Ziller for their encouragement, advice, inspiration and for what they taught me. We also thank Christine Allen-Blanchette, Arthur Azevedo de Amorim, Kostas Daniilidis, Carlos Esteves, Spyridon Leonardos, Stephen Phillips, Jo˜ ao Sedoc, Marcelo Siqueira, and Roberto Tron for reporting typos and for helpful comments. 6 Contents Contents 7 1 Introduction 11 2 Tensor Algebras 21 2.1 Linear Algebra Preliminaries: Dual Spaces and Pairings . . . . . . . . . . . 25 2.2 Tensor Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 2.3 Bases of Tensor Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 2.4 Some Useful Isomorphisms for Tensor Products . . . . . . . . . . . . . . . . 44 2.5 Duality for Tensor Products . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 2.6 Tensor Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 2.7 (r, s)-Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 2.8 (r, s)-Tensors as Multilinear Maps . . . . . . . . . . . . . . . . . . . . . . . 62 2.9 Contraction Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 2.10 Symmetric Tensor Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 2.11 Bases of Symmetric Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 2.12 Duality for Symmetric Powers . . . . . . . . . . . . . . . . . . . . . . . . . . 78 2.13 Symmetric Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 2.14 Tensor Products of Modules over a Commmutative Ring . . . . . . . . . . . 85 2.15 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 3 Exterior Tensor Powers and Exterior Algebras 91 3.1 Exterior Tensor Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 3.2 Bases of Exterior Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 3.3 Duality for Exterior Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 3.4 Exterior Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 3.5 The Hodge ∗-Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 3.6 Left and Right Hooks ⊛. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 3.7 Testing Decomposability ⊛. . . . . . . . . . . . . . . . . . . . . . . . . . . 124 3.8 The Grassmann-Pl¨ ucker’s Equations and Grassmannians ⊛ . . . . . . . . . 127 3.9 Vector-Valued Alternating Forms . . . . . . . . . . . . . . . . . . . . . . . . 130 3.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 4 Differential Forms 137 7 8 CONTENTS 4.1 Differential Forms on Rn and de Rham Cohomology . . . . . . . . . . . . . 141 4.2 Pull-Back of Differential Forms . . . . . . . . . . . . . . . . . . . . . . . . . 152 4.3 Differential Forms on Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . 159 4.4 Lie Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 4.5 Vector-Valued Differential Forms . . . . . . . . . . . . . . . . . . . . . . . . 179 4.6 Differential Forms on Lie Groups . . . . . . . . . . . . . . . . . . . . . . . . 186 4.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 5 Tensor Fields 195 5.1 Tensor Fields as Sections of the Tensor Bundle T r,s(M) . . . . . . . . . . . 195 5.2 Tensor Fields as C∞(M)-Multilinear Maps . . . . . . . . . . . . . . . . . . . 198 5.3 Tensor Field Derivations and Wilmore’s Theorem . . . . . . . . . . . . . . . 203 5.4 Using Contractions Instead of Kronecker’s delta . . . . . . . . . . . . . . . . 210 5.5 Extension of the Covariant Derivative to Tensor Fields . . . . . . . . . . . . 213 5.6 Lie Derivative of Tensor Fields . . . . . . . . . . . . . . . . . . . . . . . . . 215 6 Distributions and the Frobenius Theorem 217 6.1 Tangential Distributions, Involutive Distributions . . . . . . . . . . . . . . . 218 6.2 Frobenius Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 6.3 Differential Ideals and Frobenius Theorem . . . . . . . . . . . . . . . . . . . 225 6.4 A Glimpse at Foliations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 6.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 7 Integration on Manifolds 231 7.1 Orientation of Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 7.2 Volume Forms on Riemannian Manifolds and Lie Groups . . . . . . . . . . . 239 7.3 Integration in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 7.4 Integration on Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 7.5 Densities ⊛. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 7.6 Manifolds With Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 7.7 Integration on Regular Domains and Stokes’ Theorem . . . . . . . . . . . . 256 7.8 Integration on Riemannian Manifolds and Lie Groups . . . . . . . . . . . . 271 7.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 8 Spherical Harmonics and Linear Representations 281 8.1 Hilbert Spaces and Hilbert Sums . . . . . . . . . . . . . . . . . . . . . . . . 284 8.2 Spherical Harmonics on the Circle . . . . . . . . . . . . . . . . . . . . . . . 298 8.3 Spherical Harmonics on the 2-Sphere . . . . . . . . . . . . . . . . . . . . . . 301 8.4 The Laplace-Beltrami Operator . . . . . . . . . . . . . . . . . . . . . . . . . 309 8.5 Harmonic Polynomials, Spherical Harmonics and L2(Sn) . . . . . . . . . . . 319 8.6 Zonal Spherical Functions and Gegenbauer Polynomials . . . . . . . . . . . 328 8.7 More on the Gegenbauer Polynomials . . . . . . . . . . . . . . . . . . . . . 339 8.8 The Funk–Hecke Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 CONTENTS 9 8.9 Linear Representations of Compact Lie Groups ⊛. . . . . . . . . . . . . . . 348 8.10 Consequences of The Peter–Weyl Theorem . . . . . . . . . . . . . . . . . . . 356 8.11 Gelfand Pairs, Spherical Functions, Fourier Transform ⊛. . . . . . . . . . . 360 8.12 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364 9 Operators on Riemannian Manifolds 369 9.1 The Gradient and Hessian Operators . . . . . . . . . . . . . . . . . . . . . . 370 9.2 The Hodge ∗Operator on Riemannian Manifolds . . . . . . . . . . . . . . . 380 9.3 The Hodge Laplacian and the Hodge Divergence Operators . . . . . . . . . 382 9.4 The Hodge and Laplace–Beltrami Laplacians of Functions . . . . . . . . . . 386 9.5 Divergence and Lie Derivative of the Volume Form . . . . . . . . . . . . . . 389 9.6 Harmonic Forms, the Hodge Theorem, Poincar´ e Duality . . . . . . . . . . . 393 9.7 The Bochner Laplacian and the Bochner Technique . . . . . . . . . . . . . . 395 9.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403 10 Bundles, Metrics on Bundles, Homogeneous Spaces 405 10.1 Fibre Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 10.2 Bundle Morphisms, Equivalent and Isomorphic Bundles . . . . . . . . . . . 417 10.3 Bundle Constructions Via the Cocycle Condition . . . . . . . . . . . . . . . 424 10.4 Vector Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429 10.5 Operations on Vector Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . 438 10.6 Properties of Vector Bundle Sections . . . . . . . . . . . . . . . . . . . . . . 442 10.7 Covariant Derivatives of Tensor Fields . . . . . . . . . . . . . . . . . . . . . 444 10.8 Metrics on Bundles, Reduction, Orientation . . . . . . . . . . . . . . . . . . 446 10.9 Principal Fibre Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450 10.10 Proper and Free Actions, Homogeneous Spaces . . . . . . . . . . . . . . . . 458 10.11 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462 11 Connections and Curvature in Vector Bundles 467 11.1 Introduction to Connections in Vector Bundles . . . . . . . . . . . . . . . . 467 11.2 Connections in Vector Bundles and Riemannian Manifolds . . . . . . . . . . 471 11.3 Connection Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476 11.4 Parallel Transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481 11.5 Curvature and Curvature Form . . . . . . . . . . . . . . . . . . . . . . . . . 484 11.6 Structure Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490 11.7 A Formula for d∇◦d∇. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494 11.8 Connections Compatible with a Metric . . . . . . . . . . . . . . . . . . . . . 495 11.9 Connections on the Dual Bundle . . . . . . . . . . . . . . . . . . . . . . . . 499 11.10 The Levi-Civita Connection on TM Revisited . . . . . . . . . . . . . . . . . 501 11.11 Pontrjagin Classes and Chern Classes, a Glimpse . . . . . . . . . . . . . . . 505 11.12 The Pfaffian Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513 11.13 Euler Classes and The Generalized Gauss-Bonnet Theorem . . . . . . . . . 517 11.14 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 10 CONTENTS 12 Connections and Curvature in Principal Bundles 525 12.1 Connections and Connection Forms in Principal Bundes . . . . . . . . . . . 526 12.2 Curvature Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536 13 Clifford Algebras, Clifford Groups, Pin and Spin 541 13.1 Introduction: Rotations As Group Actions . . . . . . . . . . . . . . . . . . . 541 13.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548 13.3 Clifford Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 551 13.4 Clifford Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557 13.5 The Groups Pin(n) and Spin(n) . . . . . . . . . . . . . . . . . . . . . . . . 565 13.6 The Groups Pin(p, q) and Spin(p, q) . . . . . . . . . . . . . . . . . . . . . . 572 13.7 The Groups Pin(p, q) and Spin(p, q) as double covers . . . . . . . . . . . . 577 13.8 Periodicity of the Clifford Algebras Clp,q . . . . . . . . . . . . . . . . . . . . 584 13.9 The Complex Clifford Algebras Cl(n, C) . . . . . . . . . . . . . . . . . . . . 588 13.10 Clifford Groups Over a Field K . . . . . . . . . . . . . . . . . . . . . . . . . 589 13.11 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 596 Bibliography 605 Symbol Index 613 Index 619 Chapter 1 Introduction This book covers topics which belong to a second course in differential geometry. Differential forms constitute the main tool needed to understand and prove many of the results presented in this book. Thus one need have a solid understanding of differential forms, which turn out to be certain kinds of skew-symmetric (also called alternating) tensors. Differential forms have two main roles: (1) Describe various systems of partial differential equations on manifolds. (2) To define various geometric invariants reflecting the global structure of manifolds or bundles. Such invariants are obtained by integrating certain differential forms. Differential forms can be combined using a notion of product called the wedge product, but what really gives power to the formalism of differential forms is the magical operation d of exterior differentiation. Given a form ω, we obtain another form dω, and remarkably, the following equation holds ddω = 0. As silly as it looks, the above equation lies at the core of the notion of cohomology, a powerful algebraic tool to understanding the topology of manifolds, and more generally of topological spaces. ´ Elie Cartan had many of the intuitions that led to the cohomology of differential forms, but it was Georges de Rham who defined it rigorously and proved some important theorems about it. It turns out that the notion of Laplacian can also be defined on differential forms using a device due to Hodge, and some important theorems can be obtained: the Hodge decomposition theorem, and Hodge’s theorem about the isomorphism between the de Rham cohomology groups and the spaces of harmonic forms. Differential forms can also be used to define the notion of curvature of a connection on a certain type of manifold called a vector bundle. Because differential forms are such a fundamental tool we made the (perhaps painful) decision to provide a fairly detailed exposition of tensors, starting with arbitrary tensors, 11 12 CHAPTER 1. INTRODUCTION and then specializing to symmetric and alternating tensors. In particular, we explain rather carefully the process of taking the dual of a tensor (of all three flavors). Tensors, symmetric tensors, tensor algebras, and symmetrc algebras are discussed on Chapter 2. Alternating tensors, and exterior algebras are discussed in Chapter 3. The Hodge ∗operator is intro-duced, we discuss criteria for the decomposablity of an alternating tensor in terms of hook operators, and we present the Grassmann-Pl¨ ucker’s equations. We now give a preview of the topics discussed in this book. Chapter 4 is devoted to a thorough presentation of differential forms, including vector-valued differential forms, differential forms on Lie Groups, and Maurer-Cartan forms. We also introduce de Rham cohomology. Chapter 6 is a short chapter devoted to distributions and the Frobenius theorem. Distri-butions are a generalization of vector fields, and the issue is to understand when a distribution is integrable. The Frobenius theorem gives a necessary and sufficient condition for a distri-bution to have an integral manifold at every point. One version of the Frobenius theorem is stated in terms of vector fields, the second version in terms of differential forms. The theory of integration on manifolds and Lie groups is presented in Chapter 7. We introduce the notion of orientation of a smooth manifold (of dimension n), volume forms, and then explain how to integrate a smooth n-form with compact support. We define densities which allow integrating n-forms even if the manifold is not orientable, but we do not go into the details of this theory. We define manifolds with boundary, and explain how to integrate forms on certain kinds of manifolds with boundaries called regular domains. We state and prove a version of the famous result known as Stokes’ theorem. In the last section we discuss integrating functions on Riemannian manifolds or Lie groups. The main theme of Chapter 8 is to generalize Fourier analysis on the circle to higher dimensional spheres. One of our goals is to understand the structure of the space L2(Sn) of real-valued square integrable functions on the sphere Sn, and its complex analog L2 C(Sn). Both are Hilbert spaces if we equip them with suitable inner products. It turns out that each of L2(Sn) and L2 C(Sn) contains a countable family of very nice finite dimensional subspaces Hk(Sn) (and HC k (Sn)), where Hk(Sn) is the space of (real) spherical harmonics on Sn, that is, the restrictions of the harmonic homogeneous polynomials of degree k (in n + 1 real variables) to Sn (and similarly for HC k (Sn)); these polynomials satisfy the Laplace equation ∆P = 0, where the operator ∆is the (Euclidean) Laplacian, ∆= ∂2 ∂x2 1 + · · · + ∂2 ∂x2 n+1 . Remarkably, each space Hk(Sn) (resp. HC k (Sn)) is the eigenspace of the Laplace-Beltrami operator ∆Sn on Sn, a generalization to Riemannian manifolds of the standard Laplacian 13 (in fact, Hk(Sn) is the eigenspace for the eigenvalue −k(n + k −1)). As a consequence, the spaces Hk(Sn) (resp. HC k (Sn)) are pairwise orthogonal. Furthermore (and this is where analysis comes in), the set of all finite linear combinations of elements in S∞ k=0 Hk(Sn) (resp. S∞ k=0 HC k (Sn)) is is dense in L2(Sn) (resp. dense in L2 C(Sn)). These two facts imply the following fundamental result about the structure of the spaces L2(Sn) and L2 C(Sn). The family of spaces Hk(Sn) (resp. HC k (Sn)) yields a Hilbert space direct sum decompo-sition L2(Sn) = ∞ M k=0 Hk(Sn) (resp. L2 C(Sn) = ∞ M k=0 HC k (Sn)), which means that the summands are closed, pairwise orthogonal, and that every f ∈L2(Sn) (resp. f ∈L2 C(Sn)) is the sum of a converging series f = ∞ X k=0 fk in the L2-norm, where the fk ∈Hk(Sn) (resp. fk ∈HC k (Sn)) are uniquely determined functions. Furthermore, given any orthonormal basis (Y 1 k , . . . , Y ak,n+1 k ) of Hk(Sn), we have fk = ak,n+1 X mk=1 ck,mkY mk k , with ck,mk = ⟨f, Y mk k ⟩Sn. The coefficients ck,mk are “generalized” Fourier coefficients with respect to the Hilbert basis {Y mk k | 1 ≤mk ≤ak,n+1, k ≥0}; see Theorems 8.18 and 8.19. When n = 2, the functions Y mk k correspond to the spherical harmonics, which are defined in terms of the Legendre functions. Along the way, we prove the famous Funk–Hecke formula. The purpose of Section 8.9 is to generalize the results about the structure of the space of functions L2 C(Sn) defined on the sphere Sn, especially the results of Sections 8.5 and 8.6 (such as Theorem 8.19, except part (3)), to homogeneous spaces G/K where G is a compact Lie group and K is a closed subgroup of G. The first step is to consider the Hilbert space L2 C(G) where G is a compact Lie group and to find a Hilbert sum decomposition of this space. The key to this generalization is the notion of (unitary) linear representation of the group G. The result that we are alluding to is a famous theorem known as the Peter–Weyl theorem about unitary representations of compact Lie groups. The Peter–Weyl theorem can be generalized to any representation V : G →Aut(E) of G into a separable Hilbert space E, and we obtain a Hilbert sum decomposition of E in terms of subspaces Eρ of E. The next step is to consider the subspace L2 C(G/K) of L2 C(G) consisting of the functions that are right-invariant under the action of K. These can be viewed as functions on the 14 CHAPTER 1. INTRODUCTION homogeneous space G/K. Again we obtain a Hilbert sum decomposition. It is also interest-ing to consider the subspace L2 C(K\G/K) of functions in L2 C(G) consisting of the functions that are both left and right-invariant under the action of K. The functions in L2 C(K\G/K) can be viewed as functions on the homogeneous space G/K that are invariant under the left action of K. Convolution makes the space L2 C(G) into a non-commutative algebra. Remarkably, it is possible to characterize when L2 C(K\G/K) is commutative (under convolution) in terms of a simple criterion about the irreducible representations of G. In this situation, (G, K) is a called a Gelfand pair. When (G, K) is a Gelfand pair, it is possible to define a well-behaved notion of Fourier transform on L2 C(K\G/K). Gelfand pairs and the Fourier transform are briefly considered in Section 8.11. Chapter 9 deals with various generalizations of the Lapacian to manifolds. The Laplacian is a very important operator because it shows up in many of the equations used in physics to describe natural phenomena such as heat diffusion or wave propagation. Therefore, it is highly desirable to generalize the Laplacian to functions defined on a manifold. Furthermore, in the late 1930’s, Georges de Rham (inspired by ´ Elie Cartan) realized that it was fruitful to define a version of the Laplacian operating on differential forms, because of a fundamental and almost miraculous relationship between harmonics forms (those in the kernel of the Laplacian) and the de Rham cohomology groups on a (compact, orientable) smooth manifold. Indeed, as we will see in Section 9.6, for every cohomology group Hk DR(M), every cohomology class [ω] ∈Hk DR(M) is represented by a unique harmonic k-form ω; this is the Hodge theorem. The connection between analysis and topology lies deep and has many important consequences. For example, Poincar´ e duality follows as an “easy” consequence of the Hodge theorem. Technically, the Hodge Laplacian can be defined on differential forms using the Hodge ∗ operator (Section 3.5). On functions, there is an alternate and equivalent definition of the Laplacian using only the covariant derivative and obtained by generalizing the notions of gradient and divergence to functions on manifolds. Another version of the Laplacian on k-forms can be defined in terms of a generalization of the Levi-Civita connection ∇: X(M) × X(M) →X(M) to k-forms viewed as a linear map ∇: Ak(M) →HomC∞(M)(X(M), Ak(M)), and in terms of a certain adjoint ∇∗of ∇, a linear map ∇∗: HomC∞(M)(X(M), Ak(M)) →Ak(M). We obtain the Bochner Laplacian (or connection Laplacian ) ∇∗∇. Then it is natural to wonder how the Hodge Laplacian ∆differs from the connection Laplacian ∇∗∇? 15 Remarkably, there is a formula known as Weitzenb¨ ock’s formula (or Bochner’s formula) of the form ∆= ∇∗∇+ C(R∇), where C(R∇) is a contraction of a version of the curvature tensor on differential forms (a fairly complicated term). In the case of one-forms, ∆= ∇∗∇+ Ric, where Ric is a suitable version of the Ricci curvature operating on one-forms. Weitzenb¨ ock-type formulae are at the root of the so-called “Bochner technique,” which consists in exploiting curvature information to deduce topological information. Chapter 10 is an introduction to bundle theory; we discuss fibre bundles, vector bundles, and principal bundles. Intuitively, a fibre bundle over B is a family E = (Eb)b∈B of spaces Eb (fibres) indexed by B and varying smoothly as b moves in B, such that every Eb is diffeomorphic to some prespecified space F. The space E is called the total space, B the base space, and F the fibre. A way to define such a family is to specify a surjective map π: E →B. We will assume that E, B, F are smooth manifolds and that π is a smooth map. The type of bundles that we just described is too general and to develop a useful theory it is necessary to assume that locally, a bundle looks likes a product. Technically, this is achieved by assuming that there is some open cover U = (Uα)α∈I of B and that there is a family (ϕα)α∈I of diffeomorphisms ϕα : π−1(Uα) →Uα × F. Intuitively, above Uα, the open subset π−1(Uα) looks like a product. The maps ϕα are called local trivializations. The last important ingredient in the notion of a fibre bundle is the specifiction of the “twisting” of the bundle; that is, how the fibre Eb = π−1(b) gets twisted as b moves in the base space B. Technically, such twisting manifests itself on overlaps Uα ∩Uβ ̸= ∅. It turns out that we can write ϕα ◦ϕ−1 β (b, x) = (b, gαβ(b)(x)) for all b ∈Uα∩Uβ and all x ∈F. The term gαβ(b) is a diffeomorphism of F. Then we require that the family of diffeomorphisms gαβ(b) belongs to a Lie group G, which is expressed by specifying that the maps gαβ, called transitions maps, are maps gαβ : Uα ∩Uβ →G. The purpose of the group G, called the structure group, is to specify the “twisting” of the bundle. Fibre bundles are defined in Section 10.1. The family of transition maps gαβ satisfies an important condition on nonempty overlaps Uα ∩Uβ ∩Uγ called the cocycle condition: gαβ(b)gβγ(b) = gαγ(b) 16 CHAPTER 1. INTRODUCTION (where gαβ(b), gβγ(b), gαγ(b) ∈G), for all α, β, γ such that Uα ∩Uβ ∩Uγ ̸= ∅and all b ∈ Uα ∩Uβ ∩Uγ. In Section 10.2, following Hirzebruch and Chern , we define bundle morphisms and the notion of equivalence of bundles over the same base. We show that two bundles (over the same base) are equivalent if and only if they are isomorphic. In Section 10.3 we describe the construction of a fibre bundle with prescribed fibre F and structure group G from a base manifold, B, an open cover U = (Uα)α∈I of B, and a family of maps gαβ : Uα ∩Uβ →G satisfying the cocycle condition, called a cocycle. This construction is the basic tool for constructing new bundles from old ones. Section 10.4 is devoted to a special kind of fibre bundle called vector bundles. A vector bundle is a fibre bundle for which the fibre is a finite-dimensional vector space V , and the structure group is a subgroup of the group of linear isomorphisms (GL(n, R) or GL(n, C), where n = dim V ). Typical examples of vector bundles are the tangent bundle TM and the cotangent bundle T ∗M of a manifold M. We define maps of vector bundles and equivalence of vector bundles. In Section 10.5 we describe various operations on vector bundles: Whitney sums, ten-sor products, tensor powers, exterior powers, symmetric powers, dual bundles, and Hom bundles. We also define the complexification of a real vector bundle. In Section 10.6 we discuss properties of the sections of a vector bundle ξ. We prove that the space of sections Γ(ξ) is finitely generated projective C∞(B)-module. Section 10.7 is devoted to the the covariant derivative of tensor fields and to the duality between vector fields and differential forms. In Section 10.8 we explain how to give a vector bundle a Riemannian metric. This is achieved by supplying a smooth family (⟨−, −⟩b)b∈B of inner products on each fibre π−1(b) above b ∈B. We describe the notion of reduction of the structure group and define orientable vector bundles. In Section 10.9 we consider the special case of fibre bundles for which the fibre coincides with the structure group G, which acts on itself by left translations. Such fibre bundles are called principal bundles. It turns out that a principal bundle can be defined in terms of a free right action of Lie group on a smooth manifold. When principal bundles are defined in terms of free right actions, the notion of bundle morphism is also defined in terms of equivariant maps. There are two constructions that allow us to reduce the study of fibre bundles to the study of principal bundles. Given a fibre bundle ξ with fibre F, we can construct a principal bundle P(ξ) obtained by replacing the fibre F by the group G. Conversely, given a principal bundle ξ and an effective action of G on a manifold F, we can construct the fibre bundle ξ[F] obtained by replacing G by F. The maps ξ 7→ξ[F] and ξ 7→P(ξ) 17 induce a bijection between equivalence classes of principal G-bundles and fibre bundles (with structure group G). Furthermore, ξ is a trivial bundle iffP(ξ) is a trivial bundle. Section 10.10 is devoted to principal bundles that arise from proper and free actions of a Lie group. When the base space is a homogenous space, which means that it arises from a transitive action of a Lie group, then the total space is a principal bundle. There are many illustrations of this situation involving SO(n + 1) and SU(n + 1). In Chapter 11 we discuss connections and curvature in vector bundles. In Section 11.2 we define connections on a vector bundle. This can be done in two equivalent ways. One of the two definitions is more abstract than the other because it involves a tensor product, but it is technically more convenient. This definition states that a connection on a vector bundle ξ, as an R-linear map ∇: Γ(ξ) →A1(B) ⊗C∞(B) Γ(ξ) (∗) that satisfies the “Leibniz rule” ∇(fs) = d f ⊗s + f∇s, with s ∈Γ(ξ) and f ∈C∞(B), where Γ(ξ) and A1(B) are treated as C∞(B)-modules. Here, A1(B) = Γ(T ∗B) is the space of 1-forms on B. Since there is an isomorphism A1(B) ⊗C∞(B) Γ(ξ) ∼ = Γ(T ∗B ⊗ξ), a connection can be defined equivalently as an R-linear map ∇: Γ(ξ) →Γ(T ∗B ⊗ξ) satisfying the Leibniz rule. In Section 11.3 we show how a connection can be represented in a chart in terms of a certain matrix called a connection matrix. We prove that every vector bundle possesses a connection, and we give a formula describing how a connection matrix changes if we switch from one chart to another. In Section 11.4 we define the notion of covariant derivative along a curve and parallel transport. Section 11.5 is devoted to the very important concept of curvature form R∇of a connec-tion ∇on a vector bundle ξ. We show that the curvature form is a vector-valued two-form with values in Γ(Hom(ξ, ξ)). We also establish the relationhip between R∇and the more familiar definition of the Riemannian curvature in terms of vector fields. In Section 11.6 we show how the curvature form can be expressed in a chart in terms of a matrix of two-forms called a curvature matrix. The connection matrix and the curvature matrix are related by the structure equation. We also give a formula describing how a curvature matrix changes if we switch from one chart to another. Bianchi’s identity gives 18 CHAPTER 1. INTRODUCTION an expression for the exterior derivative of the curvature matrix in terms of the curvature matrix itself and the connection matrix. Section 11.8 deals with connections compatible with a metric and the Levi-Civita con-nection, which arise in the Riemannian geometry of manifolds. One way of characterizing the Levi-Civita connection involves defining the notion of connection on the dual bundle. This is achieved in Section 11.9. Levi-Civita connections on the tangent bundle of a manifold are investigated in Section 11.10. The purpose of Section 11.11 is to introduce the reader to Pontrjagin Classes and Chern Classes, which are fundamental invariants of real (resp. complex) vector bundles. Here we are dealing with one of the most sophisticated and beautiful parts of differential geometry. A masterly exposition of the theory of characteristic classes is given in the classic book by Milnor and Stasheff. Amazingly, the method of Chern and Weil using differential forms is quite accessible for someone who has reasonably good knowledge of differential forms and de Rham cohomology, as long as one is willing to gloss over various technical details. We give an introduction to characteristic classes using the method of Chern and Weil. If ξ is a real orientable vector bundle of rank 2m, and if ∇is a metric connection on ξ, then it is possible to define a closed global form eu(R∇), and its cohomology class e(ξ) is called the Euler class of ξ. This is shown in Section 11.13. The Euler class e(ξ) turns out to be a square root of the top Pontrjagin class pm(ξ) of ξ. A complex rank m vector bundle can be viewed as a real vector bundle of rank 2m, which is always orientable. The Euler class e(ξ) of this real vector bundle is equal to the top Chern class cm(ξ) of the complex vector bundle ξ. The global form eu(R∇) is defined in terms of a certain polynomial Pf(A) associated with a real skew-symmetric matrix A, which is a kind of square root of the determinant det(A). The polynomial Pf(A), called the Pfaffian, is defined in Section 11.12. The culmination of this chapter is a statement of the generalization due to Chern of a classical theorem of Gauss and Bonnet. This theorem known as the generalized Gauss– Bonnet formula expresses the Euler characteristic χ(M) of an orientable, compact smooth manifold M of dimension 2m as χ(M) = Z M eu(R∇), where eu(R∇) is the Euler form associated with the curvature form R∇of a metric connection ∇on M. The goal of Chapter 13 is to explain how rotations in Rn are induced by the action of a certain group Spin(n) on Rn, in a way that generalizes the action of the unit complex numbers U(1) on R2, and the action of the unit quaternions SU(2) on R3 (i.e., the action is defined in terms of multiplication in a larger algebra containing both the group Spin(n) and 19 Rn). The group Spin(n), called a spinor group, is defined as a certain subgroup of units of an algebra Cln, the Clifford algebra associated with Rn. Since the spinor groups are certain well chosen subgroups of units of Clifford algebras, it is necessary to investigate Clifford algebras to get a firm understanding of spinor groups. This chapter provides a tutorial on Clifford algebra and the groups Spin and Pin, including a study of the structure of the Clifford algebra Clp,q associated with a nondegenerate symmetric bilinear form of signature (p, q) and culminating in the beautiful 8-periodicity theorem of ´ Elie Cartan and Raoul Bott (with proofs). We also explain when Spin(p, q) is a double-cover of SO(p, q). Some preliminaries on algebras and tensor algebras are reviewed in Section 13.2. In Section 13.3 we define Clifford algebras over the field K = R. The Clifford groups (over K = R) are defined in Section 13.4. In the second half of this section we restrict our attention to the real quadratic form Φ(x1, . . . , xn) = −(x2 1 + · · · + x2 n). The corresponding Clifford algebras are denoted Cln and the corresponding Clifford groups as Γn. In Section 13.5 we define the groups Pin(n) and Spin(n) associated with the real quadratic form Φ(x1, . . . , xn) = −(x2 1+· · ·+x2 n). We prove that the maps ρ: Pin(n) →O(n) and ρ: Spin(n) →SO(n) are surjective with kernel {−1, 1}. We determine the groups Spin(n) for n = 2, 3, 4. Section 13.6 is devoted to the Spin and Pin groups associated with the real nondegenerate quadratic form Φ(x1, . . . , xp+q) = x2 1 + · · · + x2 p −(x2 p+1 + · · · + x2 p+q). We obtain Clifford algebras Clp,q, Clifford groups Γp,q, and groups Pin(p, q) and Spin(p, q). We show that the maps ρ: Pin(p, q) →O(p, q) and ρ: Spin(p, q) →SO(p, q) are surjective with kernel {−1, 1}. In Section 13.7 we show that the Lie groups Pin(p, q) and Spin(p, q) are double covers of O(p, q) and SO(p, q). In Section 13.8 we prove an amazing result due to ´ Elie Cartan and Raoul Bott, namely the 8-periodicity of the Clifford algebras Clp,q. This result says that: for all n ≥0, we have the following isomorphisms: Cl0,n+8 ∼ = Cl0,n ⊗Cl0,8 Cln+8,0 ∼ = Cln,0 ⊗Cl8,0. Furthermore, Cl0,8 = Cl8,0 = R(16), the real algebra of 16 × 16 matrices. Section 13.9 is devoted to the complex Clifford algebras Cl(n, C). In this case, we have a 2-periodicity, Cl(n + 2, C) ∼ = Cl(n, C) ⊗C Cl(2, C), 20 CHAPTER 1. INTRODUCTION with Cl(2, C) = C(2), the complex algebra of 2 × 2 matrices. Finally, in the last section, Section 13.10 we outline the theory of Clifford groups and of the Pin and Spin groups over any field K of characteristic ̸= 2. Chapter 2 Tensor Algebras and Symmetric Algebras Tensors are creatures that we would prefer did not exist but keep showing up whenever multilinearity manifests itself. One of the goals of differential geometry is to be able to generalize “calculus on Rn” to spaces more general than Rn, namely manifolds. We would like to differentiate functions f : M →R defined on a manifold, optimize functions (find their minima or maxima), but also to integrate such functions, as well as compute areas and volumes of subspaces of our manifold. he suitable notion of differentiation is the notion of tangent map, a linear notion. One of the main discoveries made at the beginning of the twentieth century by Poincar´ e and ´ Elie Cartan, is that the “right” approach to integration is to integrate differential forms, and not functions. To integrate a function f, we integrate the form fω, where ω is a volume form on the manifold M. The formalism of differential forms takes care of the process of the change of variables quite automatically, and allows for a very clean statement of Stokes’ formula. Differential forms can be combined using a notion of product called the wedge product, but what really gives power to the formalism of differential forms is the magical operation d of exterior differentiation. Given a form ω, we obtain another form dω, and remarkably, the following equation holds ddω = 0. As silly as it looks, the above equation lies at the core of the notion of cohomology, a powerful algebraic tool to understanding the topology of manifolds, and more generally of topological spaces. ´ Elie Cartan had many of the intuitions that led to the cohomology of differential forms, but it was Georges de Rham who defined it rigorously and proved some important theorems about it. It turns out that the notion of Laplacian can also be defined on differential forms using a device due to Hodge, and some important theorems can be obtained: the Hodge 21 22 CHAPTER 2. TENSOR ALGEBRAS decomposition theorem, and Hodge’s theorem about the isomorphism between the de Rham cohomology groups and the spaces of harmonic forms. To understand all this, one needs to learn about differential forms, which turn out to be certain kinds of skew-symmetric (also called alternating) tensors. If one’s only goal is to define differential forms, then it is possible to take some short cuts and to avoid introducing the general notion of a tensor. However, tensors that are not necessarily skew-symmetric arise naturally, such as the curvature tensor, and in the theory of vector bundles, general tensor products are needed. Consequently, we made the (perhaps painful) decision to provide a fairly detailed ex-position of tensors, starting with arbitrary tensors, and then specializing to symmetric and alternating tensors. In particular, we explain rather carefully the process of taking the dual of a tensor (of all three flavors). We refrained from following the approach in which a tensor is defined as a multilinear map defined on a product of dual spaces, because it seems very artificial and confusing (certainly to us). This approach relies on duality results that only hold in finite dimension, and consequently unecessarily restricts the theory of tensors to finite dimensional spaces. We also feel that it is important to begin with a coordinate-free approach. Bases can be chosen for computations, but tensor algebra should not be reduced to raising or lowering indices. Readers who feel that they are familiar with tensors could skip this chapter and the next. They can come back to them “by need.” We begin by defining tensor products of vector spaces over a field and then we investigate some basic properties of these tensors, in particular the existence of bases and duality. After this we investigate special kinds of tensors, namely symmetric tensors and skew-symmetric tensors. Tensor products of modules over a commutative ring with identity will be discussed very briefly. They show up naturally when we consider the space of sections of a tensor product of vector bundles. Given a linear map f : E →F (where E and F are two vector spaces over a field K), we know that if we have a basis (ui)i∈I for E, then f is completely determined by its values f(ui) on the basis vectors. For a multilinear map f : En →F, we don’t know if there is such a nice property but it would certainly be very useful. In many respects tensor products allow us to define multilinear maps in terms of their action on a suitable basis. The crucial idea is to linearize, that is, to create a new vector space E⊗n such that the multilinear map f : En →F is turned into a linear map f⊗: E⊗n →F which is equivalent to f in a strong sense. If in addition, f is symmetric, then we can define a symmetric tensor power Symn(E), and every symmetric multilinear map f : En →F is turned into a linear map f⊙: Symn(E) →F which is equivalent to f in a strong sense. Similarly, if f is alternating, then we can define a skew-symmetric tensor power Vn(E), and every alternating multilinear map is turned into a linear map f∧: Vn(E) →F which is equivalent to f in a strong sense. 23 Tensor products can be defined in various ways, some more abstract than others. We try to stay down to earth, without excess. In Section 2.1, we review some facts about dual spaces and pairings. In particular, we show that an inner product on a finite-dimensional vector space E induces a canonical isomorphism between E and its dual space E∗. Pairings will be used to deal with dual spaces of tensors. We also show that there is a canonical isomorphism between the vector space of bilinear forms on E and the vector space of linear maps from E to itself. Tensor products are defined in Section 2.2. Given two vector spaces E1 and E2 over a field K, the tensor product E1 ⊗E2 is defined by a universal mapping property: it is a vector space with an injection i⊗: E1 ×E2 →E1 ⊗E2, such that for every vector space F and every bilinear map f : E1 × E2 →F, there is a unique linear map f⊗: E1 ⊗E2 →F such that f = f⊗◦i⊗, as illustrated in the following diagram: E1 × E2 f & i⊗/ E1 ⊗E2 f⊗  F We prove that the above universal mapping property defines E1 ⊗E2 up to isomorphism, and then we prove its existence by constructing a suitable quotient of the free vector space K(E1⊗E2) generated by E1 ⊗E2. The generalization to any finite number of vector spaces E1, . . . , En is immediate. The universal mapping property of the tensor product yields an isomorphism between the vector space of linear maps Hom(E1 ⊗· · · ⊗En, F) and the vector space of multilinear maps Hom(E1, . . . , En; F). We show that tensor product is functorial, which means that given two linear maps f : E →F and g: E′ →F ′, there is a linear map f ⊗g: E ⊗F →E′ ⊗F ′. In Section 2.3, we show how to construct a basis for the tensor product E1 ⊗· · · ⊗En from bases for the spaces E1, . . . , En. In Section 2.4, we prove some basic isomorphisms involving tensor products. One of these isomorphisms states that Hom(E ⊗F, G) is isomorphic to Hom(E, Hom(F, G)). Section 2.5 deals with duality for tensor products. It is a very important section which needs to be throroughly understood in order to study vector bundles. The main isomorphisms state that if E1, . . . , En are finite dimensional, then (E1 ⊗· · · ⊗En)∗∼ = E∗ 1 ⊗· · · ⊗E∗ n ∼ = Hom(E1, . . . , En; K). The second isomorphism arises from the pairing µ defined on generators by µ(v∗ 1 ⊗· · · ⊗v∗ n)(u1, . . . , un) = v∗ 1(u1) · · · v∗ n(un). 24 CHAPTER 2. TENSOR ALGEBRAS We also prove that if either E or F is finite dimensional then there is a canonical isomorphism between E∗⊗F and Hom(E, F). In Section 2.6 we define the tensor algebra T(V ). This is the direct sum of the tensor powers V ⊗m = V ⊗· · · ⊗V | {z } m , T(V ) = M m≥0 V ⊗m. In addition to being a vector space, T(V ) is equipped with an associative multiplication, ⊗. The tensor algebra T(V ) satisfies a universal mapping property with respect to (associative) algebras. We also define the tensor algebras T r,s(V ) and the tensor algebra T •,•(V ). We define contraction operations. If E and F are algebras, we show how to make E ⊗F into an algebra. In Section 2.10 to turn to the special case of symmetric tensor powers, which correspond to symmetric multilinear maps ϕ: En →F. There are multilinear maps that are invariant under permutation of its arguments. Given a vector space E over a field K, for any n ≥1, the symmetric tensor power Sn(E) is defined by a universal mapping property: it is a vector space with an injection i⊙: En →Sn(E), such that for every vector space F and every symmetric multilinear map f : En →F, there is a unique linear map f⊙: Sn(E) →F such that f = f⊙◦i⊙, as illustrated in the following diagram: En f # i⊙/ Sn(E) f⊙  F We prove that the above universal mapping property defines Sn(E) up to isomorphism, and then we prove its existence by constructing the quotient of the tensor power E⊗n by the subspace C of E⊗n generated by the vectors of the form u1 ⊗· · · ⊗un −uσ(1) ⊗· · · ⊗uσ(n), for all ui ∈E, and all permutations σ: {1, . . . , n} →{1, . . . , n}. As a corollary, there is an isomorphism between the vector space of linear maps Hom(Sn(E), F) and the vector space of symmetric multilinear maps Symn(E; F). We also show that given two linear maps f, g: E →E′, there is a linear map f ⊙g: S2(E) →S2(E′). A basic isomorphism involving the symmetric power of a direct sum is shown at the end of this section. 2.1. LINEAR ALGEBRA PRELIMINARIES: DUAL SPACES AND PAIRINGS 25 In Section 2.11, we show how to construct a basis of the tensor power Sn(E) from a basis of E. This involves multisets. Section 2.12 is devoted to duality in symmetric powers. There is a nondegenerate pairing Sn(E∗) × Sn(E) − →K defined on generators as follows: (v∗ 1 ⊙· · · ⊙v∗ n, u1 ⊙· · · ⊙un) 7→ X σ∈Sn v∗ σ(1)(u1) · · · v∗ σ(n)(un). As a consequence, if E is finite dimensional and if K is a field of charactersistic 0, we have canonical isomorphisms (Sn(E))∗∼ = Sn(E∗) ∼ = Symn(E; K). The symmetric tensor power Sn(E) is also naturally embedded in E⊗n. In Section 2.13 we define symmetric tensor algebras. As in the case of tensors, we can pack together all the symmetric powers Sn(V ) into an algebra. Given a vector space V , the space S(V ) = M m≥0 Sm(V ), is called the symmetric tensor algebra of V . The symmetric tensor algebra S(V ) satisfies a universal mapping property with respect to commutative algebras. We conclude with Section 2.14 which gives a quick introduction to tensor products of modules over a commutative ring. Such tensor products arise because vector fields and dif-ferential forms on a smooth manifold are modules over the ring of smooth functions C∞(M). Except for the results about bases and duality, most other results still hold for these more general tensors. We introduce projective modules, which behave better under duality. Pro-jective modules will show up when dealing with vector bundles. 2.1 Linear Algebra Preliminaries: Dual Spaces and Pairings We assume that we are dealing with vector spaces over a field K. As usual the dual space E∗ of a vector space E is defined by E∗= Hom(E, K). The dual space E∗is the vector space consisting of all linear maps ω: E →K with values in the field K. A problem that comes up often is to decide when a space E is isomorphic to the dual F ∗of some other space F (possibly equal to E). The notion of pairing due to Pontrjagin provides a very clean criterion. Definition 2.1. Given two vector spaces E and F over a field K, a map ⟨−, −⟩: E×F →K is a nondegenerate pairing iffit is bilinear and iff⟨u, v⟩= 0 for all v ∈F implies u = 0, and ⟨u, v⟩= 0 for all u ∈E implies v = 0. A nondegenerate pairing induces two linear maps 26 CHAPTER 2. TENSOR ALGEBRAS ϕ: E →F ∗and ψ: F →E∗defined such that for all for all u ∈E and all v ∈F, ϕ(u) is the linear form in F ∗and ψ(v) is the linear form in E∗given by ϕ(u)(y) = ⟨u, y⟩ for all y ∈F ψ(v)(x) = ⟨x, v⟩ for all x ∈E. Schematically, ϕ(u) = ⟨u, −⟩and ψ(v) = ⟨−, v⟩. Proposition 2.1. For every nondegenerate pairing ⟨−, −⟩: E × F →K, the induced maps ϕ: E →F ∗and ψ: F →E∗are linear and injective. Furthermore, if E and F are finite dimensional, then ϕ: E →F ∗and ψ: F →E∗are bijective. Proof. The maps ϕ: E →F ∗and ψ: F →E∗are linear because u, v 7→⟨u, v⟩is bilinear. Assume that ϕ(u) = 0. This means that ϕ(u)(y) = ⟨u, y⟩= 0 for all y ∈F, and as our pairing is nondegenerate, we must have u = 0. Similarly, ψ is injective. If E and F are finite dimensional, then dim(E) = dim(E∗) and dim(F) = dim(F ∗). However, the injectivity of ϕ and ψ implies that that dim(E) ≤dim(F ∗) and dim(F) ≤dim(E∗). Consequently dim(E) ≤ dim(F) and dim(F) ≤dim(E), so dim(E) = dim(F). Therefore, dim(E) = dim(F ∗) and ϕ is bijective (and similarly dim(F) = dim(E∗) and ψ is bijective). Proposition 2.1 shows that when E and F are finite dimensional, a nondegenerate pairing induces canonical isomorphisms ϕ: E →F ∗and ψ: F →E∗; that is, isomorphisms that do not depend on the choice of bases. An important special case is the case where E = F and we have an inner product (a symmetric, positive definite bilinear form) on E. Remark: When we use the term “canonical isomorphism,” we mean that such an isomor-phism is defined independently of any choice of bases. For example, if E is a finite dimen-sional vector space and (e1, . . . , en) is any basis of E, we have the dual basis (e∗ 1, . . . , e∗ n) of E∗(where, e∗ i (ej) = δi j), and thus the map ei 7→e∗ i is an isomorphism between E and E∗. This isomorphism is not canonical. On the other hand, there are two important cases where we have canonical isomorphisms for a vector space E of finite dimension. (1) The pairing ⟨−, −⟩: E∗× E →K given by ⟨ω, u⟩= ω(u), ω ∈E∗, u ∈E, namely evaluation at u. This pairing induces the isomorphism ψ: E →E∗∗given by ψ(u)(ω) = ⟨ω, u⟩= ω(u), ω ∈E∗, u ∈E. Definition 2.2. The bilinear map eval: E∗× E →K, also denoted δ1 1 : E∗× E →K, given by eval(ω, u) = δ1 1(ω, u) = ω(u), ω ∈E∗, u ∈E, is the evaluation map or Kronecker delta, and for every u ∈E, the linear form evalu ∈ E∗∗is given by evalu(ω) = eval(ω, u) = ω(u), ω ∈E∗. (eval) 2.1. LINEAR ALGEBRA PRELIMINARIES: DUAL SPACES AND PAIRINGS 27 Thus, we see that that the isomorphism ψ: E →E∗∗is given by ψ(u) = evalu, u ∈E. (2) If ⟨−, −⟩is an inner product on E, then Proposition 2.1 shows that the nondegenerate pairing ⟨−, −⟩on E × E induces a canonical isomorphism between E and E∗. This isomorphism is often denoted ♭: E →E∗, and we usually write u♭for ♭(u), with u ∈E. Schematically, u♭= ⟨u, −⟩. The inverse of ♭is denoted ♯: E∗→E, and given any linear form ω ∈E∗, we usually write ω♯for ♯(ω). Schematically, ω = ⟨ω♯, −⟩. Given any basis, (e1, . . . , en) of E (not necessarily orthonormal), let (gij) be the n × n-matrix given by gij = ⟨ei, ej⟩(the Gram matrix of the inner product). Recall that the dual basis (e∗ 1, . . . , e∗ n) of E∗consists of the coordinate forms e∗ i ∈E∗, which are characterized by the following properties: e∗ i (ej) = δij, 1 ≤i, j ≤n. The inverse of the Gram matrix (gij) is often denoted by (gij) (by raising the indices). The tradition of raising and lowering indices is pervasive in the literature on tensors. It is indeed useful to have some notational convention to distinguish between vectors and linear forms (also called one-forms or covectors). The usual convention is that coordinates of vectors are written using superscripts, as in u = Pn i=1 uiei, and coordinates of one-forms are written using subscripts, as in ω = Pn i=1 ωie∗ i . Actually, since vectors are indexed with subscripts, one-forms are indexed with superscripts, so e∗ i should be written as ei. The motivation is that summation signs can then be omitted, according to the Einstein summation convention. According to this convention, whenever a summation variable (such as i) appears both as a subscript and a superscript in an expression, it is assumed that it is involved in a summation. For example the sum Pn i=1 uiei is abbreviated as uiei, and the sum Pn i=1 ωiei is abbreviated as ωiei. In this text we will typically only use the Einstein summation convention for tensors. The maps ♭and ♯can be described explicitly in terms of the Gram matrix of the inner product and its inverse. Proposition 2.2. For any vector space E, given a basis (e1, . . . , en) for E and its dual basis (e∗ 1, . . . , e∗ n) for E∗, for any inner product ⟨−, −⟩on E, if (gij) is its Gram matrix, with 28 CHAPTER 2. TENSOR ALGEBRAS gij = ⟨ei, ej⟩, and (gij) is its inverse, then for every vector u = Pn j=1 ujej ∈E and every one-form ω = Pn i=1 ωie∗ i ∈E∗, we have u♭= n X i=1 ωie∗ i , with ωi = n X j=1 gijuj, and ω♯= n X j=1 (ω♯)jej, with (ω♯)i = n X j=1 gijωj. Proof. For every u = Pn j=1 ujej, since u♭(v) = ⟨u, v⟩for all v ∈E and the Gram matrix is symmetric, we have u♭(ei) = ⟨u, ei⟩= n X j=1 ujej, ei = n X j=1 uj⟨ej, ei⟩= n X j=1 gijuj, so we get u♭= n X i=1 ωie∗ i , with ωi = n X j=1 gijuj. If we write ω ∈E∗as ω = Pn i=1 ωie∗ i and ω♯∈E as ω♯= Pn j=1(ω♯)jej, since ωi = ω(ei) = ⟨ω♯, ei⟩= n X j=1 (ω♯)jgij = n X j=1 gij(ω♯)j, 1 ≤i ≤n, we get (ω♯)i = n X j=1 gijωj, where (gij) is the inverse of the matrix (gij). The map ♭has the effect of lowering (flattening!) indices, and the map ♯has the effect of raising (sharpening!) indices. Here is an explicit example of Proposition 2.2. Let (e1, e2) be a basis of E such that ⟨e1, e1⟩= 1, ⟨e1, e2⟩= 2, ⟨e2, e2⟩= 5. Then g = 1 2 2 5  , g−1 =  5 −2 −2 1  . 2.1. LINEAR ALGEBRA PRELIMINARIES: DUAL SPACES AND PAIRINGS 29 Set u = u1e1 + u2e2 and observe that u♭(e1) = ⟨u1e1 + u2e2, e1⟩= ⟨e1, e1⟩u1 + ⟨e2, e1⟩u2 = g11u1 + g12u2 = u1 + 2u2 u♭(e2) = ⟨u1e1 + u2e2, e2⟩= ⟨e1, e2⟩u1 + ⟨e2, e2⟩u2 = g21u1 + g22u2 = 2u1 + 5u2, which in turn implies that u♭= ω1e∗ 1 + ω2e∗ 2 = u♭(e1)e∗ 1 + u♭(e2)e∗ 2 = (u1 + 2u2)e∗ 1 + (2u1 + 5u2)e∗ 2. Given ω = ω1e∗ 1 + ω2e∗ 2, we calculate ω♯= (ω♯)1e1 + (ω♯)2e2 from the following two linear equalities: ω1 = ω(e1) = ⟨ω♯, e1⟩= ⟨(ω♯)1e1 + (ω♯)2e2, e1⟩ = ⟨e1, e1⟩(ω♯)1 + ⟨e2, e1⟩(ω♯)2 = (ω♯)1 + 2(ω♯)2 = g11(ω♯)1 + g12(ω♯)2 ω2 = ω(e2) = ⟨ω♯, e2⟩= ⟨(ω♯)1e1 + (ω♯)2e2, e2⟩ = ⟨e1, e2⟩(ω♯)1 + ⟨e2, e2⟩(ω♯)2 = 2(ω♯)1 + 5(ω♯)2 = g21(ω♯)1 + g22(ω♯)2. These equalities are concisely written as ω1 ω2  = 1 2 2 5  (ω♯)1 (ω♯)2  = g (ω♯)1 (ω♯)2  . Then (ω♯)1 (ω♯)2  = g−1 ω1 ω2  =  5 −2 −2 1  ω1 ω2  , which in turn implies (ω♯)1 = 5ω1 −2ω2, (ω♯)2 = −2ω1 + ω2, i.e. ω♯= (5ω1 −2ω2)e1 + (−2ω1 + ω2)e2. The inner product ⟨−, −⟩on E induces an inner product on E∗denoted ⟨−, −⟩E∗, and given by ⟨ω1, ω2⟩E∗= ⟨ω♯ 1, ω♯ 2⟩, for all ω1, ω2 ∈E∗. Then we have ⟨u♭, v♭⟩E∗= ⟨(u♭)♯, (v♭)♯⟩= ⟨u, v⟩ for all u, v ∈E. If (e1, . . . , en) is a basis of E and gij = ⟨ei, ej⟩, as (e∗ i )♯= n X k=1 gikek, 30 CHAPTER 2. TENSOR ALGEBRAS an easy computation shows that ⟨e∗ i , e∗ j⟩E∗= ⟨(e∗ i )♯, (e∗ j)♯⟩= gji = gij; that is, in the basis (e∗ 1, . . . , e∗ n), the inner product on E∗is represented by the matrix (gij), the inverse of the matrix (gij). The inner product on a finite dimensional vector space also yields a canonical isomorphism between the space Hom(E, E; K) of bilinear forms on E, and the space Hom(E, E) of linear maps from E to itself. Using this isomorphism, we can define the trace of a bilinear form in an intrinsic manner. This technique is used in differential geometry, for example, to define the divergence of a differential one-form. Proposition 2.3. If ⟨−, −⟩is an inner product on a finite dimensional vector space E (over a field, K), then for every bilinear form f : E × E →K, there is a unique linear map f ♮: E →E such that f(u, v) = ⟨f ♮(u), v⟩, for all u, v ∈E. The map f 7→f ♮is a linear isomorphism between Hom(E, E; K) and Hom(E, E). Proof. For every g ∈Hom(E, E), the map given by f(u, v) = ⟨g(u), v⟩, u, v ∈E, is clearly bilinear. It is also clear that the above defines a linear map from Hom(E, E) to Hom(E, E; K). This map is injective, because if f(u, v) = 0 for all u, v ∈E, as ⟨−, −⟩is an inner product, we get g(u) = 0 for all u ∈E. Furthermore, both spaces Hom(E, E) and Hom(E, E; K) have the same dimension, so our linear map is an isomorphism. If (e1, . . . , en) is an orthonormal basis of E, then we check immediately that the trace of a linear map g (which is independent of the choice of a basis) is given by tr(g) = n X i=1 ⟨g(ei), ei⟩, where n = dim(E). Definition 2.3. We define the trace of the bilinear form f by tr(f) = tr(f ♮). From Proposition 2.3, tr(f) is given by tr(f) = n X i=1 f(ei, ei), 2.2. TENSOR PRODUCTS 31 for any orthonormal basis (e1, . . . , en) of E. We can also check directly that the above expression is independent of the choice of an orthonormal basis. We demonstrate how to calculate tr(f) where f : R2×R2 →R with f((x1, y1), (x2, y2)) = x1x2+2x2y1+3x1y2−y1y2. Under the standard basis for R2, the bilinear form f is represented as x1 y1  1 3 2 −1  x2 y2  . This matrix representation shows that f ♮= 1 3 2 −1 ⊤ = 1 2 3 −1  , and hence tr(f) = tr(f ♮) = tr 1 2 3 −1  = 0. We will also need the following proposition to show that various families are linearly independent. Proposition 2.4. Let E and F be two nontrivial vector spaces and let (ui)i∈I be any family of vectors ui ∈E. The family (ui)i∈I is linearly independent ifffor every family (vi)i∈I of vectors vi ∈F, there is some linear map f : E →F so that f(ui) = vi for all i ∈I. Proof. Left as an exercise. 2.2 Tensor Products First we define tensor products, and then we prove their existence and uniqueness up to isomorphism. Definition 2.4. Let K be a given field, and let E1, . . . , En be n ≥2 given vector spaces. For any vector space F, a map f : E1 × · · · × En →F is multilinear iffit is linear in each of its argument; that is, f(u1, . . . ui1, v + w, ui+1, . . . , un) = f(u1, . . . ui1, v, ui+1, . . . , un) + f(u1, . . . ui1, w, ui+1, . . . , un) f(u1, . . . ui1, λv, ui+1, . . . , un) = λf(u1, . . . ui1, v, ui+1, . . . , un), for all uj ∈Ej (j ̸= i), all v, w ∈Ei and all λ ∈K, for i = 1 . . . , n. 32 CHAPTER 2. TENSOR ALGEBRAS The set of multilinear maps as above forms a vector space denoted L(E1, . . . , En; F) or Hom(E1, . . . , En; F). When n = 1, we have the vector space of linear maps L(E, F) (also denoted Hom(E, F)). (To be very precise, we write HomK(E1, . . . , En; F) and HomK(E, F).) Definition 2.5. A tensor product of n ≥2 vector spaces E1, . . . , En is a vector space T together with a multilinear map ϕ: E1 × · · · × En →T, such that for every vector space F and for every multilinear map f : E1×· · ·×En →F, there is a unique linear map f⊗: T →F with f(u1, . . . , un) = f⊗(ϕ(u1, . . . , un)), for all u1 ∈E1, . . . , un ∈En, or for short f = f⊗◦ϕ. Equivalently, there is a unique linear map f⊗such that the following diagram commutes. E1 × · · · × En f & ϕ / T f⊗  F The above property is called the universal mapping property of the tensor product (T, ϕ). We show that any two tensor products (T1, ϕ1) and (T2, ϕ2) for E1, . . . , En, are isomorphic. Proposition 2.5. Given any two tensor products (T1, ϕ1) and (T2, ϕ2) for E1, . . . , En, there is an isomorphism h: T1 →T2 such that ϕ2 = h ◦ϕ1. Proof. Focusing on (T1, ϕ1), we have a multilinear map ϕ2 : E1 × · · · × En →T2, and thus there is a unique linear map (ϕ2)⊗: T1 →T2 with ϕ2 = (ϕ2)⊗◦ϕ1 as illustrated by the following commutative diagram. E1 × · · · × En ϕ2 & ϕ1 / T1 (ϕ2)⊗  T2 Similarly, focusing now on on (T2, ϕ2), we have a multilinear map ϕ1 : E1 × · · · × En →T1, and thus there is a unique linear map (ϕ1)⊗: T2 →T1 with ϕ1 = (ϕ1)⊗◦ϕ2 2.2. TENSOR PRODUCTS 33 as illustrated by the following commutative diagram. E1 × · · · × En ϕ1 & ϕ2 / T2 (ϕ1)⊗  T1 Putting these diagrams together, we obtain the commutative diagrams T1 (ϕ2)⊗  E1 × · · · × En ϕ1 & ϕ1 8 ϕ2 / T2 (ϕ1)⊗  T1 and T2 (ϕ1)⊗  E1 × · · · × En ϕ2 & ϕ2 8 ϕ1 / T1 (ϕ2)⊗  T2, which means that ϕ1 = (ϕ1)⊗◦(ϕ2)⊗◦ϕ1 and ϕ2 = (ϕ2)⊗◦(ϕ1)⊗◦ϕ2. On the other hand, focusing on (T1, ϕ1), we have a multilinear map ϕ1 : E1 × · · · × En →T1, but the unique linear map h: T1 →T1 with ϕ1 = h ◦ϕ1 is h = id, as illustrated by the following commutative diagram E1 × · · · × En ϕ1 & ϕ1 / T1 id  T1, and since (ϕ1)⊗◦(ϕ2)⊗is linear as a composition of linear maps, we must have (ϕ1)⊗◦(ϕ2)⊗= id. 34 CHAPTER 2. TENSOR ALGEBRAS Similarly, we have the commutative diagram E1 × · · · × En ϕ2 & ϕ2 / T2 id  T2, and we must have (ϕ2)⊗◦(ϕ1)⊗= id. This shows that (ϕ1)⊗and (ϕ2)⊗are inverse linear maps, and thus, (ϕ2)⊗: T1 →T2 is an isomorphism between T1 and T2. Now that we have shown that tensor products are unique up to isomorphism, we give a construction that produces them. Tensor products are obtained from free vector spaces by a quotient process, so let us begin by describing the construction of the free vector space generated by a set. For simplicity assume that our set I is finite, say I = {♥, ♦, ♠, ♣}. The construction works for any field K (and in fact for any commutative ring A, in which case we obtain the free A-module generated by I). Assume that K = R. The free vector space generated by I is the set of all formal linear combinations of the form a♥+ b♦+ c♠+ d♣, with a, b, c, d ∈R. It is assumed that the order of the terms does not matter. For example, 2♥−5♦+ 3♠= −5♦+ 2♥+ 3♠. Addition and multiplication by a scalar are are defined as follows: (a1♥+ b1♦+ c1♠+ d1♣) + (a2♥+ b2♦+ c2♠+ d2♣) = (a1 + a2)♥+ (b1 + b2)♦+ (c1 + c2)♠+ (d1 + d2)♣, and α · (a♥+ b♦+ c♠+ d♣) = αa♥+ αb♦+ αc♠+ αd♣, for all a, b, c, d, α ∈R. With these operations, it is immediately verified that we obtain a vector space denoted R(I). The set I can be viewed as embedded in R(I) by the injection ι given by ι(♥) = 1♥, ι(♦) = 1♦, ι(♠) = 1♠, ι(♣) = 1♣. Thus, R(I) can be viewed as the vector space with the special basis I = {♥, ♦, ♠, ♣}. In our case, R(I) is isomorophic to R4. 2.2. TENSOR PRODUCTS 35 The exact same construction works for any field K, and we obtain a vector space denoted by K(I) and an injection ι: I →K(I). The main reason why the free vector space K(I) over a set I is interesting is that it satisfies a universal mapping property. This means that for every vector space F (over the field K), any function h: I →F, where F is considered just a set, has a unique linear extension h: K(I) →F. By extension, we mean that h(i) = h(i) for all i ∈I, or more rigorously that h = h ◦ι. For example, if I = {♥, ♦, ♠, ♣}, K = R, and F = R3, the function h given by h(♥) = (1, 1, 1), h(♦) = (1, 1, 0), h(♠) = (1, 0, 0), h(♣) = (0, 0 −1) has a unique linear extension h: R(I) →R3 to the free vector space R(I), given by h(a♥+ b♦+ c♠+ d♣) = ah(♥) + bh(♦) + ch(♠) + dh(♣) = ah(♥) + bh(♦) + ch(♠) + dh(♣) = a(1, 1, 1) + b(1, 1, 0) + c(1, 0, 0) + d(0, 0, −1) = (a + b + c, a + b, a −d). To generalize the construction of a free vector space to infinite sets I, we observe that the formal linear combination a♥+ b♦+ c♠+ d♣can be viewed as the function f : I →R given by f(♥) = a, f(♦) = b, f(♠) = c, f(♣) = d, where a, b, c, d ∈R. More generally, we can replace R by any field K. If I is finite, then the set of all such functions is a vector space under pointwise addition and pointwise scalar multiplication. If I is infinite, since addition and scalar multiplication only makes sense for finite vectors, we require that our functions f : I →K take the value 0 except for possibly finitely many arguments. We can think of such functions as an infinite sequences (fi)i∈I of elements fi of K indexed by I, with only finitely many nonzero fi. The formalization of this construction goes as follows. Given any set I viewed as an index set, let K(I) be the set of all functions f : I →K such that f(i) ̸= 0 only for finitely many i ∈I. As usual, denote such a function by (fi)i∈I; it is a family of finite support. We make K(I) into a vector space by defining addition and scalar multiplication by (fi) + (gi) = (fi + gi) λ(fi) = (λfi). The family (ei)i∈I is defined such that (ei)j = 0 if j ̸= i and (ei)i = 1. It is a basis of the vector space K(I), so that every w ∈K(I) can be uniquely written as a finite linear combination of the ei. There is also an injection ι: I →K(I) such that ι(i) = ei for every i ∈I. Furthermore, it is easy to show that for any vector space F, and for any function 36 CHAPTER 2. TENSOR ALGEBRAS h: I →F, there is a unique linear map h: K(I) →F such that h = h ◦ι, as in the following diagram. I h ! ι / K(I) h  F Definition 2.6. The vector space (K(I), ι) constructed as above from a set I is called the free vector space generated by I (or over I). The commutativity of the above diagram is called the universal mapping property of the free vector space (K(I), ι) over I. Using the proof technique of Proposition 2.5, it is not hard to prove that any two vector spaces satisfying the above universal mapping property are isomorphic. We can now return to the construction of tensor products. For simplicity consider two vector spaces E1 and E2. Whatever E1 ⊗E2 and ϕ: E1 × E2 →E1 ⊗E2 are, since ϕ is supposed to be bilinear, we must have ϕ(u1 + u2, v1) = ϕ(u1, v1) + ϕ(u2, v1) ϕ(u1, v1 + v2) = ϕ(u1, v1) + ϕ(u1, v2) ϕ(λu1, v1) = λϕ(u1, v1) ϕ(u1, µv1) = µϕ(u1, v1) for all u1, u2 ∈E1, all v1, v2 ∈E2, and all λ, µ ∈K. Since E1 ⊗E2 must satisfy the universal mapping property of Definition 2.5, we may want to define E1 ⊗E2 as the free vector space K(E1×E2) generated by I = E1 × E2 and let ϕ be the injection of E1 × E2 into K(E1×E2). The problem is that in K(E1×E2), vectors such that (u1 + u2, v1) and (u1, v1) + (u2, v2) are different, when they should really be the same, since ϕ is bilinear. Since K(E1×E2) is free, there are no relations among the generators and this vector space is too big for our purpose. The remedy is simple: take the quotient of the free vector space K(E1×E2) by the subspace N generated by the vectors of the form (u1 + u2, v1) −(u1, v1) −(u2, v1) (u1, v1 + v2) −(u1, v1) −(u1, v2) (λu1, v1) −λ(u1, v1) (u1, µv1) −µ(u1, v1). Then, if we let E1 ⊗E2 be the quotient space K(E1×E2)/N and let ϕ be the quotient map, this forces ϕ to be bilinear. Checking that (K(E1×E2)/N, ϕ) satisfies the universal mapping property is straightforward. Here is the detailed construction. 2.2. TENSOR PRODUCTS 37 Theorem 2.6. Given n ≥2 vector spaces E1, . . . , En, a tensor product (E1 ⊗· · · ⊗En, ϕ) for E1, . . . , En can be constructed. Furthermore, denoting ϕ(u1, . . . , un) as u1 ⊗· · · ⊗un, the tensor product E1 ⊗· · · ⊗En is generated by the vectors u1 ⊗· · · ⊗un, where u1 ∈E1, . . . , un ∈En, and for every multilinear map f : E1 × · · · × En →F, the unique linear map f⊗: E1 ⊗· · · ⊗En →F such that f = f⊗◦ϕ is defined by f⊗(u1 ⊗· · · ⊗un) = f(u1, . . . , un) on the generators u1 ⊗· · · ⊗un of E1 ⊗· · · ⊗En. Proof. First we apply the construction of a free vector space to the cartesian product I = E1×· · ·×En, obtaining the free vector space M = K(I) on I = E1×· · ·×En. Since every basis generator ei ∈M is uniquely associated with some n-tuple i = (u1, . . . , un) ∈E1 × · · · × En, we denote ei by (u1, . . . , un). Next let N be the subspace of M generated by the vectors of the following type: (u1, . . . , ui + vi, . . . , un) −(u1, . . . , ui, . . . , un) −(u1, . . . , vi, . . . , un), (u1, . . . , λui, . . . , un) −λ(u1, . . . , ui, . . . , un). We let E1 ⊗· · · ⊗En be the quotient M/N of the free vector space M by N, π: M →M/N be the quotient map, and set ϕ = π ◦ι. By construction, ϕ is multilinear, and since π is surjective and the ι(i) = ei generate M, the fact that each i is of the form i = (u1, . . . , un) ∈E1 × · · · × En implies that ϕ(u1, . . . , un) generate M/N. Thus, if we denote ϕ(u1, . . . , un) as u1 ⊗· · · ⊗un, the space E1 ⊗· · · ⊗En is generated by the vectors u1 ⊗· · · ⊗un, with ui ∈Ei. It remains to show that (E1 ⊗· · · ⊗En, ϕ) satisfies the universal mapping property. To this end, we begin by proving there is a map h such that f = h ◦ϕ. Since M = K(E1×···×En) is free on I = E1 × · · · × En, there is a unique linear map f : K(E1×···×En) →F, such that f = f ◦ι, as in the diagram below. E1 × · · · × En f ) ι / K(E1×···×En) = M f  F Because f is multilinear, note that we must have f(w) = 0 for every w ∈N; for example, on the generator (u1, . . . , ui + vi, . . . , un) −(u1, . . . , ui, . . . , un) −(u1, . . . , vi, . . . , un) 38 CHAPTER 2. TENSOR ALGEBRAS we have f((u1, . . . , ui + vi, . . . , un) −(u1, . . . , ui, . . . , un) −(u1, . . . , vi, . . . , un)) = f(u1, . . . , ui + vi, . . . , un) −f(u1, . . . , ui, . . . , un) −f(u1, . . . , vi, . . . , un) = f(u1, . . . , ui, . . . , un) + f(u1, . . . , vi, . . . , un) −f(u1, . . . , ui, . . . , un) −f(u1, . . . , vi, . . . , un) = 0. But then, f : M →F factors through M/N, which means that there is a unique linear map h: M/N →F such that f = h ◦π making the following diagram commute M f " π / M/N h  F, by defining h([z]) = f(z) for every z ∈M, where [z] denotes the equivalence class in M/N of z ∈M. Indeed, the fact that f vanishes on N insures that h is well defined on M/N, and it is clearly linear by definition. Since f = f ◦ι, from the equation f = h ◦π, by composing on the right with ι, we obtain f = f ◦ι = h ◦π ◦ι = h ◦ϕ, as in the following commutative diagram. K(E1×···×En) π ( f  E1 × · · · × En f ( ι 6 K(E1×···×En)/N h u F We now prove the uniqueness of h. For any linear map f⊗: E1 ⊗· · · ⊗En →F such that f = f⊗◦ϕ, since the vectors u1 ⊗· · · ⊗un generate E1 ⊗· · · ⊗En and since ϕ(u1, . . . , un) = u1 ⊗· · · ⊗un, the map f⊗is uniquely defined by f⊗(u1 ⊗· · · ⊗un) = f(u1, . . . , un). Since f = h ◦ϕ, the map h is unique, and we let f⊗= h. The map ϕ from E1 × · · · × En to E1 ⊗· · · ⊗En is often denoted by ι⊗, so that ι⊗(u1, . . . , un) = u1 ⊗· · · ⊗un. 2.2. TENSOR PRODUCTS 39 What is important about Theorem 2.6 is not so much the construction itself but the fact that it produces a tensor product with the universal mapping property with respect to multilinear maps. Indeed, Theorem 2.6 yields a canonical isomorphism L(E1 ⊗· · · ⊗En, F) ∼ = L(E1, . . . , En; F) between the vector space of linear maps L(E1 ⊗· · · ⊗En, F), and the vector space of multi-linear maps L(E1, . . . , En; F), via the linear map −◦ϕ defined by h 7→h ◦ϕ, where h ∈L(E1 ⊗· · · ⊗En, F). Indeed, h ◦ϕ is clearly multilinear, and since by Theorem 2.6, for every multilinear map f ∈L(E1, . . . , En; F), there is a unique linear map f⊗∈ L(E1 ⊗· · · ⊗En, F) such that f = f⊗◦ϕ, the map −◦ϕ is bijective. As a matter of fact, its inverse is the map f 7→f⊗. We record this fact as the following proposition. Proposition 2.7. Given a tensor product (E1 ⊗· · · ⊗En, ϕ), the linear map h 7→h ◦ϕ is a canonical isomorphism L(E1 ⊗· · · ⊗En, F) ∼ = L(E1, . . . , En; F) between the vector space of linear maps L(E1⊗· · ·⊗En, F), and the vector space of multilinear maps L(E1, . . . , En; F). Using the “Hom” notation, the above canonical isomorphism is written Hom(E1 ⊗· · · ⊗En, F) ∼ = Hom(E1, . . . , En; F). Remarks: (1) To be very precise, since the tensor product depends on the field K, we should subscript the symbol ⊗with K and write E1 ⊗K · · · ⊗K En. However, we often omit the subscript K unless confusion may arise. (2) For F = K, the base field, Proposition 2.7 yields a canonical isomorphism between the vector space L(E1 ⊗· · · ⊗En, K), and the vector space of multilinear forms L(E1, . . . , En; K). However, L(E1 ⊗· · ·⊗En, K) is the dual space (E1 ⊗· · ·⊗En)∗, and thus the vector space of multilinear forms L(E1, . . . , En; K) is canonically isomorphic to (E1 ⊗· · · ⊗En)∗. 40 CHAPTER 2. TENSOR ALGEBRAS Since this isomorphism is used often, we record it as the following proposition. Proposition 2.8. Given a tensor product E1 ⊗· · · ⊗En,, there is a canonical isomorphism L(E1, . . . , En; K) ∼ = (E1 ⊗· · · ⊗En)∗ between the vector space of multilinear maps L(E1, . . . , En; K) and the dual (E1 ⊗· · · ⊗En)∗ of the tensor product E1 ⊗· · · ⊗En. The fact that the map ϕ: E1 × · · · × En →E1 ⊗· · · ⊗En is multilinear, can also be expressed as follows: u1 ⊗· · · ⊗(vi + wi) ⊗· · · ⊗un = (u1 ⊗· · · ⊗vi ⊗· · · ⊗un) + (u1 ⊗· · · ⊗wi ⊗· · · ⊗un), u1 ⊗· · · ⊗(λui) ⊗· · · ⊗un = λ(u1 ⊗· · · ⊗ui ⊗· · · ⊗un). Of course, this is just what we wanted! Definition 2.7. Tensors in E1 ⊗· · · ⊗En are called n-tensors, and tensors of the form u1 ⊗· · · ⊗un, where ui ∈Ei are called simple (or decomposable) n-tensors. Those n-tensors that are not simple are often called compound n-tensors. Not only do tensor products act on spaces, but they also act on linear maps (they are functors). Proposition 2.9. Given two linear maps f : E →E′ and g: F →F ′, there is a unique linear map f ⊗g: E ⊗F →E′ ⊗F ′ such that (f ⊗g)(u ⊗v) = f(u) ⊗g(v), for all u ∈E and all v ∈F. Proof. We can define h: E × F →E′ ⊗F ′ by h(u, v) = f(u) ⊗g(v). It is immediately verified that h is bilinear, and thus it induces a unique linear map f ⊗g: E ⊗F →E′ ⊗F ′ making the following diagram commutes E × F h & ι⊗ / E ⊗F f⊗g  E′ ⊗F ′, such that (f ⊗g)(u ⊗v) = f(u) ⊗g(v), for all u ∈E and all v ∈F. 2.2. TENSOR PRODUCTS 41 Definition 2.8. The linear map f ⊗g: E ⊗F →E′ ⊗F ′ given by Proposition 2.9 is called the tensor product of f : E →E′ and g: F →F ′. Another way to define f ⊗g proceeds as follows. Given two linear maps f : E →E′ and g: F →F ′, the map f × g is the linear map from E × F to E′ × F ′ given by (f × g)(u, v) = (f(u), g(v)), for all u ∈E and all v ∈F. Then the map h in the proof of Proposition 2.9 is given by h = ι′ ⊗◦(f × g), and f ⊗g is the unique linear map making the following diagram commute. E × F f×g  ι⊗ / E ⊗F f⊗g  E′ × F ′ ι′ ⊗ / E′ ⊗F ′ Remark: The notation f ⊗g is potentially ambiguous, because Hom(E, F) and Hom(E′, F ′) are vector spaces, so we can form the tensor product Hom(E, F)⊗Hom(E′, F ′) which contains elements also denoted f ⊗g. To avoid confusion, the first kind of tensor product of linear maps defined in Proposition 2.9 (which yields a linear map in Hom(E ⊗F, E′ ⊗F ′)) can be denoted by T(f, g). If we denote the tensor product E ⊗F by T(E, F), this notation makes it clearer that T is a bifunctor. If E, E′ and F, F ′ are finite dimensional, by picking bases it is not hard to show that the map induced by f ⊗g 7→T(f, g) is an isomorphism Hom(E, F) ⊗Hom(E′, F ′) ∼ = Hom(E ⊗F, E′ ⊗F ′). Proposition 2.10. Suppose we have linear maps f : E →E′, g: F →F ′, f ′ : E′ →E′′ and g′ : F ′ →F ′′. Then the following identity holds: (f ′ ◦f) ⊗(g′ ◦g) = (f ′ ⊗g′) ◦(f ⊗g). (∗) Proof. We have the commutative diagram E × F f×g  ι⊗ / E ⊗F f⊗g  E′ × F ′ f′×g′  ι′ ⊗ / E′ ⊗F ′ f′⊗g′  E′′ × F ′′ ι′′ ⊗ / E′′ ⊗F ′′, and thus the commutative diagram. E × F (f′×g′)◦(f×g)  ι⊗ / E ⊗F (f′⊗g′)◦(f⊗g)  E′′ × F ′′ ι′′ ⊗ / E′′ ⊗F ′′ 42 CHAPTER 2. TENSOR ALGEBRAS We also have the commutative diagram. E × F (f′◦f)×(g′◦g)  ι⊗ / E ⊗F (f′◦f)⊗(g′◦g)  E′′ × F ′′ ι′′ ⊗ / E′′ ⊗F ′′. Since we immediately verify that (f ′ ◦f) × (g′ ◦g) = (f ′ × g′) ◦(f × g), by uniqueness of the map between E ⊗F and E′′ ⊗F ′′ in the above diagram, we conclude that (f ′ ◦f) ⊗(g′ ◦g) = (f ′ ⊗g′) ◦(f ⊗g), as claimed. The above formula (∗) yields the following useful fact. Proposition 2.11. If f : E →E′ and g: F →F ′ are isomorphisms, then f ⊗g: E ⊗F → E′ ⊗F ′ is also an isomorphism. Proof. If f −1 : E′ →E is the inverse of f : E →E′ and g−1 : F ′ →F is the inverse of g: F →F ′ , then f −1 ⊗g−1 : E′ ⊗F ′ →E ⊗F is the inverse of f ⊗g: E ⊗F →E′ ⊗F ′, which is shown as follows: (f ⊗g) ◦(f −1 ⊗g−1) = (f ◦f −1) ⊗(g ◦g−1) = idE′ ⊗idF ′ = idE′⊗F ′, and (f −1 ⊗g−1) ◦(f ⊗g) = (f −1 ◦f) ⊗(g−1 ◦g) = idE ⊗idF = idE⊗F. Therefore, f ⊗g: E ⊗F →E′ ⊗F ′ is an isomorphism. The generalization to the tensor product f1 ⊗· · · ⊗fn of n ≥3 linear maps fi : Ei →Fi is immediate, and left to the reader. 2.3. BASES OF TENSOR PRODUCTS 43 2.3 Bases of Tensor Products We showed that E1 ⊗· · ·⊗En is generated by the vectors of the form u1 ⊗· · ·⊗un. However, these vectors are not linearly independent. This situation can be fixed when considering bases. To explain the idea of the proof, consider the case when we have two spaces E and F both of dimension 3. Given a basis (e1, e2, e3) of E and a basis (f1, f2, f3) of F, we would like to prove that e1 ⊗f1, e1 ⊗f2, e1 ⊗f3, e2 ⊗f1, e2 ⊗f2, e2 ⊗f3, e3 ⊗f1, e3 ⊗f2, e3 ⊗f3 are linearly independent. To prove this, it suffices to show that for any vector space G, if w11, w12, w13, w21, w22, w23, w31, w32, w33 are any vectors in G, then there is a bilinear map h: E × F →G such that h(ei, ej) = wij, 1 ≤i, j ≤3. Because h yields a unique linear map h⊗: E ⊗F →G such that h⊗(ei ⊗ej) = wij, 1 ≤i, j ≤3, and by Proposition 2.4, the vectors e1 ⊗f1, e1 ⊗f2, e1 ⊗f3, e2 ⊗f1, e2 ⊗f2, e2 ⊗f3, e3 ⊗f1, e3 ⊗f2, e3 ⊗f3 are linearly independent. This suggests understanding how a bilinear function f : E×F →G is expressed in terms of its values f(ei, fj) on the basis vectors (e1, e2, e3) and (f1, f2, f3), and this can be done easily. Using bilinearity we obtain f(u1e1 + u2e2 + u3e3, v1f1 + v2f2 + v3f3) = u1v1f(e1, f1) + u1v2f(e1, f2) + u1v3f(e1, f3) + u2v1f(e2, f1) + u2v2f(e2, f2) + u2v3f(e2, f3) + u3v1f(e3, f1) + u3v2f(e3, f2) + u3v3f(e3, f3). Therefore, given w11, w12, w13, w21, w22, w23, w31, w32, w33 ∈G, the function h given by h(u1e1 + u2e2 + u3e3, v1f1 + v2f2 + v3f3) = u1v1w11 + u1v2w12 + u1v3w13 + u2v1w21 + u2v2w22 + u2v3w23 + u3v1w31 + u3v2w33 + u3v3w33 is clearly bilinear, and by construction h(ei, fj) = wij, so it does the job. The generalization of this argument to any number of vector spaces of any dimension (even infinite) is straightforward. Proposition 2.12. Given n ≥2 vector spaces E1, . . . , En, if (uk i )i∈Ik is a basis for Ek, 1 ≤k ≤n, then the family of vectors (u1 i1 ⊗· · · ⊗un in)(i1,...,in)∈I1×...×In is a basis of the tensor product E1 ⊗· · · ⊗En. 44 CHAPTER 2. TENSOR ALGEBRAS Proof. For each k, 1 ≤k ≤n, every vk ∈Ek can be written uniquely as vk = X j∈Ik vk j uk j, for some family of scalars (vk j )j∈Ik. Let F be any nontrivial vector space. We show that for every family (wi1,...,in)(i1,...,in)∈I1×...×In, of vectors in F, there is some linear map h: E1 ⊗· · · ⊗En →F such that h(u1 i1 ⊗· · · ⊗un in) = wi1,...,in. Then by Proposition 2.4, it follows that (u1 i1 ⊗· · · ⊗un in)(i1,...,in)∈I1×...×In is linearly independent. However, since (uk i )i∈Ik is a basis for Ek, the u1 i1 ⊗· · · ⊗un in also generate E1 ⊗· · · ⊗En, and thus, they form a basis of E1 ⊗· · · ⊗En. We define the function f : E1 × · · · × En →F as follows: For any n nonempty finite subsets J1, . . . , Jn such that Jk ⊆Ik for k = 1, . . . , n, f( X j1∈J1 v1 j1u1 j1, . . . , X jn∈Jn vn jnun jn) = X j1∈J1,...,jn∈Jn v1 j1 · · · vn jn wj1,...,jn. It is immediately verified that f is multilinear. By the universal mapping property of the tensor product, the linear map f⊗: E1 ⊗· · · ⊗En →F such that f = f⊗◦ϕ, is the desired map h. In particular, when each Ik is finite and of size mk = dim(Ek), we see that the dimension of the tensor product E1⊗· · ·⊗En is m1 · · · mn. As a corollary of Proposition 2.12, if (uk i )i∈Ik is a basis for Ek, 1 ≤k ≤n, then every tensor z ∈E1 ⊗· · · ⊗En can be written in a unique way as z = X (i1,...,in) ∈I1×...×In λi1,...,in u1 i1 ⊗· · · ⊗un in, for some unique family of scalars λi1,...,in ∈K, all zero except for a finite number. 2.4 Some Useful Isomorphisms for Tensor Products Proposition 2.13. Given three vector spaces E, F, G, there exists unique canonical isomor-phisms (1) E ⊗F ∼ = F ⊗E 2.4. SOME USEFUL ISOMORPHISMS FOR TENSOR PRODUCTS 45 (2) (E ⊗F) ⊗G ∼ = E ⊗(F ⊗G) ∼ = E ⊗F ⊗G (3) (E ⊕F) ⊗G ∼ = (E ⊗G) ⊕(F ⊗G) (4) K ⊗E ∼ = E such that respectively (a) u ⊗v 7→v ⊗u (b) (u ⊗v) ⊗w 7→u ⊗(v ⊗w) 7→u ⊗v ⊗w (c) (u, v) ⊗w 7→(u ⊗w, v ⊗w) (d) λ ⊗u 7→λu. Proof. Except for (3), these isomorphisms are proved using the universal mapping property of tensor products. (1) The map from E × F to F ⊗E given by (u, v) 7→v ⊗u is clearly bilinear, thus it induces a unique linear α: E ⊗F →F ⊗E making the following diagram commute E × F % ι⊗ / E ⊗F α  F ⊗E, such that α(u ⊗v) = v ⊗u, for all u ∈E and all v ∈F. Similarly, the map from F × E to E ⊗F given by (v, u) 7→u ⊗v is clearly bilinear, thus it induces a unique linear β : F ⊗E →E ⊗F making the following diagram commute F × E % ι⊗ / F ⊗E β  E ⊗F, such that β(v ⊗u) = u ⊗v, for all u ∈E and all v ∈F. It is immediately verified that (β ◦α)(u ⊗v) = u ⊗v and (α ◦β)(v ⊗u) = v ⊗u for all u ∈E and all v ∈F. Since the tensors of the form u ⊗v span E ⊗F and similarly the tensors of the form v ⊗u span F ⊗E, the map β ◦α is actually the identity on E ⊗F, and similarly α ◦β is the identity on F ⊗E, so α and β are isomorphisms. 46 CHAPTER 2. TENSOR ALGEBRAS (2) Fix some w ∈G. The map (u, v) 7→u ⊗v ⊗w from E ×F to E ⊗F ⊗G is bilinear, and thus there is a linear map fw : E ⊗F →E ⊗F ⊗G making the following diagram commute E × F ' ι⊗ / E ⊗F fw  E ⊗F ⊗G, with fw(u ⊗v) = u ⊗v ⊗w. Next consider the map (z, w) 7→fw(z), from (E ⊗F) × G into E ⊗F ⊗G. It is easily seen to be bilinear, and thus it induces a linear map f : (E ⊗F) ⊗G →E ⊗F ⊗G making the following diagram commute (E ⊗F) × G ( ι⊗/ (E ⊗F) ⊗G f  E ⊗F ⊗G, with f((u ⊗v) ⊗w) = u ⊗v ⊗w. Also consider the map (u, v, w) 7→(u ⊗v) ⊗w from E ×F ×G to (E ⊗F)⊗G. It is trilinear, and thus there is a linear map g: E ⊗F ⊗G → (E ⊗F) ⊗G making the following diagram commute E × F × G ( ι⊗ / E ⊗F ⊗G g  (E ⊗F) ⊗G, with g(u ⊗v ⊗w) = (u ⊗v) ⊗w. Clearly, f ◦g and g ◦f are identity maps, and thus f and g are isomorphisms. The other case is similar. (3) Given a fixed vector space G, for any two vector spaces M and N and every linear map f : M →N, let τG(f) = f ⊗idG be the unique linear map making the following diagram commute. M × G f×idG  ιM⊗/ M ⊗G f⊗idG  N × G ιN⊗/ N ⊗G 2.4. SOME USEFUL ISOMORPHISMS FOR TENSOR PRODUCTS 47 The identity (∗) proved in Proposition 2.10 shows that if g: N →P is another linear map, then τG(g) ◦τG(f) = (g ⊗idG) ◦(f ⊗idG) = (g ◦f) ⊗(idG ◦idG) = (g ◦f) ⊗idG = τG(g ◦f). Clearly, τG(0) = 0, and a direct computation on generators also shows that τG(idM) = (idM ⊗idG) = idM⊗G, and that if f ′ : M →N is another linear map, then τG(f + f ′) = τG(f) + τG(f ′). In fancy terms, τG is a functor. Now, if E ⊕F is a direct sum, it is a standard fact of linear algebra that if πE : E ⊕F →E and πF : E ⊕F →F are the projection maps, then πE ◦πE = πE πF ◦πF = πF πE ◦πF = 0 πF ◦πE = 0 πE + πF = idE⊕F. If we apply τG to these identites, we get τG(πE) ◦τG(πE) = τG(πE) τG(πF) ◦τG(πF) = τG(πF) τG(πE) ◦τG(πF) = 0 τG(πF) ◦τG(πE) = 0 τG(πE) + τG(πF) = id(E⊕F)⊗G. Observe that τG(πE) = πE ⊗idG is a map from (E ⊕F) ⊗G onto E ⊗G and that τG(πF) = πF ⊗idG is a map from (E ⊕F) ⊗G onto F ⊗G, and by linear algebra, the above equations mean that we have a direct sum (E ⊗G) ⊕(F ⊗G) ∼ = (E ⊕F) ⊗G. (4) We have the linear map ϵ: E →K ⊗E given by ϵ(u) = 1 ⊗u, for all u ∈E. The map (λ, u) 7→λu from K × E to E is bilinear, so it induces a unique linear map η: K ⊗E →E making the following diagram commute K × E % ι⊗/ K ⊗E η  E, such that η(λ ⊗u) = λu, for all λ ∈K and all u ∈E. We have (η ◦ϵ)(u) = η(1 ⊗u) = 1u = u, and (ϵ ◦η)(λ ⊗u) = ϵ(λu) = 1 ⊗(λu) = λ(1 ⊗u) = λ ⊗u, which shows that both ϵ ◦η and η ◦ϵ are the identity, so ϵ and η are isomorphisms. 48 CHAPTER 2. TENSOR ALGEBRAS Remark: The isomorphism (3) can be generalized to finite and even arbitrary direct sums L i∈I Ei of vector spaces (where I is an arbitrary nonempty index set). We have an isomor-phism M i∈I Ei  ⊗G ∼ = M i∈I (Ei ⊗G). This isomorphism (with isomorphism (1)) can be used to give another proof of Proposition 2.12 (see Bertin , Chapter 4, Section 1) or Lang , Chapter XVI, Section 2). Proposition 2.14. Given any three vector spaces E, F, G, we have the canonical isomor-phism Hom(E, F; G) ∼ = Hom(E, Hom(F, G)). Proof. Any bilinear map f : E × F →G gives the linear map ϕ(f) ∈Hom(E, Hom(F, G)), where ϕ(f)(u) is the linear map in Hom(F, G) given by ϕ(f)(u)(v) = f(u, v). Conversely, given a linear map g ∈Hom(E, Hom(F, G)), we get the bilinear map ψ(g) given by ψ(g)(u, v) = g(u)(v), and it is clear that ϕ and ψ and mutual inverses. Since by Proposition 2.7 there is a canonical isomorphism Hom(E ⊗F, G) ∼ = Hom(E, F; G), together with the isomorphism Hom(E, F; G) ∼ = Hom(E, Hom(F, G)) given by Proposition 2.14, we obtain the important corollary: Proposition 2.15. For any three vector spaces E, F, G, we have the canonical isomorphism Hom(E ⊗F, G) ∼ = Hom(E, Hom(F, G)). 2.5 Duality for Tensor Products In this section all vector spaces are assumed to have finite dimension, unless specified other-wise. Let us now see how tensor products behave under duality. For this, we define a pairing between E∗ 1 ⊗· · ·⊗E∗ n and E1⊗· · ·⊗En as follows: For any fixed (v∗ 1, . . . , v∗ n) ∈E∗ 1 ×· · ·×E∗ n, we have the multilinear map lv∗ 1,...,v∗ n : (u1, . . . , un) 7→v∗ 1(u1) · · · v∗ n(un) 2.5. DUALITY FOR TENSOR PRODUCTS 49 from E1 × · · · × En to K. The map lv∗ 1,...,v∗ n extends uniquely to a linear map Lv∗ 1,...,v∗ n : E1 ⊗· · · ⊗En − →K making the following diagram commute. E1 × · · · × En lv∗ 1,...,v∗ n ) ι⊗/ E1 ⊗· · · ⊗En Lv∗ 1,...,v∗ n  K We also have the multilinear map (v∗ 1, . . . , v∗ n) 7→Lv∗ 1,...,v∗ n from E∗ 1 × · · · × E∗ n to Hom(E1 ⊗· · · ⊗En, K), which extends to a unique linear map L from E∗ 1 ⊗· · · ⊗E∗ n to Hom(E1 ⊗· · · ⊗En, K) making the following diagram commute. E∗ 1 × · · · × E∗ n Lv∗ 1,...,v∗ n ι⊗ / E∗ 1 ⊗· · · ⊗E∗ n L  Hom(E1 ⊗· · · ⊗En; K) However, in view of the isomorphism Hom(U ⊗V, W) ∼ = Hom(U, Hom(V, W)) given by Proposition 2.15, with U = E∗ 1 ⊗· · · ⊗E∗ n, V = E1 ⊗· · · ⊗En and W = K, we can view L as a linear map L: (E∗ 1 ⊗· · · ⊗E∗ n) ⊗(E1 ⊗· · · ⊗En) →K, which corresponds to a bilinear map ⟨−, −⟩: (E∗ 1 ⊗· · · ⊗E∗ n) × (E1 ⊗· · · ⊗En) − →K, (††) via the isomorphism (U ⊗V )∗∼ = Hom(U, V ; K) given by Proposition 2.8. This pairing is given explicitly on generators by ⟨v∗ 1 ⊗· · · ⊗v∗ n, u1 . . . , un⟩= v∗ 1(u1) · · · v∗ n(un). This pairing is nondegenerate, as proved below. Proof. If (e1 1, . . . , e1 m1), . . . , (en 1, . . . , en mn) are bases for E1, . . . , En, then for every basis element (e1 i1)∗⊗· · · ⊗(en in)∗of E∗ 1 ⊗· · · ⊗E∗ n, and any basis element e1 j1 ⊗· · · ⊗en jn of E1 ⊗· · · ⊗En, we have ⟨(e1 i1)∗⊗· · · ⊗(en in)∗, e1 j1 ⊗· · · ⊗en jn⟩= δi1 j1 · · · δin jn, where δi j is Kronecker delta, defined such that δi j = 1 if i = j, and 0 otherwise. Given any α ∈E∗ 1 ⊗· · · ⊗E∗ n, assume that ⟨α, β⟩= 0 for all β ∈E1 ⊗· · · ⊗En. The vector α is a finite 50 CHAPTER 2. TENSOR ALGEBRAS linear combination α = P λi1,...,in(e1 i1)∗⊗· · · ⊗(en in)∗, for some unique λi1,...,in ∈K. If we choose β = e1 i1 ⊗· · · ⊗en in, then we get 0 = ⟨α, e1 i1 ⊗· · · ⊗en in⟩= DX λi1,...,in(e1 i1)∗⊗· · · ⊗(en in)∗, e1 i1 ⊗· · · ⊗en in E = X λi1,...,in⟨(e1 i1)∗⊗· · · ⊗(en in)∗, e1 i1 ⊗· · · ⊗en in⟩ = λi1,...,in. Therefore, α = 0, Conversely, given any β ∈E1⊗· · ·⊗En, assume that ⟨α, β⟩= 0, for all α ∈E∗ 1 ⊗· · ·⊗E∗ n. The vector β is a finite linear combination β = P λi1,...,ine1 i1 ⊗· · · ⊗en in, for some unique λi1,...,in ∈K. If we choose α = (e1 i1)∗⊗· · · ⊗(en in)∗, then we get 0 = ⟨(e1 i1)∗⊗· · · ⊗(en in)∗, β⟩= D (e1 i1)∗⊗· · · ⊗(en in)∗, X λi1,...,ine1 i1 ⊗· · · ⊗en in E = X λi1,...,in⟨(e1 i1)∗⊗· · · ⊗(en in)∗, e1 i1 ⊗· · · ⊗en in⟩ = λi1,...,in. Therefore, β = 0. By Proposition 2.1,1 we have a canonical isomorphism (E1 ⊗· · · ⊗En)∗∼ = E∗ 1 ⊗· · · ⊗E∗ n. Here is our main proposition about duality of tensor products. Proposition 2.16. We have canonical isomorphisms (E1 ⊗· · · ⊗En)∗∼ = E∗ 1 ⊗· · · ⊗E∗ n, and µ: E∗ 1 ⊗· · · ⊗E∗ n ∼ = Hom(E1, . . . , En; K). Proof. The second isomorphism follows from the isomorphism (E1⊗· · ·⊗En)∗∼ = E∗ 1⊗· · ·⊗E∗ n together with the isomorphism Hom(E1, . . . , En; K) ∼ = (E1 ⊗· · ·⊗En)∗given by Proposition 2.8. The isomorphism µ: E∗ 1 ⊗· · · ⊗E∗ n ∼ = Hom(E1, . . . , En; K) can be described explicitly as the linear extension to E∗ 1 ⊗· · · ⊗E∗ n of the map given by µ(v∗ 1 ⊗· · · ⊗v∗ n)(u1 . . . , un) = v∗ 1(u1) · · · v∗ n(un). 1This is where the assumption that our spaces are finite dimensional is used. 2.5. DUALITY FOR TENSOR PRODUCTS 51 Since our spaces are all finite-dimensional, each space Ei is canonically isomorphic to E∗∗ i , where the isomorphism u 7→evalu is defined by Equation (eval) in Section 2.1, namely evalu(ω) = ω(u), ω ∈E∗ i , u ∈Ei. By replacing each Ei by its dual E∗ i , we obtain an isomorphism µ∗: E∗∗ 1 ⊗· · · ⊗E∗∗ n ∼ = Hom(E∗ 1, . . . , E∗ n; K), and using the isomorphisms Ei ∼ = E∗∗ i we also obtain an isomorphism E1 ⊗· · · ⊗En ∼ = Hom(E∗ 1, . . . , E∗ n; K). The above isomorphism can be defined explicitly. Indeed, using the same symbol µ∗with a slight abuse of notation, we define the map µ∗: E1 ⊗· · · ⊗En →Hom(E∗ 1, . . . , E∗ n; K) given by µ∗(u1 ⊗· · · ⊗un)(v∗ 1, . . . , v∗ n) = µ∗(evalu1 ⊗· · · ⊗evalun)(v∗ 1, . . . , v∗ n) = evalu1(v∗ 1) · · · evalun(v∗ n) = v∗ 1(u1) · · · v∗ n(un), for all ui ∈Ei and all v∗ i ∈E∗ i , 1 ≤i ≤n. Consequently, we have an explicit definition of the isomorphism µ∗: E1 ⊗· · · ⊗En →Hom(E∗ 1, . . . , E∗ n; K) given by µ∗(u1 ⊗· · · ⊗un)(v∗ 1, . . . , v∗ n) = v∗ 1(u1) · · · v∗ n(un), for all ui ∈Ei and all v∗ i ∈E∗ i , 1 ≤i ≤n. Observe the “duality” with the definition of the isomorphism µ: E∗ 1⊗· · ·⊗E∗ n →Hom(E1, . . . , En; K) given by µ(v∗ 1 ⊗· · · ⊗v∗ n)(u1, . . . , un) = v∗ 1(u1) · · · v∗ n(un). We record the above results in the following proposition. Proposition 2.17. We have the canonical isomorphism µ∗: E1 ⊗· · · ⊗En →Hom(E∗ 1, . . . , E∗ n; K) given by µ∗(u1 ⊗· · · ⊗un)(v∗ 1, . . . , v∗ n) = v∗ 1(u1) · · · v∗ n(un), and the canonical isomorphism µ: E∗ 1 ⊗· · · ⊗E∗ n →Hom(E1, . . . , En; K) given by µ(v∗ 1 ⊗· · · ⊗v∗ n)(u1, . . . , un) = v∗ 1(u1) · · · v∗ n(un), for all ui ∈Ei and all v∗ i ∈E∗ i , 1 ≤i ≤n. 52 CHAPTER 2. TENSOR ALGEBRAS The above isomorphisms are used by some authors to define tensor products of finite-dimensional vector spaces, for example, Dieudonn´ e , O’Neill and Sakai . They have the advantage of circumventing the quotient construction of the tensor product. The above isomorphisms are easily generalized to “mixed” tensor products E1 ⊗· · · ⊗Er ⊗F ∗ 1 ⊗· · · ⊗F ∗ s and E∗ 1 ⊗· · · ⊗E∗ r ⊗F1 ⊗· · · ⊗Fs of finite-dimensional vector spaces E1, . . . , Er and F1, . . . , Fs, with r, s ≥0. The proof of the following proposition is left as an exercise. Proposition 2.18. We have the canonical isomorphism µ∗: r O i=1 Ei ⊗ s O j=1 F ∗ j →Hom  r Y i=1 E∗ 1 × s Y j=1 Fj; K  given by µ∗(u1 ⊗· · · ⊗ur ⊗v∗ 1 ⊗· · · ⊗v∗ s)(x∗ 1, . . . , x∗ r, y1, . . . , ys) = r Y i=1 s Y j=1 x∗ i (ui)v∗ j(yj), and the canonical isomorphism µ: r O i=1 E∗ i ⊗ s O j=1 Fj →Hom  r Y i=1 E1 × s Y j=1 F ∗ j ; K  given by µ(x∗ 1 ⊗· · · ⊗x∗ r ⊗y1 ⊗· · · ⊗ys)(u1, . . . , ur, v∗ 1, . . . , v∗ s) = r Y i=1 s Y j=1 x∗ i (ui)v∗ j(yj), for all ui ∈Ei, x∗ i ∈E∗ i (1 ≤i ≤r), v∗ j ∈F ∗ j , yj ∈Fj (1 ≤j ≤s). Remark: The canonical isomorphisms of Proposition 2.16 holds under more general con-ditions. Namely, that K is a commutative ring with identity and that the Ei are finitely-generated projective K-modules (see Definition 2.28). See Bourbaki, (Chapter III, §11, Section 5, Proposition 7). In the special case where i = j = 1, we have a canonical isomorphism between E∗⊗F and Hom(E × F ∗; K). There is also a canonical isomorphism β between Hom(E × F ∗; K) 2.5. DUALITY FOR TENSOR PRODUCTS 53 and Hom(E, F ∗∗), namely for any f ∈Hom(E × F ∗; K), for any u ∈E, β(f) is the linear map from E to F ∗∗given by β(f)(u) = f(u, v∗), v∗∈F ∗. Since there is a canonical isomorphism F ∼ = F ∗∗, we also obtain canonical isomorphisms E∗⊗F ∼ = Hom(E × F ∗; K) ∼ = Hom(E, F ∗∗) ∼ = Hom(E, F). The method for deriving the isomorphism E∗⊗F ∼ = Hom(E, F) uses the canonical isomorphism F ∼ = F ∗∗, which is slightly indirect and actually unnecessary, as we now explain. Let E and F be two vector spaces and let α: E∗× F →Hom(E, F) be the map defined such that α(u∗, f)(x) = u∗(x)f, for all u∗∈E∗, f ∈F, and x ∈E. This map is clearly bilinear, and thus it induces a linear map α⊗: E∗⊗F →Hom(E, F) making the following diagram commute E∗× F α ' ι⊗ / E∗⊗F α⊗  Hom(E, F), such that α⊗(u∗⊗f)(x) = u∗(x)f. Proposition 2.19. If E and F are vector spaces (not necessarily finite dimensional), then the following properties hold: (1) The linear map α⊗: E∗⊗F →Hom(E, F) is injective. (2) If E is finite dimensional, then α⊗: E∗⊗F →Hom(E, F) is a canonical isomorphism. (3) If F is finite dimensional, then α⊗: E∗⊗F →Hom(E, F) is a canonical isomorphism. Proof. (1) Let (e∗ i )i∈I be a basis of E∗and let (fj)j∈J be a basis of F. Then we know that (e∗ i ⊗fj)i∈I,j∈J is a basis of E∗⊗F. To prove that α⊗is injective, let us show that its kernel is reduced to (0). For any vector ω = X i∈I′,j∈J′ λij e∗ i ⊗fj in E∗⊗F, with I′ and J′ some finite sets, assume that α⊗(ω) = 0. This means that for every x ∈E, we have α⊗(ω)(x) = 0; that is, X i∈I′,j∈J′ α⊗(λij e∗ i ⊗fj)(x) = X j∈J′ X i∈I′ λije∗ i (x)  fj = 0. 54 CHAPTER 2. TENSOR ALGEBRAS Since (fj)j∈J is a basis of F, for every j ∈J′, we must have X i∈I′ λije∗ i (x) = 0, for all x ∈E. But then (e∗ i )i∈I′ would be linearly dependent, contradicting the fact that (e∗ i )i∈I is a basis of E∗, so we must have λij = 0, for all i ∈I′ and all j ∈J′, which shows that ω = 0. Therefore, α⊗is injective. (2) Let (ej)1≤j≤n be a finite basis of E, and as usual, let e∗ j ∈E∗be the linear form defined by e∗ j(ek) = δj,k, where δj,k = 1 iffj = k and 0 otherwise. We know that (e∗ j)1≤j≤n is a basis of E∗(this is where we use the finite dimension of E). For any linear map f ∈Hom(E, F), for every x = x1e1 + · · · + xnen ∈E, we have f(x) = f(x1e1 + · · · + xnen) = x1f(e1) + · · · + xnf(en) = e∗ 1(x)f(e1) + · · · + e∗ n(x)f(en). Consequently, every linear map f ∈Hom(E, F) can be expressed as f(x) = e∗ 1(x)f1 + · · · + e∗ n(x)fn, for some fi ∈F. Furthermore, if we apply f to ei, we get f(ei) = fi, so the fi are unique. Observe that (α⊗(e∗ 1 ⊗f1 + · · · + e∗ n ⊗fn))(x) = n X i=1 (α⊗(e∗ i ⊗fi))(x) = n X i=1 e∗ i (x)fi. Thus, α⊗is surjective, so α⊗is a bijection. (3) Let (f1, . . . , fm) be a finite basis of F, and let (f ∗ 1, . . . , f ∗ m) be its dual basis. Given any linear map h: E →F, for all u ∈E, since f ∗ i (fj) = δij, we have h(u) = m X i=1 f ∗ i (h(u))fi. If h(u) = m X j=1 v∗ j(u)fj for all u ∈E (∗) for some linear forms (v∗ 1, . . . , v∗ m) ∈(E∗)m, then f ∗ i (h(u)) = m X j=1 v∗ j(u)f ∗ i (fj) = v∗ i (u) for all u ∈E, 2.5. DUALITY FOR TENSOR PRODUCTS 55 which shows that v∗ i = f ∗ i ◦h for i = 1, . . . , m. This means that h has a unique expression in terms of linear forms as in (∗). Define the map α from (E∗)m to Hom(E, F) by α(v∗ 1, . . . , v∗ m)(u) = m X j=1 v∗ j(u)fj for all u ∈E. This map is linear. For any h ∈Hom(E, F), we showed earlier that the expression of h in (∗) is unique, thus α is an isomorphism. Similarly, E∗⊗F is isomorphic to (E∗)m. Any tensor ω ∈E∗⊗F can be written as a linear combination p X k=1 u∗ k ⊗yk for some u∗ k ∈E∗and some yk ∈F, and since (f1, . . . , fm) is a basis of F, each yk can be written as a linear combination of (f1, . . . , fm), so ω can be expressed as ω = m X i=1 v∗ i ⊗fi, (†) for some linear forms v∗ i ∈E∗which are linear combinations of the u∗ k. If we pick a basis (w∗ i )i∈I for E∗, then we know that the family (w∗ i ⊗fj)i∈I,1≤j≤m is a basis of E∗⊗F, and this implies that the v∗ i in (†) are unique. Define the linear map β from (E∗)m to E∗⊗F by β(v∗ 1, . . . , v∗ m) = m X i=1 v∗ i ⊗fi. Since every tensor ω ∈E∗⊗F can be written in a unique way as in (†), this map is an isomorphism. Note that in Proposition 2.19, we have an isomorphism if either E or F has finite dimen-sion. The following proposition allows us to view a multilinear as a tensor product. Proposition 2.20. If the spaces E1, . . . En are finite dimensional vector spaces and F is any vector space, then we have the canonical isomorphism Hom(E1, . . . , En; F) ∼ = E∗ 1 ⊗· · · ⊗E∗ n ⊗F. Proof. In view of the canonical isomorphism Hom(E1, . . . , En; F) ∼ = Hom(E1 ⊗· · · ⊗En, F) given by Proposition 2.7 and the canonical isomorphism (E1 ⊗· · · ⊗En)∗∼ = E∗ 1 ⊗· · · ⊗E∗ n given by Proposition 2.16, if the Ei’s are finite dimensional, then Proposition 2.19 yields the canonical isomorphism Hom(E1, . . . , En; F) ∼ = E∗ 1 ⊗· · · ⊗E∗ n ⊗F, as claimed. 56 CHAPTER 2. TENSOR ALGEBRAS 2.6 Tensor Algebras Our goal is to define a vector space T(V ) obtained by taking the direct sum of the tensor products V ⊗· · · ⊗V | {z } m , and to define a multiplication operation on T(V ) which makes T(V ) into an algebraic struc-ture called an algebra. The algebra T(V ) satisfies a universal property stated in Proposition 2.21, which makes it the “free algebra” generated by the vector space V . Definition 2.9. The tensor product V ⊗· · · ⊗V | {z } m is also denoted as m O V or V ⊗m and is called the m-th tensor power of V (with V ⊗1 = V , and V ⊗0 = K). We can pack all the tensor powers of V into the “big” vector space T(V ) = M m≥0 V ⊗m, denoted T •(V ) or N V to avoid confusion with the tangent bundle. This is an interesting object because we can define a multiplication operation on it which makes it into an algebra. When V is of finite dimension n, we can pick some basis (e1 . . . , en) of V , and then every tensor ω ∈T(V ) can be expressed as a linear combination of terms of the form ei1 ⊗· · ·⊗eik, where (i1, . . . , ik) is any sequence of elements from the set {1, . . . , n}. We can think of the tensors ei1 ⊗· · ·⊗eik as monomials in the noncommuting variables e1, . . . , en. Thus the space T(V ) corresponds to the algebra of polynomials with coefficients in K in n noncommuting variables. Let us review the definition of an algebra over a field. Let K denote any (commutative) field, although for our purposes, we may assume that K = R (and occasionally, K = C). Since we will only be dealing with associative algebras with a multiplicative unit, we only define algebras of this kind. Definition 2.10. Given a field K, a K-algebra is a K-vector space A together with a bilinear operation ·: A × A →A, called multiplication, which makes A into a ring with unity 1 (or 1A, when we want to be very precise). This means that · is associative and that there is a multiplicative identity element 1 so that 1 · a = a · 1 = a, for all a ∈A. Given two 2.6. TENSOR ALGEBRAS 57 K-algebras A and B, a K-algebra homomorphism h: A →B is a linear map that is also a ring homomorphism, with h(1A) = 1B; that is, h(a1 · a2) = h(a1) · h(a2) for all a1, a2 ∈A h(1A) = 1B. The set of K-algebra homomorphisms between A and B is denoted Homalg(A, B). For example, the ring Mn(K) of all n × n matrices over a field K is a K-algebra. There is an obvious notion of ideal of a K-algebra. Definition 2.11. Let A be a K-algebra. An ideal A ⊆A is a linear subspace of A that is also a two-sided ideal with respect to multiplication in A; this means that for all a ∈A and all α, β ∈A, we have αaβ ∈A. If the field K is understood, we usually simply say an algebra instead of a K-algebra. We would like to define a multiplication operation on T(V ) which makes it into a K-algebra. As T(V ) = M i≥0 V ⊗i, for every i ≥0, there is a natural injection ιn : V ⊗n →T(V ), and in particular, an injection ι0 : K →T(V ). The multiplicative unit 1 of T(V ) is the image ι0(1) in T(V ) of the unit 1 of the field K. Since every v ∈T(V ) can be expressed as a finite sum v = ιn1(v1) + · · · + ιnk(vk), where vi ∈V ⊗ni and the ni are natural numbers with ni ̸= nj if i ̸= j, to define multiplica-tion in T(V ), using bilinearity, it is enough to define multiplication operations ·: V ⊗m × V ⊗n − →V ⊗(m+n), which, using the isomorphisms V ⊗n ∼ = ιn(V ⊗n), yield multi-plication operations ·: ιm(V ⊗m) × ιn(V ⊗n) − →ιm+n(V ⊗(m+n)). First, for ω1 ∈V ⊗m and ω2 ∈V ⊗n, we let ω1 · ω2 = ω1 ⊗ω2. This defines a bilinear map so it defines a multiplication V ⊗m × V ⊗n − →V ⊗m ⊗V ⊗n. This is not quite what we want, but there is a canonical isomorphism V ⊗m ⊗V ⊗n ∼ = V ⊗(m+n) which yields the desired multiplication ·: V ⊗m × V ⊗n − →V ⊗(m+n). The isomorphism V ⊗m ⊗V ⊗n ∼ = V ⊗(m+n) can be established by induction using the isomorphism (E ⊗F) ⊗G ∼ = E ⊗F ⊗G. First we prove by induction on m ≥2 that V ⊗(m−1) ⊗V ∼ = V ⊗m, 58 CHAPTER 2. TENSOR ALGEBRAS and then by induction on n ≥1 than V ⊗m ⊗V ⊗n ∼ = V ⊗(m+n). In summary the multiplication V ⊗m × V ⊗n − →V ⊗(m+n) is defined so that (v1 ⊗· · · ⊗vm) · (w1 ⊗· · · ⊗wn) = v1 ⊗· · · ⊗vm ⊗w1 ⊗· · · ⊗wn. (This has to be made rigorous by using isomorphisms involving the associativity of tensor products, for details, see Jacobson , Section 3.9, or Bertin , Chapter 4, Section 2.) Definition 2.12. Given a K-vector space V (not necessarily finite dimensional), the vector space T(V ) = M m≥0 V ⊗m denoted T •(V ) or N V equipped with the multiplication operations V ⊗m×V ⊗n − →V ⊗(m+n) defined above is called the tensor algebra of V . Remark: It is important to note that multiplication in T(V ) is not commutative. Also, in all rigor, the unit 1 of T(V ) is not equal to 1, the unit of the field K. However, in view of the injection ι0 : K →T(V ), for the sake of notational simplicity, we will denote 1 by 1. More generally, in view of the injections ιn : V ⊗n →T(V ), we identify elements of V ⊗n with their images in T(V ). The algebra T(V ) satisfies a universal mapping property which shows that it is unique up to isomorphism. For simplicity of notation, let i: V →T(V ) be the natural injection of V into T(V ). Proposition 2.21. Given any K-algebra A, for any linear map f : V →A, there is a unique K-algebra homomorphism f : T(V ) →A so that f = f ◦i, as in the diagram below. V i / f " T(V ) f  A Proof. Left an an exercise (use Theorem 2.6). A proof can be found in Knapp (Appendix A, Proposition A.14) or Bertin (Chapter 4, Theorem 2.4). Proposition 2.21 implies that there is a natural isomorphism Homalg(T(V ), A) ∼ = Hom(V, A), where the algebra A on the right-hand side is viewed as a vector space. Proposition 2.21 also has the following corollary. 2.6. TENSOR ALGEBRAS 59 Proposition 2.22. Given a linear map h: V1 →V2 between two vectors spaces V1, V2 over a field K, there is a unique K-algebra homomorphism ⊗h: T(V1) →T(V2) making the following diagram commute. V1 i1 / h  T(V1) ⊗h  V2 i2 / T(V2). Most algebras of interest arise as well-chosen quotients of the tensor algebra T(V ). This is true for the exterior algebra V(V ) (also called Grassmann algebra), where we take the quotient of T(V ) modulo the ideal generated by all elements of the form v ⊗v, where v ∈V ,and for the symmetric algebra Sym(V ), where we take the quotient of T(V ) modulo the ideal generated by all elements of the form v ⊗w −w ⊗v, where v, w ∈V . Algebras such as T(V ) are graded in the sense that there is a sequence of subspaces V ⊗n ⊆T(V ) such that T(V ) = M k≥0 V ⊗n, and the multiplication ⊗behaves well w.r.t. the grading, i.e., ⊗: V ⊗m × V ⊗n →V ⊗(m+n). Definition 2.13. A K-algebra E is said to be a graded algebra iffthere is a sequence of subspaces En ⊆E such that E = M k≥0 En, (with E0 = K) and the multiplication · respects the grading; that is, ·: Em × En →Em+n. Elements in En are called homogeneous elements of rank (or degree) n. If E and F are two K-algebras, we know that their tensor product E ⊗F exists as a vector space. We can make E ⊗F into an algebra as well. Indeed, we have the multilinear map E × F × E × F − →E ⊗F given by (a, b, c, d) 7→(ac) ⊗(bd), where ac is the product of a and c in E and bd is the product of b and d in F. By the universal mapping property, we get a linear map, E ⊗F ⊗E ⊗F − →E ⊗F. Using the isomorphism E ⊗F ⊗E ⊗F ∼ = (E ⊗F) ⊗(E ⊗F), we get a linear map (E ⊗F) ⊗(E ⊗F) − →E ⊗F, 60 CHAPTER 2. TENSOR ALGEBRAS and thus a bilinear map, (E ⊗F) × (E ⊗F) − →E ⊗F which is our multiplication operation in E ⊗F. This multiplication is determined by (a ⊗b) · (c ⊗d) = (ac) ⊗(bd). In summary, we have the following proposition. Proposition 2.23. Given two K-algebra E and F, the operation on E ⊗F defined on generators by (a ⊗b) · (c ⊗d) = (ac) ⊗(bd) makes E ⊗F into a K-algebra. 2.7 (r, s)-Tensors In differential geometry and in physics it is necessary to consider slightly more general tensors. Definition 2.14. Given a vector space V , for any pair of nonnegative integers (r, s), the tensor space T r,s(V ) of type (r, s) is the tensor product T r,s(V ) = V ⊗r ⊗(V ∗)⊗s = V ⊗· · · ⊗V | {z } r ⊗V ∗⊗· · · ⊗V ∗ | {z } s , with T 0,0(V ) = K. We also define the tensor algebra T •,•(V ) as the direct sum (coproduct) T •,•(V ) = M r,s≥0 T r,s(V ). Tensors in T r,s(V ) are called homogeneous of degree (r, s). Note that tensors in T r,0(V ) are just our “old tensors” in V ⊗r. We make T •,•(V ) into an algebra by defining tensor product operations as follows. Definition 2.15. The multiplication operations ⊗: T r1,s1(V ) × T r2,s2(V ) − →T r1+r2,s1+s2(V ) are defined such that if u = u1 ⊗· · · ⊗ur1 ⊗u∗ 1 ⊗· · · ⊗u∗ s1 and v = v1 ⊗· · · ⊗vr2 ⊗v∗ 1 ⊗· · · ⊗v∗ s2, then u ⊗v = u1 ⊗· · · ⊗ur1 ⊗v1 ⊗· · · ⊗vr2 ⊗u∗ 1 ⊗· · · ⊗u∗ s1 ⊗v∗ 1 ⊗· · · ⊗v∗ s2. 2.7. (r, s)-TENSORS 61 Denote by Hom(V r, (V ∗)s; W) the vector space of all multilinear maps from V r × (V ∗)s to W. Then we have the universal mapping property which asserts that there is a canonical isomorphism Hom(T r,s(V ), W) ∼ = Hom(V r, (V ∗)s; W). In particular, (T r,s(V ))∗∼ = Hom(V r, (V ∗)s; K). For finite dimensional vector spaces, the duality of Section 2.5 is also easily extended to the tensor spaces T r,s(V ). We define the pairing T r,s(V ∗) × T r,s(V ) − →K as follows: if v∗= v∗ 1 ⊗· · · ⊗v∗ r ⊗ur+1 ⊗· · · ⊗ur+s ∈T r,s(V ∗) and u = u1 ⊗· · · ⊗ur ⊗v∗ r+1 ⊗· · · ⊗v∗ r+s ∈T r,s(V ), then (v∗, u) = v∗ 1(u1) · · · v∗ r+s(ur+s). This is a nondegenerate pairing, and thus we get a canonical isomorphism (T r,s(V ))∗∼ = T r,s(V ∗). Consequently, we get a canonical isomorphism T r,s(V ∗) ∼ = Hom(V r, (V ∗)s; K). We summarize these results in the following proposition. Proposition 2.24. Let V be a vector space and let T r,s(V ) = V ⊗r ⊗(V ∗)⊗s = V ⊗· · · ⊗V | {z } r ⊗V ∗⊗· · · ⊗V ∗ | {z } s . We have the canonical isomorphisms (T r,s(V ))∗∼ = T r,s(V ∗), and T r,s(V ∗) ∼ = Hom(V r, (V ∗)s; K). Proposition 2.18 specializes to the case where Ei = Fj = V for i = 1, . . . , r and j = 1, . . . , s, and this yields a characterization of (r, s)-tensors in terms of multilinear maps. 62 CHAPTER 2. TENSOR ALGEBRAS Proposition 2.25. Let V be a finite-dimensional vector space. We have the canonical isomorphism µr,s : T r,s(V ) →Hom((V ∗)r, V s; K) given by µr,s(u1 ⊗· · · ⊗ur ⊗v∗ 1 ⊗· · · ⊗v∗ s)(x∗ 1, . . . , x∗ r, y1, . . . , ys) = r Y i=1 s Y j=1 x∗ i (ui)v∗ j(yj), for all ui ∈V, x∗ i ∈V ∗(1 ≤i ≤r), and all yj ∈V, v∗ j ∈V ∗(1 ≤j ≤s). At first glance, this view of tensors as multilinear maps may appear somewhat contorted, but we will see in Chapter 5 that it is more convenient to define and manipulate tensor fields. For this reason, such a defintion of tensors as multilinear maps is often used in the literature 2.8 (r, s)-Tensors as Multilinear Maps As we just observed, Proposition 2.25 allows us to view the space of (r, s)-tensors T r,s(V ) as the space of multilinear maps T : (V ∗)r × V s →K, a point of view that turns out to be technically very fruitful in differential geometry. Definition 2.16. The space Hom((V ∗)r, V s; K) of multilinear maps T : (V ∗)r × V s →K is denoted T r,s(V ). For example, Dieudonn´ e , O’Neill and Sakai use this definition. Proposition 2.25 shows that the map µr,s : T r,s(V ) →T r,s(V ) is an isomorphism. For this reason, the elements of T r,s(V ) are also called tensors. Note that T 0,1(V ) is V ∗, and T 1,0(V ) is V ∗∗∼ = V , using the canonical isomorphic u 7→evalu. Remark: The tensor spaces, T r,s(V ) (resp, T r,s(V )) are also denoted T r s (V ) (resp. T r s (V )). A tensor α ∈T r,s(V ) is said to be contravariant in the first r arguments and covariant in the last s arguments. This terminology refers to the way tensors behave under coordinate changes. Given a basis (e1, . . . , en) of V , if (e∗ 1, . . . , e∗ n) denotes the dual basis, then the tensors ei1 ⊗· · · ⊗eir ⊗e∗ j1 ⊗· · · ⊗e∗ js form a basis of T r,s(V ), so every tensor α ∈T r,s(V ) is given by an expression of the form α = X i1,...,ir j1,...,js ai1...ir j1...jsei1 ⊗· · · ⊗eir ⊗e∗ j1 ⊗· · · ⊗e∗ js. 2.8. (r, s)-TENSORS AS MULTILINEAR MAPS 63 Using the isomorphism µr,s, the multilinear maps µr,s(ei1 ⊗· · · ⊗eir ⊗e∗ j1 ⊗· · · ⊗e∗ js) form a basis of T r,s(V ), so every tensor T ∈T r,s(V ) can be uniquely written as T = X i1,...,ir j1,...,js ai1...ir j1...jsµr,s(ei1 ⊗· · · ⊗eir ⊗e∗ j1 ⊗· · · ⊗e∗ js). Since e∗ j(ei) = δij, observe that T(e∗ i1, . . . , e∗ ir, ej1, . . . , ejs) = ai1...ir j1...js. It is customary to drop µr,s and view ei1 ⊗· · · ⊗eir ⊗e∗ j1 ⊗· · · ⊗e∗ js as the corresponding multilinear map in T r,s(V ). To simplify notation, we often write µ instead of µr,s. At first, the practice of dropping µ can be quite confusing, but one finds rather quickly that using the space T r,s(V ) instead of the space T r,s(V ) is more convenient to deal with tensor fields. The tradition in classical tensor notation is to use lower indices on vectors and upper indices on linear forms, and in accordance to Einstein summation convention (or Einstein notation), the position of the indices on the coefficients is reversed. Einstein summation convention (already encountered in Section 2.1) is to assume that a summation is performed for all values of every index that appears simultaneously once as an upper index and once as a lower index. According to this convention, the linear form e∗ i dual to the basis vector ei is denoted by ei, and (suppressing the map µr,s), a tensor α ∈T r,s(V ) is written α = ai1...ir j1...jsei1 ⊗· · · ⊗eir ⊗ej1 ⊗· · · ⊗ejs. An older view of tensors is that they are multidimensional arrays of coefficients, ai1...ir j1...js  , subject to the rules for changes of bases (see Proposition 2.28). From now on until Section 2.10 we will use the Einstein summation convention. It does reduce considerably the number of summation signs. Proposition 2.26. The space T 1,1(V ) is isomorphic to the space of linear maps in Hom(V, V ∗∗), and in view of the canonical isomorphism V ∼ = V ∗∗, it is also isomorphic to the space Hom(V, V ) of linear maps from V to itself. 64 CHAPTER 2. TENSOR ALGEBRAS Proof. Indeed, if T ∈T 1,1(V ), we obtain the linear map f : V →V ∗∗defined such that for every u ∈V , f(u) = T(θ, u), θ ∈V ∗. If we pick a basis (e1, . . . , en) in V and if T is expressed as T = ai jei ⊗ej, then for any basis vector ej, the linear form T(−, ej) on V ∗is the (1, 0)-tensor µ(ai jei), namely, evaluation of a linar form in V ∗at ai jei ∈V . Thus the (1, 1)-tensor T defines the linear map f : V →V given by f(ej) = ai jei, and we see that the matrix representing f over the basis (e1, . . . , en) is (ai j). Note the double interpretation of ai jei. We can view it as a vector in V (a linear combi-nation of the vectors ei in V ), or as a (multi)linear map in T 1,0(V ), where this time, ei is really µ(ei), namely the linear form v 7→v(ei), with v ∈V ∗. For any vector u = ujej ∈V , we have f(u) = f(ujej) = ujf(ej) = ujai jei = ai jujei ∈V, so the ith coordinate f(u)i of f(u) is given by f(u)i = ai juj, confirming that (ai j) is the matrix representing f.  Note that in the above computation, the (1, 1)-tensor (ai j) denotes a matrix A where the upper index i is the row index and the lower index j is the column index, because we implicitly assumed that the coordinates uj of the vector u constitute a column vector and that f(u) is also represented by a column vector, so that f(u)i = ai juj is the product f(u)i = ai 1 · · · ai j · · · ai n         u1 . . . uj . . . un        of the ith row of the matrix A by the column vector whose jth entry is uj. The equation f(ej) = ai jei can be symbolically written as f(ej) = e1 · · · ei · · · en         a1 j . . . ai j . . . an j        , 2.8. (r, s)-TENSORS AS MULTILINEAR MAPS 65 where the column vector above is the jth column of the matrix A. To recap,    f(u)1 . . . f(u)n   = A    u1 . . . un   , f(e1) · · · f(en)  = e1 · · · en  A. A dual interpretation of the (1, 1)-tensor (ai j) consists in viewing the coordinates uj of the vector u as forming a row vector, in which case f(u) is also represented by a row vector, so that f(u)i = ai juj is the product f(u)i = u1 · · · uj · · · un        ai 1 . . . ai j . . . ai n        , where the column vector above is the transpose of the ith row of A, which means that the matrix involved is the transpose A⊤of the matrix A. In this point of view, the equation f(ej) = ai jei can be symbolically written as f(ej) = a1 j · · · ai j · · · an j         e1 . . . ei . . . en        , where the row vector above is the jth row of the matrix A. Again, the matrix involved is the transpose A⊤of A. To recap, f(u)1 · · · f(u)n = u1 · · · un A⊤,    f(e1) . . . f(en)   = A⊤    e1 . . . en   . Thus, in the dual interpretation, given a (1, 1) tensor (ai j), it makes sense to view the lower index j as the row index and the upper index i as the column index. We usually swap i and j so that the expression (aj i) denotes a matrix where the lower index i is the row index and the upper index j is the column index. The dual interpretation is used below in Proposition 2.28. Using the isomorphism µ: T r,s(V ) →T r,s(V ), it is not hard to show that the tensor product T r1,s1(V ) × T r2,s2(V ) − →T r1+r2,s1+s2(V ) is defined as follows. 66 CHAPTER 2. TENSOR ALGEBRAS Definition 2.17. For two tensors T1 ∈T r1,s1(V ) and T2 ∈T r2,s2(V ), their tensor product T1 ⊗T2 is the multilinear map in T r1+r2,s1+s2(V ) defined as follows: (T1 ⊗T2)(ω1, . . . , ωr1, θ1, . . . , θr2, v1, . . . , vs1, y1, . . . , ys2) = T1(ω1, . . . , ωr1, v1, . . . , vs1)T2(θ1, . . . , θr2, y1, . . . , ys2), for all ωi1 ∈V ∗(1 ≤i1 ≤r1), vj1 ∈V (1 ≤j1 ≤s1), θi2 ∈V ∗(1 ≤i2 ≤r2), yj2 ∈V (1 ≤j2 ≤s2). We check immediately that if T1 ∈T r1,s1(V ) and T2 ∈T r2,s2(V ), then µr1+r2,s1+s2(T1 ⊗T2) = µr1,s1(T1) ⊗µr2,s2(T2), where the tensor product on the left is the tensor product on T r1,s1(V ) × T r2,s2(V ) from Definition 2.15, and the tensor product on the right is the tensor product on T r1,s1(V ) × T r2,s2(V ) from Definition 2.17. If V and W are two vector spaces, for any linear map ϕ: V →W, recall that the transpose ϕ⊤of ϕ is the linear map ϕ⊤: W ∗→V ∗given by ϕ⊤(ω) = ω ◦ϕ, ω ∈W ∗. If ϕ: V →W is a linear isomorphism, it induces a linear map ϕr s : T r,s(V ) →T r,s(W) defined as follows. Definition 2.18. Let ϕ: V →W be a linear isomorphism of finite-dimensional vector spaces. The map ϕr s : T r,s(V ) →T r,s(W) is defined as follows. For every T ∈T r,s(V ), (ϕr sT)(θ1, . . . , θr, v1, . . . , vs) = T(ϕ⊤(θ1), . . . , ϕ⊤(θr), ϕ−1(v1), . . . , ϕ−1(vs)), for all θ1, . . . , θr ∈W ∗and all v1, . . . , vs ∈W. Since ϕ−1 and ϕ⊤are linear, ϕr sT is multilinear, thus in T r,s(W). The following proposition is shown in Abraham and Marsden (Proposition 1.7.4). Proposition 2.27. Let ϕ: E →F and ψ: F →G be two linear isomorphisms of finite-dimensional vector spaces. The following properties hold. (1) (ψ ◦ϕ)r s = ψr s ◦ϕr s. (2) If id: E →E is the identity, then idr s : T r,s(E) →T r,s(E) is also the identity. (3) The map ϕr s : T r,s(E) →T r,s(F) is a linear isomorphisms and (ϕr s)−1 = (ϕ−1)r s. 2.8. (r, s)-TENSORS AS MULTILINEAR MAPS 67 The next proposition gives formulae for the components of ϕr sT in terms of the compo-nents of T with respect to bases of E and F. Let (e1, . . . , en) be a basis of E, (f1, . . . , fn) be a basis of F, and let (e1, . . . , en) be the dual basis of E∗and (f 1, . . . , f n) be the dual basis of F ∗. If A = (aj i) is the matrix representing ϕ in the dual interpretation given (in Einstein notation) by ϕ(ei) = aj ifj, namely,    ϕ(e1) . . . ϕ(en)   = A    f1 . . . fn   , then it known that the matrix representing ϕ⊤(also in the dual interpretation) with respect to the dual bases (f 1, . . . , f n) and (e1, . . . , en) is the transpose A⊤of the matrix A (see Gallier and Quaintance , Section 10.6). This means that ϕ⊤(f i) = ai jej, where transposition corresponds to the fact that the indices i and j are raised when passing from vectors to their dual. If B = (bj i) denotes the inverse of A, then we have ϕ−1(fi) = bj iej. We easily obtain the following result. Proposition 2.28. Let ϕ: E →F be a linear isomorphism, (e1, . . . , en) be a basis of E, and (f1, . . . , fn) be a basis of F. If A = (aj i) is the matrix representing ϕ in the dual interpretation as above and B = (bj i) is its inverse, for any T ∈T r,s(E) with components ti1...ir j1...js with respect to the basis ei1 ⊗· · · ⊗eir ⊗ej1 · · · ⊗ejs, the components of ϕr sT with respect to the basis fi′ 1 ⊗· · · ⊗fi′ r ⊗f j′ 1 · · · ⊗f j′ s are given by (t′) i′ 1...i′ r j′ 1...j′ s = ti1...ir j1...jsa i′ 1 i1 · · · ai′ r ir · · · bj1 j′ 1 · · · bjs j′ s. Observe that the Einstein summation convention pays offin avoiding many summation symbols (P). In the special case where E = F and ϕ is the identity, we obtain a formula expressing how the components of a tensor change under a change of basis. This makes a connection with the old-fashion method for defining tensors. Another operation on general tensors, contraction, is useful in differential geometry. 68 CHAPTER 2. TENSOR ALGEBRAS 2.9 Contraction Operators Definition 2.19. For all r, s ≥1, the contraction ci,j : T r,s(V ) →T r−1,s−1(V ), with 1 ≤i ≤r and 1 ≤j ≤s, is the linear map defined on generators by ci,j(u1 ⊗· · · ⊗ur ⊗ω1 ⊗· · · ⊗ωs) = ωj(ui) u1 ⊗· · · ⊗b ui ⊗· · · ⊗ur ⊗ω1 ⊗· · · ⊗c ωj ⊗· · · ⊗ωs, ui ∈V , ωj ∈V ∗, where the hat over an argument means that it should be omitted. Note that since T r,s(V ) is the tensor product T r,s(V ) = V ⊗r ⊗(V ∗)⊗s, the map ci,j is the unique linear extension of the multilinear map from V r × (V ∗)s to V ⊗r ⊗(V ∗)⊗s given by ci,j(u1, . . . , ur, ω1, . . . , ωs) = ωj(ui) u1 ⊗· · · ⊗b ui ⊗· · · ⊗ur ⊗ω1 ⊗· · · ⊗c ωj ⊗· · · ⊗ωs. Therefore this definition is intrinsic, that is, does not depend on any choice of basis. Remark: Note that the notation ci,j is ambiguous, since technically we have an operator cr,s i,j : T r,s(V ) →T r−1,s−1(V ) for every pair r, s ≥1, with 1 ≤i ≤r and 1 ≤j ≤s, and cs,r i,j ̸= cr′,s′ i,j if (r, s) ̸= (r′, s′), 1 ≤i ≤min(r, r′) and 1 ≤j ≤min(s, s′). However it is customary to omit the superscripts r and s and rely on context to infer r and s. The notation ci,j is also used by Gallot, Hulin, Lafontaine , but Sakai uses ci j, and O’Neill uses cj i. Let us figure our what is c1,1 : T 1,1(V ) →R, that is c1,1 : V ⊗V ∗→R. If (e1, . . . , en) is a basis of V and (e1, . . . , en) is the dual basis, by Proposition 2.12, every h ∈V ⊗V ∗can be expressed as h = ai j ei ⊗ej. As c1,1(ei ⊗ej) = δi,j, we get c1,1(h) = ai i. Remark: By Proposition 2.25, V ⊗V ∗is isomorphic to T 1,1(V ), and by Proposition 2.26, T 1,1(V ) is isomorphic to Hom(V, V ). Thus we can view h ∈V ⊗V ∗as linear map. The matrix of h over the basis (e1, . . . , en) is also (ai j), so tr(h) = ai i is its trace. Since c1,1 is defined independently of any basis, c1,1 provides an intrinsic definition of the trace of a linear map h ∈Hom(V, V ). The notion of contraction transfers to the space T r,s(V ), but at first glance, we don’t have an intrinsic definition since it appears that we have to resort to bases. There is a way to circumvent this problem as explained in O’Neill (Chapter 2, Lemma 6). The first step is to show that the contraction c1,1 from T 1,1(V ) to K is intrinsically defined. 2.9. CONTRACTION OPERATORS 69 Proposition 2.29. There is a unique linear map c1,1 : T 1,1(V ) →K such that c1,1(µ(u ⊗θ)) = θ(u), u ∈V, θ ∈V ∗. Proof. If we pick a basis (e1, . . . , en) of V , we see immediately that for any T ∈T 1,1(V ) expressed as T = ti j ei ⊗ej, (again, writing ei ⊗ej instead of µ(ei ⊗ej)), since c1,1(ei ⊗ej) = ej(ei) = δij, we must have c1,1(T) = ti i. Given another basis (f1, . . . , fn), if this basis is expressed in terms of the basis (e1, . . . , en) as ei = aj ifj, then by Proposition 2.28), we have f i = ai jej, so if (bj i) is the inverse of (aj i), we obtain T = ti jak i bj l fk ⊗f l, and thus the contracted tensor over the new bases is given by c′ 1,1(T) = ti jak i bj k. But (bj i) is the inverse of (aj i), so ak i bj k = δij, which implies by Proposition 2.28 that c′ 1,1(T) = ti i = c1,1(T). Therefore, c1,1(T) is independent of the choice of basis. In the general case, we have the following result. Proposition 2.30. There is a unique linear map ci,j : T r,s(V ) →T r−1,s−1(V ), with 1 ≤i ≤r and 1 ≤j ≤s, such that (omitting the map µ) ci,j(u1 ⊗· · · ⊗ur ⊗ω1 ⊗· · · ⊗ωs) = ωj(ui) u1 ⊗· · · ⊗b ui ⊗· · · ⊗ur ⊗ω1 ⊗· · · ⊗c ωj ⊗· · · ⊗ωs, ui ∈V , ωj ∈V ∗. 70 CHAPTER 2. TENSOR ALGEBRAS Proof. If we pick a basis (e1 . . . , en) in V , using the Einstein summation convention, (omitting the map µ), for any α ∈T r,s(V ), we can write α = ai1...ir j1...js ei1 ⊗· · · ⊗eir ⊗ej1 ⊗· · · ⊗ejs, and since ej(ei) = δij, we must have ck,l(α) = ai1...ik−1mik+1...ir j1...jl−1mjl+1...js ei1 ⊗· · · ⊗c eik ⊗· · · ⊗eir ⊗ej1 ⊗· · · ⊗c ejl ⊗· · · ⊗ejs, where the repeated index m appears as a superscript in position k and as a subscript in position l, so according to the Einstein summation convention, ai1...ik−1mik+1...ir j1...jl−1mjl+1...,js denotes the sum n X m=1 ai1...ik−1mik+1...ir j1...jl−1mjl+1...js . Given another basis (f1, . . . , fn), if this basis is expressed in terms of the basis (e1, . . . , en) as ei = aj ifj, then by Proposition 2.28, we have α = ti1...ir j1...jsa i′ 1 i1 · · · ai′ r ir · · · bj1 j′ 1 · · · bjs j′ s fi′ 1 ⊗· · · ⊗fi′ r ⊗f j′ 1 ⊗· · · ⊗f j′ s. Consequently, contracting on the pair of indices (i′ k, j′ l), the contracted tensor over the new bases is given by c′ k,l(α) = ti1...m...ir j1...m...jsa i′ 1 i1 · · · c am ik · · · ai′ r ir bj1 j′ 1 · · · c bjl m · · · bjs j′ s fi′ 1 ⊗· · · ⊗c fi′ k ⊗· · · ⊗fi′ r ⊗f j′ 1 ⊗· · · ⊗c f j′ l ⊗· · · ⊗f j′ s. Since (bj i) is the inverse of (aj i), we have am ikbjl m = δikjl, which implies that c′ k,l(α) = ti1...m...ir j1...m...jsa i′ 1 i1 · · · c am ik · · · ai′ r ir bj1 j′ 1 · · · c bjl m · · · bjs j′ s fi′ 1 ⊗· · · ⊗c fi′ k ⊗· · · ⊗fi′ r ⊗f j′ 1 ⊗· · · ⊗c f j′ l ⊗· · · ⊗f j′ s = ti1...m...ir j1...m...js ei1 ⊗· · · ⊗c eik ⊗· · · ⊗eir ⊗ej1 ⊗· · · ⊗c ejl ⊗· · · ⊗ejs = ck,l(α). Therefore, the definition of ck,l is intrinsic; it does not depend on the choice of basis. We now turn to symmetric tensors. 2.10. SYMMETRIC TENSOR POWERS 71 2.10 Symmetric Tensor Powers Our goal is to come up with a notion of tensor product that will allow us to treat symmetric multilinear maps as linear maps. Note that we have to restrict ourselves to a single vector space E, rather then n vector spaces E1, . . . , En, so that symmetry makes sense. Definition 2.20. A multilinear map f : En →F is symmetric iff f(uσ(1), . . . , uσ(n)) = f(u1, . . . , un), for all ui ∈E and all permutations, σ: {1, . . . , n} →{1, . . . , n}. The group of permutations on {1, . . . , n} (the symmetric group) is denoted Sn. The vector space of all symmetric multilinear maps f : En →F is denoted by Symn(E; F) or Homsymlin(En, F). Note that Sym1(E; F) = Hom(E, F). We could proceed directly as in Theorem 2.6 and construct symmetric tensor products from scratch. However, since we already have the notion of a tensor product, there is a more economical method. First we define symmetric tensor powers. Definition 2.21. An n-th symmetric tensor power of a vector space E, where n ≥1, is a vector space S together with a symmetric multilinear map ϕ: En →S such that, for every vector space F and for every symmetric multilinear map f : En →F, there is a unique linear map f⊙: S →F, with f(u1, . . . , un) = f⊙(ϕ(u1, . . . , un)), for all u1, . . . , un ∈E, or for short f = f⊙◦ϕ. Equivalently, there is a unique linear map f⊙such that the following diagram commutes. En f ! ϕ / S f⊙  F The above property is called the universal mapping property of the symmetric tensor power (S, ϕ). We next show that any two symmetric n-th tensor powers (S1, ϕ1) and (S2, ϕ2) for E are isomorphic. Proposition 2.31. Given any two symmetric n-th tensor powers (S1, ϕ1) and (S2, ϕ2) for E, there is an isomorphism h: S1 →S2 such that ϕ2 = h ◦ϕ1. 72 CHAPTER 2. TENSOR ALGEBRAS Proof. Replace tensor product by n-th symmetric tensor power in the proof of Proposition 2.5. We now give a construction that produces a symmetric n-th tensor power of a vector space E. Theorem 2.32. Given a vector space E, a symmetric n-th tensor power (Sn(E), ϕ) for E can be constructed (n ≥1). Furthermore, denoting ϕ(u1, . . . , un) as u1 ⊙· · · ⊙un, the symmetric tensor power Sn(E) is generated by the vectors u1 ⊙· · · ⊙un, where u1, . . . , un ∈E, and for every symmetric multilinear map f : En →F, the unique linear map f⊙: Sn(E) →F such that f = f⊙◦ϕ is defined by f⊙(u1 ⊙· · · ⊙un) = f(u1, . . . , un) on the generators u1 ⊙· · · ⊙un of Sn(E). Proof. The tensor power E⊗n is too big, and thus we define an appropriate quotient. Let C be the subspace of E⊗n generated by the vectors of the form u1 ⊗· · · ⊗un −uσ(1) ⊗· · · ⊗uσ(n), for all ui ∈E, and all permutations σ: {1, . . . , n} →{1, . . . , n}. We claim that the quotient space (E⊗n)/C does the job. Let p: E⊗n →(E⊗n)/C be the quotient map, and let ϕ: En →(E⊗n)/C be the map given by ϕ = p ◦ϕ0, where ϕ0 : En →E⊗n is the injection given by ϕ0(u1, . . . , un) = u1 ⊗· · · ⊗un. Let us denote ϕ(u1, . . . , un) as u1 ⊙· · · ⊙un. It is clear that ϕ is symmetric. Since the vectors u1 ⊗· · · ⊗un generate E⊗n, and p is surjective, the vectors u1 ⊙· · · ⊙un generate (E⊗n)/C. It remains to show that ((E⊗n)/C, ϕ) satisfies the universal mapping property. To this end we begin by proving that there is a map h such that f = h ◦ϕ. Given any symmetric multilinear map f : En →F, by Theorem 2.6 there is a linear map f⊗: E⊗n →F such that f = f⊗◦ϕ0, as in the diagram below. En f # ϕ0 / E⊗n f⊗  F 2.10. SYMMETRIC TENSOR POWERS 73 However, since f is symmetric, we have f⊗(z) = 0 for every z ∈C. Thus, we get an induced linear map h: (E⊗n)/C →F making the following diagram commute. E⊗n p % f⊗  En f " ϕ0 < (E⊗n)/C h y F If we define h([z]) = f⊗(z) for every z ∈E⊗n, where [z] is the equivalence class in (E⊗n)/C of z ∈E⊗n, the above diagram shows that f = h ◦p ◦ϕ0 = h ◦ϕ. We now prove the uniqueness of h. For any linear map f⊙: (E⊗n)/C →F such that f = f⊙◦ϕ, since ϕ(u1, . . . , un) = u1 ⊙· · · ⊙un and the vectors u1 ⊙· · · ⊙un generate (E⊗n)/C, the map f⊙ is uniquely defined by f⊙(u1 ⊙· · · ⊙un) = f(u1, . . . , un). Since f = h ◦ϕ, the map h is unique, and we let f⊙= h. Thus, Sn(E) = (E⊗n)/C and ϕ constitute a symmetric n-th tensor power of E. The map ϕ from En to Sn(E) is often denoted ι⊙, so that ι⊙(u1, . . . , un) = u1 ⊙· · · ⊙un. Again, the actual construction is not important. What is important is that the symmetric n-th power has the universal mapping property with respect to symmetric multilinear maps. Remark: The notation ⊙for the commutative multiplication of symmetric tensor powers is not standard. Another notation commonly used is ·. We often abbreviate “symmetric tensor power” as “symmetric power.” The symmetric power Sn(E) is also denoted SymnE but we prefer to use the notation Sym to denote spaces of symmetric multilinear maps. To be consistent with the use of ⊙, we could have used the notation Jn E. Clearly, S1(E) ∼ = E and it is convenient to set S0(E) = K. The fact that the map ϕ: En →Sn(E) is symmetric and multilinear can also be expressed as follows: u1 ⊙· · · ⊙(vi + wi) ⊙· · · ⊙un = (u1 ⊙· · · ⊙vi ⊙· · · ⊙un) + (u1 ⊙· · · ⊙wi ⊙· · · ⊙un), u1 ⊙· · · ⊙(λui) ⊙· · · ⊙un = λ(u1 ⊙· · · ⊙ui ⊙· · · ⊙un), uσ(1) ⊙· · · ⊙uσ(n) = u1 ⊙· · · ⊙un, for all permutations σ ∈Sn. The last identity shows that the “operation” ⊙is commutative. This allows us to view the symmetric tensor u1 ⊙· · · ⊙un as an object called a multiset. 74 CHAPTER 2. TENSOR ALGEBRAS Given a set A, a multiset with elements from A is a generalization of the concept of a set that allows multiple instances of elements from A to occur. For example, if A = {a, b, c, d}, the following are multisets: M1 = {a, a, b}, M2 = {a, a, b, b, c}, M3 = {a, a, b, b, c, d, d, d}. Here is another way to represent multisets as tables showing the multiplicities of the elements in the multiset: M1 = a b c d 2 1 0 0  , M2 = a b c d 2 2 1 0  , M3 = a b c d 2 2 1 3  . The above are just graphs of functions from the set A = {a, b, c, d} to N. This suggests the following definition. Definition 2.22. A finite multiset M over a set A is a function M : A →N such that M(a) ̸= 0 for finitely many a ∈A. The multiplicity of an element a ∈A in M is M(a). The set of all multisets over A is denoted by N(A), and we let dom(M) = {a ∈A | M(a) ̸= 0}, which is a finite set. The set dom(M) is the set of elements in A that actually occur in M. For any multiset M ∈N(A), note that P a∈A M(a) makes sense, since P a∈A M(a) = P a∈dom(A) M(a), and dom(M) is finite; this sum is the total number of elements in the multiset A and is called the size of M. Let |M| = P a∈A M(a). Going back to our symmetric tensors, we can view the tensors of the form u1 ⊙· · · ⊙un as multisets of size n over the set E. Theorem 2.32 implies the following proposition. Proposition 2.33. There is a canonical isomorphism Hom(Sn(E), F) ∼ = Symn(E; F), between the vector space of linear maps Hom(Sn(E), F) and the vector space of symmetric multilinear maps Symn(E; F) given by the linear map −◦ϕ defined by h 7→h ◦ϕ, with h ∈Hom(Sn(E), F). Proof. The map h◦ϕ is clearly symmetric multilinear. By Theorem 2.32, for every symmetric multilinear map f ∈Symn(E; F) there is a unique linear map f⊙∈Hom(Sn(E), F) such that f = f⊙◦ϕ, so the map −◦ϕ is bijective. Its inverse is the map f 7→f⊙. In particular, when F = K, we get the following important fact. Proposition 2.34. There is a canonical isomorphism (Sn(E))∗∼ = Symn(E; K). 2.11. BASES OF SYMMETRIC POWERS 75 Definition 2.23. Symmetric tensors in Sn(E) are called symmetric n-tensors, and tensors of the form u1 ⊙· · · ⊙un, where ui ∈E, are called simple (or decomposable) symmetric n-tensors. Those symmetric n-tensors that are not simple are often called compound symmetric n-tensors. Given a linear maps f : E →E′, since the map ι′ ⊙◦(f × f) is bilinear and symmetric, there is a unique linear map f ⊙f : S2(E) →S2(E′) making the following diagram commute. E2 f×f  ι⊙ / S2(E) f⊙f  (E′)2 ι′ ⊙ / S2(E′). Observe that f ⊙f is determined by (f ⊙f)(u ⊙v) = f(u) ⊙f(v). Proposition 2.35. Given any two linear maps f : E →E′ and f ′ : E′ →E′′,we have (f ′ ◦f) ⊙(f ′ ◦f) = (f ′ ⊙f ′) ◦(f ⊙f). By using the proof techniques of Proposition 2.13 (3), we can show the following property of symmetric tensor products. Proposition 2.36. We have the following isomorphism: Sn(E ⊕F) ∼ = n M k=0 Sk(E) ⊗Sn−k(F). The generalization to the symmetric tensor product f ⊙· · · ⊙f of n ≥3 copies of the linear map f : E →E′ is immediate, and left to the reader. 2.11 Bases of Symmetric Powers The vectors u1 ⊙· · · ⊙um where u1, . . . , um ∈E generate Sm(E), but they are not linearly independent. We will prove a version of Proposition 2.12 for symmetric tensor powers using multisets. Recall that a (finite) multiset over a set I is a function M : I →N, such that M(i) ̸= 0 for finitely many i ∈I. The set of all multisets over I is denoted as N(I) and we let dom(M) = {i ∈I | M(i) ̸= 0}, the finite set of elements in I that actually occur in M. The size of the multiset M is |M| = P a∈A M(a). 76 CHAPTER 2. TENSOR ALGEBRAS To explain the idea of the proof, consider the case when m = 2 and E has dimension 3. Given a basis (e1, e2, e3) of E, we would like to prove that e1 ⊙e1, e1 ⊙e2, e1 ⊙e3, e2 ⊙e2, e2 ⊙e3, e3 ⊙e3 are linearly independent. To prove this, it suffices to show that for any vector space F, if w11, w12, w13, w22, w23, w33 are any vectors in F, then there is a symmetric bilinear map h: E2 →F such that h(ei, ej) = wij, 1 ≤i ≤j ≤3. Because h yields a unique linear map h⊙: S2(E) →F such that h⊙(ei ⊙ej) = wij, 1 ≤i ≤j ≤3, by Proposition 2.4, the vectors e1 ⊙e1, e1 ⊙e2, e1 ⊙e3, e2 ⊙e2, e2 ⊙e3, e3 ⊙e3 are linearly independent. This suggests understanding how a symmetric bilinear function f : E2 →F is expressed in terms of its values f(ei, ej) on the basis vectors (e1, e2, e3), and this can be done easily. Using bilinearity and symmetry, we obtain f(u1e1 + u2e2 + u3e3, v1e1 + v2e2 + v3e3) = u1v1f(e1, e1) + (u1v2 + u2v1)f(e1, e2) + (u1v3 + u3v1)f(e1, e3) + u2v2f(e2, e2) + (u2v3 + u3v2)f(e2, e3) + u3v3f(e3, e3). Therefore, given w11, w12, w13, w22, w23, w33 ∈F, the function h given by h(u1e1 + u2e2 + u3e3, v1e1 + v2e2 + v3e3) = u1v1w11 + (u1v2 + u2v1)w12 + (u1v3 + u3v1)w13 + u2v2w22 + (u2v3 + u3v2)w23 + u3v3w33 is clearly bilinear symmetric, and by construction h(ei, ej) = wij, so it does the job. The generalization of this argument to any m ≥2 and to a space E of any dimension (even infinite) is conceptually clear, but notationally messy. If dim(E) = n and if (e1, . . . , en) is a basis of E, for any m vectors vj = Pn i=1 ui,jei in E, for any symmetric multilinear map f : Em →F, we have f(v1, . . . , vm) = X k1+···+kn=m X I1∪···∪In={1,...,m} Ii∩Ij=∅, i̸=j, |Ij|=kj Y i1∈I1 u1,i1 ! · · · Y in∈In un,in !! f(e1, . . . , e1 | {z } k1 , . . . , en, . . . , en | {z } kn ). 2.11. BASES OF SYMMETRIC POWERS 77 Definition 2.24. Given any set J of n ≥1 elements, say J = {j1, . . . , jn}, and given any m ≥2, for any sequence (k1 . . . , kn) of natural numbers ki ∈N such that k1 + · · · + kn = m, the multiset M of size m M = {j1, . . . , j1 | {z } k1 , j2, . . . , j2 | {z } k2 , . . . , jn, . . . , jn | {z } kn } is denoted by M(m, J, k1, . . . , kn). Note that M(ji) = ki, for i = 1, . . . , n. Given any k ≥1, and any u ∈E, we denote u ⊙· · · ⊙u | {z } k as u⊙k. We can now prove the following proposition. Proposition 2.37. Given a vector space E, if (ei)i∈I is a basis for E, then the family of vectors  e⊙M(i1) i1 ⊙· · · ⊙e⊙M(ik) ik  M∈N(I), |M|=m, {i1,...,ik}=dom(M) is a basis of the symmetric m-th tensor power Sm(E). Proof. The proof is very similar to that of Proposition 2.12. First assume that E has finite dimension n. In this case I = {1, . . . , n}, and any multiset M ∈N(I) of size |M| = m is of the form M(m, {1, . . . , n}, k1, . . . , kn), with ki = M(i) and k1 + · · · + kn = m. For any nontrivial vector space F, for any family of vectors (wM)M∈N(I), |M|=m, we show the existence of a symmetric multilinear map h: Sm(E) →F, such that for every M ∈N(I) with |M| = m, we have h(e⊙M(i1) i1 ⊙· · · ⊙e⊙M(ik) ik ) = wM, where {i1, . . . , ik} = dom(M). We define the map f : Em →F as follows: for any m vectors v1, . . . , vm ∈E we can write vk = Pn i=1 ui,kei for k = 1, . . . , m and we set f(v1, . . . , vm) = X k1+···+kn=m X I1∪···∪In={1,...,m} Ii∩Ij=∅, i̸=j, |Ij|=kj Y i1∈I1 u1,i1 ! · · · Y in∈In un,in !! wM(m,{1,...,n},k1,...,kn). It is not difficult to verify that f is symmetric and multilinear. By the universal mapping property of the symmetric tensor product, the linear map f⊙: Sm(E) →F such that f = f⊙◦ϕ, is the desired map h. Then by Proposition 2.4, it follows that the family  e⊙M(i1) i1 ⊙· · · ⊙e⊙M(ik) ik  M∈N(I), |M|=m, {i1,...,ik}=dom(M) 78 CHAPTER 2. TENSOR ALGEBRAS is linearly independent. Using the commutativity of ⊙, we can also show that these vectors generate Sm(E), and thus, they form a basis for Sm(E). If I is infinite dimensional, then for any m vectors v1, . . . , vm ∈F there is a finite subset J of I such that vk = P j∈J uj,kej for k = 1, . . . , m, and if we write n = |J|, then the formula for f(v1, . . . , vm) is obtained by replacing the set {1, . . . , n} by J. The details are left as an exercise. As a consequence, when I is finite, say of size p = dim(E), the dimension of Sm(E) is the number of finite multisets (j1, . . . , jp), such that j1 + · · · + jp = m, jk ≥0. We leave as an exercise to show that this number is p+m−1 m  . Thus, if dim(E) = p, then the dimension of Sm(E) is p+m−1 m  . Compare with the dimension of E⊗m, which is pm. In particular, when p = 2, the dimension of Sm(E) is m + 1. This can also be seen directly. Remark: The number p+m−1 m  is also the number of homogeneous monomials Xj1 1 · · · Xjp p of total degree m in p variables (we have j1 +· · ·+jp = m). This is not a coincidence! Given a vector space E and a basis (ei)i∈I for E, Proposition 2.37 shows that every symmetric tensor z ∈Sm(E) can be written in a unique way as z = X M∈N(I) P i∈I M(i)=m {i1,...,ik}=dom(M) λM e⊙M(i1) i1 ⊙· · · ⊙e⊙M(ik) ik , for some unique family of scalars λM ∈K, all zero except for a finite number. This looks like a homogeneous polynomial of total degree m, where the monomials of total degree m are the symmetric tensors e⊙M(i1) i1 ⊙· · · ⊙e⊙M(ik) ik in the “indeterminates” ei, where i ∈I (recall that M(i1) + · · · + M(ik) = m) and implies that polynomials can be defined in terms of symmetric tensors. 2.12 Duality for Symmetric Powers In this section all vector spaces are assumed to have finite dimension over a field of charac-teristic zero. We define a nondegenerate pairing Sn(E∗)×Sn(E) − →K as follows: Consider the multilinear map (E∗)n × En − →K given by (v∗ 1, . . . , v∗ n, u1, . . . , un) 7→ X σ∈Sn v∗ σ(1)(u1) · · · v∗ σ(n)(un). 2.12. DUALITY FOR SYMMETRIC POWERS 79 Note that the expression on the right-hand side is “almost” the determinant det(v∗ j(ui)), except that the sign sgn(σ) is missing (where sgn(σ) is the signature of the permutation σ; that is, the parity of the number of transpositions into which σ can be factored). Such an expression is called a permanent. It can be verified that this expression is symmetric w.r.t. the ui’s and also w.r.t. the v∗ j. For any fixed (v∗ 1, . . . , v∗ n) ∈(E∗)n, we get a symmetric multilinear map lv∗ 1,...,v∗ n : (u1, . . . , un) 7→ X σ∈Sn v∗ σ(1)(u1) · · · v∗ σ(n)(un) from En to K. The map lv∗ 1,...,v∗ n extends uniquely to a linear map Lv∗ 1,...,v∗ n : Sn(E) →K making the following diagram commute: En lv∗ 1,...,v∗ n # ι⊙/ Sn(E) Lv∗ 1,...,v∗ n  K. We also have the symmetric multilinear map (v∗ 1, . . . , v∗ n) 7→Lv∗ 1,...,v∗ n from (E∗)n to Hom(Sn(E), K), which extends to a linear map L from Sn(E∗) to Hom(Sn(E), K) making the following diagram commute: (E∗)n ' ι⊙∗ / Sn(E∗) L  Hom(Sn(E), K). However, in view of the isomorphism Hom(U ⊗V, W) ∼ = Hom(U, Hom(V, W)), with U = Sn(E∗), V = Sn(E) and W = K, we can view L as a linear map L: Sn(E∗) ⊗Sn(E) − →K, which by Proposition 2.8 corresponds to a bilinear map ⟨−, −⟩: Sn(E∗) × Sn(E) − →K. (∗) This pairing is given explicitly on generators by ⟨v∗ 1 ⊙· · · ⊙v∗ n, u1 ⊙· · · ⊙un⟩= X σ∈Sn v∗ σ(1)(u1) · · · v∗ σ(n)(un). 80 CHAPTER 2. TENSOR ALGEBRAS Now this pairing in nondegenerate. This can be shown using bases.2 If (e1, . . . , em) is a basis of E, then for every basis element (e∗ i1)⊙n1 ⊙· · ·⊙(e∗ ik)⊙nk of Sn(E∗), with n1+· · ·+nk = n, we have ⟨(e∗ i1)⊙n1 ⊙· · · ⊙(e∗ ik)⊙nk, e⊙n1 i1 ⊙· · · ⊙e⊙nk ik ⟩= n1! · · · nk!, and ⟨(e∗ i1)⊙n1 ⊙· · · ⊙(e∗ ik)⊙nk, ej1 ⊙· · · ⊙ejn⟩= 0 if (j1 . . . , jn) ̸= (i1, . . . , i1 | {z } n1 , . . . , ik, . . . , ik | {z } nk ). If the field K has characteristic zero, then n1! · · · nk! ̸= 0. We leave the details as an exercise to the reader. Therefore we get a canonical isomorphism (Sn(E))∗∼ = Sn(E∗). The following proposition summarizes the duality properties of symmetric powers. Proposition 2.38. Assume the field K has characteristic zero. We have the canonical isomorphisms (Sn(E))∗∼ = Sn(E∗) and Sn(E∗) ∼ = Symn(E; K) = Homsymlin(En, K), which allows us to interpret symmetric tensors over E∗as symmetric multilinear maps. Proof. The isomorphism µ: Sn(E∗) ∼ = Symn(E; K) follows from the isomorphisms (Sn(E))∗∼ = Sn(E∗) and (Sn(E))∗∼ = Symn(E; K) given by Proposition 2.34. Remarks: 1. The isomorphism µ: Sn(E∗) ∼ = Symn(E; K) discussed above can be described explicitly as the linear extension of the map given by µ(v∗ 1 ⊙· · · ⊙v∗ n)(u1 ⊙· · · ⊙un) = X σ∈Sn v∗ σ(1)(u1) · · · v∗ σ(n)(un). If (e1, . . . , em) is a basis of E, then for every basis element (e∗ i1)⊙n1 ⊙· · · ⊙(e∗ ik)⊙nk of Sn(E∗), with n1 + · · · + nk = n, we have µ((e∗ i1)⊙n1 ⊙· · · ⊙(e∗ ik)⊙nk)(ei1, . . . , ei1 | {z } n1 . . . , eik, . . . , eik | {z } nk ) = n1! · · · nk!, 2This is where the assumption that we are in finite dimension and that the field has characteristic zero are used. 2.12. DUALITY FOR SYMMETRIC POWERS 81 If the field K has positive characteristic, then it is possible that n1! · · · nk! = 0, and this is why we required K to be of characteristic 0 in order for Proposition 2.38 to hold. 2. The canonical isomorphism of Proposition 2.38 holds under more general conditions. Namely, that K is a commutative algebra with identity over Q, and that the E is a finitely-generated projective K-module (see Definition 2.28). See Bourbaki, (Chapter III, §11, Section 5, Proposition 8). The map from En to Sn(E) given by (u1, . . . , un) 7→u1 ⊙· · · ⊙un yields a surjection π: E⊗n →Sn(E). Because we are dealing with vector spaces, this map has some section; that is, there is some injection β : Sn(E) →E⊗n with π ◦β = id. Since our field K has char-acteristic 0, there is a special injection having a natural definition involving a symmetrization process defined as follows: For every permutation σ, we have the map rσ : En →E⊗n given by rσ(u1, . . . , un) = uσ(1) ⊗· · · ⊗uσ(n). As rσ is clearly multilinear, rσ extends to a linear map (rσ)⊗: E⊗n →E⊗n making the following diagram commute En rσ " ι⊗ / E⊗n (rσ)⊗  E⊗n, and we get a map Sn × E⊗n − →E⊗n, namely σ · z = (rσ)⊗(z). It is immediately checked that this is a left action of the symmetric group Sn on E⊗n, and the tensors z ∈E⊗n such that σ · z = z, for all σ ∈Sn are called symmetrized tensors. We define the map η: En →E⊗n by η(u1, . . . , un) = 1 n! X σ∈Sn σ · (u1 ⊗· · · ⊗un) = 1 n! X σ∈Sn uσ(1) ⊗· · · ⊗uσ(n). As the right hand side is clearly symmetric, we get a linear map η⊙: Sn(E) →E⊗n making the following diagram commute. En η # ι⊙/ Sn(E) η⊙  E⊗n 82 CHAPTER 2. TENSOR ALGEBRAS Clearly, η⊙(Sn(E)) is the set of symmetrized tensors in E⊗n. If we consider the map S = η⊙◦π: E⊗n − →E⊗n where π is the surjection π: E⊗n →Sn(E), it is easy to check that S ◦S = S. Therefore, S is a projection, and by linear algebra, we know that E⊗n = S(E⊗n) ⊕Ker S = η⊙(Sn(E)) ⊕Ker S. It turns out that Ker S = E⊗n ∩I = Ker π, where I is the two-sided ideal of T(E) generated by all tensors of the form u ⊗v −v ⊗u ∈E⊗2 (for example, see Knapp , Appendix A). Therefore, η⊙is injective, E⊗n = η⊙(Sn(E)) ⊕(E⊗n ∩I) = η⊙(Sn(E)) ⊕Ker π, and the symmetric tensor power Sn(E) is naturally embedded into E⊗n. 2.13 Symmetric Algebras As in the case of tensors, we can pack together all the symmetric powers Sn(V ) into an algebra. Definition 2.25. Given a vector space V , the space S(V ) = M m≥0 Sm(V ), is called the symmetric tensor algebra of V . We could adapt what we did in Section 2.6 for general tensor powers to symmetric tensors but since we already have the algebra T(V ), we can proceed faster. If I is the two-sided ideal generated by all tensors of the form u ⊗v −v ⊗u ∈V ⊗2, we set S•(V ) = T(V )/I. Observe that since the ideal I is generated by elements in V ⊗2, every tensor in I is a linear combination of tensors of the form ω1 ⊗(u ⊗v −v ⊗u) ⊗ω2, with ω1 ∈V ⊗n1 and ω2 ∈V ⊗n2 for some n1, n2 ∈N, which implies that I = M m≥0 (I ∩V ⊗m). Then, S•(V ) automatically inherits a multiplication operation which is commutative, and since T(V ) is graded, that is T(V ) = M m≥0 V ⊗m, we have S•(V ) = M m≥0 V ⊗m/(I ∩V ⊗m). 2.13. SYMMETRIC ALGEBRAS 83 However, it is easy to check that Sm(V ) ∼ = V ⊗m/(I ∩V ⊗m), so S•(V ) ∼ = S(V ). When V is of finite dimension n, S(V ) corresponds to the algebra of polynomials with coefficients in K in n variables (this can be seen from Proposition 2.37). When V is of infinite dimension and (ui)i∈I is a basis of V , the algebra S(V ) corresponds to the algebra of polynomials in infinitely many variables in I. What’s nice about the symmetric tensor algebra S(V ) is that it provides an intrinsic definition of a polynomial algebra in any set of I variables. It is also easy to see that S(V ) satisfies the following universal mapping property. Proposition 2.39. Given any commutative K-algebra A, for any linear map f : V →A, there is a unique K-algebra homomorphism f : S(V ) →A so that f = f ◦i, as in the diagram below. V i / f " S(V ) f  A Remark: If E is finite dimensional, recall the isomorphism µ: Sn(E∗) − →Symn(E; K) defined as the linear extension of the map given by µ(v∗ 1 ⊙· · · ⊙v∗ n)(u1, . . . , un) = X σ∈Sn v∗ σ(1)(u1) · · · v∗ σ(n)(un). Now we have also a multiplication operation Sm(E∗) × Sn(E∗) − →Sm+n(E∗). The following question then arises: Can we define a multiplication Symm(E; K) × Symn(E; K) − →Symm+n(E; K) directly on symmetric multilinear forms, so that the following diagram commutes? Sm(E∗) × Sn(E∗) µm×µn  ⊙ / Sm+n(E∗) µm+n  Symm(E; K) × Symn(E; K) · / Symm+n(E; K) 84 CHAPTER 2. TENSOR ALGEBRAS The answer is yes! The solution is to define this multiplication such that for f ∈Symm(E; K) and g ∈Symn(E; K), (f · g)(u1, . . . , um+n) = X σ∈shuffle(m,n) f(uσ(1), . . . , uσ(m))g(uσ(m+1), . . . , uσ(m+n)), (∗) where shuffle(m, n) consists of all (m, n)-“shuffles;” that is, permutations σ of {1, . . . m + n} such that σ(1) < · · · < σ(m) and σ(m + 1) < · · · < σ(m + n). Observe that a (m, n)-shuffle is completely determined by the sequence σ(1) < · · · < σ(m). For example, suppose m = 2 and n = 1. Given v∗ 1, v∗ 2, v∗ 3 ∈E∗, the multiplication structure on S(E∗) implies that (v∗ 1 ⊙v∗ 2) · v∗ 3 = v∗ 1 ⊙v∗ 2 ⊙v∗ 3 ∈S3(E∗). Furthermore, for u1, u2, u3, ∈E, µ3(v∗ 1 ⊙v∗ 2 ⊙v∗ 3)(u1, u2, u3) = X σ∈S3 v∗ σ(1)(u1)v∗ σ(2)(u2)v∗ σ(3)(u3) = v∗ 1(u1)v∗ 2(u2)v∗ 3(u3) + v∗ 1(u1)v∗ 3(u2)v∗ 2(u3) + v∗ 2(u1)v∗ 1(u2)v∗ 3(u3) + v∗ 2(u1)v∗ 3(u2)v∗ 1(u3) + v∗ 3(u1)v∗ 1(u2)v∗ 2(u3) + v∗ 3(u1)v∗ 2(u2)v∗ 1(u3). Now the (2, 1)- shuffles of {1, 2, 3} are the following three permutations, namely 1 2 3 1 2 3  , 1 2 3 1 3 2  , 1 2 3 2 3 1  . If f ∼ = µ2(v∗ 1 ⊙v∗ 2) and g ∼ = µ1(v∗ 3), then (∗) implies that (f · g)(u1, u2, u3) = X σ∈shuffle(2,1) f(uσ(1), uσ(2))g(uσ(3)) = f(u1, u2)g(u3) + f(u1, u3)g(u2) + f(u2, u3)g(u1) = µ2(v∗ 1 ⊙v∗ 2)(u1, u2)µ1(v∗ 3)(u3) + µ2(v∗ 1 ⊙v∗ 2)(u1, u3)µ1(v∗ 3)(u2) + µ2(v∗ 1 ⊙v∗ 2)(u2, u3)µ1(v∗ 3)(u1) = (v∗ 1(u1)v∗ 2(u2) + v∗ 2(u1)v∗ 1(u2))v∗ 3(u3) + (v∗ 1(u1)v∗ 2(u3) + v∗ 2(u1)v∗ 1(u3))v∗ 3(u2) + (v∗ 1(u2)v∗ 2(u3) + v∗ 2(u2)v∗ 1(u3))v∗ 3(u1) = µ3(v∗ 1 ⊙v∗ 2 ⊙v∗ 3)(u1, u2, u3). We leave it as an exercise for the reader to verify Equation (∗) for arbitrary nonnegative integers m and n. Another useful canonical isomorphism (of K-algebras) is given below. Proposition 2.40. For any two vector spaces E and F, there is a canonical isomorphism (of K-algebras) S(E ⊕F) ∼ = S(E) ⊗S(F). 2.14. TENSOR PRODUCTS OF MODULES OVER A COMMMUTATIVE RING 85 2.14 Tensor Products of Modules over a Commmutative Ring This section provides some background on modules which is needed for Section 10.8 about metrics on vector bundles and for Chapter 11 on connections and curvature on vector bundles. What happens is that given a manifold M, the space X(M) of vector fields on M and the space Ap(M) of differential p-forms on M are vector spaces, but vector fields and p-forms can also be multiplied by smooth functions in C∞(M). This operation is a left action of C∞(M) which satisfies all the axioms of the scalar multiplication in a vector space, but since C∞(M) is not a field, the resulting structure is not a vector space. Instead it is a module, a more general notion. Definition 2.26. If R is a commutative ring with identity (say 1), a module over R (or R-module) is an abelian group M with a scalar multiplication ·: R × M →M such that all the axioms of a vector space are satisfied. At first glance, a module does not seem any different from a vector space, but the lack of multiplicative inverses in R has drastic consequences, one being that unlike vector spaces, modules are generally not free; that is, have no bases. Furthermore, a module may have torsion elements, that is, elements m ∈M such that λ · m = 0, even though m ̸= 0 and λ ̸= 0. For example, for any nonzero integer n ∈Z, the Z-module Z/nZ has no basis and n · m = 0 for all m ∈Z/nZ. Similarly, Q as a Z-module has no basis. In fact, any two distinct nonzero elements p1/q1 and p2/q2 are linearly dependent, since (p2q1) p1 q1  −(p1q2) p2 q2  = 0. Nevertheless, it is possible to define tensor products of modules over a ring, just as in Section 2.2, and the results of that section continue to hold. The results of Section 2.4 also continue to hold since they are based on the universal mapping property. However, the results of Section 2.3 on bases generally fail, except for free modules. Similarly, the results of Section 2.5 on duality generally fail. Tensor algebras can be defined for modules, as in Section 2.6. Symmetric tensor and alternating tensors can be defined for modules, but again, results involving bases generally fail. Tensor products of modules have some unexpected properties. For example, if p and q are relatively prime integers, then Z/pZ ⊗Z Z/qZ = (0). This is because, by Bezout’s identity, there are a, b ∈Z such that ap + bq = 1, 86 CHAPTER 2. TENSOR ALGEBRAS so, for all x ∈Z/pZ and all y ∈Z/qZ, we have x ⊗y = ap(x ⊗y) + bq(x ⊗y) = a(px ⊗y) + b(x ⊗qy) = a(0 ⊗y) + b(x ⊗0) = 0. It is possible to salvage certain properties of tensor products holding for vector spaces by restricting the class of modules under consideration. For example, projective modules have a pretty good behavior w.r.t. tensor products. Definition 2.27. A free R-module F is a module that has a basis (i.e., there is a family (ei)i∈I of linearly independent vectors in F that span F). Projective modules generalize free modules. They have many equivalent characteriza-tions. Here is one that is best suited for our needs. Definition 2.28. An R-module P is projective if it is a summand of a free module; that is, if there is a free R-module F, and some R-module Q, so that F = P ⊕Q. For example, we show in Section 10.8 that the space Γ(ξ) of global sections of a vector bundle ξ over a base manifold B is a finitely generated C∞(B)-projective module. Given any R-module M, we let M ∗= HomR(M, R) be its dual. We have the following proposition. Proposition 2.41. For any finitely-generated projective R-module P and any R-module Q, we have the isomorphisms: P ∗∗ ∼ = P HomR(P, Q) ∼ = P ∗⊗R Q. Proof sketch. We only consider the second isomorphism. Since P is projective, we have some R-modules P1, F with P ⊕P1 = F, where F is some free module. We know that for any R-modules U, V, W, we have HomR(U ⊕V, W) ∼ = HomR(U, W) Y HomR(V, W) ∼ = HomR(U, W) ⊕HomR(V, W), so P ∗⊕P ∗ 1 ∼ = F ∗, HomR(P, Q) ⊕HomR(P1, Q) ∼ = HomR(F, Q). 2.14. TENSOR PRODUCTS OF MODULES OVER A COMMMUTATIVE RING 87 By tensoring with Q and using the fact that tensor distributes w.r.t. coproducts, we get (P ∗⊗R Q) ⊕(P ∗ 1 ⊗Q) ∼ = (P ∗⊕P ∗ 1 ) ⊗R Q ∼ = F ∗⊗R Q. Now, the proof of Proposition 2.19 goes through because F is free and finitely generated. This implies F ∗⊗Q ∼ = Hom(F, Q), so α⊗: (P ∗⊗R Q) ⊕(P ∗ 1 ⊗Q) ∼ = F ∗⊗R Q − →HomR(F, Q) ∼ = HomR(P, Q) ⊕HomR(P1, Q) is an isomorphism, and as α⊗maps P ∗⊗R Q to HomR(P, Q), it yields an isomorphism between these two spaces. The isomorphism α⊗: P ∗⊗R Q ∼ = HomR(P, Q) of Proposition 2.41 is still given by α⊗(u∗⊗f)(x) = u∗(x)f, u∗∈P ∗, f ∈Q, x ∈P. It is convenient to introduce the evaluation map Evx : P ∗⊗R Q →Q defined for every x ∈P by Evx(u∗⊗f) = u∗(x)f, u∗∈P ∗, f ∈Q. In Section 11.5 we will need to consider a slightly weaker version of the universal mapping property of tensor products. The situation is this: We have a commutative R-algebra S, where R is a field (or even a commutative ring), we have two R-modules U and V , and moreover, U is a right S-module and V is a left S-module. In Section 11.5, this corresponds to R = R, S = C∞(B), U = Ai(B) and V = Γ(ξ), where ξ is a vector bundle. Then we can form the tensor product U ⊗R V , and we let U ⊗S V be the quotient module (U ⊗R V )/W, where W is the submodule of U ⊗R V generated by the elements of the form us ⊗R v −u ⊗R sv. As S is commutative, we can make U ⊗S V into an S-module by defining the action of S via s(u ⊗S v) = us ⊗S v. It is verified that this S-module is isomorphic to the tensor product of U and V as S-modules, and the following universal mapping property holds: Proposition 2.42. For every R-bilinear map f : U × V →Z, if f satisfies the property f(us, v) = f(u, sv), for all u ∈U, v ∈V, s ∈S, then f induces a unique R-linear map b f : U ⊗S V →Z such that f(u, v) = b f(u ⊗S v), for all u ∈U, v ∈V. Note that the linear map b f : U ⊗S V →Z is only R-linear; it is not S-linear in general. 88 CHAPTER 2. TENSOR ALGEBRAS 2.15 Problems Problem 2.1. Prove Proposition 2.4. Problem 2.2. Given two linear maps f : E →E′ and g: F →F ′, we defined the unique linear map f ⊗g: E ⊗F →E′ ⊗F ′ by (f ⊗g)(u ⊗v) = f(u) ⊗g(v), for all u ∈E and all v ∈F. See Proposition 2.9. Thus f ⊗g ∈Hom(E ⊗F, E′ ⊗F ′). If we denote the tensor product E ⊗F by T(E, F), and we assume that E, E′ and F, F ′ are finite dimensional, pick bases and show that the map induced by f ⊗g 7→T(f, g) is an isomorphism Hom(E, F) ⊗Hom(E′, F ′) ∼ = Hom(E ⊗F, E′ ⊗F ′). Problem 2.3. Adjust the proof of Proposition 2.13 (2) to show that E ⊗(F ⊗G) ∼ = E ⊗F ⊗G, whenever E, F, and G are arbitrary vector spaces. Problem 2.4. Given a fixed vector space G, for any two vector spaces M and N and every linear map f : M →N, we defined τG(f) = f ⊗idG to be the unique linear map making the following diagram commute. M × G f×idG  ιM⊗/ M ⊗G f⊗idG  N × G ιN⊗/ N ⊗G See the proof of Proposition 2.13 (3). Show that (1) τG(0) = 0, (2) τG(idM) = (idM ⊗idG) = idM⊗G, (3) If f ′ : M →N is another linear map, then τG(f + f ′) = τG(f) + τG(f ′). Problem 2.5. Induct on m ≥2 to prove the canonical isomorphism V ⊗m ⊗V ⊗n ∼ = V ⊗(m+n). Use this isomorphism to show that ·: V ⊗m × V ⊗n − →V ⊗(m+n) defined as (v1 ⊗· · · ⊗vm) · (w1 ⊗· · · ⊗wn) = v1 ⊗· · · ⊗vm ⊗w1 ⊗· · · ⊗wn. induces a multiplication on T(V ). Hint. See Jacobson , Section 3.9, or Bertin , Chapter 4, Section 2.). 2.15. PROBLEMS 89 Problem 2.6. Prove Proposition 2.21. Hint. See Knapp (Appendix A, Proposition A.14) or Bertin (Chapter 4, Theorem 2.4). Problem 2.7. Given linear maps f : E →E′ and f ′ : E′ →E′′, show that (f ′ ◦f) ⊙(f ′ ◦f) = (f ′ ⊙f ′) ◦(f ⊙f). Problem 2.8. Complete the proof of Proposition 2.37 for the case of an infinite dimensional vector space E. Problem 2.9. Let I be a finite index set of cardinality p. Let m be a nonnegative integer. Show that the number of multisets over I with cardinality m is p+m−1 m  . Problem 2.10. Prove Proposition 2.36. Problem 2.11. Using bases, show that the bilinear map at (∗) in Section 2.12 produces a nondegenerate pairing. Problem 2.12. Let I be the two-sided ideal generated by all tensors of the form u⊗v−v⊗u ∈ V ⊗2. Prove that Sm(V ) ∼ = V ⊗m/(I ∩V ⊗m). Problem 2.13. Verify Equation (∗) of Section 2.13 for arbitrary nonnegative integers m and n. Problem 2.14. Let P be a finitely generated projective R-module. Recall that P ∗= HomR(P, R). Show that P ∗∗∼ = P. Problem 2.15. Let S be a commutative R-algebra, where R is a commutative ring. Suppose we have R-modules U and V , where U is a right S-module and V is a left S-module. We form the tensor product U ⊗R V , and we let U ⊗S V be the quotient module (U ⊗R V )/W, where W is the submodule of U ⊗R V generated by the elements of the form us ⊗R v −u ⊗R sv. As S is commutative, we can make U ⊗S V into an S-module by defining the action of S via s(u ⊗S v) = us ⊗S v. Verify this S-module is isomorphic to the tensor product of U and V as S-modules. Problem 2.16. Prove Proposition 2.42. 90 CHAPTER 2. TENSOR ALGEBRAS Chapter 3 Exterior Tensor Powers and Exterior Algebras In this chapter we consider alternating (also called skew-symmetric) multilinear maps and exterior tensor powers (also called alternating tensor powers), denoted Vn(E). In many respects alternating multilinear maps and exterior tensor powers can be treated much like symmetric tensor powers, except that sgn(σ) needs to be inserted in front of the formulae valid for symmetric powers. Roughly speaking, we are now in the world of determinants rather than in the world of permanents. However, there are also some fundamental differences, one of which being that the exterior tensor power Vn(E) is the trivial vector space (0) when E is finite dimensional and n > dim(E). This chapter provides the firm foundations for understanding differential forms. In Section 3.1 we define the exterior powers of a vector space E. This time, instead of dealing with symmetric multilinear maps, we deal with alternating multilinear maps, which are multilinear maps f : En →F such that f(u1, . . . , un) = 0 whenever two adjacent arguments are identical. This implies that f(u1, . . . , un) = 0 whenever any two arguments are identical, and that f(. . . , ui, ui+1, . . .) = −f(. . . , ui+1, ui, . . .). Given a vector space E over a field K, for any n ≥1, the exterior tensor power Vn(E) is defined by a universal mapping property: it is a vector space with an injection i∧: En → Vn(E), such that for every vector space F and every alternating multilinear map f : En →F, there is a unique linear map f∧: Vn(E) →F such that f = f∧◦i∧, as illustrated in the following diagram: En f $ i∧ / Vn(E) f∧  F 91 92 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS We prove that the above universal mapping property defines Vn(E) up to isomorphism, and then we prove its existence by constructing the quotient n ^ (E) = E⊗n/(Ia ∩E⊗n) where Ia is the two-sided ideal of the tensor algebra T(E) generated by all tensors of the form u ⊗u ∈E⊗2. As a corollary, there is an isomorphism Hom( n ^ (E), F) ∼ = Altn(E; F) between the vector space of linear maps Hom(Vn(E), F) and the vector space of alternating multilinear maps Altn(E; F). A new phenomenon that arises with exterior tensor powers is that if E has dimension n, then Vk(E) = (0) for all k > n. Given any two linear maps f, g: E →E′, there is a linear map f ∧g: V2(E) →V2(E′). A basic isomorphism involving the exterior power of a direct sum is shown at the end of this Section. In Section 3.2 we show how to construct a basis of Vk(E) from a basis of E (1 ≤k ≤n). If E has dimension n and if (e1, . . . , en) is a basis of E, for any finite sequence I = (i1, . . . , ik) with 1 ≤i1 < i2 < · · · < ik ≤n, if we write eI = ei1 ∧· · · ∧eik, then the family of all the eI is a basis of Vk(E). Thus Vk(E) has dimension n k  . Section 3.3 is devoted to duality in exterior powers. There is a nondegenerate pairing ⟨−, −⟩: n ^ (E∗) × n ^ (E) − →K defined in terms of generators by ⟨v∗ 1 ∧· · · ∧v∗ n, u1 ∧· · · ∧un⟩= det(v∗ j(ui)). As a consequence, if E is finite dimensional, we have canonical isomorphisms ( n ^ (E))∗∼ = n ^ (E∗) ∼ = Altn(E; K). The exterior tensor power Vk(E) is naturally embedded in E⊗n (if K has characteristic 0). In Section 3.4 we define exterior algebras (or Grassmann algebras). As in the case of symmetric tensors, we can pack together all the exterior powers Vn(V ) into an algebra. Given any vector space V , the vector space ^ (V ) = M m≥0 m ^ (V ) 93 is an algebra called the exterior algebra (or Grassmann algebra) of V . The exterior algebra satisfies a universal mapping condition. If we define Alt(E) = M n≥0 Altn(E; K), then this is an algebra under a combinatorial definition of the wedge operation, and this algebra is isomorphic to V(E∗). In Section 3.5, we introduce the Hodge ∗-operator. Given a vector space V of dimension n with an inner product ⟨−, −⟩, for some chosen orientation of V , for each k such that 1 ≤k ≤n, there is an isomorphism ∗from Vk(V ) to Vn−k(V ). The Hodge ∗operator can be extended to an isomorphism of V(V ). It is the main tool used to define a generalization of the Laplacian (the Hodge Laplacian) to a smooth manifold. The next three sections are somewhat more technical. They deal with some contraction operators called left hooks and right hooks. The motivation comes from the problem of understanding when a tensor α ∈Vk(E) is decomposable. An arbitrary tensor α ∈Vk(E) is a linear combination of tensors of the form u1 ∧· · · ∧uk, called decomposable. The issue is to find criteria for decomposability. This is not as obvious as it looks. For example, we have e1 ∧e2 + e1 ∧e3 + e2 ∧e3 = (e1 + e2) ∧(e2 + e3), where the tensor on the right is clearly decomposable, but the tensor on the left does not look decomposable at first glance. Criteria for testing decomposability using left hooks are given in Section 3.7. Say dim(E) = n. Using our nonsingular pairing ⟨−, −⟩: p ^ E∗× p ^ E − →K (1 ≤p ≤n) defined on generators by ⟨u∗ 1 ∧· · · ∧u∗ p, v1 ∧· · · ∧up⟩= det(u∗ i (vj)), in Section 3.6 we define various contraction operations (partial evaluation operators) ⌟: p ^ E × p+q ^ E∗− → q ^ E∗, ⌟: p ^ E∗× p+q ^ E − → q ^ E left hook and ⌞: p+q ^ E∗× p ^ E − → q ^ E∗, ⌞: p+q ^ E × p ^ E∗− → q ^ E right hook. These left and right hooks also have combinatorial definitions in terms of the basis vectors eI and e∗ J. The right hooks can be expressed in terms of the left hooks. Left and right hooks induce isomorphisms γ : Vp E →Vn−p E∗and δ: Vp E∗→Vn−p E. 94 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS A criterion for testing decomposabiliity in terms of left hooks is presented in Section 3.7. In Section 3.8, based on the criterion established in Section 3.7 , we derive a criterion for testing decomposabilty in terms of equations known as the Grassmann-Pl¨ ucker’s equations. We also show that the Grassmannian manifold G(k, n) can be embedded as an algebraic variety into RP(n k)−1 defined by equations of degree 2. Section 3.9 discusses vector-valued alternating forms. The purpose of this section is to present the technical background needed for Sections 4.5 and 4.6 on vector-valued differential forms, in particular in the case of Lie groups where differential forms taking their values in a Lie algebra arise naturally. Given a finite dimensional vector space E and any vector space F, there is an isomorphism µF : n ^ (E∗) ! ⊗F − →Altn(E; F) defined on generators by µF((v∗ 1 ∧· · · ∧v∗ n) ⊗f)(u1, . . . , un) = (det(v∗ j(ui))f, with v∗ 1, . . . , v∗ n ∈E∗, u1, . . . , un ∈E, and f ∈F. We also discuss a generalization of the wedge product. 3.1 Exterior Tensor Powers As in the case of symmetric tensor powers, since we already have the tensor algebra T(V ), we can proceed rather quickly. But first let us review some basic definitions and facts. Definition 3.1. Let f : En →F be a multilinear map. We say that f alternating ifffor all ui ∈E, f(u1, . . . , un) = 0 whenever ui = ui+1, for some i with 1 ≤i ≤n −1; that is, f(u1, . . . , un) = 0 whenever two adjacent arguments are identical. We say that f is skew-symmetric (or anti-symmetric) iff f(uσ(1), . . . , uσ(n)) = sgn(σ)f(u1, . . . , un), for every permutation σ ∈Sn, and all ui ∈E. For n = 1, we agree that every linear map f : E →F is alternating. The vector space of all multilinear alternating maps f : En →F is denoted Altn(E; F). Note that Alt1(E; F) = Hom(E, F). The following basic proposition shows the relationship between alternation and skew-symmetry. Proposition 3.1. Let f : En →F be a multilinear map. If f is alternating, then the following properties hold: 3.1. EXTERIOR TENSOR POWERS 95 (1) For all i, with 1 ≤i ≤n −1, f(. . . , ui, ui+1, . . .) = −f(. . . , ui+1, ui, . . .). (2) For every permutation σ ∈Sn, f(uσ(1), . . . , uσ(n)) = sgn(σ)f(u1, . . . , un). (3) For all i, j, with 1 ≤i < j ≤n, f(. . . , ui, . . . uj, . . .) = 0 whenever ui = uj. Moreover, if our field K has characteristic different from 2, then every skew-symmetric multilinear map is alternating. Proof. (1) By multilinearity applied twice, we have f(. . . , ui + ui+1, ui + ui+1, . . .) = f(. . . , ui, ui, . . .) + f(. . . , ui, ui+1, . . .) + f(. . . , ui+1, ui, . . .) + f(. . . , ui+1, ui+1, . . .). Since f is alternating, we get 0 = f(. . . , ui, ui+1, . . .) + f(. . . , ui+1, ui, . . .); that is, f(. . . , ui, ui+1, . . .) = −f(. . . , ui+1, ui, . . .). (2) Clearly, the symmetric group, Sn, acts on Altn(E; F) on the left, via σ · f(u1, . . . , un) = f(uσ(1), . . . , uσ(n)). Consequently, as Sn is generated by the transpositions (permutations that swap exactly two elements), since for a transposition, (2) is simply (1), we deduce (2) by induction on the number of transpositions in σ. (3) There is a permutation σ that sends 1 and 2 respectively to i and j. By hypothesis ui = uj, so we have uσ(1) = ui = uj = uσ(2), and as f is alternating we have f(uσ(1), . . . , uσ(n)) = 0. However, by (2), f(u1, . . . , un) = sgn(σ)f(uσ(1), . . . , uσ(n)) = 0. Now when f is skew-symmetric, if σ is the transposition swapping ui and ui+1 = ui, as sgn(σ) = −1, we get f(. . . , ui, ui, . . .) = −f(. . . , ui, ui, . . .), so that 2f(. . . , ui, ui, . . .) = 0, and in every characteristic except 2, we conclude that f(. . . , ui, ui, . . .) = 0, namely f is alternating. 96 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS Proposition 3.1 shows that in every field of characteristic different from 2, alternating and skew-symmetric multilinear maps are identical. Using Proposition 3.1 we easily deduce the following crucial fact. Proposition 3.2. Let f : En →F be an alternating multilinear map. For any families of vectors, (u1, . . . , un) and (v1, . . . , vn), with ui, vi ∈E, if vj = n X i=1 aijui, 1 ≤j ≤n, then f(v1, . . . , vn) = X σ∈Sn sgn(σ) aσ(1),1 · · · aσ(n),n ! f(u1, . . . , un) = det(A)f(u1, . . . , un), where A is the n × n matrix, A = (aij). Proof. Use Property (ii) of Proposition 3.1. We are now ready to define and construct exterior tensor powers. Definition 3.2. An n-th exterior tensor power of a vector space E, where n ≥1, is a vector space A together with an alternating multilinear map ϕ: En →A, such that for every vector space F and for every alternating multilinear map f : En →F, there is a unique linear map f∧: A →F with f(u1, . . . , un) = f∧(ϕ(u1, . . . , un)), for all u1, . . . , un ∈E, or for short f = f∧◦ϕ. Equivalently, there is a unique linear map f∧such that the following diagram commutes: En f ! ϕ / A f∧  F. The above property is called the universal mapping property of the exterior tensor power (A, ϕ). We now show that any two n-th exterior tensor powers (A1, ϕ1) and (A2, ϕ2) for E are isomorphic. Proposition 3.3. Given any two n-th exterior tensor powers (A1, ϕ1) and (A2, ϕ2) for E, there is an isomorphism h: A1 →A2 such that ϕ2 = h ◦ϕ1. 3.1. EXTERIOR TENSOR POWERS 97 Proof. Replace tensor product by n-th exterior tensor power in the proof of Proposition 2.5. We next give a construction that produces an n-th exterior tensor power of a vector space E. Theorem 3.4. Given a vector space E, an n-th exterior tensor power (Vn(E), ϕ) for E can be constructed (n ≥1). Furthermore, denoting ϕ(u1, . . . , un) as u1 ∧· · · ∧un, the exterior tensor power Vn(E) is generated by the vectors u1 ∧· · · ∧un, where u1, . . . , un ∈E, and for every alternating multilinear map f : En →F, the unique linear map f∧: Vn(E) →F such that f = f∧◦ϕ is defined by f∧(u1 ∧· · · ∧un) = f(u1, . . . , un) on the generators u1 ∧· · · ∧un of Vn(E). Proof sketch. We can give a quick proof using the tensor algebra T(E). Let Ia be the two-sided ideal of T(E) generated by all tensors of the form u ⊗u ∈E⊗2. Then let n ^ (E) = E⊗n/(Ia ∩E⊗n) and let π be the projection π: E⊗n →Vn(E). If we let u1 ∧· · · ∧un = π(u1 ⊗· · · ⊗un), it is easy to check that (Vn(E), ∧) satisfies the conditions of Theorem 3.4. Remark: We can also define ^ (E) = T(E)/Ia = M n≥0 n ^ (E), the exterior algebra of E. This is the skew-symmetric counterpart of S(E), and we will study it a little later. For simplicity of notation, we may write Vn E for Vn(E). We also abbreviate “exterior tensor power” as “exterior power.” Clearly, V1(E) ∼ = E, and it is convenient to set V0(E) = K. The fact that the map ϕ: En →Vn(E) is alternating and multilinear can also be ex-pressed as follows: u1 ∧· · · ∧(ui + vi) ∧· · · ∧un = (u1 ∧· · · ∧ui ∧· · · ∧un) + (u1 ∧· · · ∧vi ∧· · · ∧un), u1 ∧· · · ∧(λui) ∧· · · ∧un = λ(u1 ∧· · · ∧ui ∧· · · ∧un), uσ(1) ∧· · · ∧uσ(n) = sgn(σ) u1 ∧· · · ∧un, 98 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS for all σ ∈Sn. The map ϕ from En to Vn(E) is often denoted ι∧, so that ι∧(u1, . . . , un) = u1 ∧· · · ∧un. Theorem 3.4 implies the following result. Proposition 3.5. There is a canonical isomorphism Hom( n ^ (E), F) ∼ = Altn(E; F) between the vector space of linear maps Hom(Vn(E), F) and the vector space of alternating multilinear maps Altn(E; F), given by the linear map −◦ϕ defined by 7→h ◦ϕ, with h ∈ Hom(Vn(E), F). In particular, when F = K, we get a canonical isomorphism n ^ (E) !∗ ∼ = Altn(E; K). Definition 3.3. Tensors α ∈Vn(E) are called alternating n-tensors or alternating tensors of degree n and we write deg(α) = n. Tensors of the form u1 ∧· · ·∧un, where ui ∈E, are called simple (or decomposable) alternating n-tensors. Those alternating n-tensors that are not simple are often called compound alternating n-tensors. Simple tensors u1∧· · ·∧un ∈Vn(E) are also called n-vectors and tensors in Vn(E∗) are often called (alternating) n-forms. Given a linear map f : E →E′, since the map ι′ ∧◦(f ×f) is bilinear and alternating, there is a unique linear map f ∧f : V2(E) →V2(E′) making the following diagram commute: E2 f×f  ι∧ / V2(E) f∧f  (E′)2 ι′ ∧ / V2(E′). The map f ∧f : V2(E) →V2(E′) is determined by (f ∧f)(u ∧v) = f(u) ∧f(v). Proposition 3.6. Given any two linear maps f : E →E′ and f ′ : E′ →E′′, we have (f ′ ◦f) ∧(f ′ ◦f) = (f ′ ∧f ′) ◦(f ∧f). 3.2. BASES OF EXTERIOR POWERS 99 The generalization to the alternating product f ∧· · · ∧f of n ≥3 copies of the linear map f : E →E′ is immediate, and left to the reader. We can show the following property of the exterior tensor product, using the proof technique of Proposition 2.13. Proposition 3.7. We have the following isomorphism: n ^ (E ⊕F) ∼ = n M k=0 k ^ (E) ⊗ n−k ^ (F). 3.2 Bases of Exterior Powers Definition 3.4. Let E be any vector space. For any basis (ui)i∈Σ for E, we assume that some total ordering ≤on the index set Σ has been chosen. Call the pair ((ui)i∈Σ, ≤) an ordered basis. Then for any nonempty finite subset I ⊆Σ, let uI = ui1 ∧· · · ∧uim, where I = {i1, . . . , im}, with i1 < · · · < im. Since Vn(E) is generated by the tensors of the form v1 ∧· · · ∧vn, with vi ∈E, in view of skew-symmetry, it is clear that the tensors uI with |I| = n generate Vn(E) (where ((ui)i∈Σ, ≤) is an ordered basis). Actually they form a basis. To gain an intuitive understanding of this statement, let m = 2 and E be a 3-dimensional vector space lexicographically ordered basis {e1, e2, e3}. We claim that e1 ∧e2, e1 ∧e3, e2 ∧e3 form a basis for V2(E) since they not only generate V2(E) but are linearly independent. The linear independence is argued as follows: given any vector space F, if w12, w13, w23 are any vectors in F, there is an alternating bilinear map h: E2 →F such that h(e1, e2) = w12, h(e1, e3) = w13, h(e2, e3) = w23. Because h yields a unique linear map h∧: V2 E →F such that h∧(ei ∧ej) = wij, 1 ≤i < j ≤3, by Proposition 2.4, the vectors e1 ∧e2, e1 ∧e3, e2 ∧e3 100 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS are linearly independent. This suggests understanding how an alternating bilinear function f : E2 →F is expressed in terms of its values f(ei, ej) on the basis vectors (e1, e2, e3). Using bilinearity and alternation, we obtain f(u1e1 + u2e2 + u3e3, v1e1 + v2e2 + v3e3) = (u1v2 −u2v1)f(e1, e2) + (u1v3 −u3v1)f(e1, e3) + (u2v3 −u3v2)f(e2, e3). Therefore, given w12, w13, w23 ∈F, the function h given by h(u1e1 + u2e2 + u3e3, v1e1 + v2e2 + v3e3) = (u1v2 −u2v1)w12 + (u1v3 −u3v1)w13 + (u2v3 −u3v2)w23 is clearly bilinear and alternating, and by construction h(ei, ej) = wij, with 1 ≤i < j ≤3 does the job. We now prove the assertion that tensors uI with |I| = n generate Vn(E) for arbitrary n. Proposition 3.8. Given any vector space E, if E has finite dimension d = dim(E), then for all n > d, the exterior power Vn(E) is trivial; that is Vn(E) = (0). If n ≤d or if E is infinite dimensional, then for every ordered basis ((ui)i∈Σ, ≤), the family (uI) is basis of Vn(E), where I ranges over finite nonempty subsets of Σ of size |I| = n. Proof. First assume that E has finite dimension d = dim(E) and that n > d. We know that Vn(E) is generated by the tensors of the form v1 ∧· · · ∧vn, with vi ∈E. If u1, . . . , ud is a basis of E, as every vi is a linear combination of the uj, when we expand v1 ∧· · · ∧vn using multilinearity, we get a linear combination of the form v1 ∧· · · ∧vn = X (j1,...,jn) λ(j1,...,jn) uj1 ∧· · · ∧ujn, where each (j1, . . . , jn) is some sequence of integers jk ∈{1, . . . , d}. As n > d, each sequence (j1, . . . , jn) must contain two identical elements. By alternation, uj1 ∧· · · ∧ujn = 0, and so v1 ∧· · · ∧vn = 0. It follows that Vn(E) = (0). Now assume that either dim(E) = d and n ≤d, or that E is infinite dimensional. The argument below shows that the uI are nonzero and linearly independent. As usual, let u∗ i ∈E∗be the linear form given by u∗ i (uj) = δij. For any nonempty subset I = {i1, . . . , in} ⊆Σ with i1 < · · · < in, for any n vectors v1, . . . , vn ∈E, let lI(v1, . . . , vn) = det(u∗ ij(vk)) = u∗ i1(v1) · · · u∗ i1(vn) . . . ... . . . u∗ in(v1) · · · u∗ in(vn) . 3.2. BASES OF EXTERIOR POWERS 101 If we let the n-tuple (v1, . . . , vn) vary we obtain a map lI from En to K, and it is easy to check that this map is alternating multilinear. Thus lI induces a unique linear map LI : Vn(E) →K making the following diagram commute. En lI $ ι∧ / Vn(E) LI  K Observe that for any nonempty finite subset J ⊆Σ with |J| = n, we have LI(uJ) =  1 if I = J 0 if I ̸= J. Note that when dim(E) = d and n ≤d, or when E is infinite-dimensional, the forms u∗ i1, . . . , u∗ in are all distinct, so the above does hold. Since LI(uI) = 1, we conclude that uI ̸= 0. If we have a linear combination X I λIuI = 0, where the above sum is finite and involves nonempty finite subset I ⊆Σ with |I| = n, for every such I, when we apply LI we get λI = 0, proving linear independence. As a corollary, if E is finite dimensional, say dim(E) = d, and if 1 ≤n ≤d, then we have dim( n ^ (E)) = d n  , and if n > d, then dim(Vn(E)) = 0. Remark: When n = 0, if we set u∅= 1, then (u∅) = (1) is a basis of V0(V ) = K. It follows from Proposition 3.8 that the family (uI)I where I ⊆Σ ranges over finite subsets of Σ is a basis of V(V ) = L n≥0 Vn(V ). As a corollary of Proposition 3.8 we obtain the following useful criterion for linear inde-pendence. Proposition 3.9. For any vector space E, the vectors u1, . . . , un ∈E are linearly indepen-dent iffu1 ∧· · · ∧un ̸= 0. Proof. If u1 ∧· · · ∧un ̸= 0, then u1, . . . , un must be linearly independent. Otherwise, some ui would be a linear combination of the other uj’s (with j ̸= i), and then, as in the proof of Proposition 3.8, u1 ∧· · · ∧un would be a linear combination of wedges in which two vectors are identical, and thus zero. 102 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS Conversely, assume that u1, . . . , un are linearly independent. Then we have the linear forms u∗ i ∈E∗such that u∗ i (uj) = δi,j 1 ≤i, j ≤n. As in the proof of Proposition 3.8, we have a linear map Lu1,...,un : Vn(E) →K given by Lu1,...,un(v1 ∧· · · ∧vn) = det(u∗ j(vi)) = u∗ 1(v1) · · · u∗ 1(vn) . . . ... . . . u∗ n(v1) · · · u∗ n(vn) , for all v1 ∧· · · ∧vn ∈Vn(E). As Lu1,...,un(u1 ∧· · · ∧un) = 1, we conclude that u1 ∧· · · ∧un ̸= 0. Proposition 3.9 shows that geometrically every nonzero wedge u1 ∧· · · ∧un corresponds to some oriented version of an n-dimensional subspace of E. 3.3 Duality for Exterior Powers In this section all vector spaces are assumed to have finite dimension. We define a nonde-generate pairing Vn(E∗) × Vn(E) − →K as follows: Consider the multilinear map (E∗)n × En − →K given by (v∗ 1, . . . , v∗ n, u1, . . . , un) 7→ X σ∈Sn sgn(σ) v∗ σ(1)(u1) · · · v∗ σ(n)(un) = det(v∗ j(ui)) = v∗ 1(u1) · · · v∗ 1(un) . . . ... . . . v∗ n(u1) · · · v∗ n(un) . It is easily checked that this expression is alternating w.r.t. the ui’s and also w.r.t. the v∗ j. For any fixed (v∗ 1, . . . , v∗ n) ∈(E∗)n, we get an alternating multilinear map lv∗ 1,...,v∗ n : (u1, . . . , un) 7→det(v∗ j(ui)) from En to K. The map lv∗ 1,...,v∗ n extends uniquely to a linear map Lv∗ 1,...,v∗ n : Vn(E) →K making the following diagram commute: En lv∗ 1,...,v∗ n $ ι∧ / Vn(E) Lv∗ 1,...,v∗ n  K. 3.3. DUALITY FOR EXTERIOR POWERS 103 We also have the alternating multilinear map (v∗ 1, . . . , v∗ n) 7→Lv∗ 1,...,v∗ n from (E∗)n to Hom(Vn(E), K), which extends to a linear map L from Vn(E∗) to Hom(Vn(E), K) making the following diagram commute: (E∗)n ' ι∧∗ / Vn(E∗) L  Hom(Vn(E), K). However, in view of the isomorphism Hom(U ⊗V, W) ∼ = Hom(U, Hom(V, W)), with U = Vn(E∗), V = Vn(E) and W = K, we can view L as a linear map L: n ^ (E∗) ⊗ n ^ (E) − →K, which by Proposition 2.8 corresponds to a bilinear map ⟨−, −⟩: n ^ (E∗) × n ^ (E) − →K. (∗) This pairing is given explicitly in terms of generators by ⟨v∗ 1 ∧· · · ∧v∗ n, u1 ∧· · · ∧un⟩= det(v∗ j(ui)). Now this pairing in nondegenerate. This can be shown using bases. Given any basis (e1, . . . , em) of E, for every basis element e∗ i1∧· · ·∧e∗ in of Vn(E∗) (with 1 ≤i1 < · · · < in ≤m), we have ⟨e∗ i1 ∧· · · ∧e∗ in, ej1 ∧· · · ∧ejn⟩= ( 1 if (j1, . . . , jn) = (i1, . . . , in) 0 otherwise. We leave the details as an exercise to the reader. As a consequence we get the following canonical isomorphisms. Proposition 3.10. There is a canonical isomorphism ( n ^ (E))∗∼ = n ^ (E∗). There is also a canonical isomorphism µ: n ^ (E∗) ∼ = Altn(E; K) which allows us to interpret alternating tensors over E∗as alternating multilinear maps. 104 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS Proof. The second isomorphism follows from the canonical isomorphism (Vn(E))∗∼ = Vn(E∗) and the canonical isomorphism (Vn(E))∗∼ = Altn(E; K) given by Proposition 3.5. Remarks: 1. The isomorphism µ: Vn(E∗) ∼ = Altn(E; K) discussed above can be described explicitly as the linear extension of the map given by µ(v∗ 1 ∧· · · ∧v∗ n)(u1, . . . , un) = det(v∗ j(ui)). 2. The canonical isomorphism of Proposition 3.10 holds under more general conditions. Namely, that K is a commutative ring with identity and that E is a finitely-generated projective K-module (see Definition 2.27). See Bourbaki, (Chapter III, §11, Section 5, Proposition 7). 3. Variants of our isomorphism µ are found in the literature. For example, there is a version µ′, where µ′ = 1 n!µ, with the factor 1 n! added in front of the determinant. Each version has its its own merits and inconveniences. Morita uses µ′ because it is more convenient than µ when dealing with characteristic classes. On the other hand, µ′ may not be defined for a field with positive characteristic, and when using µ′, some extra factor is needed in defining the wedge operation of alternating multilinear forms (see Section 3.4) and for exterior differentiation. The version µ is the one adopted by Warner , Knapp , Fulton and Harris , and Cartan [21, 22]. If f : E →F is any linear map, by transposition we get a linear map f ⊤: F ∗→E∗given by f ⊤(v∗) = v∗◦f, v∗∈F ∗. Consequently, we have f ⊤(v∗)(u) = v∗(f(u)), for all u ∈E and all v∗∈F ∗. For any p ≥1, the map (u1, . . . , up) 7→f(u1) ∧· · · ∧f(up) from Ep to Vp F is multilinear alternating, so it induces a unique linear map Vp f : Vp E → Vp F making the following diagram commute Ep " ι∧ / Vp E Vp f  Vp F, 3.3. DUALITY FOR EXTERIOR POWERS 105 and defined on generators by  p ^ f  (u1 ∧· · · ∧up) = f(u1) ∧· · · ∧f(up). Combining Vp and duality, we get a linear map Vp f ⊤: Vp F ∗→Vp E∗defined on generators by  p ^ f ⊤ (v∗ 1 ∧· · · ∧v∗ p) = f ⊤(v∗ 1) ∧· · · ∧f ⊤(v∗ p). Proposition 3.11. If f : E →F is any linear map between two finite dimensional vector spaces E and F, then µ  p ^ f ⊤ (ω)  (u1, . . . , up) = µ(ω)(f(u1), . . . , f(up)), ω ∈ p ^ F ∗, u1, . . . , up ∈E. Proof. It is enough to prove the formula on generators. By definition of µ, we have µ  p ^ f ⊤ (v∗ 1 ∧· · · ∧v∗ p)  (u1, . . . , up) = µ(f ⊤(v∗ 1) ∧· · · ∧f ⊤(v∗ p))(u1, . . . , up) = det(f ⊤(v∗ j)(ui)) = det(v∗ j(f(ui))) = µ(v∗ 1 ∧· · · ∧v∗ p)(f(u1), . . . , f(up)), as claimed. Remark: The map Vp f ⊤is often denoted f ∗, although this is an ambiguous notation since p is dropped. Proposition 3.11 gives us the behavior of Vp f ⊤under the identification of Vp E∗and Altp(E; K) via the isomorphism µ. As in the case of symmetric powers, the map from En to Vn(E) given by (u1, . . . , un) 7→ u1 ∧· · · ∧un yields a surjection π: E⊗n →Vn(E). Now this map has some section, so there is some injection β : Vn(E) →E⊗n with π ◦β = id. As we saw in Proposition 3.10 there is a canonical isomorphism ( n ^ (E))∗∼ = n ^ (E∗) for any field K, even of positive characteristic. However, if our field K has characteristic 0, then there is a special injection having a natural definition involving an antisymmetrization process. Recall, from Section 2.12 that we have a left action of the symmetric group Sn on E⊗n. The tensors z ∈E⊗n such that σ · z = sgn(σ) z, for all σ ∈Sn are called antisymmetrized tensors. We define the map η: En →E⊗n by η(u1, . . . , un) = 1 n! X σ∈Sn sgn(σ) uσ(1) ⊗· · · ⊗uσ(n).1 1It is the division by n! that requires the field to have characteristic zero. 106 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS As the right hand side is an alternating map, we get a unique linear map Vn η: Vn(E) → E⊗n making the following diagram commute. En η # ι∧ / Vn(E) Vn η  E⊗n. Clearly, Vn η(Vn(E)) is the set of antisymmetrized tensors in E⊗n. If we consider the map A = (Vn η) ◦π: E⊗n − →E⊗n, it is easy to check that A ◦A = A. Therefore, A is a projection, and by linear algebra, we know that E⊗n = A(E⊗n) ⊕Ker A = n ^ η( n ^ (E)) ⊕Ker A. It turns out that Ker A = E⊗n ∩Ia = Ker π, where Ia is the two-sided ideal of T(E) generated by all tensors of the form u ⊗u ∈E⊗2 (for example, see Knapp , Appendix A). Therefore, Vn η is injective, E⊗n = n ^ η( n ^ (E)) ⊕(E⊗n ∩Ia) = n ^ η( n ^ (E)) ⊕Ker π, and the exterior tensor power Vn(E) is naturally embedded into E⊗n. 3.4 Exterior Algebras As in the case of symmetric tensors, we can pack together all the exterior powers Vn(V ) into an algebra. Definition 3.5. Given any vector space V , the vector space ^ (V ) = M m≥0 m ^ (V ) is called the exterior algebra (or Grassmann algebra) of V . To make V(V ) into an algebra, we mimic the procedure used for symmetric powers. If Ia is the two-sided ideal generated by all tensors of the form u ⊗u ∈V ⊗2, we set • ^ (V ) = T(V )/Ia. Then V•(V ) automatically inherits a multiplication operation, called wedge product, and since T(V ) is graded, that is T(V ) = M m≥0 V ⊗m, 3.4. EXTERIOR ALGEBRAS 107 we have • ^ (V ) = M m≥0 V ⊗m/(Ia ∩V ⊗m). However, it is easy to check that m ^ (V ) ∼ = V ⊗m/(Ia ∩V ⊗m), so • ^ (V ) ∼ = ^ (V ). When V has finite dimension d, we actually have a finite direct sum (coproduct) ^ (V ) = d M m=0 m ^ (V ), and since each Vm(V ) has dimension d m  , we deduce that dim( ^ (V )) = 2d = 2dim(V ). The multiplication, ∧: Vm(V )×Vn(V ) →Vm+n(V ), is skew-symmetric in the following precise sense: Proposition 3.12. For all α ∈Vm(V ) and all β ∈Vn(V ), we have β ∧α = (−1)mnα ∧β. Proof. Since v ∧u = −u ∧v for all u, v ∈V , Proposition 3.12 follows by induction. Since α ∧α = 0 for every simple (also called decomposable) tensor α = u1 ∧· · · ∧un, it seems natural to infer that α ∧α = 0 for every tensor α ∈V(V ). If we consider the case where dim(V ) ≤3, we can indeed prove the above assertion. However, if dim(V ) ≥4, the above fact is generally false! For example, when dim(V ) = 4, if (u1, u2, u3, u4) is a basis for V , for α = u1 ∧u2 + u3 ∧u4, we check that α ∧α = 2u1 ∧u2 ∧u3 ∧u4, which is nonzero. However, if α ∈Vm E with m odd, since m2 is also odd, we have α ∧α = (−1)m2α ∧α = −α ∧α, so indeed α ∧α = 0 (if K is not a field of characteristic 2). 108 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS The above discussion suggests that it might be useful to know when an alternating tensor is simple (decomposable). We will show in Section 3.6 that for tensors α ∈V2(V ), α∧α = 0 iffα is simple. A general criterion for decomposability can be given in terms of some operations known as left hook and right hook (also called interior products); see Section 3.6. It is easy to see that V(V ) satisfies the following universal mapping property. Proposition 3.13. Given any K-algebra A, for any linear map f : V →A, if (f(v))2 = 0 for all v ∈V , then there is a unique K-algebra homomorphism f : V(V ) →A so that f = f ◦i, as in the diagram below. V i / f " V(V ) f  A When E is finite dimensional, recall the isomorphism µ: Vn(E∗) − →Altn(E; K), defined as the linear extension of the map given by µ(v∗ 1 ∧· · · ∧v∗ n)(u1, . . . , un) = det(v∗ j(ui)). Now, we have also a multiplication operation Vm(E∗) × Vn(E∗) − →Vm+n(E∗). The fol-lowing question then arises: Can we define a multiplication Altm(E; K) × Altn(E; K) − →Altm+n(E; K) directly on alternating multilinear forms, so that the following diagram commutes? Vm(E∗) × Vn(E∗) µm×µn  ∧ / Vm+n(E∗) µm+n  Altm(E; K) × Altn(E; K) ∧ / Altm+n(E; K) As in the symmetric case, the answer is yes! The solution is to define this multiplication such that, for f ∈Altm(E; K) and g ∈Altn(E; K), (f ∧g)(u1, . . . , um+n) = X σ∈shuffle(m,n) sgn(σ) f(uσ(1), . . . , uσ(m))g(uσ(m+1), . . . , uσ(m+n)), (∗∗) where shuffle(m, n) consists of all (m, n)-“shuffles;” that is, permutations σ of {1, . . . m + n} such that σ(1) < · · · < σ(m) and σ(m+1) < · · · < σ(m+n). For example, when m = n = 1, we have (f ∧g)(u, v) = f(u)g(v) −g(u)f(v). 3.4. EXTERIOR ALGEBRAS 109 When m = 1 and n ≥2, check that (f ∧g)(u1, . . . , um+1) = m+1 X i=1 (−1)i−1f(ui)g(u1, . . . , b ui, . . . , um+1), where the hat over the argument ui means that it should be omitted. Here is another explicit example. Suppose m = 2 and n = 1. Given v∗ 1, v∗ 2, v∗ 3 ∈E∗, the multiplication structure on V(E∗) implies that (v∗ 1 ∧v∗ 2) · v∗ 3 = v∗ 1 ∧v∗ 2 ∧v∗ 3 ∈V3(E∗). Furthermore, for u1, u2, u3, ∈E, µ3(v∗ 1 ∧v∗ 2 ∧v∗ 3)(u1, u2, u3) = X σ∈S3 sgn(σ)v∗ σ(1)(u1)v∗ σ(2)(u2)v∗ σ(3)(u3) = v∗ 1(u1)v∗ 2(u2)v∗ 3(u3) −v∗ 1(u1)v∗ 3(u2)v∗ 2(u3) −v∗ 2(u1)v∗ 1(u2)v∗ 3(u3) + v∗ 2(u1)v∗ 3(u2)v∗ 1(u3) + v∗ 3(u1)v∗ 1(u2)v∗ 2(u3) −v∗ 3(u1)v∗ 2(u2)v∗ 1(u3). Now the (2, 1)- shuffles of {1, 2, 3} are the following three permutations, namely 1 2 3 1 2 3  , 1 2 3 1 3 2  , 1 2 3 2 3 1  . If f ∼ = µ2(v∗ 1 ∧v∗ 2) and g ∼ = µ1(v∗ 3), then (∗∗) implies that (f · g)(u1, u2, u3) = X σ∈shuffle(2,1) sgn(σ)f(uσ(1), uσ(2))g(uσ(3)) = f(u1, u2)g(u3) −f(u1, u3)g(u2) + f(u2, u3)g(u1) = µ2(v∗ 1 ∧v∗ 2)(u1, u2)µ1(v∗ 3)(u3) −µ2(v∗ 1 ∧v∗ 2)(u1, u3)µ1(v∗ 3)(u2) + µ2(v∗ 1 ∧v∗ 2)(u2, u3)µ1(v∗ 3)(u1) = (v∗ 1(u1)v∗ 2(u2) −v∗ 2(u1)v∗ 1(u2))v∗ 3(u3) −(v∗ 1(u1)v∗ 2(u3) −v∗ 2(u1)v∗ 1(u3))v∗ 3(u2) + (v∗ 1(u2)v∗ 2(u3) −v∗ 2(u2)v∗ 1(u3))v∗ 3(u1) = µ3(v∗ 1 ∧v∗ 2 ∧v∗ 3)(u1, u2, u3). As a result of all this, the direct sum Alt(E) = M n≥0 Altn(E; K) is an algebra under the above multiplication, and this algebra is isomorphic to V(E∗). For the record we state 110 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS Proposition 3.14. When E is finite dimensional, the maps µ: Vn(E∗) − →Altn(E; K) induced by the linear extensions of the maps given by µ(v∗ 1 ∧· · · ∧v∗ n)(u1, . . . , un) = det(v∗ j(ui)) yield a canonical isomorphism of algebras µ: V(E∗) − →Alt(E), where the multiplication in Alt(E) is defined by the maps ∧: Altm(E; K) × Altn(E; K) − →Altm+n(E; K), with (f ∧g)(u1, . . . , um+n) = X σ∈shuffle(m,n) sgn(σ) f(uσ(1), . . . , uσ(m))g(uσ(m+1), . . . , uσ(m+n)), where shuffle(m, n) consists of all (m, n)-“shuffles,” that is, permutations σ of {1, . . . m+n} such that σ(1) < · · · < σ(m) and σ(m + 1) < · · · < σ(m + n). Remark: The algebra V(E) is a graded algebra. Given two graded algebras E and F, we can make a new tensor product E b ⊗F, where E b ⊗F is equal to E ⊗F as a vector space, but with a skew-commutative multiplication given by (a ⊗b) ∧(c ⊗d) = (−1)deg(b)deg(c)(ac) ⊗(bd), where a ∈Em, b ∈F p, c ∈En, d ∈F q. Then, it can be shown that ^ (E ⊕F) ∼ = ^ (E) b ⊗ ^ (F). 3.5 The Hodge ∗-Operator In order to define a generalization of the Laplacian that applies to differential forms on a Riemannian manifold, we need to define isomorphisms k ^ V − → n−k ^ V, for any Euclidean vector space V of dimension n and any k, with 0 ≤k ≤n. If ⟨−, −⟩ denotes the inner product on V , we define an inner product on Vk V , denoted ⟨−, −⟩∧, by setting ⟨u1 ∧· · · ∧uk, v1 ∧· · · ∧vk⟩∧= det(⟨ui, vj⟩), for all ui, vi ∈V , and extending ⟨−, −⟩∧by bilinearity. It is easy to show that if (e1, . . . , en) is an orthonormal basis of V , then the basis of Vk V consisting of the eI (where I = {i1, . . . , ik}, with 1 ≤i1 < · · · < ik ≤n) is an orthonormal basis of Vk V . Since the inner product on V induces an inner product on V ∗(recall that ⟨ω1, ω2⟩= ⟨ω♯ 1, ω♯ 2⟩, for all ω1, ω2 ∈V ∗), we also get an inner product on Vk V ∗. 3.5. THE HODGE ∗-OPERATOR 111 Definition 3.6. An orientation of a vector space V of dimension n is given by the choice of some basis (e1, . . . , en). We say that a basis (u1, . . . , un) of V is positively oriented iff det(u1, . . . , un) > 0 (where det(u1, . . . , un) denotes the determinant of the matrix whose jth column consists of the coordinates of uj over the basis (e1, . . . , en)), otherwise it is negatively oriented. An oriented vector space is a vector space V together with an orientation of V . If V is oriented by the basis (e1, . . . , en), then V ∗is oriented by the dual basis (e∗ 1, . . . , e∗ n). If σ is any permutation of {1, . . . , n}, then the basis (eσ(1), . . . , eσ(n)) has positive orientation iffthe signature sgn(σ) of the permutation σ is even. If V is an oriented vector space of dimension n, then we can define a linear isomorphism ∗: k ^ V → n−k ^ V, called the Hodge ∗-operator. The existence of this operator is guaranteed by the following proposition. Proposition 3.15. Let V be any oriented Euclidean vector space whose orientation is given by some chosen orthonormal basis (e1, . . . , en). For any alternating tensor α ∈Vk V , there is a unique alternating tensor ∗α ∈Vn−k V such that α ∧β = ⟨∗α, β⟩∧e1 ∧· · · ∧en for all β ∈Vn−k V . The alternating tensor ∗α is independent of the choice of the positive orthonormal basis (e1, . . . , en). Proof. Since Vn V has dimension 1, the alternating tensor e1 ∧· · · ∧en is a basis of Vn V . It follows that for any fixed α ∈Vk V , the linear map λα from Vn−k V to Vn V given by λα(β) = α ∧β is of the form λα(β) = fα(β) e1 ∧· · · ∧en for some linear form fα ∈ Vn−k V ∗. But then, by the duality induced by the inner product ⟨−, −⟩on Vn−k V , there is a unique vector ∗α ∈Vn−k V such that fλ(β) = ⟨∗α, β⟩∧ for all β ∈Vn−k V , which implies that α ∧β = λα(β) = fα(β) e1 ∧· · · ∧en = ⟨∗α, β⟩∧e1 ∧· · · ∧en, as claimed. If (e′ 1, . . . , e′ n) is any other positively oriented orthonormal basis, by Proposition 3.2, e′ 1 ∧· · ·∧e′ n = det(P) e1 ∧· · ·∧en = e1 ∧· · ·∧en, since det(P) = 1 where P is the change of basis from (e1, . . . , en) to (e′ 1, . . . , e′ n) and both bases are positively oriented. 112 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS Definition 3.7. The operator ∗from Vk V to Vn−k V defined by Proposition 3.15 is called the Hodge ∗-operator. Obseve that the Hodge ∗-operator is linear. The Hodge ∗-operator is defined in terms of the orthonormal basis elements of V V as follows: For any increasing sequence (i1, . . . , ik) of elements ip ∈{1, . . . , n}, if (j1, . . . , jn−k) is the increasing sequence of elements jq ∈{1, . . . , n} such that {i1, . . . , ik} ∪{j1, . . . , jn−k} = {1, . . . , n}, then ∗(ei1 ∧· · · ∧eik) = sign(i1, . . . ik, j1, . . . , jn−k) ej1 ∧· · · ∧ejn−k. In particular, for k = 0 and k = n, we have ∗(1) = e1 ∧· · · ∧en ∗(e1 ∧· · · ∧en) = 1. For example, if n = 3, we have ∗e1 = e2 ∧e3 ∗e2 = −e1 ∧e3 ∗e3 = e1 ∧e2 ∗(e1 ∧e2) = e3 ∗(e1 ∧e3) = −e2 ∗(e2 ∧e3) = e1. The Hodge ∗-operators ∗: Vk V →Vn−k V induce a linear map ∗: V(V ) →V(V ). We also have Hodge ∗-operators ∗: Vk V ∗→Vn−k V ∗. The following proposition shows that the linear map ∗: V(V ) →V(V ) is an isomorphism. Proposition 3.16. If V is any oriented vector space of dimension n, for every k with 0 ≤k ≤n, we have (i) ∗∗= (−id)k(n−k). (ii) ⟨x, y⟩∧= ∗(x ∧∗y) = ∗(y ∧∗x), for all x, y ∈Vk V . Proof. (1) Let (ei)n i=1 is an orthonormal basis of V . It is enough to check the identity on basis elements. We have ∗(ei1 ∧· · · ∧eik) = sign(i1, . . . ik, j1, . . . , jn−k) ej1 ∧· · · ∧ejn−k 3.5. THE HODGE ∗-OPERATOR 113 and ∗∗(ei1 ∧· · · ∧eik) = sign(i1, . . . ik, j1, . . . , jn−k) ∗(ej1 ∧· · · ∧ejn−k) = sign(i1, . . . ik, j1, . . . , jn−k) sign(j1, . . . , jn−k, i1, . . . ik) ei1 ∧· · · ∧eik. It is easy to see that sign(i1, . . . ik, j1, . . . , jn−k) sign(j1, . . . , jn−k, i1, . . . ik) = (−1)k(n−k), which yields ∗∗(ei1 ∧· · · ∧eik) = (−1)k(n−k) ei1 ∧· · · ∧eik, as claimed. (ii) These identities are easily checked on basis elements; see Jost , Chapter 2, Lemma 2.1.1. In particular let x = ei1 ∧· · · ∧eik, y = eij ∧· · · ∧eij, x, y ∈ k ^ V, where (ei)n i=1 is an orthonormal basis of V . If x ̸= y, ⟨x, y⟩∧= 0 since there is some eip of x not equal to any ejq of y by the orthonormality of the basis, this means the pth row of (⟨eil, ejs⟩) consists entirely of zeroes. Also x ̸= y implies that y ∧∗x = 0 since ∗x = sign(i1, . . . ik, l1, . . . , ln−k)el1 ∧· · · ∧eln−k, where els is the same as some ep in y. A similar argument shows that if x ̸= y, x ∧∗y = 0. So now assume x = y. Then ∗(ei1 ∧· · · ∧eik ∧∗(ei1 ∧· · · ∧eik)) = ∗(e1 ∧e2 · · · ∧en) = 1 = ⟨x, x⟩∧. In Section 9.2 we will need to express ∗(1) in terms of any basis (not necessarily orthonor-mal) of V . Proposition 3.17. If V is any finite dimensional oriented vector space, for any basis (v!, . . ., vn) of V , we have ∗(1) = 1 p det(⟨vi, vj⟩) v1 ∧· · · ∧vn. Proof. If (e1, . . . , en) is an orthonormal basis of V and (v1, . . . , vn) is any other basis of V , then ⟨v1 ∧· · · ∧vn, v1 ∧· · · ∧vn⟩∧= det(⟨vi, vj⟩), and since v1 ∧· · · ∧vn = det(A) e1 ∧· · · ∧en 114 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS where A is the matrix expressing the vj in terms of the ei, we have ⟨v1 ∧· · · ∧vn, v1 ∧· · · ∧vn⟩∧= det(A)2⟨e1 ∧· · · ∧en, e1 ∧· · · ∧en⟩= det(A)2. As a consequence, det(A) = p det(⟨vi, vj⟩), and v1 ∧· · · ∧vn = q det(⟨vi, vj⟩) e1 ∧· · · ∧en, from which it follows that ∗(1) = 1 p det(⟨vi, vj⟩) v1 ∧· · · ∧vn (see Jost , Chapter 2, Lemma 2.1.3). 3.6 Left and Right Hooks ⊛ The motivation for defining left hooks and right hook comes from the problem of under-standing when a tensor α ∈Vk(E) is decomposable. An arbitrary tensor α ∈Vk(E) is a linear combination of tensors of the form u1 ∧· · · ∧uk, called decomposable. The issue is to find criteria for decomposability. Criteria for testing decomposability using left hooks are given in Section 3.7. In this section all vector spaces are assumed to have finite dimension. Say dim(E) = n. Using our nonsingular pairing ⟨−, −⟩: p ^ E∗× p ^ E − →K (1 ≤p ≤n) defined on generators by ⟨u∗ 1 ∧· · · ∧u∗ p, v1 ∧· · · ∧up⟩= det(u∗ i (vj)), we define various contraction operations (partial evaluation operators) ⌟: p ^ E × p+q ^ E∗− → q ^ E∗ (left hook) and ⌞: p+q ^ E∗× p ^ E − → q ^ E∗ (right hook), as well as the versions obtained by replacing E by E∗and E∗∗by E. We begin with the left interior product or left hook, ⌟. Let u ∈Vp E. For any q such that p + q ≤n, multiplication on the right by u is a linear map ∧R(u): q ^ E − → p+q ^ E 3.6. LEFT AND RIGHT HOOKS ⊛ 115 given by v 7→v ∧u where v ∈Vq E. The transpose of ∧R(u) yields a linear map (∧R(u))⊤: p+q ^ E ∗ − →  q ^ E ∗ , which, using the isomorphisms Vp+q E ∗∼ = Vp+q E∗and Vq E ∗∼ = Vq E∗, can be viewed as a map (∧R(u))⊤: p+q ^ E∗− → q ^ E∗ given by z∗7→z∗◦∧R(u), where z∗∈Vp+q E∗. We denote z∗◦∧R(u) by u ⌟z∗. In terms of our pairing, the adjoint u ⌟of ∧R(u) defined by ⟨u ⌟z∗, v⟩= ⟨z∗, ∧R(u)(v)⟩; this in turn leads to the following definition. Definition 3.8. Let u ∈Vp E and z∗∈Vp+q E∗. We define u ⌟z∗∈Vq E∗to be q-vector uniquely determined by ⟨u ⌟z∗, v⟩= ⟨z∗, v ∧u⟩, for all v ∈Vq E. Remark: Note that to be precise the operator ⌟: p ^ E × p+q ^ E∗− → q ^ E∗ depends of p, q, so we really defined a family of operators ⌟p,q. This family of operators ⌟p,q induces a map ⌟: ^ E × ^ E∗− → ^ E∗, with ⌟p,q : p ^ E × p+q ^ E∗− → q ^ E∗ as defined before. The common practice is to omit the subscripts of ⌟. It is immediately verified that (u ∧v) ⌟z∗= u ⌟(v ⌟z∗), for all u ∈Vk E, v ∈Vp−k E, z∗∈Vp+q E∗since ⟨(u ∧v) ⌟z∗, w⟩= ⟨z∗, w ∧u ∧v⟩= ⟨v ⌟z∗, w ∧u⟩= ⟨u ⌟(v ⌟z∗), w⟩, 116 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS whenever w ∈Vq E. This means that ⌟: ^ E × ^ E∗− → ^ E∗ is a left action of the (noncommutative) ring V E with multiplication ∧on V E∗, which makes V E∗into a left V E-module. By interchanging E and E∗and using the isomorphism  k ^ F ∗ ∼ = k ^ F ∗, we can also define some maps ⌟: p ^ E∗× p+q ^ E − → q ^ E, and make the following definition. Definition 3.9. Let u∗∈Vp E∗, and z ∈Vp+q E. We define u∗⌟z ∈Vq as the q-vector uniquely defined by ⟨v∗∧u∗, z⟩= ⟨v∗, u∗⌟z⟩, for all v∗∈Vq E∗. As for the previous version, we have a family of operators ⌟p,q which define an operator ⌟: ^ E∗× ^ E − → ^ E. We easily verify that (u∗∧v∗) ⌟z = u∗⌟(v∗⌟z), whenever u∗∈Vk E∗, v∗∈Vp−k E∗, and z ∈Vp+q E; so this version of ⌟is a left action of the ring V E∗on V E which makes V E into a left V E∗-module. In order to proceed any further we need some combinatorial properties of the basis of Vp E constructed from a basis (e1, . . . , en) of E. Recall that for any (nonempty) subset I ⊆{1, . . . , n}, we let eI = ei1 ∧· · · ∧eip, where I = {i1, . . . , ip} with i1 < · · · < ip. We also let e∅= 1. Given any two nonempty subsets H, L ⊆{1, . . . , n} both listed in increasing order, say H = {h1 < . . . < hp} and L = {ℓ1 < . . . < ℓq}, if H and L are disjoint, let H ∪L be union of H and L considered as the ordered sequence (h1, . . . , hp, ℓ1, . . . , ℓq). 3.6. LEFT AND RIGHT HOOKS ⊛ 117 Then let ρH,L =  0 if H ∩L ̸= ∅, (−1)ν if H ∩L = ∅, where ν = |{(h, l) | (h, l) ∈H × L, h > l}|. Observe that when H ∩L = ∅, |H| = p and |L| = q, the number ν is the number of inversions of the sequence (h1, · · · , hp, ℓ1, · · · , ℓq), where an inversion is a pair (hi, ℓj) such that hi > ℓj.  Unless p + q = n, the function whose graph is given by  1 · · · p p + 1 · · · p + q h1 · · · hp ℓ1 · · · ℓq  is not a permutation of {1, . . . , n}. We can view ν as a slight generalization of the notion of the number of inversions of a permutation. Proposition 3.18. For any basis (e1, . . . , en) of E the following properties hold: (1) If H ∩L = ∅, |H| = p, and |L| = q, then ρH,LρL,H = (−1)ν(−1)pq−ν = (−1)pq. (2) For H, L ⊆{1, . . . , m} listed in increasing order, we have eH ∧eL = ρH,LeH∪L. Similarly, e∗ H ∧e∗ L = ρH,Le∗ H∪L. (3) For the left hook ⌟: p ^ E × p+q ^ E∗− → q ^ E∗, we have eH ⌟e∗ L = 0 if H ̸⊆L eH ⌟e∗ L = ρL−H,He∗ L−H if H ⊆L. (4) For the left hook ⌟: p ^ E∗× p+q ^ E − → q ^ E, we have e∗ H ⌟eL = 0 if H ̸⊆L e∗ H ⌟eL = ρL−H,HeL−H if H ⊆L. 118 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS Proof. These are proved in Bourbaki (Chapter III, §11, Section 11), but the proofs of (3) and (4) are very concise. We elaborate on the proofs of (2) and (4), the proof of (3) being similar. In (2) if H∩L ̸= ∅, then eH∧eL contains some vector twice and so eH∧eL = 0. Otherwise, eH ∧eL consists of eh1 ∧· · · ∧ehp ∧eℓ1 ∧· · · ∧eℓq, and to order the sequence of indices in increasing order we need to transpose any two indices (hi, ℓj) corresponding to an inversion, which yields ρH,LeH∪L. Let us now consider (4). We have |L| = p + q and |H| = p, and the q-vector e∗ H ⌟eL is characterized by ⟨v∗, e∗ H ⌟eL⟩= ⟨v∗∧e∗ H, eL⟩ for all v∗∈Vq E∗. There are two cases. Case 1: H ̸⊆L. If so, no matter what v∗∈Vq E∗is, since H contains some index h not in L, the hth row (e∗ h(eℓ1), . . . , e∗ h(eℓp+q)) of the determinant ⟨v∗∧e∗ H, eL⟩must be zero, so ⟨v∗∧e∗ H, eL⟩= 0 for all v∗∈Vq E∗, and since the pairing is nongenerate, we must have e∗ H ⌟eL = 0. Case 2: H ⊆L. In this case, for v∗= e∗ L−H, by (2) we have ⟨e∗ L−H, e∗ H ⌟eL⟩= ⟨e∗ L−H ∧e∗ H, eL⟩= ⟨ρL−H,He∗ L, eL⟩= ρL−H,H, which yields ⟨e∗ L−H, e∗ H ⌟eL⟩= ρL−H,H. The q-vector e∗ H ⌟eL can be written as a linear combination e∗ H ⌟eL = P J λJeJ with |J| = q so ⟨e∗ L−H, e∗ H ⌟eL⟩= X J λJ⟨e∗ L−H, eJ⟩. By definition of the pairing, ⟨e∗ L−H, eJ⟩= 0 unless J = L −H, which means that ⟨e∗ L−H, e∗ H ⌟eL⟩= λL−H⟨e∗ L−H, eL−H⟩= λL−H, so λL−H = ρL−H,H, as claimed. Using Proposition 3.18, we have the Proposition 3.19. For the left hook ⌟: E × q+1 ^ E∗− → q ^ E∗, for every u ∈E, x∗∈Vq+1−s E∗, and y∗∈Vs E∗, we have u ⌟(x∗∧y∗) = (−1)s(u ⌟x∗) ∧y∗+ x∗∧(u ⌟y∗). 3.6. LEFT AND RIGHT HOOKS ⊛ 119 Proof. We can prove the above identity assuming that x∗and y∗are of the form e∗ I and e∗ J using Proposition 3.18 and leave the details as an exercise for the reader. Thus, ⌟: E ×Vq+1 E∗− →Vq E∗is almost an anti-derivation, except that the sign (−1)s is applied to the wrong factor. We have a similar identity for the other version of the left hook ⌟: E∗× q+1 ^ E − → q ^ E, namely u∗⌟(x ∧y) = (−1)s(u∗⌟x) ∧y + x ∧(u∗⌟y) for every u∗∈E∗, x ∈Vq+1−s E, and y ∈Vs E. An application of this formula when q = 3 and s = 2 yields an interesting equation. In this case, u∗∈E∗and x, y ∈V2 E, so we get u∗⌟(x ∧y) = (u∗⌟x) ∧y + x ∧(u∗⌟y). In particular, for x = y, since x ∈V2 E and u∗⌟x ∈E, Proposition 3.12 implies that (u∗⌟x) ∧x = x ∧(u∗⌟x), and we obtain u∗⌟(x ∧x) = 2((u∗⌟x) ∧x). (†) As a consequence, (u∗⌟x) ∧x = 0 iffu∗⌟(x ∧x) = 0. We will use this identity together with Proposition 3.25 to prove that a 2-vector x ∈V2 E is decomposable iffx ∧x = 0. It is also possible to define a right interior product or right hook ⌞, using multiplication on the left rather than multiplication on the right. Then we use the maps ⌞: p+q ^ E∗× p ^ E − → q ^ E∗ to make the following definition. Definition 3.10. Let u ∈Vp E and z∗∈Vp+q E∗. We define z∗⌞u ∈Vq E∗to be the q-vector uniquely defined as ⟨z∗⌞u, v⟩= ⟨z∗, u ∧v⟩, for all v ∈Vq E. This time we can prove that z∗⌞(u ∧v) = (z∗⌞u) ⌞v, so the family of operators ⌞p,q defines a right action ⌞: ^ E∗× ^ E − → ^ E∗ 120 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS of the ring V E on V E∗which makes V E∗into a right V E-module. Similarly, we have maps ⌞: p+q ^ E × p ^ E∗− → q ^ E which in turn leads to the following dual formation of the right hook. Definition 3.11. Let u∗∈Vp E∗and z ∈Vp+q E. We define z ⌞u∗∈Vq to be the q-vector uniquely defined by ⟨u∗∧v∗, z⟩= ⟨v∗, z ⌞u∗⟩, for all v∗∈Vq E∗. We can prove that z ⌞(u∗∧v∗) = (z ⌞u∗) ⌞v∗, so the family of operators ⌞p,q defines a right action ⌞: ^ E × ^ E∗− → ^ E of the ring V E∗on V E which makes V E into a right V E∗-module. Since the left hook ⌟: Vp E × Vp+q E∗− →Vq E∗is defined by ⟨u ⌟z∗, v⟩= ⟨z∗, v ∧u⟩, for all u ∈Vp E, v ∈Vq E and z∗∈Vp+q E∗, the right hook ⌞: p+q ^ E∗× p ^ E − → q ^ E∗ by ⟨z∗⌞u, v⟩= ⟨z∗, u ∧v⟩, for all u ∈Vp E, v ∈Vq E, and z∗∈Vp+q E∗, and v ∧u = (−1)pqu ∧v, we conclude that z∗⌞u = (−1)pq u ⌟z∗. Similarly, since ⟨v∗∧u∗, z⟩= ⟨v∗, u∗⌟z⟩, for all u∗∈Vp E∗, v∗∈Vq E∗and z ∈Vp+q E ⟨u∗∧v∗, z⟩= ⟨v∗, z ⌞u∗⟩, for all u∗∈Vp E∗, v∗∈Vq E∗, and z ∈Vp+q E, and v∗∧u∗= (−1)pqu∗∧v∗, we have z ⌞u∗= (−1)pq u∗⌟z. We summarize the above facts in the following proposition. 3.6. LEFT AND RIGHT HOOKS ⊛ 121 Proposition 3.20. The following identities hold: z∗⌞u = (−1)pq u ⌟z∗ for all u ∈Vp E and all z∗∈Vp+q E∗ z ⌞u∗= (−1)pq u∗⌟z for all u∗∈Vp E∗and all z ∈Vp+q E. Therefore the left and right hooks are not independent, and in fact each one determines the other. As a consequence, we can restrict our attention to only one of the hooks, for example the left hook, but there are a few situations where it is nice to use both, for example in Proposition 3.23. A version of Proposition 3.18 holds for right hooks, but beware that the indices in ρL−H,H are permuted. This permutation has to do with the fact that the left hook and the right hook are related via a sign factor. Proposition 3.21. For any basis (e1, . . . , en) of E the following properties hold: (1) For the right hook ⌞: p+q ^ E × p ^ E∗− → q ^ E we have eL ⌞e∗ H = 0 if H ̸⊆L eL ⌞e∗ H = ρH,L−HeL−H if H ⊆L. (2) For the right hook ⌞: p+q ^ E∗× p ^ E − → q ^ E∗ we have e∗ L ⌞eH = 0 if H ̸⊆L e∗ L ⌞eH = ρH,L−He∗ L−H if H ⊆L. Remark: Our definition of left hooks as left actions ⌟: Vp E × Vp+q E∗− →Vq E∗and ⌟: Vp E∗×Vp+q E − →Vq E and right hooks as right actions ⌞: Vp+q E∗×Vp E − →Vq E∗ and ⌞: Vp+q E×Vp E∗− →Vq E is identical to the definition found in Fulton and Harris (Appendix B). However, the reader should be aware that this is not a universally accepted notation. In fact, the left hook u∗⌟z defined in Bourbaki is our right hook z ⌞u∗, up to the sign (−1)p(p−1)/2. This has to do with the fact that Bourbaki uses a different pairing which also involves an extra sign, namely ⟨v∗, u∗⌟z⟩= (−1)p(p−1)/2⟨u∗∧v∗, z⟩. 122 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS One of the side-effects of this choice is that Bourbaki’s version of Formula (4) of Proposition 3.18 (Bourbaki , Chapter III, page 168) is e∗ H ⌟eL = 0 if H ̸⊆L e∗ H ⌟eL = (−1)p(p−1)/2ρH,L−HeL−H if H ⊆L, where |H| = p and |L| = p + q. This correspond to Formula (1) of Proposition 3.21 up to the sign factor (−1)p(p−1)/2, which we find horribly confusing. Curiously, an older edition of Bourbaki (1958) uses the same pairing as Fulton and Harris . The reason (and the advantage) for this change of sign convention is not clear to us. We also have the following version of Proposition 3.19 for the right hook. Proposition 3.22. For the right hook ⌞: q+1 ^ E∗× E − → q ^ E∗, for every u ∈E, x∗∈Vr E∗, and y∗∈Vq+1−r E∗, we have (x∗∧y∗) ⌞u = (x∗⌞u) ∧y∗+ (−1)rx∗∧(y∗⌞u). Proof. A proof involving determinants can be found in Warner , Chapter 2. Thus, ⌞: Vq+1 E∗× E − →Vq E∗is an anti-derivation. A similar formula holds for the the right hook ⌞: Vq+1 E × E∗− →Vq E, namely (x ∧y) ⌞u∗= (x ⌞u∗) ∧y + (−1)rx ∧(y ⌞u∗), for every u∗∈E, ∈Vr E, and y ∈Vq+1−r E. This formula is used by Shafarevitch to define a hook, but beware that Shafarevitch use the left hook notation u∗⌟x rather than the right hook notation. Shafarevitch uses the terminology convolution, which seems very unfortunate. For u ∈E, the right hook z∗⌞u is also denoted i(u)z∗, and called insertion operator or interior product. This operator plays an important role in differential geometry. Definition 3.12. Let u ∈E and z∗∈Vn+1(E∗). If we view z∗as an alternating multilinear map in Altn+1(E; K), then we define i(u)z∗∈Altn(E; K) as given by (i(u)z∗)(v1, . . . , vn) = z∗(u, v1, . . . , vn). Using the left hook ⌟and the right hook ⌞we can define two linear maps γ : Vp E → Vn−p E∗and δ: Vp E∗→Vn−p E as follows: 3.6. LEFT AND RIGHT HOOKS ⊛ 123 Definition 3.13. For any basis (e1, . . . , en) of E, if we let M = {1, . . . , n}, e = e1 ∧· · · ∧en, and e∗= e∗ 1 ∧· · · ∧e∗ n, define γ : Vp E →Vn−p E∗and δ: Vp E∗→Vn−p E as γ(u) = u ⌟e∗ and δ(v∗) = e ⌞v∗, for all u ∈Vp E and all v∗∈Vp E∗. Proposition 3.23. The linear maps γ : Vp E →Vn−p E∗and δ: Vp E∗→Vn−p E are isomorphims, and γ−1 = δ. The isomorphisms γ and δ map decomposable vectors to de-composable vectors. Furthermore, if z ∈Vp E is decomposable, say z = u1 ∧· · · ∧up for some ui ∈E, then γ(z) = v∗ 1 ∧· · · ∧v∗ n−p for some v∗ j ∈E∗, and v∗ j(ui) = 0 for all i, j. A similar property holds for v∗∈Vp E∗and δ(v∗). If (e′ 1, . . . , e′ n) is any other basis of E and γ′ : Vp E →Vn−p E∗and δ′ : Vp E∗→Vn−p E are the corresponding isomorphisms, then γ′ = λγ and δ′ = λ−1δ for some nonzero λ ∈K. Proof. Using Propositions 3.18 and 3.21, for any subset J ⊆{1, . . . , n} = M such that |J| = p, we have γ(eJ) = eJ ⌟e∗= ρM−J,Je∗ M−J and δ(e∗ M−J) = e ⌞e∗ M−J = ρM−J,JeJ. Thus, δ ◦γ(eJ) = ρM−J,JρM−J,JeJ = eJ, since ρM−J,J = ±1. A similar result holds for γ ◦δ. This implies that δ ◦γ = id and γ ◦δ = id. Thus, γ and δ are inverse isomorphisms. If z ∈Vp E is decomposable, then z = u1 ∧· · · ∧up where u1, . . . , up are linearly inde-pendent since z ̸= 0, and we can pick a basis of E of the form (u1, . . . , un). Then the above formulae show that γ(z) = ±u∗ p+1 ∧· · · ∧u∗ n. Since (u∗ 1, . . . , u∗ n) is the dual basis of (u1, . . . , un), we have u∗ i (uj) = δij, If (e′ 1, . . . , e′ n) is any other basis of E, because Vn E has dimension 1, we have e′ 1 ∧· · · ∧e′ n = λe1 ∧· · · ∧en for some nonzero λ ∈K, and the rest is trivial. Applying Proposition 3.23 to the case where p = n −1, the isomorphism γ : Vn−1 E → V1 E∗maps indecomposable vectors in Vn−1 E to indecomposable vectors in V1 E∗= E∗. But every vector in E∗is decomposable, so every vector in Vn−1 E is decomposable. Corollary 3.24. If E is a finite dimensional vector space, then every vector in Vn−1 E is decomposable. 124 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS 3.7 Testing Decomposability ⊛ We are now ready to tackle the problem of finding criteria for decomposability. Such criteria will use the left hook. Once again, in this section all vector spaces are assumed to have finite dimension. But before stating our criteria, we need a few preliminary results. Proposition 3.25. Given z ∈Vp E with z ̸= 0, the smallest vector space W ⊆E such that z ∈Vp W is generated by the vectors of the form u∗⌟z, with u∗∈Vp−1 E∗. Proof. First let W be any subspace such that z ∈Vp(W) and let (e1, . . . , er, er+1, . . . , en) be a basis of E such that (e1, . . . , er) is a basis of W. Then, u∗= P I λIe∗ I, where I ⊆{1, . . . , n} and |I| = p −1, and z = P J µJeJ, where J ⊆{1, . . . , r} and |J| = p ≤r. It follows immediately from the formula of Proposition 3.18 (4), namely e∗ I ⌟eJ = ρJ−I,JeJ−I, that u∗⌟z ∈W, since J −I ⊆{1, . . . , r}. Next we prove that if W is the smallest subspace of E such that z ∈Vp(W), then W is generated by the vectors of the form u∗⌟z, where u∗∈Vp−1 E∗. Suppose not. Then the vectors u∗⌟z with u∗∈Vp−1 E∗span a proper subspace U of W. We prove that for every subspace W ′ of W with dim(W ′) = dim(W) −1 = r −1, it is not possible that u∗⌟z ∈W ′ for all u∗∈Vp−1 E∗. But then, as U is a proper subspace of W, it is contained in some subspace W ′ with dim(W ′) = r −1, and we have a contradiction. Let w ∈W −W ′ and pick a basis of W formed by a basis (e1, . . . , er−1) of W ′ and w. Any z ∈Vp(W) can be written as z = z′ + w ∧z′′, where z′ ∈Vp W ′ and z′′ ∈Vp−1 W ′, and since W is the smallest subspace containing z, we have z′′ ̸= 0. Consequently, if we write z′′ = P I λIeI in terms of the basis (e1, . . . , er−1) of W ′, there is some eI, with I ⊆ {1, . . . , r −1} and |I| = p −1, so that the coefficient λI is nonzero. Now, using any basis of E containing (e1, . . . , er−1, w), by Proposition 3.18 (4), we see that e∗ I ⌟(w ∧eI) = λw, λ = ±1. It follows that e∗ I ⌟z = e∗ I ⌟(z′ + w ∧z′′) = e∗ I ⌟z′ + e∗ I ⌟(w ∧z′′) = e∗ I ⌟z′ + λλIw, with e∗ I ⌟z′ ∈W ′, which shows that e∗ I ⌟z / ∈W ′. Therefore, W is indeed generated by the vectors of the form u∗⌟z, where u∗∈Vp−1 E∗. To help understand Proposition 3.25, let E be the vector space with basis {e1, e2, e3, e4} and z = e1 ∧e2 + e2 ∧e3. Note that z ∈V2 E. To find the smallest vector space W ⊆E 3.7. TESTING DECOMPOSABILITY ⊛ 125 such that z ∈V2 W, we calculate u∗⌟z, where u∗∈V1 E∗. The multilinearity of ⌟implies it is enough to calculate u∗⌟z for u∗∈{e∗ 1, e∗ 2, e∗ 3, e∗ 4}. Proposition 3.18 (4) implies that e∗ 1 ⌟z = e∗ 1 ⌟(e1 ∧e2 + e2 ∧e3) = e∗ 1 ⌟e1 ∧e2 = −e2 e∗ 2 ⌟z = e∗ 2 ⌟(e1 ∧e2 + e2 ∧e3) = e1 −e3 e∗ 3 ⌟z = e∗ 3 ⌟(e1 ∧e2 + e2 ∧e3) = e∗ 3 ⌟e2 ∧e3 = e2 e∗ 4 ⌟z = e∗ 4 ⌟(e1 ∧e2 + e2 ∧e3) = 0. Thus W is the two-dimensional vector space generated by the basis {e2, e1 −e3}. This is not surprising since z = −e2 ∧(e1 −e3) and is in fact decomposable. As this example demonstrates, the action of the left hook provides a way of extracting a basis of W from z. Proposition 3.25 implies the following corollary. Corollary 3.26. Any nonzero z ∈Vp E is decomposable iffthe smallest subspace W of E such that z ∈Vp W has dimension p. Furthermore, if z = u1 ∧· · ·∧up is decomposable, then (u1, . . . , up) is a basis of the smallest subspace W of E such that z ∈Vp W Proof. If dim(W) = p, then for any basis (e1, . . . , ep) of W we know that Vp W has e1∧· · ·∧ep has a basis, and thus has dimension 1. Since z ∈Vp W, we have z = λe1 ∧· · · ∧ep for some nonzero λ, so z is decomposable. Conversely assume that z ∈Vp W is nonzero and decomposable. Then, z = u1 ∧· · ·∧up, and since z ̸= 0, by Proposition 3.9 (u1, . . . , up) are linearly independent. Then for any v∗ i = u∗ 1 ∧· · · u∗ i−1 ∧u∗ i+1 ∧· · · ∧u∗ p (where u∗ i is omitted), we have v∗ i ⌟z = (u∗ 1 ∧· · · u∗ i−1 ∧u∗ i+1 ∧· · · ∧u∗ p) ⌟(u1 ∧· · · ∧up) = ±ui, so by Proposition 3.25 we have ui ∈W for i = 1, . . . , p. This shows that dim(W) ≥p, but since z = u1 ∧· · · ∧up, we have dim(W) = p, which means that (u1, . . . , up) is a basis of W. Finally we are ready to state and prove the criterion for decomposability with respect to left hooks. Proposition 3.27. Any nonzero z ∈Vp E is decomposable iff (u∗⌟z) ∧z = 0, for all u∗∈Vp−1 E∗. Proof. First assume that z ∈Vp E is decomposable. If so, by Corollary 3.26, the smallest subspace W of E such that z ∈Vp W has dimension p, so we have z = e1 ∧· · · ∧ep where e1, . . . , ep form a basis of W. By Proposition 3.25, for every u∗∈Vp−1 E∗, we have u∗⌟z ∈W, so each u∗⌟z is a linear combination of the ei’s, say u∗⌟z = α1e1 + · · · + αpep, 126 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS and (u∗⌟z) ∧z = p X i=1 αiei ∧e1 ∧· · · ∧ei ∧· · · ∧ep = 0. Now assume that (u∗⌟z) ∧z = 0 for all u∗∈Vp−1 E∗, and that dim(W) = m > p, where W is the smallest subspace of E such that z ∈Vp W If e1, . . . , em is a basis of W, then we have z = P I λIeI, where I ⊆{1, . . . , m} and |I| = p. Recall that z ̸= 0, and so, some λI is nonzero. By Proposition 3.25, each ei can be written as u∗⌟z for some u∗∈Vp−1 E∗, and since (u∗⌟z) ∧z = 0 for all u∗∈Vp−1 E∗, we get ej ∧z = 0 for j = 1, . . . , m. By wedging z = P I λIeI with each ej, as m > p, we deduce λI = 0 for all I, so z = 0, a contradiction. Therefore, m = p and Corollary 3.26 implies that z is decomposable. As a corollary of Proposition 3.27 we obtain the following fact that we stated earlier without proof. Proposition 3.28. Given any vector space E of dimension n, a vector x ∈V2 E is decom-posable iffx ∧x = 0. Proof. Recall that as an application of Proposition 3.19 we proved the formula (†), namely u∗⌟(x ∧x) = 2((u∗⌟x) ∧x) for all x ∈V2 E and all u∗∈E∗. As a consequence, (u∗⌟x) ∧x = 0 iffu∗⌟(x ∧x) = 0. By Proposition 3.27, the 2-vector x is decomposable iffu∗⌟(x ∧x) = 0 for all u∗∈E∗iff x ∧x = 0. Therefore, a 2-vector x is decomposable iffx ∧x = 0. As an application of Proposition 3.28, assume that dim(E) = 3 and that (e1, e2, e3) is a basis of E. Then any 2-vector x ∈V2 E is of the form x = αe1 ∧e2 + βe1 ∧e3 + γe2 ∧e3. We have x ∧x = (αe1 ∧e2 + βe1 ∧e3 + γe2 ∧e3) ∧(αe1 ∧e2 + βe1 ∧e3 + γe2 ∧e3) = 0, because all the terms involved are of the form c ei1 ∧ei2 ∧ei3 ∧ei4 with i1, i2, i3, i4 ∈{1, 2, 3}, and so at least two of these indices are identical. Therefore, every 2-vector x = αe1 ∧e2 + βe1 ∧e3 + γe2 ∧e3 is decomposable, although this not obvious at first glance. For example, e1 ∧e2 + e1 ∧e3 + e2 ∧e3 = (e1 + e2) ∧(e2 + e3). We now show that Proposition 3.27 yields an equational criterion for the decomposability of an alternating tensor z ∈Vp E. 3.8. THE GRASSMANN-PL ¨ UCKER’S EQUATIONS AND GRASSMANNIANS ⊛ 127 3.8 The Grassmann-Pl¨ ucker’s Equations and Grassmannian Manifolds ⊛ Let E be a vector space of dimensions n, let (e1, . . . , en) be a basis of E, and let (e∗ 1, . . . , e∗ n) be its dual basis. Our objective is to determine whether a nonzero vector z ∈Vp E is decomposable, in terms of equations. We follow an argument adapted from Bourbaki (Chapter III, §11, Section 13). By Proposition 3.27, the vector z is decomposable iff(u∗⌟z) ∧z = 0 for all u∗∈Vp−1 E∗. We can let u∗range over a basis of Vp−1 E∗, and then the conditions are (e∗ H ⌟z) ∧z = 0 for all H ⊆{1, . . . , n}, with |H| = p −1. Since (e∗ H ⌟z) ∧z ∈Vp+1 E, this is equivalent to ⟨e∗ J, (e∗ H ⌟z) ∧z⟩= 0 for all H, J ⊆{1, . . . , n}, with |H| = p −1 and |J| = p + 1. Then, for all I, I′ ⊆{1, . . . , n} with |I| = |I′| = p, Formulae (2) and (4) of Proposition 3.18 show that ⟨e∗ J, (e∗ H ⌟eI) ∧eI′⟩= 0, unless there is some i ∈{1, . . . , n} such that I −H = {i}, J −I′ = {i}. In this case, I = H ∪{i} and I′ = J −{i}, and using Formulae (2) and (4) of Proposition 3.18, we have ⟨e∗ J, (e∗ H ⌟eH∪{i}) ∧eJ−{i}⟩= ⟨e∗ J, ρ{i},Hei ∧eJ−{i}⟩= ⟨e∗ J, ρ{i},Hρ{i},J−{i}eJ⟩= ρ{i},Hρ{i},J−{i}. If we let ϵi,J,H = ρ{i},Hρ{i},J−{i}, we have ϵi,J,H = +1 if the parity of the number of j ∈J such that j < i is the same as the parity of the number of h ∈H such that h < i, and ϵi,J,H = −1 otherwise. Finally we obtain the following criterion in terms of quadratic equations (Pl¨ ucker’s equa-tions) for the decomposability of an alternating tensor. Proposition 3.29. (Grassmann-Pl¨ ucker’s Equations) For z = P I λIeI ∈Vp E, the condi-tions for z ̸= 0 to be decomposable are X i∈J−H ϵi,J,HλH∪{i}λJ−{i} = 0, with ϵi,J,H = ρ{i},Hρ{i},J−{i}, for all H, J ⊆{1, . . . , n} such that |H| = p −1, |J| = p + 1, and all i ∈J −H. 128 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS Using the above criterion, it is a good exercise to reprove that if dim(E) = n, then every tensor in Vn−1(E) is decomposable. We already proved this fact as a corollary of Proposition 3.23. Given any z = P I λIeI ∈Vp E where dim(E) = n, the family of scalars (λI) (with I = {i1 < · · · < ip} ⊆{1, . . . , n} listed in increasing order) is called the Pl¨ ucker coordinates of z. The Grassmann-Pl¨ ucker’s equations give necessary and sufficient conditions for any nonzero z to be decomposable. For example, when dim(E) = n = 4 and p = 2, these equations reduce to the single equation λ12λ34 −λ13λ24 + λ14λ23 = 0. However, it should be noted that the equations given by Proposition 3.29 are not independent in general. We are now in the position to prove that the Grassmannian G(p, n) can be embedded in the projective space RP(n p)−1. For any n ≥1 and any k with 1 ≤p ≤n, recall that the Grassmannian G(p, n) is the set of all linear p-dimensional subspaces of Rn (also called p-planes). Any p-dimensional subspace U of Rn is spanned by p linearly independent vectors u1, . . . , up in Rn; write U = span(u1, . . . , uk). By Proposition 3.9, (u1, . . . , up) are linearly independent iffu1∧· · ·∧up ̸= 0. If (v1, . . . , vp) are any other linearly independent vectors spanning U, then we have vj = p X i=1 aijui, 1 ≤j ≤p, for some aij ∈R, and by Proposition 3.2 v1 ∧· · · ∧vp = det(A) u1 ∧· · · ∧up, where A = (aij). As a consequence, we can define a map iG : G(p, n) →RP(n p)−1 such that for any k-plane U, for any basis (u1, . . . , up) of U, iG(U) = [u1 ∧· · · ∧up], the point of RP(n p)−1 given by the one-dimensional subspace of R(n p) spanned by u1 ∧· · ·∧up. Proposition 3.30. The map iG : G(p, n) →RP(n p)−1 is injective. Proof. Let U and V be any two p-planes and assume that iG(U) = iG(V ). This means that there is a basis (u1, . . . , up) of U and a basis (v1, . . . , vp) of V such that v1 ∧· · · ∧vp = c u1 ∧· · · ∧up for some nonzero c ∈R. The above implies that the smallest subspaces W and W ′ of Rn such that u1 ∧· · · ∧up ∈Vp W and v1 ∧· · · ∧vp ∈Vp W ′ are identical, so W = W ′. By Corollary 3.26, this smallest subspace W has both (u1, . . . , up) and (v1, . . . , vp) as bases, so the vj are linear combinations of the ui (and vice-versa), and U = V . 3.8. THE GRASSMANN-PL ¨ UCKER’S EQUATIONS AND GRASSMANNIANS ⊛ 129 Since any nonzero z ∈Vp Rn can be uniquely written as z = X I λIeI in terms of its Pl¨ ucker coordinates (λI), every point of RP(n p)−1 is defined by the Pl¨ ucker coordinates (λI) viewed as homogeneous coordinates. The points of RP(n p)−1 corresponding to one-dimensional spaces associated with decomposable alternating p-tensors are the points whose coordinates satisfy the Grassmann-Pl¨ ucker’s equations of Proposition 3.29. Therefore, the map iG embeds the Grassmannian G(p, n) as an algebraic variety in RP(n p)−1 defined by equations of degree 2. We can replace the field R by C in the above reasoning and we obtain an embedding of the complex Grassmannian GC(p, n) as an algebraic variety in CP(n p)−1 defined by equations of degree 2. In particular, if n = 4 and p = 2, the equation λ12λ34 −λ13λ24 + λ14λ23 = 0 is the homogeneous equation of a quadric in CP5 known as the Klein quadric. The points on this quadric are in one-to-one correspondence with the lines in CP3. There is also a simple algebraic criterion to decide whether the smallest subspaces U and V associated with two nonzero decomposable vectors u1 ∧· · · ∧up and v1 ∧· · · ∧vq have a nontrivial intersection. Proposition 3.31. Let E be any n-dimensional vector space over a field K, and let U and V be the smallest subspaces of E associated with two nonzero decomposable vectors u = u1 ∧· · · ∧up ∈Vp U and v = v1 ∧· · · ∧vq ∈Vq V . The following properties hold: (1) We have U ∩V = (0) iffu ∧v ̸= 0. (2) If U ∩V = (0), then U + V is the least subspace associated with u ∧v. Proof. Assume U ∩V = (0). We know by Corollary 3.26 that (u1, . . . , up) is a basis of U and (v1, . . . , vq) is a basis of V . Since U ∩V = (0), (u1, . . . , up, v1, . . . , vq) is a basis of U +V , and by Proposition 3.9, we have u ∧v = u1 ∧· · · ∧up ∧v1 ∧· · · ∧vq ̸= 0. This also proves (2). Conversely, assume that dim(U ∩V ) ≥1. Pick a basis (w1, . . . , wr) of W = U ∩V , and ex-tend this basis to a basis (w1, . . . , wr, wr+1, . . . , wp) of U and to a basis (w1, . . . , wr, wp+1, . . ., wp+q−r) of V . By Corollary 3.26, (u1, . . . , up) is also basis of U, so u1 ∧· · · ∧up = a w1 ∧· · · ∧wr ∧wr+1 ∧· · · ∧wp 130 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS for some a ∈K, and (v1, . . . , vq) is also basis of V , so v1 ∧· · · ∧vq = b w1 · · · ∧wr ∧wp+1 ∧· · · ∧wp+q−r for some b ∈K, and thus u ∧v = u1 ∧· · · ∧up ∧v1 ∧· · · ∧vq = 0 since it contains some repeated wi, with 1 ≤i ≤r. As an application of Proposition 3.31, consider two projective lines D1 and D2 in RP3, which means that D1 and D2 correspond to two 2-planes in R4, and thus by Proposition 3.30, to two points in RP(4 2)−1 = RP5. These two points correspond to the 2-vectors z = a1,2e1 ∧e2 + a1,3e1 ∧e3 + a1,4e1 ∧e4 + a2,3e2 ∧e3 + a2,4e2 ∧e4 + a3,4e3 ∧e4 and z′ = a′ 1,2e1 ∧e2 + a′ 1,3e1 ∧e3 + a′ 1,4e1 ∧e4 + a′ 2,3e2 ∧e3 + a′ 2,4e2 ∧e4 + a′ 3,4e3 ∧e4 whose Pl¨ ucker coordinates, (where ai,j = λij), satisfy the equation λ12λ34 −λ13λ24 + λ14λ23 = 0 of the Klein quadric, and D1 and D2 intersect iffz ∧z′ = 0 iff a1,2a′ 3,4 −a1,3a′ 3,4 + a1,4a′ 2,3 + a2,3a′ 1,4 −a2,4a′ 1,3 + a3,4a′ 1,2 = 0. Observe that for D1 fixed, this is a linear condition. This fact is very helpful for solving problems involving intersections of lines. A famous problem is to find how many lines in RP3 meet four given lines in general position. The answer is at most 2. 3.9 Vector-Valued Alternating Forms The purpose of this section is to present the technical background needed for Sections 4.5 and 4.6 on vector-valued differential forms, in particular in the case of Lie groups where differential forms taking their values in a Lie algebra arise naturally. In this section the vector space E is assumed to have finite dimension. We know that there is a canonical isomorphism Vn(E∗) ∼ = Altn(E; K) between alternating n-forms and alternating multilinear maps. As in the case of general tensors, the isomorphisms provided by Propositions 3.5, 2.19, and 3.10, namely Altn(E; F) ∼ = Hom  n ^ (E), F  Hom  n ^ (E), F  ∼ =  n ^ (E) ∗ ⊗F  n ^ (E) ∗ ∼ = n ^ (E∗) 3.9. VECTOR-VALUED ALTERNATING FORMS 131 yield a canonical isomorphism Altn(E; F) ∼ =  n ^ (E∗)  ⊗F which we record as a corollary. Corollary 3.32. For any finite dimensional vector space E and any vector space F, we have a canonical isomorphism Altn(E; F) ∼ =  n ^ (E∗)  ⊗F. Note that F may have infinite dimension. This isomorphism allows us to view the tensors in Vn(E∗)⊗F as vector-valued alternating forms, a point of view that is useful in differential geometry. If (f1, . . . , fr) is a basis of F, every tensor ω ∈Vn(E∗) ⊗F can be written as some linear combination ω = r X i=1 αi ⊗fi, with αi ∈Vn(E∗).We also let ^ (E; F) = M n=0 n ^ (E∗) ! ⊗F = ^ (E)  ⊗F. Given three vector spaces, F, G, H, if we have some bilinear map Φ: F × G →H, then we can define a multiplication operation ∧Φ : ^ (E; F) × ^ (E; G) → ^ (E; H) as follows: For every pair (m, n), we define the multiplication ∧Φ :  m ^ (E∗)  ⊗F ! ×  n ^ (E∗)  ⊗G ! − → m+n ^ (E∗)  ⊗H by ω ∧Φ η = (α ⊗f) ∧Φ (β ⊗g) = (α ∧β) ⊗Φ(f, g). As in Section 3.4 (following H. Cartan ), we can also define a multiplication ∧Φ : Altm(E; F) × Altn(E; G) − →Altm+n(E; H) directly on alternating multilinear maps as follows: For f ∈Altm(E; F) and g ∈Altn(E; G), (f ∧Φ g)(u1, . . . , um+n) = X σ∈shuffle(m,n) sgn(σ) Φ  f(uσ(1), . . . , uσ(m)), g(uσ(m+1), . . . , uσ(m+n))  , 132 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS where shuffle(m, n) consists of all (m, n)-“shuffles;” that is, permutations σ of {1, . . . m + n} such that σ(1) < · · · < σ(m) and σ(m + 1) < · · · < σ(m + n). A special case of interest is the case where F = G = H is a Lie algebra and Φ(a, b) = [a, b] is the Lie bracket of F. In this case, using a basis (f1, . . . , fr) of F, if we write ω = P i αi⊗fi and η = P j βj ⊗fj, we have ω ∧Φ η = [ω, η] = X i,j αi ∧βj ⊗[fi, fj]. It is customary to denote ω∧Φη by [ω, η] (unfortunately, the bracket notation is overloaded). Consequently, [η, ω] = (−1)mn+1[ω, η]. In general not much can be said about ∧Φ, unless Φ has some additional properties. In particular, ∧Φ is generally not associative. We now use vector-valued alternating forms to generalize both the µ map of Proposition 3.14 and generalize Proposition 2.19 by defining the map µF : n ^ (E∗) ! ⊗F − →Altn(E; F) on generators by µF((v∗ 1 ∧· · · ∧v∗ n) ⊗f)(u1, . . . , un) = (det(v∗ j(ui))f, with v∗ 1, . . . , v∗ n ∈E∗, u1, . . . , un ∈E, and f ∈F. Proposition 3.33. The map µF : n ^ (E∗) ! ⊗F − →Altn(E; F) defined as above is a canonical isomorphism for every n ≥0. Furthermore, given any three vector spaces, F, G, H, and any bilinear map Φ: F × G →H, for all ω ∈(Vn(E∗)) ⊗F and all η ∈(Vn(E∗)) ⊗G, µH(ω ∧Φ η) = µF(ω) ∧Φ µG(η). Proof. Since we already know that (Vn(E∗))⊗F and Altn(E; F) are isomorphic, it is enough to show that µF maps some basis of (Vn(E∗)) ⊗F to linearly independent elements. Pick some bases (e1, . . . , ep) in E and (fj)j∈J in F. Then we know that the vectors e∗ I ⊗fj, where I ⊆{1, . . . , p} and |I| = n, form a basis of (Vn(E∗)) ⊗F. If we have a linear dependence X I,j λI,jµF(e∗ I ⊗fj) = 0, 3.9. VECTOR-VALUED ALTERNATING FORMS 133 applying the above combination to each (ei1, . . . , ein) (I = {i1, . . . , in}, i1 < · · · < in), we get the linear combination X j λI,jfj = 0, and by linear independence of the fj’s, we get λI,j = 0 for all I and all j. Therefore, the µF(e∗ I ⊗fj) are linearly independent, and we are done. The second part of the proposition is checked using a simple computation. The following proposition will be useful in dealing with vector-valued differential forms. Proposition 3.34. If (e1, . . . , ep) is any basis of E, then every element ω ∈(Vn(E∗)) ⊗F can be written in a unique way as ω = X I e∗ I ⊗fI, fI ∈F, where the e∗ I are defined as in Section 3.2. Proof. Since, by Proposition 3.8, the e∗ I form a basis of Vn(E∗), elements of the form e∗ I ⊗f span (Vn(E∗)) ⊗F. Now if we apply µF(ω) to (ei1, . . . , ein), where I = {i1, . . . , in} ⊆ {1, . . . , p}, we get µF(ω)(ei1, . . . , ein) = µF(e∗ I ⊗fI)(ei1, . . . , ein) = fI. Therefore, the fI are uniquely determined by f. Proposition 3.34 can also be formulated in terms of alternating multilinear maps, a fact that will be useful to deal with differential forms. Corollary 3.35. Define the product ·: Altn(E; R) × F →Altn(E; F) as follows: For all ω ∈Altn(E; R) and all f ∈F, (ω · f)(u1, . . . , un) = ω(u1, . . . , un)f, for all u1, . . . , un ∈E. Then for every ω ∈(Vn(E∗)) ⊗F of the form ω = u∗ 1 ∧· · · ∧u∗ n ⊗f, we have µF(u∗ 1 ∧· · · ∧u∗ n ⊗f) = µF(u∗ 1 ∧· · · ∧u∗ n) · f. Then Proposition 3.34 yields the following result. Proposition 3.36. If (e1, . . . , ep) is any basis of E, then every element ω ∈Altn(E; F) can be written in a unique way as ω = X I e∗ I · fI, fI ∈F, where the e∗ I are defined as in Section 3.2. 134 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS 3.10 Problems Problem 3.1. Complete the induction argument used in the proof of Proposition 3.1 (2). Problem 3.2. Prove Proposition 3.2. Problem 3.3. Prove Proposition 3.7. Problem 3.4. Show that the pairing given by (∗) in Section 3.3 is nondegenerate. Problem 3.5. Let Ia be the two-sided ideal generated by all tensors of the form u⊗u ∈V ⊗2. Prove that m ^ (V ) ∼ = V ⊗m/(Ia ∩V ⊗m). Problem 3.6. Complete the induction proof of Proposition 3.12. Problem 3.7. Prove the following lemma: If V is a vector space with dim(V ) ≤3, then α ∧α = 0 whenever α ∈V(V ). Problem 3.8. Prove Proposition 3.13. Problem 3.9. Given two graded algebras E and F, define E b ⊗F to be the vector space E ⊗F, but with a skew-commutative multiplication given by (a ⊗b) ∧(c ⊗d) = (−1)deg(b)deg(c)(ac) ⊗(bd), where a ∈Em, b ∈F p, c ∈En, d ∈F q. Show that ^ (E ⊕F) ∼ = ^ (E) b ⊗ ^ (F). Problem 3.10. If ⟨−, −⟩denotes the inner product on V , recall that we defined an inner product on Vk V , also denoted ⟨−, −⟩, by setting ⟨u1 ∧· · · ∧uk, v1 ∧· · · ∧vk⟩= det(⟨ui, vj⟩), for all ui, vi ∈V , and extending ⟨−, −⟩by bilinearity. Show that if (e1, . . . , en) is an orthonormal basis of V , then the basis of Vk V consisting of the eI (where I = {i1, . . . , ik}, with 1 ≤i1 < · · · < ik ≤n) is also an orthonormal basis of Vk V . Problem 3.11. Show that (u∗∧v∗) ⌟z = u∗⌟(v∗⌟z), whenever u∗∈Vk E∗, v∗∈Vp−k E∗, and z ∈Vp+q E. Problem 3.12. Prove Statement (3) of Proposition 3.18. 3.10. PROBLEMS 135 Problem 3.13. Prove Proposition 3.19. Also prove the identity u∗⌟(x ∧y) = (−1)s(u∗⌟x) ∧y + x ∧(u∗⌟y), where u∗∈E∗, x ∈Vq+1−s E, and y ∈Vs E. Problem 3.14. Use the Grassmann-Pl¨ ucker’s equations prove that if dim(E) = n, then every tensor in Vn−1(E) is decomposable. Problem 3.15. Recall that the map µF : n ^ (E∗) ! ⊗F − →Altn(E; F) is defined on generators by µF((v∗ 1 ∧· · · ∧v∗ n) ⊗f)(u1, . . . , un) = (det(v∗ j(ui))f, with v∗ 1, . . . , v∗ n ∈E∗, u1, . . . , un ∈E, and f ∈F. Given any three vector spaces, F, G, H, and any bilinear map Φ: F × G →H, for all ω ∈(Vn(E∗)) ⊗F and all η ∈(Vn(E∗)) ⊗G prove that µH(ω ∧Φ η) = µF(ω) ∧Φ µG(η). 136 CHAPTER 3. EXTERIOR TENSOR POWERS AND EXTERIOR ALGEBRAS Chapter 4 Differential Forms The theory of differential forms is one of the main tools in geometry and topology. This theory has a surprisingly large range of applications, and it also provides a relatively easy access to more advanced theories such as cohomology. For all these reasons, it is really an indispensable theory, and anyone with more than a passable interest in geometry should be familiar with it. The theory of differential forms was initiated by Poincar´ e and further elaborated by ´ Elie Cartan at the end of the nineteenth century. Differential forms have two main roles: (1) Describe various systems of partial differential equations on manifolds. (2) To define various geometric invariants reflecting the global structure of manifolds or bundles. Such invariants are obtained by integrating certain differential forms. As we will see shortly, as soon as one tries to define integration on higher-dimensional objects, such as manifolds, one realizes that it is not functions that are integrated, but instead differential forms. Furthermore, as by magic, the algebra of differential forms handles changes of variables automatically and yields a neat form of “Stokes formula.” We begin with differential forms defined on an open subset U of Rn. A p-form is any smooth function ω: U →Vp(Rn)∗taking as values alternating tensors in the exterior power Vp(Rn)∗. The set of all p-forms on U is a vector space denoted Ap(U). The vector space A∗(U) = L p≥0 Ap(U) is the set of differential forms on U. Proposition 3.14 shows that for every finite-dimensional vector space E, there are iso-morphisms µ: n ^ (E∗) − →Altn(E; R), and these yield a canonical isomorphism of algebras µ: V(E∗) − →Alt(E), where Alt(E) = M n≥0 Altn(E; R), 137 138 CHAPTER 4. DIFFERENTIAL FORMS and where Altn(E; R) is the vector space of real valued alternating multilinear maps on En. In view of these isomorphisms, we will identify ω and µ(ω) for any ω ∈Vn(E∗), and we will write ω(u1, . . . , un) as an abbreviation for µ(ω)(u1, . . . , un). Because Alt(Rn) is an algebra under the wedge product, differential forms also have a wedge product, and thus A∗(U) is an algebra with the wedge product ∧on forms. However, the power of differential forms stems from the exterior differential d: Ap(U) →Ap+1(U), which is a skew-symmetric version of the usual differentiation operator. In Section 4.1 we prove some basic properties of the wedge product and of the exterior differential d. One of the most crucial properties of d is that the composition Ap(U) d − →Ap+1(U) d − →Ap+2(U) is identically zero; that is d ◦d = 0, which is an abbreviation for dp+1 ◦dp = 0. We explain that in R3, the notions of gradient, curl, and divergence, arise naturally from the exterior differential d. When is there a smooth field (P, Q, R) (in R3) whose curl is given by a prescribed smooth field (A, B, C)? Equivalently, when is there a 1-form ω = Pdx + Qdy + Rdz such that dω = η = Ady ∧dz + Bdz ∧dx + Cdx ∧dy? Because d ◦d = 0, it is necessary that dη = 0; that is, (A, B, C) must have zero divergence. However, this condition is not sufficient in general; it depends on the topology of U. More generally, we say that a differential p-form ω is closed if dω = 0 and exact if ω = dη for some (p −1)-form η. Since d ◦d = 0, every exact form is closed, but the converse is false in general. The purpose of de Rham cohomology is to measure the failure of a differential forms to be exact in terms of certain abelian groups (in fact, algebras). The diagram (a cochain complex) A0(U) d − →A1(U) − →· · · − →Ap−1(U) d − →Ap(U) d − →Ap+1(U) − →· · · is called the de Rham complex of U. For every p ≥0, let Zp(U) = {ω ∈Ap(U) | dω = 0} = Ker d: Ap(U) − →Ap+1(U) be the vector space of closed p-forms, also called p-cocycles, and for every p ≥1, let Bp(U) = {ω ∈Ap(U) | ∃η ∈Ap−1(U), ω = dη} = Im d: Ap−1(U) − →Ap(U) 139 be the vector space of exact p-forms, also called p-coboundaries. Set B0(U) = (0). Forms in Ap(U) are also called p-cochains. As Bp(U) ⊆Zp(U) for every p ≥0, we define the pth de Rham cohomology group of U as the quotient space Hp DR(U) = Zp(U)/Bp(U); The real vector space H• DR(U) = L p≥0 Hp DR(U) is called the de Rham cohomology algebra of U. The de Rham cohomology groups will be generalized to smooth manidolds in Section 4.3. They are important invariants of a manifold (which means that diffeomorphic manifolds have isomorphic cohomology groups). In Section 4.2 we consider the behavior of differential forms under smooth maps ϕ: U → V . Any such map induces a map ϕ∗: Ap(V ) →Ap(U) on differential p-forms called a pullback (notice the reversal of U and V ). Note that ϕ need not be a diffeomorphism, which is one of the technical advantages of forms over vector fields. We state various properties of the behavior of wedge products and the exterior differential d under pullback. In particular, dϕ∗(ω) = ϕ∗(dω). This property shows that a map ϕ: U →V induces a map H• DR(ϕ): H• DR(V ) →H• DR(U) on cohomology. We state a fundamental result known as the Poincar´ e lemma, which says that the de Rham cohomology of a star-shaped open subset of Rn vanishes for p ≥1, and that H0(U) = R. Thus every closed p-form on such a domain is exact (p ≥1). In Section 4.3 we generalize differential forms to smooth manifolds. Having defined differential forms on open subsets of Rn, this is not a difficult task. Technically, the set Ak(M) of smooth differential k-forms on M is the set of smooth sections Γ(M, Vk T ∗M) of the bundle Vk T ∗M, and the set A∗(M) of all smooth differential forms on M is the set of smooth sections Γ(M, V T ∗M) of the bundle V T ∗M. These definitions are quite abstract, so we explain how p-forms are defined locally in terms of charts. Wedge products, pullbacks, and the exterior differential d: Ak(M) →Ak+1(M). are defined. As in the case of open subsets of Rn, we have d ◦d = 0, and d commutes with pullbacks. As a consequence, we have the de Rham complex A0(M) d − →A1(M) − →· · · − →Ak−1(M) d − →Ak(M) d − →Ak+1(M) − →· · · , 140 CHAPTER 4. DIFFERENTIAL FORMS and we can define the cohomology groups Hk DR(M) and the graded cohomology algebra H• DR(M). Another important property of the exterior differential d is that it is a local operator, which means that the value of dω at p only depends of the values of ω near p. As a consequence, we obtain a characterization of the the operator d; see Theorem 4.14. Smooth differential forms can also be defined in terms of alternating C∞(M)-multilinear maps on smooth vector fields. This approach also yields a global formula for the exterior derivative dω(X1, . . . , Xk+1) of a k-form ω applied to k + 1 vector fields X1, . . . , Xk+1. This formula is not very useful for computing dω at a given point p since it requires vector fields as input, but it is quite useful in theoretical investigations. Let ω ∈Ak(M) be any smooth k-form on M. Then ω induces an alternating multilinear map ω: X(M) × · · · × X(M) | {z } k − →C∞(M) as follows: for any k smooth vector fields X1, . . . , Xk ∈X(M), ω(X1, . . . , Xk)(p) = ωp(X1(p), . . . , Xk(p)). This map is obviously alternating and R-linear, but it is also C∞(M)-linear. Let M be a smooth manifold. It is shown in Proposition 4.15 that for every k ≥0, there is an isomorphism between the space of k-forms Ak(M) and the space Altk C∞(M)(X(M)) of alternating C∞(M)-multilinear maps on smooth vector fields. That is, Ak(M) ∼ = Altk C∞(M)(X(M)), viewed as C∞(M)-modules. Then Proposition 4.16 gives an expression for dω(X1, . . . , Xk+1) (where X1, . . . , Xk+1 are vector fields) in terms of the Xi and some of their Lie brackets. Section 4.4 is a technical section devoted to Lie derivatives of differential forms. We prove various properties about the interaction of Lie derivatives with the wedge operator and the exterior differential d. In particular, we prove Cartan’s formula, which expresses the Lie derivative of a differential form in terms of d and an operator i(X): Ak(M) →Ak−1(M) called an insertion operator, where X is a vector field. We also generalize Lie derivatives to tensors. In Section 4.5 we show how differential forms can be generalized so that they take values in any vector space F, rather than just R. Vector-valued differential forms are needed in the theory of Lie groups and to define connections and curvature on vector bundles; see Chapter 11. For simplicity, assume that U is an open subset of Rn. Then it is natural to define dif-ferential forms with values in F as smooth maps ω: U →Altp(Rn; F), where Altp(Rn; F) 4.1. DIFFERENTIAL FORMS ON RN AND DE RHAM COHOMOLOGY 141 denotes the vector space of alternating multilinear linear maps with values in F. The vec-tor space of all p-forms on U with values in F is denoted Ap(U; F), and the vector space A∗(U; F) = L p≥0 Ap(U; F) is the set of differential forms on U with values in F. There is no difficulty in defining the exterior differential d: Ap(U; F) →Ap+1(U; F), and it can be shown that d ◦d = 0. The pullback of a form in Ap(V ; F) along a smooth map ϕ: U →V is defined as before. The major difference is that there is no longer an obvious notion of wedge product. To define such an operation we need a bilinear form Φ: F ×G →H, where F, G, H are some vector spaces. Then we can define a wedge product ∧Φ : Ap(U; F) × Aq(U; G) →Ap+q(U; H). Such a wedge product is not associative in general, and not much can be said about it unless Φ has some additional properties. In general, unlike the case where F = R, there is no nice formula for d(ω∧Φη), unless F, G, H are finite-dimensional. The case where F = H = G = g where g is a Lie algebra and Φ(a, b) = [a, b] is of particular interest. The generalization of vector-valued differential forms to manifolds is no problem, except that some results involving the wedge product fail for the same reason that they fail in the case of forms on open subsets of Rn. In Section 4.6 we discuss left-invariant one-forms on a Lie group G. They form a space isomorphic to the dual g∗of the Lie algebra g of G. We prove the Maurer–Cartan equations in two versions, the second one involving a g-valued one-form ωMC called the Maurer–Cartan form. Our main goal is to define differential forms on manifolds, but we begin with differential forms on open subsets of Rn in order to build up intuition. 4.1 Differential Forms on Subsets of Rn and de Rham Cohomology Differential forms are smooth functions on open subsets U of Rn, taking as values alternating tensors in some exterior power Vp(Rn)∗. Definition 4.1. Given any open subset U of Rn, a smooth differential p-form on U, for short a p-form on U, is any smooth function ω: U →Vp(Rn)∗. The vector space of all p-forms on U is denoted Ap(U). The vector space A∗(U) = L p≥0 Ap(U) is the set of differential forms on U. Observe that A0(U) = C∞(U, R), the vector space of smooth functions on U, and A1(U) = C∞(U, (Rn)∗), the set of smooth functions from U to the set of linear forms on Rn. Also, Ap(U) = (0) for p > n. 142 CHAPTER 4. DIFFERENTIAL FORMS Remark: The space A∗(U) is also denoted A•(U). Other authors use Ωp(U) instead of Ap(U), but we prefer to reserve Ωp for holomorphic forms. Recall from Sections 3.3 and 3.4, in particular Proposition 3.14, that for every finite-dimensional vector space E, the isomorphisms µ: Vn(E∗) − →Altn(E; R) induced by the linear extensions of the maps given by µ(v∗ 1 ∧· · · ∧v∗ n)(u1, . . . , un) = v∗ 1(u1) · · · v∗ 1(un) . . . ... . . . v∗ n(u1) · · · v∗ n(un) = det(v∗ j(ui)) yield a canonical isomorphism of algebras µ: V(E∗) − →Alt(E), where Alt(E) = M n≥0 Altn(E; R), and where Altn(E; R) is the vector space of real valued alternating multilinear maps on En. Recall that multiplication on alternating multilinear forms is defined such that, for f ∈Altm(E; K) and g ∈Altn(E; K), (f ∧g)(u1, . . . , um+n) = X σ∈shuffle(m,n) sgn(σ) f(uσ(1), . . . , uσ(m))g(uσ(m+1), . . . , uσ(m+n)), (∗∗) where shuffle(m, n) consists of all (m, n)-“shuffles;” that is, permutations σ of {1, . . . m + n} such that σ(1) < · · · < σ(m) and σ(m + 1) < · · · < σ(m + n). The isomorphism µ has the property that µ(ω ∧η) = µ(ω) ∧µ(η), ω, η ∈ ^ (E∗), where the wedge operation on the left is the wedge on the exterior algebra V(E∗), and the wedge on the right is the multiplication on Alt(E) defined in (∗∗). In view of these isomorphisms, we will identify ω and µ(ω) for any ω ∈Vn(E∗), and we will write ω(u1, . . . , un) as an abbreviation for µ(ω)(u1, . . . , un). Because Alt(Rn) is an algebra under the wedge product, differential forms also have a wedge product. However, the power of differential forms stems from the exterior differential d, which is a skew-symmetric version of the usual differentiation operator. Recall from Section 3.2 that if (e1, . . . , en) is any basis of Rn and (e∗ 1, . . . , e∗ n) is its dual basis, then the alternating tensors e∗ I = e∗ i1 ∧· · · ∧e∗ ip form basis of Vp(Rn)∗, where I = {i1, . . . , ip} ⊆{1, . . . , n}, with i1 < · · · < ip. Thus, with respect to the basis (e1, . . . , en), every p-form ω can be uniquely written ω(x) = X I fI(x) e∗ i1 ∧· · · ∧e∗ ip = X I fI(x) e∗ I x ∈U, 4.1. DIFFERENTIAL FORMS ON RN AND DE RHAM COHOMOLOGY 143 where each fI is a smooth function on U. For example, if U = R2 −{0}, then ω(x, y) = −y x2 + y2 e∗ 1 + x x2 + y2 e∗ 2 is a 1-form on U (with e1 = (1, 0) and e2 = (0, 1)). We often write ωx instead of ω(x). Now, not only is A∗(U) a vector space, it is also an algebra. Definition 4.2. The wedge product on A∗(U) is defined as follows: For all p, q ≥0, the wedge product ∧: Ap(U) × Aq(U) →Ap+q(U) is given by (ω ∧η)x = ωx ∧ηx, x ∈U. For example, if ω and η are one-forms, then (ω ∧η)x(u, v) = ωx(u)ηx(v) −ωx(v)ηx(u). In particular, if U ⊆R3 and ωx = a1e∗ 1 + a3e∗ 3 and ηx = b1e∗ 1 + b2e∗ 2, for u = (u1, u2, u3) ∈R3 and v = (v1, v2, v3) ∈R3, the preceding line implies ωx(u)ηx(v) −ωx(v)ηx(u) = a1e∗ 1(u) + a3e∗ 3(u) b1e∗ 1(v) + b2e∗ 2(v)  − a1e∗ 1(v) + a3e∗ 3(v) b1e∗ 1(u) + b2e∗ 2(u)  = (a1u1 + a3u3)(b1v1 + b2v2) −(a1v1 + a3v3)(b1u1 + b2u2) = a1b2(u1v2 −v1u2) −a3b1(u1v3 −v1u3) −a3b2(u2v3 −u3v2) = a1b2 e∗ 1(u) e∗ 1(v) e∗ 2(u) e∗ 2(v) −a3b1 e∗ 1(u) e∗ 1(v) e∗ 3(u) e∗ 3(v) −a3b2 e∗ 2(u) e∗ 2(v) e∗ 3(u) e∗ 3(v) = (a1b2e∗ 1 ∧e∗ 2 −a3b1e∗ 1 ∧e∗ 3 −a3b2e∗ 2 ∧e∗ 3)(u, v) = (a1b1e∗ 1 ∧e∗ 1 + a1b2e∗ 1 ∧e∗ 2 + a3b1e∗ 3 ∧e∗ 1 + a3b2e∗ 3 ∧e∗ 2)(u, v) = (a1e∗ 1 + a3e∗ 3) ∧(b1e∗ 1 + b2e∗ 2)  (u, v) = (ω ∧η)x(u, v), since e∗ i ∧e∗ i = 0 and e∗ i ∧e∗ j = −e∗ j ∧e∗ i for all 1 ≤i < j ≤3. For f ∈A0(U) = C∞(U, R) and ω ∈Ap(U), we have f ∧ω = fω. Thus, the algebra A∗(U) is also a C∞(U, R)-module, Proposition 3.12 immediately yields Proposition 4.1. For all forms ω ∈Ap(U) and η ∈Aq(U), we have η ∧ω = (−1)pqω ∧η. 144 CHAPTER 4. DIFFERENTIAL FORMS We now come to the crucial operation of exterior differentiation. First recall that if f : U →V is a smooth function from U ⊆Rn to a (finite-dimensional) normed vector space V , the derivative f ′ : U →Hom(Rn, V ) of f (also denoted Df) is a function with domain U, with f ′(x) a linear map in Hom(Rn, V ) for every x ∈U, such that if (e1, . . . , en) is the canonical basis of Rn, (u1, . . . , um) is a basis of V , and if f(x) = f1(x)u1 + · · · + fm(x)um, then f ′(x)(y1e1 + · · · + ynen) = m X i=1 n X j=1 ∂fi ∂xj (x) yj ! ui. The m × n matrix  ∂fi ∂xj (x)  is the Jacobian matrix of f at x, and if we write z1u1 + · · · + zmum = f ′(x)(y1e1 + · · · + ynen), then in matrix form, we have    z1 . . . zm   =    ∂f1 ∂x1(x) · · · ∂f1 ∂xn(x) . . . ... . . . ∂fm ∂x1 (x) · · · ∂fm ∂xn (x)       y1 . . . yn   . We also write f ′ x(u) for f ′(x)(u). Observe that since a p-form is a smooth map ω: U → Vp(Rn)∗, its derivative is a map ω′ : U →Hom  Rn, p ^ (Rn)∗  such that ω′ x is a linear map from Rn to Vp(Rn)∗for every x ∈U. By the isomorphism Vp(Rn)∗∼ = Altp(Rn; R), we can view ω′ x as a linear map ωx : Rn →Altp(Rn; R), or equiva-lently as a multilinear form ω′ x : (Rn)p+1 →R which is alternating in its last p arguments. The exterior derivative (dω)x is obtained by making ω′ x into an alternating map in all of its p + 1 arguments. To make things more concrete, let us pick a basis (e1, . . . , en) of Rn, so that the n p  tensors e∗ I form a basis of Vp(Rn)∗, where I is any subset I = {i1, . . . , ip} ⊆{1, . . . , n} such that i1 < · · · < ip. Then every p-form ω can be uniquely written as ωx = X I fI(x) e∗ I x ∈U, where each fI is a smooth function on U, and for any v = (v1, . . . , vn) ∈Rn, ω′ x(v) = X I f ′ I(x)(v) e∗ I = X I n X j=1 ∂fI ∂xj (x) vj e∗ I = X I (grad(fI)x · v)e∗ I, 4.1. DIFFERENTIAL FORMS ON RN AND DE RHAM COHOMOLOGY 145 where · is the standard Euclidean inner product. Remark: Observe that ω′ x is given by the n p  × n Jacobian matrix ∂fI ∂xj (x)  and that the product of the Ith row of the above matrix by v ∂fI ∂x1 (x) · · · ∂fI ∂xn (x)     v1 . . . vn    gives the coefficient grad(fI)x · v of e∗ I. Definition 4.3. For every p ≥0, the exterior differential d: Ap(U) →Ap+1(U) is given by (dω)x(u1, . . . , up+1) = p+1 X i=1 (−1)i−1ω′ x(ui)(u1, . . . , b ui, . . . , up+1), for all ω ∈Ap(U), all x ∈U, and all u1, . . . , up+1 ∈Rn, where the hat over the argument ui means that it should be omitted. In terms of a basis (e1, . . . , en) of Rn, if ωx = P I fI(x) e∗ I, then (dω)x(u1, . . . , up+1) = p+1 X i=1 (−1)i−1 X I f ′ I(x)(ui) e∗ I(u1, . . . , b ui, . . . , up+1) = p+1 X i=1 (−1)i−1 X I (grad(fI)x · ui)e∗ I(u1, . . . , b ui, . . . , up+1). One should check that (dω)x is indeed alternating, but this is easy. If necessary to avoid confusion, we write dp : Ap(U) →Ap+1(U) instead of d: Ap(U) →Ap+1(U). Remark: Definition 4.3 is the definition adopted by Cartan [21, 22]1 and Madsen and Tornehave . Some authors use a different approach often using Propositions 4.2 and 4.3 as a starting point, but we find the approach using Definition 4.3 more direct. Furthermore, this approach extends immediately to the case of vector-valued forms. For any smooth function, f ∈A0(U) = C∞(U, R), we get d fx(u) = f ′ x(u). 1We warn the reader that a few typos have crept up in the English translation, Cartan , of the orginal version Cartan . 146 CHAPTER 4. DIFFERENTIAL FORMS Therefore, for smooth functions, the exterior differential d f coincides with the usual derivative f ′ (we identify V1(Rn)∗and (Rn)∗). For any 1-form ω ∈A1(U), we have dωx(u, v) = ω′ x(u)(v) −ω′ x(v)(u). It follows that the map (u, v) 7→ω′ x(u)(v) is symmetric iffdω = 0. For a concrete example of exterior differentiation, consider ω(x,y) = −y x2 + y2 e∗ 1 + x x2 + y2 e∗ 2 = f1(x, y)e∗ 1 + f2(x, y)e∗ 2. Since grad(f1)⊤ (x,y) =  2xy (x2 + y2)2 y2 −x2 (x2 + y2)2  grad(f2)⊤ (x,y) =  y2 −x2 (x2 + y2)2 −2xy (x2 + y2)2  , if we write u1 = u11 u12  and u2 = u21 u22  , then we have ω′ (x,y)(u1)(u2) = (grad(f1)(x,y) · u1)e∗ 1(u2) + (grad(f2)(x,y) · u1)e∗ 2(u2) =  2xy (x2 + y2)2 y2 −x2 (x2 + y2)2  u11 u12  e∗ 1 u21 u22  +  y2 −x2 (x2 + y2)2 −2xy (x2 + y2)2  u11 u12  e∗ 2 u21 u22  = 2xy(u11u21 −u12u22) + (y2 −x2)(u12u21 + u11u22) (x2 + y2)2 . A similar computation shows that ω′ (x,y)(u2)(u1) = 2xy(u11u21 −u12u22) + (y2 −x2)(u12u21 + u11u22) (x2 + y2)2 = ω′ (x,y)(u1)(u2), and so dω(x,y)(u1, u2) = ω′ (x,y)(u1)(u2) −ω′ (x,y)(u2)(u1) = 0. Therefore dω(x,y) = 0 for all (x, y) ∈U, that is, dω = 0. The following observation is quite trivial but it will simplify notation: On Rn, we have the projection function pri : Rn →R with pri(u1, . . . , un) = ui. Note that pri = e∗ i , where 4.1. DIFFERENTIAL FORMS ON RN AND DE RHAM COHOMOLOGY 147 (e1, . . . , en) is the canonical basis of Rn. Let xi : U →R be the restriction of pri to U. Then note that x′ i is the constant map given by x′ i(x) = pri, x ∈U. It follows that dxi = x′ i is the constant function with value pri = e∗ i . Now, since every p-form ω can be uniquely expressed as ωx = X I fI(x) e∗ i1 ∧· · · ∧e∗ ip = X I fI(x)e∗ I, x ∈U, using Definition 4.2, we see immediately that ω can be uniquely written in the form ω = X I fI(x) dxi1 ∧· · · ∧dxip, (∗1) where the fI are smooth functions on U. Observe that for f ∈A0(U) = C∞(U, R), we have d fx = n X i=1 ∂f ∂xi (x) e∗ i and d f = n X i=1 ∂f ∂xi dxi. Proposition 4.2. For every p form ω ∈Ap(U) with ω = fdxi1 ∧· · · ∧dxip, we have dω = d f ∧dxi1 ∧· · · ∧dxip. Proof. Recall that ωx = fe∗ i1 ∧· · · ∧e∗ ip = fe∗ I, so ω′ x(u) = f ′ x(u)e∗ I = d fx(u)e∗ I, and by Definition 4.3, we get dωx(u1, . . . , up+1) = p+1 X i=1 (−1)i−1d fx(ui)e∗ I(u1, . . . , b ui, . . . , up+1) = (d fx ∧e∗ I)(u1, . . . , up+1), where the last equation is an instance of the equation stated just before Proposition 3.14. In practice we use Proposition 4.2 to compute dω. For example, if we take the previous example of ω = −y x2 + y2dx + x x2 + y2dy, 148 CHAPTER 4. DIFFERENTIAL FORMS Proposition 4.2 implies that dω = d  −y x2 + y2  ∧dx + d  x x2 + y2  ∧dy =  2xy (x2 + y2)2dx + y2 −x2 (x2 + y2)2dy  ∧dx +  y2 −x2 (x2 + y2)2dx − 2xy (x2 + y2)2dy  ∧dy = y2 −x2 (x2 + y2)2dy ∧dx + y2 −x2 (x2 + y2)2dx ∧dy = 0. We can now prove Proposition 4.3. For all ω ∈Ap(U) and all η ∈Aq(U), d(ω ∧η) = dω ∧η + (−1)pω ∧dη. Proof. In view of the unique representation (∗), it is enough to prove the proposition when ω = fe∗ I and η = ge∗ J. In this case, as ω ∧η = fg e∗ I ∧e∗ J, by Proposition 4.2 we have d(ω ∧η) = d(fg) ∧e∗ I ∧e∗ J = ((d f)g + f(dg)) ∧e∗ I ∧e∗ J = (d f)g ∧e∗ I ∧e∗ J + f(dg) ∧e∗ I ∧e∗ J = d f ∧e∗ I ∧ge∗ J + (−1)pfe∗ I ∧dg ∧e∗ J = dω ∧η + (−1)pω ∧dη since by Proposition 4.2, dω = d f ∧e∗ I and dη = gJ ∧e∗ J. We say that d is an anti-derivation of degree −1. Finally, here is the crucial and almost magical property of d. Proposition 4.4. For every p ≥0, the composition Ap(U) d − →Ap+1(U) d − →Ap+2(U) is identically zero; that is d ◦d = 0, which is an abbreviation for dp+1 ◦dp = 0. Proof. It is enough to prove the proposition when ω = fe∗ I. We have dωx = d fx ∧e∗ I = ∂f ∂x1 (x) e∗ 1 ∧e∗ I + · · · + ∂f ∂xn (x) e∗ n ∧e∗ I. As e∗ i ∧e∗ j = −e∗ j ∧e∗ i and e∗ i ∧e∗ i = 0, we get (d ◦d)ω = n X i,j=1 ∂2f ∂xi∂xj (x) e∗ i ∧e∗ j ∧e∗ I = X i<j  ∂2f ∂xi∂xj (x) − ∂2f ∂xj∂xi (x)  e∗ i ∧e∗ j ∧e∗ I = 0, since partial derivatives commute (as f is smooth). 4.1. DIFFERENTIAL FORMS ON RN AND DE RHAM COHOMOLOGY 149 It turns out that Propositions 4.3 and 4.4 together with the fact that d coincides with the derivative on A0(U) characterize the differential d. Theorem 4.5. There is a unique linear map d: A∗(U) →A∗(U) with d = (dp) and dp : Ap(U) →Ap+1(U) for every p ≥0, such that (1) d f = f ′, for every f ∈A0(U) = C∞(U, R). (2) d ◦d = 0. (3) For every ω ∈Ap(U) and every η ∈Aq(U), d(ω ∧η) = dω ∧η + (−1)pω ∧dη. Proof. Existence has already been shown, so we only have to prove uniqueness. Let δ be another linear map satisfying Conditions (1)–(3). By (1), d f = δf = f ′ if f ∈A0(U). In particular, this hold when f = xi, with xi : U →R the restriction of pri to U. In this case, we know that δxi = e∗ i , the constant function e∗ i = pri. By (2), δe∗ i = 0. Using (3), we get δe∗ I = 0 for every nonempty subset I ⊆{1, . . . , n}. If ω = fe∗ I, by (3), we get δω = δf ∧e∗ I + f ∧δe∗ I = δf ∧e∗ I = d f ∧e∗ I = dω. Finally, since every differential form is a linear combination of special forms fIe∗ I, we conclude that δ = d. Propositions 4.2, 4.3 and 4.4 can be summarized by saying that A∗(U) together with the product ∧and the differential d is a differential graded algebra. As A∗(U) = L p≥0 Ap(U) and dp : Ap(U) →Ap+1(U), we can view d = (dp) as a linear map d: A∗(U) →A∗(U) such that d ◦d = 0. Let us consider one more example. Assume n = 3 and consider any function f ∈A0(U). We have d f = ∂f ∂x dx + ∂f ∂y dy + ∂f ∂z dz, and the vector ∂f ∂x, ∂f ∂y , ∂f ∂z  is the gradient of f. Next let ω = Pdx + Qdy + Rdz 150 CHAPTER 4. DIFFERENTIAL FORMS be a 1-form on some open U ⊆R3. An easy calculation yields dω = dP ∧dx + dQ ∧dy + dR ∧dz = ∂P ∂x dx + ∂P ∂y dy + ∂P ∂z dz  ∧dx + ∂Q ∂x dx + ∂Q ∂y dy + ∂Q ∂z dz  ∧dy + ∂R ∂x dx + ∂R ∂y dy + ∂R ∂z dz  ∧dz = ∂P ∂y dy ∧dx + ∂P ∂z dz ∧dx + ∂Q ∂x dx ∧dy + ∂Q ∂z dz ∧dy + ∂R ∂x dx ∧dz + ∂R ∂y dy ∧dz = ∂R ∂y −∂Q ∂z  dy ∧dz + ∂P ∂z −∂R ∂x  dz ∧dx + ∂Q ∂x −∂P ∂y  dx ∧dy. The vector field given by ∂R ∂y −∂Q ∂z , ∂P ∂z −∂R ∂x , ∂Q ∂x −∂P ∂y  is the curl of the vector field given by (P, Q, R). Now if η = Ady ∧dz + Bdz ∧dx + Cdx ∧dy is a 2-form on R3, we get dη = dA ∧dy ∧dz + dB ∧dz ∧dx + dC ∧dx ∧dy = ∂A ∂x dx + ∂A ∂y dy + ∂A ∂z dz  ∧dy ∧dz + ∂B ∂x dx + ∂B ∂y dy + ∂B ∂z dz  ∧dz ∧dx + ∂C ∂x dx + ∂C ∂y dy + ∂C ∂z dz  ∧dx ∧dy = ∂A ∂x dx ∧dy ∧dz + ∂B ∂y dy ∧dz ∧dx + ∂C ∂z dz ∧dx ∧dy = ∂A ∂x + ∂B ∂y + ∂C ∂z  dx ∧dy ∧dz. The real number ∂A ∂x + ∂B ∂y + ∂C ∂z is called the divergence of the vector field (A, B, C). When is there a smooth field (P, Q, R) whose curl is given by a prescribed smooth field (A, B, C)? Equivalently, when is there a 1-form ω = Pdx + Qdy + Rdz such that dω = η = Ady ∧dz + Bdz ∧dx + Cdx ∧dy? 4.1. DIFFERENTIAL FORMS ON RN AND DE RHAM COHOMOLOGY 151 By Proposition 4.4 it is necessary that dη = 0; that is, (A, B, C) has zero divergence. However, this condition is not sufficient in general; it depends on the topology of U. If U is star-like, Poincar´ e’s Lemma (to be considered shortly) says that this condition is sufficient. Definition 4.4. The diagram A0(U) d − →A1(U) − →· · · − →Ap−1(U) d − →Ap(U) d − →Ap+1(U) − →· · · is called the de Rham complex of U. It is a cochain complex. Definition 4.5. A differential form ω is closed iffdω = 0; exact iffω = dη for some differential form η. For every p ≥0, let Zp(U) = {ω ∈Ap(U) | dω = 0} = Ker d: Ap(U) − →Ap+1(U) be the vector space of closed p-forms, also called p-cocycles, and for every p ≥1, let Bp(U) = {ω ∈Ap(U) | ∃η ∈Ap−1(U), ω = dη} = Im d: Ap−1(U) − →Ap(U) be the vector space of exact p-forms, also called p-coboundaries. Set B0(U) = (0). Forms in Ap(U) are also called p-cochains. As Bp(U) ⊆Zp(U) (by Proposition 4.4), for every p ≥0, we define the pth de Rham cohomology group of U as the quotient space Hp DR(U) = Zp(U)/Bp(U); This is an abelian group under addition of cosets. An element of Hp DR(U) is called a co-homology class and is denoted [ω], where ω ∈Zp(U) is a cocycle. The real vector space H• DR(U) = L p≥0 Hp DR(U) is called the de Rham cohomology algebra of U. We also we define the vector spaces Z∗(U) and B∗(U) by Z∗(U) = M p≥0 Zp(U) and B∗(U) = M p≥0 Bp(U). We often drop the subscript DR and write Hp(U) for Hp DR(U) (resp. H•(U) for H• DR(U)), when no confusion arises. Proposition 4.4 shows that every exact form is closed, but the converse is false in general. Measuring the extent to which closed forms are not exact is the object of de Rham cohomology. For example, if we consider the form ω(x,y) = −y x2 + y2 dx + x x2 + y2 dy, on U = R2 −{0}, we have dω = 0. Yet, it is not hard to show (using integration, see Madsen and Tornehave , Chapter 1) that there is no smooth function f on U such that d f = ω. Thus, ω is a closed form which is not exact. This is because U is punctured. 152 CHAPTER 4. DIFFERENTIAL FORMS Observe that H0(U) = Z0(U) = {f ∈C∞(U, R) | d f = 0}; that is, H0(U) is the space of locally constant functions on U, equivalently, the space of functions that are constant on the connected components of U. Thus, the cardinality of H0(U) gives the number of connected components of U. For a large class of open sets (for example, open sets that can be covered by finitely many convex sets), the cohomology groups Hp(U) are finite dimensional. Now, A∗(U) is a graded algebra with multiplication ∧. Proposition 4.6. The vector space Z∗(U) is a subalgebra of A∗(U), and B∗(U) is an ideal in Z∗(U). Proof. The vector space Z∗(U) is a subalgebra of A∗(U), because d(ω ∧η) = dω ∧η + (−1)pω ∧dη, so dω = 0 and dη = 0 implies d(ω ∧η) = 0. The vector space B∗(U) is an ideal in Z∗(U), because if ω = dη and dτ = 0, then d(η ∧τ) = dη ∧τ + (−1)p−1η ∧dτ = ω ∧τ, with η ∈Ap−1(U). Therefore, H• DR = Z∗(U)/B∗(U) inherits a graded algebra structure from A∗(U). Ex-plicitly, the multiplication in H• DR is given by [ω] [η] = [ω ∧η]. We now consider the action of smooth maps ϕ: U →U ′ on differential forms in A∗(U ′). We will see that ϕ induces a map from A∗(U ′) to A∗(U) called a pull-back map. This corresponds to a change of variables. 4.2 Pull-Back of Differential Forms Recall Proposition 3.11 which states that if f : E →F is any linear map between two finite-dimensional vector spaces E and F, then µ  p ^ f ⊤ (ω)  (u1, . . . , up) = µ(ω)(f(u1), . . . , f(up)), ω ∈ p ^ F ∗, u1, . . . , up ∈E. We apply this proposition with E = Rn, F = Rm, and f = ϕ′ x (x ∈U), and get µ  p ^ (ϕ′ x)⊤ (ωϕ(x))  (u1, . . . , up) = µ(ωϕ(x))(ϕ′ x(u1), . . . , ϕ′ x(up)), ω ∈Ap(V ), ui ∈Rn. This gives us the behavior of Vp(ϕ′ x)⊤under the identification of Vp(R)∗and Altn(Rn; R) via the isomorphism µ. Consequently, denoting Vp(ϕ′ x)⊤by ϕ∗, we make the following definition: 4.2. PULL-BACK OF DIFFERENTIAL FORMS 153 Definition 4.6. Let U ⊆Rn and V ⊆Rm be two open subsets. For every smooth map ϕ: U →V , for every p ≥0, we define the map ϕ∗: Ap(V ) →Ap(U) by ϕ∗(ω)x(u1, . . . , up) = ωϕ(x)(ϕ′ x(u1), . . . , ϕ′ x(up)), for all ω ∈Ap(V ), all x ∈U, and all u1, . . . , up ∈Rn. We say that ϕ∗(ω) (for short, ϕ∗ω) is the pull-back of ω by ϕ. As ϕ is smooth, ϕ∗ω is a smooth p-form on U. The maps ϕ∗: Ap(V ) →Ap(U) induce a map also denoted ϕ∗: A∗(V ) →A∗(U). Using the chain rule we obtain the following result. Proposition 4.7. The following identities hold: id∗= id, (ψ ◦ϕ)∗= ϕ∗◦ψ∗. Here is an example of Definition 4.6. Let U = [0, 1] × [0, 1] ⊂R2 and let V = R3. Define ϕ : Q →R3 as ϕ(u, v) = (ϕ1(u, v), ϕ2(u, v), ϕ3(u, v)) = (x, y, z) where x = u + v, y = u −v, z = uv. Let w = xdy ∧dz + ydx ∧dz be a 2-form in V . Clearly ϕ′ (u,v) =   1 1 1 −1 v u  . Set u1 = u11 u12  and u2 = u21 u22  . Definition 4.6 implies that the pull back of ω into U is ϕ∗(ω)(u,v)(u1, u2) = ωϕ(u,v)(ϕ′ (u,v)(u1), ϕ′ (u,v)(u2)) = ωϕ(u,v)     1 1 1 −1 v u   u11 u12  ,   1 1 1 −1 v u   u21 u22   = ωϕ(u,v)     u11 + u12 u11 −u12 vu11 + uu12  ,   u21 + u22 u21 −u22 vu21 + uu22     = (u + v)dy ∧dz     u11 + u12 u11 −u12 vu11 + uu12  ,   u21 + u22 u21 −u22 vu21 + uu22     + (u −v)dx ∧dz     u11 + u12 u11 −u12 vu11 + uu12  ,   u21 + u22 u21 −u22 vu21 + uu22     = (u + v) u11 −u12 u21 −u22 vu11 + uu12 vu21 + uu22 + (u −v) u11 + u12 u21 + u22 vu11 + uu12 vu21 + uu22 = (u + v)(u + v)(u11u22 −u21u12) + (u −v)(u −v)(u11u22 −u21u12) 154 CHAPTER 4. DIFFERENTIAL FORMS = (u + v)(u + v) u11 u21 u12 u22 + (u −v)(u −v) u11 u21 u12 u22 = (u + v)(u + v)du ∧dv(u1, u2) + (u −v)(u −v)du ∧dv(u1, u2) = 2(u2 + v2)du ∧dv(u1, u2). As the preceding example demonstrates, Definition 4.6 is not convenient for computa-tions, so it is desirable to derive rules that yield a recursive definition of the pull-back. The first rule has to do with the constant form ω = e∗ i . Proposition 4.8. We have ϕ∗e∗ i = dϕi, with ϕi = pri ◦ϕ. Proof. We have ϕx = (ϕ1)xe1 + · · · + (ϕm)xem for all x ∈U, ϕ′ x(u) = (ϕ1)′ x(u)e1 + · · · + (ϕm)′ x(u)em, and (ϕi)′ x(u) = n X l=1 ∂ϕi ∂xl (x) ul = n X l=1 ∂ϕi ∂xl (x) e∗ l (u), so ϕ∗(e∗ i )x(u) = e∗ i (ϕ′ x(u)) = e∗ i ((ϕ1)′ x(u)e1 + · · · + (ϕm)′ x(u)em) = (ϕi)′ x(u) = n X l=1 ∂ϕi ∂xl (x) e∗ l (u) = d(ϕi)x(u), as claimed. The next proposition shows that the pull-back behaves well with respect to the wedge and the exterior derivative and provides the rest of the computational rules necessary for efficiently computing a pull-back. Proposition 4.9. Let U ⊆Rn and V ⊆Rm be two open sets and let ϕ: U →V be a smooth map. Then (i) ϕ∗(ω ∧η) = ϕ∗ω ∧ϕ∗η, for all ω ∈Ap(V ) and all η ∈Aq(V ). (ii) ϕ∗(f) = f ◦ϕ, for all f ∈A0(V ). (iii) dϕ∗(ω) = ϕ∗(dω), for all ω ∈Ap(V ); that is, the following diagram commutes for all p ≥0: Ap(V ) ϕ∗ / d  Ap(U) d  Ap+1(V ) ϕ∗/ Ap+1(U). 4.2. PULL-BACK OF DIFFERENTIAL FORMS 155 Proof. (i) (See Madsen and Tornehave , Chapter 3). For any x ∈U and any vectors u1, . . . , up+q ∈Rn (with p, q ≥1), we have ϕ∗(ω ∧η)x(u1, . . . , up+q) = (ω ∧η)ϕ(x)(ϕ′ x(u1), . . . , ϕ′ x(up+q)) = X σ∈shuffle(p,q) sgn(σ) ωϕ(x)(ϕ′ x(uσ(1)), . . . , ϕ′ x(uσ(p))) ηϕ(x)(ϕ′ x(uσ(p+1)), . . . , ϕ′ x(uσ(p+q))) = X σ∈shuffle(p,q) sgn(σ) ϕ∗(ω)x(uσ(1), . . . , uσ(p)) ϕ∗(η)x(uσ(p+1), . . . , uσ(p+q)) = (ϕ∗(ω)x ∧ϕ∗(η)x)(u1, . . . , up+q). If p = 0 or q = 0, the proof is similar but simpler. We leave it as exercise to the reader. (ii) If f ∈A0(V ) = C∞(V ), by definition ϕ∗(f)x = f(ϕ(x)), which means that ϕ∗(f) = f ◦ϕ. First we prove (iii) in the case ω ∈A0(V ). Using (i) and (ii) and the fact that ϕ∗e∗ i = dϕi, since d f = m X k=1 ∂f ∂xk e∗ k, we have ϕ∗(d f) = m X k=1 ϕ∗  ∂f ∂xk  ∧ϕ∗(e∗ k) = m X k=1  ∂f ∂xk ◦ϕ  ∧ n X l=1 ∂ϕk ∂xl e∗ l ! = m X k=1 n X l=1  ∂f ∂xk ◦ϕ  ∂ϕk ∂xl  e∗ l = n X l=1 m X k=1  ∂f ∂xk ◦ϕ  ∂ϕk ∂xl ! e∗ l = n X l=1 ∂(f ◦ϕ) ∂xl e∗ l = d(f ◦ϕ) = d(ϕ∗(f)). For the case where ω = fe∗ I, we know by Proposition 4.2 that dω = d f ∧e∗ I. We claim that dϕ∗(e∗ I) = 0. 156 CHAPTER 4. DIFFERENTIAL FORMS To prove this first we show by induction on p that dϕ∗(e∗ I) = d(ϕ∗(e∗ i1 ∧· · · ∧e∗ ip)) = d(ϕ∗(e∗ i1) ∧· · · ∧ϕ∗(e∗ ip)) = p X k=1 (−1)k−1ϕ∗(e∗ i1) ∧· · · ∧d(ϕ∗(e∗ ik)) ∧· · · ∧ϕ∗(e∗ ip). The base case p = 1 is trivial. Assuming that the induction hypothesis holds for any p ≥1, with I = {i1 < i2 < · · · < ip+1}, using Proposition 4.3, we have dϕ∗(e∗ I) = d(ϕ∗(e∗ i1) ∧ϕ∗(e∗ i2) ∧· · · ∧ϕ∗(e∗ ip+1)) = d(ϕ∗(e∗ i1)) ∧ϕ∗(e∗ i2) ∧· · · ∧ϕ∗(e∗ ip+1) + (−1)1ϕ∗(e∗ i1) ∧d(ϕ∗(e∗ i2) ∧· · · ∧ϕ∗(e∗ ip+1)) = d(ϕ∗(e∗ i1)) ∧ϕ∗(e∗ i2) ∧· · · ∧ϕ∗(e∗ ip+1) −ϕ∗(e∗ i1) ∧ p+1 X k=2 (−1)k−2ϕ∗(e∗ i2) ∧· · · ∧d(ϕ∗(e∗ ik)) ∧· · · ∧ϕ∗(e∗ ip+1)  = d(ϕ∗(e∗ i1)) ∧ϕ∗(e∗ i2) ∧· · · ∧ϕ∗(e∗ ip+1) + p+1 X k=2 (−1)k−1ϕ∗(e∗ i1) ∧ϕ∗(e∗ i2) ∧· · · ∧d(ϕ∗(e∗ ik)) ∧· · · ∧ϕ∗(e∗ ip+1) = p+1 X k=1 (−1)k−1ϕ∗(e∗ i1) ∧· · · ∧d(ϕ∗(e∗ ik)) ∧· · · ∧ϕ∗(e∗ ip+1), establishing the induction hypothesis. As a consequence of the above equation, we have dϕ∗(e∗ I) = d(ϕ∗(e∗ i1) ∧· · · ∧ϕ∗(e∗ ip)) = p X k=1 (−1)k−1ϕ∗(e∗ i1) ∧· · · ∧d(ϕ∗(e∗ ik)) ∧· · · ∧ϕ∗(e∗ ip) = 0, since ϕ∗(e∗ ik) = dϕik and d ◦d = 0. Consequently, Proposition 4.3 implies that d(ϕ∗(f) ∧ϕ∗(e∗ I)) = d(ϕ∗f) ∧ϕ∗(e∗ I). Then we have ϕ∗(dω) = ϕ∗(d f) ∧ϕ∗(e∗ I) = d(ϕ∗f) ∧ϕ∗(e∗ I) = d(ϕ∗(f) ∧ϕ∗(e∗ I)) = d(ϕ∗(fe∗ I)) = d(ϕ∗ω). Since every differential form is a linear combination of special forms fe∗ I, we are done. 4.2. PULL-BACK OF DIFFERENTIAL FORMS 157 We use Proposition 4.9 to recompute the pull-back of w = x dy ∧dz + y dx ∧dz. Recall Q = [0, 1] × [0, 1] ⊂R2 and ϕ : U →R3 was defined via x = u + v, y = u −v, z = uv. Proposition 4.9 implies that ϕ∗(ω) = (u + v)ϕ∗(dy) ∧ϕ∗(dz) + (u −v)ϕ∗(dx) ∧ϕ∗(dz) = (u + v)d(ϕ∗y) ∧d(ϕ∗z) + (u −v)d(ϕ∗x) ∧d(ϕ∗z) = (u + v)d(u −v) ∧d(uv) + (u −v)d(u + v) ∧d(uv) = (u + v)(du −dv) ∧(vdu + udv) + (u −v)(du + dv) ∧(vdu + udv) = 2(u2 + v2)du ∧dv. We may generalize the techniques of the preceding calculation by using Proposition 4.9 to compute ϕ∗ω where ϕ: U →V is a smooth map between two open subsets U and V of Rn and ω = fdy1 ∧· · · ∧dyn is a p-form on V . We can write ϕ = (ϕ1, . . . , ϕn) with ϕi : U →R. By Proposition 4.9, we have ϕ∗ω = ϕ∗(f)ϕ∗(dy1) ∧· · · ∧ϕ∗(dyn) = ϕ∗(f)d(ϕ∗y1) ∧· · · ∧d(ϕ∗yn) = (f ◦ϕ)d(ϕ∗y1) ∧· · · ∧d(ϕ∗yn). However, ϕ∗yi = ϕi so we have ϕ∗ω = (f ◦ϕ)dϕ1 ∧· · · ∧dϕn. For any x ∈U, since d(ϕi)x = n X j=1 ∂ϕi ∂xj (x) dxj we get dϕ1 ∧· · · ∧dϕn = det ∂ϕi ∂xj (x)  dx1 ∧· · · ∧dxn = J(ϕ)x dx1 ∧· · · ∧dxn where J(ϕ)x = det ∂ϕi ∂xj (x)  is the Jacobian of ϕ at x ∈U. It follows that (ϕ∗ω)x = ϕ∗(fdy1 ∧· · · ∧dyn)x = f(ϕ(x))J(ϕ)x dx1 ∧· · · ∧dxn. The fact that d and pull-back commutes is an important fact. It allows us to show that a map ϕ: U →V induces a map H•(ϕ): H•(V ) →H•(U) on cohomology, and it is crucial in generalizing the exterior differential to manifolds. 158 CHAPTER 4. DIFFERENTIAL FORMS To a smooth map ϕ: U →V , we associate the map Hp(ϕ): Hp(V ) →Hp(U) given by Hp(ϕ)([ω]) = [ϕ∗(ω)]. This map is well defined, because if we pick any representative ω + dη in the cohomology class [ω] specified by the closed form ω, then dϕ∗ω = ϕ∗dω = 0, so ϕ∗ω is closed, and ϕ∗(ω + dη) = ϕ∗ω + ϕ∗(dη) = ϕ∗ω + dϕ∗η, which shows that Hp(ϕ)([ω]) is well defined. It is also clear that Hp+q(ϕ)([ω][η]) = Hp(ϕ)([ω])Hq(ϕ)([η]), which means that H•(ϕ) is a homomorphism of graded algebras. We often denote H•(ϕ) by ϕ∗. We conclude this section by stating without proof an important result known as the Poincar´ e Lemma. Recall that a subset S ⊆Rn is star-shaped iffthere is some point c ∈S such that for every point x ∈S, the closed line segment [c, x] joining c and x is entirely contained in S. Theorem 4.10. (Poincar´ e’s Lemma) If U ⊆Rn is any star-shaped open set, then we have Hp(U) = (0) for p > 0 and H0(U) = R. Thus, for every p ≥1, every closed form ω ∈Ap(U) is exact. Sketch of proof. Pick c so that U is star-shaped w.r.t. c and let g: U →U be the constant function with value c. Then we see that g∗ω =  0 if ω ∈Ap(U), with p ≥1, ω(c) if ω ∈A0(U), where ω(c) denotes the constant function with value ω(c). The trick is to find a family of linear maps hp : Ap(U) →Ap−1(U), for p ≥1, with h0 = 0, such that d ◦hp + hp+1 ◦d = id −g∗, p > 0, called a chain homotopy. Indeed, if ω ∈Ap(U) is closed and p ≥1, we get dhpω = ω, so ω is exact, and if p = 0 we get h1dω = 0 = ω −ω(c), so ω is constant. It remains to find the hp, which is not obvious. A construction of these maps can be found in Madsen and Tornehave (Chapter 3), Warner (Chapter 4), Cartan (Section 2) Morita (Chapter 3). 4.3. DIFFERENTIAL FORMS ON MANIFOLDS 159 In Section 4.3, we promote differential forms to manifolds. As preparation, note that every open subset U ⊆Rn is a manifold, and that for every x ∈U, the tangent space TxU to U at x is canonically isomorphic to Rn. It follows that the tangent bundle TU and the cotangent bundle T ∗U are trivial, namely TU ∼ = U ×Rn and T ∗U ∼ = U ×(Rn)∗, so the bundle k ^ T ∗U ∼ = U × k ^ (Rn)∗ is also trivial. Consequently, we can view Ak(U) as the set of smooth sections of the vector bundle Vk T ∗(U). The generalization to manifolds is then to define the space of differential p-forms on a manifold M as the space of smooth sections of the bundle Vk T ∗M. 4.3 Differential Forms on Manifolds Let M be any smooth manifold of dimension n. We define the vector bundle V T ∗M as the direct sum bundle ^ T ∗M = n M k=0 k ^ T ∗M; see Section 10.5 for details. Recall that a smooth section of the bundle Vk T ∗M is a smooth function ω: M → Vk T ∗M such that ω(p) ∈Vk T ∗ p M for all p ∈M. Definition 4.7. Let M be any smooth manifold of dimension n. The set Ak(M) of smooth differential k-forms on M is the set of smooth sections Γ(M, Vk T ∗M) of the bundle Vk T ∗M, and the set A∗(M) of all smooth differential forms on M is the set of smooth sections Γ(M, V T ∗M) of the bundle V T ∗M. Observe that A0(M) ∼ = C∞(M, R), the set of smooth functions on M, since the bundle V0 T ∗M is isomorphic to M × R, and smooth sections of M × R are just graphs of smooth functions on M. We also write C∞(M) for C∞(M, R). If ω ∈A∗(M), we often write ωp for ω(p). Definition 4.7 is quite abstract, and it is important to get a more down-to-earth feeling by taking a local view of differential forms, namely with respect to a chart. So let (U, ϕ) be a local chart on M, with ϕ: U →Rn, and let xi = pri ◦ϕ, the ith local coordinate (1 ≤i ≤n); see Tu (Chapter 3, §8) or Gallier and Quaintance . Recall that for any p ∈U, the vectors  ∂ ∂x1  p , . . . ,  ∂ ∂xn  p form a basis of the tangent space TpM. Furthermore, the linear forms (dx1)p, . . . , (dxn)p form a basis of T ∗ p M, (where (dxi)p, the differential of xi at p, is identified with the linear form such that d fp(v) = v(f), for every smooth function f on U and every v ∈TpM). The 160 CHAPTER 4. DIFFERENTIAL FORMS basis ((dx1)p, . . . , (dxn)p) of (TpM)∗is the dual of the basis  ∂ ∂x1  p , . . . ,  ∂ ∂xn  p  of TpM. Indeed, since xi = pri ◦ϕ, we have (dxi)p  ∂ ∂xj  p ! =  ∂ ∂xj  p xi = ∂(pri ◦ϕ ◦ϕ−1) ∂Xj p = ∂pri ∂Xj p = δij. Consequently, locally on U, every k-form ω ∈Ak(M) can be written uniquely as ωp = X I fI(p)dxi1 ∧· · · ∧dxik = X I fI(p)dxI, p ∈U, where I = {i1, . . . , ik} ⊆{1, . . . , n}, with i1 < . . . < ik and dxI = dxi1 ∧· · · ∧dxik. Furthermore, each fI is a smooth function on U. Remark: We define the set of smooth (r, s)-tensor fields as the set Γ(M, T r,s(M)) of smooth sections of the tensor bundle T r,s(M) = T ⊗rM ⊗(T ∗M)⊗s. Tensor fields are discussed quite extensively in Chapter 5. The operations on the algebra V T ∗M yield operations on differential forms using point-wise definitions. If ω, η ∈A∗(M) and λ ∈R, then for every x ∈M, (ω + η)x = ωx + ηx (λω)x = λωx (ω ∧η)x = ωx ∧ηx. Actually, it is necessary to check that the resulting forms are smooth, but this is easily done using charts. When f ∈A0(M), we write fω instead of f ∧ω. It follows that A∗(M) is a graded real algebra and a C∞(M)-module. Proposition 4.1 generalizes immediately to manifolds. Proposition 4.11. For all forms ω ∈Ar(M) and η ∈As(M), we have η ∧ω = (−1)pqω ∧η. For any smooth map ϕ: M →N between two manifolds M and N, we have the differential map dϕ: TM →TN, also a smooth map, and for every p ∈M, the map dϕp : TpM →Tϕ(p)N is linear. As in Section 4.1, Proposition 3.11 gives us the formula µ  k ^ (dϕp)⊤ (ωϕ(p))  (u1, . . . , uk) = µ(ωϕ(p))(dϕp(u1), . . . , dϕp(uk)), ω ∈Ak(N), for all u1, . . . , uk ∈TpM. This gives us the behavior of Vk(dϕp)⊤under the identification of Vk T ∗ p M and Altk(TpM; R) via the isomorphism µ. Here is the extension of Definition 4.6 to differential forms on a manifold. 4.3. DIFFERENTIAL FORMS ON MANIFOLDS 161 Definition 4.8. For any smooth map ϕ: M →N between two smooth manifolds M and N, for every k ≥0, we define the map ϕ∗: Ak(N) →Ak(M) by ϕ∗(ω)p(u1, . . . , uk) = ωϕ(p)(dϕp(u1), . . . , dϕp(uk)), for all ω ∈Ak(N), all p ∈M, and all u1, . . . , uk ∈TpM. We say that ϕ∗(ω) (for short, ϕ∗ω) is the pull-back of ω by ϕ. The maps ϕ∗: Ak(N) →Ak(M) induce a map also denoted ϕ∗: A∗(N) →A∗(M). Using the chain rule, we check immediately that id∗ = id, (ψ ◦ϕ)∗ = ϕ∗◦ψ∗. We need to check that ϕ∗ω is smooth, and for this it is enough to check it locally on a chart (U, ψ). For any chart (V, θ) on N such that ϕ(U) ⊆V , on V we know that ω ∈Ak(N) can be written uniquely as ω = X I fIdxi1 ∧· · · ∧dxik, with fI smooth on V , and it is easy to see (using the definition) that locally on U we have ϕ∗ω = X I (fI ◦ϕ)d(xi1 ◦ϕ) ∧· · · ∧d(xik ◦ϕ), (†) which is smooth. In the special case of M = Rn, ϕ: M →N is a parametrization of N, and (†) is what we use to efficiently calculate the pull-back of ω on the embedded manifold N. For example, let M = {(θ, ϕ) : 0 < θ < π, 0 < ϕ < 2π} ⊂R2, N = S2 and ψ : M →N the parametrization of S2 given by x = sin θ cos ϕ, y = sin θ sin ϕ, z = cos θ. See Figure 4.1. Let w = x dy be a form on S2. The pull-back of ω into M is calculated via (†) as ψ∗w = sin θ cos ϕ d(sin θ sin ϕ) = sin θ cos ϕ(cos θ sin ϕ dθ + sin θ cos ϕ dϕ), where we applied Proposition 4.2 since M ⊂R2. Remark: The fact that the pull-back of differential forms makes sense for arbitrary smooth maps ϕ: M →N, and not just diffeomorphisms, is a major technical superiority of forms over vector fields. 162 CHAPTER 4. DIFFERENTIAL FORMS z x y Θ φ Figure 4.1: The spherical coordinates of S2. The next step is to define d on A∗(M). There are several ways to proceed, but since we already considered the special case where M is an open subset of Rn, we proceed using charts. Given a smooth manifold M of dimension n, let (U, ϕ) be any chart on M. For any ω ∈Ak(M) and any p ∈U, define (dω)p as follows: If k = 0, that is ω ∈C∞(M), let (dω)p = dωp, the differential of ω at p, and if k ≥1, let (dω)p = ϕ∗d((ϕ−1)∗ω)ϕ(p)  p, where d is the exterior differential on Ak(ϕ(U)). More explicitly, (dω)p is given by (dω)p(u1, . . . , uk+1) = d((ϕ−1)∗ω)ϕ(p)(dϕp(u1), . . . , dϕp(uk+1)), (∗∗) for every p ∈U and all u1, . . . , uk+1 ∈TpM. Observe that the above formula is still valid when k = 0 if we interpret the symbold d in d((ϕ−1)∗ω)ϕ(p) = d(ω◦ϕ−1)ϕ(p) as the differential. Since ϕ−1 : ϕ(U) →U is map whose domain is an open subset W = ϕ(U) of Rn, the form (ϕ−1)∗ω is a differential form in A∗(W), so d((ϕ−1)∗ω) is well-defined. The formula at Line (∗∗) encapsulates the following “natural” three step procedure: Step 1: Take the form ω on the manifold M and precompose ω with the parameterization ϕ−1 so that (ϕ−1)∗ω is now a form in U, a subset of Rm, where m is the dimension of M. Step 2: Differentiate (ϕ−1)∗ω via Proposition 4.2. Step 3: Compose the result of Step 2 with the chart map ϕ and pull the differential form on U back into M. 4.3. DIFFERENTIAL FORMS ON MANIFOLDS 163 We need to check that the definition at Line (∗∗) does not depend on the chart (U, ϕ). Proof. For any other chart (V, ψ), with U ∩V ̸= ∅, the map θ = ψ ◦ϕ−1 is a diffeomorphism between the two open subsets ϕ(U ∩V ) and ψ(U ∩V ), and ψ = θ ◦ϕ. Let x = ϕ(p) and y = ψ(p). We need to check that d((ϕ−1)∗ω)x(dϕp(u1), . . . , dϕp(uk+1)) = d((ψ−1)∗ω)y(dψp(u1), . . . , dψp(uk+1)), for every p ∈U ∩V and all u1, . . . , uk+1 ∈TpM. However, y = ψ(p) = θ(ϕ(p)) = θ(x), so d((ψ−1)∗ω)y(dψp(u1), . . . , dψp(uk+1)) = d((ϕ−1◦θ−1)∗ω)θ(x)(d(θ◦ϕ)p(u1), . . . , d(θ◦ϕ)p(uk+1)). Since (ϕ−1 ◦θ−1)∗= (θ−1)∗◦(ϕ−1)∗ and, by Proposition 4.9 (iii), d(((θ−1)∗◦(ϕ−1)∗)ω) = d((θ−1)∗((ϕ−1)∗ω)) = (θ−1)∗(d((ϕ−1)∗ω)), we get d((ϕ−1 ◦θ−1)∗ω)θ(x)(d(θ ◦ϕ)p(u1), . . . , d(θ ◦ϕ)p(uk+1)) = (θ−1)∗(d((ϕ−1)∗ω))θ(x)(d(θ ◦ϕ)p(u1), . . . , d(θ ◦ϕ)p(uk+1)). Then by Definition 4.8, we obtain (θ−1)∗(d((ϕ−1)∗ω))θ(x)(d(θ ◦ϕ)p(u1), . . . , d(θ ◦ϕ)p(uk+1)) = d((ϕ−1)∗ω)x((dθ−1)θ(x)(d(θ ◦ϕ)p(u1)), . . . , (dθ−1)θ(x)(d(θ ◦ϕ)p(uk+1))). As (dθ−1)θ(x)(d(θ ◦ϕ)p(ui)) = d(θ−1 ◦(θ ◦ϕ))p(ui) = dϕp(ui), by the chain rule, we obtain d((ψ−1)∗ω)θ(x)(dψp(u1), . . . , dψp(uk+1)) = d((ϕ−1)∗ω)x(dϕp(u1), . . . , dϕp(uk+1)), as desired. Observe that (dω)p is smooth on U, and as our definition of (dω)p does not depend on the choice of a chart, the forms (dω) ↾U agree on overlaps and yield a differential form dω defined on the whole of M. Thus we can make the following definition: Definition 4.9. If M is any smooth manifold, there is a linear map d: Ak(M) →Ak+1(M) for every k ≥0, such that for every ω ∈Ak(M), for every chart (U, ϕ), for every p ∈U, if k = 0, that is ω ∈C∞(M), then (dω)p = dωp, the differential of ω at p, else if k ≥1, then (dω)p = ϕ∗d((ϕ−1)∗ω)ϕ(p)  p, where d is the exterior differential on Ak(ϕ(U)) from Definition 4.3. We obtain a linear map d: A∗(M) →A∗(M) called exterior differentiation. 164 CHAPTER 4. DIFFERENTIAL FORMS To explicitly demonstrate Definition 4.9, we return to our previous example of ψ : M →S2 and ω = xdy considered as a one form on S2. Note that ψ(θ, ϕ) = (sin θ cos ϕ, sin θ sin ϕ, cos θ) is a parameterization of the S2 and hence ψ−1(x, y, z) = (cos−1(z), tan−1(y/x)) provides the structure of a chart on S2. We already found that the pull-back of ω into M is ψ∗ω = sin θ cos ϕ cos θ sin ϕ dθ + sin θ cos ϕ sin θ cos ϕ dϕ. Proposition 4.2 is now applied ψ∗ω to give us dψ∗ω = d(sin θ cos ϕ cos θ sin ϕ dθ) + d(sin θ cos ϕ sin θ cos ϕ dϕ) = d(sin θ cos ϕ cos θ sin ϕ) ∧dθ + d(sin θ cos ϕ sin θ cos ϕ) ∧dϕ = ∂ ∂ϕ(sin θ cos ϕ cos θ sin ϕ)dϕ ∧dθ + ∂ ∂θ(sin θ cos ϕ sin θ cos ϕ)dθ ∧dϕ = sin θ cos θ(−sin2 ϕ + cos2 ϕ)dϕ ∧dθ + 2 sin θ cos θ cos2 ϕ dθ ∧dϕ = sin θ cos θ(sin2 ϕ + cos2 ϕ)dθ ∧dϕ = sin θ cos θ dθ ∧dϕ. It just remains to compose dψ∗ω with ψ−1 to obtain dω = (ψ−1)∗(dψ∗ω) = z √ 1 −z2d(cos−1 z) ∧d(tan−1 y/x) = z √ 1 −z2  − 1 √ 1 −z2dz  ∧ −y x2 1 + y2 x2 dx + 1 x 1 + y2 x2 dy ! = −z dz ∧  − y x2 + y2 dx + x x2 + y2 dy  = zy x2 + y2 dz ∧dx − zx x2 + y2 dz ∧dy. Since x2 + y2 + z2 = 1, we obtain the constraint x dx + y dy + z dz = 0, which implies that −z dz = x dx + y dy. Then we find that dω is equivalent to dω = −z dz ∧  − y x2 + y2 dx + x x2 + y2 dy  = (x dx + y dy) ∧  − y x2 + y2 dx + x x2 + y2 dy  = x2 x2 + y2 dx ∧dy + y2 x2 + y2 dx ∧dy = dx ∧dy, 4.3. DIFFERENTIAL FORMS ON MANIFOLDS 165 where we interpret dx ∧dy as the restriction of 2-form in R3 to S2, i.e. dx ∧dy|S2 is defined as (dx ∧dy|S2)p(v) = (dx ∧dy)p(v), p ∈S2, v ∈TpS2. Propositions 4.3, 4.4 and 4.9 generalize to manifolds. Proposition 4.12. Let M and N be smooth manifolds and let ϕ: M →N be a smooth map. (1) For all ω ∈Ar(M) and all η ∈As(M), d(ω ∧η) = dω ∧η + (−1)rω ∧dη. (2) For every k ≥0, the composition Ak(M) d − →Ak+1(M) d − →Ak+2(M) is identically zero; that is, d ◦d = 0. (3) ϕ∗(ω ∧η) = ϕ∗ω ∧ϕ∗η, for all ω ∈Ar(N) and all η ∈As(N). (4) ϕ∗(f) = f ◦ϕ, for all f ∈A0(N). (5) dϕ∗(ω) = ϕ∗(dω), for all ω ∈Ak(N); that is, the following diagram commutes for all k ≥0. Ak(N) ϕ∗ / d  Ak(M) d  Ak+1(N) ϕ∗/ Ak+1(M) Proof. It is enough to prove these properties in a chart (U, ϕ), which is easy. We only check (2). We have (d(dω))p = d ϕ∗d((ϕ−1)∗ω)ϕ(p)  p = ϕ∗h d  (ϕ−1)∗ ϕ∗d((ϕ−1)∗ω)ϕ(p)  ϕ(p) i p = ϕ∗h d d((ϕ−1)∗ω)ϕ(p)  ϕ(p) i p = 0, as (ϕ−1)∗◦ϕ∗= (ϕ ◦ϕ−1)∗= id∗= id and d ◦d = 0 on forms in Ak(ϕ(U)), with ϕ(U) ⊆ Rn. As a consequence, Definition 4.5 of the de Rham cohomology generalizes to manifolds. 166 CHAPTER 4. DIFFERENTIAL FORMS Definition 4.10. For every manifold M, we have the de Rham complex A0(M) d − →A1(M) − →· · · − →Ak−1(M) d − →Ak(M) d − →Ak+1(M) − →· · · , and we can define the cohomology groups Hk DR(M) and the graded cohomology algebra H• DR(M). For every k ≥0, let Zk(M) = {ω ∈Ak(M) | dω = 0} = Ker d: Ak(M) − →Ak+1(M) be the vector space of closed k-forms, and for every k ≥1, let Bk(M) = {ω ∈Ak(M) | ∃η ∈Ak−1(M), ω = dη} = Im d: Ak−1(M) − →Ak(M) be the vector space of exact k-forms, with B0(M) = (0). Then, for every k ≥0, we define the kth de Rham cohomology group of M as the quotient space Hk DR(M) = Zk(M)/Bk(M). This is an abelian group under addition of cosets. The real vector space H• DR(M) = L k≥0 Hk DR(M) is called the de Rham cohomology algebra of M. We often drop the sub-script DR when no confusion arises. Every smooth map ϕ: M →N between two manifolds induces an algebra map ϕ∗: H•(N) →H•(M). Another important property of the exterior differential is that it is a local operator, which means that the value of dω at p only depends of the values of ω near p. Not all operators are local. For example, the operator I : C∞([a, b]) →C∞([a, b]) given by I(f) = Z b a f(t) dt, where I(f) is the constant function on [a, b], is not local since for any point p ∈[a, b], the calculation of I(f) requires evaluating f over [a, b]. More generally, we have the following definition. Definition 4.11. A linear map D: A∗(M) →A∗(M) is a local operator if for all k ≥0, for any nonempty open subset U ⊆M and for any two k-forms ω, η ∈Ak(M), if ω ↾U = η ↾U, then (Dω) ↾U = (Dη) ↾U. Since D is linear, the above condition is equivalent to saying that for any k-form ω ∈Ak(M), if ω ↾U = 0, then (Dω) ↾U = 0. Since Property (1) of Proposition 4.12 comes up a lot, we introduce the following defini-tion. Definition 4.12. Given any smooth manifold M, a linear map D: A∗(M) →A∗(M) is called an antiderivation if for all r, s ≥0, for all ω ∈Ar(M) and all η ∈As(M), D(ω ∧η) = Dω ∧η + (−1)rω ∧Dη. The antiderivation is of degree m ∈Z if D: Ap(M) →Ap+m(M) for all p such that p+m ≥0. 4.3. DIFFERENTIAL FORMS ON MANIFOLDS 167 By Proposition 4.12, exterior differentiation d: A∗(M) →A∗(M) is an antiderivation of degree 1. Proposition 4.13. Let M be a smooth manifold. Any linear antiderivation D: A∗(M) → A∗(M) is a local operator. Proof. By linearity, it is enough to show that if ω ↾U = 0, then (Dω) ↾U = 0. There is an apparent problem, which is that although ω is zero on U, it may not be zero outside U, so it is not obvious that we can conclude that Dω is zero on U. The crucial ingredient to circumvent this difficulty is the existence of “bump functions;” see Tu (Chapter 3, §8) or Morita (Chapter 1, Section 13(b)). By Lemma 1.28 of Morita applied to the constant function with value 1, for every p ∈U, there some open subset V ⊆U containing p and a smooth function f : M →R such that supp f ⊆U and f ≡1 on V . Consequently, fω is a smooth differential form which is identically zero, and since D is an antiderivation D(fω) = Df ∧ω + fDω, which, evaluated at p yields 0 = Dfp ∧ωp + 1Dωp = Dfp ∧0 + 1Dωp = Dωp; that is, Dωp = 0, as claimed. Remark: If D: A∗(M) →A∗(M) is a linear map which is a derivation, which means that D(ω ∧η) = Dω ∧η + ω ∧Dη for all ω ∈Ar(M) and all η ∈As(M), then the proof of Proposition 4.13 still works and shows that D is also a local operator. By Proposition 4.13, exterior differentiation d: A∗(M) →A∗(M) is a local operator. As in the case of differential forms on Rn, the operator d is uniquely determined by the properties of Theorem 4.5. Theorem 4.14. Let M be a smooth manifold. There is a unique linear operator d: A∗(M) →A∗(M), with d = (dk) and dk : Ak(M) →Ak+1(M) for every k ≥0, such that (1) (d f)p = d fp, where d fp is the differential of f at p ∈M for every f ∈A0(M) = C∞(M). (2) d ◦d = 0. (3) For every ω ∈Ar(M) and every η ∈As(M), d(ω ∧η) = dω ∧η + (−1)rω ∧dη. Furthermore, any linear operator d satisfying (1)–(3) is a local operator. 168 CHAPTER 4. DIFFERENTIAL FORMS Proof. Existence has already been established. Let D: A∗(M) →A∗(M) be any linear operator satisfying (1)–(3). We need to prove that D = d where d is defined in Definition 4.9. For any k ≥0, pick any ω ∈Ak(M). For every p ∈M, we need to prove that (Dω)p = (dω)p. Let (U, ϕ) be any chart with p ∈U, and let xi = pri ◦ϕ be the corresponding local coordinate maps. We know that ω ∈Ak(M) can be written uniquely as ωq = X I fI(q)dxi1 ∧· · · ∧dxik q ∈U. Using a bump function, there is some open subset V of U with p ∈V and some functions e fI, and e xi1, . . . , e xik defined on M and agreeing with fI, xi1, . . . , xik on V . If we define e ω = X I e fIde xi1 ∧· · · ∧de xik then e ω is defined for all p ∈M and ω ↾V = e ω ↾V. By Proposition 4.13, since D is a linear map satisfying (3), it is a local operator so Dω ↾V = De ω ↾V. Since D satisfies (1), we have De xij = de xij. Then at p, by linearity, we have (Dω)p = (De ω)p =  D X I e fIde xi1 ∧· · · ∧de xik  p =  D X I e fIDe xi1 ∧· · · ∧De xik  p = X I D  e fIDe xi1 ∧· · · ∧De xik  p . As in the proof of Proposition 4.9(iii), we can show by induction that D(De xi1 ∧· · · ∧De xik) = k X j=1 (−1)j−1De xi1 ∧· · · ∧DDe xij ∧· · · ∧De xik and since by (2) DDe xij = 0, we have D(De xi1 ∧· · · ∧De xik) = 0. 4.3. DIFFERENTIAL FORMS ON MANIFOLDS 169 Then, using the above, by (3) and (1), we get X I D  e fIDe xi1 ∧· · · ∧De xik  p = X I D e fI ∧De xi1 ∧· · · ∧De xik  p + X I e fI ∧D(De xi1 ∧· · · ∧De xik)  p = X I d e fI ∧de xi1 ∧· · · ∧de xik  p = X I d fI ∧dxi1 ∧· · · ∧dxik  p = (dω)p. Therefore (Dω)p = (dω)p, which proves that D = d. Remark: A closer look at the proof of Theorem 4.14 shows that it is enough to assume DDω = 0 on forms ω ∈A0(M) = C∞(M). Smooth differential forms can also be defined in terms of alternating C∞(M)-multilinear maps on smooth vector fields. This approach also yields a global formula for the exterior derivative dω(X1, . . . , Xk+1) of a k-form ω applied to k + 1 vector fields X1, . . . , Xk+1. This formula is not very useful for computing dω at a given point p since it requires vector fields as input, but it is quite useful in theoretical investigations. Let ω ∈Ak(M) be any smooth k-form on M. Then ω induces an alternating multilinear map ω: X(M) × · · · × X(M) | {z } k − →C∞(M) as follows: For any k smooth vector fields X1, . . . , Xk ∈X(M), ω(X1, . . . , Xk)(p) = ωp(X1(p), . . . , Xk(p)). This map is obviously alternating and R-linear, but it is also C∞(M)-linear, since for every f ∈C∞(M), ω(X1, . . . , fXi, . . . Xk)(p) = ωp(X1(p), . . . , f(p)Xi(p), . . . , Xk(p)) = f(p)ωp(X1(p), . . . , Xi(p), . . . , Xk(p)) = (fω)p(X1(p), . . . , Xi(p), . . . , Xk(p)). (Recall, that the set of smooth vector fields X(M) is a real vector space and a C∞(M)-module.) Interestingly, every alternating C∞(M)-multilinear map on smooth vector fields deter-mines a differential form. This is because ω(X1, . . . , Xk)(p) only depends on the values of X1, . . . , Xk at p. 170 CHAPTER 4. DIFFERENTIAL FORMS Proposition 4.15. Let M be a smooth manifold. For every k ≥0, there is an isomorphism between the space of k-forms Ak(M) and the space Altk C∞(M)(X(M)) of alternating C∞(M)-multilinear maps on smooth vector fields. That is, Ak(M) ∼ = Altk C∞(M)(X(M)), viewed as C∞(M)-modules. Proof. We follow the proof in O’Neill (Chapter 2, Lemma 3 and Proposition 2).¿ Let Φ: X(M) × · · · × X(M) | {z } k − →C∞(M) be an alternating C∞(M)-multilinear map. First we prove that for any vector fields X1, . . . , Xk and Y1, . . . , Yk, for every p ∈M, if Xi(p) = Yi(p), then Φ(X1, . . . , Xk)(p) = Φ(Y1, . . . , Yk)(p). Observe that Φ(X1, . . . , Xk) −Φ(Y1, . . . , Yk) = Φ(X1 −Y1, X2, . . . , Xk) + Φ(Y1, X2 −Y2, X3, . . . , Xk) + Φ(Y1, Y2, X3 −Y3, . . . , Xk) + · · · + Φ(Y1, . . . , Yk−2, Xk−1 −Yk−1, Xk) + Φ(Y1, . . . , Yk−1, Xk −Yk). As a consequence, it is enough to prove that if Xi(p) = 0 for some i, then Φ(X1, . . . , Xk)(p) = 0. Without loss of generality, assume i = 1. In any local chart (U, ϕ) near p, we can write X1 = n X i=1 fi ∂ ∂xi , and as Xi(p) = 0, we have fi(p) = 0 for i = 1, . . . , n. Since the expression on the right-hand side is only defined on U, we extend it using a bump function once again. There is some open subset V ⊆U containing p and a smooth function h: M →R such that supp h ⊆U and h ≡1 on V . Then we let hi = hfi, a smooth function on M, Yi = h ∂ ∂xi, a smooth vector field on M, so that h2X1 = n X i=1 hiYi, and since fi(p) = 0 we have hi(p) = 0. Since Φ is C∞(M)-multilinear, we have h2Φ(X1, X2, . . . , Xk)(q) = Φ(h2X1, X2, . . . , Xk) = Φ  n X i=1 hiYi, X2, . . . , Xk  = n X i=1 hiΦ(Yi, X2, . . . , Xk), 4.3. DIFFERENTIAL FORMS ON MANIFOLDS 171 and since hi(p) = 0 and h(p) = 1, we get Φ(X1, X2, . . . , Xk)(p) = 0, as claimed. Next we show that Φ induces a smooth differential form. For every p ∈M, for any u1, . . . , uk ∈TpM, we can pick a smooth function f equal to 1 on some open set V containing p and 0 outside V , so that we get smooth vector fields X1, . . . , Xk with Xk(p) = uk. We set ωp(u1, . . . , uk) = Φ(X1, . . . , Xk)(p). As we proved that Φ(X1, . . . , Xk)(p) only depends on X1(p) = u1, . . . , Xk(p) = uk, the function ωp is well defined, and it is easy to check that it is smooth. Therefore, the map Φ 7→ω just defined is indeed an isomorphism. Remark: The space HomC∞(M)(X(M), C∞(M)) of all C∞(M)-linear maps from X(M) to C∞(M) is also a C∞(M)-module, called the dual of X(M). Proposition 4.15 shows that as C∞(M)-modules, A1(M) ∼ = HomC∞(M)(X(M), C∞(M)). A result analogous to Proposition 4.15 holds for tensor fields. This matter is discussed in Chapter 5. Recall that for any function f ∈C∞(M) and every vector field X ∈X(M), the Lie derivative X[f] (or X(f)) of f w.r.t. X is defined so that X[f]p = d fp(X(p)); Also recall the notion of the Lie bracket [X, Y ] of two vector fields; see Warner (Chapter 1), Morita (Chapter 1, Section 1.4), Gallier and Qaintance . The interpretation of differential forms as C∞(M)-multilinear maps given by Proposition 4.15 yields the following formula for (dω)(X1, . . . , Xk+1), where the Xi are vector fields: Proposition 4.16. Let M be a smooth manifold. For every k-form ω ∈Ak(M), we have (dω)(X1, . . . , Xk+1) = k+1 X i=1 (−1)i−1Xi[ω(X1, . . . , c Xi, . . . , Xk+1)] + X i<j (−1)i+jω([Xi, Xj], X1, . . . , c Xi, . . . , c Xj, . . . , Xk+1)], for all vector fields, X1, . . . , Xk+1 ∈X(M): Proof sketch. First one checks that the right-hand side of the formula in Proposition 4.16 is alternating and C∞(M)-multilinear. For this, use Proposition 5.3 from Chapter 0 of Do Carmo , or see Gallier and Quaintance . Consequently, by Proposition 4.15, this 172 CHAPTER 4. DIFFERENTIAL FORMS expression defines a (k + 1)-form. Secondly, it is enough to check that both sides of the equation agree on charts (U, ϕ). We know in a chart that ω can be written uniquely as ωp = X I fI(p)dxi1 ∧· · · ∧dxik p ∈U. Also, as differential forms are C∞(M)-multilinear, it is enough to consider vector fields of the form Xi = ∂ ∂xji . However, for such vector fields, [Xi, Xj] = 0, and then it is a simple matter to check that the equation holds. For more details, see Morita (Chapter 2). In particular, when k = 1, Proposition 4.16 yields the often used formula: Corollary 4.17. The following formula holds: dω(X, Y ) = X[ω(Y )] −Y [ω(X)] −ω([X, Y ]). There are other ways of proving the formula of Proposition 4.16, for instance, using Lie derivatives. Before considering the Lie derivative LXω of differential forms, we define interior multi-plication by a vector field, i(X)(ω). We will see shortly that there is a relationship between LX, i(X), and d, known as Cartan’s Formula. Definition 4.13. Let M be a smooth manifold. For every vector field X ∈X(M), for all k ≥1, there is a linear map i(X): Ak(M) →Ak−1(M) defined so that, for all ω ∈Ak(M), for all p ∈M, for all u1, . . . , uk−1 ∈TpM, (i(X)ω)p(u1, . . . , uk−1) = ωp(Xp, u1, . . . , uk−1). Obviously, i(X) is C∞(M)-linear in X, namely i(fX)ω = fi(X)ω, i(X)(fω) = fi(X)ω, f ∈C∞(M), ω ∈Ak(M), and it is easy to check that i(X)ω is indeed a smooth (k −1)-form. When k = 0, we set i(X)ω = 0. Observe that i(X)ω is also given by (i(X)ω)p = i(Xp)ωp, p ∈M, where i(Xp) is the interior product (or insertion operator) defined in Section 3.6 (with i(Xp)ωp equal to our right hook, ωp ⌞Xp). As a consequence, by Proposition 3.22, the operator i(X) is an anti-derivation of degree −1; that is, we have i(X)(ω ∧η) = (i(X)ω) ∧η + (−1)rω ∧(i(X)η), for all ω ∈Ar(M) and all η ∈As(M). Remark: Other authors, including Marsden , use a left hook instead of a right hook, and denote i(X)ω as X ⌟ω. 4.4. LIE DERIVATIVES 173 4.4 Lie Derivatives We just saw in Section 4.3 that for any function f ∈C∞(M) and every vector field X ∈ X(M), the Lie derivative X[f] (or X(f)) of f w.r.t. X is defined so that X[f]p = d fp(Xp). recall that for any manifold M, given any two vector fields X, Y ∈X(M), the Lie derivative of X with respect to Y is given by (LX Y )p = lim t− →0 Φ∗ tY  p −Yp t = d dt Φ∗ tY  p t=0 , where Φt is the local one-parameter group associated with X (Φ is the global flow associated with X; see Warner , Chapters 1 and 2, or Gallier and Quaintance ), and Φ∗ t is the pull-back of the diffeomorphism Φt given by (Φ∗ tY )p = d(Φ−1 t )Φt(p)(YΦt(p)). Furthermore, to calculate LXY recall that LXY = [X, Y ]. Proposition 4.18. The following identity holds: Xp[f] = lim t− →0 (Φ∗ tf)(p) −f(p) t = d dt(Φ∗ tf)(p) t=0 , with Φ∗ tf = f ◦Φt (as usual for functions). Proof. Recall that if Φ is the flow of X, then for every p ∈M, the map t 7→Φt(p) is an integral curve of X through p; that is ˙ Φt(p) = X(Φt(p)), Φ0(p) = p, in some open set containing p. In particular, ˙ Φ0(p) = Xp. Then we have lim t− →0 (Φ∗ tf)(p) −f(p) t = lim t− →0 f(Φt(p)) −f(Φ0(p)) t = d dt(f ◦Φt(p)) t=0 = d fp( ˙ Φ0(p)) = d fp(Xp) = Xp[f]. We would like to define the Lie derivative of differential forms. This can be done alge-braically or in terms of flows; the two approaches are equivalent, but it seems more natural to give a definition using flows. 174 CHAPTER 4. DIFFERENTIAL FORMS Definition 4.14. Let M be a smooth manifold. For every vector field X ∈X(M), for every k-form ω ∈Ak(M), the Lie derivative of ω with respect to X, denoted LXω, is given by (LXω)p = lim t− →0 Φ∗ tω  p −ωp t = d dt Φ∗ tω  p t=0 , where Φ∗ tω is the pull-back of ω along Φt (see Definition 4.8). Obviously, LX : Ak(M) →Ak(M) is a linear map, but it has many other interesting properties. The generalization of the Lie derivative to tensor fields is discussed in Chapter 5. We now state, mostly without proofs, a number of properties of Lie derivatives. Most of these proofs are fairly straightforward computations, often tedious, and can be found in most texts, including Warner , Morita , and Gallot, Hullin and Lafontaine . Proposition 4.19. Let M be a smooth manifold. For every vector field X ∈X(M), the following properties hold: (1) For all ω ∈Ar(M) and all η ∈As(M), LX(ω ∧η) = (LXω) ∧η + ω ∧(LXη); that is, LX is a derivation. (2) For all ω ∈Ak(M), for all Y1, . . . , Yk ∈X(M), LX(ω(Y1, . . . , Yk)) = (LXω)(Y1, . . . , Yk) + k X i=1 ω(Y1, . . . , Yi−1, LXYi, Yi+1, . . . , Yk). (3) The Lie derivative commutes with d: LX ◦d = d ◦LX. Proof. We only prove (2). First we claim that if ϕ: M →M is a diffeomorphism, then for every ω ∈Ak(M), for all X1, . . . , Xk ∈X(M), (ϕ∗ω)(X1, . . . , Xk) = ϕ∗(ω((ϕ−1)∗X1, . . . , (ϕ−1)∗Xk)), (∗) where (ϕ−1)∗Xi is the pull-back of the vector field Xi (also equal to the push-forward ϕ∗Xi of Xi). Recall that ((ϕ−1)∗Y )p = dϕϕ−1(p)(Yϕ−1(p)), 4.4. LIE DERIVATIVES 175 for any vector field Y . Then by Definition 4.8, for every p ∈M, we have (ϕ∗ω(X1, . . . , Xk))(p) = ωϕ(p)(dϕp(X1(p)), . . . , dϕp(Xk(p))) = ωϕ(p) dϕϕ−1(ϕ(p))(X1(ϕ−1(ϕ(p)))), . . . , dϕϕ−1(ϕ(p))(Xk(ϕ−1(ϕ(p))))  = ωϕ(p)(((ϕ−1)∗X1)ϕ(p), . . . , ((ϕ−1)∗Xk)ϕ(p)) = (ω((ϕ−1)∗X1, . . . , (ϕ−1)∗Xk)) ◦ϕ)(p) = ϕ∗(ω((ϕ−1)∗X1, . . . , (ϕ−1)∗Xk))(p), since for any function g ∈C∞(M), we have ϕ∗g = g ◦ϕ. We know that Xp[f] = lim t− →0 (Φ∗ tf)(p) −f(p) t and for any vector field Y , [X, Y ]p = (LXY )p = lim t− →0 Φ∗ tY  p −Yp t . Since the one-parameter group associated with −X is Φ−t, (this follows from Φ−t ◦Φt = id), we have lim t− →0 Φ∗ −tY  p −Yp t = −[X, Y ]p. Now, using Φ−1 t = Φ−t and (∗), we have (LXω)(Y1, . . . , Yk) = lim t− →0 (Φ∗ tω)(Y1, . . . , Yk) −ω(Y1, . . . , Yk) t = lim t− →0 Φ∗ t(ω(Φ∗ −tY1, . . . , Φ∗ −tYk)) −ω(Y1, . . . , Yk) t = lim t− →0 Φ∗ t(ω(Φ∗ −tY1, . . . , Φ∗ −tYk)) −Φ∗ t(ω(Y1, . . . , Yk)) t + lim t− →0 Φ∗ t(ω(Y1, . . . , Yk)) −ω(Y1, . . . , Yk) t . Call the first term A and the second term B. Then as Xp[f] = lim t− →0 (Φ∗ tf)(p) −f(p) t , we have B = X[ω(Y1, . . . , Yk)] = LX(ω(Y1, . . . , Yk)). 176 CHAPTER 4. DIFFERENTIAL FORMS As to A, we have A = lim t− →0 Φ∗ t(ω(Φ∗ −tY1, . . . , Φ∗ −tYk)) −Φ∗ t(ω(Y1, . . . , Yk)) t = lim t− →0 Φ∗ t ω(Φ∗ −tY1, . . . , Φ∗ −tYk) −ω(Y1, . . . , Yk) t  = lim t− →0 Φ∗ t ω(Φ∗ −tY1, . . . , Φ∗ −tYk) −ω(Y1, Φ∗ −tY2, . . . , Φ∗ −tYk) t  + lim t− →0 Φ∗ t ω(Y1, Φ∗ −tY2, . . . , Φ∗ −tYk) −ω(Y1, Y2, Φ∗ −tY3, . . . , Φ∗ −tYk) t  + · · · + lim t− →0 Φ∗ t ω(Y1, . . . , Yk−1, Φ∗ −tYk) −ω(Y1, . . . , Yk) t  = lim t− →0 Φ∗ t ω(Φ∗ −tY1 −Y1, Φ∗ −tY2, . . . , Φ∗ −tYk) t  + lim t− →0 Φ∗ t ω(Y1, Φ∗ −tY2 −Y2, . . . , Φ∗ −tYk) t  + · · · + lim t− →0 Φ∗ t ω(Y1, . . . , Yk−1, Φ∗ −tYk −Yk) t  = lim t− →0 Φ∗ t  ω Φ∗ −tY1 −Y1 t , . . . , Φ∗ −tYk  + lim t− →0 Φ∗ t  ω  Y1, Φ∗ −tY2 −Y2 t , . . . , Φ∗ −tYk  + · · · + lim t− →0 Φ∗ t  ω  Y1, . . . , Yk−1, Φ∗ −tYk −Yk t  = k X i=1 ω(Y1, . . . , −[X, Yi], . . . , Yk), since limt− →0 Φ∗ t = id. When we add up A and B, we get A + B = LX(ω(Y1, . . . , Yk)) − k X i=1 ω(Y1, . . . , [X, Yi], . . . , Yk) = (LXω)(Y1, . . . , Yk), which finishes the proof. Part (2) of Proposition 4.19 shows that the Lie derivative of a differential form can be defined in terms of the Lie derivatives of functions and vector fields: (LXω)(Y1, . . . , Yk) = LX(ω(Y1, . . . , Yk)) − k X i=1 ω(Y1, . . . , Yi−1, LXYi, Yi+1, . . . , Yk) = X[ω(Y1, . . . , Yk)] − k X i=1 ω(Y1, . . . , Yi−1, [X, Yi], Yi+1, . . . , Yk). However, to best calculate LXω, we use Cartan’s formula. Recall the definition of i(X): Ak(M) →Ak−1(M) given in Definition 4.13. 4.4. LIE DERIVATIVES 177 Proposition 4.20. (Cartan’s Formula) Let M be a smooth manifold. For every vector field X ∈X(M), for every ω ∈Ak(M), we have LXω = i(X)dω + d(i(X)ω), that is, LX = i(X) ◦d + d ◦i(X). Proof. If k = 0, then LXf = X[f] = d f(X) for a function f, and on the other hand, i(X)f = 0 and i(X)d f = d f(X), so the equation holds. If k ≥1, then by Proposition 4.16, we have (i(X)dω)(X1, . . . , Xk) = dω(X, X1, . . . , Xk) = X[ω(X1, . . . , Xk)] + k X i=1 (−1)iXi[ω(X, X1, . . . , c Xi, . . . , Xk)] + k X j=1 (−1)jω([X, Xj], X1, . . . , c Xj, . . . , Xk) + X i<j (−1)i+jω([Xi, Xj], X, X1, . . . , c Xi, . . . , c Xj, . . . , Xk). On the other hand, again by Proposition 4.16, we have d(i(X)ω)(X1, . . . , Xk) = k X i=1 (−1)i−1Xi[ω(X, X1, . . . , c Xi, . . . , Xk)] + X i n. Of course we would like to have a “good” notion of exterior differential, and we would like as many properties of “ordinary” differential forms as possible to remain valid. As will see in our somewhat sketchy presentation, these goals can be achieved, except for some properties of the exterior product. Using the isomorphism µF : p ^ (Rn)∗ ! ⊗F − →Altp(Rn; F) and Proposition 3.34, we obtain a convenient expression for differential forms in A∗(U; F). 180 CHAPTER 4. DIFFERENTIAL FORMS Proposition 4.22. If (e1, . . . , en) is any basis of Rn and (e∗ 1, . . . , e∗ n) is its dual basis, then every differential p-form ω ∈Ap(U; F) can be written uniquely as ω(x) = X I e∗ i1 ∧· · · ∧e∗ ip ⊗fI(x) = X I e∗ I ⊗fI(x) x ∈U, (∗2) where each fI : U →F is a smooth function on U. As explained in Section 3.9, to express the above formula directly on alternating multi-linear maps, define the product ·: Altp(Rn; R) × F →Altp(Rn; F) as follows. Definition 4.16. For all ω ∈Altp(Rn; R) and all f ∈F, define ω · f by (ω · f)(u1, . . . , up) = ω(u1, . . . , up)f, for all u1, . . . , up ∈Rn. Then it is immediately verified that for every ω ∈(Vp(Rn)∗) ⊗F of the form ω = u∗ 1 ∧· · · ∧u∗ p ⊗f, we have µF(u∗ 1 ∧· · · ∧u∗ p ⊗f) = µF(u∗ 1 ∧· · · ∧u∗ p) · f. By Proposition 3.36, the above property can be restated as the fact that for any basis (e1, . . . , en) of Rn, every differential p-form ω ∈Ap(U; F) can be written uniquely as ω(x) = X I e∗ i1 ∧· · · ∧e∗ ip · fI(x), x ∈U, where each fI : U →F is a smooth function on U. In order to define a multiplication on differential forms, we use a bilinear form Φ: F ×G → H; see Section 3.9. Definition 4.17. For every pair (p, q), we define the multiplication ∧Φ :  p ^ (Rn)∗ ⊗F ! ×  q ^ (Rn)∗ ⊗G ! − → p+q ^ (Rn)∗ ⊗H by (α ⊗f) ∧Φ (β ⊗g) = (α ∧β) ⊗Φ(f, g). We can also define a multiplication ∧Φ directly on alternating multilinear maps as follows: For f ∈Altp(Rn; F) and g ∈Altq(Rn; G), (f ∧Φ g)(u1, . . . , up+q) = X σ∈shuffle(p,q) sgn(σ) Φ(f(uσ(1), . . . , uσ(p)), g(uσ(p+1), . . . , uσ(p+q))), where shuffle(p, q) consists of all (m, n)-“shuffles;” that is, permutations σ of {1, . . . p + q} such that σ(1) < · · · < σ(p) and σ(p + 1) < · · · < σ(p + q). 4.5. VECTOR-VALUED DIFFERENTIAL FORMS 181 Then, we obtain a multiplication ∧Φ : Ap(U; F) × Aq(U; G) →Ap+q(U; H), defined so that for any differential forms ω ∈Ap(U; F) and η ∈Aq(U; G), (ω ∧Φ η)x = ωx ∧Φ ηx, x ∈U. In general, not much can be said about ∧Φ, unless Φ has some additional properties. In particular, ∧Φ is generally not associative, and there is no analog of Proposition 4.1. For simplicity of notation, we write ∧for ∧Φ. Using Φ, we can also define a multiplication ·: Ap(U; F) × A0(U; G) →Ap(U; H). Definition 4.18. The multiplication ·: Ap(U; F) × A0(U; G) →Ap(U; H) is given by (ω · f)x(u1, . . . , up) = Φ(ωx(u1, . . . , up), f(x)), for all x ∈U, all f ∈A0(U; G) = C∞(U, G), and all u1, . . . , up ∈Rn. This multiplication will be used in the case where F = R and G = H to obtain a normal form for differential forms. Generalizing d is no problem. Observe that since a differential p-form is a smooth map ω: U →Altp(Rn; F), its derivative is a map ω′ : U →Hom(Rn, Altp(Rn; F)) such that ω′ x is a linear map from Rn to Altp(Rn; F) for every x ∈U. We can view ω′ x as a multilinear map ω′ x : (Rn)p+1 →F which is alternating in its last p arguments. As in Section 4.1, the exterior derivative (dω)x is obtained by making ω′ x into an alternating map in all of its p + 1 arguments. Definition 4.19. For every p ≥0, the exterior differential d: Ap(U; F) →Ap+1(U; F) is given by (dω)x(u1, . . . , up+1) = p+1 X i=1 (−1)i−1ω′ x(ui)(u1, . . . , b ui, . . . , up+1), for all ω ∈Ap(U; F) and all u1, . . . , up+1 ∈Rn, where the hat over the argument ui means that it should be omitted. Note that d depends on the vector space F, so to be very precise we should denote d as dF. To keep notation simple, it is customary to drop the subscript F. For any smooth function f ∈A0(U; F) = C∞(U, F), we get d fx(u) = f ′ x(u). 182 CHAPTER 4. DIFFERENTIAL FORMS Therefore, for smooth functions, the exterior differential d f coincides with the usual derivative f ′. The important observation following Definition 4.3 also applies here. If xi : U →R is the restriction of pri to U, then x′ i is the constant map given by x′ i(x) = pri, x ∈U. It follows that dxi = x′ i is the constant function with value pri = e∗ i . As a consequence, every p-form ω can be uniquely written as ωx = X I dxi1 ∧· · · ∧dxip ⊗fI(x), (∗3) where each fI : U →F is a smooth function on U. Using the multiplication · induced by the scalar multiplication in F (Φ(λ, f) = λf, with λ ∈R and f ∈F), we see that every p-form ω can be uniquely written as ω = X I dxi1 ∧· · · ∧dxip · fI. (∗4) As for real-valued functions, for any f ∈A0(U; F) = C∞(U, F), we have d fx(u) = n X i=1 ui ∂f ∂xi (x) = n X i=1 e∗ i (u) ∂f ∂xi (x), and so, d f = n X i=1 dxi · ∂f ∂xi . (∗5) In general, Proposition 4.3 fails, unless F is finite-dimensional (see below). However for any arbitrary F, a weak form of Proposition 4.3 can be salvaged. Again, let Φ: F × G →H be a bilinear form, let ·: Ap(U; F) × A0(U; G) →Ap(U; H) be as defined before Definition 4.19, and let ∧Φ be the wedge product associated with Φ. The following fact is proved in Cartan (Section 2.4): Proposition 4.23. For all ω ∈Ap(U; F) and all f ∈A0(U; G), we have d(ω · f) = (dω) · f + ω ∧Φ d f. Fortunately, d ◦d still vanishes, but this requires a completely different proof, since we can’t rely on Proposition 4.2 (see Cartan , Section 2.5). Similarly, Proposition 4.2 holds, but a different proof is needed. 4.5. VECTOR-VALUED DIFFERENTIAL FORMS 183 Proposition 4.24. The composition Ap(U; F) d − →Ap+1(U; F) d − →Ap+2(U; F) is identi-cally zero for every p ≥0; that is d ◦d = 0, which is an abbreviation for dp+1 ◦dp = 0. To generalize Proposition 4.2, we use Proposition 4.23 with the product · and the wedge product ∧Φ induced by the bilinear form Φ given by scalar multiplication in F; that, is Φ(λ, f) = λf, for all λ ∈R and all f ∈F. Proposition 4.25. For every p form ω ∈Ap(U; F) with ω = dxi1 ∧· · · ∧dxip · f, we have dω = dxi1 ∧· · · ∧dxip ∧F d f, where ∧is the usual wedge product on real-valued forms and ∧F is the wedge product asso-ciated with scalar multiplication in F. More explicitly, for a p form ω = dxi1 ∧· · ·∧dxip ·f, for every x ∈U, for all u1, . . . , up+1 ∈ Rn, we have (dωx)(u1, . . . , up+1) = p+1 X i=1 (−1)i−1(dxi1 ∧· · · ∧dxip)x(u1, . . . , b ui, . . . , up+1)d fx(ui). If we use the fact that d f = n X i=1 dxi · ∂f ∂xi , we see easily that dω = n X j=1 dxi1 ∧· · · ∧dxip ∧dxj · ∂f ∂xj , (∗6) the direct generalization of the real-valued case, except that the “coefficients” are functions with values in F. The pull-back of forms in A∗(V, F) is defined as before. Luckily, Proposition 4.9 holds (see Cartan , Section 2.8). Proposition 4.26. Let U ⊆Rn and V ⊆Rm be two open sets and let ϕ: U →V be a smooth map. Then (i) ϕ∗(ω ∧η) = ϕ∗ω ∧ϕ∗η, for all ω ∈Ap(V ; F) and all η ∈Aq(V ; F). (ii) ϕ∗(f) = f ◦ϕ, for all f ∈A0(V ; F). 184 CHAPTER 4. DIFFERENTIAL FORMS (iii) dϕ∗(ω) = ϕ∗(dω), for all ω ∈Ap(V ; F); that is, the following diagram commutes for all p ≥0. Ap(V ; F) ϕ∗ / d  Ap(U; F) d  Ap+1(V ; F) ϕ∗/ Ap+1(U; F) Let us now consider the special case where F has finite dimension m. Pick any basis (f1, . . . , fm) of F. Then as every differential p-form ω ∈Ap(U; F) can be written uniquely as ω(x) = X I e∗ i1 ∧· · · ∧e∗ ip · fI(x), x ∈U, where each fI : U →F is a smooth function on U, by expressing the fI over the basis (f1, . . . , fm), we see that ω can be written uniquely as ω = m X i=1 ωi · fi, (∗7) where ω1, . . . , ωm are smooth real-valued differential forms in Ap(U; R), and we view fi as the constant map with value fi from U to F. Then as ω′ x(u) = m X i=1 (ω′ i)x(u)fi, for all u ∈Rn, we see that dω = m X i=1 dωi · fi. (∗8) Actually, because dω is defined independently of bases, the fi do not need to be linearly independent; any choice of vectors and forms such that ω = k X i=1 ωi · fi will do. Given a bilinear map Φ: F ×G →H, a simple calculation shows that for all ω ∈Ap(U; F) and all η ∈Ap(U; G), we have ω ∧Φ η = m X i=1 m′ X j=1 ωi ∧ηj · Φ(fi, gj), with ω = Pm i=1 ωi ·fi and η = Pm′ j=1 ηj ·gj, where (f1, . . . , fm) is a basis of F and (g1, . . . , gm′) is a basis of G. From this and Proposition 4.23, it follows that Proposition 4.3 holds for finite-dimensional spaces. 4.5. VECTOR-VALUED DIFFERENTIAL FORMS 185 Proposition 4.27. If F, G, H are finite dimensional and Φ: F × G →H is a bilinear map, then or all ω ∈Ap(U; F) and all η ∈Aq(U; G), d(ω ∧Φ η) = dω ∧Φ η + (−1)pω ∧Φ dη. On the negative side, in general, Proposition 4.1 still fails. A special case of interest is the case where F = G = H = g is a Lie algebra, and Φ(a, b) = [a, b] is the Lie bracket of g. In this case, using a basis (f1, . . . , fr) of g, if we write ω = P i αifi and η = P j βjfj, we have ω ∧Φ η = [ω, η] = X i,j αi ∧βj[fi, fj], where for simplicity of notation we dropped the subscript Φ on [ω, η] and the multiplication sign ·. Let us figure out what [ω, ω] is for a one-form ω ∈A1(U, g). By definition, [ω, ω] = X i,j ωi ∧ωj[fi, fj], so ω, ω = X i,j (ωi ∧ωj)(u, v)[fi, fj] = X i,j (ωi(u)ωj(v) −ωi(v)ωj(u))[fi, fj] = X i,j ωi(u)ωj(v)[fi, fj] − X i,j ωi(v)ωj(u)[fi, fj] = h X i ωi(u)fi, X j ωj(v)fj i − h X i ωi(v)fi, X j ωj(u)fj i = [ω(u), ω(v)] −[ω(v), ω(u)] = 2[ω(u), ω(v)]. Therefore, ω, ω = 2[ω(u), ω(v)]. Note that in general, [ω, ω] ̸= 0, because ω is vector valued. Of course, for real-valued forms, [ω, ω] = 0. Using the Jacobi identity of the Lie algebra, we easily find that [[ω, ω], ω] = 0. The generalization of vector-valued differential forms to manifolds is no problem, except that some results involving the wedge product fail for the same reason that they fail in the case of forms on open subsets of Rn. 186 CHAPTER 4. DIFFERENTIAL FORMS Definition 4.20. Given a smooth manifold M of dimension n and a vector space F, the set Ak(M; F) of differential k-forms on M with values in F is the set of maps p 7→ωp with ωp ∈ Vk T ∗ p M  ⊗F ∼ = Altk(TpM; F), which vary smoothly in p ∈M. This means that the map p 7→ωp(X1(p), . . . , Xk(p)) is smooth for all vector fields X1, . . . , Xk ∈X(M). It can be shown (see Proposition 10.12) that Ak(M; F) ∼ = Ak(M) ⊗C∞(M) C∞(M; F) ∼ = Altk C∞(M)(X(M); C∞(M; F)), which reduces to Proposition 4.15 when F = R. The reader may want to carry out the verification that the theory generalizes to manifolds on her/his own. The following result will be used in the next section. Proposition 4.28. If ω ∈A1(M; F) is a vector valued one-form, then for any two vector fields X, Y on M, we have dω(X, Y ) = X(ω(Y )) −Y (ω(X)) −ω([X, Y ]). In the next section we consider some properties of differential forms on Lie groups. 4.6 Differential Forms on Lie Groups and Maurer-Cartan Forms Given a Lie group G, it is well known that the set of left-invariant vector fields on G is isomorphic to the Lie algebra g = T1G of G (where 1 denotes the identity element of G); see Warner (Chapter 4) or Gallier and Quaintance . Recall that a vector field X on G is left-invariant iff d(La)b(Xb) = XLab = Xab, for all a, b ∈G. In particular, for b = 1, we get Xa = d(La)1(X1). which shows that X is completely determined by its value at 1. The map X 7→X(1) is an isomorphism between left-invariant vector fields on G and g. The above suggests looking at left-invariant differential forms on G. We will see that the set of left-invariant one-forms on G is isomorphic to g∗, the dual of g as a vector space. Definition 4.21. Given a Lie group G, a differential form ω ∈Ak(G) is left-invariant iff L∗ aω = ω, for all a ∈G, where L∗ aω is the pull-back of ω by La (left multiplication by a). The left-invariant one-forms ω ∈A1(G) are also called Maurer-Cartan forms. 4.6. DIFFERENTIAL FORMS ON LIE GROUPS 187 Here is a simple example of a left-invariant one-form on S1. Let g = (cos t, sin t) ∈S1. Then Lg : S1 →S1 is given by Lg(u, v) = (cos t u −sin t v, sin t u + cos t v) = (x, y). Let ω = −y dx + x dy. Then L∗ gω = (−sin t u −cos t v)d(cos t u −sin t v) + (cos t u −sin t v)d(sin t u + cos t v) = (−sin t u −cos t v)(cos t du −sin t dv) + (cos t u −sin t v)(sin t du + cos t dv) = −(sin2 t + cos2 t)v du + (sin2 t + cos2 t)u dv = −v du + u dv, which (by setting u →x and v →y) shows that L∗ gω = ω. For a one-form ω ∈A1(G) left-invariance means that (L∗ aω)g(u) = ωLag(d(La)gu) = ωag(d(La)gu) = ωg(u), for all a, g ∈G and all u ∈TgG. For a = g−1, we get ωg(u) = ω1(d(Lg−1)gu) = ω1(d(L−1 g )gu), which shows that ωg is completely determined by ω1 (the value of ωg at g = 1). Proposition 4.29. The map ω 7→ω1 is an isomorphism between the set of left-invariant one-forms on G and g∗. Proof. First, for any linear form α ∈g∗, the one-form αL given by αL g (u) = α(d(L−1 g )gu), u ∈Tg(G) is left-invariant, because (L∗ hαL)g(u) = αL hg(d(Lh)g(u)) = α(d(L−1 hg )hg(d(Lh)g(u))) = α(d(L−1 hg ◦Lh)g(u)) = α(d(L−1 g )g(u)) = αL g (u). Secondly, we saw that for every one-form ω ∈A1(G), ωg(u) = ω1(d(L−1 g )gu), so ω1 ∈g∗is the unique element such that ω = ωL 1 , which shows that the map α 7→αL is an isomorphism whose inverse is the map ω 7→ω1. 188 CHAPTER 4. DIFFERENTIAL FORMS Now, since every left-invariant vector field is of the form X = uL for some unique u ∈g, where uL is the vector field given by uL(a) = d(La)1u, and since the left-invariance of ω implies that ωag(d(La)gu) = ωg(u), for g = 1, we get ωa(d(La)1u) = ω1(u); that is ωa(X) = ω1(u), a ∈G, which shows that ω(X) is a constant function on G. It follows that for every vector field Y (not necessarily left-invariant), Y [ω(X)] = 0. Recall that by Corollary 4.17, we have dω(X, Y ) = X[ω(Y )] −Y [ω(X)] −ω([X, Y ]). Consequently, for all left-invariant vector fields X, Y on G, for every left-invariant one-form ω, we have dω(X, Y ) = −ω([X, Y ]). If we identify the set of left-invariant vector fields on G with g and the set of left-invariant one-forms on G with g∗, we have dω(X, Y ) = −ω([X, Y ]), ω ∈g∗, X, Y ∈g. We summarize these facts in the following proposition. Proposition 4.30. Let G be any Lie group. (1) The set of left-invariant one-forms on G is isomorphic to g∗, the dual of the Lie algebra g of G, via the isomorphism ω 7→ω1. (2) For every left-invariant one form ω and every left-invariant vector field X, the value of the function ω(X) is constant and equal to ω1(X1). (3) If we identify the set of left-invariant vector fields on G with g and the set of left-invariant one-forms on G with g∗, then dω(X, Y ) = −ω([X, Y ]), ω ∈g∗, X, Y ∈g. Pick any basis X1, . . . , Xr of the Lie algebra g, and let ω1, . . . , ωr be the dual basis of g∗. There are some constants ck ij such that [Xi, Xj] = r X k=1 ck ijXk. 4.6. DIFFERENTIAL FORMS ON LIE GROUPS 189 The constants ck ij are called the structure constants of the Lie algebra g. Observe that ck ji = −ck ij. As ωi([Xp, Xq]) = ci pq and dωi(X, Y ) = −ωi([X, Y ]), we have dωi(X, Y ) = −ci pq. Proposition 4.31. The following equations known as the Maurer-Cartan equations hold: dωi = −1 2 X j,k ci jkωj ∧ωk. Proof. Since X j,k ci jkωj ∧ωk(Xp, Xq) = X j,k ci jk(ωj(Xp)ωk(Xq) −ωj(Xq)ωk(Xp)) = X j,k ci jkωj(Xp)ωk(Xq) − X j,k ci jkωj(Xq)ωk(Xp) = X j,k ci jkωj(Xp)ωk(Xq) + X j,k ci kjωj(Xq)ωk(Xp) = ci pq + ci pq = 2ci pq, we get the equations dωi = −1 2 X j,k ci jkωj ∧ωk. These equations can be neatly described if we use differential forms valued in g. Let ωMC be the one-form given by (ωMC)g(u) = d(L−1 g )gu, g ∈G, u ∈TgG. What ωMC does is to “bring back” a vector v ∈TgG to g = T1G. The same computation that showed that αL is left-invariant if α ∈g∗shows that ωMC is left-invariant, and obviously (ωMC)1 = id. Definition 4.22. Given any Lie group G, the Maurer-Cartan form on G is the g-valued differential 1-form ωMC ∈A1(G, g) given by (ωMC)g = d(L−1 g )g, g ∈G. The same argument that we used to prove Property (2) of Proposition 4.30 shows that for every left-invariant one-form ω ∈A1(G, g) and every left-invariant vector field X ∈X(G), the value of the function ω(X) is constant and equal to ω1(X1). In particular, this holds for the Maurer-Cartan form ωMC. As in Section 4.5, the Lie bracket on g induces a multiplication [−, −]: Ap(G; g) × Aq(G; g) →Ap+q(G; g) 190 CHAPTER 4. DIFFERENTIAL FORMS given by [ω, η] = X ij αi ∧βj · [fi, fj], where (f1, . . . , fr) is a basis of g and where ω = P i αi ·fi and η = P j βj ·fj. Using the same proof, we obtain the equation ω, ω = 2[ω(X), ω(Y )]. Recall that for every g ∈G, conjugation by g is the map given by a 7→gag−1; that is, a 7→(Lg ◦Rg−1)a, and the adjoint map Ad(g): g →g associated with g is the derivative of Adg = Lg ◦Rg−1 at 1; that is, we have Ad(g)(u) = d(Adg)1(u), u ∈g. Furthermore, it is obvious that Lg and Rh commute. Proposition 4.32. Given any Lie group G, for all g ∈G, the Maurer-Cartan form ωMC has the following properties: (1) (ωMC)1 = idg. (2) For all g ∈G, R∗ gωMC = Ad(g−1) ◦ωMC. (3) The 2-form dωMC ∈A2(G, g) satisfies the Maurer-Cartan equation dωMC = −1 2[ωMC, ωMC]. Proof. Property (1) is obvious. (2) For simplicity of notation, if we write ω = ωMC, then (R∗ gω)h = ωhg ◦d(Rg)h = d(L−1 hg )hg ◦d(Rg)h = d(L−1 hg ◦Rg)h = d((Lh ◦Lg)−1 ◦Rg)h = d(L−1 g ◦L−1 h ◦Rg)h = d(L−1 g ◦Rg ◦L−1 h )h = d(Lg−1 ◦Rg)1 ◦d(L−1 h )h = Ad(g−1) ◦ωh, as claimed. 4.6. DIFFERENTIAL FORMS ON LIE GROUPS 191 (3) We can easily express ωMC in terms of a basis of g. If X1, . . . , Xr is a basis of g and ω1, . . . , ωr is the dual basis, then by Proposition 4.30 (2) and Part (1) of Proposition 4.32, we have ωMC(Xi) = (ωMC)1(Xi) = Xi, for i = 1, . . . , r, so ωMC is given by ωMC = ω1 · X1 + · · · + ωr · Xr, (∗9) under the usual identification of left-invariant vector fields (resp. left-invariant one forms) with elements of g (resp. elements of g∗). Then we have dωMC = dω1 · X1 + · · · + dωr · Xr. (∗10) We will use the Maurer-Cartan equations dωi = −1 2 X j,k ci jkωj ∧ωk to obtain the desired equation. Using the fact that the ci jk are skew-symmetric in j, k, for all u, v ∈g, we have ωMC, ωMC = h X j ωj(u) · Xj, X k ωj(v) · Xk i = X i,j,k ωj(u)ωk(v)ci jk · Xi = X i,j,k ci jk(ωj ∧ωk)(u, v) · Xi = −2 X i dωi(u, v) · Xi = −2dωMC(u, v), namely dωMC = −1 2[ωMC, ωMC], as claimed. In the case of a matrix group G ⊆GL(n, R), it is easy to see that the Maurer-Cartan form is given explicitly by ωMC(v) = g−1v, v ∈TgG, g ∈G. Since TgG is isomorphic to gg, we have ωMC(gv) = v, v ∈g. 192 CHAPTER 4. DIFFERENTIAL FORMS The above expression suggests that, with some abuse of notation, ωMC may be denoted by g−1dg, where g = (gij) and where dg is an abbreviation for the n × n matrix (dgij). Thus, ωMC is a kind of logarithmic derivative of the identity. For n = 2, if we write g = α β γ δ  , we get ωMC = 1 αδ −βγ  δdα −βdγ δdβ −βdδ −γdα + αdγ −γdβ + αdδ  . Remarks: (1) The quantity, dωMC + 1 2[ωMC, ωMC] is the curvature of the connection ωMC on G. The Maurer-Cartan equation says that the curvature of the Maurer-Cartan connection is zero. We also say that ωMC is a flat connection. (2) As dωMC = −1 2[ωMC, ωMC], we get d[ωMC, ωMC] = 0, which yields [[ωMC, ωMC], ωMC] = 0. It is easy to show that the above expresses the Jacobi identity in g. (3) As in the case of real-valued one-forms, for every left-invariant one-form ω ∈A1(G, g), we have ωg(u) = ω1(d(L−1 g )gu) = ω1((ωMC)gu), for all g ∈G and all u ∈TgG, and where ω1 : g →g is a linear map. Consequently, there is a bijection between the set of left-invariant one-forms in A1(G, g) and Hom(g, g). (4) The Maurer-Cartan form can be used to define the Darboux derivative of a map f : M →G, where M is a manifold and G is a Lie group. The Darboux derivative of f is the g-valued one-form ωf ∈A1(M, g) on M given by ωf = f ∗ωMC. Then it can be shown that when M is connected, if f1 and f2 have the same Darboux derivative ωf1 = ωf2, then f2 = Lg ◦f1, for some g ∈G. Elie Cartan also characterized which g-valued one-forms on M are Darboux derivatives (dω + 1 2[ω, ω] = 0 must hold). For more on Darboux derivatives, see Sharpe (Chapter 3) and Malliavin (Chapter III). 4.7. PROBLEMS 193 4.7 Problems Problem 4.1. Recall that d: Ap(U) →Ap+1(U) is given by (dω)x(u1, . . . , up+1) = p+1 X i=1 (−1)i−1ω′ x(ui)(u1, . . . , b ui, . . . , up+1), for all ω ∈Ap(U), all x ∈U, and all u1, . . . , up+1 ∈Rn. Show that (dω)x is alternating in its p + 1 arguments. Problem 4.2. Given the 1-form ω(x,y) = −y x2 + y2 dx + x x2 + y2 dy, on U = R2 −{0}, we have dω = 0. Show that there is no smooth function f on U such that d f = ω. Hint. See Madsen and Tornehave , Chapter 1. Problem 4.3. Let U ⊆Rn, V ⊆Rm, W ⊆Rp be three open subsets. Let ϕ: U →V , ψ: V →W, and id: U →U be smooth maps. Show that id∗= id, (ψ ◦ϕ)∗= ϕ∗◦ψ∗. Problem 4.4. Complete the proof of Proposition 4.12. Problem 4.5. Complete the proof details for Proposition 4.16. Hint. See Morita (Chapter 2). Problem 4.6. Let M be a smooth manifold. For every vector field X ∈X(M), for all k ≥1, let i(X): Ak(M) →Ak−1(M) the the linear insertion operator defined in Definition 4.13. Show that i(fX)ω = fi(X)ω, i(X)(fω) = fi(X)ω, f ∈C∞(M), ω ∈Ak(M), Also prove that i(X)(ω ∧η) = (i(X)ω) ∧η + (−1)rω ∧(i(X)η), for all ω ∈Ar(M) and all η ∈As(M). Problem 4.7. Let M be a smooth manifold. For every vector field X ∈X(M), prove the following: (1) For all ω ∈Ar(M) and all η ∈As(M), LX(ω ∧η) = (LXω) ∧η + ω ∧(LXη); 194 CHAPTER 4. DIFFERENTIAL FORMS (2) The Lie derivative commutes with d: LX ◦d = d ◦LX. Hint. See Warner , Morita , and Gallot, Hullin and Lafontaine . Problem 4.8. Prove Proposition 4.21. Problem 4.9. Prove Proposition 5.12. Hint. See Gallot, Hullin and Lafontaine (Chapter 1). Problem 4.10. Show that the composition Ap(U; F) d − →Ap+1(U; F) d − →Ap+2(U; F) is identically zero for every p ≥0; that is d ◦d = 0. Hint. See Cartan , Section 2.5. Problem 4.11. (i) Prove Proposition 4.23. Hint. See Cartan , Section 2.4. (ii) Use Proposition 4.23 with the product · and the wedge product ∧Φ induced by the bilinear form Φ given by scalar multiplication in F to prove Proposition 4.25. Problem 4.12. Prove Proposition 4.26. Hint. See Cartan , Section 2.8. Problem 4.13. Let F and G be finite dimensional vector spaces with (f1, . . . , fm) a basis of F and (g1, . . . , gm′) a basis of G. Show that ω ∧Φ η = m X i=1 m′ X j=1 ωi ∧ηj · Φ(fi, gj), where ω = Pm i=1 ωi · fi and η = Pm′ j=1 ηj · gj. Problem 4.14. Use Problem 4.13 to prove Proposition 4.27. Chapter 5 Tensor Fields In this chapter we explore tensor fields in more depth. In particular, we discuss tensor derivations and the generalization of the covariant derivative and the Lie derivative to tensor fields. These have applications in physics, especially in the general theory of relativity. 5.1 Tensor Fields as Sections of the Tensor Bundle T r,s(M) A tensor field T ∈Γ(M, T r,s(M)) is a smooth section of the tensor bundle T r,s(M) = T ⊗rM ⊗(T ∗M)⊗s. Technically, since in view of Proposition 2.25 we have the canonical isomorphism µ: T r,s(TpM) →Hom((TpM ∗)r, (TpM)s; R) for every p ∈M, we prefer using the following definition. Definition 5.1. A tensor field T ∈Γ(M, T r,s(M)) is a smooth map assigning a multilinear map Tp ∈T r,s(TpM) = Hom((TpM ∗)r, TpM s; R) to every p ∈M. Following Abraham and Marsden , we denote the space of tensor fields Γ(M, T r,s(M)) as T r,s(M). Example 5.1. Observe that a tensor field in T 0,1(M) is a differential one-form, so T 0,1(M) = A1(M). A tensor field T in T 1,0(M) is a vector field such that Tp ∈(TpM)∗∗for every p, so using the isomorphism between TpM and (TpM)∗∗, T defines a vector field in X(M). Typically, we view a vector field X ∈X(M) as the tensor field TX ∈T 1,0(M) given by (TX)p(θ) = θ(Xp), θ ∈(TpM)∗, and we identify X(M) and T 1,0(M) using the isomorphism X 7→TX. To simplify notation, we write X instead of TX, and by abuse of notation, we write X ∈T 1,0(M). We also identify T 0,0(M) with C∞(M). 195 196 CHAPTER 5. TENSOR FIELDS Tensor products and contractions of tensor fields in T r,s(M) are defined fibrewise. Definition 5.2. The tensor product ⊗: T r1,s1(M) × T r2,s2(M) − →T r1+r2,s1+s2(M) is defined such that for any tensor fields T1 ∈T r1,s1(M) and T2 ∈T r2,s2(M), (T1 ⊗T2)p = (T1)p ⊗(T2)p for all p ∈M, where (T1)p ⊗(T2)p is the tensor product of the tensors (T1)p ∈T r1,s1(TpM) and (T2)p ∈T r2,s2(TpM) as in Definition 2.17. Definition 5.3. The contraction operator ci,j : T r,s(M) →T r−1,s−1(M) is given by (ci,jT)p = ci,j(Tp), p ∈M, for any tensor field T ∈T r,s(M), where ci,j(Tp) on the right-hand side is given by Propositions 2.29 and 2.30. For any function f ∈C∞(M) and any tensor field T ∈T r,s(M), the tensor field fT ∈T r,s(M) is given by (fT)p = f(p)Tp, p ∈M. The Kronecker tensor field δ ∈T 1,1(M) is defined such that for all p ∈M, δp(θ, u) = (δp)1 1(θ, u) = θ(u), θ ∈(TpM)∗, u ∈TpM. Observe that Kronecker’s δ is related to the contraction c1,1 by the equation (c1,1)p(u ⊗θ) = δp(θ, u) = θ(u), θ ∈(TpM)∗, u ∈TpM, but δ ∈T 1,1(M) and c1,1 is a map c1,1 : T 1,1(M) →C∞(M). Recall that u ⊗θ is really the bilinear map µ(u ⊗θ) ∈T 1,1(TpM). The operation (f, T) 7→fT from C∞(M) × T r,s(M) to T r,s(M) makes T r,s(M) into a C∞(M)-module. It is immediately verified that the tensor product is C∞(M)-bilinear and associative. Given r vector fields Xi ∈X(M) ∼ = T 1,0(M) and s differential one-forms ωj ∈A1(M) = T 0,1(M), by taking their tensor product as in Definition 5.2, we obtain a tensor field X1 ⊗· · · ⊗Xr ⊗ω1 ⊗· · · ⊗ωs in T r,s(M) given explicitly by the formula ((X1 ⊗· · · ⊗Xr ⊗ω1 ⊗· · · ⊗ωs)p((θ1)p, . . . , (θr)p, (Y1)p, . . . , (Ys)p) = r Y i=1 (θi)p((Xi)p) s Y j=1 (ωj)p((Yj)p), (†1) 5.1. TENSOR FIELDS AS SECTIONS OF THE TENSOR BUNDLE T r,s(M) 197 (Yj)p ∈TpM (1 ≤j ≤s), (θi)p ∈(TpM)∗(1 ≤i ≤r). Formula (†1) will allow us to express a tensor field in a local chart. In this chapter it pays offto adopt Einstein summation convention; see Section 2.8. Let (U, ϕ) be a local chart on M, with ϕ: U →Rn, and let xi = pri ◦ϕ, the ith local coordinate (1 ≤i ≤n); see Tu (Chapter 3, §8) or Gallier and Quaintance . Note carefully that we now denote the local ith coordinate as xi with a superscript, in order to be able to use the Einstein summation convention. Recall that the vector fields in X(U), ∂ ∂x1, . . . , ∂ ∂xn form a basis of the C∞(M)-module X(U). Furthermore, the differential one-forms dx1, . . . , dxn form a basis of the C∞(M)-module A1(U), (where (dxi)p, the differential of xi at p, is iden-tified with the linear form such that d fp(v) = v(f), for every smooth function f on U and every v ∈TpM). The basis dx1, . . . , dxn of A1(U) is the dual of the basis ∂ ∂x1, . . . , ∂ ∂xn  of X(U), which means that for all p ∈U, (dxi)p  ∂ ∂xj  p ! = δi,j, 1 ≤i, j ≤n. For all (i1, . . . , ir) and all (j1, . . . , js), with 1 ≤ik, jl ≤n, the tensor products ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs are tensor fields in T r,s(U). Proposition 5.1. The tensor fields ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs (†2) form a basis of the C∞(M)-module T r,s(U). Consequently, locally on U, every tensor field T ∈T r,s(M) can be written uniquely as Tp = ai1...ir j1...js(p)  ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs  (p), p ∈U, (†3) where ai1,...,ir j1,...,js is a smooth function on U given by ai1...ir j1...js(p) = Tp (dxi1)p, . . . , (dxir)p,  ∂ ∂xj1  p , . . . ,  ∂ ∂xjs  p ! . (†4) 198 CHAPTER 5. TENSOR FIELDS Proof. For any vectors (Xl)p = Xjl l (p) ∂ ∂xjl  p in TpM and any one-forms (θk)p = θk ik(p)(dxik)p in (TpM)∗, with 1 ≤l ≤s and 1 ≤k ≤r, by multilinearity, we have Tp((θ1)p, . . . , (θr)p, (X1)p, . . . , (Xs)p) = Tp (dxi1)p, . . . , (dxir)p,  ∂ ∂xj1  p , . . . ,  ∂ ∂xjs  p ! θ1 i1(p) · · · θr ir(p)Xj1 1 (p) · · · Xjs s (p), which shows that the tensor fields in (†2) span the C∞(M)-module T r,s(U), with coefficients given by (†4). The tensor fields in (†2) are linearly independent, because since (dxi)p  ∂ ∂xj  p ! = δi,j, 1 ≤i, j ≤n, we have  ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs  p (dxi′ 1)p, . . . , (dxi′ r)p,  ∂ ∂xj′ 1  p , . . . ,  ∂ ∂xj′ s  p ! = δi1,i′ 1 · · · δir,i′ r · · · δj1,j′ 1 · · · δjs,j′ s, so if we have a C∞(M)-linear combination of the tensor fields in (†2) with coefficents λi1...ir j1...js ∈ C∞(U), evaluating this linear combination at p on (dxi1)p, . . . , (dxir)p,  ∂ ∂xj1  p , . . . ,  ∂ ∂xjs  p yields λi1...ir j1...js(p) = 0. Proposition 5.1 implies that if we have a family T α of tensor fields in T r,s(Uα) each locally defined on the domain Uα of a chart of a manifold M, and if the tensor fields T α and T β agree on the overlap Uα ∩Uβ, then the T α patch to form a tensor field T ∈T r,s(M). 5.2 Tensor Fields as C∞(M)-Multilinear Maps Generalizing what we did just before Proposition 4.15, tensor fields can be viewed as C∞(M)-multilinear maps. Definition 5.4. Given a tensor field T ∈T r,s(M) = Γ(M, T r,s(M)), we define a C∞(M)-multilinear map µr,s(T): A1(M) × · · · × A1(M) | {z } r × X(M) × · · · × X(M) | {z } s − →C∞(M), 5.2. TENSOR FIELDS AS C∞(M)-MULTILINEAR MAPS 199 given by µr,s(T)(ω1, . . . , ωr, X1, . . . Xr)(p) = Tp((ω1)p, . . . , (ωr)p, (X1)p, . . . (Xs)p), for all differential one-forms ωi ∈A1(M), all vector fields Xj ∈X(M), and all p ∈M. Note that we are commiting an abuse of notation since the notation µr,s was already used to denote the map on tensors from T r,s(V ) to T r,s(V ). The context should make it clear whether we are dealing with tensors or tensor fields. We also usually suppress the superscripts r and s. Conversely, a C∞(M)-multilinear map Φ: A1(M) × · · · × A1(M) | {z } r × X(M) × · · · × X(M) | {z } s − →C∞(M) induces a family of multilinear linear maps Tp ∈T r,s(TpM) for every p ∈M which constitute a tensor field. This fact relies on the following two propositions (see O’Neill , Chapter 2, Lemma 3 and Proposition 2). Proposition 5.2. Given a C∞(M)-multilinear map Φ: A1(M) × · · · × A1(M) | {z } r × X(M) × · · · × X(M) | {z } s − →C∞(M), for any p ∈M, for any one forms ωi ∈A1(M) (1 ≤i ≤r) and any vector fields Xj ∈X(M) (1 ≤j ≤s), if either (ωi)p = 0 for some i or (Xj)p = 0 for some j, then Φ(ω1, . . . , ωr, X1, . . . Xs)(p) = 0. Proposition 5.2 is used to prove the next proposition. Proposition 5.3. Given a C∞(M)-multilinear map Φ: A1(M) × · · · × A1(M) | {z } r × X(M) × · · · × X(M) | {z } s − →C∞(M), for any p ∈M, for any one forms ωi, θi ∈A1(M) (1 ≤i ≤r) and any vector fields Xj, Yj ∈X(M) (1 ≤j ≤s), if (ωi)p = (θi)p for i = 1, . . . r and (Xj)p = (Yj)p for j = 1, . . . s, then Φ(ω1, . . . , ωr, X1, . . . Xs)(p) = Φ(θ1, . . . , θr, Y1, . . . Xs)(p). A C∞(M)-multilinear map Φ induces a tensor Tp ∈T r,s(TpM) for every p ∈M as follows. For any linear forms αi ∈(TpM)∗(1 ≤i ≤r) and any tangent vectors uj ∈TpM (1 ≤j ≤s), we can pick a smooth function f equal to 1 on some open set U containing p 200 CHAPTER 5. TENSOR FIELDS and 0 outside U (a bump function), so that we get smooth one-forms ωi and vector fields Xj with (ωi)p = αi for i = 1, . . . , r and (Xj)p = uj for j = 1, . . . , s, and we set Tp(α1, . . . , αr, u1, . . . , us) = Φ(ω1, . . . , ωr, X1, . . . , Xs)(p). By Proposition 5.3, the above definition does not depend on the one-forms ωi and the vector fields Xj chosen such that (ωi)p = αi and (Xj)p = uj. As a cororally we obtain the following isomorphism. Theorem 5.4. Let M be a smooth manifold. There is an isomorphism between the space of of tensor fields T r,s(M) and the space HomC∞(M)((A1(M))r, (X(M))s); C∞(M)) of C∞(M)-multilinear maps Φ: A1(M) × · · · × A1(M) | {z } r × X(M) × · · · × X(M) | {z } s − →C∞(M). Definition 5.5. The space HomC∞(M)((A1(M))r, (X(M))s); C∞(M)) is denoted as Tr,s(M) (see Abraham and Marsden ). With a slight abuse of language, we also call the multilinear maps in Tr,s(M) tensor fields. Multiplication of a tensor field in Tr,s(M) by a smooth function and tensor products of tensor fields (in Tr,s(M) for some r, s) are defined as follows. Recall that given two functions f, g: M →R, their product fg is defined pointwise by (fg)(p) = f(p)g(p), p ∈M. Definition 5.6. The tensor product ⊗: Tr1,s1(M) × Tr2,s2(M) − →Tr1+r2,s1+s2(M) is defined such that for any two tensors T1 ∈Tr1,s1(M) and T2 ∈Tr2,s2(M), (T1 ⊗T2)(ω1, . . . , ωr1, θ1, . . . , θr2, X1, . . . , Xs1, Y1, . . . , Ys2) = T1(ω1, . . . , ωr1, X1, . . . , Xs1)T2(θ1, . . . , θr2, Y1, . . . , Ys2), for all ωi1 ∈A1(M) (1 ≤i1 ≤r1), Xj1 ∈X(M) (1 ≤j1 ≤s1), θi2 ∈A1(M) (1 ≤i2 ≤r2), Yj2 ∈X(M) (1 ≤j2 ≤s2), where T1(ω1, . . . , ωr1, X1, . . . , Xs1)T2(θ1, . . . , θr2, Y1, . . . , Ys2) is the product of the functions T1(ω1, . . . , ωr1, X1, . . . , Xs1) and T2(θ1, . . . , θr2, Y1, . . . , Ys2). In the special case where r1 = s1 = 0, T1 = f is a function in C∞(M) and f ⊗T2 is also denoted by fT2. Similarly, in the special case where r2 = s2 = 0, T2 = g is a function in C∞(M) and T1 ⊗g is also denoted by T1g. It is immediately verified that the tensor product is C∞(M)-bilinear and associative. However, it is not commutative (find a counter-example). 5.2. TENSOR FIELDS AS C∞(M)-MULTILINEAR MAPS 201 Remark: We admit that the notation ⊗for tensor products is heavily overloaded, since it has at least four different meanings: (1)-(2) for tensors in the space T r,s(V ) and in the space of multilimear forms in T r,s(V ), where V is a vector space; (3) for tensor fields in T r,s(M); (4) for C∞(M)-multilinear maps in the space Tr,s(M), where M is a manifold. This is quite confusing, but hopefully context should make it clear which tensor product is used. Example 5.2. Proposition 4.15 shows that as C∞(M)-modules, A1(M) ∼ = T0,1(M). The isomorphism is defined as follows. Given a differential one-form θ ∈A1(M) and a vector field X ∈X(M), we have the C∞(M)-linear map Tθ from X(M) to C∞(M) given by (Tθ(X))(p) = θp(Xp), p ∈M. Example 5.3. Given a vector field X ∈X(M), we obtain a map TX from A1(M) to C∞(M) given by (TX(θ))(p) = θp(Xp), p ∈M. We check immediately that this map is C∞(M)-linear, so TX ∈T1,0(M). It can be shown that the map X 7→TX is a bijection (see O’Neill , Chapter 2). Therefore, X(M) is isomorphic to T1,0(M). As in Section 5.1, given r vector fields Xi ∈X(M) ∼ = T1,0(M) and s differential one-forms ωj ∈A1(M) ∼ = T0,1(M), by taking their tensor product as in Definition 5.6, we obtain a tensor field X1 ⊗· · · ⊗Xr ⊗ω1 ⊗· · · ⊗ωs in Tr,s(M) given explicitly by the formula  (X1 ⊗· · · ⊗Xr ⊗ω1 ⊗· · · ⊗ωs)(θ1, . . . , θr, Y1, . . . , Ys)  (p) = r Y i=1 (θi)p((Xi)p) s Y j=1 (ωj)p((Yj)p), (†5) Yj ∈X(M) (1 ≤j ≤s), θi ∈A1(M) (1 ≤i ≤r). This fact leads to a version of Proposition 5.1 for tensor fields in Tr,s(M). Let (U, ϕ) be a local chart on M, with ϕ: U →Rn, and let xi = pri ◦ϕ, the ith local coordinate (1 ≤i ≤n). Recall that the vector fields in X(U), ∂ ∂x1, . . . , ∂ ∂xn form a basis of the C∞(M)-module X(U). Furthermore, the differential one-forms dx1, . . . , dxn form a basis of the C∞(M)-module A1(U). For all (i1, . . . , ir) and all (j1, . . . , js), with 1 ≤ik, jl ≤n, the tensor products ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs are tensor fields in Tr,s(U). 202 CHAPTER 5. TENSOR FIELDS Proposition 5.5. The tensor fields ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs (†6) form a basis of the C∞(M)-module Tr,s(U). Consequently, locally on U, every tensor field T ∈Tr,s(M) can be written uniquely as T = ai1...ir j1...js ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs, (†7) where ai1,...,ir j1,...,js is a smooth function on U given by ai1...ir j1...js = T  dxi1, . . . , dxir, ∂ ∂xj1 , . . . , ∂ ∂xjs  . (†8) There is also a version of the contraction operators ci,j : Tr,s(M) →Tr−1,s−1(M). The method for defining these contractions is analogous to the method used in Section 2.9 to define contractions on the space T r,s(V ). Following O’Neill (Chapter 2), we first define contraction operators locally in terms of charts (U, ϕ) on the tensors in Tr,s(U). Since these operators agree on overlaps, they patch to give contraction operators globally defined. We sketch how to proceed with the contraction c1,1 : T1,1(M) →C∞(M), leaving the case of the more general contractions ci,j as as exercise. The proof mimicks the proof of Proposition 2.29. We need to prove that there is a unique C∞(M)-linear map c1,1 : T1,1(M) →C∞(M) such that c1,1(X ⊗θ) = θ(X), X ∈X(M), θ ∈A1(M). (†9) It suffices to prove that c1,1 is uniquely defined on T1,1(U), where (U, ϕ) is a chart. As usual, we have the local coordinates xi = pri ◦ϕ. By Proposition 5.5, a tensor field T ∈T1,1(U) is written uniquely as T = ti j ∂ ∂xi ⊗dxj, so (†9) implies that c1,1(T) = ti i. If (V, ψ) is another chart with local coordinates yi = pri ◦ψ and U ∩V ̸= ∅, the map ψ ◦ϕ−1 has the Jacobian matrix  ∂yi ∂xj  1≤i,j≤n, and its inverse ϕ ◦ψ−1 has the Jacobian matrix  ∂xi ∂yj  1≤i,j≤n, which is the inverse of the previous matrix. Then on the overlap U ∩V we have  ∂ ∂xi  p = ∂yj ∂xi (ϕ(p))  ∂ ∂yj  p . 5.3. TENSOR FIELD DERIVATIONS AND WILMORE’S THEOREM 203 At this point, we have the change of basis matrix (aj i) =  ∂yj ∂xi  1≤i,j≤n expressing the basis  ∂ ∂xi  (1 ≤i ≤n) in terms of the basis  ∂ ∂yj  (1 ≤j ≤n), and we also have (dyi)p = ∂yi ∂xj (ϕ(p))(dxj)p, so we can repeat the computation in the proof of Proposition 2.29 and deduce that c1,1(T) is independent of the change of basis. We are now ready to tackle the important topic of tensor derivations. 5.3 Tensor Field Derivations and Wilmore’s Theorem There are two parallel approaches to the topic of tensor field derivations depending on which version of tensor fields is used; as the space T r,s(M) of sections of the tensor bundle T r,s(M), or as C∞(M)-multilinear maps in Tr,s(M). The first approach is discussed in Gallot, Hulin, Lafontaine and Abraham and Marsden , and the second in O’Neill and Sakai . We present the first approach, following quite closely Abraham and Marsden , because it is more fundamental since it does not rely on the smoothness of the manifold M. We also find it a bit more elegant since it does not involve the contractions but instead the Kronecker tensor field delta (see Definition 5.2). Wilmore’s theorem asserts the existence of differential operators on tensor fields that satisfy a Leibniz-type property with respect to tensor products. Given any (smooth) manifold M and any open subset U of M, the subspace of T r,s(M) consisting of the restrictions of tensor fields in T r,s(M) to U is denoted T r,s(U). Definition 5.7. A differential operator on tensor fields on a (smooth or Ck, k ≥1) manifold M is a family of D of maps Dr sU : T r,s(U) →T r,s(U) for every nonempty open subset U of M and all pairs r, s ≥0, such that the following properties hold. (DO1) If we write Dr s = Dr sM, then each Dr s is a tensor derivation, which means that Dr s is R-linear and that Dr1+r2 s1+s2(T1 ⊗T2) = Dr1 s1(T1) ⊗T2 + T1 ⊗Dr2 s2(T2), for all T1 ∈T r1,s1(M) and all T2 ∈T r2,s2(M). (DO2) The family (Dr sU) is local, or natural with respect to restriction, that is, for all open subsets U, V of M such that U ⊆V and all T ∈T r,s(V ), ((Dr sV )T)|U = (Dr sU)(T|U) ∈T r,s(U), 204 CHAPTER 5. TENSOR FIELDS where |U means restriction to U, or equivalently the following diagram commutes. T r,s(V ) |U / Dr sV  T r,s(U) Dr sU  T r,s(V ) |U / T r,s(U). (DO3) (D1 1U)δ = 0, where δ1 1 ∈T 1,1(U) is Kronecker’s delta. Even though this is an abuse of notation, the indices r and s in Dr s are often omitted. The following fundamental theorem shows that given a family of maps EU : C∞(U) → C∞(U) and FU : X(U) →X(U) for each open subset U of M, if these maps are local tensor derivations, then they have a unique extension to a differential operator on the tensor field algebras T r,s(U). Theorem 5.6. (Wilmore) Assume that for every open subset U of M, we have a family of maps EU : C∞(U) →C∞(U) and FU : X(U) →X(U), which are R-linear tensor derivations and natural with respect to restriction. More explicitly, this means that (1) EU(fg) = (EU(f))g + fEU(g), f, g ∈C∞(U). (2) EU(f|U) = (EMf)|U, f ∈C∞(M). (3) FU(fX) = (EU(f))X + fFU(X), f ∈C∞(U), X ∈X(U). (4) FU(X|U) = (FMX)|U, X ∈X(M). Then there is a unique family of differential operators Dr sU : T r,s(U) →T r,s(U) that coincides with EU on C∞(U) ∼ = T 0,0(U) and with FU on X(U) ∼ = T 1,0(U). Furthermore, define the product rule as follows. (DO4) For all T ∈T r,s(M), ω1, . . . , ωr ∈A1(M) and X1, . . . , Xs ∈X(M), D0 0(T(ω1, . . . , ωr, X1, . . . , Xs)) = (Dr sT)(ω1, . . . , ωr, X1, . . . , Xs) + r X i=1 T(ω1, . . . , D0 1ωi, . . . , ωr, X1, . . . , Xs) + s X j=1 T(ω1, . . . ωr, X1, . . . , D1 0Xj, . . . , Xs). In the above equations we are using implicitly Definition 5.4 involving µr,s to view tensor fields in T r,s(M) as tensor fields in Tr,s(M). Under the hypothesis of (DO1) and (DO2), (DO3) is equivalent to (DO4). If M is smooth, (DO4) determines the operators Dr sU. In the special case r = 0, s = 1, T is a differential one-form θ ∈A1(M) = T 0,1(M), and we have (D0 1θ)(X) = D0 0(θ(X)) −θ(D1 0X) = E(θ(X)) −θ(FX), X ∈X(M). (Dθ) 5.3. TENSOR FIELD DERIVATIONS AND WILMORE’S THEOREM 205 Proof. Suppose a family of operators Dr sU satisfying the conditions of the theorem exists. Consider any chart (U, ϕ). By (DO2) and (2), (4) above we may restrict our attention to tensor fields in T r,s(U). By Proposition 5.1, every tensor field T ∈T r,s(U) can be written uniquely as Tp = ai1...ir j1...js(p)  ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs  (p), p ∈U, (†10) where ai1,...,ir j1,...,js is a smooth function on U. Using R-linearity and (DO1), we obtain (Dr sU)T = (D0 0U) ai1...ir j1...js  ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs  + r X k=1 ai1...ir j1...js  ∂ ∂xi1 ⊗· · · ⊗(D1 0U) ∂ ∂xik ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs  + s X l=1 ai1...ir j1...js  ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗(D0 1U)dxjl ⊗· · · ⊗dxjs  , and since D0 0U agrees with EU on C∞(M) and D1 0U agrees with FU on X(U), we have (D0 0U) ai1...ir j1...js = EUai1...ir j1...js and (D1 0U) ∂ ∂xik = FU ∂ ∂xik , so (Dr sU)T is uniquely determined if the expressions (D0 1(U)dxjl are uniquely determined. This is where (DO3) comes in. The trick is that locally on U, the Kronecker’s tensor field δ is given by the expression (using Einstein’s sum convention) δp =  ∂ ∂xj  p ⊗(dxj)p, p ∈U, (δo) as we check by applying both sides to  (dxl)p, ∂ ∂xk  p  , since (dxj)p  ∂ ∂xk  p ! = δj,k, (dxl)p  ∂ ∂xj  p ! = δl,j, so δp (dxl)p,  ∂ ∂xk  p ! = (dxl)p  ∂ ∂xk  p = δl,k and  ∂ ∂xj  p ⊗(dxj)p (dxl)p,  ∂ ∂xk  p ! = (dxj)p  ∂ ∂xk  p ! (dxl)p  ∂ ∂xj  p ! = n X j=1 δj,kδl,j = δl,k. 206 CHAPTER 5. TENSOR FIELDS Note that if we denote δj,k as δj k and δl,j as δl j, then according to the Einstein summation convention n X j=1 δj,kδl,j = δj kδl j. Now if we apply D1 1U to both sides of (δp), since by (DO3) we have (D1 1U)δ = 0, by (DO1), we obtain (D1 1U)  ∂ ∂xj ⊗dxj  = (D1 0U) ∂ ∂xj ⊗dxj + ∂ ∂xj ⊗(D0 1U)dxj = 0. Appying the above to  (dxk)p, ∂ ∂xi  p  , we obtain (dxk)p  (D1 0U) ∂ ∂xj  p ! (dxj)p  ∂ ∂xi  p ! + (dxk)p  ∂ ∂xj  p ! ((D0 1U)dxj)p  ∂ ∂xi  p ! = (dxk)p  FU ∂ ∂xi  p ! + ((D0 1U)dxk)p  ∂ ∂xi  p ! = 0. Therefore, we obtain ((D0 1U)dxk)p  ∂ ∂xi  p ! = −(dxk)p  FU ∂ ∂xi  p ! , p ∈U, in other words, ((D0 1U)dxk)  ∂ ∂xi  = −dxk  FU ∂ ∂xi  , on U, (†11) which shows that (D0 1U)dxk is uniquely defined on U. In summary, we proved that if the family of operators Dr sU exists, then it is unique. We now use the equations (Dr sU)T = EU ai1...ir j1...js  ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs  + r X k=1 ai1...ir j1...js  ∂ ∂xi1 ⊗· · · ⊗FU ∂ ∂xik ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs  + s X l=1 ai1...ir j1...js  ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗(D0 1U)dxjl ⊗· · · ⊗dxjs  , together with (†11) to define the operators Dr sU. Then we check that the resulting family is well defined (DO1), (DO2), (DO3) are satisfied, which is a tedious task. 5.3. TENSOR FIELD DERIVATIONS AND WILMORE’S THEOREM 207 To prove (DO4), we use (†10) applied to one-forms θk ∈A1(U) and vector fields Xl ∈X(U) witten locally as θk = θk i dxi and Xl = Xj l ∂ ∂xj for some functions θk i , Xj l ∈C∞(U). We obtain T(θ1, . . . , θr, X1, . . . , Xs) = ai1...ir j1...js θ1 i1 · · · θr irXj1 1 · · · Xjs s . By (DO1) for functions, (D0 0U)(T(θ1, . . . , θr, X1, . . . , Xs) = (EU ai1...ir j1...js)θ1 i1 · · · θr irXj1 1 · · · Xjs s + r X k=1 ai1...ir j1...js θ1 i1 · · · EUθk ik · · · θr irXj1 1 · · · Xjs s + s X l=1 ai1...ir j1...js θ1 i1 · · · θr irXj1 1 · · · EUXjl l · · · Xjs s . (†12) From (Dr sU)T = EU ai1...ir j1...js  ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs  + r X k=1 ai1...ir j1...js  ∂ ∂xi1 ⊗· · · ⊗FU ∂ ∂xik ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs  + s X l=1 ai1...ir j1...js  ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗(D0 1U)dxjl ⊗· · · ⊗dxjs  and θk = θk i dxi, Xl = Xj l ∂ ∂xj , we obtain (Dr sU)T  ((θ1, . . . , θr, X1, . . . , Xs) = EU ai1...ir j1...js θ1 i1 · · · θr irXj1 1 · · · Xjs s + r X k=1 ai1...ir j1...js θ1 i1 · · · θk  FU ∂ ∂xik  · · · θr irXj1 1 · · · Xjs s + s X l=1 ai1...ir j1...js θ1 i1 · · · θr irXj1 1 · · · ((D0 1U)dxjl)(Xl) · · · Xjs. s (†13) From (†3), we also have T(θ1, . . . , (D0 1U)θk, . . . θr, X1, . . . , Xs) = ai1...ir j1...js θ1 i1 · · · (D0 1U)θk  ∂ ∂xik  · · · θr irXj1 1 · · · Xjs s , (†14) 208 CHAPTER 5. TENSOR FIELDS and T θ1, . . . , θr, X1, . . . , (D1 0U)(Xl), . . . , Xs  = ai1...ir j1...js θi1 · · · θr irXj1 1 · · · dxjl ((D1 0U)(Xl)  · · · Xjs s . (†15) By adding up (†13), (†14) and (†15) for k = 1, . . . , r and l = 1, . . . , s, we obtain (Dr sU)T  ((θ1, . . . , θr, X1, . . . , Xs) + r X k=1 T(θ1, . . . , (D0 1U)θk, . . . θr, X1, . . . , Xs) + s X l=1 T  θ1, . . . , θr, X1, . . . , FU ∂ ∂xjl , . . . , Xs  = EU ai1...ir j1...js θ1 i1 · · · θr irXj1 1 · · · Xjs s + r X k=1 ai1...ir j1...js θ1 i1 · · ·  θk  FU ∂ ∂xik  + (D0 1U)θk  ∂ ∂xik  · · · θr irXj1 1 · · · Xjs s + s X l=1 ai1...ir j1...js θ1 i1 · · · θr irXj1 1 · · · ((D0 1U)dxjl)(Xl) + dxjl ((D1 0U)(Xl)  · · · Xjs s . To conclude that the above expression on the right-hand side is equal to the right-hand side of (†12), we need to prove the equations θk  FU ∂ ∂xik  + (D0 1U)θk  ∂ ∂xik  = EUθk ik (†16) ((D0 1U)dxjl)(Xl) + dxjl ((D1 0U)(Xl)  = EUXjl l . (†17) Rrenaming ik as i, we have θk  FU ∂ ∂xi  = (θk j dxj)  FU ∂ ∂xi  . (†18) By (DO1), we also have (D0 1U)(θk) = (D0 1U)(θk j dxj) = (EUθk j )dxj + θk j (D0 1U)dxj. Applying the above to ∂ ∂xi and using (†11), we obtain EUθk i + θk j ((D0 1U)dxj)  ∂ ∂xi  = EUθk i −θk j dxj  FU ∂ ∂xi  , (†19) and so θk  FU ∂ ∂xi  + (D0 1U)θk  ∂ ∂xi  = θk j dxj  FU ∂ ∂xi  + EUθk i −θk j dxj  FU ∂ ∂xi  = EUθk i , 5.3. TENSOR FIELD DERIVATIONS AND WILMORE’S THEOREM 209 proving (†16). Renaming jl as j, by (DO3), we have (D1 0U)(Xl) = (D1 0U)  Xi l ∂ ∂xi  = (EUXi l ) ∂ ∂xi + Xi l FU ∂ ∂xi. Applying dxj to the above (remembering that dxj is C∞(M)-linear) and using (†11), we obtain dxj (D1 0U)(Xl)  = EUXj l + Xi l dxj  FU ∂ ∂xi  = EUXj l −Xi l ((D0 1U)dxj)  ∂ ∂xi  . (†20) As a consequence, we have ((D0 1U)dxj)(Xl) + dxj ((D1 0U)(Xl)  = Xi l ((D0 1U)dxj)  ∂ ∂xi  + EUXj l −Xi l ((D0 1U)dxj)  ∂ ∂xi  = EUXj l , proving (†17). Finally, this proves that (DO4) holds locally, and since this equation holds on overlaps, it also holds globally. To prove that (DO4) implies (DO3), observe that by (DO4) and the definition of δ, we have E(θ(X)) = E(δ(θ, X)) = D1 1(δ)(θ, X) + δ(D0 1θ, X) + δ(θ, FX) = D1 1(δ)(θ, X) + (D0 1θ)(X) + θ(FX), and by (Dθ), E(θ(X)) = (D0 1θ)(X) + θ(FX), so we get D1 1(δ)(θ, X) = 0 for all θ, X, which means that D1 1δ = 0. For any chart (U, ϕ), since we have ((D0 1U)dxj)  ∂ ∂xi  = −dxj  FU ∂ ∂xi  , on U, (†11) if we have the local expression (D1 0U) ∂ ∂xi = FU ∂ ∂xi = Γj i ∂ ∂xj , (†21) then ((D0 1U)dxj)  ∂ ∂xi  = −dxj  FU ∂ ∂xi  = −dxj  Γk i ∂ ∂xk  = −Γj i, so we obtain the local expression (D0 1U)dxj = −Γj idxi. (†22) Observe the sort of duality between (†21) and (†22) 210 CHAPTER 5. TENSOR FIELDS 5.4 Using Contractions Instead of Kronecker’s delta There is another approach to tensor field derivations using the contraction ci,j : T r,s(M) → T r−1,s−1(M) instead of Kronecker’s δ. The reason is that the contraction operators commute with D, and this fact implies (DO4). Proposition 5.7. If D is a differential operator (a family of maps Dr sU as in Definition 5.7), then the contractions operators ci,j : T r,s(M) →T r−1,s−1(M) commute with D, which means that ci,j(Dr sT) = Dr−1 s−1(ci,jT), T ∈T r,s(M). (DO5) Proof. We begin with the case r = s = 1. It suffices to prove (DO5) for T = X ⊗θ, with X ∈X(M) and θ ∈A1(M). In this case, c1,1(X ⊗θ) = δ(θ, X) = θ(X), and by (Dθ) and (DO1), D0 0(c1,1(X ⊗θ)) = D0 0(θ(X)) = (D0 1θ)(X) + θ(D1 0X) = c1,1(X ⊗D0 1θ) + c1,1(D1 0X ⊗θ) = c1,1(X ⊗D0 1θ + D1 0X ⊗θ) = c1,1(D1 1(X ⊗θ)), proving (DO5). Recall that the proof of (Dθ) uses (DO3), namely, D1 1δ = 0. Let us now consider the general case. It suffices to prove (DO5) for T = X1 ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗θs, with Xi ∈X(M) and θj ∈A1(M). Recall that ci,j(X1 ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗θs) = θj(Xi)X1 ⊗· · · ⊗c Xi ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗b θj ⊗· · · ⊗θs. Using (DO1) and (Dθ), we obtain Dr−1 s−1(ci,j(X1 ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗θs)) = Dr−1 s−1(θj(Xi)X1 ⊗· · · ⊗c Xi ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗b θj ⊗· · · ⊗θs) = D0 0(θj(Xi))X1 ⊗· · · ⊗c Xi ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗b θj ⊗· · · ⊗θs + r X k=1,k̸=i θj(Xi) X1 ⊗· · · ⊗D1 0Xk ⊗· · · ⊗c Xi ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗b θj ⊗· · · ⊗θs + s X l=1,l̸=j θj(Xi) X1 ⊗· · · ⊗c Xi ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗D0 1θl ⊗· · · ⊗b θj ⊗· · · ⊗θs 5.4. USING CONTRACTIONS INSTEAD OF KRONECKER’S DELTA 211 = θj(D1 0Xi) X1 ⊗· · · ⊗c Xi ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗b θj ⊗· · · ⊗θs + (D0 1θj)(Xi) X1 ⊗· · · ⊗c Xi ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗b θj ⊗· · · ⊗θs + r X k=1,k̸=i θj(Xi) X1 ⊗· · · ⊗D1 0Xk ⊗· · · ⊗c Xi ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗b θj ⊗· · · ⊗θs + s X l=1,l̸=j θj(Xi) X1 ⊗· · · ⊗c Xi ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗D0 1θl ⊗· · · ⊗b θj ⊗· · · ⊗θs = ci,j(X1 ⊗· · · ⊗D1 0Xi ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗θs) + ci,j(X1 ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗D0 1θj ⊗· · · ⊗θs) + r X k=1,k̸=i ci,j(X1 ⊗· · · ⊗D1 0Xk ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗θs) + s X l=1,l̸=j ci,j(X1 ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗D0 1θl ⊗· · · ⊗θs) = ci,j(Dr s(X1 ⊗· · · ⊗Xr ⊗θ1 ⊗· · · ⊗θs)), proving (DO5). Proposition 5.7 proves that assuming (DO1) and (DO2), (DO3) implies (DO5). The converse holds. Proposition 5.8. If D is a differential operator (a family of maps Dr sU as in Definition 5.7) satisfying Conditions (DO1), (DO2) and (DO5), then (DO3) holds (and thus (DO4) holds). Proof. The proof that (DO4) implies (DO3) only uses the fact that (DO4) holds for T ∈ T 0,1(M) and T ∈T 1,1(M), so we prove that (DO5) implies that (DO4) holds in these two cases. Since c1,1(X ⊗θ) = θ(X), X ∈X(M), θ ∈A1(M), by (DO5) and (DO1), D0 0(θ(X)) = D0 0(c1,1(X ⊗θ)) = c1,1(D0 0(X ⊗θ)) = c1,1(D1 0X ⊗θ + X ⊗D0 1θ) = c1,1(D1 0X ⊗θ) + c1,1(X ⊗D0 1θ) = (D0 1θ)(X) + θ(D1 0X), which is (DO4). If T ∈T 1,1(M), we use the following trick from O’Neill (Chapter 2, Proposition 13): T(θ, X) = c1,1 c1,1(θ ⊗T ⊗X)  , X ∈X(M), θ ∈A1(M). (†23) Observe that the inner c1,1 is an operator on T 2,2(M) but the outer c1,1 is an operator on T 1,1(M). 212 CHAPTER 5. TENSOR FIELDS To prove the above equation, we may express T, X, θ locally in a chart (U, ϕ). If T = ai j ∂ ∂xi ⊗dxj, θ = θkdxk, X = Xl ∂ ∂xl, then we have θ ⊗T ⊗X = ai jθkXl ∂ ∂xi ⊗∂ ∂xl ⊗dxk ⊗dxj, and so c1,1(θ ⊗T ⊗X) = ai jθiXl ∂ ∂xl ⊗dxj and c1,1 c1,1(θ ⊗T ⊗X)  = ai jθiXj = T(θ, X). Now by (†23), (DO5) twice and (DO1), we have D0 0(T(θ, X)) = D0 0 c1,1(c1,1(θ ⊗T ⊗X))  = c1,1 c1,1(D2 2(θ ⊗T ⊗X))  = c1,1 c1,1(D0 1θ ⊗T ⊗X + θ ⊗D1 1T ⊗X + θ ⊗T ⊗D1 0X)  = c1,1 c1,1(D0 1θ ⊗T ⊗X)  + c1,1 c1,1(θ ⊗D1 1T ⊗X)  + c1,1 c1,1(θ ⊗T ⊗D1 0X)  = T(D0 1θ, X) + (D1 1T)(θ, X) + T(θ, D1 0X), which is (DO4). Finally, as in the proof of Theorem 5.6, we can prove that the two above instances of (DO4) imply (DO3). In view of Theorem 5.6 and Propositions 5.7 and 5.8, assuming (DO1) and (DO2), Con-ditions (DO3), (DO4) and (DO5) are equivalent. We also obtain a version of Theorem 5.6 with Conditions (DO1), (DO2) and (DO5). Remark: In general, if T ∈T r,s(M), we can check that in (U, ϕ), T(θ1, . . . , θr, X1, . . . , Xs) = c1,1 · · · (c1,1(θ1 ⊗· · · ⊗θr ⊗T ⊗X1 ⊗· · · ⊗Xs) · · · )  . (†24) In the sequence of c1,1, the inner c1,1 is an operator on T r,s(M), ..., and the outer c1,1 is an operator on T 1,1(M). Also, the above equation and (DO5) can be used to derive (DO4). There is also a version of Wilmore’s theorem for tensor fields viewed as C∞(M)-multilinear maps in Tr,s(M). This approach is discussed in O’Neill (Chapter 2, Tensor Derivations), and we now summarize it. We begin with the following definition from O’Neill (Chapter 2, Definition 11). Definition 5.8. A tensor derivation on a smooth manifold M is a family D of R-linear maps Dr s : Tr,s(M) →Tr,s(M), such that the following conditions hold: (1) for all T1 ∈Tr1,s1(M) and all T2 ∈Tr2,s2(M), Dr1+r2 s1+s2(T1 ⊗T2) = Dr1 s1(T1) ⊗T2 + T1 ⊗Dr2 s2(T2). 5.5. EXTENSION OF THE COVARIANT DERIVATIVE TO TENSOR FIELDS 213 (2) For all T ∈Tr,s(M), D commutes with the contractions, that is Dr−1 s−1(ci,j(T)) = ci,j(Dr s(T)), (r, s ≥1). In the special case r = s = 0, D0 0 is a derivation on C∞(M), and it can be shown that there is a vector field V ∈X(M) such that D0 0f = V [f], f ∈C∞(M). See O’Neill (Chapter 1). The following result that uses the fact that tensor fields in Tr,s(M) are C∞(M)-multilinear shows how (DO2) can be recovered. Proposition 5.9. If D is a tensor derivation on M, for any open subset U of M, there is a unique tensor derivation D|U on U such that (Dr s|U)(T|U) = (Dr sT)|U, T ∈Tr,s(M). The proof of Proposition 5.9 is given in O’Neill (Chapter 2, Proposition 12). The next step is to prove that a tensor derivation satisfies Property (DO4). The proof uses the trick (†23) and its generalization (†24); see O’Neill (Chapter 2, Proposition 13). Finally, a version of Wilmore’s theorem is obtained. Theorem 5.10. Given a vector field V and an R-linear map F : X(M) →X(M) such that F(fX) = V [f]X + fF(X) X ∈X(M), f ∈C∞(M), there exists a unique tensor derivation D on M such that D0 0 agrees with V , which means that D0 0f = V [f], f ∈C∞(M), and D1 0 agrees with F. The above theorem is proven in O’Neill (Chapter 2, Theorem 15). The tensor derivation D satisfies (DO4), where E : C∞(M) →C∞(M) is given by E(f) = V [f]. We have (D0 1θ)(X) = V [θ(X)] −θ(F(X)), X ∈X(M), θ ∈A1(M), and more generally, D is given by (DO4). A nice application of Theorem 5.6 is the extension of the covariant derivative defined by a connection to tensor fields. 5.5 Extension of the Covariant Derivative to Tensor Fields Let M be a smooth manifold. Recall that a connection on M is a R-bilinear map ∇: X(M) × X(M) →X(M), 214 CHAPTER 5. TENSOR FIELDS where we write ∇XY for ∇(X, Y ), such that the following two conditions hold: ∇fXY = f∇XY ∇X(fY ) = X[f]Y + f∇XY, for all X, Y ∈X(M) and all f ∈C∞(M), where Xf = d fp(Xp). The vector field ∇XY is called the covariant derivative of Y with respect to X. Given a fixed vector field X ∈X(M), if we set EU(f) = (X|U)[f], f ∈C∞(U) FU(Y ) = ∇X|UY, Y ∈X(U), we check immediately that the hypotheses of Theorem 5.6 are satisfied, and thus the covariant derivative ∇X has a unique extension to tensor field as a differential operator. In particular, this extension ∇X satisfies Condition (DO4), in particular, (∇Xθ)(Y ) = X[θ(Y )] −θ(∇XY ), Y ∈X(M), θ ∈A1(M). More generally, if T ∈T r,s(M), we have (∇XT)(ω1, . . . , ωr, X1, . . . , Xs) = X[(T(ω1, . . . , ωr, X1, . . . , Xs)] − r X i=1 T(ω1, . . . , ∇Xωi, . . . , ωr, X1, . . . , Xs) − s X j=1 T(ω1, . . . ωr, X1, . . . , ∇XXj, . . . , Xs). We leave it as an exercise to prove that in a chart (U, ϕ), if ∇ ∂ ∂xi  ∂ ∂xj  = Γk ij ∂ ∂xk in terms of the Christoffel symbols Γk ij and T = ai1...ir j1...js ∂ ∂xi1 ⊗· · · ⊗ ∂ ∂xir ⊗dxj1 ⊗· · · ⊗dxjs, then  ∇ ∂ ∂xi T i1...ir j1...js = ∂ ∂xi(ai1...ir j1...js) + r X α=1 Γiα i kai1...k...ir j1...js − s X β=1 Γl i jβai1...ir j1...l...js. (†25) In the special case where T = dxj, we have a0 j = 1 and a0 k = 0 if k ̸= j, we get  ∇ ∂ ∂xi dxj0 k = −Γj ik, and so ∇ ∂ ∂xi dxj = −Γj ikdxk. (†26) Another application of Theorem 5.6 is the extension of the Lie derivative to tensor fields. 5.6. LIE DERIVATIVE OF TENSOR FIELDS 215 5.6 Lie Derivative of Tensor Fields In order to define the Lie derivative of a tensor field we need to generalize the notion of pullback to tensor fields. For this definition it is more convenient to use the original definition of tensor fields as smooth sections, namely as elements of T r,s(M). Definition 5.9. Given a diffeomorphism ϕ: M →N, the pullback (ϕr s)∗T ∈T r,s(M) of a tensor field T ∈T r,s(N) is defined such that for every p ∈M, ((ϕr s)∗T)p(θ1, . . . , θr, u1, . . . , us) = Tp(θ1 ◦dϕ−1 ϕ(p), . . . , θr ◦dϕ−1 ϕ(p), dϕp(u1), . . . , dϕp(us)), for all θi ∈(TpM)∗(1 ≤i ≤r) and all uj ∈TpM (1 ≤j ≤s). The following proposition is left as an exercise. Proposition 5.11. The following properties hold. (1) If ϕ: M →N is a diffeomorphism, then the map (ϕr s)∗is a linear isomorphism from T r,s(N) to T r,s(M). (2) For any two diffeomorphisms ϕ: M1 →M2 and ψ: M2 →M3, we have ((ψ ◦ϕ)r s)∗= (ϕr s)∗◦(ψr s)∗. Observe that if r = 0, the pullback (ϕ0 s)∗T is well-defined for any smooth map ϕ, even if ϕ is not a diffeomorphism. This corresponds to the fact that the pullback of a differential form is defined for any smooth map ϕ. To define the Lie derivative of a tensor field we follow Abraham and Marsden ) (Section 2.2, Theorem 2.2.20). Given any vector field X on a smooth manifold M, recall that by Proposition 10.12 of Gallier and Quaintance (see also Theorem 2.1.8 of Abraham and Marsden ) that for every p ∈M, there is a local flow ϕ: Ia × U →M (with Ia = (−a, a), a > 0) for X at p such that if Φt : U →M is the map given by Φt(q) = ϕ(t, q), then Ut ⊆U, Ut = Φt(U) is open for all t ∈Ia, and each Φt : U →Ut is a diffeomorphism onto Ut. Definition 5.10. Given any vector field X on a smooth manifold M, for any p ∈M, if Φt : U →Ut is the diffeomorphism defined above, then for any tensor field T ∈T r,s(M), the pullback ((Φt)r s)∗(T|Ut) of the restriction T|Ut ∈T r,s(Ut) of T to Ut is a tensor field in T r,s(U), and we define the Lie derivative (LXT)p of T at p as (LXT)p = lim t− →0 ((Φt)r s)∗T  p −Tp t = d dt ((Φt)r s)∗T  p t=0 . 216 CHAPTER 5. TENSOR FIELDS So, as long we can define the “right” notion of pull-back, the formula giving the Lie derivative of a function, a vector field, a differential form, and more generally a tensor field, is basically the same. The Lie derivative of tensors is used in most areas of mechanics, for example in elasticity (the rate of strain tensor) and in fluid dynamics. Finally here is a proposition about the Lie derivative of tensor fields. Proposition 5.12. Let M be a smooth manifold. For every vector field X ∈X(M), the Lie derivative LX : T •,•(M) →T •,•(M) is the unique linear local operator satisfying the following properties: (1) LXf = X[f] = d f(X), for all f ∈C∞(M). (2) LXY = [X, Y ], for all Y ∈X(M). (3) LX(α ⊗β) = (LXα) ⊗β + α ⊗(LXβ), for all tensor fields α ∈T r1,s1(M) and β ∈ T r2,s2(M); that is, LX is a derivation. (4) For all tensor fields α ∈T r,s(M), with r, s > 0, for every contraction operator ci,j, LX(ci,j(α)) = ci,j(LXα). The proof of Proposition 5.12 is an immediate application of Theorem 5.6 with (DO5) instead of (DO3), by taking EU and FU to be LX|U. See also Theorem 2.1.8 of Abraham and Marsden ). A proof can also be found in Gallot, Hullin and Lafontaine (Chapter 1). The following proposition is also useful: Proposition 5.13. For every (0, q)-tensor S ∈T 0,q(M), we have (LXS)(X1, . . . , Xq) = X[S(X1, . . . , Xq)] − q X i=1 S(X1, . . . , [X, Xi], . . . , Xq), for all X1, . . . , Xq, X ∈X(M). Chapter 6 Distributions and the Frobenius Theorem Given any smooth manifold M (of dimension n), for any smooth vector field X on M, it is known that for every point p ∈M, there is a unique maximal integral curve through p; see Warner (Chapter 1, Theorem 1.48) or Gallier and Quaintance . Furthermore, any two distinct integral curves do not intersect each other, and the union of all the integral curves is M itself. A nonvanishing vector field X can be viewed as the smooth assignment of a one-dimensional vector space to every point of M, namely p 7→RXp ⊆TpM, where RXp denotes the line spanned by Xp. Thus, it is natural to consider the more general situation where we fix some integer r, with 1 ≤r ≤n, and we have an assignment p 7→D(p) ⊆TpM, where D(p) is some r-dimensional subspace of TpM such that D(p) “varies smoothly” with p ∈M. Is there a notion of integral manifold for such assignments? Do they always exist? It is indeed possible to generalize the notion of integral curve and to define integral manifolds, but unlike the situation for vector fields (r = 1), not every assignment D as above possess an integral manifold. However, there is a necessary and sufficient condition for the existence of integral manifolds given by the Frobenius theorem. This theorem has several equivalent formulations. First we will present a formulation in terms of vector fields. Then we show that there are advantages in reformulating the notion of involutivity in terms of differential ideals, and we state a differential form version of the Frobenius theorem. The above versions of the Frobenius theorem are “local.” We will briefly discuss the notion of foliation and state a global version of the Frobenius theorem. Since Frobenius’ theorem is a standard result of differential geometry, we will omit most proofs, and instead refer the reader to the literature. A complete treatment of Frobenius’ theorem can be found in Warner , Morita , and Lee . 217 218 CHAPTER 6. DISTRIBUTIONS AND THE FROBENIUS THEOREM 6.1 Tangential Distributions, Involutive Distributions Our first task is to define precisely what we mean by a smooth assignment p 7→D(p) ⊆TpM, where D(p) is an r-dimensional subspace. Recall the definition of an immersed submanifold given in Warner , Chapter 1, Definition 1.2, namely a pair (M, ψ) where ψ: M →N is a smooth injective immersion (which means that dψp is injective for all p ∈M). Definition 6.1. Let M be a smooth manifold of dimension n. For any integer r, with 1 ≤r ≤n, an r-dimensional tangential distribution (for short, a distribution) is a map D: M →TM, such that (a) D(p) ⊆TpM is an r-dimensional subspace for all p ∈M. (b) For every p ∈M, there is some open subset U with p ∈U, and r smooth vector fields X1, . . . , Xr defined on U, such that (X1(q), . . . , Xr(q)) is a basis of D(q) for all q ∈U. We say that D is locally spanned by X1, . . . , Xr. An immersed submanifold N of M is an integral manifold of D iffD(p) = TpN for all p ∈N. We say that D is completely integrable iffthere exists an integral manifold of D through every point of M. We also write Dp for D(p). Remarks: (1) An r-dimensional distribution D is just a smooth subbundle of TM. (2) An integral manifold is only an immersed submanifold, not necessarily an embedded submanifold. (3) Some authors (such as Lee) reserve the locution “completely integrable” to a seemingly strongly condition (see Lee , Chapter 19, page 500). This condition is in fact equivalent to “our” definition (which seems the most commonly adopted). (4) Morita uses a stronger notion of integral manifold. Namely, an integral manifold is actually an embedded manifold. Most of the results including Frobenius theorem still hold, but maximal integral manifolds are immersed but not embedded manifolds, and this is why most authors prefer to use the weaker definition (immersed manifolds). Here is an example of a distribution which does not have any integral manifolds. This is the two-dimensional distribution in R3 spanned by the vector fields X = ∂ ∂x + y ∂ ∂z, Y = ∂ ∂y. To show why this distribution is not integrable, we will need an involutivity condition. Here is the definition. 6.2. FROBENIUS THEOREM 219 Definition 6.2. Let M be a smooth manifold of dimension n and let D be an r-dimensional distribution on M. For any smooth vector field X, we say that X belongs to D (or lies in D) iffXp ∈Dp for all p ∈M. We say that D is involutive ifffor any two smooth vector fields X, Y on M, if X and Y belong to D, then [X, Y ] also belongs to D. Proposition 6.1. Let M be a smooth manifold of dimension n. If an r-dimensional distri-bution D is completely integrable, then D is involutive. Proof. A proof can be found in in Warner (Chapter 1), and Lee (Proposition 19.3). Another proof is given in Morita (Section 2.3), but beware that Morita defines an integral manifold to be an embedded manifold. In the example before Definition 6.1, we have [X, Y ] = −∂ ∂z, so this distribution is not involutive. Therefore, by Proposition 6.1, this distribution is not completely integrable. 6.2 Frobenius Theorem Frobenius’ theorem asserts that the converse of Proposition 6.1 holds. Although we do not intend to prove it in full, we would like to explain the main idea of the proof of Frobenius’ theorem. It turns out that the involutivity condition of two vector fields is equivalent to the commutativity of their corresponding flows, and this is the crucial fact used in the proof. Definition 6.3. Given a manifold, M, we say that two vector fields X and Y are mutually commutative iff[X, Y ] = 0. For example, on R2, the vector fields ∂ ∂x and ∂ ∂y are commutative since ∂2f ∂x∂y = ∂2f ∂y∂x. On the other hand, the vector fields ∂ ∂x and x ∂ ∂y are not since  ∂ ∂x, x ∂ ∂y  f = ∂ ∂x  x∂f ∂y  −x ∂ ∂y ∂f ∂x  = ∂f ∂y + x ∂2f ∂x∂y −x ∂2f ∂y∂x = ∂f ∂y , which in turn implies h ∂ ∂x, x ∂ ∂y i = ∂ ∂y. 220 CHAPTER 6. DISTRIBUTIONS AND THE FROBENIUS THEOREM Recall that we denote by ΦX the (global) flow of the vector field X. For every p ∈M, the map t 7→ΦX(t, p) = γp(t) is the maximal integral curve through p. We also write Φt(p) for ΦX(t, p) (dropping X). Recall that the map p 7→Φt(p) is a diffeomorphism on its domain (an open subset of M). For the next proposition, given two vector fields X and Y , we write Φ for the flow associated with X and Ψ for the flow associated with Y . Proposition 6.2. Given a manifold M, for any two smooth vector fields X and Y , the following conditions are equivalent: (1) X and Y are mutually commutative (i.e. [X, Y ] = 0). (2) Y is invariant under Φt; that is, (Φt)∗Y = Y , whenever the left hand side is defined. (3) X is invariant under Ψs; that is, (Ψs)∗X = X, whenever the left hand side is defined. (4) The maps Φt and Ψt are mutually commutative. This means that Φt ◦Ψs = Ψs ◦Φt, for all s, t such that both sides are defined. (5) LXY = [X, Y ] = 0. (6) LY X = [Y, X] = 0. (In (5) LXY is the Lie derivative and similarly in (6).) Proof. A proof can be found in Lee (Chapter 18, Proposition 18.5) and in Morita (Chapter 2, Proposition 2.18). For example, to prove the implication (2) = ⇒(4), we observe that if ϕ is a diffeomorphism on some open subset U of M, then the integral curves of ϕ∗Y through a point p ∈M are of the form ϕ◦γ, where γ is the integral curve of Y through ϕ−1(p). Consequently, the local one-parameter group generated by ϕ∗Y is {ϕ◦Ψs ◦ϕ−1}. If we apply this to ϕ = Φt, as (Φt)∗Y = Y , we get Φt ◦Ψs ◦Φ−1 t = Ψs, and hence Φt ◦Ψs = Ψs ◦Φt. In order to state our first version of the Frobenius theorem we make the following defini-tion. Definition 6.4. Let M be a smooth manifold of dimension n. Given any smooth r-dimensional distribution D on M, a chart (U, ϕ) is flat for D iff ϕ(U) ∼ = U ′ × U ′′ ⊆Rr × Rn−r, where U ′ and U ′′ are connected open subsets such that for every p ∈U, the distribution D is spanned by the vector fields ∂ ∂x1 , . . . , ∂ ∂xr . 6.2. FROBENIUS THEOREM 221 U φ 1 1 1 1 x x 1 2 φ (U) φ Figure 6.1: A flat chart for the solid ball B3. Each slice in ϕ(U) is parallel to the xy-plane and turns into a cap shape inside of B3. If (U, ϕ) is flat for D, then each slice of (U, ϕ) Sc = {q ∈U | xr+1 = cr+1, . . . , xn = cn}, is an integral manifold of D, where xi = pri ◦ϕ is the ith-coordinate function on U and c = (cr+1, . . . , cn) ∈Rn−r is a fixed vector, as illustrated in Figure 6.1. Theorem 6.3. (Frobenius) Let M be a smooth manifold of dimension n. A smooth r-dimensional distribution D on M is completely integrable iffit is involutive. Furthermore, for every p ∈U, there is flat chart (U, ϕ) for D with p ∈U so that every slice of (U, ϕ) is an integral manifold of D. Proof. A proof of Theorem 6.3 can be found in Warner (Chapter 1, Theorem 1.60), Lee (Chapter 19, Theorem 19.10), and Morita (Chapter 2, Theorem 2.17). Since we already have Proposition 6.1, it is only necessary to prove that if a distribution is involutive, then it is completely integrable. Here is a sketch of the proof, following Morita. 222 CHAPTER 6. DISTRIBUTIONS AND THE FROBENIUS THEOREM Pick any p ∈M. As D is a smooth distribution, we can find some chart (U, ϕ) with p ∈U, and some vector fields Y1, . . . , Yr so that Y1(q), . . . , Yr(q) are linearly independent and span Dq for all q ∈U. Locally, we can write Yi = n X j=1 aij ∂ ∂xj , i = 1, . . . , r. (†) Since Y1, . . . , Yr are linearly independent, the r×n matrix (aij) has rank r, so by renumbering the coordinates if necessary, we may assume that the first r columns are linearly independent in which case the r × r matrix A(q) = (aij(q)), 1 ≤i, j ≤r. q ∈U is invertible. Then the inverse matrix B(q) = A−1(q) defines r × r functions bij(q), and let Xi = r X j=1 bijYj, j = 1, . . . , r. (††) Now in matrix form Line (†) becomes    Y1 . . . Yr   = A R     ∂ ∂x1 . . . ∂ ∂xn   , for some r × (n −r) matrix R, and Line (††) becomes    X1 . . . Xr   = B    Y1 . . . Yr   , so we get    X1 . . . Xr   = I BR     ∂ ∂x1 . . . ∂ ∂xn   , that is, Xi = ∂ ∂xi + n X j=r+1 cij ∂ ∂xj , i = 1, . . . , r, (∗) where the cij are functions defined on U. Obviously, X1, . . . , Xr are linearly independent and they span Dq for all q ∈U. Since D is involutive, there are some functions fk defined on U so that [Xi, Xj] = r X k=1 fkXk. 6.2. FROBENIUS THEOREM 223 On the other hand, by (∗), each [Xi, Xj] is a linear combination of ∂ ∂xr+1, . . . , ∂ ∂xn. Using (∗), we obtain [Xi, Xj] = r X k=1 fkXk = r X k=1 fk ∂ ∂xk + r X k=1 n X j=r+1 fkckj ∂ ∂xj , and since this is supposed to be a linear combination of ∂ ∂xr+1, . . . , ∂ ∂xn, we must have fk = 0 for k = 1, . . . , r, which shows that [Xi, Xj] = 0, 1 ≤i, j ≤r; that is, the vector fields X1, . . . , Xr are mutually commutative. Let Φi t be the local one-parameter group associated with Xi. By Proposition 6.2 (4), the Φi t commute; that is, Φi t ◦Φj s = Φj s ◦Φi t 1 ≤i, j ≤r, whenever both sides are defined. We can pick a sufficiently small open subset V in Rr containing the origin and define the map Φ: V →U by Φ(t1, . . . , tr) = Φ1 t1 ◦· · · ◦Φr tr(p). Clearly, Φ is smooth, and using the fact that each Xi is invariant under each Φj s for j ̸= i, and dΦi p  ∂ ∂ti  = Xi(p), we get dΦp  ∂ ∂ti  = Xi(p). As X1, . . . , Xr are linearly independent, we deduce that dΦp : T0Rr →TpM is an injection, and thus we may assume by shrinking V if necessary that our map Φ: V →M is an embedding. But then, N = Φ(V ) is a an immersed submanifold of M, and it only remains to prove that N is an integral manifold of D through p. Obviously, TpN = Dp, so we just have to prove that TqN = Dq for all q ∈N. Now for every q ∈N, we can write q = Φ(t1, . . . , tr) = Φ1 t1 ◦· · · ◦Φr tr(p), for some (t1, . . . , tr) ∈V . Since the Φi t commute for any i, with 1 ≤i ≤r, we can write q = Φi ti ◦Φ1 t1 ◦· · · ◦Φi−1 ti−1 ◦Φi+1 ti+1 ◦· · · ◦Φr tr(p). If we fix all the tj but ti and vary ti by a small amount, we obtain a curve in N through q, and this is an orbit of Φi t. Therefore, this curve is an integral curve of Xi through q whose velocity vector at q is equal to Xi(q), and so Xi(q) ∈TqN. Since the above reasoning holds for all i, we get TqN = Dq, as claimed. Therefore, N is an integral manifold of D through p. 224 CHAPTER 6. DISTRIBUTIONS AND THE FROBENIUS THEOREM To best understand how the proof of Theorem 6.3 constructs the integral manifold N, we provide the following example found in Chapter 19 of Lee . Let D ⊂TR3 be the distribution Y1 := V = x ∂ ∂x + ∂ ∂y + x(y + 1) ∂ ∂z Y2 := W = ∂ ∂x + y ∂ ∂z. Given f ∈C∞(R3), observe that V, W = V (W(f)) −W(V (f)) =  x ∂ ∂x + ∂ ∂y + x(y + 1) ∂ ∂z  ∂f ∂x + y∂f ∂z  −  ∂ ∂x + y ∂ ∂z   x∂f ∂x + ∂f ∂y + x(y + 1)∂f ∂z  = ∂f ∂z −∂f ∂x −(y + 1)∂f ∂z = −∂f ∂x −y∂f ∂z = −W(f). Thus D is involutive and Theorem 6.3 is applicable. Our goal is to find a flat chart around the origin. In order to construct this chart, we note that ∂ ∂z is not in the span of V and W since if ∂ ∂z = aV + bW, then ∂ ∂z = (ax + b) ∂ ∂x + a ∂ ∂y + (a(x + 1) + by) ∂ ∂z, which in turn implies a = 0 = b, a contradiction. This means we may rewrite a basis for D in terms of Line (∗) and find that X1 := X = W = ∂ ∂x + y ∂ ∂z X2 := Y = V −xW = ∂ ∂y + x ∂ ∂z. Alternatively we may obtain X1, X2 from the matrix form of Line (†), Y1 Y2  = x 1 x(y + 1) 1 0 y      ∂ ∂x ∂ ∂y ∂ ∂z    , 6.3. DIFFERENTIAL IDEALS AND FROBENIUS THEOREM 225 (with A = x 1 1 0  ), and the matrix form of Line (††), namely X1 X2  = 0 1 1 −x  Y1 Y2  . The flow of X is αu(x, y, z) := Φ1 u(x, y, z) = (x + u, y, z + uy), while the flow of Y is βv(x, y, z) := Φ2 v(x, y, z) = (x, y + v, z + vx). For a fixed point on the z-axis near the origin, say (0, 0, w), we define Φ: R3 →R3 as a composition of the flows, namely Φ(u, v)(0, 0, w) = αu ◦βv(0, 0, w) = αu(0, v, w) = (u, v, w + uv). In other words Φ(u, v)(0, 0, w) provides the parameterization of R3 given by x = u, y = v, z = w + uv, and thus the flat chart is given by Φ−1(x, y, z) = (u, v, z −xy). By the paragraph immediately preceding Theorem 6.3, we conclude that the N, the integral manifolds of D, are given by the level sets of w(x, y, z) = z −xy. In preparation for a global version of Frobenius theorem in terms of foliations, we state the following proposition proved in Lee (Chapter 19, Proposition 19.12): Proposition 6.4. Let M be a smooth manifold of dimension n and let D be an involutive r-dimensional distribution on M. For every flat chart (U, ϕ) for D, for every integral manifold N of D, the set N ∩U is a countable disjoint union of open parallel r-dimensional slices of U, each of which is open in N and embedded in M. We now describe an alternative method for describing involutivity in terms of differential forms. 6.3 Differential Ideals and Frobenius Theorem First, we give a smoothness criterion for distributions in terms of one-forms. 226 CHAPTER 6. DISTRIBUTIONS AND THE FROBENIUS THEOREM Proposition 6.5. Let M be a smooth manifold of dimension n and let D be an assignment p 7→Dp ⊆TpM of some r-dimensional subspace of TpM, for all p ∈M. Then D is a smooth distribution ifffor every p ∈U, there is some open subset U with p ∈U, and some linearly independent one-forms ω1, . . . , ωn−r defined on U, so that Dq = {u ∈TqM | (ω1)q(u) = · · · = (ωn−r)q(u) = 0}, for all q ∈U. Proof. Proposition 6.5 is proved in Lee (Chapter 19, Lemma 19.5). The idea is to either extend a set of linearly independent differential one-forms to a coframe and then consider the dual frame, or to extend some linearly independent vector fields to a frame and then take the dual basis. Proposition 6.5 suggests the following definitions. Definition 6.5. Let M be a smooth manifold of dimension n and let D be an r-dimensional distribution on M. 1. Some linearly independent one-forms ω1, . . . , ωn−r defined on some open subset U ⊆M are called local defining one-forms for D if Dq = {u ∈TqM | (ω1)q(u) = · · · = (ωn−r)q(u) = 0}, for all q ∈U. 2. We say that a k-form ω ∈Ak(M) annihilates D iff ωq(X1(q), . . . , Xk(q)) = 0, for all q ∈M and for all vector fields X1, . . . , Xk belonging to D. We write Ik(D) = {ω ∈Ak(M) | ωq(X1(q), . . . , Xk(q)) = 0}, for all q ∈M and for all vector fields X1, . . . , Xk belonging to D, and we let I(D) = n M k=1 Ik(D). Thus, I(D) is the collection of differential forms that “vanish on D.” In the classical terminology, a system of local defining one-forms as above is called a system of Pfaffian equations. It turns out that I(D) is not only a vector space, but also an ideal of A•(M). Recall that a subspace I of A•(M) is an ideal ifffor every ω ∈I, we have θ ∧ω ∈I for every θ ∈A•(M). Proposition 6.6. Let M be a smooth n-dimensional manifold and D be an r-dimensional distribution. If I(D) is the space of forms annihilating D, then the following hold: 6.3. DIFFERENTIAL IDEALS AND FROBENIUS THEOREM 227 (a) I(D) is an ideal in A•(M). (b) I(D) is locally generated by n −r linearly independent one-forms, which means for every p ∈U, there is some open subset U ⊆M with p ∈U and a set of linearly independent one-forms ω1, . . . , ωn−r defined on U, so that (i) If ω ∈Ik(D), then ω ↾U belongs to the ideal in A•(U) generated by ω1, . . . , ωn−r; that is, ω = n−r X i=1 θi ∧ωi, on U, for some (k −1)-forms θi ∈Ak−1(U). (ii) If ω ∈Ak(M) and if there is an open cover by subsets U (as above) such that for every U in the cover, ω ↾U belongs to the ideal generated by ω1, . . . , ωn−r, then ω ∈I(D). (c) If I ⊆A•(M) is an ideal locally generated by n −r linearly independent one-forms, then there exists a unique smooth r-dimensional distribution D for which I = I(D). Proof. Proposition 6.6 is proved in Warner (Chapter 2, Proposition 2.28); see also Morita (Chapter 2, Lemma 2.19), and Lee (Chapter 19, pages 498-500). In order to characterize involutive distributions, we need the notion of a differential ideal. Definition 6.6. Let M be a smooth manifold of dimension n. An ideal I ⊆A•(M) is a differential ideal iffit is closed under exterior differentiation; that is, dω ∈I whenever ω ∈I, which we also express by dI ⊆I. Here is the differential ideal criterion for involutivity. Proposition 6.7. Let M be a smooth manifold of dimension n. A smooth r-dimensional distribution D is involutive iffthe ideal I(D) is a differential ideal. Proof. Proposition 6.7 is proved in Warner (Chapter 2, Proposition 2.30), Morita (Chapter 2, Proposition 2.20), and Lee (Chapter 19, Proposition 19.19). Assume D is involutive. Let ω ∈Ak(M) be any k form on M and let X0, . . . , Xk be k + 1 smooth vector fields lying in D. Then by Proposition 4.16 and the fact that D is involutive, we deduce that dω(X0, . . . , Xk) = 0. Hence, dω ∈I(D), which means that I(D) is a differential ideal. For the converse, assume I(D) is a differential ideal. We know that for any one-form ω, dω(X, Y ) = X(ω(Y )) −Y (ω(X)) −ω([X, Y ]), 228 CHAPTER 6. DISTRIBUTIONS AND THE FROBENIUS THEOREM for any vector fields X, Y . Now, if ω1, . . . , ωn−r are linearly independent one-forms that define D locally on U, using a bump function, we can extend ω1, . . . , ωn−r to M, and then using the above equation, for any vector fields X, Y belonging to D, we get ωi([X, Y ]) = X(ωi(Y )) −Y (ωi(X)) −dωi(X, Y ), and since ωi(X) = ωi(Y ) = dωi(X, Y ) = 0,( because I(D) is a differential ideal and ωi ∈ I(D)), we get ωi([X, Y ]) = 0 for i = 1, . . . , n−r, which means that [X, Y ] belongs to D. Using Proposition 6.6, we can give a more concrete criterion. Proposition 6.8. A distibution D is involutive ifffor every local defining one-forms ω1, . . ., ωn−r for D (on some open subset U), there are some one-forms ωij ∈A1(U) so that dωi = n−r X j=1 ωij ∧ωj (i = 1, . . . , n −r). The above conditions are often called the integrability conditions. Definition 6.7. Let M be a smooth manifold of dimension n. Given any ideal I ⊆A•(M), an immersed manifold N = (M, ψ) of M, (where ψ: N →M), is an integral manifold of I iff ψ∗ω = 0, for all ω ∈I. A connected integral manifold of the ideal I is maximal iffits image is not a proper subset of the image of any other connected integral manifold of I. Finally, here is the differential form version of the Frobenius theorem. Theorem 6.9. (Frobenius Theorem, Differential Ideal Version) Let M be a smooth manifold of dimension n. If I ⊆A•(M) is a differential ideal locally generated by n −r linearly independent one-forms, then for every p ∈M, there exists a unique maximal, connected, integral manifold of I through p, and this integral manifold has dimension r. Proof. Theorem 6.9 is proved in Warner . This theorem follows immediately from Theorem 1.64 in Warner . Another version of the Frobenius theorem goes as follows; see Morita (Chapter 2, Theorem 2.21). Theorem 6.10. (Frobenius Theorem, Integrability Conditions Version) Let M be a smooth manifold of dimension n. An r-dimensional distribution D on M is completely integrable iff for every local defining one-forms ω1, . . . , ωn−r for D (on some open subset, U), there are some one-forms ωij ∈A1(U) so that we have the integrability conditions dωi = n−r X j=1 ωij ∧ωj (i = 1, . . . , n −r). 6.4. A GLIMPSE AT FOLIATIONS 229 There are applications of Frobenius theorem (in its various forms) to systems of partial differential equations, but we will not deal with this subject. The reader is advised to consult Lee , Chapter 19, and the references there. 6.4 A Glimpse at Foliations and a Global Version of Frobenius Theorem All the maximal integral manifolds of an r-dimensional involutive distribution on a manifold M yield a decomposition of M with some nice properties, those of a foliation. Definition 6.8. Let M be a smooth manifold of dimension n. A family F = {Fα}α of subsets of M is a k-dimensional foliation iffit is a family of pairwise disjoint, connected, immersed k-dimensional submanifolds of M called the leaves of the foliation, whose union is M, and such that for every p ∈M, there is a chart (U, ϕ) with p ∈U called a flat chart for the foliation, and the following property holds: ϕ(U) ∼ = U ′ × U ′′ ⊆Rr × Rn−r, where U ′ and U ′′ are some connected open subsets, and for every leaf Fα of the foliation, if Fα ∩U ̸= ∅, then Fα ∩U is a countable union of k-dimensional slices given by xr+1 = cr+1, . . . , xn = cn, for some constants cr+1, . . . , cn ∈R. The structure of a foliation can be very complicated. For instance, the leaves can be dense in M. For example, there are spirals on a torus that form the leaves of a foliation (see Lee , Example 19.9). Foliations are in one-to-one correspondence with involutive distributions. Proposition 6.11. Let M be a smooth manifold of dimension n. For any foliation F on M, the family of tangent spaces to the leaves of F forms an involutive distribution on M. The converse to the above proposition may be viewed as a global version of Frobenius theorem. Theorem 6.12. Let M be a smooth manifold of dimension n. For every r-dimensional smooth, involutive distribution D on M, the family of all maximal, connected, integral man-ifolds of D forms a foliation of M. Proof. The proof of Theorem 6.12 can be found in Lee (Theorem 19.21). 230 CHAPTER 6. DISTRIBUTIONS AND THE FROBENIUS THEOREM 6.5 Problems Problem 6.1. Prove Proposition 6.1. Hint. See Warner , Chapter 1, and Lee , Proposition 19.3. Problem 6.2. Prove Proposition 6.2. Hint. See Lee , Chapter 18, Proposition 18.5, and Morita , Chapter 2, Proposition 2.18. Problem 6.3. Prove Proposition 6.4. Hint. See , Chapter 19, Proposition 19.12. Problem 6.4. Prove Proposition 6.5. Hint. See Lee , Chapter 19, Lemma 19.5. Problem 6.5. Prove Proposition 6.6. Hint. See Warner, Chapter 2, Proposition 2.28; see also Morita , Chapter 2, Lemma 2.19, and Lee , Chapter 19, pages 498-500. Chapter 7 Integration on Manifolds The purpose of this chapter is to generalize the theory of integration known for functions defined on open subsets of Rn to manifolds. As a first step, we explain how differential forms defined on an open subset of Rn are integrated. Then, if M is a smooth manifold of dimension n, and if ω is an n-form on M (with compact support), the integral R M ω is defined by patching together the integrals defined on small-enough open subsets covering M using a partition of unity. If (U, ϕ) is a chart such that the support of ω is contained in U, then the pullback (ϕ−1)∗ω of ω is an n-form on Rn, so we know how to compute its integral R ϕ(U)(ϕ−1)∗ω. To ensure that these integrals have a consistent value on overlapping charts, we need for M to be orientable. Actually, there is a more general notion of integration on a manifold that uses densities instead differential forms, but we do not need such generality. In Section 7.1 we define the notion of orientation of a manifold. First we define an orientation of a vector space. Then we define an oriented smooth manifold as a manifold that has an atlas consisting of charts such that the transition maps all have positive Jacobian determinants. Technically, a more convenient criterion for orientability is the existence of a differential n-form (where n is the dimension of the manifold) which is nowhere-vanishing. Such an n-form is called a volume form. We prove that a smooth manifold (Hausdorffand second-countable) is orientable if and only if it possesses a volume form. We also define orientable diffeomorphisms. In Section 7.2 we consider the special case of Riemannian manifolds. An orientable Riemannian manifold has a special volume form expressible in a chart in terms of the square root of the determinant of the matrix expressing the metric. Lie groups are always orientable and posssess a left-invariant volume form. In Section 7.3 we explain how to integrate differential forms with compact support defined on an open subset U of Rn. Since a differential n-form on U can be expressed as ωx = f(x)dx1 ∧· · · ∧dxn, where f : U →R is a smooth function with compact support contained in U, we can define R U ω as R U f(x)dx1 · · · dxn, the standard Riemann integral of f. We also give a formula for the change of variable induced by a diffeomorphism ϕ: U →V . 231 232 CHAPTER 7. INTEGRATION ON MANIFOLDS In Section 7.4, we promote the definition of the integral of a differential form defined on an open subset of Rn to smooth oriented manifolds. For any n-form ω with compact support on a smooth n-dimensional oriented manifold M, the integral R M ω is computed by patching together the integrals defined on small-enough open subsets covering M using a partition of unity. The orientability of M is needed to ensure that the above integrals have a consistent value on overlapping charts In preparation for discussing Stokes’ theorem, we need to define manifolds with bound-aries, which is the object of Section 7.6. The idea is to allow a class of manifolds that can be covered with open subsets homeomorhic to open subset of the half space Hn = {(x1, . . . , xn) ∈Rn | xn ≥0}. In Section 7.7 we define a class of manifolds with boundaries called regular domains, and we prove Stokes’ theorem, which roughly speaking is stated as Z ∂N ω = Z N dω, where N is an oriented domain with smooth boundary ∂N. We also present the classical versions of Stokes’ theorem in R3 as well as the divergence theoerem. We also mention the class of manifolds with corners, which is more general than the class of manifolds with boundaries, for which a version of Stokes’ theorem holds. In Section 7.8 we define the integral of a smooth function f with compact support defined on an orientable Riemannian manifold M. For this we use the canonical volume form VolM induced by the Riemannian metric, and let R M f = R M fVolM. Since a Lie group G is orientable, we can pick a left invariant volume form ω and define the integral of a function f with compact support as R G f = R G fω. Such an integral is left-invariant. Roughly speaking this means that the integral does not change if the variable t in f(t) is replaced by st. In general it is not right-invariant (the integral is right invariant if the integral does not change when the variable t in f(t) is replaced by ts). The failure of right-invariance of the integral is measured by the modular function of the group. Technically, R G fω = ∆(g) R G f(tg)ω. Lie groups for which ∆≡1 are particularly nice, and are called unimodular. Compact Lie groups are unimodular, and so are semisimple Lie groups. 7.1 Orientation of Manifolds Although the notion of orientation of a manifold is quite intuitive it is technically rather subtle. We restrict our discussion to smooth manifolds (the notion of orientation can also be defined for topological manifolds, but more work is involved). Intuitively, a manifold M is orientable if it is possible to give a consistent orientation to its tangent space TpM at every point p ∈M. So, if we go around a closed curve starting at p ∈M, when we come back to p, the orientation of TpM should be the same as when we 7.1. ORIENTATION OF MANIFOLDS 233 started. For example, if we travel on a M¨ obius strip (a manifold with boundary) dragging a coin with us, we will come back to our point of departure with the coin flipped. Try it; see Figure 7.3 for an illustration. To be rigorous, we have to say what it means to orient TpM (a vector space) and what consistency of orientation means. We begin by quickly reviewing the notion of orientation of a vector space. Let E be a vector space of dimension n. If u1, . . . , un and v1, . . . , vn are two bases of E, a basic and crucial fact of linear algebra says that there is a unique linear map g mapping each ui to the corresponding vi (i.e., g(ui) = vi, i = 1, . . . , n). Then look at the determinant det(g) of this map. We know that det(g) = det(P), where P is the matrix whose j-th column consists of the coordinates of vj over the basis u1, . . . , un. Either det(g) is negative, or it is positive. This leads to the following definition. Definition 7.1. Let E be a vector space of dimension n with bases u1, . . . , un and v1, . . . , vn. Let g be the unique linear map such that g(ui) = vi, i = 1, . . . , n. We say u1, . . . , un and v1, . . . , vn have the same orientation iffdet(g) is positive. Definition 7.1 defines an equivalence relation on bases where two bases are equivalent iff they have the same orientation. Definition 7.2. Let E be a vector space of dimension n. An orientation of E is the choice of one of the two equivalence classes, which amounts to picking some basis as an orientation frame. For E = R, an orientation is given by e1 or −e1. Such an orientation is visualized as either right or left translation from the origin. For E = R2, an orientation is given by (e1, e2) or (e2, e1), i.e. either counterclockwise or clockwise rotation about the origin. For E = R3, the orientation is represented by (e1, e2, e3) or (e2, e1, e3), namely the right-handed or left handed orientation of the i, j, k axis system. See Figure 7.1. Definition 7.2 is perfectly fine, but it turns out that it is more convenient, in the long term, to use a definition of orientation in terms of differential forms and the exterior algebra Vn E∗. This approach is especially useful when defining the notion of integration on a manifold. We observe that two bases u1, . . . , un and v1, . . . , vn have the same orientation iff ω(u1, . . . , un) and ω(v1, . . . , vn) have the same sign for all ω ∈Vn E∗−{0} (where 0 denotes the zero n-form). As Vn E∗is one-dimensional, picking an orientation of E is equivalent to picking a generator (a one-element basis) ω of Vn E∗, and to say that u1, . . . , un has positive orientation iffω(u1, . . . , un) > 0. Definition 7.3. Let E be a vector space of dimension n. Given an orientation (say, given by ω ∈Vn E∗) of E, a linear map f : E →E is orientation preserving iffω(f(u1), . . . , f(un)) > 0 whenever ω(u1, . . . , un) > 0 (or equivalently, iffdet(f) > 0). 234 CHAPTER 7. INTEGRATION ON MANIFOLDS 0 0 e e 1 1 -e1 e2 e1 e2 e1 e2 e3 e1 e2 e3 Figure 7.1: The two orientations of R, R2, and R3. To define the orientation of an n-dimensional manifold M we use charts. Given any p ∈M, for any chart (U, ϕ) at p, the tangent map dϕ−1 ϕ(p) : Rn →TpM makes sense. If (e1, . . . , en) is the standard basis of Rn, as it gives an orientation to Rn, we can orient TpM by giving it the orientation induced by the basis dϕ−1 ϕ(p)(e1), . . . , dϕ−1 ϕ(p)(en). The consistency of orientations of the TpM’s is given by the overlapping of charts. See Figure 7.2. We require that the Jacobian determinants of all ϕj ◦ϕ−1 i have the same sign whenever (Ui, ϕi) and (Uj, ϕj) are any two overlapping charts. Thus, we are led to the definition below. All definitions and results stated in the rest of this section apply to manifolds with or without boundary. Definition 7.4. Given a smooth manifold M of dimension n, an orientation atlas of M is any atlas so that the transition maps ϕj i = ϕj ◦ϕ−1 i (from ϕi(Ui ∩Uj) to ϕj(Ui ∩Uj)) all have a positive Jacobian determinant for every point in ϕi(Ui ∩Uj). A manifold is orientable iff its has some orientation atlas. We should mention that not every manifold is orientable. The open Mobius strip, i.e. the Mobius strip with circle boundary removed, is not orientable, as demonstrated in Figure 7.3. Definition 7.4 can be hard to check in practice and there is an equivalent criterion is terms of n-forms which is often more convenient. The idea is that a manifold of dimension 7.1. ORIENTATION OF MANIFOLDS 235 p φ φ(p) φ(p) i j = 0 = 0 e e e e 1 1 2 2 d φ (p) i φ d φ (p) i φ d φ (p) j φ d φ (p) j ( e ) 1 ( e ) 1 2 ( e ) 2 ( e ) -1 -1 -1 -1 M φ φi j Figure 7.2: The sphere S2 with consistent orientation on two overlapping charts. n is orientable iffthere is a map p 7→ωp, assigning to every point p ∈M a nonzero n-form ωp ∈Vn T ∗ p M, so that this map is smooth. Definition 7.5. If M is an n-dimensional manifold, recall that a smooth section ω ∈ Γ(M, Vn T ∗M) is called a (smooth) n-form. An n-form ω is a nowhere-vanishing n-form on M or volume form on M iffωp is a nonzero form for every p ∈M. This is equivalent to saying that ωp(u1, . . . , un) ̸= 0, for all p ∈M and all bases u1, . . . , un, of TpM. The determinant function (u1, . . . , un) 7→det(u1, . . . , un) where the ui are expressed over the canonical basis (e1, . . . , en) of Rn, is a volume form on Rn. We will denote this volume form by ωRn = dx1 ∧· · · ∧dxn. Observe the justification for the term volume form: the quantity det(u1, . . . , un) is indeed the (signed) volume of the parallelepiped {λ1u1 + · · · + λnun | 0 ≤λi ≤1, 1 ≤i ≤n}. A volume form on the sphere Sn ⊆Rn+1 is obtained as follows: ωSn(u1, . . . un) = det(p, u1, . . . un), 236 CHAPTER 7. INTEGRATION ON MANIFOLDS 1 1’ Figure 7.3: The Mobius strip does not have a consistent orientation. The frame starting at 1 is reversed when traveling around the loop to 1′. where p ∈Sn and u1, . . . un ∈TpSn. As the ui are orthogonal to p, this is indeed a volume form. Observe that if f is a smooth function on M and ω is any n-form, then fω is also an n-form. More interesting is the following proposition. Proposition 7.1. (a) If h: M →N is a local diffeomorphism of manifolds, where dim M = dim N = n, and ω ∈An(N) is a volume form on N, then h∗ω is a volume form on M. (b) Assume M has a volume form ω. For every n-form η ∈An(M), there is a unique smooth function f ∈C∞(M) so that η = fω. If η is a volume form, then f(p) ̸= 0 for all p ∈M. Proof. (a) By definition, h∗ωp(u1, . . . , un) = ωh(p)(dhp(u1), . . . , dhp(un)), for all p ∈M and all u1, . . . , un ∈TpM. As h is a local diffeomorphism, dph is a bijection for every p. Thus, if u1, . . . , un is a basis, then so is dhp(u1), . . . , dhp(un), and as ω is nonzero at every point for every basis, h∗ωp(u1, . . . , un) ̸= 0. (b) Pick any p ∈M and let (U, ϕ) be any chart at p. As ϕ is a diffeomorphism, by (a), we see that ϕ−1∗ω is a volume form on ϕ(U). But then, it is easy to see that ϕ−1∗η = gϕ−1∗ω, for some unique smooth function g on ϕ(U), and so η = fUω, for some unique smooth function fU on U. For any two overlapping charts (Ui, ϕi) and (Uj, ϕj), for every p ∈Ui ∩Uj, for every basis u1, . . . , un of TpM, we have ηp(u1, . . . , un) = fi(p)ωp(u1, . . . , un) = fj(p)ωp(u1, . . . , un), 7.1. ORIENTATION OF MANIFOLDS 237 and as ωp(u1, . . . , un) ̸= 0, we deduce that fi and fj agree on Ui ∩Uj. But then the fi’s patch on the overlaps of the cover {Ui} of M, and so there is a smooth function f defined on the whole of M and such that f ↾Ui = fi. As the fi’s are unique, so is f. If η is a volume form, then ηp does not vanish for all p ∈M, and since ωp is also a volume form, ωp does not vanish for all p ∈M, so f(p) ̸= 0 for all p ∈M. Remark: If h1 and h2 are smooth maps of manifolds, it is easy to prove that (h2 ◦h1)∗= h∗ 1 ◦h∗ 2, and that for any smooth map h: M →N, h∗(fω) = (f ◦h)h∗ω, where f is any smooth function on N and ω is any n-form on N. The connection between Definition 7.4 and volume forms is given by the following im-portant theorem whose proof contains a wonderful use of partitions of unity. Theorem 7.2. A smooth manifold (Hausdorffand second-countable) is orientable iffit pos-sesses a volume form. Proof. First assume that a volume form ω exists on M, and say n = dim M. For any atlas {(Ui, ϕi)}i of M, by Proposition 7.1, each n-form ϕ−1 i ∗ω is a volume form on ϕi(Ui) ⊆Rn, and ϕ−1 i ∗ω = fiωRn, for some smooth function fi never zero on ϕi(Ui), where ωRn is the volume form on Rn. By composing ϕi with an orientation-reversing linear map if necessary, we may assume that for this new atlas, fi > 0 on ϕi(Ui). We claim that the family (Ui, ϕi)i is an orientation atlas. This is because, on any (nonempty) overlap Ui ∩Uj, as ω = ϕ∗ j(fjωRn) and (ϕj ◦ϕ−1 i )∗= (ϕ−1 i )∗◦ϕ∗ j, we have (ϕj ◦ϕ−1 i )∗(fjωRn) = (ϕ−1 i )∗◦ϕ∗ j(fjωRn) = (ϕ−1 i )∗ω = fiωRn, and by the definition of pull-backs, we see that for every x ∈ϕi(Ui ∩Uj), if we let y = ϕj ◦ϕ−1 i (x), then fi(x)(ωRn)x(e1, . . . , en) = (ϕj ◦ϕ−1 i )∗ x(fjωRn)(e1, . . . , en) = fj(y)(ωRn)y(d(ϕj ◦ϕ−1 i )x(e1), . . . , d(ϕj ◦ϕ−1 i )x(en)) = fj(y)J((ϕj ◦ϕ−1 i )x)(ωRn)y(e1, . . . , en), where e1, . . . , en is the standard basis of Rn and J((ϕj ◦ϕ−1 i )x) is the Jacobian determinant of ϕj ◦ϕ−1 i at x. As both fj(y) > 0 and fi(x) > 0, we have J((ϕj ◦ϕ−1 i )x) > 0, as desired. 238 CHAPTER 7. INTEGRATION ON MANIFOLDS Conversely, assume that J((ϕj ◦ϕ−1 i )x) > 0, for all x ∈ϕi(Ui ∩Uj), whenever Ui ∩Uj ̸= ∅. We need to make a volume form on M. For each Ui, let ωi = ϕ∗ i ωRn, where ωRn is the volume form on Rn. As ϕi is a diffeomorphism, by Proposition 7.1, we see that ωi is a volume form on Ui. Then if we apply Theorem 1.11 from Chapter 1 of Warner , we can find a partition of unity {fi} subordinate to the cover {Ui}, with the same index set. Let, ω = X i fiωi. We claim that ω is a volume form on M. It is clear that ω is an n-form on M. Now since every p ∈M belongs to some Ui, check that on ϕi(Ui), we have ϕ−1 i ∗ω = X j∈finite set ϕ−1 i ∗(fjωj) = X j∈finite set ϕ−1 i ∗(fjϕ∗ jωRn) = X j∈finite set (fj ◦ϕ−1 i )(ϕ−1 i ∗◦ϕ∗ j)ωRn = X j∈finite set (fj ◦ϕ−1 i )(ϕj ◦ϕ−1 i )∗ωRn = X j∈finite set (fj ◦ϕ−1 i )J(ϕj ◦ϕ−1 i ) ! ωRn, and this sum is strictly positive because the Jacobian determinants are positive, and as P j fj = 1 and fj ≥0, some term must be strictly positive. Therefore, ϕ−1 i ∗ω is a volume form on ϕi(Ui), so ϕ∗ i ϕ−1 i ∗ω = ω is a volume form on Ui. As this holds for all Ui, we conclude that ω is a volume form on M. Since we showed there is a volume form on the sphere Sn, by Theorem 7.2, the sphere Sn is orientable. It can be shown that the projective spaces RPn are non-orientable iffn is even, and thus orientable iffn is odd. In particular, RP2 is not orientable. Also, even though M may not be orientable, its tangent bundle T(M) is always orientable! (Prove it). It is also easy to show that if f : Rn+1 →R is a smooth submersion, then M = f −1(0) is a smooth orientable manifold. Another nice fact is that every Lie group is orientable. By Proposition 7.1 (b), given any two volume forms ω1 and ω2 on a manifold M, there is a function f : M →R never 0 on M, such that ω2 = fω1. This fact suggests the following definition. 7.2. VOLUME FORMS ON RIEMANNIAN MANIFOLDS AND LIE GROUPS 239 Definition 7.6. Given an orientable manifold M, two volume forms ω1 and ω2 on M are equivalent iffω2 = fω1 for some smooth function f : M →R, such that f(p) > 0 for all p ∈M. An orientation of M is the choice of some equivalence class of volume forms on M, and an oriented manifold is a manifold together with a choice of orientation. If M is a manifold oriented by the volume form ω, for every p ∈M, a basis (b1, . . . , bn) of TpM is posively oriented iffωp(b1, . . . , bn) > 0, else it is negatively oriented (where n = dim(M)). If M is an orientable manifold, for any two volume forms ω1 and ω2 on M, as ω2 = fω1 for some function f on M which is never zero, f has a constant sign on every connected component of M. Consequently, a connected orientable manifold has two orientations. We will also need the notion of orientation-preserving diffeomorphism. Definition 7.7. Let h: M →N be a diffeomorphism of oriented manifolds M and N, of dimension n, and say the orientation on M is given by the volume form ω1 while the orientation on N is given by the volume form ω2. We say that h is orientation preserving iff h∗ω2 determines the same orientation of M as ω1. Using Definition 7.7, we can define the notion of a positive atlas. Definition 7.8. If M is a manifold oriented by the volume form ω, an atlas for M is positive ifffor every chart (U, ϕ), the diffeomorphism ϕ: U →ϕ(U) is orientation preserving, where U has the orientation induced by M and ϕ(U) ⊆Rn has the orientation induced by the standard orientation on Rn (with dim(M) = n). The proof of Theorem 7.2 shows Proposition 7.3. If a manifold M has an orientation atlas, then there is a uniquely deter-mined orientation on M such that this atlas is positive. 7.2 Volume Forms on Riemannian Manifolds and Lie Groups Recall that a smooth manifold M is a Riemannian manifold iffthe vector bundle TM has a Euclidean metric. This means that there is a family (⟨−, −⟩p)p∈M of inner products on the tangent spaces TpM, such that ⟨−, −⟩p depends smoothly on p, which can be expressed by saying that that the maps x 7→⟨dϕ−1 x (ei), dϕ−1 x (ej)⟩ϕ−1(x), x ∈ϕ(U), 1 ≤i, j ≤n are smooth, for every chart (U, ϕ) of M, where (e1, . . . , en) is the canonical basis of Rn. We let gij(x) = ⟨dϕ−1 x (ei), dϕ−1 x (ej)⟩ϕ−1(x), 240 CHAPTER 7. INTEGRATION ON MANIFOLDS and we say that the n × n matrix (gij(x)) is the local expression of the Riemannian metric on M at x in the coordinate patch (U, ϕ). If a Riemannian manifold M is orientable, then there is a volume form on M with some special properties. Proposition 7.4. Let M be a Riemannian manifold with dim(M) = n. If M is orientable, then there is a uniquely determined volume form VolM on M with the following property: For every p ∈M, for every positively oriented orthonormal basis (b1, . . . , bn) of TpM, we have (VolM)p(b1, . . . , bn) = 1. Furthermore, if the above equation holds then in every orientation preserving local chart (U, ϕ), we have ((ϕ−1)∗VolM)q = q det(gij(q)) dx1 ∧· · · ∧dxn, q ∈ϕ(U). Proof. Say the orientation of M is given by ω ∈An(M). For any two positively oriented orthonormal bases (b1, . . . , bn) and (b′ 1, . . . , b′ n) in TpM, by expressing the second basis over the first, there is an orthogonal matrix C = (cij) so that b′ i = n X j=1 cijbj. We have ωp(b′ 1, . . . , b′ n) = det(C)ωp(b1, . . . , bn), and as these bases are positively oriented, we conclude that det(C) = 1 (as C is orthogonal, det(C) = ±1). As a consequence, we have a well-defined function ρ: M →R with ρ(p) > 0 for all p ∈M, such that ρ(p) = ωp(b1, . . . , bn) for every positively oriented orthonormal basis (b1, . . . , bn) of TpM. If we can show that ρ is smooth, then (VolM)p = 1 ρ(p)ωp is the required volume form. Let (U, ϕ) be a positively oriented chart and consider the vector fields Xj on ϕ(U) given by Xj(q) = dϕ−1 q (ej), q ∈ϕ(U), 1 ≤j ≤n. Then (X1(q), . . . , Xn(q)) is a positively oriented basis of Tϕ−1(q). If we apply Gram-Schmidt orthogonalization, we get an upper triangular matrix A(q) = (aij(q)) of smooth functions on ϕ(U) with aii(q) > 0, such that bi(q) = n X j=1 aij(q)Xj(q), 1 ≤i ≤n, 7.2. VOLUME FORMS ON RIEMANNIAN MANIFOLDS AND LIE GROUPS 241 and (b1(q), . . . , bn(q)) is a positively oriented orthonormal basis of Tϕ−1(q). We have ρ(ϕ−1(q)) = ωϕ−1(q)(b1(q), . . . , bn(q)) = det(A(q))ωϕ−1(q)(X1(q), . . . , Xn(q)) = det(A(q))(ϕ−1)∗ωq(e1, . . . , en), which shows that ρ is smooth. If we repeat the end of the proof with ω = VolM, then ρ ≡1 on M, and the above formula yields ((ϕ−1)∗VolM)q = (det(A(q)))−1dx1 ∧· · · ∧dxn. If we compute ⟨bi(q), bk(q)⟩ϕ−1(q), we get δik = ⟨bi(q), bk(q)⟩ϕ−1(q) = n X j=1 n X l=1 aij(q)gjl(q)akl(q), and so I = A(q)G(q)A(q)⊤, where G(q) = (gjl(q)). Thus, (det(A(q)))2 det(G(q)) = 1, and since det(A(q)) = Q i aii(q) > 0, we conclude that (det(A(q)))−1 = q det(gij(q)), which proves the second formula. We saw in Section 7.1 that a volume form ωSn on the sphere Sn ⊆Rn+1 is given by (ωSn)p(u1, . . . un) = det(p, u1, . . . un), where p ∈Sn and u1, . . . un ∈TpSn. To be more precise, we consider the n-form ˜ ωRn ∈An(Rn+1) given by the above formula. As (˜ ωRn)p(e1, . . . , b ei, . . . , en+1) = det(p, e1, . . . , b ei, . . . , en+1) = (−1)i−1pi, where p = (p1, . . . , pn+1), we have (˜ ωRn)p = n+1 X i=1 (−1)i−1pi dx1 ∧· · · ∧c dxi ∧· · · ∧dxn+1. (∗) Let i: Sn →Rn+1 be the inclusion map. For every p ∈Sn and every basis (u1, . . . , un) of TpSn, the (n + 1)-tuple (p, u1, . . . , un) is a basis of Rn+1, and so (˜ ωRn)p ̸= 0. Hence, ˜ ωRn ↾Sn = i∗˜ ωRn is a volume form on Sn. If we give Sn the Riemannian structure induced by Rn+1, then the discussion above shows that VolSn = ˜ ωRn ↾Sn. 242 CHAPTER 7. INTEGRATION ON MANIFOLDS To obtain another representation for VolSn, let r: Rn+1 −{0} →Sn be the map given by r(x) = x ∥x∥, and set ω = r∗VolSn, a closed n-form on Rn+1 −{0}. Clearly, ω ↾Sn = VolSn. Furthermore ωx(u1, . . . , un) = (˜ ωRn)r(x)(drx(u1), . . . , drx(un)) = ∥x∥−1 det(x, drx(u1), . . . , drx(un)). We leave it as an exercise to prove that ω is given by ωx = 1 ∥x∥n n+1 X i=1 (−1)i−1xi dx1 ∧· · · ∧c dxi ∧· · · ∧dxn+1. The procedure used to construct VolSn can be generalized to any n-dimensional orientable manifold embedded in Rm. Let U be an open subset of Rn and ψ: U →M ⊆Rm be an orientation-preserving parametrization. Assume that x1, x2, . . . , xm are the coordinates of Rm (the ambient coordinates of M) and that u1, u2, . . . , un are the coordinates of U (the local coordinates of M). Let x = ψ(u) be a point in M. Edwards (Theorem 5.6) shows that VolM = X (i1,i2,...,in) 1≤i1<i2<···<in≤m ni1,i2,...,indxi1 ∧dxi2 ∧· · · ∧dxin, (∗∗) where ni1,i2,··· ,in(x) = 1 D ∂(ψi1, ψi2, . . . , ψin) ∂(u1, u2, . . . , un) , D = [det J⊤(ψ)(u)J(ψ)(u)  ] 1 2 and ∂(ψi1,ψi2,...,ψin) ∂(u1,u2,...,un) is the determinant of the n×n matrix obtained by selecting rows i1 through in of dψu. If M is a smooth orientable manifold of dimension m −1, Edwards’s formula for VolM reduces to VolM = m X i=1 (−1)i−1ni dx1 ∧· · · ∧dxi−1 ∧dxi+1 ∧· · · ∧dxm, (∗∗∗) where ni = ni(x) is the ith component of the unit normal vector N(x) on M given by ni(x) = (−1)i−1 D ∂(ψ1, . . . , ˆ ψi, . . . , ψm) ∂(u1, u2, . . . , um−1) . 7.2. VOLUME FORMS ON RIEMANNIAN MANIFOLDS AND LIE GROUPS 243 In particular, if M = Sn embedded in Rn+1, for p ∈Sn, N(p) = (p1, p2, . . . , pn+1) and (∗∗∗) becomes (∗). For a particular example of (∗∗), let M = S2 and ψ : U →S2 where x = sin θ cos ϕ, y = sin θ sin ϕ, z = cos θ. and U = {(θ, ϕ) : 0 < θ < π, 0 < ϕ < 2π} ⊂R2. See Figure 4.1. Clearly J(ψ)(θ, ϕ) =   cos θ cos ϕ −sin θ sin ϕ cos θ sin ϕ sin θ cos ϕ −sin θ 0  , which in turn implies D = [det J⊤(ψ)(θ, ϕ)J(ψ)(θ, ϕ)  ] 1 2 =  det 1 0 0 sin2 θ  1 2 = sin θ. Then VolS2 = n1,2 dx ∧dy + n1,3 dx ∧dz + n2,3 dy ∧dz, where n1,2 = 1 sin θ ∂(x, y) ∂(θ, ϕ) = 1 sin θdet cos θ cos ϕ −sin θ sin ϕ cos θ sin ϕ sin θ cos ϕ  = cos θ sin θ sin θ = cos θ = z n1,3 = 1 sin θ ∂(x, z) ∂(θ, ϕ) = 1 sin θdet cos θ cos ϕ −sin θ sin ϕ −sin θ 0  = −sin2 θ sin ϕ sin θ = −sin θ sin ϕ = −y n2,3 = 1 sin θ ∂(y, z) ∂(θ, ϕ) = 1 sin θdet cos θ sin ϕ sin θ cos ϕ −sin θ 0  = sin2 θ cos ϕ sin θ = sin θ cos ϕ = x. Thus VolS2 = n1,2 dx ∧dy + n1,3 dx ∧dz + n2,3 dy ∧dz = z dx ∧dy −y dx ∧dz + x dy ∧dz, which agrees with (∗) when n = 2. We mention that the orientation of Sn provides a way of orienting projective spaces of odd dimension. We know that there is a map π: Sn →RPn such that π−1([p]) consists of two antipodal points for every [p] ∈RPn. It can be shown that there is a volume form on RPn iffn is odd, in which case π∗(VolRPn) = VolSn. 244 CHAPTER 7. INTEGRATION ON MANIFOLDS Thus, RPn is orientable iffn is odd. We end this section with an important result regarding orientability of Lie groups. Let G be a Lie group of dimension n. For any basis (ω1, . . . , ωn) of the dual g∗of the Lie algebra g of G, we have the left-invariant one-forms defined by the ωi, also denoted ωi, and obviously (ω1, . . . , ωn) is a frame for T ∗G. Therefore, ω = ω1 ∧· · · ∧ωn is an n-form on G that is never zero; that is, a volume form. Since pull-back commutes with ∧, the n-form ω is left-invariant. We summarize this as Proposition 7.5. Every Lie group G possesses a left-invariant volume form. Therefore, every Lie group is orientable. 7.3 Integration in Rn As we said in Section 4.1, one of the raison d’ˆ etre for differential forms is that they are the objects that can be integrated on manifolds. We will be integrating differential forms that are at least continuous (in most cases, smooth) and with compact support. In the case of forms ω on Rn, this means that the closure of the set {x ∈Rn | ωx ̸= 0} is compact. Similarly, for a form ω ∈A∗(M) where M is a manifold, the support suppM(ω) of ω is the closure of the set {p ∈M | ωp ̸= 0}. We let A∗ c(M) denote the set of differential forms with compact support on M. If M is a smooth manifold of dimension n, our ultimate goal is to define a linear operator Z M : An c (M) − →R which generalizes in a natural way the usual integral on Rn. In this section we assume that M = Rn or M = U for some open subset U of Rn. Now every n-form (with compact support) on Rn is given by ωx = f(x) dx1 ∧· · · ∧dxn, where f is a smooth function with compact support. Thus, we set Z Rn ω = Z Rn f(x)dx1 · · · dxn, where the expression on the right-hand side is the usual Riemann integral of f on Rn. For the reader who would like to review the definition of the Riemann integral, we suggest Sections 23.1 to 23.3 of and Sections 4.1 to 4.3 of . Actually we will need to integrate smooth forms ω ∈An c (U) with compact support defined on some open subset U ⊆Rn (with supp(ω) ⊆U). However, this is no problem since we still have ωx = f(x) dx1 ∧· · · ∧dxn, where f : U →R is a smooth function with compact support contained in U, and f can be smoothly extended to Rn by setting it to 0 on Rn −supp(f). We write R U ω for this integral and make the following definition. 7.3. INTEGRATION IN RN 245 Definition 7.9. Let U be an open subset of Rn and let An c (U) denote the set of smooth n-forms with compact support contained in U. In other words ω ∈An c (U) if and only if ωx = f(x) dx1 ∧· · · ∧dxn for some smooth function f : U →R and the closure of {x ∈Rn | ωx ̸= 0} is a compact set of Rn contained in U. For ω ∈An c (U), the expression R U ω is defined as Z U ω = Z U f(x)dx1 · · · dxn, () where the right side of () is interpreted as the Riemann integral. In Definition 7.9, the n-form must be represented as dx1 ∧· · · ∧dxn. This is not a problem since Proposition 4.1 says that we may switch order within the wedge product by adjusting the functional coefficient with the appropriate negative signs. For example, if ωx = f(x) dx1 ∧dx3 ∧dx2, Definition 7.9 implies that Z U ω = Z U f(x) dx1 ∧dx3 ∧dx2 = − Z U f(x) dx1 dx2 dx3. For this reason, R U ω is often called a “signed” integral. It is crucial for the generalization of the integral to manifolds to see what the change of variable formula looks like in terms of differential forms. Proposition 7.6. Let ϕ: U →V be a diffeomorphism between two open subsets of Rn. If the Jacobian determinant J(ϕ)(x) has a constant sign δ = ±1 on U, then for every ω ∈An c (V ), we have Z U ϕ∗ω = δ Z V ω. Proof. We know that ω can be written as ωx = f(x) dx1 ∧· · · ∧dxn, x ∈V, where f : V →R has compact support. From the example after Proposition 4.9 we have (ϕ∗ω)y = f(ϕ(y))J(ϕ)y dy1 ∧· · · ∧dyn = δf(ϕ(y))|J(ϕ)y| dy1 ∧· · · ∧dyn. On the other hand, the change of variable formula (using ϕ) is Z ϕ(U) f(x) dx1 · · · dxn = Z U f(ϕ(y)) |J(ϕ)y| dy1 · · · dyn, so the formula follows. We will promote the integral on open subsets of Rn to manifolds using partitions of unity. 246 CHAPTER 7. INTEGRATION ON MANIFOLDS 7.4 Integration on Manifolds Definition 7.10. Let M be an oriented manifold of dimension n. We say ω is a smooth n-form on M with compact support if the closure of {p ∈M | ωp ̸= 0} is compact in M. We denote {p ∈M | ωp ̸= 0} by supp(ω). The set of smooth n-forms on M with compact support is denoted An c (M) while A∗ c(M) is the set of all smooth differential forms on M with compact support. Intuitively, for any n-form ω ∈An c (M) on a smooth n-dimensional oriented manifold M, the integral R M ω is computed by patching together the integrals defined on small-enough open subsets covering M using a partition of unity. If (U, ϕ) is a chart such that supp(ω) ⊆U, then the form (ϕ−1)∗ω is an n-form on Rn, and the integral R ϕ(U)(ϕ−1)∗ω makes sense. The orientability of M is needed to ensure that the above integrals have a consistent value on overlapping charts. Proposition 7.7. Let M be a smooth oriented manifold of dimension n. There exists a unique linear operator Z M : An c (M) − →R with the following property: For any ω ∈An c (M), if supp(ω) ⊆U, where (U, ϕ) is a positively oriented chart, then Z M ω = Z ϕ(U) (ϕ−1)∗ω. (†) Proof. First, assume that supp(ω) ⊆U, where (U, ϕ) is a positively oriented chart. Then we wish to set Z M ω = Z ϕ(U) (ϕ−1)∗ω. However, we need to prove that the above expression does not depend on the choice of the chart. Let (V, ψ) be another chart such that supp(ω) ⊆V , so that supp(ω) ⊆U ∩V . The map θ = ψ ◦ϕ−1 is a diffeomorphism from W = ϕ(U ∩V ) to W ′ = ψ(U ∩V ), and by hypothesis, its Jacobian determinant is positive on W. Since suppϕ(U)((ϕ−1)∗ω) ⊆W, suppψ(V )((ψ−1)∗ω) ⊆W ′, and θ∗◦(ψ−1)∗ω = (ϕ−1)∗◦ψ∗◦(ψ−1)∗ω = (ϕ−1)∗ω, Proposition 7.6 yields Z W ′(ψ−1)∗ω = Z W θ∗((ψ−1)∗ω) = Z W (ϕ−1)∗ω, as claimed. In the general case, using a partition of unity, for every open cover of M by positively oriented charts (Ui, ϕi), we have a partition of unity (ρi)i∈I subordinate to this cover. Recall that supp(ρi) ⊆Ui, i ∈I. 7.4. INTEGRATION ON MANIFOLDS 247 Thus, ρiω is an n-form whose support is a subset of Ui. Furthermore, as P i ρi = 1, ω = X i ρiω. Define I(ω) = X i Z Ui ρiω, where each term in the sum is defined by Z Ui ρiω = Z ϕi(Ui) (ϕ−1 i )∗ρiω, where (Ui, ϕi) is the chart associated with i ∈I. It remains to show that I(ω) does not depend on the choice of open cover and on the choice of partition of unity. Let (Vj, ψj) be another open cover by positively oriented charts, and let (θj)j∈J be a partition of unity subordinate to the open cover (Vj). Note that Z Ui ρiθjω = Z Vj ρiθjω, since supp(ρiθjω) ⊆Ui ∩Vj, and as P i ρi = 1 and P j θj = 1, we have X i Z Ui ρiω = X i,j Z Ui ρiθjω = X i,j Z Vj ρiθjω = X j Z Vj θjω, proving that I(ω) is indeed independent of the open cover and of the partition of unity. The uniqueness assertion is easily proved using a partition of unity. Since the integral at (†) is well-defined we are able to make the following definition. Definition 7.11. Let M be a smooth oriented manifold of dimension n. For ω ∈An c (M), if (U, ϕ) is a positively oriented chart, and supp(ω) ⊆U, we define R M ω by Z M ω = Z ϕ(U) (ϕ−1)∗ω. Given an embedded manifold M in Rn, Definition 7.11 shows that integration of a form over a manifold reduces, after a change of variables, to an appropriate Riemann integral over the parameter space. We will demonstrate the meaning of this sentence by explicitly calculating R S2 VolS2. In Section 7.2 we described a parametrization of S2 by ψ : U →S2 where x = sin θ cos ϕ, y = sin θ sin ϕ, z = cos θ, 248 CHAPTER 7. INTEGRATION ON MANIFOLDS and U = {(θ, ϕ) : 0 < θ < π, 0 < ϕ < 2π} ⊂R2. See Figure 4.1. We then showed that VolS2 = z dx ∧dy −y dx ∧dz + x dy ∧dz. To calculate R S2 VolS2, we first use (†) of Section 4.3 to calculate ψ∗(z dx ∧dy) = cos θ(d(sin θ cos ϕ) ∧d(sin θ sin ϕ)) = cos θ ((cos θ cos ϕ dθ −sin θ sin ϕ dϕ) ∧(cos θ sin ϕ dθ + sin θ cos ϕ dϕ)) = cos θ(cos2 ϕ cos θ sin θ + sin2 ϕ sin θ cos θ)dθ ∧dϕ = cos2 θ sin θ dθ ∧dϕ ψ∗(−y dx ∧dz) = −sin θ sin ϕ (d(sin θ cos ϕ) ∧d(cos θ)) = −sin θ sin ϕ ((cos θ cos ϕ dθ −sin θ sin ϕ dϕ) ∧−sin θ dθ) = sin3 θ sin2 ϕ dθ ∧dϕ ψ∗(x dy ∧dz) = sin θ cos ϕ (d(sin θ sin ϕ) ∧d(cos θ)) = sin θ cos ϕ ((cos θ sin ϕ dθ + sin θ cos ϕ dϕ) ∧−sin θ dθ) = sin3 θ cos2 ϕ dθ ∧dϕ. Then ϕ∗(VolS2) = (cos2 θ sin θ + sin3 θ sin2 ϕ + sin3 θ cos2 ϕ)dθ ∧dϕ = (cos2 θ sin θ + sin3 θ)dθ ∧dϕ = sin θ(cos2 θ + sin2 θ)dθ ∧dϕ = sin θdθ ∧dϕ, and Line (†) implies that Z S2(VolS2) = Z 2π 0 Z π 0 ϕ∗(VolS2) = Z 2π 0 Z π 0 sin θ dθ dϕ = 2π [−cos θ]π 0 = 4π. Observe that 4π is indeed the surface area of S2, a result we should have expected since we were integrating the volume form. The integral of Definition 7.11 has the following properties: Proposition 7.8. Let M be an oriented manifold of dimension n. The following properties hold: (1) If M is connected, then for every n-form ω ∈An c (M), the sign of R M ω changes when the orientation of M is reversed. (2) For every n-form ω ∈An c (M), if supp(ω) ⊆W for some open subset W of M, then Z M ω = Z W ω, where W is given the orientation induced by M. 7.5. DENSITIES ⊛ 249 (3) If ϕ: M →N is an orientation-preserving diffeomorphism, then for every ω ∈An c (N), we have Z N ω = Z M ϕ∗ω. Proof. Use a partition of unity to reduce to the case where supp(ω) is contained in the domain of a chart, and then use Proposition 7.6 and (†) from Proposition 7.7. It is also possible to define integration on non-orientable manifolds using densities. The next section will not be used anywhere else in this book and can be omitted. 7.5 Densities ⊛ Definition 7.12. Given a vector space V of dimension n ≥1, a density on V is a function µ: V n →R such that for every linear map f : V →V , we have µ(f(v1), . . . , f(vn)) = | det(f)|µ(v1, . . . , vn) for all v1, . . . , vn ∈V . If (v1, . . . , vn) are linearly dependent, then for any basis (e1, . . . , en) of V there is a unique linear map f such that f(ei) = vi for i = 1, . . . , n, and since (v1, . . . , vn) are linearly dependent, f is singular so det(f) = 0, which implies that µ(v1, . . . , vn) = | det(f)|µ(e1, . . . , en) = 0 for any linearly dependent vectors v1, . . . , vn ∈V . In view of this fact, a density is sometimes defined as a function µ: Vn V →R such that for every automorphism f ∈GL(V ), µ(f(v1) ∧· · · ∧f(vn)) = | det(f)|µ(v1 ∧· · · ∧vn) (††) for all v1 ∧· · · ∧vn ∈V (with µ(0) = 0). For any nonzero v1 ∧· · · ∧vn, w1 ∧· · · ∧wn ∈Vn V , because w1 ∧· · · ∧wn = det(P)v1 ∧· · · ∧vn where P is the matrix whose jth column consists of the coefficients of wj over the basis (v1, . . . , vn), it is not hard to show that Condition (††) is equivalent to µ(cw) = |c|µ(w), w ∈ n ^ V, c ∈R. Densities are not multilinear, but it is not hard to show that for any fixed n, they form a vector space of dimension 1 which is spanned by the absolute value |ω| of any nonzero n-form ω ∈Vn V ∗. Let den(V ) be the set of all densities on V . We have the following proposition from Lee (Chapter 14, Proposition 14.26). 250 CHAPTER 7. INTEGRATION ON MANIFOLDS Proposition 7.9. Let V be any vector space of dimension n ≥1. The following properties hold: (a) The set den(V ) is a vector space. (b) For any two densities µ1, µ2 ∈den(V ) and for any basis (e1, . . . , en) of V , if µ1(e1, . . . , en) = µ2(e1, . . . , en), then µ1 = µ2. (c) For any n-form ω ∈Vn V ∗, the function |ω| given by |ω|(v1, . . . , vn) = |ω(v1, . . . , vn)| is a density. (d) The vector space den(V ) is a one-dimensional space spanned by |ω| for any nonzero ω ∈Vn V ∗. Proof. (a) That den(V ) is a vector space is immediate from the definition. (b) Pick any n vectors (v1, . . . , vn) ∈V n. Since (e1, . . . , en) is a basis of V , there is a unique linear map f : V →V such that f(ei) = vi for i = 1, . . . , n, and since by hypothesis µ1(e1, . . . , en) = µ2(e1, . . . , en), we have µ1(v1, . . . , vn) = µ1(f(e1), . . . , f(en)) = | det(f)|µ1(e1, . . . , en) = | det(f)|µ2(e1, . . . , en) = µ2(f(e1), . . . , f(en)) = µ2(v1, . . . , vn), which proves that µ1 = µ2. (c) If ω ∈Vn V ∗, then |ω|(f(v1), . . . , f(vn)) = |ω(f(v1), . . . , f(vn))| = | det(f)ω(v1, . . . , vn)| = | det(f)| |ω|(v1, . . . , vn), which shows that |ω| is a density. (d) Let (e1, . . . , en) be any basis of V , and let ω ∈Vn V ∗be any nonzero n-form. For any density µ, we need to show that µ = c|ω| for some c ∈R. Let a = |ω|(e1, . . . , en) = |ω(e1, . . . , en)| b = µ(e1, . . . , en). 7.5. DENSITIES ⊛ 251 Since ω ̸= 0 and (e1, . . . , en) is a basis, ω(e1, . . . , en) ̸= 0 so a ̸= 0, and by Condition (c) (b/a)|ω| is a density. Since (b/a)|ω|(e1, . . . , en) = b = µ(e1, . . . , en), by Condition (b) µ = (b/a)|ω|, as desired. If we denote the vector space of densities on V by den(V ), then given a manifold M, we can form the density bundle den(M) whose underlying set is the disjoint union of the vector spaces den(TpM) for all p ∈M. This set can be made into a smooth bundle, and a density on M is a smooth section of the density bundle. The main property of densities is that every smooth manifold admits a global smooth (positive) density, without any orientability assumptions. Then it is possible to carry out the theory of integration on manifolds using densities instead of volume forms, as we did in this section. This development can be found in Lee (Chapter 14), but we have no need for this extra generality. It turns out that orientations can be defined as certain functions satisfying a variant of the condition used in Definition 7.12, and this definition clarifies the relationship between volume forms and densities. The sign function is defined such that for any λ ∈R, sign(λ) =      +1 if λ > 0 −1 if λ < 0 0 if λ = 0. Definition 7.13. Given a vector space V of dimension n ≥1, an orientation on V is a function o: V n →R such that for every linear map f : V →V , we have o(f(v1), . . . , f(vn)) = sign(det(f))o(v1, . . . , vn) for all v1, . . . , vn ∈V . If (v1, . . . , vn) are linearly dependent, then for any basis (e1, . . . , en) of V there is a unique linear map f such that f(ei) = vi for i = 1, . . . , n, and since (v1, . . . , vn) are linearly dependent, f is singular so det(f) = 0, which implies that o(v1, . . . , vn) = sign(det(f))o(e1, . . . , en) = 0 for any linearly dependent vectors v1, . . . , vn ∈V . For any two bases (u1, . . . , un) and (v1, . . . , vn), there is a unique linear map f such that f(ui) = vi for i = 1, . . . , n, and o(u1, . . . , un) = o(v1, . . . , vn) iffdet(f) > 0, which is indeed the condition for (u1, . . . , un) and (v1, . . . , vn) to have the same orientation. There are exactly two orientations o such that |o(u1, . . . , un)| = 1. Let Or(V ) be the set of all orientations on V . We have the following proposition. 252 CHAPTER 7. INTEGRATION ON MANIFOLDS Proposition 7.10. Let V be any vector space of dimension n ≥1. The following properties hold: (a) The set Or(V ) is a vector space. (b) For any two orientations o1, o2 ∈Or(V ) and for any basis (e1, . . . , en) of V , if o1(e1, . . . , en) = o2(e1, . . . , en), then o1 = o2. (c) For any nonzero n-form ω ∈Vn V ∗, the function o(ω) given by o(ω)(v1, . . . , vn) = 0 if (v1, . . . , vn) are linearly dependent and o(ω)(v1, . . . , vn) = ω(v1, . . . , vn) |ω(v1, . . . , vn)|, if (v1, . . . , vn) are linearly independent, is an orientation. (d) The vector space Or(V ) is a one-dimensional space, and it is spanned by o(ω) for any nonzero ω ∈Vn V ∗. Proof. (a) That Or(V ) is a vector space is immediate from the definition. (b) Pick any n vectors (v1, . . . , vn) ∈V n. Since (e1, . . . , en) is a basis of V , there is a unique linear map f : V →V such that f(ei) = vi for i = 1, . . . , n, and since by hypothesis o1(e1, . . . , en) = o2(e1, . . . , en), we have o1(v1, . . . , vn) = o1(f(e1), . . . , f(en)) = sign(det(f))o1(e1, . . . , en) = sign(det(f))o2(e1, . . . , en) = o2(f(e1), . . . , f(en)) = o2(v1, . . . , vn), which proves that o1 = o2. (c) Let ω ∈Vn V ∗be any nonzero form. If (v1, . . . , vn) are linearly independent, then we know that ω(f(v1), . . . , f(vn)) = det(f) ω(v1, . . . , vn) |ω|(f(v1), . . . , f(vn)) = | det(f)| |ω|(v1, . . . , vn). We know that det(f) = 0 ifff is singular, but then (f(v1), . . . , f(vn)) are linearly dependent so o(ω)(f(v1), . . . , f(vn)) = 0 = sign(det(f))o(ω)(v1, . . . , vn). 7.5. DENSITIES ⊛ 253 If det(f) ̸= 0, then o(ω)(f(v1), . . . , f(vn)) = ω(f(v1), . . . , f(vn)) |ω(f(v1), . . . , f(vn))| = det(f) | det(f)| ω(f(v1), . . . , f(vn)) |ω(f(v1), . . . , f(vn))| = sign(det(f))o(ω)(v1, . . . , vn), which shows that o(ω) is an orientation. (d) Let (e1, . . . , en) be any basis of V , and let ω ∈Vn V ∗be any nonzero n-form. For any orientation o, we need to show that o = co(ω) for some c ∈R. Let a = o(ω)(e1, . . . , en) = ω(e1, . . . , ven) |ω(e1, . . . , en)| b = o(e1, . . . , en). Since ω ̸= 0 and (e1, . . . , en) is a basis, ω(e1, . . . , en) ̸= 0 so a ̸= 0, and by Condition (c) (b/a)o(ω) is an orientation. Since (b/a)o(ω)(e1, . . . , en) = b = o(e1, . . . , en), by Condition (b) o = (b/a)o(ω), as desired. Part (c) of Proposition 7.10 implies that for every nonzero n-form ω ∈Vn V ∗, there exists some density |ω| and some orientation o(ω) such that o(ω)|ω| = ω. This shows that orientations are just normalized volume forms that take exactly two values c and −c on linearly independent vectors (with c > 0), whereas densities are absolute values of volume forms. We have the following results showing the relationship between the spaces Vn V ∗, Or(V ), and den(V ). Proposition 7.11. Let V be any vector space of dimension n ≥1. For any nonzero n-form ω ∈Vn V ∗, the bilinear map Φ: Or(V ) × den(V ) →Vn V ∗given by Φ(αo(ω), β|ω|) = αβω, α, β ∈R induces an isomorphism Or(V ) ⊗den(V ) ∼ = Vn V ∗. Proof. The spaces Vn V ∗, Or(V ), and den(V ) are all one-dimensional, and if ω ̸= 0, then ω is a basis of Vn V ∗and Propositions 7.9 and 7.10 show that o(ω) is a basis of Or(V ) and |ω| is a basis of den(V ), so the map Φ defines a bilinear map from Or(V ) × den(V ) to Vn V ∗. 254 CHAPTER 7. INTEGRATION ON MANIFOLDS Therefore, by the universal mapping property, we obtain a linear map Φ⊗: Or(V ) ⊗den(V ) →Vn V ∗. Since ω ̸= 0, we have o(ω)|ω| = ω, which shows that Φ is surjective, and thus Φ⊗is surjective. Since all the spaces involved are one-dimensional, Or(V ) ⊗den(V ) is also one-dimensional, so Φ⊗is bijective. Given a manifold M, we can form the orientation bundle Or(M) whose underlying set is the disjoint union of the vector spaces Or(TpM) for all p ∈M. This set can be made into a smooth bundle, and an orientation of M is a smooth global section of the orientation bundle. Then it can be shown that there is a bundle isomorphism Or(M) ⊗den(M) ∼ = n ^ T ∗M. and since den(M) always has global sections, we see that there is a global volume form iff Or(M) has a global section iffM is orientable. The theory or integration developed so far deals with domains that are not general enough. Indeed, for many applications, we need to integrate over domains with boundaries. 7.6 Manifolds With Boundary Up to now we have defined manifolds locally diffeomorphic to an open subset of Rm. This excludes many natural spaces such as a closed disk, whose boundary is a circle, a closed ball B(1), whose boundary is the sphere Sm−1, a compact cylinder S1 × [0, 1], whose boundary consist of two circles, a M¨ obius strip, etc. These spaces fail to be manifolds because they have a boundary; that is, neighborhoods of points on their boundaries are not diffeomorphic to open sets in Rm. Perhaps the simplest example is the (closed) upper half space Hm = {(x1, . . . , xm) ∈Rm | xm ≥0}. Under the natural embedding Rm−1 ∼ = Rm−1 × {0} , →Rm, the subset ∂Hm of Hm defined by ∂Hm = {x ∈Hm | xm = 0} is isomorphic to Rm−1, and is called the boundary of Hm. When m = 0 we have H0 = ∅and ∂H0 = ∅. We also define the interior of Hm as Int(Hm) = Hm −∂Hm. Now if U and V are open subsets of Hm, where Hm ⊆Rm has the subset topology, and if f : U →V is a continuous function, we need to explain what we mean by f being smooth. 7.6. MANIFOLDS WITH BOUNDARY 255 Definition 7.14. Let U and V be open subsets of Hm. We say that f : U →V as above is smooth if it has an extension e f : e U →e V , where e U and e V are open subsets of Rm with U ⊆e U and V ⊆e V , and with e f a smooth function. We say that f is a (smooth) diffeomorphism iff f −1 exists and if both f and f −1 are smooth, as just defined. To define a manifold with boundary, we replace everywhere R by H in the defintion of a chart and in the definition of an atlas (see Tu , Chapter 6, §22, or Gallier and Quaintance ). So, for instance, given a topological space M, a chart is now pair (U, ϕ), where U is an open subset of M and ϕ: U →Ωis a homeomorphism onto an open subset Ω= ϕ(U) of Hnϕ (for some nϕ ≥1), etc. Thus, we obtain Definition 7.15. Given some integer n ≥1 and given some k such that k is either an integer k ≥1 or k = ∞, a Ck-manifold of dimension n with boundary consists of a topological space M together with an equivalence class A of Ck n-atlases on M (where the charts are now defined in terms of open subsets of Hn). Any atlas A in the equivalence class A is called a differentiable structure of class Ck (and dimension n) on M. We say that M is modeled on Hn. When k = ∞, we say that M is a smooth manifold with boundary. It remains to define what is the boundary of a manifold with boundary. Definition 7.16. Let M be a manifold with boundary as defined by Definition 7.15. The boundary ∂M of M is the set of all points p ∈M, such that there is some chart (Uα, ϕα), with p ∈Uα and ϕα(p) ∈∂Hn. We also let Int(M) = M −∂M and call it the interior of M. p q 0 0 H H 2 2 φ φ(p) Ψ Ψ (q) Figure 7.4: A two dimensional manifold with red boundary. 256 CHAPTER 7. INTEGRATION ON MANIFOLDS  Do not confuse the boundary ∂M and the interior Int(M) of a manifold with bound-ary embedded in RN with the topological notions of boundary and interior of M as a topological space. In general, they are different. For example, if M is the subset [0, 1) ∪{2} of the real line, then its manifold boundary is ∂M = {0}, and its topological boundary is Bd(M) = {0, 1, 2}. Note that manifolds are also manifolds with boundary: their boundary is just empty. We shall still reserve the word “manifold” for these, but for emphasis, we will sometimes call them “boundaryless.” The definition of tangent spaces, tangent maps, etc., are easily extended to manifolds with boundary. The reader should note that if M is a manifold with boundary of dimension n, the tangent space TpM is defined for all p ∈M and has dimension n, even for boundary points p ∈∂M. The only notion that requires more care is that of a submanifold. For more on this, see Hirsch , Chapter 1, Section 4. One should also beware that the product of two manifolds with boundary is generally not a manifold with boundary (consider the product [0, 1]×[0, 1] of two line segments). There is a generalization of the notion of a manifold with boundary called manifold with corners, and such manifolds are closed under products (see Hirsch , Chapter 1, Section 4, Exercise 12). If M is a manifold with boundary, we see that Int(M) is a manifold without boundary. What about ∂M? Interestingly, the boundary ∂M of a manifold with boundary M of dimension n is a manifold of dimension n −1. For this we need the following proposition. Proposition 7.12. If M is a manifold with boundary of dimension n, for any p ∈∂M on the boundary on M, for any chart (U, ϕ) with p ∈M, we have ϕ(p) ∈∂Hn. Proof. Since p ∈∂M, by definition, there is some chart (V, ψ) with p ∈V and ψ(p) ∈∂Hn. Let (U, ϕ) be any other chart, with p ∈M, and assume that q = ϕ(p) ∈Int(Hn). The transition map ψ ◦ϕ−1 : ϕ(U ∩V ) →ψ(U ∩V ) is a diffeomorphism, and q = ϕ(p) ∈Int(Hn). By the inverse function theorem, there is some open W ⊆ϕ(U ∩V ) ∩Int(Hn) ⊆Rn, with q ∈W, so that ψ ◦ϕ−1 maps W homeomorphically onto some subset Ωopen in Int(Hn), with ψ(p) ∈Ω, contradicting the hypothesis, ψ(p) ∈∂Hn. Using Proposition 7.12, we immediately derive the fact that ∂M is a manifold of dimen-sion n −1. We obtain charts on ∂M by considering the charts (U ∩∂M, L ◦ϕ), where (U, ϕ) is a chart on M such that U ∩∂M = ϕ−1(∂Hn) ̸= ∅and L: ∂Hn →Rn−1 is the natural isomorphism (see see Hirsch , Chapter 1, Section 4). 7.7 Integration on Regular Domains and Stokes’ Theorem Given a manifold M, we define a class of subsets with boundaries that can be integrated on, and for which Stokes’ theorem holds. In Warner (Chapter 4), such subsets are called 7.7. INTEGRATION ON REGULAR DOMAINS AND STOKES’ THEOREM 257 regular domains, and in Madsen and Tornehave (Chapter 10), they are called domains with smooth boundary. Definition 7.17. Let M be a smooth manifold of dimension n. A subset N ⊆M is called a domain with smooth boundary (or codimension zero submanifold with boundary) ifffor every p ∈M, there is a chart (U, ϕ) with p ∈U such that ϕ(U ∩N) = ϕ(U) ∩Hn, (∗) where Hn is the closed upper-half space Hn = {(x1, . . . , xn) ∈Rn | xn ≥0}. Note that (∗) is automatically satisfied when p is an interior or an exterior point of N, since we can pick a chart such that ϕ(U) is contained in an open half space of Rn defined by either xn > 0 or xn < 0. If p is a boundary point of N, then ϕ(p) has its last coordinate equal to 0; see Figure 7.5. p U φ φ(p) φ( U ) H X 2 Figure 7.5: The subset N, the peach region of the torus M, is a domain with smooth boundary. If M is orientable, then any orientation of M induces an orientation of ∂N, the boundary of N. This follows from the following proposition: 258 CHAPTER 7. INTEGRATION ON MANIFOLDS Proposition 7.13. Let ϕ: Hn →Hn be a diffeomorphism with everywhere positive Jaco-bian determinant. Then ϕ induces a diffeomorphism Φ: ∂Hn →∂Hn, which viewed as a diffeomorphism of Rn−1, also has everywhere positive Jacobian determinant. Proof. By the inverse function theorem, every interior point of Hn is the image of an interior point, so ϕ maps the boundary to itself. If ϕ = (ϕ1, . . . , ϕn), then Φ = (ϕ1(x1, . . . , xn−1, 0), . . . , ϕn−1(x1, . . . , xn−1, 0)), since ϕn(x1, . . . , xn−1, 0) = 0. It follows that ∂ϕn ∂xi (x1, . . . , xn−1, 0) = 0 for i = 1, . . . , n −1, and as ϕ maps Hn to itself, ∂ϕn ∂xn (x1, . . . , xn−1, 0) > 0. Now the Jacobian matrix of ϕ at q = ϕ(p) ∈∂Hn is of the form J(ϕ)(q) =      ∗ dΦq . . . ∗ 0 · · · 0 ∂ϕn ∂xn (q)      and since ∂ϕn ∂xn (q) > 0 and by hypothesis det(J(ϕ)q) > 0, we have det(J(Φ)q) > 0, as claimed. In order to make Stokes’ formula sign free, if Rn has the orientation given by dx1∧· · ·∧dxn, then ∂Hn is given the orientation given by (−1)ndx1 ∧· · ·∧dxn−1 if n ≥2, and −1 for n = 1. In particular ∂H2 is oriented by e1 while ∂H3 is oriented by −e1 ∧e2 = e2 ∧e1. See Figure 7.6. Definition 7.18. Given any domain with smooth boundary N ⊆M, a tangent vector w ∈TpM at a boundary point p ∈∂N is outward directed iffthere is a chart (U, ϕ) with p ∈U, ϕ(U ∩N) = ϕ(U) ∩Hn, and dϕp(w) has a negative nth coordinate prn(dϕp(w)); see Figure 7.7. Let (V, ψ) be another chart with p ∈V . The transition map θ = ψ ◦ϕ−1 : ϕ(U ∩V ) →ψ(U ∩V ) induces a map ϕ(U ∩V ) ∩Hn − →ψ(U ∩V ) ∩Hn which restricts to a diffeomorphism Θ: ϕ(U ∩V ) ∩∂Hn →ψ(U ∩V ) ∩∂Hn. 7.7. INTEGRATION ON REGULAR DOMAINS AND STOKES’ THEOREM 259 e2 e2 -e1 e1 e3 -e2 -^ e1 = e1 ^ e2 e3 -^ e2 ^ e1 = e1 ^ e2 ^ e3 Figure 7.6: The boundary orientations of ∂H2 and ∂H3. The proof of Proposition 7.13 shows that the Jacobian matrix of dθq at q = ϕ(p) ∈∂Hn is of the form J(θ)(q) =      ∗ J(Θ)q . . . ∗ 0 · · · 0 ∂θn ∂xn(q)      with θ = (θ1, . . . , θn), and that ∂θn ∂xn(q) > 0. As dψp = d(ψ ◦ϕ−1)q ◦dϕp, we see that for any w ∈TpM with prn(dϕp(w)) < 0, since prn(dψp(w)) = ∂θn ∂xn(q) prn(dϕp(w)), we also have prn(dψp(w)) < 0 (recall that θ = ψ ◦ϕ−1). Therefore, the negativity condition of Definition 7.18 does not depend on the chart at p. The following proposition is then easy to show. Proposition 7.14. Let N ⊆M be a domain with smooth boundary, where M is a smooth manifold of dimension n. (1) The boundary ∂N of N is a smooth manifold of dimension n −1. (2) Assume M is oriented. If n ≥2, there is an induced orientation on ∂N determined as follows: For every p ∈∂N, if v1 ∈TpM is an outward directed tangent vector, then a basis (v2, . . . , vn) for Tp∂N is positively oriented iffthe basis (v1, v2, . . . , vn) for TpM is positively oriented. When n = 1, every p ∈∂N has the orientation +1 ifffor every 260 CHAPTER 7. INTEGRATION ON MANIFOLDS p U U φ φ( U ) H X 2 w φ (w) p d Figure 7.7: An example of an outward directed tangent vector to N. Notice this red tangent vector points away from N. outward directed tangent vector v1 ∈TpM, the vector v1 is a positively oriented basis of TpM. Part (2) of Proposition 7.14 is summarized as “outward pointing vector first.” When M is an n-dimensional embedded manifold in Rm with an orientation preserving parametrization ψ : U →Rm, for any point p = ψ(q) ∈∂N, let v1 be a tangent vector pointing away from N. This means dψq(−en) = v1. To complete the basis of TpM in a manner consistent with the positive orientation of U given by dx1 ∧· · · ∧dxn, we choose an ordered basis (v2, · · · , vn) of Tp∂N such that dψq((−1)ne1) = v2 and dψq(ei) = vi+1 whenever 2 ≤i ≤n −1. Intuitively, ψ maps the positive orientation of U to a positive orientation of TpM with the condition that the first vector in the orientation frame of TpM points away from N. See Figure 7.8. Another way to describe the induced orientation of ∂N is through the insertion operator; see Definition 4.13. Let ω be a volume form on M, let p ∈∂N, and let v1 ∈TpM be an outward directed tangent vector. The volume form on ∂N is given by iv1ω where iv1ω(v2, · · · , vn) = ω(v1, v2, · · · , vn). If M is oriented, then for every n-form ω ∈An c (M), the integral R N ω is well-defined. More precisely, Proposition 7.7 can be generalized to domains with a smooth boundary. This 7.7. INTEGRATION ON REGULAR DOMAINS AND STOKES’ THEOREM 261 p U Ψ Ψ Ψ Ψ 1 2 3 4 Figure 7.8: The orientation of Tp∂N consistent with the positive orientation of R2. can be shown in various ways. The most natural way to proceed is to prove an extension of Proposition 7.6 using a slight generalization of the change of variable formula. Proposition 7.15. Let ϕ: U →V be a diffeomorphism between two open subsets of Rn, and assume that ϕ maps U ∩Hn to V ∩Hn. Then for every smooth function f : V →R with compact support, Z V ∩Hn f(x)dx1 · · · dxn = Z U∩Hn f(ϕ(y)) |J(ϕ)y| dy1 · · · dyn. One alternative way to define R N ω involves covering N with special kinds of open subsets arising from regular simplices (see Warner , Chapter 4). Remark: Another alternative way to proceed is to apply techniques of measure theory. In Madsen and Tornehave it is argued that integration theory goes through for continuous n-forms with compact support. If σ is a volume form on M, then for every continuous 262 CHAPTER 7. INTEGRATION ON MANIFOLDS function with compact support f, the map f 7→Iσ(f) = Z M fσ is a linear positive operator1 (which means that I(f) ≥0 for f ≥0). By Riesz’ representation theorem (see Rudin , Chapter 2), Iσ determines a positive Borel measure µσ which satisfies Z M fdµσ = Z M fσ for all continuous functions f with compact support. Since any C1 n-form ω can be written uniquely as ω = fσ for some C1 function f, we can set Z N ω = Z M 1Nfσ, where 1N is the function with value 1 on N and 0 outside N. We now have all the ingredient to state and prove Stokes’s formula. Our proof is based on the proof found in Section 23.5 of Tu . Alternative proofs can be found in many places (for example, Warner (Chapter 4), Bott and Tu (Chapter 1), and Madsen and Tornehave (Chapter 10). Theorem 7.16. (Stokes’ Theorem) Let N ⊆M be a domain with smooth boundary, where M is a smooth oriented manifold of dimension n, give ∂N the orientation induced by M, and let i: ∂N →M be the inclusion map. For every differential form with compact support ω ∈An−1 c (M) with N ∩supp(ω) compact, we have Z ∂N i∗ω = Z N dω. In particular, if N = M is a smooth oriented manifold with boundary, then Z ∂M i∗ω = Z M dω, (∗∗∗) and if M is a smooth oriented manifold without boundary, then Z M dω = 0. Of course, i∗ω is the restriction of ω to ∂N, and for simplicity of notation i∗ω is usually written ω, and Stokes’ formula is written Z ∂N ω = Z N dω. 1In fact, a Radon measure. 7.7. INTEGRATION ON REGULAR DOMAINS AND STOKES’ THEOREM 263 Proof based on Tu . We select a covering {(Ui, ϕi)}i∈I of M and we restrict to those charts (Ui, ϕi) such that ϕi(Ui ∩N) = ϕi(Ui)∩Hn is diffeomorphic to either Rn or Hn via an orientation preserving diffeomorphism. Note that each Ui has a nonempty intersection with N. Let (ρi)i∈I be a partition of unity subordinate to this cover. An adaptation of the proof of Proposition 7.7 shows that ρiω is an (n −1)-form on M with compact support in Ui. Assume that Stokes’ theorem is true for Rn and Hn. Then Stokes’ theorem will hold for all Ui which are diffeomorphic to either Rn or Hn. Observe that the paragraph preceding Proposition 7.14 implies that ∂N ∩Ui = ∂Ui. Since P i ρi = 1, we have Z ∂N ω = Z ∂N X i ρiω = X i Z ∂N ρiω, since X i ρiω is finite = X i Z ∂Ui ρiω, since supp(ρiω) ⊆Ui = X i Z Ui d(ρiω), by assumption that Stokes’ is true for Ui = X i Z N d(ρω), since supp(d(ρiω)) ⊆Ui ∩N = Z N d X i ρiω ! = Z N dω. Thus it remains to prove Stokes’ theorem for Rn and Hn. Since ω is now assumed to be an (n −1)-form on Rn or Hn with compact support, ω = n X i=1 fi dx1 ∧· · · ∧c dxi ∧· · · ∧dxn, where each fi is a smooth function with compact support in Rn or Hn. By using the R-linearity of the exterior derivative and the integral operator, we may assume that ω has only one term, namely ω = f dx1 ∧· · · ∧c dxi ∧· · · ∧dxn, and d ω = n X j=1 ∂f ∂xj dxj ∧dx1 ∧· · · ∧c dxi ∧· · · ∧dxn = (−1)i−1 ∂f ∂xi dx1 ∧· · · ∧dxi ∧· · · ∧dxn. 264 CHAPTER 7. INTEGRATION ON MANIFOLDS where f is smooth function on Rn such that supp(f) is contained in the interior of the n-cube [−a, a]n for some fixed a > 0. To verify Stokes’ theorem for Rn, we evaluate R Rn d ω as an iterated integral via Fubini’s theorem. (See Edwards , Theorem 4.1.) In particular, we find that Z Rn d ω = Z Rn(−1)i−1 ∂f ∂xi dx1 · · · dxi · · · dxn = (−1)i−1 Z Rn−1 Z ∞ −∞ ∂f ∂xi dxi  dx1 · · · c dxi · · · dxn = (−1)i−1 Z Rn−1 Z a −a ∂f ∂xi dxi  dx1 · · · c dxi · · · dxn = (−1)i−1 Z Rn−1 0 dx1 · · · c dxi · · · dxn since supp(f) ⊂[−a, a]n = 0 = Z ∅ ω = Z ∂Rn ω. The verification of Stokes’ theorem for Hn involves the analysis of two cases. For the first case assume i ̸= n. Since ∂Hn is given by xn = 0, then dxn ≡0 on ∂Hn. An application of Fubini’s theorem shows that Z Hn d ω = Z Hn(−1)i−1 ∂f ∂xi dx1 · · · dxi · · · dxn = (−1)i−1 Z Hn−1 Z ∞ −∞ ∂f ∂xi dxi  dx1 · · · c dxi · · · dxn = (−1)i−1 Z Hn−1 Z a −a ∂f ∂xi dxi  dx1 · · · c dxi · · · dxn = (−1)i−1 Z Hn−1 0 dx1 · · · c dxi · · · dxn since supp(f) ⊂[−a, a]n = 0 = Z ∂Hn f dx1 · · · c dxi · · · dxn, since dxn ≡0 on ∂Hn. It remains to analyze the case i = n. Fubini’s theorem implies Z Hn d ω = Z Hn(−1)n−1 ∂f ∂xn dx1 · · · dxn = (−1)n−1 Z Rn−1 Z ∞ 0 ∂f ∂xn dxn  dx1, · · · dxn−1 = (−1)n−1 Z Rn−1 Z a 0 ∂f ∂xn dxn  dx1 · · · dxn−1 = (−1)n Z Rn−1 f(x1, · · · , xn−1, 0) dx1 · · · dxn−1, since supp(f) ⊂[−a, a]n = Z ∂Hn ω, 7.7. INTEGRATION ON REGULAR DOMAINS AND STOKES’ THEOREM 265 where the last equality follows from the fact that (−1)nRn−1 is the induced boundary orien-tation of ∂Hn. Stokes’ theorem, as presented in Theorem 7.16, unifies the integral theorems of vector calculus since the classical integral theorems of vector calculus are particular examples of (∗∗∗) when M is an n-dimensional manifold embedded in R3. If n = 3, ω ∈A2 c(M), and (∗∗∗) becomes the Divergence theorem. Given a smooth function F : R3 →R3, recall that the divergence of F is the smooth real-valued function divF : R3 →R where divF = ∂F1 ∂x1 + ∂F2 ∂x2 + ∂F3 ∂x3 , and (x1, x2, x3) are the standard coordinates of R3 (often represented as (x, y, z)). The Divergence theorem is as follows: Proposition 7.17. (Divergence Theorem) Let F : R3 →R3 be a smooth vector field defined on a neighborhood of M, a compact oriented smooth 3-dimensional manifold with boundary. Then Z M divF VolM = Z ∂M F · N Vol∂M, (1) where N(x) = (n1(x), n2(x), n3(x)) is the unit outer normal vector field on ∂M, Vol∂M = n1 dx2 ∧dx3 −n2 dx1 ∧dx3 + n3 dx1 ∧dx2 and ∂M is positively oriented as the boundary of M ⊆R3. In calculus books (1) is often written as Z Z Z divF dx dy dz = Z F · N dS, (2) where dS is the surface area differential. In particular if ∂M is parametrized by ϕ(x, y) = (x, y, f(x, y)), Z F · N dS = Z Z F ·  −∂f ∂x, −∂f ∂y , 1  dx dy. (3) The verification of (3) is an application of Equation (∗∗) from Section 7.2. In particular J(ϕ)(x, y) =   1 0 0 1 ∂f ∂x ∂f ∂y  , which in turn implies D = det J(ϕ)⊤(x, y)J(ϕ)(x, y) 1 2 = s 1 + ∂f ∂x 2 + ∂f ∂y 2 . 266 CHAPTER 7. INTEGRATION ON MANIFOLDS Hence n1,2 = det 1 0 0 1  D = 1 r 1 + ∂f ∂x 2 +  ∂f ∂y 2 n1,3 = det  1 0 ∂f ∂x ∂f ∂y  D = ∂f ∂y r 1 + ∂f ∂x 2 +  ∂f ∂y 2 n2,3 = det  0 1 ∂f ∂x ∂f ∂y  D = −∂f ∂x r 1 + ∂f ∂x 2 +  ∂f ∂y 2 and dS = n1,2 dx ∧dy + n1,3 dx ∧dz + n2,3 dy ∧dz = dx ∧dy r 1 + ∂f ∂x 2 +  ∂f ∂y 2 + ∂f ∂y dx ∧dz r 1 + ∂f ∂x 2 +  ∂f ∂y 2 + −∂f ∂x dy ∧dz r 1 + ∂f ∂x 2 +  ∂f ∂y 2. Since z = f(x, y), ϕ∗(dS) = dx ∧dy r 1 + ∂f ∂x 2 +  ∂f ∂y 2 + ∂f ∂y dx ∧( ∂f ∂x dx + ∂f ∂y dy) r 1 + ∂f ∂x 2 +  ∂f ∂y 2 + −∂f ∂x dy ∧( ∂f ∂x dx + ∂f ∂y dy) r 1 + ∂f ∂x 2 +  ∂f ∂y 2 = s 1 + ∂f ∂x 2 + ∂f ∂y 2 dx ∧dy. Furthermore, N = ∂ϕ ∂x × ∂ϕ ∂y ∂ϕ ∂x × ∂ϕ ∂y = 1 r 1 + ∂f ∂x 2 +  ∂f ∂y 2  −∂f ∂x, −∂f ∂y , 1  . Substituting the expressions for N and ϕ∗(dS) into R F · N dS give the right side of (3). If n = 2, ω ∈A1 c(M), and (∗∗∗) becomes the classical Stokes’ theorem. Given a smooth function F : R3 →R3, recall that the curl of F is the smooth function curlF : R3 →R3 curlF = ∂F3 ∂x2 −∂F2 ∂x3 , ∂F1 ∂x3 −∂F3 ∂x1 , ∂F2 ∂x1 −∂F1 ∂x2  . The classical Stokes’ theorem is as follows: 7.7. INTEGRATION ON REGULAR DOMAINS AND STOKES’ THEOREM 267 Proposition 7.18. Let M be an oriented compact 2-dimensional manifold with boundary locally parametrized in R3 by the orientation-preserving local diffeomorphism ψ : U →R3 such that ψ(u, v) = (x1, x2, x3) ∈M. Define N = ∂ψ ∂u × ∂ψ ∂v ∂ψ ∂u × ∂ψ ∂v to be the smooth outward unit normal vector field on M. Let n be the outward directed tangent vector field on ∂M. Let T = N × n. Given F : R3 →R3, a smooth vector field defined on a open subset of R3 containing M, Z M curlF · N VolM = Z ∂M F · T Vol∂M, (4) where VolM is defined as in Vol∂M of Proposition 7.17 and Vol∂M = ds, the line integral form. If M is parametrized by ϕ(x, y) = (x, y, f(x, y)), we have shown that the left side of (4) may be written as Z M curlF · N VolM = Z curlF · N dS = Z Z curlF ·  −∂f ∂x, −∂f ∂y , 1  dx dy. Many calculus books represent the right side of (4) as Z ∂M F · T ds = Z F · dr, (5) where dr = (dx, dy, dz). Once again the verification of (5) is an application of Equation (∗∗) from Section 7.2. Let ψ(x) = (x, y(x), z(x)) be a parameterization of ∂M. Then J(ψ)(x) = (1, yx, zx)⊤, where yx = dy dx and zx = dz dx. Then D = det J(ψ)⊤(x)J(ψ)(x) 1 2 = p 1 + y2 x + z2 x, ds = dx + yx dy + zx dz p 1 + y2 x + z2 x , and ψ∗ds = p 1 + y2 x + z2 x dx. Furthermore T = J(ψ)(x) p 1 + y2 x + z2 x = (1, yx, zx)⊤ p 1 + y2 x + z2 x . 268 CHAPTER 7. INTEGRATION ON MANIFOLDS Substituting the expressions for T and ψ∗ds into the left side of (5) gives Z ∂M F · T ds = Z F ·  1, dy dx, dz dx  dx = Z F · (dx, dy, dz) = Z F · dr. Thus the classical form of Stokes’ theorem often appears as Z Z curlF ·  −∂f ∂x, −∂f ∂y , 1  dx dy = Z F ·  1, dy dx, dz dx  dx = Z F · dr, where M is parametrized via ϕ(x, y) = (x, y, f(x, y)). The orientation frame (n, T, N) given in Proposition 7.18 provides the standard orien-tation of R3 given by (e1, e2, e3) and is visualized as follows. Pick a preferred side of the surface. This choice is represented by N. At each boundary point, draw the outward point-ing tangent vector n which is locally perpendicular (in the tangent plane) to the boundary curve. To determine T, pretend you are a bug on the side of the surface selected by N. You must walk along the boundary curve in the direction that keeps the boundary of the surface your right. Then T = N × n and (n, T, N) is oriented via the right-hand rule in the same manner as (e1, e2, e3); see Figure 7.9. N N n T M Figure 7.9: The orientation frame (n, T, N) for the bell shaped surface M. Notice the bug must walk along the boundary in a counter clockwise direction. For those readers who wish to learn more about the connections between the classical integration theorems of vector calculus and Stokes’ theorem, we refer them to Edwards (Chapter 5, Section 7). The version of Stokes’ theorem that we have presented applies to domains with smooth boundaries, but there are many situations where it is necessary to deal with domains with 7.7. INTEGRATION ON REGULAR DOMAINS AND STOKES’ THEOREM 269 singularities, for example corners (as a cube, a tetrahedron, etc.). Manifolds with corners form a nice class of manifolds that allow such a generalization of Stokes’ theorem. To model corners, we adapt the idea that we used when we defined charts of manifolds with boundaries but instead of using the closed half space Hm, we use the closed convex cone Rm + = {(x1, . . . , xm) ∈Rm | x1 ≥0, . . . , xm ≥0}. The boundary ∂Rm + of Rm + is the space ∂Rm + = {(x1, . . . , xm) ∈Rm | x1 ≥0, . . . , xm ≥0, xi = 0 for some i}, which can also be written as ∂Rm + = H1 ∪· · · ∪Hm, with Hi = {(x1, . . . , xm) ∈Rm + | xi = 0}. The set of corner points of Rm + is the subset {(x1, . . . , xm) ∈Rm + | ∃i∃j(i ̸= j), xi = 0 and xj = 0}. Equivalently, the set of corner points is the union of all intersections Hi1 ∩· · · ∩Hik for all finite subsets {i1, . . . , ik} of {1, . . . , m} with k ≥2. See Figure 7.10. R R R 1 2 3 + + + H H 1 2 H1 H1 H3 H2 Figure 7.10: The closed convex cones R1 +, R2 +, and R3 +. Corner points are in red. Definition 7.19. Given a topological space M, a chart with corners is a pair (U, ϕ) where U is some open subset of M and ϕ is a homeomorphism of U onto some open subset of Rm + (with the subspace topology of Rm). Compatible charts, atlases, equivalent atlases are defined as usual, and a smooth manifold with corners is a topological space together with an equivalence class of atlases of charts with corners. A point p ∈M is a corner point if there is a chart (U, ϕ) with p ∈U such that ϕ(p) is a corner point of Rm +. 270 CHAPTER 7. INTEGRATION ON MANIFOLDS p1 p2 p3 φ φ φ1 2 3 φ1 ( p1 ) φ ( p ) 2 2 φ ( p ) 3 3 M Figure 7.11: The three types of charts on M, a manifold with corners. Note that p2 is a corner point of M. It is not hard to show that the definition of corner point does not depend on the chart (U, ϕ) with p ∈U. See Figure 7.11. Now, in general, the boundary of a smooth manifold with corners is not a smooth manifold with corners. For example, ∂Rm + is not a smooth manifold with corners, but it is the union of smooth manifolds with corners, since ∂Rm + = H1 ∪· · · ∪Hm, and each Hi is a smooth manifold with corners. We can use this fact to define R ∂M ω where ω is an (n−1)-form whose support in contained in the domain of a chart with corners (U, ϕ) by setting Z ∂M ω = m X i=1 Z Hi (ϕ−1)∗ω, where each Hi is given a suitable orientation. Then it is not hard to prove a version of Stokes’ theorem for manifolds with corners. For a detailed exposition, see Lee , Chapter 14. An even more general class of manifolds with singularities (in RN) for which Stokes’ theorem is valid is discussed in Lang (Chapter XVII. §3). 7.8. INTEGRATION ON RIEMANNIAN MANIFOLDS AND LIE GROUPS 271 7.8 Integration on Riemannian Manifolds and Lie Groups We saw in Section 7.2 that every orientable Riemannian manifold has a uniquely defined volume form VolM (see Proposition 7.4). Definition 7.20. Given any smooth real-valued function f with compact support on M, we define the integral of f over M by Z M f = Z M f VolM. Actually it is possible to define the integral R M f using densities even if M is not ori-entable, but we do not need this extra generality. If M is compact, then R M 1M = R M VolM is the volume of M (where 1M is the constant function with value 1). If M and N are Riemannian manifolds, then we have the following version of Proposition 7.8 (3). Proposition 7.19. If M and N are oriented Riemannian manifolds and if ϕ: M →N is an orientation preserving diffeomorphism, then for every function f ∈C∞(N) with compact support, we have Z N f VolN = Z M f ◦ϕ | det(dϕ)| VolM, where f ◦ϕ | det(dϕ)| denotes the function p 7→f(ϕ(p))| det(dϕp)|, with dϕp : TpM →Tϕ(p)N. In particular, if ϕ is an orientation preserving isometry (see Definition 6 in Chapter 3 of O’Neill , or Gallier and Quaintance ), then Z N f VolN = Z M f ◦ϕ VolM. We often denote R M f VolM by R M f(t)dt. If f : M →C is a smooth complex valued-function then we can write f = u + iv for two real-valued functions u: M →R and v: M →R with u(p) = ℜ(f(p)) and v(p) = ℑ(f(p)) for all p ∈M. Then, if f has compact support so do u and v, and we define R M f VolM by Z M f VolM = Z M u VolM + i Z M v VolM. Remark: A volume form on an orientable Riemannian manifold (M, g) yields a linear form vg : C0(M) →R with domain the vector space C0(M) of continuous real-valued functions on M with compact support, given by vg(f) = Z M f VolM. 272 CHAPTER 7. INTEGRATION ON MANIFOLDS This linear form turns out to a Radon measure (see Sakai (Chapter II, Section 5). This Radon measure can be used to define an analog of the Lebesgue integral on the Riemannian manifold M, and to define measurable sets and measurable functions on the manifold M (see Sakai (Chapter II, Section 5). Given a diffeomorphism Φ: M →N between two Riemannnian manifolds (M, g) and (N, h), the Radon measure vh on N can be pulled back to a Radon measure Φ∗vh on M given by (Φ∗vh)(f) = vh(f ◦Φ−1). If e g denotes the canonical Riemannian metric on TpM given by gp, then we can estab-lish a relationship between exp∗ p vg and ve g that involves the Ricci curvature at p, where expp : TpM →M is the exponential map at p. Actually, in general expp is only defined in an open ball Br(0) in TpM centered at the origin, and it is a diffeomorphism on this open ball. The following result can be shown; see Sakai (Chapter 5, Section Lemma 5.4). We use polar coordinates in TpM, which means that every nonzero x ∈TpM is expressed at x = tu with t = ∥x∥and u = x/ ∥x∥. Proposition 7.20. Given a Riemannian manifold (M, g), for every p ∈M, in normal coordinates, near p, we have exp∗ p vg = θ ve g, where θ is the function given by θ(t, u) = tn−1q det(gij(expp tu)). Furthermore, we have θ(t, u) = tn−1 −1 6Ricp(u, u) tn+1 + o(tn+2). Recall from Proposition 7.4 that if (M, g) is an oriented Riemannian manifold, then the volume form VolM satisfies the following property: in every orientation preserving local chart (U, ϕ), we have ((ϕ−1)∗VolM)q = q det(gij(q)) dx1 ∧· · · ∧dxn, q ∈ϕ(U). In particular, if ϕ = exp−1 p is a chart specified by the inverse of the exponential map on some small enough open subset containing p, then near p we obtain (exp∗ p VolM)tu =  1 −1 6Ricp(u, u) t2 + o(t3)  dx1 ∧· · · ∧dxn. Thus we can think of the Ricci curvature as a measure of the deviation of the volume form on M (at tu) from the Euclidean volume form. If G is a Lie group, we know from Section 7.2 that G is always orientable and that G possesses left-invariant volume forms. Since dim(Vn g∗) = 1 if dim(G) = n, and since every left-invariant volume form is determined by its value at the identity, the space of left-invariant volume forms on G has dimension 1. If we pick some left-invariant volume form ω defining the orientation of G, then every other left-invariant volume form is proportional to ω. 7.8. INTEGRATION ON RIEMANNIAN MANIFOLDS AND LIE GROUPS 273 Definition 7.21. Let G be a Lie group and ω be a left invariant volume form. Given any smooth real-valued function f with compact support on G, we define the integral of f over G (w.r.t. ω) by Z G f = Z G fω. This integral depends on ω, but since ω is defined up to some positive constant, so is the integral. When G is compact, we usually pick ω so that Z G ω = 1. If f : G →C is a smooth complex valued-function then we can write f = u + iv for two real-valued functions u: G →R and v: G →R as before and we define Z G f ω = Z G u ω + i Z G v ω. For every g ∈G, as ω is left-invariant, L∗ gω = ω, so L∗ g is an orientation-preserving diffeomorphism, and by Proposition 7.8 (3), Z G fω = Z G L∗ g(fω), so using Proposition 4.12, we get Z G f = Z G fω = Z G L∗ g(fω) = Z G L∗ gf L∗ gω = Z G L∗ gf ω = Z G (f ◦Lg)ω = Z G f ◦Lg. Thus we proved the following proposition. Proposition 7.21. Given any left-invariant volume form ω on a Lie group G, for any smooth function f with compact support, we have Z G f = Z G f ◦Lg, a property called left-invariance. It is then natural to ask when our integral is right-invariant; that is, when Z G f = Z G f ◦Rg. Observe that R∗ gω is left-invariant, since L∗ hR∗ gω = R∗ gL∗ hω = R∗ gω. 274 CHAPTER 7. INTEGRATION ON MANIFOLDS It follows that R∗ gω is some constant multiple of ω, and so there is a function ∆: G →R such that R∗ gω = ∆(g)ω. One can check that ∆is smooth, and we let ∆(g) = |∆(g)|. Since ∆(gh)ω = R∗ ghω = (Rh ◦Rg)∗ω = R∗ g ◦R∗ hω = ∆(g)∆(h)ω, we deduce that ∆(gh) = ∆(g)∆(h), so ∆is a homomorphism of G into R+. Definition 7.22. The function ∆defined above is called the modular function of G. Proposition 7.22. Given any left-invariant volume form ω on a Lie group G, for any smooth function f with compact support, we have Z G fω = ∆(g) Z G (f ◦Rg)ω. Proof. By Proposition 7.8 (3), for a fixed g ∈G, as R∗ g is an orientation-preserving diffeo-morphism, Z G fω = Z G R∗ g(fω) = Z G R∗ gf R∗ gω = Z G (f ◦Rg)∆(g)ω, or equivalently, Z G fω = ∆(g) Z G (f ◦Rg)ω, which is the desired formula. Consequently, our integral is right-invariant iff∆≡1 on G. Thus, our integral is not always right-invariant. When it is, i.e. when ∆≡1 on G, we say that G is unimodular. Proposition 7.23. Any compact Lie group G is unimodular. Proof. In this case, 1 = Z G ω = Z G 1Gω = Z G ∆(g)ω = ∆(g) Z G ω = ∆(g), for all g ∈G. 7.8. INTEGRATION ON RIEMANNIAN MANIFOLDS AND LIE GROUPS 275 Therefore, for a compact Lie group G, our integral is both left and right invariant. We say that our integral is bi-invariant. As a matter of notation, the integral R G f = R G fω is often written R G f(g)dg. Then left-invariance can be expressed as Z G f(g)dg = Z G f(hg)dg, and right-invariance as Z G f(g)dg = Z G f(gh)dg, for all h ∈G. If ω is left-invariant, then it can be shown (see Dieudonn´ e , Chapter XIV, Section 3) that Z G f(g−1)∆(g−1)dg = Z G f(g)dg. Consequently, if G is unimodular, then Z G f(g−1)dg = Z G f(g)dg. Proposition 7.24. If ωl is any left-invariant volume form on G and if ωr is any right-invariant volume form on G, then ωr(g) = c∆(g−1)ωl(g), for some constant c ̸= 0. Proof. Indeed, define the form ω by ω(g) = ∆(g−1)ωl(g). By Proposition 4.12 we have (R∗ hω)g = ∆((gh)−1)(R∗ hωl)g = ∆(h−1)∆(g−1)∆(h)(ωl)g = ∆(g−1)(ωl)g, which shows that ω is right-invariant, and thus ωr(g) = c∆(g−1)ωl(g), as claimed (since ∆(g−1) = ±∆(g−1)). Actually, the following property holds. Proposition 7.25. For any Lie group G, for any g ∈G, we have ∆(g) = | det(Ad(g−1))|. 276 CHAPTER 7. INTEGRATION ON MANIFOLDS Proof. For this recall that Ad(g) = d(Lg ◦Rg−1)1. For any left-invariant n-form ω ∈Vn g∗, we claim that (R∗ gω)h = det(Ad(g−1))ωh, which shows that ∆(g) = | det(Ad(g−1))|. Indeed, for all v1, . . . , vn ∈ThG, we have (R∗ gω)h(v1, . . . , vn) = ωhg(d(Rg)h(v1), . . . , d(Rg)h(vn)) = ωhg(d(Lg ◦Lg−1 ◦Rg ◦Lh ◦Lh−1)h(v1), . . . , d(Lg ◦Lg−1 ◦Rg ◦Lh ◦Lh−1)h(vn)) = ωhg(d(Lh ◦Lg ◦Lg−1 ◦Rg ◦Lh−1)h(v1), . . . , d(Lh ◦Lg ◦Lg−1 ◦Rg ◦Lh−1)h(vn)) = ωhg(d(Lhg ◦Lg−1 ◦Rg ◦Lh−1)h(v1), . . . , d(Lhg ◦Lg−1 ◦Rg ◦Lh−1)h(vn)) = ωhg d(Lhg)1(Ad(g−1)(d(Lh−1)h(v1))), . . . , d(Lhg)1(Ad(g−1)(d(Lh−1)h(vn)))  = (L∗ hgω)1 Ad(g−1)(d(Lh−1)h(v1)), . . . , Ad(g−1)(d(Lh−1)h(vn))  = ω1 Ad(g−1)(d(Lh−1)h(v1)), . . . , Ad(g−1)(d(Lh−1)h(vn))  = det(Ad(g−1)) ω1 d(Lh−1)h(v1), . . . , d(Lh−1)h(vn)  = det(Ad(g−1)) (L∗ h−1ω)h(v1, . . . , vn) = det(Ad(g−1)) ωh(v1, . . . , vn), where we used the left-invariance of ω twice. In general, if G is not unimodular then ωl ̸= ωr. A simple example provided by Vinroot is the group G of direct affine transformations of the real line, which can be viewed as the group of matrices of the form g = x y 0 1  , x, y, ∈R, x > 0. Let A = a b 0 1  ∈G and define T : G →G as T(g) = Ag = a b 0 1  x y 0 1  = ax ay + b 0 1  . Since G is homeomorphic to R+×R, T(g) is also represented by T(x, y) = (ax, ay+b). Then the Jacobian matrix of T is given by J(T)(x, y) = a 0 0 a  , which implies that det(J(T)(x, y)) = a2. Let F : G →R+ be a smooth function on G with compact support. Furthermore assume that Θ(x, y) = F(x, y)x−2 is also smooth on G with 7.8. INTEGRATION ON RIEMANNIAN MANIFOLDS AND LIE GROUPS 277 compact support. Since Θ ◦T(x, y) = Θ(ax, ay + b) = F(ax, ay + b)(ax)−2, Proposition 7.19 implies that Z G F(x, y)x−2dx dy = Z G Θ(x, y) ◦T | det(J(T)(x, y))|dx dy = Z G F(ax, ay + b)(ax)−2a2 dx dy = Z G F ◦Tx−2dx dy. In summary we have shown for g = x y 0 1  , we have Z G F(Ag)x−2 dx dy = Z G F(g)x−2dx dy which implies that the left-invariant volume form on G is ωl = dx dy x2 . To define a right-invariant volume form on G, define S : G →G as S(g) = gA = x y 0 1  a b 0 1  = ax bx + y 0 1  , which is represented by S(x, y) = (ax, bx + y). Then the Jacobian matrix of S is J(S)(x, y) = a 0 b 1  , and det(J(S)(x, y)) = a. Using F(x, y) as above and Θ(x, y) = F(x, y)x−1, we find that Z G F(x, y)x−1dx dy = Z G Θ(x, y) ◦S | det(J(S)(x, y))|dx dy = Z G F(ax, bx + y)(ax)−1a dx dy = Z G F ◦Sx−1dx dy, which implies that ωr = dx dy x . Note that ∆(g) = |x−1|. Observe that ∆(A) = |a−1|, F ◦RA = F(ax, bx + y), and that 1 |a| Z G F ◦RA wl = 1 |a| Z G F(ax, bx + y) dx dy x2 = 1 |a| Z G F(u, v) du a (dv −b a du) u a 2 , u = ax, v = bx + y = Z G F(u, v) dudv u2 = Z G F(u, v) wl, 278 CHAPTER 7. INTEGRATION ON MANIFOLDS which is a special case of Proposition 7.22. Remark: By the Riesz’ representation theorem, ω defines a positive measure µω which satisfies Z G fdµω = Z G fω. Using what we have shown, this measure is left-invariant. Such measures are called left Haar measures, and similarly we have right Haar measures. It can be shown that every two left Haar measures on a Lie group are proportional (see Knapp, , Chapter VIII). Given a left Haar measure µ, the function ∆such that µ(Rgh) = ∆(g)µ(h) for all g, h ∈G is the modular function of G. However, beware that some authors, including Knapp, use ∆(g−1) instead of ∆(g). As above, we have ∆(g) = | det(Ad(g−1))|. Beware that authors who use ∆(g−1) instead of ∆(g) give a formula where Ad(g) appears instead of Ad(g−1). Again, G is unimodular iff∆≡1. It can be shown that compact, semisimple, reductive, and nilpotent Lie groups are uni-modular (for instance, see Knapp, , Chapter VIII). On such groups, left Haar measures are also right Haar measures (and vice versa). In this case, we can speak of Haar measures on G. For more details on Haar measures on locally compact groups and Lie groups, we refer the reader to Folland (Chapter 2), Helgason (Chapter 1), and Dieudonn´ e (Chapter XIV). 7.9 Problems Problem 7.1. Prove the following: if f : Rn+1 →R is a smooth submersion, then M = f −1(0) is a smooth orientable manifold. Problem 7.2. Let r: Rn+1 −{0} →Sn be the map given by r(x) = x ∥x∥, and set ω = r∗VolSn. a closed n-form on Rn+1 −{0}. Clearly, ω ↾Sn = VolSn. Show that ω is given by ωx = 1 ∥x∥n n+1 X i=1 (−1)i−1xi dx1 ∧· · · ∧c dxi ∧· · · ∧dxn+1. 7.9. PROBLEMS 279 Problem 7.3. Recall that π: Sn →RPn is the map such that π−1([p]) consists of two antipodal points for every [p] ∈RPn. Show there is a volume form on RPn iffn is odd, given by π∗(VolRPn) = VolSn. Problem 7.4. Complete the proof of Proposition 7.7 by using a partition of unity argument to show the uniqueness of the linear operator Z M : An c (M) − →R which satisfies the following property: For any ω ∈An c (M), if supp(ω) ⊆U, where (U, ϕ) is a positively oriented chart, then Z M ω = Z ϕ(U) (ϕ−1)∗ω. Problem 7.5. Complete the proof sketch details of Proposition 7.8. Problem 7.6. Recall that a density may be defined as a function µ: Vn V →R such that for every automorphism f ∈GL(V ), µ(f(v1) ∧· · · ∧f(vn)) = | det(f)|µ(v1 ∧· · · ∧vn) (††) for all v1 ∧· · · ∧vn ∈V (with µ(0) = 0). Show that Condition (††) is equivalent to µ(cw) = |c|µ(w), w ∈ n ^ V, c ∈R. Problem 7.7. Prove Proposition 7.14. Problem 7.8. Prove Proposition 7.15. Problem 7.9. Prove Proposition 7.19. Problem 7.10. Prove Proposition 7.20. Hint. See Sakai , Chapter 5, Section Lemma 5.4. Problem 7.11. Let G be a Lie group with a left invariant volume form ω. Show that Z G f(g−1)∆(g−1)dg = Z G f(g)dg. Hint. See Dieudonn´ e , Chapter XIV, Section 3. 280 CHAPTER 7. INTEGRATION ON MANIFOLDS Chapter 8 Spherical Harmonics and Linear Representations of Lie Groups This chapter and the next focus on topics that are somewhat different from the more geomet-ric and algebraic topics discussed in previous chapters. Indeed, the focus of this chapter is on the types of functions that can be defined on a manifold, the sphere Sn in particular, and this involves some analysis. A main theme of this chapter is to generalize Fourier analysis on the circle to higher dimensional spheres. One of our goals is to understand the structure of the space L2(Sn) of real-valued square integrable functions on the sphere Sn, and its complex analog L2 C(Sn). Both are Hilbert spaces if we equip them with the inner product ⟨f, g⟩Sn = Z Sn f(t)g(t) dt = Z Sn fg VolSn, and in the complex case with the Hermitian inner product ⟨f, g⟩Sn = Z Sn f(t)g(t) dt = Z Sn fg VolSn. This means that if we define the L2-norm associated with the above inner product as ∥f∥= p ⟨f, f⟩, then L2(Sn) and L2 C(Sn) are complete normed vector spaces (see Section 8.1 for a review of Hilbert spaces). It turns out that each of L2(Sn) and L2 C(Sn) contains a countable family of very nice finite dimensional subspaces Hk(Sn) (and HC k (Sn)), where Hk(Sn) is the space of (real) spherical harmonics on Sn, that is, the restrictions of the harmonic homogeneous polynomials of degree k (in n + 1 real variables) to Sn (and similarly for HC k (Sn)); these polynomials satisfy the Laplace equation ∆P = 0, where the operator ∆is the (Euclidean) Laplacian, ∆= ∂2 ∂x2 1 + · · · + ∂2 ∂x2 n+1 . 281 282 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Remarkably, each space Hk(Sn) (resp. HC k (Sn)) is the eigenspace of the Laplace-Beltrami operator ∆Sn on Sn, a generalization to Riemannian manifolds of the standard Laplacian (in fact, Hk(Sn) is the eigenspace for the eigenvalue −k(n + k −1)). As a consequence, the spaces Hk(Sn) (resp. HC k (Sn)) are pairwise orthogonal. Furthermore (and this is where analysis comes in), the set of all finite linear combinations of elements in S∞ k=0 Hk(Sn) (resp. S∞ k=0 HC k (Sn)) is is dense in L2(Sn) (resp. dense in L2 C(Sn)). These two facts imply the following fundamental result about the structure of the spaces L2(Sn) and L2 C(Sn). The family of spaces Hk(Sn) (resp. HC k (Sn)) yields a Hilbert space direct sum decompo-sition L2(Sn) = ∞ M k=0 Hk(Sn) (resp. L2 C(Sn) = ∞ M k=0 HC k (Sn)), which means that the summands are closed, pairwise orthogonal, and that every f ∈L2(Sn) (resp. f ∈L2 C(Sn)) is the sum of a converging series f = ∞ X k=0 fk in the L2-norm, where the fk ∈Hk(Sn) (resp. fk ∈HC k (Sn)) are uniquely determined functions. Furthermore, given any orthonormal basis (Y 1 k , . . . , Y ak,n+1 k ) of Hk(Sn), we have fk = ak,n+1 X mk=1 ck,mkY mk k , with ck,mk = ⟨f, Y mk k ⟩Sn. The coefficients ck,mk are “generalized” Fourier coefficients with respect to the Hilbert basis {Y mk k | 1 ≤mk ≤ak,n+1, k ≥0}; see Theorems 8.18 and 8.19. In Section 8.2 we begin by reviewing the simple case n = 1, where S1 is a circle, which corresponds to standard Fourier analysis. In this case, there is a simple expression in polar coordinates for the Laplacian ∆S1 on the circle, and we are led to the equation ∆S1g = −k2g. We find that H0(S1) = R, and Hk(S1) is the two-dimensional space spanned by cos kθ and sin kθ for k ≥1. We also determine explicitly the harmonic polynomials in two variables. In Section 8.3 we consider the sphere S2. This time we need to find the Laplacian ∆S2 on the sphere. This is an old story, and we use the formula in terms of spherical coordinates. Then we need to solve the equation ∆S2g = −k(k −1)g. This is a classical problem that was solved in the early 1780s by the separation of variables method. After some labor, we are led to the general Legendre equation. The solutions 283 are the associated Legendre functions P m k (t), which are defined in terms of the Legendre polynomials. The upshot is that the functions cos mϕ P m k (cos θ), sin mϕ P m k (cos θ) are eigenfunctions of the Laplacian ∆S2 on the sphere for the eigenvalue −k(k + 1). For k fixed, as 0 ≤m ≤k, we get 2k + 1 linearly independent functions, so Hk(S2) has dimension 2k + 1. These functions are the spherical harmonics, but they are usually expressed in a different notation (ym l (θ, ϕ) with −l ≤m ≤l). Expressed in Cartesian coordinates, these are the homogeneous harmonic polynomials. In order to generalize the above cases to n ≥3, we need to define the Laplace-Beltrami operator on a manifold, which is done in Section 8.4. We also find a formula relating the Laplacian on Rn+1 to the Laplacian ∆Sn on Sn. The Hilbert sum decomposition of L2(Sn) is accomplished in Section 8.5. In Section 8.6 we describe the zonal spherical functions Zτ k on Sn and show that they essentially come from certain polynomials generalizing the Legendre polynomials known as the Gegenbauer polynomials. For any fixed point τ on Sn and any constant c ∈C, the zonal spherical function Zτ k is the unique homogeneous harmonic polynomial of degree k such that Zτ k(τ) = c, and Zτ k is invariant under any rotation fixing τ. An interesting property of the zonal spherical functions is a formula for obtaining the kth spherical harmonic component of a function f ∈L2 C(Sn); see Proposition 8.27. Another important property of the zonal spherical functions Zτ k is that they generate HC k (Sn). A closer look at the Gegenbauer polynomials is taken in Section 8.7. In Section 8.8 we prove the Funk-Hecke formula. This formula basically allows one to perform a sort of convolution of a “kernel function” with a spherical function in a convenient way. The Funk-Hecke formula was used in a ground-breaking paper by Basri and Jacobs to compute the reflectance function r from the lighting function ℓas a pseudo-convolution K ⋆ℓ(over S2) with the Lambertian kernel K. The final Sections 8.9 and 8.11 are devoted to more advanced material which is presented without proofs. The purpose of Section 8.9 is to generalize the results about the structure of the space of functions L2 C(Sn) defined on the sphere Sn, especially the results of Sections 8.5 and 8.6 (such as Theorem 8.19, except part (3)), to homogeneous spaces G/K where G is a compact Lie group and K is a closed subgroup of G. The first step is to consider the Hilbert space L2 C(G) where G is a compact Lie group and to find a Hilbert sum decomposition of this space. The key to this generalization is the notion of (unitary) linear representation of the group G. The result that we are alluding to is a famous theorem known as the Peter–Weyl theorem about unitary representations of compact Lie groups (Herman Klauss Hugo Weyl, 1885-1955). 284 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS The Peter–Weyl theorem can be generalized to any representation V : G →Aut(E) of G into a separable Hilbert space E, and we obtain a Hilbert sum decomposition of E in terms of subspaces Eρ of E. The next step is to consider the subspace L2 C(G/K) of L2 C(G) consisting of the functions that are right-invariant under the action of K. These can be viewed as functions on the homogeneous space G/K. Again we obtain a Hilbert sum decomposition. It is also interest-ing to consider the subspace L2 C(K\G/K) of functions in L2 C(G) consisting of the functions that are both left and right-invariant under the action of K. The functions in L2 C(K\G/K) can be viewed as functions on the homogeneous space G/K that are invariant under the left action of K. Convolution makes the space L2 C(G) into a non-commutative algebra. Remarkably, it is possible to characterize when L2 C(K\G/K) is commutative (under convolution) in terms of a simple criterion about the irreducible representations of G. In this situation, (G, K) is a called a Gelfand pair. When (G, K) is a Gelfand pair, it is possible to define a well-behaved notion of Fourier transform on L2 C(K\G/K). Gelfand pairs and the Fourier transform are briefly considered in Section 8.11. 8.1 Hilbert Spaces and Hilbert Sums The material in this chapter assumes that the reader has some familiarity with the concepts of a Hilbert space and a Hilbert basis. We present this section to review these important concepts. Many of the proofs are omitted and are found in traditional sources such as Bourbaki , Dixmier , Lang [73, 74], and Rudin . The special case of separable Hilbert spaces is treated very nicely in Deitmar . We begin our review by recalling the definition of a Hermitian space. To do this we need to define the notion of a Hermitian form. Definition 8.1. Given two vector spaces E and F over C, a function f : E →F is semilinear if f(u + v) = f(u) + f(v) f(λu) = λu, for all u, v ∈E and λ ∈C. Definition 8.2. Given a complex vector space E, a function ϕ: E ×E →C is a sesquilinear form if it is linear in its first argument and semilinear in its second argument, which means 8.1. HILBERT SPACES AND HILBERT SUMS 285 that ϕ(u1 + u2, v) = ϕ(u1, v) + ϕ(u2, v) ϕ(u, v1 + v2) = ϕ(u, v1) + ϕ(u, v2) ϕ(λu, v) = λϕ(u, v) ϕ(u, λv) = λϕ(u, v), for all u, v, u1, u2, v1, v2 ∈E and λ ∈C. A function ϕ: E × E →C is a Hermitian form if it is sesquilinear and if ϕ(u, v) = ϕ(v, u), for all u, v ∈E. Definition 8.3. Given a complex vector space E, a Hermitian form ϕ: E×E →C is positive definite if ϕ(u, u) > 0 for all u ̸= 0. A pair ⟨E, ϕ⟩where E is a complex vector space and ϕ is a Hermitian form on E is called a Hermitian (or unitary) space if ϕ is positive definite. The standard example of a Hermitian form on Cn is the map ϕ defined such that ϕ((x1, . . . , xn), (y1, . . . , yn)) = x1y1 + x2y2 + · · · + xnyn. This map is also positive definite and makes Cn into a Hermitian space. Given a Hermitian space ⟨E, ϕ⟩, we can readily show that the function ∥∥: E →R defined such that ⟨u, u⟩= ∥u∥= ϕ(u, u), is a norm on E. Thus, E is a normed vector space. If E is also complete, then it is a very interesting space. Recall that completeness has to do with the convergence of Cauchy sequences. A normed vector space ⟨E, ∥∥⟩is automatically a metric space under the metric d defined such that d(u, v) = ∥v −u∥. This leads us to the following definition. Definition 8.4. Given a metric space E with metric d, a sequence (an)n≥1 of elements an ∈E is a Cauchy sequence ifffor every ϵ > 0, there is some N ≥1 such that d(am, an) < ϵ for all m, n ≥N. We say that E is complete iffevery Cauchy sequence converges to a limit (which is unique, since a metric space is Hausdorff). Every finite dimensional vector space over R or C is complete. One can show by induction that given any basis (e1, . . . , en) of E, the linear map h: Cn →E defined such that h((z1, . . . , zn)) = z1e1 + · · · + znen 286 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS is a homeomorphism (using the sup-norm on Cn). One can also use the fact that any two norms on a finite dimensional vector space over R or C are equivalent (see Lang , Dixmier , or Schwartz ). However, if E has infinite dimension, it may not be complete. When a Hermitian space is complete, a number of the properties that hold for finite dimensional Hermitian spaces also hold for infinite dimensional spaces. For example, any closed subspace has an orthogonal complement, and in particular, a finite dimensional subspace has an orthogonal complement. Hermitian spaces that are also complete play an important role in analysis. Since they were first studied by Hilbert, they are called Hilbert spaces. Definition 8.5. A (complex) Hermitian space ⟨E, ϕ⟩which is a complete normed vector space under the norm ∥∥induced by ϕ is called a Hilbert space. A real Euclidean space ⟨E, ϕ⟩which is complete under the norm ∥∥induced by ϕ is called a real Hilbert space. All the results in this section hold for complex Hilbert spaces as well as for real Hilbert spaces. We state all results for the complex case only, since they also apply to the real case, and since the proofs in the complex case need a little more care. Example 8.1. The space l2 of all countably infinite sequences x = (xi)i∈N of complex numbers such that P∞ i=0 |xi|2 < ∞is a Hilbert space. It will be shown later that the map ϕ: l2 × l2 →C defined such that ϕ ((xi)i∈N, (yi)i∈N) = ∞ X i=0 xiyi is well defined, and that l2 is a Hilbert space under ϕ. In fact, we will prove a more general result (Proposition 8.3). Example 8.2. The set C∞[a, b] of smooth functions f : [a, b] →C is a Hermitian space under the Hermitian form ⟨f, g⟩= Z b a f(x)g(x)dx, but it is not a Hilbert space because it is not complete (see Section 7.8 for the definition of the integral of a complex-valued function). It is possible to construct its completion L2([a, b]), which turns out to be the space of Lebesgue square-integrable functions on [a, b]. One of the most important facts about finite-dimensional Hermitian (and Euclidean) spaces is that they have orthonormal bases. This implies that, up to isomorphism, every finite-dimensional Hermitian space is isomorphic to Cn (for some n ∈N) and that the inner product is given by ⟨(x1, . . . , xn), (y1, . . . , yn)⟩= n X i=1 xiyi. 8.1. HILBERT SPACES AND HILBERT SUMS 287 Furthermore, every subspace W has an orthogonal complement W ⊥, and the inner product induces a natural duality between E and E∗, where E∗is the space of linear forms on E. When E is a Hilbert space, E may be infinite dimensional, often of uncountable dimen-sion. Thus, we can’t expect that E always have an orthonormal basis. However, if we modify the notion of basis so that a “Hilbert basis” is an orthogonal family that is also dense in E, i.e., every v ∈E is the limit of a sequence of finite combinations of vectors from the Hilbert basis, then we can recover most of the “nice” properties of finite-dimensional Hermitian spaces. For instance, if (uk)k∈K is a Hilbert basis, for every v ∈E, we can define the Fourier coefficients ck = ⟨v, uk⟩/∥uk∥, and then, v is the “sum” of its Fourier series P k∈K ckuk. However, the cardinality of the index set K can be very large, and it is necessary to define what it means for a family of vectors indexed by K to be summable. It turns out that every Hilbert space is isomorphic to a space of the form l2(K), where l2(K) is a generalization of the space of Example 8.1 (see Theorem 8.7, usually called the Riesz-Fischer theorem). Definition 8.6. Given a Hilbert space E, a family (uk)k∈K of nonnull vectors is an orthogonal family iffthe uk are pairwise orthogonal, i.e., ⟨ui, uj⟩= 0 for all i ̸= j (i, j ∈K), and an orthonormal family iff⟨ui, uj⟩= δi, j, for all i, j ∈K. A total orthogonal family (or system) or Hilbert basis is an orthogonal family that is dense in E. This means that for every v ∈E, for every ϵ > 0, there is some finite subset I ⊆K and some family (λi)i∈I of complex numbers, such that v − X i∈I λiui < ϵ. Given an orthogonal family (uk)k∈K, for every v ∈E, for every k ∈K, the scalar ck = ⟨v, uk⟩/∥uk∥2 is called the k-th Fourier coefficient of v over (uk)k∈K. Remark: The terminology Hilbert basis is misleading, because a Hilbert basis (uk)k∈K is not necessarily a basis in the algebraic sense. Indeed, in general, (uk)k∈K does not span E. Intuitively, it takes linear combinations of the uk’s with infinitely many nonnull coefficients to span E. Technically, this is achieved in terms of limits. In order to avoid the confusion between bases in the algebraic sense and Hilbert bases, some authors refer to algebraic bases as Hamel bases and to total orthogonal families (or Hilbert bases) as Schauder bases. Definition 8.7. Given an orthogonal family (uk)k∈K of a Hilbert space E, for any finite subset I of K, and for any family (λi)i∈I of complex numbers, we call sums of the form P i∈I λiui partial sums of Fourier series, and if these partial sums converge to a limit denoted as P k∈K ckuk, we call P k∈K ckuk a Fourier series. However, we have to make sense of such sums! Indeed, when K is unordered or uncount-able, the notion of limit or sum has not been defined. This can be done as follows (for more details, see Dixmier ). 288 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Definition 8.8. Given a normed vector space E (say, a Hilbert space), for any nonempty index set K, we say that a family (uk)k∈K of vectors in E is summable with sum v ∈E iff for every ϵ > 0, there is some finite subset I of K, such that, v − X j∈J uj < ϵ for every finite subset J with I ⊆J ⊆K. We say that the family (uk)k∈K is summable iffthere is some v ∈E such that (uk)k∈K is summable with sum v. A family (uk)k∈K is a Cauchy family ifffor every ϵ > 0, there is a finite subset I of K, such that, X j∈J uj < ϵ for every finite subset J of K with I ∩J = ∅. If (uk)k∈K is summable with sum v, we usually denote v as P k∈K uk. Remark: The notion of summability implies that the sum of a family (uk)k∈K is independent of any order on K. In this sense, it is a kind of “commutative summability”. More precisely, it is easy to show that for every bijection ϕ: K →K (intuitively, a reordering of K), the family (uk)k∈K is summable iffthe family (ul)l∈ϕ(K) is summable, and if so, they have the same sum. To state some important properties of Fourier coefficients the following technical propo-sition, whose proof is found in Bourbaki , will be needed. Proposition 8.1. Let E be a complete normed vector space (say, a Hilbert space). (1) For any nonempty index set K, a family (uk)k∈K is summable iffit is a Cauchy family. (2) Given a family (rk)k∈K of nonnegative reals rk ≥0, if there is some real number B > 0 such that P i∈I ri < B for every finite subset I of K, then (rk)k∈K is summable and P k∈K rk = r, where r is least upper bound of the set of finite sums P i∈I ri (I ⊆K). The following proposition gives some of the main properties of Fourier coefficients. Among other things, at most countably many of the Fourier coefficient may be nonnull, and the partial sums of a Fourier series converge. Given an orthogonal family (uk)k∈K, we let Uk = Cuk. Proposition 8.2. Let E be a Hilbert space, (uk)k∈K an orthogonal family in E, and V the closure of the subspace generated by (uk)k∈K. The following properties hold: (1) For every v ∈E, for every finite subset I ⊆K, we have X i∈I |ci|2 ≤∥v∥2, where the ck = ⟨v, uk⟩/∥uk∥2 are the Fourier coefficients of v. 8.1. HILBERT SPACES AND HILBERT SUMS 289 (2) For every vector v ∈E, if (ck)k∈K are the Fourier coefficients of v, the following conditions are equivalent: (2a) v ∈V (2b) The family (ckuk)k∈K is summable and v = P k∈K ckuk. (2c) The family (|ck|2)k∈K is summable and ∥v∥2 = P k∈K |ck|2; (3) The family (|ck|2)k∈K is summable, and we have the Bessel inequality: X k∈K |ck|2 ≤∥v∥2. As a consequence, at most countably many of the ck may be nonzero. The family (ckuk)k∈K forms a Cauchy family, and thus, the Fourier series P k∈K ckuk converges in E to some vector u = P k∈K ckuk. See Figure 8.1. Proof. (1) Let uI = X i∈I ciui for any finite subset I of K. We claim that v −uI is orthogonal to ui for every i ∈I. Indeed, ⟨v −uI, ui⟩= v − X j∈I cjuj, ui + = ⟨v, ui⟩− X j∈I cj ⟨uj, ui⟩ = ⟨v, ui⟩−ci∥ui∥2 = ⟨v, ui⟩−⟨v, ui⟩= 0, since ⟨uj, ui⟩= 0 for all i ̸= j and ci = ⟨v, ui⟩/∥ui∥2. As a consequence, we have ∥v∥2 = v − X i∈I ciui + X i∈I ciui 2 = v − X i∈I ciui 2 + X i∈I ciui 2 = v − X i∈I ciui 2 + X i∈I |ci|2, since the ui are pairwise orthogonal; that is, ∥v∥2 = v − X i∈I ciui 2 + X i∈I |ci|2, 290 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS E V = span(u ) k v form c = k v, uk uk 2 u = ck uk k K Σ e E V = span(u ) k v form c = k v, uk uk 2 ck uk k K Σ e = (i.) (ii.) Figure 8.1: A schematic illustration of Proposition 8.2. Figure (i.) illustrates Condition (2b), while Figure (ii.) illustrates Condition (3). Note E is the purple oval and V is the magenta oval. In both cases, take a vector of E, form the Fourier coefficients ck, then form the Fourier series P k∈K ckuk. Condition (2b) ensures v equals its Fourier series since v ∈V . However, if v / ∈V , the Fourier series does not equal v. Eventually, we will discover that V = E, which implies that that Fourier series converges to its vector v. which in turn implies X i∈I |ci|2 ≤∥v∥2, as claimed. (2) We prove the chain of implications (a) ⇒(b) ⇒(c) ⇒(a). (a) ⇒(b) If v ∈V , since V is the closure of the subspace spanned by (uk)k∈K, for every ϵ > 0, there is some finite subset I of K and some family (λi)i∈I of complex numbers, such that v − X i∈I λiui < ϵ. 8.1. HILBERT SPACES AND HILBERT SUMS 291 Now for every finite subset J of K such that I ⊆J, we have v − X i∈I λiui 2 = v − X j∈J cjuj + X j∈J cjuj − X i∈I λiui 2 = v − X j∈J cjuj 2 + X j∈J cjuj − X i∈I λiui 2 , since I ⊆J and the uj (with j ∈J) are orthogonal to v −P j∈J cjuj by the argument in (1), which shows that v − X j∈J cjuj ≤ v − X i∈I λiui < ϵ, and thus, that the family (ckuk)k∈K is summable with sum v, so that v = X k∈K ckuk. (b) ⇒(c) If v = P k∈K ckuk, then for every ϵ > 0, there some finite subset I of K, such that v − X j∈J cjuj < √ϵ, for every finite subset J of K such that I ⊆J, and since we proved in (1) that ∥v∥2 = v − X j∈J cjuj 2 + X j∈J |cj|2, we get ∥v∥2 − X j∈J |cj|2 < ϵ, which proves that (|ck|2)k∈K is summable with sum ∥v∥2. (c) ⇒(a) Finally, if (|ck|2)k∈K is summable with sum ∥v∥2, for every ϵ > 0, there is some finite subset I of K such that ∥v∥2 − X j∈J |cj|2 < ϵ2 for every finite subset J of K such that I ⊆J, and again, using the fact that ∥v∥2 = v − X j∈J cjuj 2 + X j∈J |cj|2, we get v − X j∈J cjuj < ϵ, 292 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS which proves that (ckuk)k∈K is summable with sum P k∈K ckuk = v, and v ∈V . (3) Since Part (1) implies P i∈I |ci|2 ≤∥v∥2 for every finite subset I of K, by Proposition 8.1 (2), the family (|ck|2)k∈K is summable. The Bessel inequality X k∈K |ck|2 ≤∥v∥2 is an obvious consequence of the inequality P i∈I |ci|2 ≤∥v∥2 (for every finite I ⊆K). Now, for every natural number n ≥1, if Kn is the subset of K consisting of all ck such that |ck| ≥1/n, (i.e. n|ck| ≥1 whenever ck ∈Kn), the number of elements in each Kn is finite since X k∈Kn |nck|2 ≤n2 X k∈K |ck|2 ≤n2∥v∥2 converges. Hence, at most a countable number of the ck may be nonzero. Since (|ck|2)k∈K is summable with sum c, Proposition 8.1 (1) shows that for every ϵ > 0, there is some finite subset I of K such that X j∈J |cj|2 < ϵ2 for every finite subset J of K such that I ∩J = ∅. Since X j∈J cjuj 2 = X j∈J |cj|2, we get X j∈J cjuj < ϵ. This proves that (ckuk)k∈K is a Cauchy family, which, by Proposition 8.1 (1), implies that (ckuk)k∈K is summable, since E is complete. Thus, the Fourier series P k∈K ckuk is summable, with its sum denoted u ∈V . Proposition 8.2 suggests looking at the space of sequences (zk)k∈K (where zk ∈C) such that (|zk|2)k∈K is summable. Indeed, such spaces are Hilbert spaces, and it turns out that every Hilbert space is isomorphic to one of those. Such spaces are the infinite-dimensional version of the spaces Cn under the usual Euclidean norm. Definition 8.9. Given any nonempty index set K, the space l2(K) is the set of all sequences (zk)k∈K, where zk ∈C, such that (|zk|2)k∈K is summable, i.e., P k∈K |zk|2 < ∞. Remarks: (1) When K is a finite set of cardinality n, l2(K) is isomorphic to Cn. 8.1. HILBERT SPACES AND HILBERT SUMS 293 (2) When K = N, the space l2(N) corresponds to the space l2 of Example 8.1. In that example we claimed that l2 was a Hermitian space, and in fact, a Hilbert space. We now state this fact for any index set K. For a proof of Proposition 8.3 we refer the reader to Schwartz ). Proposition 8.3. Given any nonempty index set K, the space l2(K) is a Hilbert space under the Hermitian product ⟨(xk)k∈K, (yk)k∈K⟩= X k∈K xkyk. The subspace consisting of sequences (zk)k∈K such that zk = 0, except perhaps for finitely many k, is a dense subspace of l2(K). We just need two more propositions before being able to prove that every Hilbert space is isomorphic to some l2(K). Proposition 8.4. Let E be a Hilbert space, and (uk)k∈K an orthogonal family in E. The following properties hold: (1) For every family (λk)k∈K ∈l2(K), the family (λkuk)k∈K is summable. Furthermore, v = P k∈K λkuk is the only vector such that ck = λk for all k ∈K, where the ck are the Fourier coefficients of v. (2) For any two families (λk)k∈K ∈l2(K) and (µk)k∈K ∈l2(K), if v = P k∈K λkuk and w = P k∈K µkuk, we have the following equation, also called Parseval identity: ⟨v, w⟩= X k∈K λkµk. Proof. (1) The fact that (λk)k∈K ∈l2(K) means that (|λk|2)k∈K is summable. The proof given in Proposition 8.2 (3) applies to the family (|λk|2)k∈K (instead of (|ck|2)k∈K), and yields the fact that (λkuk)k∈K is summable. Letting v = P k∈K λkuk, recall that ck = ⟨v, uk⟩/∥uk∥2. Pick some k ∈K. Since ⟨−, −⟩is continuous, for every ϵ > 0, there is some η > 0 such that | ⟨v, uk⟩−⟨w, uk⟩| = | ⟨v −w, uk⟩| < ϵ∥uk∥2 whenever ∥v −w∥< η. However, since for every η > 0, there is some finite subset I of K such that v − X j∈J λjuj < η 294 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS for every finite subset J of K such that I ⊆J, we can pick J = I ∪{k}, and letting w = P j∈J λjuj, we get ⟨v, uk⟩− X j∈J λjuj, uk + < ϵ∥uk∥2. However, ⟨v, uk⟩= ck∥uk∥2 and X j∈J λjuj, uk + = λk∥uk∥2, and thus, the above proves that |ck −λk| < ϵ for every ϵ > 0, and thus, that ck = λk. (2) Since ⟨−, −⟩is continuous, for every ϵ > 0, there are some η1 > 0 and η2 > 0, such that | ⟨x, y⟩| < ϵ whenever ∥x∥< η1 and ∥y∥< η2. Since v = P k∈K λkuk and w = P k∈K µkuk, there is some finite subset I1 of K such that v − X i∈I λiui < η1 for every finite subset I of K such that I1 ⊆I, and there is some finite subset I2 of K such that w − X i∈I µiui < η2 for every finite subset I of K such that I2 ⊆I. Letting I = I1 ∪I2, we get v − X i∈I λiui, w − X i∈I µiui + < ϵ. Furthermore, ⟨v, w⟩= v − X i∈I λiui + X i∈I λiui, w − X i∈I µiui + X i∈I µiui + = v − X i∈I λiui, w − X i∈I µiui + + X i∈I λiµi, since the ui are orthogonal to v −P i∈I λiui and w −P i∈I µiui for all i ∈I. This proves that for every ϵ > 0, there is some finite subset I of K such that ⟨v, w⟩− X i∈I λiµi < ϵ. 8.1. HILBERT SPACES AND HILBERT SUMS 295 We already know from Proposition 8.3 that (λkµk)k∈K is summable, and since ϵ > 0 is arbitrary, we get ⟨v, w⟩= X k∈K λkµk. The next proposition states properties characterizing Hilbert bases (total orthogonal families). Proposition 8.5. Let E be a Hilbert space, and let (uk)k∈K be an orthogonal family in E. The following properties are equivalent: (1) The family (uk)k∈K is a total orthogonal family. (2) For every vector v ∈E, if (ck)k∈K are the Fourier coefficients of v, then the family (ckuk)k∈K is summable and v = P k∈K ckuk. (3) For every vector v ∈E, we have the Parseval identity: ∥v∥2 = X k∈K |ck|2. (4) For every vector u ∈E, if ⟨u, uk⟩= 0 for all k ∈K, then u = 0. See Figure 8.2. E V = span(u ) k v form c = k v, uk uk 2 ck uk k K Σ e = = Figure 8.2: A schematic illustration of Proposition 8.5. Since (uk)k∈K is a Hilbert basis, V = E. Then given a vector of E, if we form the Fourier coefficients ck, then form the Fourier series P k∈K ckuk, we are ensured that v is equal to its Fourier series. 296 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Proof. The equivalence of (1), (2), and (3), is an immediate consequence of Proposition 8.2 and Proposition 8.4. It remains to show that (1) and (4) are equivalent. (1) ⇒(4) If (uk)k∈K is a total orthogonal family and ⟨u, uk⟩= 0 for all k ∈K, since u = P k∈K ckuk where ck = ⟨u, uk⟩/∥uk∥2, we have ck = 0 for all k ∈K, and u = 0. (4) ⇒(1) Conversely, assume that the closure V , where V is the subspace generated by (uk)k∈K, is different from E. Then we have E = V ⊕V ⊥, where V ⊥is the orthogonal complement of V , and V ⊥is nontrivial since V ̸= E. As a consequence, there is some nonnull vector u ∈V ⊥. But then, u is orthogonal to every vector in V , and in particular, ⟨u, uk⟩= 0 for all k ∈K, which, by assumption, implies that u = 0, contradicting the fact that u ̸= 0. At last, we can prove that every Hilbert space is isomorphic to some Hilbert space l2(K) for some suitable K. First, we need the fact that every Hilbert space has some Hilbert basis. This proof uses Zorn’s Lemma (see Rudin ). Proposition 8.6. Let E be a Hilbert space. Given any orthogonal family (uk)k∈K in E, there is a total orthogonal family (ul)l∈L containing (uk)k∈K. All Hilbert bases for a Hilbert space E have index sets K of the same cardinality. For a proof, see Bourbaki . Definition 8.10. A Hilbert space E is separable if its Hilbert bases are countable. Theorem 8.7. (Riesz-Fischer) For every Hilbert space E, there is some nonempty set K such that E is isomorphic to the Hilbert space l2(K). More specifically, for any Hilbert basis (uk)k∈K of E, the maps f : l2(K) →E and g: E →l2(K) defined such that f ((λk)k∈K) = X k∈K λkuk and g(u) = ⟨u, uk⟩/∥uk∥2 k∈K = (ck)k∈K, are bijective linear isometries such that g ◦f = id and f ◦g = id. Proof. By Proposition 8.4 (1), the map f is well defined, and it it clearly linear. By Propo-sition 8.2 (3), the map g is well defined, and it is also clearly linear. By Proposition 8.2 (2b), we have f(g(u)) = u = X k∈K ckuk, and by Proposition 8.4 (1), we have g(f ((λk)k∈K)) = (λk)k∈K, and thus g ◦f = id and f ◦g = id. By Proposition 8.4 (2), the linear map g is an isometry. Therefore, f is a linear bijection and an isometry between l2(K) and E, with inverse g. 8.1. HILBERT SPACES AND HILBERT SUMS 297 Remark: The surjectivity of the map g: E →l2(K) is known as the Riesz-Fischer theorem. Having done all this hard work, we sketch how these results apply to Fourier series. Again, we refer the readers to Rudin or Lang [73, 74] for a comprehensive exposition. Let C(T) denote the set of all periodic continuous functions f : [−π, π] →C with period 2π. There is a Hilbert space L2(T) containing C(T) and such that C(T) is dense in L2(T), whose inner product is given by ⟨f, g⟩= Z π −π f(x)g(x)dx. The Hilbert space L2(T) is the space of Lebesgue square-integrable periodic functions (of period 2π). It turns out that the family (eikx)k∈Z is a total orthogonal family in L2(T), because it is already dense in C(T) (for instance, see Rudin ). Then the Riesz-Fischer theorem says that for every family (ck)k∈Z of complex numbers such that X k∈Z |ck|2 < ∞, there is a unique function f ∈L2(T) such that f is equal to its Fourier series f(x) = X k∈Z ckeikx, where the Fourier coefficients ck of f are given by the formula ck = 1 2π Z π −π f(t)e−iktdt. The Parseval theorem says that +∞ X k=−∞ ckdk = 1 2π Z π −π f(t)g(t)dt for all f, g ∈L2(T), where ck and dk are the Fourier coefficients of f and g. Thus, there is an isomorphism between the two Hilbert spaces L2(T) and l2(Z), which is the deep reason why the Fourier coefficients “work”. Theorem 8.7 implies that the Fourier series P k∈Z ckeikx of a function f ∈L2(T) converges to f in the L2-sense, i.e., in the mean-square sense. This does not necessarily imply that the Fourier series converges to f pointwise! This is a subtle issue, and for more on this subject, the reader is referred to Lang [73, 74] or Schwartz [103, 104]. An alternative Hilbert basis for L2(T) is given by {cos kx, sin kx}∞ k=0. This particular Hilbert basis will play an important role representing the spherical harmonics on S1 as seen the next section. 298 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS 8.2 Spherical Harmonics on the Circle For the remainder of this chapter we discuss spherical harmonics and take a glimpse at the linear representation of Lie groups. Spherical harmonics on the sphere S2 have interesting applications in computer graphics and computer vision so this material is not only important for theoretical reasons but also for practical reasons. Joseph Fourier (1768-1830) invented Fourier series in order to solve the heat equation . Using Fourier series, every square-integrable periodic function f (of period 2π) can be expressed uniquely as the sum of a power series of the form f(θ) = a0 + ∞ X k=1 (ak cos kθ + bk cos kθ), where the Fourier coefficients ak, bk of f are given by the formulae a0 = 1 2π Z π −π f(θ) dθ, ak = 1 π Z π −π f(θ) cos kθ dθ, bk = 1 π Z π −π f(θ) sin kθ dθ, for k ≥1. The reader will find the above formulae in Fourier’s famous book in Chapter III, Section 233, page 256, essentially using the notation that we use today. This remarkable discovery has many theoretical and practical applications in physics, signal processing, engineering, etc. We can describe Fourier series in a more conceptual manner if we introduce the following inner product on square-integrable functions: ⟨f, g⟩= Z π −π f(θ)g(θ) dθ, which we will also denote by ⟨f, g⟩= Z S1 f(θ)g(θ) dθ, where S1 denotes the unit circle. After all, periodic functions of (period 2π) can be viewed as functions on the circle. With this inner product, the space L2(S1) is a complete normed vector space, that is, a Hilbert space. Furthermore, if we define the subspaces Hk(S1) of L2(S1) so that H0(S1) (= R) is the set of constant functions and Hk(S1) is the two-dimensional space spanned by the functions cos kθ and sin kθ, then it turns out that we have a Hilbert sum decomposition L2(S1) = ∞ M k=0 Hk(S1) into pairwise orthogonal subspaces, where S∞ k=0 Hk(S1) is dense in L2(S1). The functions cos kθ and sin kθ are also orthogonal in Hk(S1). Now it turns out that the spaces Hk(S1) arise naturally when we look for homogeneous solutions of the Laplace equation ∆f = 0 in R2 (Pierre-Simon Laplace, 1749-1827). Roughly 8.2. SPHERICAL HARMONICS ON THE CIRCLE 299 speaking, a homogeneous function in R2 is a function that can be expressed in polar coordi-nates (r, θ) as f(r, θ) = rkg(θ). Recall that the Laplacian on R2 expressed in Cartesian coordinates (x, y) is given by ∆f = ∂2f ∂x2 + ∂2f ∂y2 , where f : R2 →R is a function which is at least of class C2. In polar coordinates (r, θ), where (x, y) = (r cos θ, r sin θ) and r > 0, since ∂f ∂r = cos θ∂f ∂x + sin θ∂f ∂y , ∂2f ∂r2 = cos2 θ∂2f ∂x2 + sin2 θ∂2f ∂y2 + 2 sin θ cos θ ∂2f ∂x∂y, and ∂2f ∂θ2 = −r  cos θ∂f ∂x + sin θ∂f ∂y  + r2  sin2 θ∂2f ∂x2 −2 sin θ cos θ ∂2f ∂x∂y + cos2 θ∂2f ∂y2  = −r∂f ∂r + r2  sin2 θ∂2f ∂x2 −2 sin θ cos θ ∂2f ∂x∂y + cos2 θ∂2f ∂y2  , we find that ∂2f ∂r2 + 1 r2 ∂2f ∂θ2 = ∂2f ∂x2 + ∂2f ∂y2 −1 r ∂f ∂r , which implies that the Laplacian (in polar coordinates) is given by ∆f = 1 r ∂ ∂r  r∂f ∂r  + 1 r2 ∂2f ∂θ2 . If we restrict f to the unit circle S1, then the Laplacian on S1 is given by ∆s1f = ∂2f ∂θ2 . It turns out that the space Hk(S1) is the eigenspace of ∆S1 for the eigenvalue −k2. To show this, we consider another question, namely what are the harmonic functions on R2; that is, the functions f that are solutions of the Laplace equation ∆f = 0. 300 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Our ancestors had the idea that the above equation can be solved by separation of vari-ables. This means that we write f(r, θ) = F(r)g(θ) , where F(r) and g(θ) are independent functions. To make things easier, let us assume that F(r) = rk for some integer k ≥0, which means that we assume that f is a homogeneous function of degree k. Recall that a function f : R2 →R is homogeneous of degree k iff f(tx, ty) = tkf(x, y) for all t > 0. Now, using the Laplacian in polar coordinates, we get ∆f = 1 r ∂ ∂r  r∂(rkg(θ)) ∂r  + 1 r2 ∂2(rkg(θ)) ∂θ2 = 1 r ∂ ∂r krkg  + rk−2∂2g ∂θ2 = rk−2k2g + rk−2∂2g ∂θ2 = rk−2(k2g + ∆S1g). Thus, we deduce that ∆f = 0 iff ∆S1g = −k2g; that is, g is an eigenfunction of ∆S1 for the eigenvalue −k2. But the above equation is equivalent to the second-order differential equation d2g dθ2 + k2g = 0, whose general solution is given by g(θ) = an cos kθ + bn sin kθ. In summary, we showed the following facts. Proposition 8.8. The integers 0, −1, −4, −9, . . . , −k2, . . . are eigenvalues of ∆S1, and the functions cos kθ and sin kθ are eigenfunctions for the eigenvalue −k2, with k ≥0. It looks like the dimension of the eigenspace corresponding to the eigenvalue −k2 is 1 when k = 0, and 2 when k ≥1. It can indeed be shown that ∆S1 has no other eigenvalues and that the dimensions claimed for the eigenspaces are correct. Observe that if we go back to our homogeneous harmonic functions f(r, θ) = rkg(θ), we see that this space is spanned by the functions uk = rk cos kθ, vk = rk sin kθ. 8.3. SPHERICAL HARMONICS ON THE 2-SPHERE 301 Now, (x + iy)k = rk(cos kθ + i sin kθ), and since ℜ(x + iy)k = uk and ℑ(x + iy)k = vk are homogeneous polynomials, we see that uk and vk are homogeneous polynomials called harmonic polynomials. For example, here is a list of a basis for the harmonic polynomials (in two variables) of degree k = 0, 1, 2, 3, 4, listed as ˜ uk = cos kθ, ˜ vk = sin kθ: k = 0 1 k = 1 x, y k = 2 x2 −y2, 2xy k = 3 x3 −3xy2, 3x2y −y3 k = 4 x4 −6x2y2 + y4, x3y −xy3. To derive these formulas, we simply expand (x + iy)k via the binomial theorem and take ˜ uk as the real part, and ˜ vk as the imaginary part. Therefore, the eigenfunctions of the Laplacian on S1 are the restrictions of the harmonic polynomials on R2 to S1, and we have a Hilbert sum decomposition L2(S1) = L∞ k=0 Hk(S1). It turns out that this phenomenon generalizes to the sphere Sn ⊆Rn+1 for all n ≥1. Let us take a look at next case n = 2. 8.3 Spherical Harmonics on the 2-Sphere The material of section is very classical and can be found in many places, for example An-drews, Askey and Roy (Chapter 9), Sansone (Chapter III), Hochstadt (Chapter 6), and Lebedev (Chapter ). We recommend the exposition in Lebedev because we find it particularly clear and uncluttered. We have also borrowed heavily from some lecture notes by Hermann Gluck for a course he offered in 1997-1998. Our goal is to find the homogeneous solutions of the Laplace equation ∆f = 0 in R3, and to show that they correspond to spaces Hk(S2) of eigenfunctions of the Laplacian ∆S2 on the 2-sphere S2 = {(x, y, z) ∈R3 | x2 + y2 + z2 = 1}. Then the spaces Hk(S2) will give us a Hilbert sum decomposition of the Hilbert space L2(S2) of square-integrable functions on S2. This is the generalization of Fourier series to the 2-sphere and the functions in the spaces Hk(S2) are called spherical harmonics. The Laplacian in R3 is of course given by ∆f = ∂2f ∂x2 + ∂2f ∂y2 + ∂2f ∂z2 . If we use spherical coordinates x = r sin θ cos ϕ y = r sin θ sin ϕ z = r cos θ, 302 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS in R3, where 0 ≤θ < π, 0 ≤ϕ < 2π and r > 0 (recall that ϕ is the so-called azimuthal angle in the xy-plane originating at the x-axis and θ is the so-called polar angle from the z-axis, angle defined in the plane obtained by rotating the xz-plane around the z-axis by the angle ϕ), then since ∂f ∂r = sin θ cos ϕ∂f ∂x + sin θ sin ϕ∂f ∂y + cos θ∂f ∂z , ∂2f ∂r2 = sin2 θ cos2 ϕ∂2f ∂x2 + sin2 θ sin2 ϕ∂2f ∂y2 + cos2 θ∂2f ∂z2 + 2 sin2 θ sin ϕ cos ϕ ∂2f ∂x∂y + 2 sin θ cos θ cos ϕ ∂2f ∂x∂z + 2 cos θ sin θ sin ϕ ∂2f ∂y∂z, ∂f ∂θ = r cos θ cos ϕ∂f ∂x + r cos θ sin ϕ∂f ∂y −r sin θ∂f ∂z , ∂2f ∂θ2 = −r∂f ∂r + r2 cos2 θ cos2 ϕ∂2f ∂x2 + r2 cos2 θ sin2 ϕ∂2f ∂y2 + r2 sin2 θ∂2f ∂z2 + 2r2 cos2 θ cos ϕ sin ϕ ∂2f ∂x∂y −2r2 cos θ sin θ cos ϕ ∂2f ∂x∂z −2r2 cos θ sin θ sin ϕ ∂2f ∂y∂z, ∂f ∂ϕ = −r sin θ sin ϕ∂f ∂x + r sin θ cos ϕ∂f ∂y , and ∂2f ∂ϕ2 = −r sin θ cos ϕ∂f ∂x −r sin θ sin ϕ∂f ∂y + r2 sin2 θ sin2 ϕ∂2f ∂x2 + r2 sin2 θ cos2 ϕ∂2f ∂y2 −2r2 sin2 θ cos ϕ sin ϕ ∂2f ∂x∂y, we discover that ∂2f ∂r2 + 1 r2 ∂2f ∂θ2 + 1 r2 sin2 θ ∂2f ∂ϕ2 + 2 r ∂f ∂r + cos θ r2 sin θ ∂f ∂θ = ∂2f ∂x2 + ∂2f ∂y2 + ∂2f ∂z2 , which implies that the Laplacian in spherical coordinates is given by ∆f = 1 r2 ∂ ∂r  r2∂f ∂r  + 1 r2∆S2f, 8.3. SPHERICAL HARMONICS ON THE 2-SPHERE 303 where ∆S2f = 1 sin θ ∂ ∂θ  sin θ∂f ∂θ  + 1 sin2 θ ∂2f ∂ϕ2, is the Laplacian on the sphere S2. Let us look for homogeneous harmonic functions f(r, θ, ϕ) = rkg(θ, ϕ) on R3; that is, solutions of the Laplace equation ∆f = 0. We get ∆f = 1 r2 ∂ ∂r  r2∂(rkg) ∂r  + 1 r2∆S2(rkg) = 1 r2 ∂ ∂r krk+1g  + rk−2∆S2g = rk−2k(k + 1)g + rk−2∆S2g = rk−2(k(k + 1)g + ∆S2g). Therefore, ∆f = 0 iff ∆S2g = −k(k + 1)g; that is, g is an eigenfunction of ∆S2 for the eigenvalue −k(k + 1). We can look for solutions of the above equation using the separation of variables method. If we let g(θ, ϕ) = Θ(θ)Φ(ϕ), then we get the equation Φ sin θ ∂ ∂θ  sin θ∂Θ ∂θ  + Θ sin2 θ ∂2Φ ∂ϕ2 = −k(k + 1)ΘΦ; that is, dividing by ΘΦ and multiplying by sin2 θ, sin θ Θ ∂ ∂θ  sin θ∂Θ ∂θ  + k(k + 1) sin2 θ = −1 Φ ∂2Φ ∂ϕ2 . Since Θ and Φ are independent functions, the above is possible only if both sides are equal to a constant, say µ. This leads to two equations ∂2Φ ∂ϕ2 + µΦ = 0 sin θ Θ ∂ ∂θ  sin θ∂Θ ∂θ  + k(k + 1) sin2 θ −µ = 0. However, we want Φ to be periodic in ϕ since we are considering functions on the sphere, so µ be must of the form µ = m2 for some non-negative integer m. Then we know that the space of solutions of the equation ∂2Φ ∂ϕ2 + m2Φ = 0 304 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS is two-dimensional and is spanned by the two functions Φ(ϕ) = cos mϕ, Φ(ϕ) = sin mϕ. We still have to solve the equation sin θ ∂ ∂θ  sin θ∂Θ ∂θ  + (k(k + 1) sin2 θ −m2)Θ = 0, which is equivalent to sin2 θ Θ′′ + sin θ cos θ Θ′ + (k(k + 1) sin2 θ −m2)Θ = 0. a variant of Legendre’s equation. For this, we use the change of variable t = cos θ, and we consider the function u given by u(cos θ) = Θ(θ) (recall that 0 ≤θ < π), so we get the second-order differential equation (1 −t2)u′′ −2tu′ +  k(k + 1) − m2 1 −t2  u = 0 sometimes called the general Legendre equation (Adrien-Marie Legendre, 1752-1833). The trick to solve this equation is to make the substitution u(t) = (1 −t2) m 2 v(t); see Lebedev , Chapter 7, Section 7.12. Then we get (1 −t2)v′′ −2(m + 1)tv′ + (k(k + 1) −m(m + 1))v = 0. When m = 0, we get the Legendre equation: (1 −t2)v′′ −2tv′ + k(k + 1)v = 0; see Lebedev , Chapter 7, Section 7.3. This equation has two fundamental solution Pk(t) and Qk(t) called the Legendre functions of the first and second kinds. The Pk(t) are actually polynomials and the Qk(t) are given by power series that diverge for t = 1, so we only keep the Legendre polynomials Pk(t). The Legendre polynomials can be defined in various ways. One definition is in terms of Rodrigues’ formula: Pn(t) = 1 2nn! dn dtn(t2 −1)n; see Lebedev , Chapter 4, Section 4.2. In this version of the Legendre polynomials they are normalized so that Pn(1) = 1. There is also the following recurrence relation: P0 = 1 P1 = t (n + 1)Pn+1 = (2n + 1)tPn −nPn−1 n ≥1; 8.3. SPHERICAL HARMONICS ON THE 2-SPHERE 305 see Lebedev , Chapter 4, Section 4.3. For example, the first six Legendre polynomials are 1 t 1 2(3t2 −1) 1 2(5t3 −3t) 1 8(35t4 −30t2 + 3) 1 8(63t5 −70t3 + 15t). Let us now return to our differential equation (1 −t2)v′′ −2(m + 1)tv′ + (k(k + 1) −m(m + 1))v = 0. (∗) Observe that if we differentiate with respect to t, we get the equation (1 −t2)v′′′ −2(m + 2)tv′′ + (k(k + 1) −(m + 1)(m + 2))v′ = 0. This shows that if v is a solution of our Equation (∗) for given k and m, then v′ is a solution of the same equation for k and m + 1. Thus, if Pk(t) solves (∗) for given k and m = 0, then P ′ k(t) solves (∗) for the same k and m = 1, P ′′ k (t) solves (∗) for the same k and m = 2, and in general dm/dtm(Pk(t)) solves (∗) for k and m. Therefore, our original equation (1 −t2)u′′ −2tu′ +  k(k + 1) − m2 1 −t2  u = 0 (†) has the solution u(t) = (1 −t2) m 2 dm dtm(Pk(t)) := P k m(t). The function u(t) is traditionally denoted P m k (t) and called an associated Legendre function; see Lebedev , Chapter 7, Section 7.12. The index k is often called the band index. Obviously, P m k (t) ≡0 if m > k and P 0 k (t) = Pk(t), the Legendre polynomial of degree k. An associated Legendre function is not a polynomial in general, and because of the factor (1 −t2) m 2 , it is only defined on the closed interval [−1, 1].  Certain authors add the factor (−1)m in front of the expression for the associated Leg-endre function P m k (t), as in Lebedev , Chapter 7, Section 7.12, see also Footnote 29 on Page 193. This seems to be common practice in the quantum mechanics literature where it is called the Condon Shortley phase factor. 306 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS The associated Legendre functions satisfy various recurrence relations that allows us to compute them. For example, for fixed m ≥0, we have (see Lebedev , Chapter 7, Section 7.12) the recurrence (k −m + 1)P m k+1(t) = (2k + 1)tP m k (t) −(k + m)P m k−1(t), k ≥1, and for fixed k ≥2, we have P m+2 k (t) = 2(m + 1)t (t2 −1) 1 2 P m+1 k (t) + (k −m)(k + m + 1)P m k (t), 0 ≤m ≤k −2, which can also be used to compute P m k starting from P 0 k (t) = Pk(t) P 1 k (t) = kt (t2 −1) 1 2 Pk(t) − k (t2 −1) 1 2 Pk−1(t). Observe that the recurrence relation for m fixed yields the following equation for k = m (as P m m−1 = 0): P m m+1(t) = (2m + 1)tP m m (t). It it also easy to see that P m m (t) = (2m)! 2mm! (1 −t2) m 2 . Observe that (2m)! 2mm! = (2m −1)(2m −3) · · · 5 · 3 · 1, an expression that is sometimes denoted (2m −1)!! and called the double factorial.  Beware that some papers in computer graphics adopt the definition of associated Legen-dre functions with the scale factor (−1)m added, so this factor is present in these papers, for example Green . The equation above allows us to “lift” P m m to the higher band m + 1. The computer graphics community (see Green ) uses the following three rules to compute P m k (t) where 0 ≤m ≤k: (1) Compute P m m (t) = (2m)! 2mm! (1 −t2) m 2 . If m = k, stop. Otherwise do Step 2 once. (2) Compute P m m+1(t) = (2m + 1)tP m m (t). If k = m + 1, stop. Otherwise, iterate Step 3. 8.3. SPHERICAL HARMONICS ON THE 2-SPHERE 307 (3) Starting from i = m + 1, compute (i −m + 1)P m i+1(t) = (2i + 1)tP m i (t) −(i + m)P m i−1(t) until i + 1 = k. If we recall that Equation (†) was obtained from the equation sin2 θ Θ′′ + sin θ cos θ Θ′ + (k(k + 1) sin2 θ −m2)Θ = 0 using the substitution u(cos θ) = Θ(θ), we see that Θ(θ) = P m k (cos θ) is a solution of the above equation. Putting everything together, as f(r, θ, ϕ) = rkΘ(θ)Φ(ϕ), we proved the following facts. Proposition 8.9. The homogeneous functions f(r, θ, ϕ) = rk cos mϕ P m k (cos θ), f(r, θ, ϕ) = rk sin mϕ P m k (cos θ) are solutions of the Laplacian ∆in R3, and the functions cos mϕ P m k (cos θ), sin mϕ P m k (cos θ) are eigenfunctions of the Laplacian ∆S2 on the sphere for the eigenvalue −k(k + 1). For k fixed, as 0 ≤m ≤k, we get 2k + 1 linearly independent functions. The notation for the above functions varies quite a bit, essentially because of the choice of normalization factors used in various fields (such as physics, seismology, geodesy, spectral analysis, magnetics, quantum mechanics etc.). We will adopt the notation ym l , where l is a nonnegative integer but m is allowed to be negative, with −l ≤m ≤l. Thus, we set ym l (θ, ϕ) =    N 0 l Pl(cos θ) if m = 0 √ 2N m l cos mϕ P m l (cos θ) if m > 0 √ 2N m l sin(−mϕ) P −m l (cos θ) if m < 0 for l = 0, 1, 2, . . ., and where the N m l are scaling factors. In physics and computer graphics, N m l is chosen to be N m l = s (2l + 1)(l −|m|)! 4π(l + |m|)! . Definition 8.11. The functions ym l are called the real spherical harmonics of degree l and order m. The index l is called the band index. 308 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS The functions, ym l , have some very nice properties, but to explain these we need to recall the Hilbert space structure of the space L2(S2) of square-integrable functions on the sphere. Recall that we have an inner product on L2(S2) given by ⟨f, g⟩= Z S2 fg VolS2 = Z 2π 0 Z π 0 f(θ, ϕ)g(θ, ϕ) sin θdθdϕ, where f, g ∈L2(S2) and where VolS2 is the volume form on S2 (induced by the metric on R3). With this inner product, L2(S2) is a complete normed vector space using the norm ∥f∥= p ⟨f, f⟩associated with this inner product; that is, L2(S2) is a Hilbert space. Now, it can be shown that the Laplacian ∆S2 on the sphere is a self-adjoint linear operator with respect to this inner product. As the functions ym1 l1 and ym2 l2 with l1 ̸= l2 are eigenfunctions corresponding to distinct eigenvalues (−l1(l1 + 1) and −l2(l2 + 1)), they are orthogonal; that is, ⟨ym1 l1 , ym2 l2 ⟩= 0, if l1 ̸= l2. It is also not hard to show that for a fixed l, ⟨ym1 l , ym2 l ⟩= δm1,m2; that is, the functions ym l with −l ≤m ≤l form an orthonormal system, and we denote by Hl(S2) the (2l + 1)-dimensional space spanned by these functions. It turns out that the functions ym l form a basis of the eigenspace El of ∆S2 associated with the eigenvalue −l(l + 1), so that El = Hl(S2), and that ∆S2 has no other eigenvalues. More is true. It turns out that L2(S2) is the orthogonal Hilbert sum of the eigenspaces Hl(S2). This means that the Hl(S2) are (1) mutually orthogonal (2) closed, and (3) The space L2(S2) is the Hilbert sum L∞ l=0 Hl(S2), which means that for every function f ∈L2(S2), there is a unique sequence of spherical harmonics fj ∈Hl(S2) so that f = ∞ X l=0 fl; that is, the sequence Pl j=0 fj converges to f (in the norm on L2(S2)). Observe that each fl is a unique linear combination fl = P ml aml l yml l . Therefore, (3) gives us a Fourier decomposition on the sphere generalizing the famil-iar Fourier decomposition on the circle. Furthermore, the Fourier coefficients amll can be computed using the fact that the ym l form an orthonormal Hilbert basis: aml l = ⟨f, yml l ⟩. 8.4. THE LAPLACE-BELTRAMI OPERATOR 309 We also have the corresponding homogeneous harmonic functions Hm l (r, θ, ϕ) on R3 given by Hm l (r, θ, ϕ) = rlym l (θ, ϕ). If one starts computing explicitly the Hm l for small values of l and m, one finds that it is always possible to express these functions in terms of the Cartesian coordinates x, y, z as homogeneous polynomials! This remarkable fact holds in general: The eigenfunctions of the Laplacian ∆S2, and thus the spherical harmonics, are the restrictions of homogeneous harmonic polynomials in R3. Here is a list of bases of the homogeneous harmonic polynomials of degree k in three variables up to k = 4 (thanks to Herman Gluck). k = 0 1 k = 1 x, y, z k = 2 x2 −y2, x2 −z2, xy, xz, yz k = 3 x3 −3xy2, 3x2y −y3, x3 −3xz2, 3x2z −z3, y3 −3yz2, 3y2z −z3, xyz k = 4 x4 −6x2y2 + y4, x4 −6x2z2 + z4, y4 −6y2z2 + z4, x3y −xy3, x3z −xz3, y3z −yz3, 3x2yz −yz3, 3xy2z −xz3, 3xyz2 −x3y. Subsequent sections will be devoted to a proof of the important facts stated earlier. 8.4 The Laplace-Beltrami Operator In order to define rigorously the Laplacian on the sphere Sn ⊆Rn+1 and establish its relation-ship with the Laplacian on Rn+1, we need the definition of the Laplacian on a Riemannian manifold (M, g), the Laplace-Beltrami operator (Eugenio Beltrami, 1835-1900). A more gen-eral definition of the the Laplace-Beltrami operator as an operator on differential forms is given in Section 9.3. In this chapter we only need the definition of the Laplacian on functions. Recall that a Riemannian metric g on a manifold M is a smooth family of inner products g = (gp), where gp is an inner product on the tangent space TpM for every p ∈M. The inner product gp on TpM establishes a canonical duality between TpM and T ∗ p M, namely, we have the isomorphism ♭: TpM →T ∗ p M defined such that for every u ∈TpM, the linear form u♭∈T ∗ p M is given by u♭(v) = gp(u, v), v ∈TpM. The inverse isomorphism ♯: T ∗ p M →TpM is defined such that for every ω ∈T ∗ p M, the vector ω♯is the unique vector in TpM so that gp(ω♯, v) = ω(v), v ∈TpM. 310 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS The isomorphisms ♭and ♯induce isomorphisms between vector fields X ∈X(M) and one-forms ω ∈A1(M). In particular, for every smooth function f ∈C∞(M), the vector field corresponding to the one-form d f is the gradient grad f of f. The gradient of f is uniquely determined by the condition gp((grad f)p, v) = d fp(v), v ∈TpM, p ∈M. Definition 8.12. Let (M, g) be a Riemannian manifold. If ∇X is the covariant derivative associated with the Levi-Civita connection induced by the metric g, then the divergence of a vector field X ∈X(M) is the function div X : M →R defined so that (div X)(p) = tr(Y (p) 7→(∇Y X)p); namely, for every p, (div X)(p) is the trace of the linear map Y (p) 7→(∇Y X)p. Then the Laplace-Beltrami operator, for short, Laplacian, is the linear operator ∆: C∞(M) →C∞(M) given by ∆f = div grad f. Remark: The definition just given differs from the definition given in Section 9.3 by a negative sign. We adopted this sign convention to conform with most of the literature on spherical harmonics (where the negative sign is omitted). A consequence of this choice is that the eigenvalues of the Laplacian are negative. For more details on the Laplace-Beltrami operator, we refer the reader to Chapter 9 or to Gallot, Hulin and Lafontaine (Chapter 4) or O’Neill (Chapter 3), Postnikov (Chapter 13), Helgason (Chapter 2) or Warner (Chapters 4 and 6). All this being rather abstract, it is useful to know how grad f, div X, and ∆f are expressed in a chart. If (U, ϕ) is a chart of M, with p ∈M, and if as usual  ∂ ∂x1  p , . . . ,  ∂ ∂xn  p ! denotes the basis of TpM induced by ϕ, the local expression of the metric g at p is given by the n × n matrix (gij)p, with (gij)p = gp  ∂ ∂xi  p ,  ∂ ∂xj  p ! . The matrix (gij)p is symmetric, positive definite, and its inverse is denoted (gij)p. We also let |g|p = det(gij)p. For simplicity of notation we often omit the subscript p. Then it can be 8.4. THE LAPLACE-BELTRAMI OPERATOR 311 shown that for every function f ∈C∞(M), in local coordinates given by the chart (U, ϕ), we have grad f = X ij gij ∂f ∂xj ∂ ∂xi , where as usual ∂f ∂xj (p) =  ∂ ∂xj  p f = ∂(f ◦ϕ−1) ∂uj (ϕ(p)), and (u1, . . . , un) are the coordinate functions in Rn. There are formulae for div X and ∆f involving the Christoffel symbols. Let X = n X i=1 Xi ∂ ∂xi , be a vector field expressed over a chart (U, ϕ). Recall that the Christoffel symbol Γk ij is defined as Γk ij = 1 2 n X l=1 gkl (∂igjl + ∂jgil −∂lgij) , (∗) where ∂kgij = ∂ ∂xk (gij). Then divX = n X i=1 " ∂Xi ∂xi + n X j=1 Γi ijXj # , and ∆f = X i,j gij " ∂2f ∂xi∂xj − n X k=1 Γk ij ∂f ∂xk # , whenever f ∈C∞(M); see Pages 86 and 87 of O’Neill . We take a moment to use O’Neill formula to re-derive the expression for the Laplacian on R2 in terms of polar coordinates (r, θ), where x = r cos θ and y = r sin θ. Note that ∂ ∂x1 = ∂ ∂r = (cos θ, sin θ) ∂ ∂x2 = ∂ ∂θ = (−r sin θ, r cos θ), which in turn gives gij = 1 0 0 r2  gij = 1 0 0 r−2  . Some computations show that the only nonzero Christoffel symbols are Γ2 12 = Γ2 21 = 1 r Γ1 22 = −r; 312 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS see Gallier and Quaintance . Hence ∆f = 2 X i,j=1 gij " ∂2f ∂xi∂xj − 2 X k=1 Γk ij ∂f ∂xk # = g11 " ∂2f ∂x2 1 − 2 X k=1 Γk 11 ∂f ∂xk # + g22 " ∂2f ∂x2 2 − 2 X k=1 Γk 22 ∂f ∂xk # = ∂2f ∂r2 + 1 r2 ∂2f ∂θ2 −Γ1 22 ∂f ∂r  = ∂2f ∂r2 + 1 r2 ∂2f ∂θ2 + r∂f ∂r  = 1 r2 ∂2f ∂θ2 + ∂2f ∂r2 + 1 r ∂f ∂r = 1 r2 ∂2f ∂θ2 + 1 r ∂ ∂r  r∂f ∂r  . O’Neill’s formula may also be used to re-derive the expression for the Laplacian on R3 in terms of spherical coordinates (r, θ, ϕ) where x = r sin θ cos ϕ y = r sin θ sin ϕ z = r cos θ. We have ∂ ∂x1 = ∂ ∂r = sin θ cos ϕ ∂ ∂x + sin θ sin ϕ ∂ ∂y + cos θ ∂ ∂z = b r ∂ ∂x2 = ∂ ∂θ = r  cos θ cos ϕ ∂ ∂x + cos θ sin ϕ ∂ ∂y −sin θ ∂ ∂z  = rb θ ∂ ∂x3 = ∂ ∂ϕ = r  −sin θ sin ϕ ∂ ∂x + sin θ cos ϕ ∂ ∂y  = r b ϕ. Observe that b r, b θ and b ϕ are pairwise orthogonal. Therefore, the matrix (gij) is given by (gij) =   1 0 0 0 r2 0 0 0 r2 sin2 θ   and |g| = r4 sin2 θ. The inverse of (gij) is (gij) =   1 0 0 0 r−2 0 0 0 r−2 sin−2 θ  . 8.4. THE LAPLACE-BELTRAMI OPERATOR 313 By using Line (∗), it is not hard to show that Γk ij = 0 except for Γ1 22 = −1 2g11∂1g22 = −1 2 ∂ ∂rr2 = −r Γ1 33 = −1 2g11∂1g33 = −1 2 ∂ ∂rr2 sin2 θ = −r sin2 θ Γ2 12 = Γ2 21 = 1 2g22∂1g22 = 1 2r2 ∂ ∂rr2 = 1 r Γ2 33 = −1 2g22∂2g33 = −1 2r2 ∂ ∂θr2 sin2 θ = −sin θ cos θ Γ3 13 = Γ3 31 = 1 2g33∂1g33 = 1 2r2 sin2 θ ∂ ∂rr2 sin2 θ = 1 r Γ3 23 = Γ2 32 = 1 2g33∂2g33 = 1 2r2 sin2 θ ∂ ∂θr2 sin2 θ = cot θ. Then ∆f = 3 X i,j=1 gij " ∂2f ∂xi∂xj − 3 X k=1 Γk ij ∂f ∂xk # = g11 " ∂2f ∂x2 1 − 3 X k=1 Γk 11 ∂f ∂xk # + g22 " ∂2f ∂x2 2 − 3 X k=1 Γk 22 ∂f ∂xk # + g33 " ∂2f ∂x2 3 − 3 X k=1 Γk 33 ∂f ∂xk # = ∂2f ∂r2 + 1 r2 ∂2f ∂θ2 −Γ1 22 ∂f ∂r  + 1 r2 sin2 θ ∂2f ∂ϕ2 −  Γ1 33 ∂f ∂r + Γ2 33 ∂f ∂θ  = ∂2f ∂r2 + 1 r2 ∂2f ∂θ2 + r∂f ∂r  + 1 r2 sin2 θ ∂2f ∂ϕ2 + r sin2 θ∂f ∂r + sin θ cos θ∂f ∂θ  = ∂2f ∂r2 + 2 r ∂f ∂r + 1 r2 ∂2f ∂θ2 + cos θ sin θ ∂f ∂θ  + 1 r2 sin2 θ ∂2f ∂ϕ2 = 1 r2 ∂ ∂r  r2∂f ∂r  + 1 r2 sin θ ∂ ∂θ  sin θ∂f ∂θ  + 1 r2 sin2 θ ∂2f ∂ϕ2. O’Neill’s formulae for the divergence and the Laplacian can be tedious to calculate since they involve knowing the Christoffel symbols. Fortunately there are other formulas for the the divergence and the Laplacian which only involve (gij) and (gij) and hence will be more convenient for our purposes: For every vector field X ∈X(M) expressed in local coordinates as X = n X i=1 Xi ∂ ∂xi , we have div X = 1 p |g| n X i=1 ∂ ∂xi p |g| Xi  , (†) 314 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS and for every function f ∈C∞(M), the Laplacian ∆f is given by ∆f = 1 p |g| X i,j ∂ ∂xi p |g| gij ∂f ∂xj  . (∗∗) A detailed proof of Equation (†) is given in Helgason (Chapter II, Lemma 2.5). This formula is also stated in Postnikov (Chapter 13, Section 6) and O’Neill (Chapter 7, Exercise 5). One should check that for M = Rn with its standard coordinates, the Laplacian is given by the familiar formula ∆f = ∂2f ∂x2 1 + · · · + ∂2f ∂x2 n . By using Equation (∗∗), we quickly rediscover the Laplacian in spherical coordinates, namely ∆f = 1 r2 sin θ 3 X i=1 3 X j=1 ∂ ∂xi  r2 sin θgij ∂f ∂xj  = 1 r2 sin θ  ∂ ∂r  r2 sin θ∂f ∂r  + ∂ ∂θ  r2 sin θr−2∂f ∂θ  + ∂ ∂ϕ  r2 sin θr−2 sin−2 θ ∂f ∂ϕ  = 1 r2 ∂ ∂r  r2∂f ∂r  + 1 r2 sin θ ∂ ∂θ  sin θ∂f ∂θ  + 1 r2 sin2 θ ∂2f ∂ϕ2. Since (θ, ϕ) are coordinates on the sphere S2 via x = sin θ cos ϕ y = sin θ sin ϕ z = cos θ, we see that in these coordinates, the metric (e gij) on S2 is given by the matrix (e gij) = 1 0 0 sin2 θ  , that |e g| = sin2 θ, and that the inverse of (e gij) is (e gij) = 1 0 0 sin−2 θ  . It follows immediately that ∆S2f = 1 sin θ ∂ ∂θ  sin θ∂f ∂θ  + 1 sin2 θ ∂2f ∂ϕ2, 8.4. THE LAPLACE-BELTRAMI OPERATOR 315 so we have verified that ∆f = 1 r2 ∂ ∂r  r2∂f ∂r  + 1 r2∆S2f. Let us now generalize the above formula to the Laplacian ∆on Rn+1, and the Laplacian ∆Sn on Sn, where Sn = {(x1, . . . , xn+1) ∈Rn+1 | x2 1 + · · · + x2 n+1 = 1}. Following Morimoto (Chapter 2, Section 2), let us use “polar coordinates.” The map from R+ × Sn to Rn+1 −{0} given by (r, σ) 7→rσ is clearly a diffeomorphism. Thus, for any system of coordinates (u1, . . . , un) on Sn, the tuple (u1, . . . , un, r) is a system of coordinates on Rn+1 −{0} called polar coordinates. Let us establish the relationship between the Laplacian ∆, on Rn+1 −{0} in polar coordinates and the Laplacian ∆Sn on Sn in local coordinates (u1, . . . , un). Proposition 8.10. If ∆is the Laplacian on Rn+1 −{0} in polar coordinates (u1, . . . , un, r) and ∆Sn is the Laplacian on the sphere Sn in local coordinates (u1, . . . , un), then ∆f = 1 rn ∂ ∂r  rn∂f ∂r  + 1 r2∆Snf. Proof. Let us compute the (n + 1) × (n + 1) matrix G = (gij) expressing the metric on Rn+1 in polar coordinates and the n × n matrix e G = (e gij) expressing the metric on Sn. Recall that if σ ∈Sn, then σ · σ = 1, and so ∂σ ∂ui · σ = 0, as ∂σ ∂ui · σ = 1 2 ∂(σ · σ) ∂ui = 0. If x = rσ with σ ∈Sn, we have ∂x ∂ui = r ∂σ ∂ui , 1 ≤i ≤n, and ∂x ∂r = σ. 316 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS It follows that gij = ∂x ∂ui · ∂x ∂uj = r2 ∂σ ∂ui · ∂σ ∂uj = r2e gij gin+1 = ∂x ∂ui · ∂x ∂r = r ∂σ ∂ui · σ = 0 gn+1n+1 = ∂x ∂r · ∂x ∂r = σ · σ = 1. Consequently, we get G =  r2 e G 0 0 1  , |g| = r2n|e g|, and G−1 =  r−2 e G−1 0 0 1  . Using the above equations and ∆f = 1 p |g| X i,j ∂ ∂xi p |g| gij ∂f ∂xj  , we get ∆f = 1 rnp |e g| n X i,j=1 ∂ ∂xi  rnp |e g| 1 r2e gij ∂f ∂xj  + 1 rnp |e g| ∂ ∂r  rnp |e g| ∂f ∂r  = 1 r2p |e g| n X i,j=1 ∂ ∂xi p |e g| e gij ∂f ∂xj  + 1 rn ∂ ∂r  rn ∂f ∂r  = 1 r2∆Snf + 1 rn ∂ ∂r  rn ∂f ∂r  , as claimed. It is also possible to express ∆Sn in terms of ∆Sn−1. If en+1 = (0, . . . , 0, 1) ∈Rn+1, then we can view Sn−1 as the intersection of Sn with the hyperplane xn+1 = 0; that is, as the set Sn−1 = {σ ∈Sn | σ · en+1 = 0}. If (u1, . . . , un−1) are local coordinates on Sn−1, then (u1, . . . , un−1, θ) are local coordinates on Sn, by setting σ = sin θ e σ + cos θ en+1, with e σ ∈Sn−1 and 0 ≤θ < π. 8.4. THE LAPLACE-BELTRAMI OPERATOR 317 Proposition 8.11. We have ∆Snf = 1 sinn−1 θ ∂ ∂θ  sinn−1 θ ∂f ∂θ  + 1 sin2 θ∆Sn−1f. Proof. Note that e σ · e σ = 1, which in turn implies ∂e σ ∂ui · e σ = 0. Furthermore, e σ · en+1 = 0, and hence ∂e σ ∂ui · en+1 = 0. By using these local coordinate systems, we find the relationship between ∆Sn and ∆Sn−1 as follows: First observe that ∂σ ∂ui = sin θ ∂e σ ∂ui + 0 en+1 ∂σ ∂θ = cos θ e σ −sin θ en+1. If e G = (e gij) represents the metric on Sn and b G = (b gij) is the restriction of this metric to Sn−1 as defined above then for 1 ≤i, j ≤n −1, we have e gij = ∂σ ∂ui · ∂σ ∂uj = sin2 θ ∂e σ ∂ui · ∂e σ ∂uj = sin2 θ b gij e gin = ∂σ ∂ui · ∂σ ∂θ =  sin θ ∂e σ ∂ui + 0 en+1  · (cos θ e σ −sin θ en+1) = 0 e gnn = ∂σ ∂θ · ∂σ ∂θ = (cos θ e σ −sin θ en+1) · (cos θ e σ −sin θ en+1) = cos2 θ + sin2 θ = 1. These calculations imply that e G =  sin2 θ b G 0 0 1  , |e g| = sin2n−2 θ|b g|, and that e G−1 =  sin−2 θ b G−1 0 0 1  . Hence ∆Snf = 1 sinn−1 θ p |b g| n−1 X i,j=1 ∂ ∂ui  sinn−1 θ p |b g| 1 sin2 θ ˆ gij ∂f ∂uj  + 1 sinn−1 θ p |b g| ∂ ∂θ  sinn−1 θ p |b g|∂f ∂θ  = 1 sinn−1 θ ∂ ∂θ  sinn−1 θ ∂f ∂θ  + 1 sin2 θ p |b g| n−1 X i,j=1 ∂ ∂ui p |b g|ˆ gij ∂f ∂uj  = 1 sinn−1 θ ∂ ∂θ  sinn−1 θ ∂f ∂θ  + 1 sin2 θ∆Sn−1f, 318 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS as claimed. A fundamental property of the divergence is known as Green’s formula. There are actually two Greens’ formulae, but we will only need the version for an orientable manifold without boundary given in Proposition 9.15. Recall that Green’s formula states that if M is a compact, orientable, Riemannian manifold without boundary, then, for every smooth vector field X ∈X(M), we have Z M (div X) VolM = 0, where VolM is the volume form on M induced by the metric. Definition 8.13. If M is a compact, orientable Riemannian manifold, then for any two smooth functions f, h ∈C∞(M), we define ⟨f, h⟩M by ⟨f, h⟩M = Z M fh VolM. Then, it is not hard to show that ⟨−, −⟩M is an inner product on C∞(M). An important property of the Laplacian on a compact, orientable Riemannian manifold is that it is a self-adjoint operator. This fact is proved in the more general case of an inner product on differential forms in Proposition 9.8, but it is instructive to give another proof in the special case of functions using Green’s formula. First we need the following two properties: For any two functions f, h ∈C∞(M), and any vector field X ∈X(M), we have: div(fX) = fdiv X + X(f) = fdiv X + g(grad f, X) grad f (h) = g(grad f, grad h) = grad h (f). Using the above identities, we obtain the following important result. Proposition 8.12. Let M be a compact, orientable, Riemannian manifold without boundary. The Laplacian on M is self-adjoint; that is, for any two functions f, h ∈C∞(M), we have ⟨∆f, h⟩M = ⟨f, ∆h⟩M, or equivalently Z M f∆h VolM = Z M h∆f VolM. Proof. By the two identities before Proposition 8.12, f∆h = fdiv grad h = div(fgrad h) −g(grad f, grad h) and h∆f = hdiv grad f = div(hgrad f) −g(grad h, grad f), 8.5. HARMONIC POLYNOMIALS, SPHERICAL HARMONICS AND L2(Sn) 319 so we get f∆h −h∆f = div(fgrad h −hgrad f). By Green’s formula, Z M (f∆h −h∆f)VolM = Z M div(fgrad h −hgrad f)VolM = 0, which proves that ∆is self-adjoint. The importance of Proposition 8.12 lies in the fact that as ⟨−, −⟩M is an inner product on C∞(M), the eigenspaces of ∆for distinct eigenvalues are pairwise orthogonal. We will make heavy use of this property in the next section on harmonic polynomials. 8.5 Harmonic Polynomials, Spherical Harmonics and L2(Sn) Harmonic homogeneous polynomials and their restrictions to Sn, where Sn = {(x1, . . . , xn+1) ∈Rn+1 | x2 1 + · · · + x2 n+1 = 1}, turn out to play a crucial role in understanding the structure of the eigenspaces of the Lapla-cian on Sn (with n ≥1). The results in this section appear in one form or another in Stein and Weiss (Chapter 4), Morimoto (Chapter 2), Helgason (Introduction, Section 3), Dieudonn´ e (Chapter 7), Axler, Bourdon and Ramey (Chapter 5), and Vilenkin (Chapter IX). Some of these sources assume a fair amount of mathematical background, and consequently uninitiated readers will probably find the exposition rather condensed, especially Helgason. We tried hard to make our presentation more “user-friendly.” Recall that a homogeneous polynomial P of degree k in n variables x1, . . . , xn is an expression of the form P = X α1+···+αn=k (α1,...,αn)∈Nk a(α1,...,αn) xα1 1 · · · xαn n , where the coefficients a(α1,...,αn) are either real or complex numbers. We view such a ho-mogeneous polynomial as a function P : Rn →C, or as a function P : Rn →R when the coefficients are all real. The Laplacian ∆P of P is defined by ∆P = X α1+···+αn=k (α1,...,αn)∈Nk a(α1,...,αn)  ∂2 ∂x2 1 + · · · + ∂2 ∂x2 n  (xα1 1 · · · xαn n ). 320 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Definition 8.14. Let Pk(n + 1) (resp. PC k (n + 1)) denote the space of homogeneous poly-nomials of degree k in n + 1 variables with real coefficients (resp. complex coefficients), and let Pk(Sn) (resp. PC k (Sn)) denote the restrictions of homogeneous polynomials in Pk(n+1) to Sn (resp. the restrictions of homogeneous polynomials in PC k (n+1) to Sn). Let Hk(n + 1) (resp. HC k (n + 1)) denote the space of (real) harmonic polynomials (resp. complex harmonic polynomials), with Hk(n + 1) = {P ∈Pk(n + 1) | ∆P = 0} and HC k (n + 1) = {P ∈PC k (n + 1) | ∆P = 0}. Harmonic polynomials are sometimes called solid harmonics. Finally, let Hk(Sn) (resp. HC k (Sn)) denote the space of (real) spherical harmonics (resp. complex spherical harmonics) be the set of restrictions of harmonic polynomials in Hk(n + 1) to Sn (resp. restrictions of harmonic polynomials in HC k (n + 1) to Sn). Definition 8.15. A function f : Rn →R (resp. f : Rn →C) is homogeneous of degree k iff f(tx) = tkf(x), for all x ∈Rn and t > 0. The restriction map ρ: Hk(n + 1) →Hk(Sn) is a surjective linear map. In fact, it is a bijection. Indeed, if P ∈Hk(n + 1), observe that P(x) = ∥x∥k P  x ∥x∥  , with x ∥x∥∈Sn, for all x ̸= 0. Consequently, if P ↾Sn = Q ↾Sn, that is P(σ) = Q(σ) for all σ ∈Sn, then P(x) = ∥x∥k P  x ∥x∥  = ∥x∥k Q  x ∥x∥  = Q(x) for all x ̸= 0, which implies P = Q (as P and Q are polynomials). Therefore, we have a linear isomorphism between Hk(n + 1) and Hk(Sn) (and between HC k (n + 1) and HC k (Sn)). It will be convenient to introduce some notation to deal with homogeneous polynomials. Given n ≥1 variables x1, . . . , xn, and any n-tuple of nonnegative integers α = (α1, . . . , αn), let |α| = α1+· · ·+αn, let xα = xα1 1 · · · xαn n , and let α! = α1! · · · αn!. Then every homogeneous polynomial P of degree k in the variables x1, . . . , xn can be written uniquely as P = X |α|=k cαxα, with cα ∈R or cα ∈C. It is well known that Pk(n) is a (real) vector space of dimension dk = n + k −1 k  8.5. HARMONIC POLYNOMIALS, SPHERICAL HARMONICS AND L2(Sn) 321 and PC k (n) is a complex vector space of the same dimension dk. For example, P2(3) is a vector space of dimension 6 with basis {x1x2, x1x3, x2x3, x2 1, x2 2, x2 3}. We can define an Hermitian inner product on PC k (n) whose restriction to Pk(n) is an inner product by viewing a homogeneous polynomial as a differential operator as follows. Definition 8.16. For every P = P |α|=k cαxα ∈PC k (n), let ∂(P) = X |α|=k cα ∂k ∂xα1 1 · · · ∂xαn n . Then for any two polynomials P, Q ∈PC k (n), let ⟨P, Q⟩= ∂(P)Q. Observe that ⟨xα, xβ⟩= 0 unless α = β, in which case we have ⟨xα, xα⟩= α!. For example, in P2(3), if xα = x2 1 and xβ = x1x2, then ⟨x2 1, x1x2⟩= ∂2 dx2 1 x1x2 = 0, while ⟨x2 1, x2 1⟩= ∂2 dx2 1 x2 1 = 2!. Then a simple computation shows that X |α|=k aαxα, X |α|=k bαxα + = X |α|=k α! aαbα. Therefore, ⟨P, Q⟩is indeed an inner product. Also observe that ∂(x2 1 + · · · + x2 n) = ∂2 ∂x2 1 + · · · + ∂2 ∂x2 n = ∆. Another useful property of our inner product is this: For P ∈PC k (n), Q ∈PC j (n), and R ∈PC k−j(n), ⟨P, QR⟩= ⟨∂(Q)P, R⟩. Indeed. ⟨P, QR⟩ = ⟨QR, P⟩ = ∂(QR)P = ∂(Q)(∂(R)P) = ∂(R)(∂(Q)P) = ⟨R, ∂(Q)P⟩ = ⟨∂(Q)P, R⟩. 322 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS In particular, ⟨(x2 1 + · · · + x2 n)P, Q⟩= ⟨P, ∂(x2 1 + · · · + x2 n)Q⟩= ⟨P, ∆Q⟩. Let us write ∥x∥2 for x2 1 + · · · + x2 n. Using our inner product, we can prove the following important theorem. Theorem 8.13. The map ∆: Pk(n) →Pk−2(n) is surjective for all n, k ≥2 (and simi-larly for ∆: PC k (n) →PC k−2(n)). Furthermore, we have the following orthogonal direct sum decompositions: Pk(n) = Hk(n) ⊕∥x∥2 Hk−2(n) ⊕· · · ⊕∥x∥2j Hk−2j(n) ⊕· · · ⊕∥x∥2[k/2] Hk/2 and PC k (n) = HC k (n) ⊕∥x∥2 HC k−2(n) ⊕· · · ⊕∥x∥2j HC k−2j(n) ⊕· · · ⊕∥x∥2[k/2] HC k/2, with the understanding that only the first term occurs on the right-hand side when k < 2. Proof. If the map ∆: PC k (n) →PC k−2(n) is not surjective, then some nonzero polynomial Q ∈PC k−2(n) is orthogonal to the image of ∆, i.e. = ⟨Q, ∆P⟩. Since P = ∥x∥2 Q ∈PC k (n), and 0 = ⟨Q, ∆P⟩, a fact established earlier shows that 0 = ⟨Q, ∆P⟩= ⟨∥x∥2 Q, P⟩= ⟨P, P⟩, which implies that P = ∥x∥2 Q = 0, and thus Q = 0, a contradiction. The same proof is valid in the real case. We claim that we have an orthogonal direct sum decomposition PC k (n) = HC k (n) ⊕∥x∥2 PC k−2(n), and similarly in the real case, with the understanding that the second term is missing if k < 2. If k = 0, 1, then PC k (n) = HC k (n), so this case is trivial. Assume k ≥2. Since Ker ∆= HC k (n) and ∆is surjective, dim(PC k (n)) = dim(HC k (n)) + dim(PC k−2(n)), so it is sufficient to prove that HC k (n) is orthogonal to ∥x∥2 PC k−2(n). Now, if H ∈HC k (n) and P = ∥x∥2 Q ∈ ∥x∥2 PC k−2(n), we have ⟨∥x∥2 Q, H⟩= ⟨Q, ∆H⟩= 0, so HC k (n) and ∥x∥2 PC k−2(n) are indeed orthogonal. Using induction, we immediately get the orthogonal direct sum decomposition PC k (n) = HC k (n) ⊕∥x∥2 HC k−2(n) ⊕· · · ⊕∥x∥2j HC k−2j(n) ⊕· · · ⊕∥x∥2[k/2] HC k/2 and the corresponding real version. 8.5. HARMONIC POLYNOMIALS, SPHERICAL HARMONICS AND L2(Sn) 323 Remark: Theorem 8.13 also holds for n = 1. Theorem 8.13 has some important corollaries. Since every polynomial in n + 1 variables is the sum of homogeneous polynomials, we get Corollary 8.14. The restriction to Sn of every polynomial (resp. complex polynomial) in n + 1 ≥2 variables is a sum of restrictions to Sn of harmonic polynomials (resp. complex harmonic polynomials). We can also derive a formula for the dimension of Hk(n) (and HC k (n)). Corollary 8.15. The dimension ak,n of the space of harmonic polynomials Hk(n) is given by the formula ak,n = n + k −1 k  − n + k −3 k −2  if n, k ≥2, with a0,n = 1 and a1,n = n, and similarly for HC k (n). As Hk(n + 1) is isomorphic to Hk(Sn) (and HC k (n + 1) is isomorphic to HC k (Sn)) we have dim(HC k (Sn)) = dim(Hk(Sn)) = ak,n+1 = n + k k  − n + k −2 k −2  . Proof. The cases k = 0 and k = 1 are trivial, since in this case Hk(n) = Pk(n). For k ≥2, the result follows from the direct sum decomposition Pk(n) = Hk(n) ⊕∥x∥2 Pk−2(n) proved earlier. The proof is identical in the complex case. Observe that when n = 2, we get ak,2 = 2 for k ≥1, and when n = 3, we get ak,3 = 2k+1 for all k ≥0, which we already knew from Section 8.3. The formula even applies for n = 1 and yields ak,1 = 0 for k ≥2. Remark: It is easy to show that ak,n+1 = n + k −1 n −1  + n + k −2 n −1  for k ≥2; see Morimoto (Chapter 2, Theorem 2.4) or Dieudonn´ e (Chapter 7, Formula 99), where a different proof technique is used. Definition 8.17. Let L2(Sn) be the space of (real) square-integrable functions on the sphere Sn. We have an inner product on L2(Sn) given by ⟨f, g⟩Sn = Z Sn fg VolSn, 324 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS where f, g ∈L2(Sn) and where VolSn is the volume form on Sn (induced by the metric on Rn+1). With this inner product, L2(Sn) is a complete normed vector space using the norm ∥f∥= ∥f∥2 = p ⟨f, f⟩associated with this inner product; that is, L2(Sn) is a Hilbert space. In the case of complex-valued functions, we use the Hermitian inner product ⟨f, g⟩Sn = Z Sn f g VolSn, and we get the complex Hilbert space L2 C(Sn) (see Section 7.8 for the definition of the integral of a complex-valued function). We also denote by C(Sn) the space of continuous (real) functions on Sn with the L∞ norm; that is, ∥f∥∞= sup{|f(x)|}x∈Sn, and by CC(Sn) the space of continuous complex-valued functions on Sn also with the L∞ norm. Recall that C(Sn) is dense in L2(Sn) (and CC(Sn) is dense in L2 C(Sn)); see Rudin (Chapter 3). The following proposition shows why the spherical harmonics play an important role. Proposition 8.16. The set of all finite linear combinations of elements in S∞ k=0 Hk(Sn) (resp. S∞ k=0 HC k (Sn)) is (i) dense in C(Sn) (resp. in CC(Sn)) with respect to the L∞-norm; (ii) dense in L2(Sn) (resp. dense in L2 C(Sn)). Proof. (i) As Sn is compact, by the Stone-Weierstrass approximation theorem (Lang , Chapter III, Corollary 1.3), if g is continuous on Sn, then it can be approximated uniformly by polynomials Pj restricted to Sn. By Corollary 8.14, the restriction of each Pj to Sn is a linear combination of elements in S∞ k=0 Hk(Sn). (ii) We use the fact that C(Sn) is dense in L2(Sn). Given f ∈L2(Sn), for every ϵ > 0, we can choose a continuous function g so that ∥f −g∥2 < ϵ/2. By (i), we can find a linear combination h of elements in S∞ k=0 Hk(Sn) so that ∥g −h∥∞< ϵ/(2 p vol(Sn)), where vol(Sn) is the volume of Sn (really, area). Thus we get ∥f −h∥2 ≤∥f −g∥2 + ∥g −h∥2 < ϵ/2 + p vol(Sn) ∥g −h∥∞< ϵ/2 + ϵ/2 = ϵ, which proves (ii). The proof in the complex case is identical. We need one more proposition before showing that the spaces Hk(Sn) constitute an orthogonal Hilbert space decomposition of L2(Sn). Proposition 8.17. For every harmonic polynomial P ∈Hk(n + 1) (resp. P ∈HC k (n + 1)), the restriction H ∈Hk(Sn) (resp. H ∈HC k (Sn)) of P to Sn is an eigenfunction of ∆Sn for the eigenvalue −k(n + k −1). 8.5. HARMONIC POLYNOMIALS, SPHERICAL HARMONICS AND L2(Sn) 325 Proof. We have P(rσ) = rkH(σ), r > 0, σ ∈Sn, and by Proposition 8.10, for any f ∈C∞(Rn+1), we have ∆f = 1 rn ∂ ∂r  rn∂f ∂r  + 1 r2∆Snf. Consequently, ∆P = ∆(rkH) = 1 rn ∂ ∂r  rn∂(rkH) ∂r  + 1 r2∆Sn(rkH) = 1 rn ∂ ∂r krn+k−1H  + rk−2∆SnH = 1 rnk(n + k −1)rn+k−2H + rk−2∆SnH = rk−2(k(n + k −1)H + ∆SnH). Thus, ∆P = 0 iff ∆SnH = −k(n + k −1)H, as claimed. From Proposition 8.17, we deduce that the space Hk(Sn) is a subspace of the eigenspace Ek of ∆Sn associated with the eigenvalue −k(n + k −1) (and similarly for HC k (Sn)). Re-markably, Ek = Hk(Sn), but it will take more work to prove this. What we can deduce immediately is that Hk(Sn) and Hl(Sn) are pairwise orthogonal whenever k ̸= l. This is because, by Proposition 8.12, the Laplacian is self-adjoint, and thus any two eigenspaces Ek and El are pairwise orthogonal whenever k ̸= l, and as Hk(Sn) ⊆ Ek and Hl(Sn) ⊆El, our claim is indeed true. Furthermore, by Proposition 8.15, each Hk(Sn) is finite-dimensional, and thus closed. Finally, we know from Proposition 8.16 that S∞ k=0 Hk(Sn) is dense in L2(Sn). But then we can apply a standard result from Hilbert space theory (for example, see Lang , Chapter V, Proposition 1.9) to deduce the following important result. Theorem 8.18. The family of spaces Hk(Sn) (resp. HC k (Sn)) yields a Hilbert space direct sum decomposition L2(Sn) = ∞ M k=0 Hk(Sn) (resp. L2 C(Sn) = ∞ M k=0 HC k (Sn)), which means that the summands are closed, pairwise orthogonal, and that every f ∈L2(Sn) (resp. f ∈L2 C(Sn)) is the sum of a converging series f = ∞ X k=0 fk 326 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS in the L2-norm, where the fk ∈Hk(Sn) (resp. fk ∈HC k (Sn)) are uniquely determined functions. Furthermore, given any orthonormal basis (Y 1 k , . . . , Y ak,n+1 k ) of Hk(Sn), we have fk = ak,n+1 X mk=1 ck,mkY mk k , with ck,mk = ⟨f, Y mk k ⟩Sn. The coefficients ck,mk are “generalized” Fourier coefficients with respect to the Hilbert basis {Y mk k | 1 ≤mk ≤ak,n+1, k ≥0}. We can finally prove the main theorem of this section. Theorem 8.19. (1) The eigenspaces (resp. complex eigenspaces) of the Laplacian ∆Sn on Sn are the spaces of spherical harmonics Ek = Hk(Sn) (resp. Ek = HC k (Sn)), and Ek corresponds to the eigenvalue −k(n + k −1). (2) We have the Hilbert space direct sum decompositions L2(Sn) = ∞ M k=0 Ek (resp. L2 C(Sn) = ∞ M k=0 Ek). (3) The complex polynomials of the form (c1x1 + · · · + cn+1xn+1)k, with c2 1 + · · · + c2 n+1 = 0, span the space HC k (n + 1) ∼ = HC k (Sn), for k ≥1. Proof. We follow essentially the proof in Helgason (Introduction, Theorem 3.1). In (1) and (2) we only deal with the real case, the proof in the complex case being identical. (1) We already know that the integers −k(n + k −1) are eigenvalues of ∆Sn and that Hk(Sn) ⊆Ek. We will prove that ∆Sn has no other eigenvalues and no other eigenvectors using the Hilbert basis {Y mk k | 1 ≤mk ≤ak,n+1, k ≥0} given by Theorem 8.18. Let λ be any eigenvalue of ∆Sn and let f ∈L2(Sn) be any eigenfunction associated with λ so that ∆f = ∆Snf = λ f. We have a unique series expansion f = ∞ X k=0 ak,n+1 X mk=1 ck,mkY mk k , with ck,mk = ⟨f, Y mk k ⟩Sn. Now, as ∆Sn is self-adjoint and ∆SnY mk k = −k(n + k −1)Y mk k , the Fourier coefficients dk,mk of ∆f are given by dk,mk = ⟨∆Snf, Y mk k ⟩Sn = ⟨f, ∆SnY mk k ⟩Sn = −k(n + k −1)⟨f, Y mk k ⟩Sn = −k(n + k −1)ck,mk. 8.5. HARMONIC POLYNOMIALS, SPHERICAL HARMONICS AND L2(Sn) 327 On the other hand, as ∆f = λ f, the Fourier coefficients of ∆f are given by dk,mk = λck,mk. By uniqueness of the Fourier expansion, we must have λck,mk = −k(n + k −1)ck,mk for all k ≥0. Since f ̸= 0, there some k such that ck,mk ̸= 0, and we must have λ = −k(n + k −1) for any such k. However, the function k 7→−k(n+k−1) reaches its maximum for k = −n−1 2 , and as n ≥1, it is strictly decreasing for k ≥0, which implies that k is unique and that cj,mj = 0 for all j ̸= k. Therefore f ∈Hk(Sn), and the eigenvalues of ∆Sn are exactly the integers −k(n + k −1), so Ek = Hk(Sn) as claimed. Since we just proved that Ek = Hk(Sn), (2) follows immediately from the Hilbert decom-position given by Theorem 8.18. (3) If H = (c1x1 + · · · + cn+1xn+1)k, with c2 1 + · · · + c2 n+1 = 0, then for k ≤1 it is obvious that ∆H = 0, and for k ≥2 we have ∆H = k(k −1)(c2 1 + · · · + c2 n+1)(c1x1 + · · · + cn+1xn+1)k−2 = 0, so H ∈HC k (n + 1). A simple computation shows that for every Q ∈PC k (n + 1), if c = (c1, . . . , cn+1), then we have ∂(Q)(c1x1 + · · · + cn+1xn+1)m = m(m −1) · · · (m −k + 1)Q(c)(c1x1 + · · · + cn+1xn+1)m−k, for all m ≥k ≥1. Assume that HC k (n + 1) is not spanned by the complex polynomials of the form (c1x1 + · · ·+cn+1xn+1)k, with c2 1 +· · ·+c2 n+1 = 0, for k ≥1. Then some Q ∈HC k (n+1) is orthogonal to all polynomials of the form H = (c1x1 + · · · + cn+1xn+1)k, with c2 1 + · · · + c2 n+1 = 0. Recall that ⟨P, ∂(Q)H⟩= ⟨QP, H⟩ and apply this equation to P = Q(c), H and Q. Since ∂(Q)H = ∂(Q)(c1x1 + · · · + cn+1xn+1)k = k!Q(c), and as Q is orthogonal to H, we get k!⟨Q(c), Q(c)⟩= ⟨Q(c), k!Q(c)⟩= ⟨Q(c), ∂(Q)H⟩= ⟨Q Q(c), H⟩= Q(c)⟨Q, H⟩= 0, 328 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS which implies Q(c) = 0. Consequently, Q(x1, . . . , xn+1) vanishes on the complex algebraic variety {(x1, . . . , xn+1) ∈Cn+1 | x2 1 + · · · + x2 n+1 = 0}. By the Hilbert Nullstellensatz, some power Qm belongs to the ideal (x2 1+· · ·+x2 n+1) generated by x2 1 + · · · + x2 n+1. Now, if n ≥2, it is well-known that the polynomial x2 1 + · · · + x2 n+1 is irreducible so the ideal (x2 1 + · · · + x2 n+1) is a prime ideal, and thus Q is divisible by x2 1+· · ·+x2 n+1. However, we know from the proof of Theorem 8.13 that we have an orthogonal direct sum PC k (n + 1) = HC k (n + 1) ⊕∥x∥2 PC k−2(n + 1). Since Q ∈HC k (n + 1) and Q is divisible by x2 1 + · · · + x2 n+1 , we must have Q = 0. Therefore, if n ≥2, we proved (3). However, when n = 1, we know from Section 8.2 that the complex harmonic homogeneous polynomials in two variables P(x, y) are spanned by the real and imaginary parts Uk, Vk of the polynomial (x + iy)k = Uk + iVk. Since (x −iy)k = Uk −iVk we see that Uk = 1 2 (x + iy)k + (x −iy)k , Vk = 1 2i (x + iy)k −(x −iy)k , and as 1 + i2 = 1 + (−i)2 = 0, the space HC k (R2) is spanned by (x + iy)k and (x −iy)k (for k ≥1), so (3) holds for n = 1 as well. As an illustration of Part (3) of Theorem 8.19, the polynomials (x1 +i cos θx2 +i sin θx3)k are harmonic. Of course, the real and imaginary part of a complex harmonic polynomial (c1x1 + · · · + cn+1xn+1)k are real harmonic polynomials. 8.6 Zonal Spherical Functions and Gegenbauer Polynomials In this section we describe the zonal spherical functions Zτ k on Sn and show that they essentially come from certain polynomials generalizing the Legendre polynomials known as the Gegenbauer polynomials. An interesting property of the zonal spherical functions is a formula for obtaining the kth spherical harmonic component of a function f ∈L2 C(Sn); see Proposition 8.27. Another important property of the zonal spherical functions Zτ k is that they generate HC k (Sn). Most proofs will be omitted. We refer the reader to Stein and Weiss (Chapter 4) and Morimoto (Chapter 2) for a complete exposition with proofs. In order to define zonal spherical functions we will need the following proposition. Proposition 8.20. If P is any (complex) polynomial in n variables such that P(R(x)) = P(x) for all rotations R ∈SO(n), and all x ∈Rn, 8.6. ZONAL SPHERICAL FUNCTIONS AND GEGENBAUER POLYNOMIALS 329 then P is of the form P(x) = m X j=0 cj(x2 1 + · · · + x2 n)j, for some c0, . . . , cm ∈C. Proof. Write P as the sum of its homogeneous pieces P = Pk l=0 Ql, where Ql is homogeneous of degree l. For every ϵ > 0 and every rotation R, we have k X l=0 ϵlQl(x) = P(ϵx) = P(R(ϵx)) = P(ϵR(x)) = k X l=0 ϵlQl(R(x)), which implies that Ql(R(x)) = Ql(x), l = 0, . . . , k. If we let Fl(x) = ∥x∥−l Ql(x), then Fl is a homogeneous function of degree 0 since Fl(tx) = ∥tx∥−l Ql(tx) = t−l ∥x∥tlQl(x) = Fl(x). Furthermore, Fl is invariant under all rotations since Fl(R(x)) = ∥R(x)∥−l Ql(R(x)) = ∥x∥−l Ql(x) = Fl(x). This is only possible if Fl is a constant function, thus Fl(x) = al for all x ∈Rn. But then, Ql(x) = al ∥x∥l. Since Ql is a polynomial, l must be even whenever al ̸= 0. It follows that P(x) = m X j=0 cj ∥x∥2j with cj = a2j for j = 0, . . . , m, and where m is the largest integer ≤k/2. Proposition 8.20 implies that if a polynomial function on the sphere Sn, in particular a spherical harmonic, is invariant under all rotations, then it is a constant. If we relax this condition to invariance under all rotations leaving some given point τ ∈Sn invariant, then we obtain zonal harmonics. The following theorem from Morimoto (Chapter 2, Theorem 2.24) gives the relation-ship between zonal harmonics and the Gegenbauer polynomials: Theorem 8.21. Fix any τ ∈Sn. For every constant c ∈C, there is a unique homogeneous harmonic polynomial Zτ k ∈HC k (n + 1) satisfying the following conditions: (1) Zτ k(τ) = c; (2) For every rotation R ∈SO(n+1), if Rτ = τ, then Zτ k(R(x)) = Zτ k(x) for all x ∈Rn+1. 330 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Furthermore, we have Zτ k(x) = c ∥x∥k Pk,n  x ∥x∥· τ  , for some polynomial Pk,n(t) of degree k. Remark: The proof given in Morimoto is essentially the same as the proof of Theorem 2.12 in Stein and Weiss (Chapter 4), but Morimoto makes an implicit use of Proposition 8.20 above. Also, Morimoto states Theorem 8.21 only for c = 1, but the proof goes through for any c ∈C, including c = 0, and we will need this extra generality in the proof of the Funk-Hecke formula. Proof. Let en+1 = (0, . . . , 0, 1) ∈Rn+1, and for any τ ∈Sn, let Rτ be some rotation such that Rτ(en+1) = τ. Assume Z ∈HC k (n + 1) satisfies Conditions (1) and (2), and let Z′ be given by Z′(x) = Z(Rτ(x)). As Rτ(en+1) = τ, we have Z′(en+1) = Z(τ) = c. Furthermore, for any rotation S such that S(en+1) = en+1, observe that Rτ ◦S ◦R−1 τ (τ) = Rτ ◦S(en+1) = Rτ(en+1) = τ, and so, as Z satisfies property (2) for the rotation Rτ ◦S ◦R−1 τ , we get Z′(S(x)) = Z(Rτ ◦S(x)) = Z(Rτ ◦S ◦R−1 τ ◦Rτ(x)) = Z(Rτ(x)) = Z′(x), which proves that Z′ is a harmonic polynomial satisfying Properties (1) and (2) with respect to en+1. Therefore, we may assume that τ = en+1. Write Z(x) = k X j=0 xk−j n+1Pj(x1, . . . , xn), where Pj(x1, . . . , xn) is a homogeneous polynomial of degree j. Since Z is invariant under every rotation R fixing en+1, and since the monomials xk−j n+1 are clearly invariant under such a rotation, we deduce that every Pj(x1, . . . , xn) is invariant under all rotations of Rn (clearly, there is a one-to-one correspondence between the rotations of Rn+1 fixing en+1 and the rotations of Rn). By Proposition 8.20, we conclude that Pj(x1, . . . , xn) = cj(x2 1 + · · · + x2 n) j 2, which implies that Pj = 0 if j is odd. Thus we can write Z(x) = [k/2] X i=0 cixk−2i n+1 (x2 1 + · · · + x2 n)i, (†) 8.6. ZONAL SPHERICAL FUNCTIONS AND GEGENBAUER POLYNOMIALS 331 where [k/2] is the greatest integer m such that 2m ≤k. If k < 2, then Z(x) = c0, so c0 = c and Z is uniquely determined. If k ≥2, we know that Z is a harmonic polynomial so we assert that ∆Z = 0. For i ≤j ≤n, ∂ ∂xj (x2 1 + · · · + x2 j + · · · x2 n)i = 2ixj(x2 1 + · · · + x2 n)i−1, and ∂2 ∂x2 j (x2 1 + · · · + x2 j + · · · + x2 n)i = 2i(x2 1 + · · · x2 n)i−1 + 4x2 ji(i −1)(x2 1 + · · · + x2 n)i−2 = 2i(x2 1 + · · · x2 n)i−2[x2 1 + · · · + x2 n + 2(i −1)x2 j]. Since ∆(x2 1 + · · · + x2 n)i = Pn j=1 ∂2 ∂x2 j (x2 1 + · · · + x2 j + · · · + x2 n)i, we find that ∆(x2 1 + · · · + x2 n)i = 2i(x2 1 + · · · + x2 n)i−2 n X j=1 [x2 1 + · · · + x2 n + 2(i −1)x2 j] = 2i(x2 1 + · · · + x2 n)i−2 " n(x2 1 + · · · + x2 n) + 2(i −1) n X j=1 x2 j # = 2i(x2 1 + · · · + x2 n)i−2[n(x2 1 + · · · + x2 n) + 2(i −1)(x2 1 + · · · + x2 n)] = 2i(n + 2i −2)(x2 1 + · · · + x2 n)i−1. Thus ∆xk−2i n+1 (x2 1 + · · · + x2 n)i = (k −2i)(k −2i −1)xk−2i−2 n+1 (x2 1 + · · · + x2 n)i + xk−2i n+1 ∆(x2 1 + · · · + x2 n)i = (k −2i)(k −2i −1)xk−2i−2 n+1 (x2 1 + · · · + x2 n)i + 2i(n + 2i −2)xk−2i n+1 (x2 1 + · · · + x2 n)i−1, and so we get ∆Z = [k/2]−1 X i=0 ((k −2i)(k −2i −1)ci + 2(i + 1)(n + 2i)ci+1) xk−2i−2 n+1 (x2 1 + · · · + x2 n)i. Then ∆Z = 0 yields the relations 2(i + 1)(n + 2i)ci+1 = −(k −2i)(k −2i −1)ci, i = 0, . . . , [k/2] −1, (††) which shows that Z is uniquely determined up to the constant c0. Since we are requiring Z(en+1) = c, we get c0 = c, and Z is uniquely determined. Now on Sn we have x2 1 + · · · + x2 n+1 = 1, so if we let t = xn+1, for c0 = 1, we get a polynomial in one variable Pk,n(t) = [k/2] X i=0 citk−2i(1 −t2)i. (∗) 332 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Thus we proved that when Z(en+1) = c, we have Z(x) = c ∥x∥k Pk,n xn+1 ∥x∥  = c ∥x∥k Pk,n  x ∥x∥· en+1  . When Z(τ) = c, we write Z = Z′ ◦R−1 τ with Z′ = Z ◦Rτ and where Rτ is a rotation such that Rτ(en+1) = τ. Then, as Z′(en+1) = c, using the formula above for Z′, we have Z(x) = Z′(R−1 τ (x)) = c R−1 τ (x) k Pk,n  R−1 τ (x) ∥R−1 τ (x)∥· en+1  = c ∥x∥k Pk,n  x ∥x∥· Rτ(en+1)  = c ∥x∥k Pk,n  x ∥x∥· τ  , since Rτ is an isometry. To best understand the proof of Theorem 8.21, we let n = 2, k = 3, and construct Z(x) ∈HC 3 (3) such that Z satisfies Conditions (1) and (2) with τ = e3. Line (†) implies that Z(x) = c0x3 3 + c1x3(x2 1 + x2 2). The conditions of Line (††) show that c1 = −3/2c0. Hence Z(x) = cx3 3 −3 2cx3(x2 1 + x2 2), where we let c = c0. We want to rewrite Z(x) via P3,2(t), where P3,2(t) is given by Line (∗) as P3,2(t) = t3 −3 2t(1 −t2). Then a simple verification shows that Z(x) = c ∥x∥3 P3,2  x3 ∥x∥  . Definition 8.18. The function, Zτ k, is called a zonal function and its restriction to Sn is a zonal spherical function. The polynomial Pk,n(t) is called the Gegenbauer polynomial of degree k and dimension n + 1 or ultraspherical polynomial. By definition, Pk,n(1) = 1. The proof of Theorem 8.21 shows that for k even, say k = 2m, the polynomial P2m,n is of the form P2m,n(t) = m X j=0 cm−jt2j(1 −t2)m−j, 8.6. ZONAL SPHERICAL FUNCTIONS AND GEGENBAUER POLYNOMIALS 333 and for k odd, say k = 2m + 1, the polynomial P2m+1,n is of the form P2m+1,n(t) = m X j=0 cm−jt2j+1(1 −t2)m−j. Consequently, Pk,n(−t) = (−1)kPk,n(t), for all k ≥0. The proof also shows that the “natural basis” for these polynomials consists of the polynomials, ti(1−t2) k−i 2 , with k−i even. Indeed, with this basis, there are simple recurrence equations for computing the coefficients of Pk,n(t). Remark: Morimoto calls the polynomials Pk,n(t) “Legendre polynomials.” For n = 2, they are indeed the Legendre polynomials. Stein and Weiss denotes our (and Morimoto’s) Pk,n(t) by P n−1 2 k (t) (up to a constant factor), and Dieudonn´ e (Chapter 7) by Gk,n+1(t). When n = 2, using the notation of Section 8.3, the zonal spherical functions on S2 are the spherical harmonics y0 l for which m = 0; that is (up to a constant factor), y0 l (θ, ϕ) = r (2l + 1) 4π Pl(cos θ), where Pl is the Legendre polynomial of degree l. For example, for l = 2, Pl(t) = 1 2(3t2 −1). Zonal spherical functions have many important properties. One such property is associ-ated with the reproducing kernel of HC k (Sn). Definition 8.19. Let HC k (Sn) be the space of spherical harmonics. Let ak,n+1 be the dimen-sion of HC k (Sn) where ak,n+1 = n + k k  − n + k −2 k −2  , if n ≥1 and k ≥2, with a0,n+1 = 1 and a1,n+1 = n + 1. Let (Y 1 k , . . . , Y ak,n+1 k ) be any orthonormal basis of HC k (Sn), and define Fk(σ, τ) by Fk(σ, τ) = ak,n+1 X i=1 Y i k(σ)Y i k(τ), σ, τ ∈Sn. The function Fk(σ, τ) is the reproducing kernel of HC k (Sn). The following proposition is easy to prove (see Morimoto , Chapter 2, Lemma 1.19 and Lemma 2.20). Proposition 8.22. The function Fk is independent of the choice of orthonormal basis. Fur-thermore, for every orthogonal transformation R ∈O(n + 1), we have Fk(Rσ, Rτ) = Fk(σ, τ), σ, τ ∈Sn. 334 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Clearly, Fk is a symmetric function. Since we can pick an orthonormal basis of real orthogonal functions for HC k (Sn) (pick a basis of Hk(Sn)), Proposition 8.22 shows that Fk is a real-valued function. The function Fk satisfies the following property which justifies its name as the reproducing kernel for HC k (Sn): Remark: In the proofs below, integration is performed with respect to the repeated variable. Proposition 8.23. For every spherical harmonic H ∈HC j (Sn), we have Z Sn H(τ)Fk(σ, τ) VolSn = δj kH(σ), σ, τ ∈Sn, for all j, k ≥0. Proof. When j ̸= k, since HC k (Sn) and HC j (Sn) are orthogonal and since Fk(σ, τ) = Pak,n+1 i=1 Y i k(σ)Y i k(τ), it is clear that the integral in Proposition 8.23 vanishes. When j = k, we have Z Sn H(τ)Fk(σ, τ) VolSn = Z Sn H(τ) ak,n+1 X i=1 Y i k(σ)Y i k(τ) VolSn = ak,n+1 X i=1 Y i k(σ) Z Sn H(τ)Y i k(τ) VolSn = ak,n+1 X i=1 Y i k(σ) ⟨H, Y i k⟩ = H(σ), since (Y 1 k , . . . , Y ak,n+1 k ) is an orthonormal basis. Remark: In Stein and Weiss (Chapter 4), the function Fk(σ, τ) is denoted by Z(k) σ (τ) and it is called the zonal harmonic of degree k with pole σ. Before we investigate the relationship between Fk(σ, τ) and Zτ k(σ), we need two technical propositions. Both are proven in Morimoto . The first, Morimoto (Chapter 2, Lemma 2.21), is needed to prove the second, Morimoto (Chapter 2, Lemma 2.23). Proposition 8.24. For all σ, τ, σ′, τ ′ ∈Sn, with n ≥1, the following two conditions are equivalent: (i) There is some orthogonal transformation R ∈O(n + 1) such that R(σ) = σ′ and R(τ) = τ ′. 8.6. ZONAL SPHERICAL FUNCTIONS AND GEGENBAUER POLYNOMIALS 335 (ii) σ · τ = σ′ · τ ′. Propositions 8.22 and 8.24 immediately yield Proposition 8.25. For all σ, τ, σ′, τ ′ ∈Sn, if σ · τ = σ′ · τ ′, then Fk(σ, τ) = Fk(σ′, τ ′). Consequently, there is some function ϕ: R →R such that Fk(σ, τ) = ϕ(σ · τ). We claim that the ϕ(σ · τ) of Proposition 8.25 is a zonal spherical function Zτ k(σ). To see why this is true, define Z(rkσ) := rkFk(σ, τ) for a fixed τ. By the definition of Fk(σ, τ), it is clear that Z is a homogeneous harmonic polynomial. The value Fk(τ, τ) does not depend of τ, because by transitivity of the action of SO(n + 1) on Sn, for any other σ ∈Sn, there is some rotation R so that Rτ = σ, and by Proposition 8.22, we have Fk(σ, σ) = Fk(Rτ, Rτ) = Fk(τ, τ). To compute Fk(τ, τ), since Fk(τ, τ) = ak,n+1 X i=1 Y i k(τ) 2 , and since (Y 1 k , . . . , Y ak,n+1 k ) is an orthonormal basis of HC k (Sn), observe that ak,n+1 = ak,n+1 X i=1 ⟨Y i k, Y i k⟩ = ak,n+1 X i=1 Z Sn Y i k(τ) 2 VolSn = Z Sn ak,n+1 X i=1 Y i k(τ) 2 ! VolSn = Z Sn Fk(τ, τ) VolSn = Fk(τ, τ) vol(Sn). Therefore, Fk(τ, τ) = ak,n+1 vol(Sn).  Beware that Morimoto uses the normalized measure on Sn, so the factor involving vol(Sn) does not appear. Remark: The volume of the n-sphere is given by vol(S2d) = 2d+1πd 1 · 3 · · · (2d −1) if d ≥1 and vol(S2d+1) = 2πd+1 d! if d ≥0. 336 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS These formulae will be proved in Section 8.8 just after the proof of Theorem 8.36. Now, if Rτ = τ, Proposition 8.22 shows that Z(R(rkσ)) = Z(rkR(σ)) = rkFk(Rσ, τ) = rkFk(Rσ, Rτ) = rkFk(σ, τ) = Z(rkσ). Therefore, the function Z satisfies Conditions (1) and (2) of Theorem 8.21 with c = ak,n+1 vol(Sn), and by uniqueness, we conclude that Z is the zonal function Zτ k whose restriction to Sn is the zonal spherical function Fk(σ, τ) = ak,n+1 vol(Sn) Pk,n(σ · τ). Consequently, we have obtained the so-called addition formula: Proposition 8.26. (Addition Formula) If (Y 1 k , . . . , Y ak,n+1 k ) is any orthonormal basis of HC k (Sn), then Pk,n(σ · τ) = vol(Sn) ak,n+1 ak,n+1 X i=1 Y i k(σ)Y i k(τ). Again, beware that Morimoto does not have the factor vol(Sn). For n = 1, we can write σ = (cos θ, sin θ) and τ = (cos ϕ, sin ϕ), and it is easy to see that the addition formula reduces to Pk,1(cos(θ −ϕ)) = cos kθ cos kϕ + sin kθ sin kϕ = cos k(θ −ϕ), the standard addition formula for trigonometric functions. Proposition 8.26 implies that Pk,n(t) has real coefficients. Furthermore Proposition 8.23 is reformulated as ak,n+1 vol(Sn) Z Sn Pk,n(σ · τ)H(τ) VolSn = δj kH(σ), (rk) showing that the Gengenbauer polynomials are reproducing kernels. A neat application of this formula is a formula for obtaining the kth spherical harmonic component of a function f ∈L2 C(Sn). Proposition 8.27. For every function f ∈L2 C(Sn), if f = P∞ k=0 fk is the unique decompo-sition of f over the Hilbert sum L∞ k=0 HC k (Sk), then fk is given by fk(σ) = ak,n+1 vol(Sn) Z Sn f(τ)Pk,n(σ · τ) VolSn, for all σ ∈Sn. 8.6. ZONAL SPHERICAL FUNCTIONS AND GEGENBAUER POLYNOMIALS 337 Proof. If we recall that HC k (Sk) and HC j (Sk) are orthogonal for all j ̸= k, using the Formula (rk), we have ak,n+1 vol(Sn) Z Sn f(τ)Pk,n(σ · τ) VolSn = ak,n+1 vol(Sn) Z Sn ∞ X j=0 fj(τ)Pk,n(σ · τ) VolSn = ak,n+1 vol(Sn) ∞ X j=0 Z Sn fj(τ)Pk,n(σ · τ) VolSn = ak,n+1 vol(Sn) Z Sn fk(τ)Pk,n(σ · τ) VolSn = fk(σ), as claimed. Another important property of the zonal spherical functions Zτ k is that they generate HC k (Sn). In order to prove this fact, we use the following proposition. Proposition 8.28. If H1, . . . , Hm ∈HC k (Sn) are linearly independent, then there are m points σ1, . . . , σm on Sn so that the m × m matrix (Hj(σi)) is invertible. Proof. We proceed by induction on m. The case m = 1 is trivial. For the induction step, we may assume that we found m points σ1, . . . , σm on Sn so that the m × m matrix (Hj(σi)) is invertible. Consider the function σ 7→ H1(σ) . . . Hm(σ) Hm+1(σ) H1(σ1) . . . Hm(σ1) Hm+1(σ1) . . . ... . . . . . . H1(σm) . . . Hm(σm) Hm+1(σm) . Since H1, . . . , Hm+1 are linearly independent, the above function does not vanish for all σ, since otherwise, by expanding this determinant with respect to the first row, we would get a linear dependence among the Hj’s where the coefficient of Hm+1 is nonzero. Therefore, we can find σm+1 so that the (m + 1) × (m + 1) matrix (Hj(σi)) is invertible. Definition 8.20. We say that ak,n+1 points, σ1, . . . , σak,n+1 on Sn form a fundamental system iffthe ak,n+1 × ak,n+1 matrix (Pn,k(σi · σj)) is invertible. Theorem 8.29. The following properties hold: (i) There is a fundamental system σ1, . . . , σak,n+1 for every k ≥1. (ii) Every spherical harmonic H ∈HC k (Sn) can be written as H(σ) = ak,n+1 X j=1 cj Pk,n(σj · σ), for some unique cj ∈C. 338 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Proof. (i) By the addition formula, Pk,n(σi · σj) = vol(Sn) ak,n+1 ak,n+1 X l=1 Y l k(σi)Y l k(σj) for any orthonormal basis (Y 1 k , . . . , Y ak,n+1 k ). It follows that the matrix (Pk,n(σi · σj)) can be written as (Pk,n(σi · σj)) = vol(Sn) ak,n+1 Y Y ∗, where Y = (Y l k(σi)), and by Proposition 8.28, we can find σ1, . . . , σak,n+1 ∈Sn so that Y and thus also Y ∗are invertible, and so (Pn,k(σi · σj)) is invertible. (ii) Again, by the addition formula, Pk,n(σ · σj) = vol(Sn) ak,n+1 ak,n+1 X i=1 Y i k(σ)Y i k(σj). However, as (Y 1 k , . . . , Y ak,n+1 k ) is an orthonormal basis, Part (i) proved that the matrix Y ∗is invertible, so the Y i k(σ) can be expressed uniquely in terms of the Pk,n(σ·σj), as claimed. Statement (ii) of Theorem 8.29 shows that the set of Pk,n(σ·τ) = vol(Sn) ak,n+1 Fk(σ, τ) do indeed generate HC k (Sn). We end this section with a neat geometric characterization of the zonal spherical functions is given in Stein and Weiss . For this, we need to define the notion of a parallel on Sn. A parallel of Sn orthogonal to a point τ ∈Sn is the intersection of Sn with any (affine) hyperplane orthogonal to the line through the center of Sn and τ. See Figure 8.3 Clearly, any rotation R leaving τ fixed leaves every parallel orthogonal to τ globally invariant, and for any two points σ1 and σ2, on such a parallel, there is a rotation leaving τ fixed that maps σ1 to σ2. Consequently, the zonal function Zτ k defined by τ is constant on the parallels orthogonal to τ. In fact, this property characterizes zonal harmonics, up to a constant. The theorem below is proved in Stein and Weiss (Chapter 4, Theorem 2.12). The proof uses Proposition 8.20 and it is very similar to the proof of Theorem 8.21. To save space, it is omitted. Theorem 8.30. Fix any point τ ∈Sn. A spherical harmonic Y ∈HC k (Sn) is constant on parallels orthogonal to τ iffY = cZτ k for some constant c ∈C. In the next section we show how the Gegenbauer polynomials can actually be computed. 8.7. MORE ON THE GEGENBAUER POLYNOMIALS 339 τ Figure 8.3: The purple planes are parallels of S2 orthogonal to the red point τ. Any rotation around the red axis maps each parallel to itself. 8.7 More on the Gegenbauer Polynomials The Gegenbauer polynomials are characterized by a formula generalizing the Rodrigues formula defining the Legendre polynomials (see Section 8.3). The expression  k + n −2 2   k −1 + n −2 2  · · ·  1 + n −2 2  can be expressed in terms of the Γ function as Γ k + n 2  Γ n 2  . Recall that the Γ function is a generalization of factorial that satisfies the equation Γ(z + 1) = zΓ(z). For z = x + iy with x > 0, Γ(z) is given by Γ(z) = Z ∞ 0 tz−1e−t dt, where the integral converges absolutely. If n is an integer n ≥0, then Γ(n + 1) = n!. It is proved in Morimoto (Chapter 2, Theorem 2.35) that Proposition 8.31. The Gegenbauer polynomial Pk,n is given by Rodrigues’ formula: Pk,n(t) = (−1)k 2k Γ n 2  Γ k + n 2  1 (1 −t2) n−2 2 dk dtk (1 −t2)k+ n−2 2 , with n ≥2. 340 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS The Gegenbauer polynomials satisfy the following orthogonality properties with respect to the kernel (1 −t2) n−2 2 (see Morimoto (Chapter 2, Theorem 2.34): Proposition 8.32. The Gegenbauer polynomial Pk,n have the following properties: Z −1 −1 (Pk,n(t))2(1 −t2) n−2 2 dt = vol(Sn) ak,n+1vol(Sn−1) Z −1 −1 Pk,n(t)Pl,n(t)(1 −t2) n−2 2 dt = 0, k ̸= l. The Gegenbauer polynomials satisfy a second-order differential equation generalizing the Legendre equation from Section 8.3. Proposition 8.33. The Gegenbauer polynomial Pk,n are solutions of the differential equation (1 −t2)P ′′ k,n(t) −ntP ′ k,n(t) + k(k + n −1)Pk,n(t) = 0. Proof. If we let τ = en+1, then the function H given by H(σ) = Pk,n(σ · τ) = Pk,n(cos θ) belongs to HC k (Sn), so ∆SnH = −k(k + n −1)H. Recall from Section 8.4 that ∆Snf = 1 sinn−1 θ ∂ ∂θ  sinn−1 θ ∂f ∂θ  + 1 sin2 θ∆Sn−1f, in the local coordinates where σ = sin θ e σ + cos θ en+1, with e σ ∈Sn−1 and 0 ≤θ < π. If we make the change of variable t = cos θ, then it is easy to see that the above formula becomes ∆Snf = (1 −t2)∂2f ∂t2 −nt∂f ∂t + 1 1 −t2∆Sn−1f (see Morimoto , Chapter 2, Theorem 2.9.) But H being zonal, it only depends on θ, that is on t, so ∆Sn−1H = 0, and thus −k(k + n −1)Pk,n(t) = ∆SnPk,n(t) = (1 −t2)∂2Pk,n ∂t2 −nt∂Pk,n ∂t , which yields our equation. Note that for n = 2, the differential equation of Proposition 8.33 is the Legendre equation from Section 8.3. The Gegenbauer polynomials also appear as coefficients in some simple generating func-tions. The following proposition is proved in Morimoto (Chapter 2, Theorem 2.53 and Theorem 2.55): 8.8. THE FUNK–HECKE FORMULA 341 Proposition 8.34. For all r and t such that −1 < r < 1 and −1 ≤t ≤1, for all n ≥1, we have the following generating formula: ∞ X k=0 ak,n+1 rkPk,n(t) = 1 −r2 (1 −2rt + r2) n+1 2 . Furthermore, for all r and t such that 0 ≤r < 1 and −1 ≤t ≤1, if n = 1, then ∞ X k=1 rk k Pk,1(t) = −1 2 log(1 −2rt + r2), and if n ≥2, then ∞ X k=0 n −1 2k + n −1 ak,n+1 rkPk,n(t) = 1 (1 −2rt + r2) n−1 2 . In Stein and Weiss (Chapter 4, Section 2), the polynomials P λ k (t), where λ > 0, are defined using the following generating formula: ∞ X k=0 rkP λ k (t) = 1 (1 −2rt + r2)λ. Each polynomial P λ k (t) has degree k and is called an ultraspherical polynomial of degree k associated with λ. In view of Proposition 8.34, we see that P n−1 2 k (t) = n −1 2k + n −1 ak,n+1 Pk,n(t), as we mentionned ealier. There is also an integral formula for the Gegenbauer polynomials known as Laplace representation; see Morimoto (Chapter 2, Theorem 2.52). 8.8 The Funk–Hecke Formula The Funk–Hecke formula (also known as Hecke–Funk formula) basically allows one to per-form a sort of convolution of a “kernel function” with a spherical function in a convenient way. Given a measurable function K on [−1, 1] such that the integral Z 1 −1 |K(t)|(1 −t2) n−2 2 dt < ∞, (which means the integral makes sense), given a function f ∈L2 C(Sn), we can view the expression K ⋆f(σ) = Z Sn K(σ · τ)f(τ) VolSn 342 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS as a sort of convolution of K and f. Actually, the use of the term convolution is really unfortunate because in a “true” convo-lution f ∗g, either the argument of f or the argument of g should be multiplied by the inverse of the variable of integration, which means that the integration should really be taking place over the group SO(n + 1). We will come back to this point later. For the time being, let us call the expression K ⋆f defined above a pseudo-convolution. Now, if f is expressed in terms of spherical harmonics as f = ∞ X k=0 ak,n+1 X mk=1 ck,mkY mk k , then the Funk–Hecke formula states that K ⋆Y mk k (σ) = αkY mk k (σ), for some fixed constant αk, and so K ⋆f = ∞ X k=0 ak,n+1 X mk=1 αkck,mkY mk k . Thus, if the constants αk are known, then it is “cheap” to compute the pseudo-convolution K ⋆f. This method was used in a ground-breaking paper by Basri and Jacobs to compute the reflectance function r from the lighting function ℓas a pseudo-convolution K ⋆ℓ(over S2) with the Lambertian kernel K given by K(σ · τ) = max(σ · τ, 0). Below, we give a proof of the Funk–Hecke formula due to Morimoto (Chapter 2, Theorem 2.39); see also Andrews, Askey and Roy (Chapter 9). This formula was first published by Funk in 1916 and then by Hecke in 1918. But before we get to the Funk–Hecke formula, we need the following auxiliary proposition. Proposition 8.35. Let σ ∈Sn be given by the local coordinates on Sn where σ = √ 1 −t2 e σ + t en+1, with e σ ∈Sn−1 and −1 ≤t ≤1. The volume form on Sn is given by VolSn = (1 −t2) n−2 2 VolSn−1 dt. 8.8. THE FUNK–HECKE FORMULA 343 Proof. We need to compute the determinant of the n × n matrix g = (gij) expressing the Riemannian metric on Sn in this local coordinate system. Say the local coordinates on Sn−1 are t1, . . . , tn−1. Given σ = √ 1 −t2 e σ + t en+1, we compute ∂σ ∂ti = √ 1 −t2 ∂e σ ∂ti ∂σ ∂t = − t √ 1 −t2 e σ + en+1, and then using the fact that e σ and en+1 are orthogonal unit vectors, gij = ∂σ ∂ti · ∂σ ∂tj = (1 −t2)∂e σ ∂ti · ∂e σ ∂ti 1 ≤i, j ≤n −1 gin = gni = ∂σ ∂ti · ∂σ ∂t = 0 1 ≤i ≤n −1 gnn = ∂σ ∂t · ∂σ ∂t = t2 1 −t2 + 1 = 1 1 −t2. If we let e g be the (n −1) × (n −1) matrix given by f gij = ∂e σ ∂ti · ∂e σ ∂tj , then g is the matrix g = (1 −t2)e g 0 0 1 1−t2  , and since e g is an (n −1) × (n −1) matrix, p det(g) = (1 −t2) n−2 2 p det(e g), as Proposition 7.4 implies VolSn−1 = p det(e g)dt1 ∧· · · ∧dtn−1, it follows that VolSn = (1 −t2) n−2 2 VolSn−1 dt, as claimed. Theorem 8.36. (Funk–Hecke Formula) Given any measurable function K on [−1, 1] such that the integral Z 1 −1 |K(t)|(1 −t2) n−2 2 dt makes sense, for every function H ∈HC k (Sn), we have Z Sn K(σ · ξ)H(ξ) VolSn =  vol(Sn−1) Z 1 −1 K(t)Pk,n(t)(1 −t2) n−2 2 dt  H(σ). Observe that when n = 2, the term (1 −t2) n−2 2 is missing and we are simply requiring that R 1 −1 |K(t)| dt makes sense. 344 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Proof. We first prove the formula in the case where H(ξ) = Pk,n(ξ ·τ) for some fixed τ ∈Sn, and then use the fact that the Pk,n’s are reproducing kernels (Formula (rk)). For any fixed τ ∈Sn and every σ ∈Sn, define F by F(σ, τ) = Z Sn K(σ · ξ)H(ξ) VolSn = Z Sn K(σ · ξ)Pk,n(ξ · τ) VolSn. Since the volume form on the sphere is invariant under orientation-preserving isometries, for every R ∈SO(n + 1), we have F(Rσ, Rτ) = F(σ, τ), which means that F(σ, τ) is a function of σ · τ. On the other hand, for σ fixed, it is not hard to see that as a function in τ, the function F(σ, −) is a spherical harmonic. This is because the function given by H(ξ) = Pk,n(ξ · τ) may be viewed as a function of τ, namely H(τ) = Pk,n(ξ · τ). Furthermore H ∈HC k (Sn), and H satisfies the equation ∆SnH(τ) = −k(k + n −1)H(τ), with respect to the τ coordinates. This implies ∆SnF(σ, −) = −k(k + n −1)F(σ, −), since −k(k + n −1)F(σ, τ) = −k(k + n −1) Z Sn K(σ · ξ)Pk,n(ξ · τ) VolSn = Z Sn K(σ · ξ)(−k(k + n −1)H(τ) VolSn = Z Sn K(σ · ξ)∆SnH(τ) VolSn = ∆Sn Z Sn K(σ · ξ)H(ξ) VolSn = ∆Sn Z Sn K(σ · ξ)F(σ, τ). Thus F(σ, −) ∈HC k (Sn). Now for every rotation R that fixes σ, F(σ, τ) = F(Rσ, Rτ) = F(σ, Rτ), which means that F(σ, −) satisfies Condition (2) of Theorem 8.21. By Theorem 8.21, we get F(σ, τ) = F(σ, σ)Pk,n(σ · τ), since F(σ, σ) = cPk,n(σ · σ) = cPk,n(1) = c. 8.8. THE FUNK–HECKE FORMULA 345 We now want to explicitly compute F(σ, σ) = c. In order to do so, we apply Proposition 8.35 and find that for σ = en+1, F(σ, σ) = Z Sn K(σ · ξ)Pk,n(ξ · σ) VolSn = Z Sn K(en+1 · ξ)Pk,n(ξ · en+1) VolSn = vol(Sn−1) Z 1 −1 K(t)Pk,n(t)(1 −t2) n−2 2 dt, and thus, F(σ, τ) =  vol(Sn−1) Z 1 −1 K(t)Pk,n(t)(1 −t2) n−2 2 dt  Pk,n(σ · τ), which is the Funk–Hecke formula when H(σ) = Pk,n(σ · τ). Let us now consider any function H ∈HC k (Sn). Recall that by the reproducing kernel property (rk), we have ak,n+1 vol(Sn) Z Sn Pk,n(ξ · τ)H(τ) VolSn = H(ξ). Then we can compute R Sn K(σ · ξ)H(ξ) VolSn using Fubini’s Theorem and the Funk–Hecke formula in the special case where H(σ) = Pk,n(σ · τ), as follows: Z Sn K(σ · ξ)H(ξ) VolSn = Z Sn K(σ · ξ)  ak,n+1 vol(Sn) Z Sn Pk,n(ξ · τ)H(τ) VolSn  VolSn = ak,n+1 vol(Sn) Z Sn H(τ) Z Sn K(σ · ξ)Pk,n(ξ · τ) VolSn  VolSn = ak,n+1 vol(Sn) Z Sn H(τ)  vol(Sn−1) Z 1 −1 K(t)Pk,n(t)(1 −t2) n−2 2 dt  Pk,n(σ · τ)  VolSn =  vol(Sn−1) Z 1 −1 K(t)Pk,n(t)(1 −t2) n−2 2 dt   ak,n+1 vol(Sn) Z Sn Pk,n(σ · τ)H(τ) VolSn  =  vol(Sn−1) Z 1 −1 K(t)Pk,n(t)(1 −t2) n−2 2 dt  H(σ), which proves the Funk–Hecke formula in general. Remark: The formula VolSn = (1 −t2) n−2 2 VolSn−1 dt. 346 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS can be recursively integrated to a obtain closed form for vol(Sn). We follow Morimoto and let t = √u. Then dt = 1 2u−1 2 and the integral of the previous line becomes vol(Sn) = vol(Sn−1) Z 1 −1 (1 −t2) n−2 2 dt = 2vol(Sn−1) Z 1 0 (1 −t2) n−2 2 dt = vol(Sn−1) Z 1 0 (1 −u) n−2 2 u−1 2 du = vol(Sn−1)B n 2, 1 2  , where the last equality made use of the beta function formula B(x, y) = Z 1 0 tx−1(1 −t)y−1 dt, Re x > 0, Re y > 0. Since B(x, y) = Γ(x)Γ(y) Γ(x + y) , (see Theorem 1.1.4 of Andrews, Askey and Roy ), our calculations imply that vol(Sn) = Γ( 1 2)Γ( n 2) Γ( n+1 2 ) vol(Sn−1) = √πΓ( n 2) Γ( n+1 2 ) vol(Sn−1), where the last equality used Γ 1 2  = √π. We now recursively apply this formula n−1 times to obtain vol(Sn) = (√π)nΓ( 1 2) Γ( n+1 2 ) vol(S0) = 2π n+1 2 Γ( n+1 2 ), since vol(S0) = 0. It is now a matter of evaluating Γ n+1 2  . If n is odd, say n = 2d + 1, vol(S2d+1) = 2π 2d+2 2 Γ(d + 1) = 2πd+1 d! . If n is even, say n = 2d, by using the formula Γ(x + 1) = xΓ(x), we find that Γ 2d + 1 2  = Γ 2d −1 2 + 1  = 2d −1 2  · · · 3 2  1 2  Γ 1 2  = (2d −1) · · · 3 · 1√π 2d . 8.8. THE FUNK–HECKE FORMULA 347 Then vol(S2d) = 2π 2d+1 2 Γ( 2d+1 2 ) = 2d+1πd (2d −1) · · · 3 · 1. The Funk–Hecke formula can be used to derive an “addition theorem” for the ultraspher-ical polynomials (Gegenbauer polynomials). We omit this topic and we refer the interested reader to Andrews, Askey and Roy (Chapter 9, Section 9.8). Remark: Oddly, in their computation of K ⋆ℓ, Basri and Jacobs first expand K in terms of spherical harmonics as K = ∞ X n=0 knY 0 n , and then use the Funk–Hecke formula to compute K ⋆Y m n . They get (see page 222) K ⋆Y m n = αnY m n , with αn = r 4π 2n + 1 kn, for some constant kn given on page 230 of their paper (see below). However, there is no need to expand K, as the Funk–Hecke formula yields directly K ⋆Y m n (σ) = Z S2 K(σ · ξ)Y m n (ξ) VolSn =  vol(S1) Z 1 −1 K(t)Pn(t) dt  Y m n (σ), where Pn(t) is the standard Legendre polynomial of degree n, since we are in the case of S2. By the definition of K (K(t) = max(t, 0)) and since vol(S1) = 2π, we get K ⋆Y m n =  2π Z 1 0 tPn(t) dt  Y m n , which is equivalent to Basri and Jacobs’ formula (14), since their αn on page 222 is given by αn = r 4π 2n + 1 kn, but from page 230, kn = p (2n + 1)π Z 1 0 tPn(t) dt. What remains to be done is to compute R 1 0 tPn(t) dt, which is done by using the Rodrigues Formula and integrating by parts (see Appendix A of Basri and Jacobs ). In the next section we show how spherical harmonics fit into the broader framework of linear representations of (Lie) groups. 348 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS 8.9 Linear Representations of Compact Lie Groups; A Glimpse ⊛ The purpose of this section and the next is to generalize the results about the structure of the space of functions L2 C(Sn) defined on the sphere Sn, especially the results of Sections 8.5 and 8.6 (such as Theorem 8.19, except Part (3)), to homogeneous spaces G/K where G is a compact Lie group and K is a closed subgroup of G. The first step is to consider the Hilbert space L2 C(G) where G is a compact Lie group and to find a Hilbert sum decomposition of this space. The key to this generalization is the notion of (unitary) linear representation of the group G. The space L2 C(Sn) is replaced by L2 C(G), and each subspace HC k (Sn) involved in the Hilbert sum L2 C(Sn) = ∞ M k=0 HC k (Sn) is replaced by a subspace aρ of L2 C(G) isomorphic to a finite-dimensional algebra of nρ × nρ matrices. More precisely, there is a basis of aρ consisting of n2 ρ functions m(ρ) ij (from G to C) and if for every g ∈G we form the matrix Mρ(g) = 1 nρ    m(ρ) 11 (g) . . . m(ρ) 1nρ(g) . . . ... . . . m(ρ) nρ1(g) . . . m(ρ) nρnρ(g)   , (∗) then the matrix Mρ(g) is unitary and Mρ(g1g2) = Mρ(g1)Mρ(g2) for all g1, g2 ∈G. This means that the map g 7→Mρ(g) is a unitary representation of G in the vector space Cnρ. Furthermore, this representation is irreducible. Thus, the set of indices ρ is the set of equivalence classes of irreducible unitary representations of G. The result that we are sketching is a famous theorem known as the Peter–Weyl Theorem about unitary representations of compact Lie groups (Herman, Klauss, Hugo Weyl, 1885-1955). The Peter–Weyl Theorem can be generalized to any representation V : G →Aut(E) of G into a separable Hilbert space E (see Definition 8.10), and we obtain a Hilbert sum decomposition of E in terms of subspaces Eρ of E. The corresponding subrepresentations are not irreducible but each nontrivial Eρ splits into a Hilbert sum whose subspaces correspond to irreducible representations. The next step is to consider the subspace L2 C(G/K) of L2 C(G) consisting of the functions that are right-invariant under the action of K. These can be viewed as functions on the homogeneous space G/K. Again, we obtain a Hilbert sum decomposition L2 C(G/K) = M ρ Lρ = L2 C(G/K) ∩aρ. 8.9. LINEAR REPRESENTATIONS OF COMPACT LIE GROUPS ⊛ 349 It is also interesting to consider the subspace L2 C(K\G/K) of functions in L2 C(G) consisting of the functions that are both left and right-invariant under the action of K. The functions in L2 C(K\G/K) can be viewed as functions on the homogeneous space G/K that are invariant under the left action of K. Convolution makes the space L2 C(G) into a non-commutative algebra. Remarkably, it is possible to characterize when L2 C(K\G/K) is commutative (under convolution) in terms of a simple criterion about the irreducible representations of G. In this situation, (G, K) is a called a Gelfand pair. When (G, K) is a Gelfand pair, it is possible to define a well-behaved notion of Fourier transform on L2 C(K\G/K). Gelfand pairs and the Fourier transform are briefly considered in Section 8.11. First we review the notion of a linear representation of a group. A good and easy-going introduction to representations of Lie groups can be found in Hall . We begin with finite-dimensional representations. Definition 8.21. Given a Lie group G and a vector space V of dimension n, a linear representation of G of dimension (or degree) n is a group homomorphism U : G →GL(V ) such that the map g 7→U(g)(u) is continuous for every u ∈V , where GL(V ) denotes the group of invertible linear maps from V to itself. The space V , called the representation space, may be a real or a complex vector space. If V has a Hermitian (resp Euclidean) inner product ⟨−, −⟩, we say that U : G →GL(V ) is a unitary representation iff ⟨U(g)(u), U(g)(v)⟩= ⟨u, v⟩, for all g ∈G and all u, v ∈V. Thus, a linear representation of G is a map U : G →GL(V ) satisfying the properties: U(gh) = U(g)U(h) U(g−1) = U(g)−1 U(1) = I. For simplicity of language, we usually abbreviate linear representation as representation. The representation space V is also called a G-module, since the representation U : G → GL(V ) is equivalent to the left action ·: G×V →V , with g·v = U(g)(v). The representation such that U(g) = I for all g ∈G is called the trivial representation. As an example, we describe a class of representations of SL(2, C), the group of complex matrices with determinant +1, a b c d  , ad −bc = 1. 350 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Recall that PC k (2) denotes the vector space of complex homogeneous polynomials of degree k in two variables (z1, z2). For every matrix A ∈SL(2, C), with A = a b c d  , for every homogeneous polynomial Q ∈PC k (2), we define Uk(A)(Q(z1, z2)) by Uk(A)(Q(z1, z2)) = Q(dz1 −bz2, −cz1 + az2). If we think of the homogeneous polynomial Q(z1, z2) as a function Q z1 z2  of the vector z1 z2  , then Uk(A)  Q z1 z2  = QA−1 z1 z2  = Q  d −b −c a  z1 z2  . The expression above makes it clear that Uk(AB) = Uk(A)Uk(B) for any two matrices A, B ∈SL(2, C), so Uk is indeed a representation of SL(2, C) into PC k (2). One might wonder why we considered SL(2, C) rather than SL(2, R). This is because it can be shown that SL(2, R) has no nontrivial unitary (finite-dimensional) representations! For more on representations of SL(2, R), see Dieudonn´ e (Chapter 14). Given any basis (e1, . . . , en) of V , each U(g) is represented by an n × n matrix U(g) = (Uij(g)). We may think of the scalar functions g 7→Uij(g) as special functions on G. As explained in Dieudonn´ e (see also Vilenkin ), essentially all special functions (Legendre polynomials, ultraspherical polynomials, Bessel functions etc.) arise in this way by choosing some suitable G and V . There is a natural and useful notion of equivalence of representations: Definition 8.22. Given any two representations U1 : G →GL(V1) and U2 : G →GL(V2), a G-map (or morphism of representations) ϕ: U1 →U2 is a linear map ϕ: V1 →V2 so that the following diagram commutes for every g ∈G: V1 U1(g) / ϕ  V1 ϕ  V2 U2(g) / V2, i.e. ϕ(U1(g)(v)) = U2(g)(ϕ(v)), v ∈V1. The space of all G-maps between two representations as above is denoted HomG(U1, U2). Two representations U1 : G →GL(V1) and U2 : G →GL(V2) are equivalent iffϕ: V1 →V2 8.9. LINEAR REPRESENTATIONS OF COMPACT LIE GROUPS ⊛ 351 is an invertible linear map (which implies that dim V1 = dim V2). In terms of matrices, the representations U1 : G →GL(V1) and U2 : G →GL(V2) are equivalent iffthere is some invertible n × n matrix, P, so that U2(g) = PU1(g)P −1, g ∈G. If W ⊆V is a subspace of V , then in some cases, a representation U : G →GL(V ) yields a representation U : G →GL(W). This is interesting because under certain conditions on G (e.g., G compact) every representation may be decomposed into a “sum” of so-called irreducible representations (defined below), and thus the study of all representations of G boils down to the study of irreducible representations of G; for instance, see Knapp (Chapter 4, Corollary 4.7), or Br¨ ocker and tom Dieck (Chapter 2, Proposition 1.9). Definition 8.23. Let U : G →GL(V ) be a representation of G. If W ⊆V is a subspace of V , then we say that W is invariant (or stable) under U iffU(g)(w) ∈W, for all g ∈G and all w ∈W. If W is invariant under U, then we have a homomorphism, U : G →GL(W), called a subrepresentation of G. A representation U : G →GL(V ) with V ̸= (0) is irreducible iffit only has the two subrepresentations U : G →GL(W) corresponding to W = (0) or W = V . It can be shown that the representations Uk of SL(2, C) defined earlier are irreducible, and that every representation of SL(2, C) is equivalent to one of the Uk’s (see Br¨ ocker and tom Dieck , Chapter 2, Section 5). The representations Uk are also representations of SU(2). Again, they are irreducible representations of SU(2), and they constitute all of them (up to equivalence). The reader should consult Hall for more examples of representations of Lie groups. An easy but crucial lemma about irreducible representations is “Schur’s Lemma.” Lemma 8.37. (Schur’s Lemma) Let U1 : G →GL(V ) and U2 : G →GL(W) be any two real or complex representations of a group G. If U1 and U2 are irreducible, then the following properties hold: (i) Every G-map ϕ: U1 →U2 is either the zero map or an isomorphism. (ii) If U1 is a complex representation, then every G-map ϕ: U1 →U1 is of the form ϕ = λid, for some λ ∈C. Proof. (i) Observe that the kernel Ker ϕ ⊆V of ϕ is invariant under U1. Indeed, for every v ∈Ker ϕ and every g ∈G, we have ϕ(U1(g)(v)) = U2(g)(ϕ(v)) = U2(g)(0) = 0, so U1(g)(v) ∈Ker ϕ. Thus, U1 : G →GL(Ker ϕ) is a subrepresentation of U1, and as U1 is irreducible, either Ker ϕ = (0) or Ker ϕ = V . In the second case, ϕ = 0. If Ker ϕ = (0), 352 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS then ϕ is injective. However, ϕ(V ) ⊆W is invariant under U2, since for every v ∈V and every g ∈G, U2(g)(ϕ(v)) = ϕ(U1(g)(v)) ∈ϕ(V ), and as ϕ(V ) ̸= (0) (as V ̸= (0) since U1 is irreducible) and U2 is irreducible, we must have ϕ(V ) = W; that is, ϕ is an isomorphism. (ii) Since V is a complex vector space, the linear map ϕ has some eigenvalue λ ∈C. Let Eλ ⊆V be the eigenspace associated with λ. The subspace Eλ is invariant under U1, since for every u ∈Eλ and every g ∈G, we have ϕ(U1(g)(u)) = U1(g)(ϕ(u)) = U1(g)(λu) = λU1(g)(u), so U1 : G →GL(Eλ) is a subrepresentation of U1, and as U1 is irreducible and Eλ ̸= (0), we must have Eλ = V . An interesting corollary of Schur’s Lemma is the following fact: Proposition 8.38. Every complex irreducible representation U : G →GL(V ) of a commu-tative group G is one-dimensional. Proof. Since G is abelian, we claim that for every g ∈G, the map τg : V →V given by τg(v) = U(g)(v) for all v ∈V is a G-map. This amounts to checking that the following diagram commutes V U(g1)/ τg  V τg  V U(g1)/ V for all g, g1 ∈G. This is equivalent to checking that τg(U(g1)(v)) = U(g)(U(g1)(v)) = U(gg1)(v) = U(g1)(τg(v)) = U(g1)(U(g)(v)) = U(g1g)(v) for all v ∈V , that is, U(gg1)(v) = U(g1g)(v), which holds since G is commutative (so gg1 = g1g). By Schur’s Lemma (Lemma 8.37 (ii)), τg = λgid for some λg ∈C. It follows that any subspace of V is invariant. If the representation is irreducible, we must have dim(V ) = 1 since otherwise V would contain a one-dimentional invariant subspace, contradicting the assumption that U is irreducible. Let us now restrict our attention to compact Lie groups. If G is a compact Lie group, then it is known that it has a left and right-invariant volume form ωG, so we can define the integral of a (real or complex) continuous function f defined on G by Z G f = Z G f ωG, 8.9. LINEAR REPRESENTATIONS OF COMPACT LIE GROUPS ⊛ 353 also denoted R G f dµG or simply R G f(t) dt, with ωG normalized so that R G ωG = 1. (See Section 7.8, or Knapp , Chapter 8, or Warner , Chapters 4 and 6.) Because G is compact, the Haar measure µG induced by ωG is both left and right-invariant (G is a unimodular group), and our integral has the following invariance properties: Z G f(t) dt = Z G f(st) dt = Z G f(tu) dt = Z G f(t−1) dt, for all s, u ∈G (see Section 7.8). Since G is a compact Lie group, we can use an “averaging trick” to show that every (finite-dimensional) representation is equivalent to a unitary representation; see Br¨ ocker and tom Dieck (Chapter 2, Theorem 1.7) or Knapp (Chapter 4, Proposition 4.6). If we define the Hermitian inner product ⟨f, g⟩= Z G f g ωG, then, with this inner product the space of square-integrable functions L2 C(G) is a Hilbert space (in fact, a separable Hilbert space). Definition 8.24. The convolution f ∗g of two functions f, g ∈L2 C(G) is given by (f ∗g)(x) = Z G f(xt−1)g(t)dt = Z G f(t)g(t−1x)dt. In general, f ∗g ̸= g ∗f, unless G is commutative. With the convolution product, L2 C(G) becomes an associative algebra (non-commutative in general). This leads us to consider unitary representations of G into the infinite-dimensional vector space L2 C(G), and more generally into a Hilbert space E. Given a Hilbert space E, the definition of a unitary representation U : G →Aut(E) is the same as in Definition 8.21, except that GL(E) is replaced by the group of automorphisms (unitary operators) Aut(E) of the Hilbert space E, and ⟨U(g)(u), U(g)(v)⟩= ⟨u, v⟩ with respect to the inner product on E. Also, in the definition of an irreducible representation U : G →Aut(E), we require that the only closed subrepresentations U : G →Aut(W) of the representation U : G →Aut(E) correspond to W = (0) or W = E. Here, a subrepresentation U : G →Aut(W) is closed if W is closed in E. The Peter–Weyl Theorem gives a decomposition of L2 C(G) as a Hilbert sum of spaces that correspond to all the irreducible unitary representations of G. We present a version of the Peter–Weyl Theorem found in Dieudonn´ e (Chapters 3-8) and Dieudonn´ e (Chapter XXI, Sections 1-4), which contains complete proofs. Other versions can be found in Br¨ ocker and tom Dieck (Chapter 3), Knapp (Chapter 4) or Duistermaat and Kolk (Chapter 4). A good preparation for these fairly advanced books is Deitmar . 354 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Theorem 8.39. (Peter–Weyl (1927)) Given a compact Lie group G, there is a decomposition of the associative algebra (under convolution ∗) L2 C(G) as a Hilbert sum L2 C(G) = M ρ∈R(G) aρ of countably many two-sided ideals aρ, where each aρ is isomorphic to a finite-dimensional algebra of nρ × nρ complex matrices, where the set of indices R(G) corresponds to the set of equivalence classes of irreducible representations of G. More precisely, for each ρ ∈R(G), there is a basis of aρ consisting of n2 ρ pairwise orthogonal continuous functions m(ρ) ij : G →C, that is ⟨m(ρ) ij , m(ρ′) hk ⟩= 0 unless ρ = ρ′, i = h and j = k, and satisfying the properties m(ρ) ij ∗m(ρ) hk = δjhm(ρ) ik m(ρ) ij (e) = δijnρ ⟨m(ρ) ij , m(ρ) ij ⟩= nρ m(ρ) ji (g) = m(ρ) ij (g−1), and if for any g ∈G we form the nρ × nρ matrix Mρ(g) given by Mρ(g) = 1 nρ    m(ρ) 11 (g) . . . m(ρ) 1nρ(g) . . . ... . . . m(ρ) nρ1(g) . . . m(ρ) nρnρ(g)   , then the matrix Mρ(g) is unitary, Mρ(g1g2) = Mρ(g1)Mρ(g2), and the map g 7→Mρ(g) is an irreducible unitary representation of G in the vector space Cnρ (Mρ is a group homomorphism Mρ : G →GL(Cnρ)). Furthermore, every irreducible unitary representation of G is equivalent to some Mρ. The function uρ given by uρ(g) = nρ X j=1 m(ρ) jj (g) = nρtr(Mρ(g)) is the unit of the algebra aρ, and the orthogonal projection of L2 C(G) onto aρ is the map f 7→uρ ∗f = f ∗uρ; that is, convolution with uρ. The Peter–Weyl theorem implies that all irreducible unitary representations of a compact Lie group are finite-dimensional. The constant functions on G form a one-dimensional ideal aρ0 called the trivial ideal, corresponding to the trivial representation ρ0 (such that Mρ0(g) = 1 for all g ∈G). The fact that the m(ρ) ij form an orthogonal system implies that Z G m(ρ) ij (g) dg = 0 for all ρ ̸= ρ0. 8.9. LINEAR REPRESENTATIONS OF COMPACT LIE GROUPS ⊛ 355 Theorem 8.39 implies that the countable family of functions  1 √nρ m(ρ) ij  ρ∈R(G), 1≤i,j≤nρ is a Hilbert basis of L2 C(G). Remark: We will often refer to the decomposition of the Hilbert space L2 C(G) in terms of the ideals aρ as the master decomposition of L2 C(G). A complete proof of Theorem 8.39 is given in Dieudonn´ e , Chapter XXI, Section 2, but see also Sections 3 and 4. Remark: The Peter–Weyl theorem actually holds for any compact topological metrizable group, not just for a compact Lie group. Definition 8.25. The function χρ = 1 nρ uρ = tr(Mρ) is the character of G associated with the representation Mρ. The functions χρ satisfy the following properties: χρ(e) = nρ χρ(sts−1) = χρ(t) for all s, t ∈G χρ(s−1) = χρ(s) for all s ∈G χρ ∗χρ′ = 0 if ρ ̸= ρ′ χρ ∗χρ = 1 nρ χρ. Furthermore, the characters form an orthonormal Hilbert basis of the Hilbert subspace of L2 C(G) consisting of the central functions, namely those functions f ∈L2 C(G) such that for every s ∈G, f(sts−1) = f(t) almost everywhere. So, we have Z G χρ(t)χρ′(t) dt = 0 if ρ ̸= ρ′, Z G |χρ(t)|2dt = 1, and Z g χρ(g) dg = 0 for all ρ ̸= ρ0. If G (compact) is commutative, then by Proposition 8.38 all representations Mρ are one-dimensional. Then each character s 7→χρ(s) is a continuous homomorphism of G into U(1), the group of unit complex numbers. For the torus group S1 = T = R/Z, the characters are 356 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS the homomorphisms θ 7→ek2πiθ, with k ∈N. This is the special case of Fourier analysis on the circle. An important corollary of the Peter–Weyl theorem is that every compact Lie group is isomorphic to a matrix group. Theorem 8.40. For every compact Lie group G, there is some integer N ≥1 and an isomorphism of G onto a closed subgroup of U(N). The proof of Theorem 8.40 can be found in Dieudonn´ e , Chapter XXI, Theorem 21.13.1) or Knapp (Chapter 4, Corollary 4.22). There is more to the Peter–Weyl theorem: It gives a description of all unitary represen-tations of G into a separable Hilbert space. 8.10 Consequences of The Peter–Weyl Theorem Recall that a Hilbert space is separable if it has a countable total orthogonal family, also called a Hilbert basis; see Definition 8.10. If f : G →E is function from a compact Lie group G to a Hilbert space E and if for all z ∈E the function s 7→⟨f(s), z⟩is integrable and the function s 7→∥f(s)∥is integrable, then it can be shown that the map z 7→ Z G ⟨f(s), z⟩ds for all z ∈E is a bounded linear functional on L2 C(G) (using the dominated convergence theorem). By the Riesz representation theorem for Hilbert spaces, there is a unique y ∈E such that ⟨y, z⟩= Z G ⟨f(s), z⟩ds for all z ∈E; see Dieudonn´ e (Chapter XIII, Proposition 13.10.4). Definition 8.26. If f : G →E is function from a compact Lie group G to a Hilbert space E, under the conditions on f stated above, the unique vector y ∈E such that ⟨y, z⟩= Z G ⟨f(s), z⟩ds for all z ∈E is denoted by Z G f(s) ds and is called the weak integral (for short, integral) of f. 8.10. CONSEQUENCES OF THE PETER–WEYL THEOREM 357 Theorem 8.41. Given a compact Lie group G, if V : G →Aut(E) is a unitary represen-tation of G in a separable Hilbert space E, using the notation of Theorem 8.39, for every ρ ∈R(G), for every x ∈E the map x 7→Vuρ(x) = Z G uρ(s)(V (s)(x)) ds is an orthogonal projection of E onto a closed subspace Eρ, where the expression on the right-hand side is the weak integral of the function s 7→uρ(s)(V (s)(x)). Furthermore, E is the Hilbert sum E = M ρ∈R(G) Eρ of those Eρ such that Eρ ̸= (0). Each such Eρ is invariant under V , but the subrepresentation of V in Eρ is not necessarily irreducible. However, each Eρ is a (finite or countable) Hilbert sum of closed subspaces invariant under V , and the subrepresentations of V corresponding to these subspaces of Eρ are all equivalent to Mρ, where Mρ is defined as in Theorem 8.39, and Mρ is the representation of G given by Mρ(g) = Mρ(g) for all g ∈G. These repre-sentations are all irreducible. As a consequence, every irreducible unitary representation of G is equivalent to some representation of the form Mρ. For any closed subspace F of E, if F is invariant under V , then F is the Hilbert sum of the orthogonal spaces F ∩Eρ for those ρ ∈R(G) for which F ∩Eρ is not reduced to 0, and each nontrivial subspace F ∩Eρ is itself the Hilbert sum of closed subspaces invariant under V , and such that the corresponding subrepresentations are all irreducible and equivalent to Mρ. If Eρ ̸= (0), we say that the irreducible representation Mρ is contained in the represen-tation V . Definition 8.27. If Eρ is finite-dimensional, then dim(Eρ) = dρnρ for some positive integer dρ. The integer dρ is called the multiplicity of Mρ in V . An interesting special case of Theorem 8.41 is the case of the so-called regular represen-tation of G in L2 C(G) itself, that is, E = L2 C(G). Definition 8.28. The (left) regular representation R (or λ) of G in L2 C(G) is defined by (Rs(f))(t) = λs(f)(t) = f(s−1t), f ∈L2 C(G), s, t ∈G. We have (Rs1(Rs2(f)))(t) = (Rs2(f))(s−1 1 t) = f(s−1 2 s−1 1 t) = f((s1s2)−1t) = (Rs1s2(f))(t), which shows that (Rs1 ◦Rs2)(f) = Rs1s2(f), 358 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS namely, R is a representation of G in L2 C(G). Observe that if we had defined Rs(f) as f(st) instead of f(s−1t), then would get (Rs1 ◦Rs2)(f) = Rs2s1(f), so this version of Rs(f) would not be a representation since the above composition is Rs2s1(f) rather than Rs1s2(f). This is the reason for using s−1 instead of s in the definition of Rs(f) = f(s−1t). Theorem 8.41 implies that we also get a Hilbert sum L2 C(G) = L ρ∈R(G) Eρ, and it turns out that Eρ = aρ, where aρ is the ideal occurring in the master decomposition of L2 C(G), so again we get the Hilbert sum L2 C(G) = M ρ∈R(G) aρ of the master decomposition. This time, the aρ generally do not correspond to irreducible subrepresentations of R. However, aρ splits into dρ = nρ minimal left ideals b(ρ) j , where b(ρ) j is spanned by the jth columm of Mρ, that is, aρ = nρ M j=1 b(ρ) j and b(ρ) j = nρ M k=1 Cm(ρ) kj , and all the subrepresentations R: G →GL(b(ρ) j ) of G in b(ρ) j are equivalent to Mρ, and thus are irreducible (see Dieudonn´ e , Chapter 3). Finally, assume that besides the compact Lie group G, we also have a closed subgroup K of G. Then we know that M = G/K is a manifold called a homogeneous space, and G acts on M on the left. For example, if G = SO(n+1) and K = SO(n), then Sn = SO(n+1)/SO(n) (see Warner , Chapter 3, or Gallier and Quaintance ). Definition 8.29. The subspace of L2 C(G) consisting of the functions f ∈L2 C(G) that are right-invariant under the action of K, that is, such that f(su) = f(s) for all s ∈G and all u ∈K, forms a closed subspace of L2 C(G) denoted by L2 C(G/K). Since a function as above is constant on every left coset sK (s ∈G), such a function can be viewed as a function on the homogeneous space G/K. For example, if G = SO(n + 1) and K = SO(n), then L2 C(G/K) = L2 C(Sn). It turns out that L2 C(G/K) is invariant under the regular representation R of G in L2 C(G), so we get a subrepresentation (of the regular representation) of G in L2 C(G/K). The corollary of the Peter–Weyl theorem (Theorem 8.41) gives us a Hilbert sum decom-position of L2 C(G/K) of the form L2 C(G/K) = M ρ Lρ = L2 C(G/K) ∩aρ, 8.10. CONSEQUENCES OF THE PETER–WEYL THEOREM 359 for the same ρ’s as before. However, these subrepresentations of R in Lρ are not necessarily irreducible. What happens is that there is some dρ with 0 ≤dρ ≤nρ, so that if dρ ≥1, then Lρ is the direct sum of the subspace spanned by the first dρ columns of Mρ. The number dρ can be characterized as follows. If we consider the restriction of the representation Mρ : G →GL(Cnρ) to K, then this representation is generally not irreducible, so Cnρ splits into subspaces Fσ1, . . . , Fσr such that the restriction of the subrepresentation Mρ to Fσi is an irreducible representation of K. Then dρ is the multiplicity of the trivial representation σ0 of K if it occurs. for this reason, dρ is also denoted (ρ : σ0) (see Dieudonn´ e , Chapter 6 and Dieudonn´ e , Chapter XXII, Sections 4-5). Definition 8.30. The subspace of L2 C(G) consisting of the functions f ∈L2 C(G) that are left-invariant under the action of K, that is, such that f(ts) = f(s) for all s ∈G and all t ∈K, is a closed subspace of L2 C(G) denoted L2 C(K\G). We get a Hilbert sum decomposition of L2 C(K\G) of the form L2 C(K\G) = M ρ L′ ρ = M ρ L2 C(K\G) ∩aρ, and for the same dρ as before, L′ ρ is the direct sum of the subspace spanned by the first dρ rows of Mρ. Finally, we consider the folowing algebra. Definition 8.31. The space L2 C(K\G/K) is defined by L2 C(K\G/K) = L2 C(G/K) ∩L2 C(K\G) = {f ∈L2 C(G) | f(tsu) = f(s)} for all s ∈G and all t, u ∈K. Functions in L2 C(K\G/K) can be viewed as functions on the homogeneous space G/K that are invariant under the left action of K. These functions are constant on the double cosets KsK (s ∈G). In the case where G = SO(3) and K = SO(2), these are the functions on S2 that are invariant under the action of SO(2) (more precisely, a subgroup of SO(3) leaving invariant some chosen element of S2). The functions in L2 C(K\G/K) are reminiscent of zonal spherical functions, and indeed these functions are often called spherical functions, as in Helgason (Chapter 4). From our previous discussion, we see that we have a Hilbert sum decomposition L2 C(K\G/K) = M ρ Lρ ∩L′ ρ and each Lρ ∩L′ ρ for which dρ ≥1 is a matrix algebra of dimension d2 ρ having as a basis the functions m(ρ) ij for 1 ≤i, j ≤dρ. As a consequence, the algebra L2 C(K\G/K) is commutative iffdρ ≤1 for all ρ. 360 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS 8.11 Gelfand Pairs, Spherical Functions, and Fourier Transform ⊛ In this section we investigate briefly what happens when the algebra L2 C(K\G/K) is com-mutative. In this case, the space L2 C(K\G/K) is a Hilbert sum of one-dimensional subspaces spanned by the functions ωρ = (1/nρ)m(ρ) 11 , which are called zonal spherical harmonics. It is also the case that L2 C(G/K) is a Hilbert sum of nρ-dimensional subspaces Lρ, where Lρ is spanned by the left translates of ωρ. Finally, it is possible to define a well-behaved notion of Fourier transform on L2 C(K\G/K), in the sense that the Fourier transform on L2 C(K\G/K) satisfies the fundamental relation F(f ∗g) = F(f)F(g). Observe that in order for this equation to hold, convolution has to be commutative. This is why the Fourier transform is defined on L2 C(K\G/K), where convolution is commutative, rather than the whole of L2 C(G/K). Definition 8.32. Given a compact Lie group G and a closed subgroup K, if the algebra L2 C(K\G/K) is commutative (for the convolution product), we say that (G, K) is a Gelfand pair; see Dieudonn´ e , Chapter 8 and Dieudonn´ e , Chapter XXII, Sections 6-7. In this case, the Lρ in the Hilbert sum decomposition of L2 C(G/K) are nontrivial of dimension nρ iff(ρ : σ0) = dρ = 1, and the subrepresentation U (of the regular representation R) of G into Lρ is irreducible and equivalent to Mρ. The space Lρ is generated by the functions m(ρ) 11 , . . . , m(ρ) nρ1, but the function ωρ(s) = 1 nρ m(ρ) 11 (s) plays a special role. Definition 8.33. Given a compact Lie group G and a closed subgroup K, if (G, K) is a Gelfand pair, then function ωρ = 1 nρ m(ρ) 11 is called a zonal spherical function, for short a spherical function. The set of zonal spherical functions on G/K is denoted S(G/K). Because G is compact, S(G/K) it is a countable set in bijection with the set of equivalence classes of representations ρ ∈R(G) such that (ρ : σ0) = 1. Spherical functions defined in Definition 8.33 are generalizations of the zonal functions on Sn of Definition 8.18. They have some interesting properties, some of which are listed below. In particular, they are a key ingredient in generalizing the notion of Fourier transform on the homogeneous space G/K. 8.11. GELFAND PAIRS, SPHERICAL FUNCTIONS, FOURIER TRANSFORM ⊛ 361 First, ωρ is a continuous function, even a smooth function since G is a Lie group. The function ωρ is such that ωρ(e) = 1 (where e is the identity element of the group, G), and ωρ(ust) = ωρ(s) for all s ∈G and all u, t ∈K. In addition, ωρ is of positive type. A function f : G →C is of positive type iff n X j,k=1 f(s−1 j sk)zjzk ≥0, for every finite set {s1, . . . , sn} of elements of G and every finite tuple (z1, . . . , zn) ∈Cn. When L2 C(K\G/K) is commutative, it is the Hilbert sum of all the 1-dimensional sub-spaces Cωρ for all ρ ∈R(G) such that dρ = 1. The orthogonal projection of L2 C(K\G/K) onto Cωρ is given by g 7→g ∗ωρ g ∈L2 C(K\G/K). Since Cωρ is an ideal in the algebra L2 C(K\G/K), there is some homomorphism ξρ : L2 C(K\G/K) →C such that g ∗ωρ = ξρ(g)ωρ g ∈L2 C(K\G/K). To be more precise, ξρ has the property ξρ(g1 ∗g2) = ξρ(g1)ξρ(g2) for all g1, g2 ∈L2 C(K\G/K). In other words, ξρ is a character of the algebra L2 C(K\G/K) (see below for the definition of characters). Because the subrepresentation R of G into Lρ is irreducible (if (G/K) is a Gelfand pair, all nontrivial Lρ are one-dimensional), the function ωρ generates Lρ under left translation. This means the following: If we recall that for any function f on G, λs(f)(t) = f(s−1t), s, t ∈G, then Lρ is generated by the functions λs(ωρ), as s varies in G. It can be shown that a (non-identically-zero) function ω in the set CC(K\G/K) of contin-uous complex-valued functions in L2 C(K\G/K) belongs to S(G/K) iffthe functional equation Z K ω(xsy) ds = ω(x)ω(y) (∗) holds for all s ∈K and all x, y ∈G. The space S(G/K) is also in bijection with the characters of the algebra L2 C(K\G/K). 362 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Definition 8.34. If (G, K) is a Gelfand pair a character of the commutative algebra of L2 C(K\G/K) is a non-identically-zero linear map ξ : L2 C(K\G/K) →C such that ξ(f ∗g) = ξ(f)ξ(g) for all f, g ∈L2 C(K\G/K). Let X0 denote the set of characters of L2 C(K\G/K). Then it can be shown that for every character ξ ∈X0, there is a unique spherical function ω ∈S(G/K) such that ξ(f) = Z G f(s)ω(s) ds. It follows that there is a bijection between S(G/K) and X0. All this is explained in Dieudonn´ e (Chapters 8 and 9) and Dieudonn´ e (Chapter XXII, Sections 6-9). It is remarkable that fairly general criteria (due to Gelfand) for a pair (G, K) to be a Gelfand pair exist. This is certainly the case if G is commutative and K = (e); this situation corresponds to commutative harmonic anaysis. If G is a semisimple compact connected Lie group and if σ: G →G is an involutive automorphism of G (that is, σ2 = id), if K is the subgroup of fixed points of σ K = {s ∈G | σ(s) = s}, then it can be shown that (G, K) is a Gelfand pair. Involutive automorphims as above were determined explicitly by E. Cartan. It turns out that G = SO(n + 1) and K = SO(n) form a Gelfand pair corresponding to the above situation (see Dieudonn´ e , Chapters 7-8 and Dieudonn´ e , Chapter XXIII, Section 38). In this particular case, ρ = k is any nonnegative integer and Lρ = Ek, the eigenspace of the Laplacian on Sn corresponding to the eigenvalue −k(n+k−1); all this was shown in Section 8.5. Therefore, the regular representation of SO(n+1) into Ek = HC k (Sn) is irreducible. This can be proved more directly; for example, see Helgason (Introduction, Theorem 3.1) or Br¨ ocker and tom Dieck (Chapter 2, Proposition 5.10). The zonal spherical harmonics ωk can be expressed in terms of the ultraspherical poly-nomials (also called Gegenbauer polynomials) P (n−1)/2 k (up to a constant factor); this was discussed in Sections 8.6 and 8.7. The reader should also consult Stein and Weiss (Chapter 4), Morimoto (Chapter 2) and Dieudonn´ e (Chapter 7). For n = 2, P 1 2 k is just the ordinary Legendre polynomial (up to a constant factor). Returning to arbitrary Gelfand pairs (G compact), the Fourier transform is defined as follows. For any function f ∈L2 C(K\G/K), the Fourier transform F(f) is a function defined on the space S(G/K). Definition 8.35. If (G, K) is a Gelfand pair (with G a compact group), the Fourier trans-form F(f) of a function f ∈L2 C(K\G/K) is the function F(f): S(G/K) →C given by F(f)(ω) = Z G f(s)ω(s−1) ds ω ∈S(G/K). 8.11. GELFAND PAIRS, SPHERICAL FUNCTIONS, FOURIER TRANSFORM ⊛ 363 More explicitly, because ωρ = 1 nρm(ρ) 11 and m(ρ) 11 (s−1) = m(ρ) 11 (s), the Fourier transform F(f) is the countable family ρ 7→1 nρ D f, m(ρ) 11 E = Z G f(s)ωρ(s−1) ds for all ρ ∈R(G) such that (ρ : σ0) = 1. This Fourier transform is often called the spherical Fourier transform or spherical trans-form, as in Helgason (Chapter 4). It appears that it was first introduced by Harish-Chandra around 1957. The Fourier transform on L2 C(K\G/K) satisfies the fundamental relation F(f ∗g) = F(g ∗f) = F(f)F(g). Observe that in order for this equation to hold, convolution has to be commutative. This is why the Fourier transform is defined on L2 C(K\G/K) rather than the whole of L2 C(G/K). For a Gelfand pair, convolution on L2 C(K\G/K) is commutative. The notion of Gelfand pair and of the Fourier transform can be generalized to locally-compact unimodular groups that are not necessary compact, but we will not discuss this here. Let us just say that when G is a commutative locally-compact group and K = (e), then Equation (∗) implies that ω(xy) = ω(x)ω(y), which means that the functions ω are characters of G, so S(G/K) is the Pontrjagin dual group b G of G, which is the group of characters of G (continuous homomorphisms of G into the group U(1)). In this case, the Fourier transform F(f) is defined for every function f ∈L1 C(G) as a function on the characters of G. This is the case of commutative harmonic analysis, as discussed in Folland and Deitmar . For more on Gelfand pairs, curious readers may consult Dieudonn´ e (Chapters 8 and 9) and Dieudonn´ e (Chapter XXII, Sections 6-9). Another approach to spherical functions (not using Gelfand pairs) is discussed in Helgason (Chapter 4). Helgason contains a short section on Gelfand pairs (chapter III, Section 12). The material in this section belongs to the overlapping areas of representation theory and noncommutative harmonic analysis. These are deep and vast areas. Besides the references cited earlier, for noncommutative harmonic analysis, the reader may consult Knapp , Folland , Taylor , or Varadarajan , but they may find the pace rather rapid. Another great survey on both topics is Kirillov , although it is not geared for the beginner. In a different direction, namely Fourier analysis on finite groups, Audrey Terras’s book contains some fascinating material. 364 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS 8.12 Problems Problem 8.1. Let E be a complex vector space of dimension n. (i) Show that that given any basis (e1, . . . , en) of E, the linear map h: Cn →E defined such that h((z1, . . . , zn)) = z1e1 + · · · + znen is a homeomorphism (using the sup-norm on Cn). (ii) Use Part (i.) and the fact that any two norms on a finite dimensional vector space over R or C are equivalent to prove that E is complete. Problem 8.2. Let E be a normed vector space. Let K be a nonempty index set. Show that for every bijection ϕ: K →K (intuitively, a reordering of K), the family (uk)k∈K is summable iffthe family (ul)l∈ϕ(K) is summable, and if so, they have the same sum. Problem 8.3. Prove Proposition 8.1. Problem 8.4. Let Θ be a function of the independent variable θ. Take second order differ-ential equation sin2 θ Θ′′ + sin θ cos θ Θ′ + (k(k + 1) sin2 θ −m2)Θ = 0, and use the change of variable t = cos θ to obtain (1 −t2)u′′ −2tu′ +  k(k + 1) − m2 1 −t2  u = 0. Then make the substitution u(t) = (1 −t2) m 2 v(t); to obtain (1 −t2)v′′ −2(m + 1)tv′ + (k(k + 1) −m(m + 1))v = 0. Hint. See Lebedev , Chapter 7, Section 7.12. Problem 8.5. Recall that the Legendre polynomial Pn(t) is defined as Pn(t) = 1 2nn! dn dtn(t2 −1)n. (i) Show that Pk(t) is a solution to the second order differential equation (1 −t2)v′′ −2tv′ + k(k + 1)v = 0. (ii) Show that the Legendre polynomials satisfy the following recurrence relation: P0 = 1 P1 = t (n + 1)Pn+1 = (2n + 1)tPn −nPn−1 n ≥1; 8.12. PROBLEMS 365 Hint. See Lebedev , Chapter 4, Section 4.3. Problem 8.6. Recall that the associated Legendre function P k m(t) is defined by P k m(t) = (1 −t2) m 2 dm dtm(Pk(t)), where Pk(t) is the Legendre polynomial of order k. (i) For fixed m ≥0, prove the recurrence relation (k −m + 1)P m k+1(t) = (2k + 1)tP m k (t) −(k + m)P m k−1(t), k ≥1. (ii) Fore fixed k ≥2, prove the recurrence relation P m+2 k (t) = 2(m + 1)t (t2 −1) 1 2 P m+1 k (t) + (k −m)(k + m + 1)P m k (t), 0 ≤m ≤k −2. Hint. See Lebedev , Chapter 7, Section 7.12. Problem 8.7. Let M be a n-dimensional Riemannian manifold with chart (U, ϕ). If for p ∈M  ∂ ∂x1  p , . . . ,  ∂ ∂xn  p ! denotes the basis of TpM induced by ϕ, the local expression of the metric g at p is given by the n × n matrix (gij)p, with (gij)p = gp  ∂ ∂xi  p ,  ∂ ∂xj  p ! . Its inverse is denoted (gij)p. We also let |g|p = det(gij)p. Show that for every function f ∈C∞(M), in local coordinates given by the chart (U, ϕ), we have grad f = X ij gij ∂f ∂xj ∂ ∂xi , where as usual ∂f ∂xj (p) =  ∂ ∂xj  p f = ∂(f ◦ϕ−1) ∂uj (ϕ(p)), and (u1, . . . , un) are the coordinate functions in Rn. 366 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Problem 8.8. Let M be a Riemannian manifold of dimension n with chart (U, ϕ). Let X = n X i=1 Xi ∂ ∂xi , be a vector field expressed over this chart. Recall that the Christoffel symbol Γk ij is defined as Γk ij = 1 2 n X l=1 gkl (∂igjl + ∂jgil −∂lgij) , (∗) where ∂kgij = ∂ ∂xk (gij). Show that divX = n X i=1 " ∂Xi ∂xi + n X j=1 Γi ijXj # , and that ∆f = X i,j gij " ∂2f ∂xi∂xj − n X k=1 Γk ij ∂f ∂xk # , whenever f ∈C∞(M). Hint. See Pages 86 and 87 of O’Neill . Problem 8.9. Let M be a Riemannian manifold of dimension n with chart (U, ϕ). For every vector field X ∈X(M) expressed in local coordinates as X = n X i=1 Xi ∂ ∂xi , show that div X = 1 p |g| n X i=1 ∂ ∂xi p |g| Xi  , and for every function f ∈C∞(M), show that ∆f = 1 p |g| X i,j ∂ ∂xi p |g| gij ∂f ∂xj  . Hint. See Helgason ,Chapter II, Lemma 2.5, Postnikov , Chapter 13, Section 6, and O’Neill ,Chapter 7, Exercise 5. Problem 8.10. Let M be a Riemannian manifold with metric g. For any two functions f, h ∈C∞(M), and any vector field X ∈X(M), show that div(fX) = fdiv X + X(f) = fdiv X + g(grad f, X) grad f (h) = g(grad f, grad h) = grad h (f). 8.12. PROBLEMS 367 Problem 8.11. Recall that ak,n denotes the dimension of Hk(n). Show that ak,n+1 = n + k −1 n −1  + n + k −2 n −1  for k ≥2. Hint. See Morimoto , Chapter 2, Theorem 2.4, or Dieudonn´ e , Chapter 7, Formula 99. Problem 8.12. Let C(Sn) the space of continuous (real) functions on Sn. Show that C(Sn) is dense in L2(Sn), where L2(Sn) is the space of (real) square-integrable functions on the sphere Sn. with norm given by ⟨f, f⟩Sn = Z Sn f 2 VolSn. Problem 8.13. Prove Theorem 8.18. Hint. See Lang , Chapter V, Proposition 1.9. Problem 8.14. Prove Proposition 8.22. Hint. See Morimoto , Chapter 2, Lemma 1.19 and Lemma 2.20. Problem 8.15. Prove Propositions 8.23 and 8.25. Hint. See Morimoto Chapter 2, Lemma 2.21 and Morimoto Chapter 2, Lemma 2.23. Problem 8.16. Prove Theorem 8.30. Hint. See Stein and Weiss ,Chapter 4, Theorem 2.12. Problem 8.17. Show that the Gegenbauer polynomial Pk,n is given by Rodrigues’ formula: Pk,n(t) = (−1)k 2k Γ n 2  Γ k + n 2  1 (1 −t2) n−2 2 dk dtk (1 −t2)k+ n−2 2 , with n ≥2. Hint. See Morimoto , Chapter 2, Theorem 2.35. Problem 8.18. Prove Proposition 8.32. Hint. See Morimoto , Chapter 2, Theorem 2.34. Problem 8.19. Prove Proposition 8.34. Hint. See Morimoto , Chapter 2, Theorem 2.53 and Theorem 2.55. 368 CHAPTER 8. SPHERICAL HARMONICS AND LINEAR REPRESENTATIONS Chapter 9 Operators on Riemannian Manifolds; Hodge Laplacian, Laplace-Beltrami Laplacian, The Bochner Laplacian, and Weitzenb¨ ock Formulae The Laplacian is a very important operator because it shows up in many of the equations used in physics to describe natural phenomena such as heat diffusion or wave propagation. Therefore, it is highly desirable to generalize the Laplacian to functions defined on a manifold. Furthermore, in the late 1930’s, Georges de Rham (inspired by ´ Elie Cartan) realized that it was fruitful to define a version of the Laplacian operating on differential forms, because of a fundamental and almost miraculous relationship between harmonics forms (those in the kernel of the Laplacian) and the de Rham cohomology groups on a (compact, orientable) smooth manifold. Indeed, as we will see in Section 9.6, for every cohomology group Hk DR(M), every cohomology class [ω] ∈Hk DR(M) is represented by a unique harmonic k-form ω. The connection between analysis and topology lies deep and has many important consequences. For example, Poincar´ e duality follows as an “easy” consequence of the Hodge theorem. Technically, the Hodge Laplacian can be defined on differential forms using the Hodge ∗ operator (Section 3.5). On functions, there is an alternate and equivalent definition of the Laplacian using only the covariant derivative and obtained by generalizing the notions of gradient and divergence to functions on manifolds. Another version of the Laplacian on k-forms can be defined in terms of a generalization of the Levi-Civita connection ∇: X(M) × X(M) →X(M) to k-forms viewed as a linear map ∇: Ak(M) →HomC∞(M)(X(M), Ak(M)), and in terms of a certain adjoint ∇∗of ∇, a linear map ∇∗: HomC∞(M)(X(M), Ak(M)) →Ak(M). 369 370 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS For this, we will define an inner product (−, −) on k-forms and an inner product ((−, −)) on HomC∞(M)(X(M), Ak(M)) and define ∇∗so that (∇∗A, ω) = ((A, ∇ω)) for all A ∈HomC∞(M)(X(M), Ak(M)) and all ω ∈Ak(M). We obtain the Bochner Laplacian (or connection Laplacian ) ∇∗∇. Then it is natural to wonder how the Hodge Laplacian ∆differs from the connection Laplacian ∇∗∇? Remarkably, there is a formula known as Weitzenb¨ ock’s formula (or Bochner’s formula) of the form ∆= ∇∗∇+ C(R∇), where C(R∇) is a contraction of a version of the curvature tensor on differential forms (a fairly complicated term). In the case of one-forms, ∆= ∇∗∇+ Ric, where Ric is a suitable version of the Ricci curvature operating on one-forms. Weitzenb¨ ock-type formulae are at the root of the so-called “Bochner technique,” which consists in exploiting curvature information to deduce topological information. For example, if the Ricci curvature on a compact orientable Riemannian manifold is strictly positive, then H1 DR(M) = (0), a theorem due to Bochner. 9.1 The Gradient and Hessian Operators on Rieman-nian Manifolds In preparation for defining the (Hodge) Laplacian, we define the gradient of a function on a Riemannian manifold, as well as the Hessian, which plays an important role in optimization theory. Unlike the situation where M is a vector space (M is flat), the Riemannian metric on M is critically involved in the definition of the gradient and of the Hessian. If (M, ⟨−, −⟩) is a Riemannian manifold of dimension n, then for every p ∈M, the inner product ⟨−, −⟩p on TpM yields a canonical isomorphism ♭: TpM →T ∗ p M, as explained in Sections 2.2 and 10.7. Namely, for any u ∈TpM, u♭= ♭(u) is the linear form in T ∗ p M defined by u♭(v) = ⟨u, v⟩p, v ∈TpM. Recall that the inverse of the map ♭is the map ♯: T ∗ p M →TpM. As a consequence, for every smooth function f ∈C∞(M), we get smooth vector field grad f = (d f)♯defined so that (grad f)p = (d fp)♯. 9.1. THE GRADIENT AND HESSIAN OPERATORS 371 Definition 9.1. For every smooth function f over a Riemannian manifold (M, ⟨−, −⟩), the vector field grad f defined by ⟨(grad f)p, u⟩p = d fp(u), for all u ∈TpM, and all p ∈M, is the gradient of the function f. Definition 9.2. Let (M, ⟨−, −⟩) be a Riemannian manifold. For any vector field X ∈X(M), The one-form X♭∈A1(M) is given by (X♭)p = (Xp)♭. The one-form X♭is uniquely defined by the equation (X♭)p(v) = ⟨Xp, v⟩p, for all p ∈M and all v ∈TpM. In view of this equation, the one-form X♭is an insertion operator in the sense discussed in Section 3.6 just after Proposition 3.22, so it is also denoted by iXg, where g = ⟨−, −⟩is the Riemannian metric on M. In the special case X = grad f, we have (grad f)♭ p(v) = ⟨(grad f)p, v⟩= d fp(v), and since dd = 0, we deduce that d(grad f)♭= 0. Therefore, for an arbitrary vector field X, the 2-form dX♭measures the extent to which X is a gradient field. If (U, ϕ) is a chart of M, with p ∈M, and if  ∂ ∂x1  p , . . . ,  ∂ ∂xn  p ! denotes the basis of TpM induced by ϕ, the local expression of the metric g at p is given by the n × n matrix (gij)p, with (gij)p = gp  ∂ ∂xi  p ,  ∂ ∂xj  p ! . The inverse is denoted by (gij)p. We often omit the subscript p and observe that for every function f ∈C∞(M), grad f = X ij gij ∂f ∂xj ∂ ∂xi . 372 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS It is instructive to look at the following special case of the preceding formula. Let f ∈C∞(M), where M is a two-dimensional manifold. For each p ∈M, let { ∂ ∂x1, ∂ ∂x2} be basis for the tangent space Tp(M). Let v = a ∂ ∂x1 + b ∂ ∂x2 ∈Tp(M). Then grad f =  g11 ∂f ∂x1 + g12 ∂f ∂x2  ∂ ∂x1 +  g21 ∂f ∂x1 + g22 ∂f ∂x2  ∂ ∂x2 . Since g12 = g21, g12 = g21, and g11 g12 g21 g22  g11 g12 g21 g22  = g11 g12 g21 g22  g11 g12 g21 g22  = 1 0 0 1  , we discover that ⟨grad f, v⟩=  g11 ∂f ∂x1 + g12 ∂f ∂x2  ∂ ∂x1 +  g21 ∂f ∂x1 + g22 ∂f ∂x2  ∂ ∂x2 , a ∂ ∂x1 + b ∂ ∂x2 = a  g11 ∂f ∂x1 + g12 ∂f ∂x2  g11 + b  g11 ∂f ∂x1 + g12 ∂f ∂x2  g12 + a  g21 ∂f ∂x1 + g22 ∂f ∂x2  g12 + b  g21 ∂f ∂x1 + g22 ∂f ∂x2  g22 = a(g11g11 + g21g21) + b(g11g12 + g21g22) ∂f ∂x1 + a(g12g11 + g22g21) + b(g12g21 + g22g22) ∂f ∂x2 = a ∂f ∂x1 + b ∂f ∂x2 =  ∂f ∂x1, ∂f ∂x2  a b  = d fp(v). We now define the Hessian of a function. For this we assume that ∇is the Levi-Civita connection. Definition 9.3. The Hessian Hess(f) (or ∇2(f)) of a function f ∈C∞(M) is the (0, 2)-tensor defined by Hess(f)(X, Y ) = X(Y (f)) −(∇XY )(f) = X(d f(Y )) −d f(∇XY ), for all vector fields X, Y ∈X(M). Remark: The Hessian of f is defined in various ways throughout the literature. For our purposes, Definition 9.3 is sufficient, but for completeness sake, we point out two alternative formulations of Hess(f)(X, Y ). 9.1. THE GRADIENT AND HESSIAN OPERATORS 373 The first reformulation utilizes the covariant derivative of a one-form. Let X ∈X(M) and θ ∈A1(M). The covariant derivative ∇Xθ of any one-form may be defined as (∇Xθ)(Y ) := X(θ(Y )) −θ(∇XY ). Thus the Hessian of f may be written as Hess(f)(X, Y ) = (∇Xd f)(Y ). The Hessian of f also appears in the literature as Hess(f)(X, Y ) = (∇d f)(X, Y ) = (∇Xd f)(Y ), which means that the (0, 2)-tensor Hess(f) is given by Hess(f) = ∇d f. Since by definition ∇Xf = d f(X), we can also write Hess(f) = ∇∇f, but we find this expression confusing. Proposition 9.1. The Hessian is given by the equation Hess(f)(X, Y ) = ⟨∇X(grad f), Y ⟩, X, Y ∈X(M). Proof. We have X(Y (f)) = X(d f(Y )) = X(⟨grad f, Y ⟩) = ⟨∇X(grad f), Y ⟩+ ⟨grad f, ∇XY ⟩ = ⟨∇X(grad f), Y ⟩+ (∇XY )(f) which yields ⟨∇X(grad f), Y ⟩= X(Y (f)) −(∇XY )(f) = Hess(f)(X, Y ), as claimed. The Hessian can also be defined in terms of Lie derivatives; this is the approach followed by Petersen (Chapter 2, Section 1.3). This approach utilizes the observation that the Levi-Civita connection can be defined in terms of the Lie derivative of the Riemannian metric g on M by the equation 2g(∇XY, Z) = (LY g)(X, Z) + (d(iY g))(X, Z), X, Y, Z ∈X(M). Proposition 9.2. The Hessian of f is given by Hess(f) = 1 2Lgrad f g. 374 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS Proof. To prove the above equation, we use the fact that d(igrad fg) = d(grad f)♭= 0 and Proposition 9.1. We have 2Hess(f)(X, Y ) = 2g(∇X(grad f), Y ), by Proposition 9.1 = (Lgrad f g)(X, Y ) + d(igrad fg)(X, Y ) = (Lgrad f g)(X, Y ), as claimed. Since ∇is torsion-free, we get Hess(f)(X, Y ) = X(Y (f)) −(∇XY )(f) = Y (X(f)) −(∇Y X)(f) = Hess(f)(Y, X), which means that the Hessian is a symmetric (0, 2)-tensor. Since the Hessian is a symmetric bilinear form, it is determined by the quadratic form X 7→Hess(f)(X, X), and it can be recovered by polarization from this quadratic form. There is also a way to compute Hess(f)(X, X) using geodesics. When geodesics are easily computable, this is usually the simplest way to compute the Hessian. Proposition 9.3. Given any p ∈M and any tangent vector X ∈TpM, if γ is a geodesic such that γ(0) = p and γ′(0) = X, then at p, we have Hess(f)p(X, X) = d2 dt2 f(γ(t)) t=0 . Proof. To prove the above formula, following Jost , we have X(X(f))(p) = γ′⟨(grad f)p, γ′⟩ = γ′  d dt f(γ(t)) t=0  = d2 dt2 f(γ(t)) t=0 . Furthermore, since γ is a geodesic, ∇γ′γ′ = 0, so we get Hess(f)p(X, X) = X(X(f))(p) −(∇XX)(f)(p) = X(X(f))(p), which proves our claim. Proposition 9.4. In local coordinates with respect to a chart, if we write d f = n X i=1 ∂f ∂xi dxi, 9.1. THE GRADIENT AND HESSIAN OPERATORS 375 then Hess f = n X i,j=1 ∂2f ∂xi∂xj − n X k=1 ∂f ∂xk Γk ij ! dxi ⊗dxj, where the Γk ij are the Christoffel symbols of the connection in the chart, namely Γk ij = 1 2 n X l=1 gkl (∂igjl + ∂jgil −∂lgij) , (∗) with ∂kgij = ∂ ∂xk (gij). The formula of Proposition 9.4 is shown in O’Neill . If (gij) is the standard Euclidean metric, the Christoffel symbols vanish and O’Neill’s formula becomes Hess f = n X i,j=1 ∂2f ∂xi∂xj dxi ⊗dxj. For another example of the preceding formula, take f ∈C∞(R2) and let us compute Hess f in terms of polar coordinates (r, θ), where x = r cos θ, and y = r sin θ. Note that ∂ ∂x1 = ∂ ∂r = (cos θ, sin θ) ∂ ∂x2 = ∂ ∂θ = (−r sin θ, r cos θ), which in turn gives gij = 1 0 0 r2  gij = 1 0 0 r−2  . A computation shows that that the only nonzero Christoffel symbols were Γ2 12 = Γ2 21 = 1 r Γ1 22 = −r. Hence Hess f = 2 X i,j=1 ∂2f ∂xi∂xj − 2 X k=1 ∂f ∂xk Γk ij ! dxi ⊗dxj = ∂2f ∂r2 dr ⊗dr +  ∂2f ∂r∂θ −∂f ∂θ Γ2 12  dr ⊗dθ +  ∂2f ∂θ∂r −∂f ∂θ Γ2 21  dθ ⊗dr + ∂2f ∂2θ −∂f ∂r Γ1 22  dθ ⊗dθ = ∂2f ∂r2 dr ⊗dr +  ∂2f ∂r∂θ −1 r ∂f ∂θ  dr ⊗dθ +  ∂2f ∂r∂θ −1 r ∂f ∂θ  dθ ⊗dr + ∂2f ∂2θ + r∂f ∂r  dθ ⊗dθ. 376 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS If we write X = x1 ∂ ∂r + x2 ∂ ∂θ and Y = y1 ∂ ∂r + y2 ∂ ∂θ, then Hess f(X, Y ) = x1 x2  ∂2f ∂r2 ∂2f ∂r∂θ −1 r ∂f ∂θ ∂2f ∂r∂θ −1 r ∂f ∂θ ∂2f ∂2θ + r ∂f ∂r ! y1 y2  . Definition 9.4. A function f ∈C∞(M) is convex (resp. strictly convex) iffits Hessian Hess(f) is positive semi-definite (resp. positive definite). The computation of the gradient of a function defined either on the Stiefel manifold or on the Grassmannian manifold is instructive. Let us first consider the Stiefel manifold S(k, n). Recall that S(k, n) is the set of all orthonormal k-frames, where an orthonormal k-frame is a k-tuples of orthonormal vectors (u1, . . . , uk) with ui ∈Rn. Then SO(n) acts transitively on S(k, n) via the action ·: SO(n) × S(k, n) →S(k, n) R · (u1, . . . , uk) = (Ru1, . . . , Ruk). and that the stabilizer of this action is H = I 0 0 R  R ∈SO(n −k)  . It follows (see Warner , Chapter 3, and Gallier and Quaintance ) that S(k, n) ∼ = G/H, with G = SO(n) and H ∼ = SO(n −k). Observe that the points of G/H ∼ = S(k, n) are the cosets QH, with Q ∈SO(n). If we write Q = [Y Y⊥], where Y consists of the first k columns of Q and Y⊥consists of the last n −k columns of Q, it is clear that [Q] is uniquely determined by Y . We also found that g/h ∼ = m where m = T −A⊤ A 0  T ∈so(k), A ∈Mn−k,k(R)  . The inner product on m is given by ⟨X, Y ⟩= −1 2tr(XY ) = 1 2tr(X⊤Y ), X, Y ∈m. The vector space m is the tangent space ToS(k, n) to S(k, n) at o = [H], the coset of the point corresponding to H. For any other point [Q] ∈G/H ∼ = S(k, n), the tangent space T[Q]S(k, n) is given by T[Q]S(k, n) =  Q S −A⊤ A 0  S ∈so(k), A ∈Mn−k,k(R)  . For every n × k matrix Y ∈S(k, n), this observation implies that tangent vectors to S(k, n) at Y are of the form X = Y S + Y⊥A, 9.1. THE GRADIENT AND HESSIAN OPERATORS 377 where S is any k × k skew-symmetric matrix, A is any (n −k) × k matrix, and [Y Y⊥] is an orthogonal matrix. Given any differentiable function F : S(k, n) →R, if we let FY be the n × k matrix of partial derivatives FY =  ∂F ∂Yij  , we then have dFY (X) = tr(F ⊤ Y X). The gradient grad(F)Y of F at Y is the uniquely defined tangent vector to S(k, n) at Y such that ⟨grad(F)Y , X⟩= dFY (X) = tr(F ⊤ Y X), for all X ∈TY S(k, n). For short, if write Z = grad(F)Y , then it can be shown that Z must satisfy the equation tr(F ⊤ Y X) = tr  Z⊤ I −1 2Y Y ⊤ X  , and since Z is of the form Z = Y S + Y⊥A, and since Y ⊤Y = Ik×k, Y ⊤ ⊥Y = 0, Y ⊤ ⊥Y⊥= I(n−k)×(n−k), we get tr(F ⊤ Y X) = tr  (S⊤Y ⊤+ A⊤Y ⊤ ⊥)  I −1 2Y Y ⊤ X  = tr 1 2S⊤Y ⊤+ A⊤Y ⊤ ⊥  X  for all X ∈TY S(k, n). The above equation implies that we must find Z = Y S + Y⊥A such that F ⊤ Y = 1 2S⊤Y ⊤+ A⊤Y ⊤ ⊥, which is equivalent to FY = 1 2Y S + Y⊥A. From the above equation, we deduce that Y ⊤ ⊥FY = A Y ⊤FY = 1 2S. Since S is skew-symmetric, we get F ⊤ Y Y = −1 2S, so S = Y ⊤FY −F ⊤ Y Y, 378 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS and thus, Z = Y S + Y⊥A = Y (Y ⊤FY −F ⊤ Y Y ) + Y⊥Y ⊤ ⊥FY = (Y Y ⊤+ Y⊥Y ⊤ ⊥)FY −Y F ⊤ Y Y = FY −Y F ⊤ Y Y. Therefore, we proved that the gradient of F at Y is given by grad(F)Y = FY −Y F ⊤ Y Y. Let us now turn to the Grassmannian G(k, n). Recall that G(k, n) is the set of all linear k-dimensional subspaces of Rn, where the k-dimensional subspace U of R is spanned by k linearly independent vectors u1, . . . , uk in Rn; write U = span(u1, . . . , uk). It can be shown that the action ·: SO(n) × G(k, n) →G(k, n) R · U = span(Ru1, . . . , Ruk). is well-defined, transitive, and has the property that stabilizer of U is the set of matrices in SO(n) with the form R = S 0 0 T  , where S ∈O(k), T ∈O(n −k) and det(S) det(T) = 1. We denote this group by S(O(k) × O(n −k)). Since SO(n) is a connected, compact semisimple Lie group whenever n ≥3, This implies that G(k, n) ∼ = SO(n)/S(O(k) × O(n −k)) is a naturally reductive homogeneous manifold whenever n ≥3. It can be shown that g/h ∼ = m where m = 0 −A⊤ A 0  A ∈Mn−k,k(R)  ; see Gallier and Quaintance . For any point [Q] ∈G(k, n) with Q ∈SO(n), if we write Q = [Y Y⊥], where Y denotes the first k columns of Q and Y⊥denotes the last n−k columns of Q, the tangent vectors X ∈T[Q]G(k, n) are of the form X = [Y Y⊥] 0 −A⊤ A 0  = [Y⊥A −Y A⊤], A ∈Mn−k,k(R). This implies that the tangent vectors to G(k, n) at Y are of the form X = Y⊥A, 9.1. THE GRADIENT AND HESSIAN OPERATORS 379 where A is any (n−k)×k matrix. We would like to compute the gradient at Y of a function F : G(k, n) →R. Again, if write Z = grad(F)Y , then Z must satisfy the equation tr(F ⊤ Y X) = ⟨Z, X⟩= tr(Z⊤X), for all X ∈T[Y ]G(k, n). Since Z is of the form Z = Y⊥A, we get tr(F ⊤ Y X) = tr(A⊤Y ⊤ ⊥X), for all X ∈T[Y ]G(k, n), which implies that F ⊤ Y = A⊤Y ⊤ ⊥; that is, FY = Y⊥A. The above yields A = Y ⊤ ⊥FY , so we have Z = Y⊥Y ⊤ ⊥FY = (I −Y Y ⊤)FY . Therefore, the gradient of F at Y is given by grad(F)Y = FY −Y Y ⊤FY . Since the geodesics in the Stiefel manifold and in the Grassmannian can be determined explicitly (see Gallier and Quaintance ), we can find the Hessian of a function using the formula Hess(f)p(X, X) = d2 dt2 f(γ(t)) t=0 . Let us do this for a function F defined on the Grassmannian, the computation on the Stiefel manifold being more complicated; see Edelman, Arias and Smith for details. For any two tangent vectors X1, X2 ∈TY G(k, n) to G(k, n) at Y , define FY Y (X1, X2) by FY Y (X1, X2) = X ij,kl (FY Y )ij,kl(X1)ij(X2)kl, with (FY Y )ij,kl = ∂2F ∂Yij∂Ykl . By using Proposition 9.3, Edelman, Arias and Smith find that a somewhat lengthy computation yields Hess(F)Y (X1, X2) = FY Y (X1, X2) −tr(X⊤ 1 X2Y ⊤FY ), where FY =  ∂F ∂Yij  , as above, when we found a formula for the gradient of F at Y . 380 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS 9.2 The Hodge ∗Operator on Riemannian Manifolds Let M be an n-dimensional Riemann manifold. By Section 2.1 the inner product ⟨−, −⟩p on TpM induces an inner product on T ∗ p M defined as follows. Definition 9.5. For any Riemannian manifold M, the inner product ⟨−, −⟩p on TpM induces an inner product on T ∗ p M given by ⟨w1, w2⟩:= ⟨w♯ 1, w♯ 2⟩, w1, w2 ∈T ∗ p M. This inner product on T ∗ p M defines an inner product on Vk T ∗ p M, with ⟨u1 ∧· · · ∧uk, v1 ∧· · · ∧vk⟩∧= det(⟨ui, vj⟩), for all ui, vi ∈T ∗ p M, and extending ⟨−, −⟩by bilinearity. Therefore, for any two k-forms ω, η ∈Ak(M), we get the smooth function ⟨ω, η⟩given by ⟨ω, η⟩(p) = ⟨ωp, ηp⟩p. Furthermore, if M is oriented, then we can apply the results of Section 3.5 so the vector bundle T ∗M is oriented (by giving T ∗ p M the orientation induced by the orientation of TpM, for every p ∈M), and for every p ∈M, we get a Hodge ∗-operator ∗: k ^ T ∗ p M → n−k ^ T ∗ p M. Then given any k-form ω ∈Ak(M), we can define ∗ω by (∗ω)p = ∗(ωp), p ∈M. We have to check that ∗ω is indeed a smooth form in An−k(M), but this is not hard to do in local coordinates (for help, see Morita , Chapter 4, Section 1). Therefore, if M is a Riemannian oriented manifold of dimension n, we have Hodge ∗-operators ∗: Ak(M) →An−k(M). Observe that ∗1 is just the volume form VolM induced by the metric. Indeed, we know from Section 2.2 that in local coordinates x1, . . . , xn near p, the metric on T ∗ p M is given by the inverse (gij) of the metric (gij) on TpM, and by the results of Section 3.5 (Proposition 3.17), ∗(1) = 1 p det(gij) dx1 ∧· · · ∧dxn = q det(gij) dx1 ∧· · · ∧dxn = VolM. Proposition 3.16 yields the following: 9.2. THE HODGE ∗OPERATOR ON RIEMANNIAN MANIFOLDS 381 Proposition 9.5. If M is a Riemannian oriented manifold of dimension n, then we have the following properties: (i) ∗(fω + gη) = f ∗ω + g ∗η, for all ω, η ∈Ak(M) and all f, g ∈C∞(M). (ii) ∗∗= (−id)k(n−k). (iii) ω ∧∗η = η ∧∗ω = ⟨ω, η⟩VolM, for all ω, η ∈Ak(M). (iv) ∗(ω ∧∗η) = ∗(η ∧∗ω) = ⟨ω, η⟩, for all ω, η ∈Ak(M). (v) ⟨∗ω, ∗η⟩= ⟨ω, η⟩, for all ω, η ∈Ak(M). Recall that exterior differentiation d is a map d: Ak(M) →Ak+1(M). Using the Hodge ∗-operator, we can define an operator δ: Ak(M) →Ak−1(M) that will turn out to be adjoint to d with respect to an inner product on A•(M). Definition 9.6. Let M be an oriented Riemannian manifold of dimension n. For any k, with 1 ≤k ≤n, let δ = (−1)n(k+1)+1 ∗d ∗. Clearly, δ: Ak(M) →Ak−1(M), and δ = 0 on A0(M) = C∞(M). Here is an example of Definition 9.6. Let M = R3 and ω = x dx ∧dy. Since {dx, dy, dz} is an orthonormal basis of T ∗ p R3, we apply Proposition 9.5 (i) and the calculations of Section 3.5 to discover that ∗x dx ∧dy = x ∗dx ∧dy = x dz. Then d(x dz) = d(x) ∧dz = dx ∧dz, and ∗dx ∧dz = −dy. Since n = 3 and k = 2, these calculations imply that δ x dx ∧dy = (−1)3(3)+1(−dy) = −dy. By using the definition of δ, the fact that d ◦d = 0, and Proposition 9.5 (ii), it is a easy to prove the following proposition. Proposition 9.6. Let M be an oriented Riemannian manifold of dimension n. Let d the exterior derivative as defined in Definition 4.9. Let δ be as defined in Definition 9.6. Then ∗δ = (−1)kd∗, δ∗= (−1)k+1 ∗d, δ ◦δ = 0. 382 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS 9.3 The Hodge Laplacian and the Hodge Divergence Operators on Riemannian Manifolds Using d and δ, we can generalize the Laplacian to an operator on differential forms. Definition 9.7. Let M be an oriented Riemannian manifold of dimension n. The Hodge Laplacian or Laplace-Beltrami operator, for short Laplacian, is the operator ∆: Ak(M) → Ak(M) defined by ∆= dδ + δd. A form, ω ∈Ak(M) such that ∆ω = 0, is a harmonic form. In particular, a function f ∈A0(M) = C∞(M) such that ∆f = 0 is called a harmonic function. To demonstrate the Hodge Laplacian, we let M = R3 and calculate ∆ω, where ω = f12 dx ∧dy + f13 dx ∧dz + f23 dy ∧dz. We first determine dδω. Since n = 3 and k = 2, δ = ∗d ∗. Since dx, dy, dz is an orthonormal basis, we use the calculations of Section 3.5 and Proposition 9.5 (i) to determine δω. Note that ∗ω = f12 ∗dx ∧dy + f13 ∗dx ∧dz + f23 ∗dy ∧dz = f12 dz −f13 dy + f23 dx. Then d(f12 dz −f13 dy + f23 dx) =  −∂f13 ∂x −∂f23 ∂y  dx ∧dy + ∂f12 ∂x −∂f23 ∂z  dx ∧dz + ∂f12 ∂y + ∂f13 ∂z  dy ∧dz, and δω = ∗d(f12 dz −f13 dy + f23 dx) =  −∂f13 ∂x −∂f23 ∂y  dz − ∂f12 ∂x −∂f23 ∂z  dy + ∂f12 ∂y + ∂f13 ∂z  dx. Thus dδω =  −∂2f13 ∂x2 −∂2f23 ∂x∂y −∂2f12 ∂y∂z −∂2f13 ∂z2  dx ∧dz +  −∂2f13 ∂x∂y −∂2f23 ∂y2 + ∂2f12 ∂x∂z −∂2f23 ∂z2  dy ∧dz +  −∂2f12 ∂x2 + ∂2f23 ∂x∂z −∂2f12 ∂y2 −∂2f13 ∂y∂z  dx ∧dy. 9.3. THE HODGE LAPLACIAN AND THE HODGE DIVERGENCE OPERATORS 383 It remains to compute δdω. Observe that dω = ∂f12 ∂z −∂f13 ∂y + ∂f23 ∂x  dx ∧dy ∧dz. Since dω is a three form, δ = (−1) ∗d ∗. Once again we go through a three step process to calculate δ. First ∗dω = ∂f12 ∂z −∂f13 ∂y + ∂f23 ∂x . Next d ∗dω = ∂2f12 ∂x∂z −∂2f13 ∂x∂y + ∂2f23 ∂x2  dx + ∂2f12 ∂y∂z −∂2f13 ∂y2 + ∂2f23 ∂x∂y  dy + ∂2f12 ∂z2 −∂2f13 ∂y∂z + ∂2f23 ∂x∂z  dz. Lastly δdω = (−1) ∗d ∗dω = − ∂2f12 ∂x∂z −∂2f13 ∂x∂y + ∂2f23 ∂x2  dy ∧dz + ∂2f12 ∂y∂z −∂2f13 ∂y2 + ∂2f23 ∂x∂y  dx ∧dz − ∂2f12 ∂z2 −∂2f13 ∂y∂z + ∂2f23 ∂x∂z  dx ∧dy. Finally we discover that ∆ω = dδω + δdω =  −∂2f13 ∂x2 −∂2f23 ∂x∂y −∂2f12 ∂y∂z −∂2f13 ∂z2  dx ∧dz +  −∂2f13 ∂x∂y −∂2f23 ∂y2 + ∂2f12 ∂x∂z −∂2f23 ∂z2  dy ∧dz +  −∂2f12 ∂x2 + ∂2f23 ∂x∂z −∂2f12 ∂y2 −∂2f13 ∂y∂z  dx ∧dy − ∂2f12 ∂x∂z −∂2f13 ∂x∂y + ∂2f23 ∂x2  dy ∧dz + ∂2f12 ∂y∂z −∂2f13 ∂y2 + ∂2f23 ∂x∂y  dx ∧dz − ∂2f12 ∂z2 −∂2f13 ∂y∂z + ∂2f23 ∂x∂z  dx ∧dy =  −∂2f12 ∂x2 −∂2f12 ∂y2 −∂2f12 ∂z2  dx ∧dy +  −∂2f13 ∂x2 −∂2f13 ∂y2 −∂2f13 ∂z2  dx ∧dz +  −∂2f23 ∂x2 −∂2f23 ∂y2 −∂2f23 ∂z2  dy ∧dz. 384 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS Notice that the coefficients of the two-form ∆ω are given by the negative of the harmonic operator on functions as defined in Section 8.5. In fact, if M = Rn with the Euclidean metric and f is a smooth function, a laborious computation yields ∆f = − n X i=1 ∂2f ∂x2 i ; that is, the usual Laplacian with a negative sign in front. (The computation can be found in Morita , Example 4.12, or Jost , Chapter 2, Section 2.1). By using Proposition 9.6, it is easy to see that ∆commutes with ∗; that is, ∆∗= ∗∆, We have ∆∗= (dδ + δd)∗= dδ ∗+δd∗ = (−1)k+1d ∗d + (−1)kδ ∗δ = (−1)k+1(−1)k+1 ∗δd + (−1)kδ ∗δ, since ∗acts on a k + 1 form = ∗δd + (−1)k(−1)kdδ∗, since ∗acts on a k −1 form = ∗(δd + dδ) = ∗∆. Definition 9.8. Let M be an oriented Riemannian manifold of dimension n. Given any vector field X ∈X(M), its Hodge divergence div X is defined by div X = δX♭. Now for a function f ∈C∞(M), we have δf = 0, so ∆f = δd f. However, div(grad f) = δ(grad f)♭= δ((d f)♯)♭= δd f, so ∆f = div grad f, as in the case of Rn. Remark: Since the definition of δ involves two occurrences of the Hodge ∗-operator, δ also makes sense on non-orientable manifolds by using a local definition. Therefore, the Laplacian ∆and the divergence also makes sense on non-orientable manifolds. In the rest of this section we assume that M is orientable. The relationship between δ and d can be made clearer by introducing an inner product on forms with compact support. Recall that Ak c(M) denotes the space of k-forms with compact support (an infinite dimensional vector space). Let k ≥1. 9.3. THE HODGE LAPLACIAN AND THE HODGE DIVERGENCE OPERATORS 385 Definition 9.9. For any two k-forms with compact support ω, η ∈Ak c(M), set (ω, η) = Z M ⟨ω, η⟩VolM = Z M ⟨ω, η⟩∗(1). If k = 0, then ω, η ∈C∞(M) and we define (ω, η) = Z M ω η VolM. Using Proposition 9.5 (iii), we have (ω, η) = Z M ⟨ω, η⟩VolM = Z M ω ∧∗η = Z M η ∧∗ω, so it is easy to check that (−, −) is indeed an inner product on k-forms with compact support. We can extend this inner product to forms with compact support in A• c(M) = Ln k=0 Ak c(M) by making Ah c(M) and Ak c(M) orthogonal if h ̸= k. Proposition 9.7. If M is an n-dimensional orientable Riemannian manifold, then δ is (formally) adjoint to d; that is, (dω, η) = (ω, δη), for all ω ∈Ak−1 c (M) and η ∈Ak c(M) with compact support. Proof. By linearity and orthogonality of the Ak c(M), the proof reduces to the case where ω ∈Ak−1 c (M) and η ∈Ak c(M) (both with compact support). By definition of δ and the fact that ∗∗= (−id)(k−1)(n−k+1) for ∗: Ak−1(M) →An−k+1(M), we have ∗δ = (−1)kd∗, and since d(ω ∧∗η) = dω ∧∗η + (−1)k−1ω ∧d ∗η = dω ∧∗η −ω ∧∗δη we get Z M d(ω ∧∗η) = Z M dω ∧∗η − Z M ω ∧∗δη = (dω, η) −(ω, δη). However, by Stokes’ theorem (Theorem 7.16), Z M d(ω ∧∗η) = 0, so (dω, η) −(ω, δη) = 0; that is, (dω, η) = (ω, δη), as claimed. 386 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS Corollary 9.8. If M is an n-dimensional orientable Riemannian manifold, then the Lapla-cian ∆is self-adjoint; that is, (∆ω, η) = (ω, ∆η), for all k-forms ω, η with compact support. Proof. Uisng Proposition 9.7 several times we have (∆ω, η) = (dδω + δdω, η) = (dδω, η) + (δdω, η) = (δω, δη) + (dω, dη) = (ω, dδη) + (ω, δdη) = (ω, dδη + δdη) = (ω, ∆η). We also obtain the following useful fact: Proposition 9.9. If M is an n-dimensional orientable Riemannian manifold, then for every k-form ω with compact support, ∆ω = 0 iffdω = 0 and δω = 0. Proof. Since ∆= dδ +δd, it is obvious that if dω = 0 and δω = 0, then ∆ω = 0. Conversely, (∆ω, ω) = ((dδ + δd)ω, ω) = (dδω, ω) + (δdω, ω) = (δω, δω) + (dω, dω). Thus, if ∆ω = 0, then (δω, δω) = (dω, dω) = 0, which implies dω = 0 and δω = 0. As a consequence of Proposition 9.9, if M is a connected, orientable, compact Riemannian manifold, then every harmonic function on M is a constant. Indeed, if M is compact then f is a 0-form of compact support, and if ∆f = 0 then d f = 0. Since f is connected, f is a constant function. 9.4 The Hodge and Laplace–Beltrami Laplacians of Functions For practical reasons we need a formula for the Hodge Laplacian of a function f ∈C∞(M), in local coordinates. If (U, ϕ) is a chart near p, as usual, let ∂f ∂xj (p) = ∂(f ◦ϕ−1) ∂uj (ϕ(p)), where (u1, . . . , un) are the coordinate functions in Rn. Write |g| = det(gij), where (gij) is the symmetric, positive definite matrix giving the metric in the chart (U, ϕ). 9.4. THE HODGE AND LAPLACE–BELTRAMI LAPLACIANS OF FUNCTIONS 387 Proposition 9.10. If M is an n-dimensional orientable Riemannian manifold, then for every local chart (U, ϕ), for every function f ∈C∞(M), we have ∆f = − 1 p |g| X i,j ∂ ∂xi p |g| gij ∂f ∂xj  . Proof. We follow Jost , Chapter 2, Section 1. Pick any function h ∈C∞(M) with compact support. We have Z M (∆f)h ∗(1) = (∆f, h) = (δd f, h) = (d f, dh) = Z M ⟨d f, dh⟩∗(1) = Z M X ij gij ∂f ∂xi ∂h ∂xj p |g| dx1 · · · dxn = − Z M X ij 1 p |g| ∂ ∂xj p |g| gij ∂f ∂xi  h p |g| dx1 · · · dxn = − Z M X ij 1 p |g| ∂ ∂xj p |g| gij ∂f ∂xi  h ∗(1). where we have used integration by parts in the second to last line. Since the above equation holds for all h, we get our result. It turns out that in a Riemannian manifold, the divergence of a vector field and the Laplacian of a function can be given by a definition that uses the covariant derivative instead of the Hodge ∗-operator. We did this in Section 8.4. A comparison of Proposition 9.10 with Line (∗∗) of Section 8.4, shows that the definition of the Hodge Laplacian of a function differs by a sign factor with the definition of the Laplacian provided by Definition 8.12. We reconcile the difference between these two definitions by defining the notion of connection divergence and connection Laplacian via the negation of the quantity described in Definition 8.12. Definition 9.10. Let M be a Riemannian manifold. If ∇is the Levi-Civita connection induced by the Riemannian metric, then the connection divergence (for short divergence) of a vector field X ∈X(M) is the function divC X : M →R defined so that (divC X)(p) = tr(Y (p) 7→(−∇Y X)p); namely, for every p, (divC X)(p) is the trace of the linear map Y (p) 7→(−∇Y X)p. The connection Laplacian of f ∈C∞M is defined as ∆Cf = divC grad f. 388 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS The connection divergence and the connection Laplacian make sense even if M is non-orientable. This is also true for the Hodge divergence and the Hodge Laplacian. Because of the sign change provided by Definition 9.10, the Hodge Laplacian ∆f agrees with the con-nection Laplacian ∆Cf. Thus, we will not distinguish between the two notions of Laplacian on a function. Since the connection Laplacian and the Hodge Laplacian (for functions) agree, we should expect that the two variants of the divergence operator also agree. This is indeed the case but a proof is not so easily found in the literature. We are aware of two proofs: one is found in Petersen (Chapter 7, Proposition 32) for compact orientable manifolds, and the other in Rosenberg for orientable manifolds, closer to the proof of Proposition 9.10. We present the second proof because it applies to a more general situation and yields an explicit formula. Proposition 9.11. If M is an n-dimensional orientable Riemannian manifold, then for every local chart (U, ϕ), for every vector field X ∈X(M), we have div X = − n X i=1 1 p |g| ∂ ∂xi p |g|Xi  . Proof. (Following Rosenberg .) Let (U, ϕ) be a chart for M. Within this chart, any X ∈X(M) is expressed as X = Pn i=1 Xi ∂ ∂xi. Take f ∈C∞(M) with compact support and compute (X, gradf) = Z M ⟨X, gradf⟩∗(1) = Z M d f(X) ∗(1) = Z M n X i=1 Xi ∂f ∂xi p |g| dx1 · · · dxn = − Z M n X i=1 1 p |g| ∂ ∂xi p |g|Xi  f p |g| dx1 · · · dxn, where the last equality follows from integration by parts. We claim (X, gradf) = (div X, f) since (divX, f) = (δX♭, f) = (X♭, d f), by Proposition 9.7 = ((X♭)♯, (d f)♯), definition of inner product on one forms = (X, (d f)♯) = (X, grad f), by the remark preceding Definition 9.1. 9.5. DIVERGENCE AND LIE DERIVATIVE OF THE VOLUME FORM 389 Thus we have shown (div X, f) = − Z M n X i=1 1 p |g| ∂ ∂xi p |g|Xi  f ∗(1) = − n X i=1 1 p |g| ∂ ∂xi p |g|Xi  , f + for all f ∈C∞(M) with compact support, and this concludes the proof. By comparing the expression for div X provided by Proposition 9.11 with the expression of divCX given by Line (†) of Section 8.4, we have the following proposition. Proposition 9.12. If M is an orientable Riemannian manifold, then for every vector field X ∈X(M), the connection divergence is given by divC X = δX♭= div X. Consequently, for the Laplacian, we have ∆f = δd f = div grad f. Proposition 9.12 shows there is no need to distinguish between the Hodge divergence and the connection divergence. Thus we will use the notation div X to simply denote the divergence of a vector field over T(M). Our next result shows relationship between div X and the Lie derivative of the volume form. 9.5 Divergence and Lie Derivative of the Volume Form Proposition 9.13. Let M be an n-dimensional Riemannian manifold. For any vector field X ∈X(M), we have LX VolM = −(div X)VolM, where div X is the connection divergence of X. Proof. (Following O’Neill (Chapter 7, Lemma 21).) Let X1, X2, . . . Xn be an orthonormal frame on M such that VolM(X1, . . . , Xn) = 1. Then LX(VolM(X1, . . . , Xn)) = LX(1) = X(1) = 0, and Proposition 4.19 (2) implies (LXVolM)(X1, . . . , Xn) = − n X i=1 VolM(X1, . . . , LXXi, . . . Xn). 390 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS Fix i and set LXXi = [X, Xi] = Pn j=1 fijXj. Since VolM is multilinear and skew-symmetric, we find that VolM(X1, . . . , LXXi, . . . Xn) = VolM(X1, . . . , n X j=1 fijXj, . . . Xn) = n X j=1 VolM(X1, . . . , fijXj, . . . Xn) = fiiVolM(X1, . . . Xi, . . . Xn) = fii. By varying i we discover that (LXVolM)(X1, . . . , Xn) = − n X i=1 VolM(X1, . . . , LXXi, . . . Xn) = − n X i=1 fii. On the other hand, since (div X)(p) = tr(Y (p) 7→(−∇Y X)p), X1, . . . , Xn is an orthonormal frame, and ∇is the Levi-Civita connection (which is torsion free), the equation before Definition 2.3 implies that −div X = n X i=1 ⟨∇XiX, Xi⟩ = − n X i=1 ⟨[X, Xi], Xi⟩+ n X i=1 ⟨∇XXi, Xi⟩, since ∇XiX −∇XXi = [Xi, X] = − n X i=1 ⟨[X, Xi], Xi⟩, since 0 = ∇X⟨Xi, Xi⟩= 2⟨∇XXi, Xi⟩ = − n X i=1 ⟨ n X j=1 fijXj, Xi⟩= − n X i=1 fii⟨Xi, Xi⟩= − n X i=1 fii. Thus we have shown −div X = (LXVolM)(X1, . . . , Xn), which is equivalent to the statement found in the proposition. Proposition 9.13 is interesting in its own right since it is used in the proof of Green’s theorem. But before stating and proving Green’s theorem, we reformulate Proposition 9.13 through the application of Cartan’s formula. Proposition 9.14. The following formula holds: (div X)VolM = −d(i(X)VolM). 9.5. DIVERGENCE AND LIE DERIVATIVE OF THE VOLUME FORM 391 Proof. By Cartan’s formula (Proposition 4.20), LX = i(X) ◦d + d ◦i(X); as dVolM = 0 (since VolM is a top form), Proposition 9.13 implies (div X)VolM = −d(i(X)VolM). The above formulae also holds for a local volume form (i.e. for a volume form on a local chart). Proposition 9.15. (Green’s Formula) If M is an orientable and compact Riemannian man-ifold without boundary, then for every vector field X ∈X(M), we have Z M (div X) VolM = 0. Proof. Proofs of Proposition 9.15 can be found in Gallot, Hulin and Lafontaine (Chapter 4, Proposition 4.9) and Helgason (Chapter 2, Section 2.4). Since Proposition 9.13 implies that (div X)VolM = −d(i(X)VolM), we have Z M (div X) VolM = − Z M d(i(X)VolM) = − Z ∂M i(X)VolM = 0 where the last equality follows by Stokes’ theorem, since ∂M = 0. We end this section by discussing an alternative definition for the operator δ: A1(M) → A0(M) in terms of the covariant derivative (see Gallot, Hulin and Lafontaine , Chapter 4). For any one-form ω ∈A1(M), and any X, Y ∈X(M), define (∇Xω)(Y ) := X(ω(Y )) −ω(∇XY ). It turns out that δω = −tr ∇ω, where the trace should be interpreted as the trace of the R-bilinear map X, Y 7→(∇Xω)(Y ), as in Chapter 2 (see Proposition 2.3). This means that in any chart (U, ϕ), δω = − n X i=1 (∇Eiω)(Ei), for any orthonormal frame field (E1, . . . , En) over U. By applying this trace definition of δω, it can be shown that δ(fd f) = f∆f −⟨grad f, grad f⟩. Proposition 9.16. For any orientable, compact manifold M, we have (∆f, f) = Z M f∆f VolM = Z M ⟨grad f, grad f⟩VolM. 392 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS Proof. Proposition 9.12 implies that δ(fd f) = δ((f d f)♯)♭= div(f d f)♯, and since Green’s formula implies that Z M δ(f d f) VolM = Z M div(f d f)♯VolM = 0, we conclude that (∆f, f) = Z M f∆f VolM = Z M ⟨grad f, grad f⟩VolM, for any orientable, compact manifold M. There is a generalization of the formula expressing δω over an orthonormal frame E1, . . ., En for a one-form ω that applies to any differential form. In fact, there are formulae express-ing both d and δ over an orthornormal frame and its coframe, and these are often handy in proofs. The formula for δω will be used in the proof of Theorem 9.26. Recall that for every vector field X ∈X(M), the interior product i(X): Ak+1(M) → Ak(M) is defined by (i(X)ω)(Y1, . . . , Yk) = ω(X, Y1, . . . , Yk), for all Y1, . . . , Yk ∈X(M). Proposition 9.17. Let M be a compact, orientable Riemannian manifold. For every p ∈M, for every local chart (U, ϕ) with p ∈M, if (E1, . . . , En) is an orthonormal frame over U and (θ1, . . . , θn) is its dual coframe, then for every k-form ω ∈Ak(M), we have: dω = n X i=1 θi ∧∇Eiω δω = − n X i=1 i(Ei)∇Eiω. A proof of Proposition 9.17 can be found in Petersen (Chapter 7, Proposition 37) or Jost (Chapter 3, Lemma 3.3.4). When ω is a one-form, δωp is just a number, and indeed δω = − n X i=1 i(Ei)∇Eiω = − n X i=1 (∇Eiω)(Ei), as stated earlier. 9.6. HARMONIC FORMS, THE HODGE THEOREM, POINCAR´ E DUALITY 393 9.6 Harmonic Forms, the Hodge Theorem, Poincar´ e Duality Let us now assume that M is orientable and compact. Definition 9.11. Let M be an orientable and compact Riemannian manifold of dimension n. For every k, with 0 ≤k ≤n, let Hk(M) = {ω ∈Ak(M) | ∆ω = 0}, the space of harmonic k-forms. The following proposition is left as an easy exercise: Proposition 9.18. Let M be an orientable and compact Riemannian manifold of dimension n. The Laplacian commutes with the Hodge ∗-operator, and we have a linear map ∗: Hk(M) →Hn−k(M). One of the deepest and most important theorems about manifolds is the Hodge decom-position theorem, which we now state. Theorem 9.19. (Hodge Decomposition Theorem) Let M be an orientable and compact Rie-mannian manifold of dimension n. For every k, with 0 ≤k ≤n, the space Hk(M) is finite dimensional, and we have the following orthogonal direct sum decomposition of the space of k-forms: Ak(M) = Hk(M) ⊕d(Ak−1(M)) ⊕δ(Ak+1(M)). The proof of Theorem 9.19 involves a lot of analysis and it is long and complicated. A complete proof can be found in Warner (Chapter 6). Other treatments of Hodge theory can be found in Morita (Chapter 4) and Jost (Chapter 2). The Hodge decomposition theorem has a number of important corollaries, one of which is Hodge theorem: Theorem 9.20. (Hodge Theorem) Let M be an orientable and compact Riemannian mani-fold of dimension n. For every k, with 0 ≤k ≤n, there is an isomorphism between Hk(M) and the de Rham cohomology vector space Hk DR(M): Hk DR(M) ∼ = Hk(M). Proof. Since by Proposition 9.9, every harmonic form ω ∈Hk(M) is closed, we get a linear map from Hk(M) to Hk DR(M) by assigning its cohomology class [ω] to ω. This map is injective. Indeed, if [ω] = 0 for some ω ∈Hk(M), then ω = dη for some η ∈Ak−1(M) so (ω, ω) = (dη, ω) = (η, δω). 394 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS But, as ω ∈Hk(M) we have δω = 0 by Proposition 9.9, so (ω, ω) = 0; that is, ω = 0. Our map is also surjective. This is the hard part of Hodge theorem. By the Hodge decomposition theorem, for every closed form ω ∈Ak(M), we can write ω = ωH + dη + δθ, with ωH ∈Hk(M), η ∈Ak−1(M), and θ ∈Ak+1(M). Since ω is closed and ωH ∈Hk(M), we have dω = 0 and dωH = 0, thus dδθ = 0 and so 0 = (dδθ, θ) = (δθ, δθ); that is, δθ = 0. Therefore, ω = ωH +dη, which implies [ω] = [ωH] with ωH ∈Hk(M), proving the surjectivity of our map. The Hodge theorem also implies the Poincar´ e duality theorem. If M is a compact, orientable, n-dimensional smooth manifold, for each k, with 0 ≤k ≤n, we define a bilinear map ((−, −)): Hk DR(M) × Hn−k DR (M) − →R by setting (([ω], [η])) = Z M ω ∧η. We need to check that this definition does not depend on the choice of closed forms in the cohomology classes [ω] and [η]. However, if ω+dα is another representative in [ω] and η+dβ is another representative in [η], as dω = dη = 0, we have d(α ∧η + (−1)kω ∧β + α ∧dβ) = dα ∧η + ω ∧dβ + dα ∧dβ, so by Stokes’ theorem, Z M (ω + dα) ∧(η + dβ) = Z M ω ∧η + Z M d(α ∧η + (−1)kω ∧β + α ∧dβ) = Z M ω ∧η. Theorem 9.21. (Poincar´ e Duality) If M is a compact, orientable, smooth manifold of dimension n, then the bilinear map ((−, −)): Hk DR(M) × Hn−k DR (M) − →R defined above is a nondegenerate pairing, and hence yields an isomorphism Hk DR(M) ∼ = (Hn−k DR (M))∗. 9.7. THE BOCHNER LAPLACIAN AND THE BOCHNER TECHNIQUE 395 Proof. Pick any Riemannian metric on M. It is enough to show that for every nonzero cohomology class [ω] ∈Hk DR(M), there is some [η] ∈Hn−k DR (M) such that (([ω], [η])) = Z M ω ∧η ̸= 0. By the Hodge theorem, we may assume that ω is a nonzero harmonic form. By Proposition 9.18, η = ∗ω is also harmonic and η ∈Hn−k(M). Then, we get (ω, ω) = Z M ω ∧∗ω = (([ω], [η])), and indeed, (([ω], [η])) ̸= 0, since ω ̸= 0. 9.7 The Bochner Laplacian, Weitzenb¨ ock Formula and the Bochner Technique Let M be a compact orientable Riemannian manifold.1 The goal of this section is to de-fine another notion of Laplacian on k-forms in terms of a generalization of the Levi-Civita connection ∇: X(M) × X(M) →X(M) to k-forms viewed as a linear map ∇: Ak(M) →HomC∞(M)(X(M), Ak(M)), and in terms of a certain adjoint ∇∗of ∇, a linear map ∇∗: HomC∞(M)(X(M), Ak(M)) →Ak(M). Since we already have an inner product (−, −) on k-forms as explained in Section 9.3, we will define an inner product ((−, −)) on HomC∞(M)(X(M), Ak(M)) and define ∇∗so that (∇∗A, ω) = ((A, ∇ω)) for all A ∈HomC∞(M)(X(M), Ak(M)) and all ω ∈Ak(M). Our exposition is heavily inspired by Petersen (Chapter 7, Section 3.2), but Petersen deals with the more general case of a vector bundle and we restrict our attention to the simpler case of a Riemannian manifold. The definition of the inner product ((−, −)) on HomC∞(M)(X(M), Ak(M)) is accom-plished in four steps. 1The Bochner Laplacian makes sense for noncompact manifolds as long as we consider forms with compact support, but we have no need for this more general setting. 396 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS 1. First, we define the connection ∇: Ak(M) →HomC∞(M)(X(M), Ak(M)) on k-forms. We define the covariant derivative ∇Xω of any k-form ω ∈Ak(M) as the k-form given by (∇Xω)(Y1, . . . , Yk) = X(ω(Y1, . . . , Yk)) − k X j=1 ω(Y1, . . . , ∇XYj, . . . , Yk); (†) see Proposition 10.13 for a justification. We can view ∇as a linear map ∇: Ak(M) →HomC∞(M)(X(M), Ak(M)), where ∇ω is the C∞(M)-linear map X 7→∇Xω. 2. The second step is to define the adjoint of a linear map in HomC∞(M)(X(M), Ak(M)). We use two inner products, one on differential forms and one on vector fields. The inner product ⟨−, −⟩p on TpM (with p ∈M) induces an inner product on differ-ential forms, namely (ω, η) = Z M ⟨ω, η⟩VolM = Z M ⟨ω, η⟩∗(1), as we explained in Section 9.3. We also obtain an inner product on vector fields if, for any two vector field X, Y ∈ X(M), we define (X, Y )X by (X, Y )X = Z M ⟨X, Y ⟩VolM, where ⟨X, Y ⟩is the function defined pointwise by ⟨X, Y ⟩(p) = ⟨X(p), Y (p)⟩p. Now for any linear map A ∈HomC∞(M)(X(M), Ak(M)), let A∗be the adjoint of A defined by (AX, θ) = (X, A∗θ)X, for all vector fields X ∈X(M) and all k-forms θ ∈Ak(M). It can be verified that A∗∈HomC∞(M)(Ak(M), X(M)). 3. In the third step, given A, B ∈HomC∞(M)(X(M), Ak(M)), the expression tr(A∗B) is a smooth function on M, and it can be verified that ⟨⟨A, B⟩⟩= tr(A∗B) defines a non-degenerate pairing on HomC∞(M)(X(M), Ak(M)). Using this pairing, we obtain the (R-valued) inner product on HomC∞(M)(X(M), Ak(M)) given by ((A, B)) = Z M tr(A∗B) VolM. 9.7. THE BOCHNER LAPLACIAN AND THE BOCHNER TECHNIQUE 397 4. The fourth and final step is to define the (formal) adjoint ∇∗of ∇: Ak(M) →HomC∞(M)(X(M), Ak(M)) as the linear map ∇∗: HomC∞(M)(X(M), Ak(M)) →Ak(M) defined implicitly by (∇∗A, ω) = ((A, ∇ω)); that is, (∇∗A, ω) = Z M ⟨∇∗A, ω⟩VolM = Z M ⟨⟨A, ∇ω⟩⟩VolM = Z M tr(A∗∇ω) VolM = ((A, ∇ω)), for all A ∈HomC∞(M)(X(M), Ak(M)) and all ω ∈Ak(M).  The notation ∇∗for the adjoint of ∇should not be confused with the dual connection on T ∗M of a connection ∇on TM! In the second interpretation, ∇∗denotes the connection on A∗(M) induced by the original connection ∇on TM. The argument type (differential form or vector field) should make it clear which ∇is intended, but it might have been better to use a notation such as ∇⊤instead of ∇∗. What we just did also applies to A∗(M) = Ln k=0 Ak(M) (where dim(M) = n), and so we can view the connection ∇as a linear map ∇: A∗(M) →HomC∞(M)(X(M), A∗(M)), and its adjoint as a linear map ∇∗: HomC∞(M)(X(M), A∗(M)) →A∗(M). Definition 9.12. Given a compact, orientable Riemannian manifold M, the Bochner Lapla-cian (or connection Laplacian) ∇∗∇is defined as the composition of the connection ∇: A∗(M) →HomC∞(M)(X(M), A∗(M)) with its adjoint ∇∗: HomC∞(M)(X(M), A∗(M)) → A∗(M), as defined above. Observe that (∇∗∇ω, ω) = ((∇ω, ∇ω)) = Z M ⟨⟨∇ω, ∇ω⟩⟩VolM, for all ω ∈Ak(M). Consequently, the “harmonic forms” ω with respect to ∇∗∇must satisfy ∇ω = 0, but this condition is not equivalent to the harmonicity of ω with respect to the Hodge Laplacian. Thus, in general, ∇∗∇and ∆are different operators. The relationship between the two is given by formulae involving contractions of the curvature tensor, and are known as Weitzenb¨ ock formulae. We will state such a formula in case of one-forms later on. In order to do this, we need to give another definition of the Bochner Laplacian using second covariant derivatives of forms. 398 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS If ω ∈A1(M) is a one-form, then the covariant derivative of ω defines a (0, 2)-tensor T given by T(Y, Z) = (∇Y ω)(Z). Thus, we can define the second covariant derivative ∇2 X,Y ω of ω as the covariant derivative of T (see Proposition 10.13); that is, ∇2 X,Y ω = (∇XT)(Y, Z) = X(T(Y, Z)) −T(∇XY, Z) −T(Y, ∇XZ). Proposition 9.22. The following formula holds for any one-form ω ∈A1(M): ∇2 X,Y ω = ∇X(∇Y ω) −∇∇XY ω; Proof. We have (∇2 X,Y ω)(Z) = X((∇Y ω)(Z)) −(∇∇XY ω)(Z) −(∇Y ω)(∇XZ) = X((∇Y ω)(Z)) −(∇Y ω)(∇XZ) −(∇∇XY ω)(Z) = X(β(Z)) −β(∇XZ) −(∇∇XY ω)(Z), β is the one-form ∇Y ω = ∇Xβ(Z) −(∇∇XY ω)(Z), definition of covariant derivative given by (†) = (∇X(∇Y ω))(Z) −(∇∇XY ω)(Z). Therefore, ∇2 X,Y ω = ∇X(∇Y ω) −∇∇XY ω, as claimed. Note that ∇2 X,Y ω is formally the same as the second covariant derivative ∇2 X,Y Z with ω replacing Z; see Gallier and Quaintance . It is natural to generalize the second covariant derivative ∇2 X,Y to k-forms as follows. Definition 9.13. Given any k-form ω ∈Ak(M), for any two vector fields X, Y ∈X(M), we define ∇2 X,Y ω by ∇2 X,Y ω = ∇X(∇Y ω) −∇∇XY ω. We also need the definition of the trace of ∇2 X,Y ω. Definition 9.14. Given any local chart (U, ϕ) and given any orthonormal frame (E1, . . ., En) over U, we can defined the trace tr(∇2ω) of ∇2 X,Y ω given by tr(∇2ω) = n X i=1 ∇2 Ei,Eiω. It is easily seen that tr(∇2ω) does not depend on the choice of local chart and orthonormal frame. By using the this notion of trace, may calculate the connection Laplacian as follows: 9.7. THE BOCHNER LAPLACIAN AND THE BOCHNER TECHNIQUE 399 Proposition 9.23. If is M a compact, orientable, Riemannian manifold, then the connection Laplacian ∇∗∇is given by ∇∗∇ω = −tr(∇2ω), for all differential forms ω ∈A∗(M). The proof of Proposition 9.23, which is quite technical, can be found in Petersen (Chapter 7, Proposition 34). Given any one-forms ω ∈A1(M), it is natural to ask what is the one-form ∇2 X,Y ω −∇2 Y,Xω. To answer this question, we need to first recall the definition of the curvature tensor. Given X, Y, Z ∈X(M), the curvature tensor R(X, Y )Z is the (1, 3)-tensor defined by R(X, Y )(Z) = ∇[X,Y ]Z + ∇Y ∇XZ −∇X∇Y Z. Assuming that ∇is the Levi-Civita connection, the following result can be shown. Proposition 9.24. The following equation holds: R(X, Y )Z = ∇2 Y,XZ −∇2 X,Y Z. For a proof, see Gallot, Hullin, Lafontaine or Gallier and Quaintance . We now are in a position to answer the preceding question. The answer is given by the following proposition which plays a crucial role in the proof of a version of Bochner’s formula: Proposition 9.25. For any vector fields X, Y, Z ∈X(M) and any one-form ω ∈A1(M) on a Riemannian manifold M, we have ((∇2 X,Y −∇2 Y,X)ω)(Z) = ω(R(X, Y )Z). Proof. (Adapted from Gallot, Hullin, Lafontaine , Lemma 4.13.) It is proven in Section 10.7 that (∇Xω)♯= ∇Xω♯. We claim that we also have (∇2 X,Y ω)♯= ∇2 X,Y ω♯. This is because (∇2 X,Y ω)♯= (∇X(∇Y ω))♯−(∇∇XY ω)♯ = ∇X(∇Y ω)♯−∇∇XY ω♯ = ∇X(∇Y ω♯) −∇∇XY ω♯ = ∇2 X,Y ω♯. 400 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS Thus, using Proposition 9.24 we deduce that ((∇2 X,Y −∇2 Y,X)ω)♯= (∇2 X,Y −∇2 Y,X)ω♯= R(Y, X)ω♯. Consequently, ((∇2 X,Y −∇2 Y,X)ω)(Z) = ⟨((∇2 X,Y −∇2 Y,X)ω)♯, Z⟩ = ⟨R(Y, X)ω♯, Z⟩ = R(Y, X, ω♯, Z) = R(X, Y, Z, ω♯) = ⟨R(X, Y )Z, ω♯⟩ = ω(R(X, Y )Z), using properties of the Riemann tensor; see Gallot, Hullin, Lafontaine or Gallier and Quaintance . We are now ready to prove the Weitzenb¨ ock formulae for one-forms. Theorem 9.26. (Weitzenb¨ ock–Bochner Formula) If is M a compact, orientable, Rieman-nian manifold, then for every one-form ω ∈A1(M), we have ∆ω = ∇∗∇ω + Ric(ω), where Ric(ω) is the one-form given by Ric(ω)(X) = ω(Ric♯(X)), and where Ric♯is the Ricci curvature viewed as a (1, 1)-tensor (that is, ⟨Ric♯(u), v⟩p = Ric(u, v), for all u, v ∈TpM and all p ∈M). Proof. (Adapted from Gallot, Hullin, Lafontaine , Proposition 4.36.) For any p ∈M, pick any normal local chart (U, ϕ) with p ∈U, and pick any orthonormal frame (E1, . . . , En) over U. Because (U, ϕ) is a normal chart at p, we have (∇EjEj)p = 0 for all i, j. Recall from the discussion at the end of Section 9.3 as a special case of Proposition 9.17 that for every one-form ω, we have δω = − X i ∇Eiω(Ei), where δω ∈C∞(M). Then dδ(w) is the one form defined via d(δω)(X) = − X i d(∇Eiω(Ei))(X) = − X i ∇X∇Eiω(Ei). since (∇Xf)p = d fp(Xp) for all X ∈X(M). Also recall Proposition 4.16, which states that dω(X, Y ) = X(ω(Y )) −Y (ω(X)) −ω([X, Y ]). 9.7. THE BOCHNER LAPLACIAN AND THE BOCHNER TECHNIQUE 401 By definition, ∇Xω(Y ) = X(ω(Y )) −ω(∇XY ) ∇Y ω(X) = Y (ω(X)) −ω(∇Y X). Hence dω(X, Y ) = ∇Xω(Y ) + ω(∇XY ) −∇Y ω(X) −ω(∇Y X) −ω([X, Y ]) = ∇Xω(Y ) −∇Xω(Y ) + ω(∇XY −∇Y X) −ω([X, Y ]). Since we are using the Levi-Civita connection, ∇XY −∇Y X = [X, Y ], and the preceding calculation becomes dω(X, Y ) = ∇Xω(Y ) −∇Y ω(X). Let β be the two-form dω. Note that ∇Eiβ is also a two-form. We use Proposition 9.17 to calculate the one form δβ as follows: (δβ)(X) = − X i (i(Ei)∇Eiβ) (X) = − X i ∇Eiβ(Ei, X). In other words, we found that (δdω)(X) = − X i ∇Eidω(Ei, X) = − X i ∇Ei∇Eiω(X) + X i ∇Ei∇Xω(Ei), where the last equality is an application of Proposition 4.16. Thus, we get ∆ω(X) = − X i ∇Ei∇Eiω(X) + X i (∇Ei∇X −∇X∇Ei)ω(Ei) = − X i ∇2 Ei,Eiω(X) + X i (∇2 Ei,X −∇2 X,Ei)ω(Ei) = ∇∗∇ω(X) + X i ω(R(Ei, X)Ei) = ∇∗∇ω(X) + ω(Ric♯(X)), using the fact that (∇EjEj)p = 0 for all i, j, and using Proposition 9.25 and Proposition 9.23. For simplicity of notation, we will write Ric(u) for Ric♯(u). There should be no confusion, since Ric(u, v) denotes the Ricci curvature, a (0, 2)-tensor. There is another way to express Ric(ω) which will be useful in the proof of the next theorem. Proposition 9.27. The Weitzenb¨ ock formula can be written as ∆ω = ∇∗∇ω + (Ric(ω♯))♭. 402 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS Proof. Observe that Ric(ω)(Z) = ω(Ric(Z)) = ⟨ω♯, Ric(Z)⟩ = ⟨Ric(Z), ω♯⟩ = Ric(Z, ω♯) = Ric(ω♯, Z) = ⟨Ric(ω♯), Z⟩ = ((Ric(ω♯))♭(Z), and thus, Ric(ω)(Z) = ((Ric(ω♯))♭(Z). Consequently the Weitzenb¨ ock formula can be written as ∆ω = ∇∗∇ω + (Ric(ω♯))♭. The Weitzenb¨ ock–Bochner formula implies the following theorem due to Bochner: Theorem 9.28. (Bochner) If M is a compact, orientable, connected Riemannian manifold, then the following properties hold: (i) If the Ricci curvature is non-negative, that is Ric(u, u) ≥0 for all p ∈M and all u ∈TpM, and if Ric(u, u) > 0 for some p ∈M and all u ∈TpM, then H1 DRM = (0). (ii) If the Ricci curvature is non-negative, then ∇ω = 0 for all ω ∈A1(M), and dim H1 DRM ≤dim M. Proof. (After Gallot, Hullin, Lafontaine ; Theorem 4.37.) (i) Assume H1 DRM ̸= (0). Then by the Hodge theorem, Theorem 9.20, there is some nonzero harmonic one-form ω. The Weitzenb¨ ock–Bochner formula implies that (∆ω, ω) = (∇∗∇ω, ω) + ((Ric(ω♯))♭, ω). Since ∆ω = 0, we get 0 = (∇∗∇ω, ω) + ((Ric(ω♯))♭, ω) = ((∇ω, ∇ω)) + Z M ⟨(Ric(ω♯))♭, ω⟩VolM = ((∇ω, ∇ω)) + Z M ⟨Ric(ω♯), ω♯⟩VolM = ((∇ω, ∇ω)) + Z M Ric(ω♯, ω♯) VolM. 9.8. PROBLEMS 403 However, ((∇ω, ∇ω)) ≥0, and by the assumption on the Ricci curvature, the integrand is nonnegative and strictly positive at some point, so the integral is strictly positive, a contradiction. (ii) Again, for any one-form ω, we have (∆ω, ω) = ((∇ω, ∇ω)) + Z M Ric(ω♯, ω♯) VolM, so if the Ricci curvature is non-negative, ∆ω = 0 iff∇ω = 0. This means that ω is invariant by parallel transport (see Section 11.4), and thus ω is completely determined by its value ωp at some point p ∈M, so there is an injection H1(M) − →T ∗ p M, which implies that dim H1 DRM = dim H1(M) ≤dim M. There is a version of the Weitzenb¨ ock formula for p-forms, but it involves a more com-plicated curvature term and its proof is also more complicated; see Petersen (Chapter 7). The Bochner technique can also be generalized in various ways, in particular, to spin manifolds, but these considerations are beyond the scope of these notes. Let us just say that Weitzenb¨ ock formulae involving the Dirac operator play an important role in physics and 4-manifold geometry. We refer the interested reader to Gallot, Hulin and Lafontaine (Chapter 4) Petersen (Chapter 7), Jost (Chaper 3), and Berger (Section 15.6), for more details on Weitzenb¨ ock formulae and the Bochner technique. 9.8 Problems Problem 9.1. Let M be a Riemannian manifold and let f ∈C∞(M). In local coordinates with respect to a chart, if we write d f = n X i=1 ∂f ∂xi dxi, show that Hess f = n X i,j=1 ∂2f ∂xi∂xj − n X k=1 ∂f ∂xk Γk ij ! dxi ⊗dxj, where the Γk ij are the Christoffel symbols of the connection in the chart, namely Γk ij = 1 2 n X l=1 gkl (∂igjl + ∂jgil −∂lgij) , (∗) with ∂kgij = ∂ ∂xk (gij). Hint. See O’Neill . 404 CHAPTER 9. OPERATORS ON RIEMANNIAN MANIFOLDS Problem 9.2. Let G(k, n) be the set of all linear k-dimensional subspaces of Rn. Using the notation of Section 9.1 show that for any two tangent vectors X1, X2 ∈TY G(k, n) to G(k, n) at Y , Hess(F)Y (X1, X2) = FY Y (X1, X2) −tr(X⊤ 1 X2Y ⊤FY ), where FY =  ∂F ∂Yij  , and where FY Y (X1, X2) = X ij,kl (FY Y )ij,kl(X1)ij(X2)kl, with (FY Y )ij,kl = ∂2F ∂Yij∂Ykl . Hint. Use Proposition 9.3, and Edelman, Arias and Smith . Problem 9.3. Prove Proposition 9.5. Hint. See Proposition 3.16. Problem 9.4. Prove Proposition 9.6. Problem 9.5. For any one-form ω ∈A1(M), and any X, Y ∈X(M), recall that (∇Xω)(Y ) := X(ω(Y )) −ω(∇XY ). It turns out that δω = −tr ∇ω, where the trace should be interpreted as the trace of the R-bilinear map X, Y 7→(∇Xω)(Y ), as in Chapter 2 (see Proposition 2.3). By applying this trace definition of δω, show that δ(fd f) = f∆f −⟨grad f, grad f⟩. Problem 9.6. Prove Proposition 9.17. Hint. See Petersen , Chapter 7, Proposition 37, or Jost , Chapter 3, Lemma 3.3.4. Problem 9.7. Prove Proposition 9.18. Problem 9.8. Prove Proposition 9.23. Hint. See Petersen , Chapter 7, Proposition 34. Chapter 10 Bundles, Metrics on Bundles, and Homogeneous Spaces A transitive action ·: G × X →X of a group G on a set X yields a description of X as a quotient G/Gx, where Gx is the stabilizer of any element, x ∈X (see Warner , Chapter 3, of Gallier and Quaintance ). The points of X are identified with the left cosets gGx (g ∈G). If X is a “well-behaved” topological space (say a locally compact Hausdorffspace), G is a “well-behaved” topological group (say a locally compact topological group which is countable at infinity), and the action is continuous, then G/Gx is homeomorphic to X (see Bourbaki , Chapter IX, Section 5, Proposition 6, or Gallier and Quaintance ). In particular these conditions are satisfied if G is a Lie group and X is a manifold. Intuitively, the above theorem says that G can be viewed as a family of “fibres” gGx, all isomorphic to Gx, these fibres being parametrized by the “base space” X, and varying smoothly when the point corresponding to the coset gGx moves in X. We have an example of what is called a fibre bundle, in fact, a principal fibre bundle. This view of G as a family of fibres gGx, with x ∈X, is a special case of the notion of a fibre bunlde. Intuitively, a fibre bundle over B is a family E = (Eb)b∈B of spaces Eb (fibres) indexed by B and varying smoothly as b moves in B, such that every Eb is diffeomorphic to some prespecified space F. The space E is called the total space, B the base space, and F the fibre. A way to define such a family is to specify a surjective map π: E →B. We will assume that E, B, F are smooth manifolds and that π is a smooth map. The theory of bundles can be developed for topological spaces but we do need such generality. The type of bundles that we just described is too general and to develop a useful theory it is necessary to assume that locally, a bundle looks likes a product. Technically, this is achieved by assuming that there is some open cover U = (Uα)α∈I of B and that there is a family (ϕα)α∈I of diffeomorphisms ϕα : π−1(Uα) →Uα × F. Intuitively, above Uα, the open subset π−1(Uα) looks like a product. The maps ϕα are called local trivializations. 405 406 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES The last important ingredient in the notion of a fibre bundle is the specifiction of the “twisting” of the bundle; that is, how the fibre Eb = π−1(b) gets twisted as b moves in the base space B. Technically, such twisting manifests itself on overlaps Uα ∩Uβ ̸= ∅. It turns out that we can write ϕα ◦ϕ−1 β (b, x) = (b, gαβ(b)(x)) for all b ∈Uα∩Uβ and all x ∈F. The term gαβ(b) is a diffeomorphism of F. Then we require that the family of diffeomorphisms gαβ(b) belongs to a Lie group G, which is expressed by specifying that the maps gαβ, called transitions maps, are maps gαβ : Uα ∩Uβ →G. The purpose of the group G, called the structure group, is to specify the “twisting” of the bundle. Fibre bundles are defined in Section 10.1. The family of transition maps gαβ satisfies an important condition on nonempty overlaps Uα ∩Uβ ∩Uγ called the cocycle condition: gαβ(b)gβγ(b) = gαγ(b) (where gαβ(b), gβγ(b), gαγ(b) ∈G), for all α, β, γ such that Uα ∩Uβ ∩Uγ ̸= ∅and all b ∈ Uα ∩Uβ ∩Uγ. In Section 10.2, we define bundle morphisms, and the notion of equivalence of bundles over the same base, following Hirzebruch and Chern . We show that two bundles (over the same base) are equivalent if and only if they are isomorphic. In Section 10.3, we describe the construction of a fibre bundle with prescribed fibre F and structure group G from a base manifold, B, an open cover U = (Uα)α∈I of B, and a family of maps gαβ : Uα ∩Uβ →G satisfying the cocycle condition, called a cocycle. This construction is the basic tool for constructing new bundles from old ones. This construction is applied to define the notion of pullback bundle. Section 10.4 is devoted to a special kind of fibre bundle called vector bundles. A vector bundle is a fibre bundle for which the fibre is a finite-dimensional vector space V , and the structure group is a subgroup of the group of linear isomorphisms (GL(n, R) or GL(n, C), where n = dim V ). Typical examples of vector bundles are the tangent bundle TM and the cotangent bundle T ∗M of a manifold M. We define maps of vector bundles, and equivalence of vector bundles. The constuction of a vector bundle in terms of a cocycle also applies to vector bundles. We give a criterion for a vector bundle to be trivial (isomorphic to B × V ) in terms of the existence of a frame of global sections. In Section 10.5 we describe various operations on vector bundles: Whitney sums, ten-sor products, tensor powers, exterior powers, symmetric powers, dual bundles, and Hom bundles. We also define the complexification of a real vector bundle. In Section 10.6 we discuss properties of the sections of a vector bundle ξ. We prove that the space of sections Γ(ξ) is a finitely generated projective C∞(B)-module. We also prove various useful isomorphisms. 10.1. FIBRE BUNDLES 407 Section 10.7 is devoted to the covariant derivative of tensor fields, and to the duality between vector fields and differential forms. In Section 10.8 we explain how to give a vector bundle a Riemannian metric. This is achieved by supplying a smooth family (⟨−, −⟩b)b∈B of inner products on each fibre π−1(b) above b ∈B. We describe the notion of reduction of the structure group and define orientable vector bundles. In Section 10.9 we consider the special case of fibre bundles for which the fibre coincides with the structure group G, which acts on itself by left translations. Such fibre bundles are called principal bundles. It turns out that a principal bundle can be defined in terms of a free right action of a Lie group on a smooth manifold. When principal bundles are defined in terms of free right actions, the notion of bundle morphism is also defined in terms of equivariant maps. There are two constructions that allow us to reduce the study of fibre bundles to the study of principal bundles. Given a fibre bundle ξ with fibre F, we can construct a principal bundle P(ξ) obtained by replacing the fibre F by the group G. Conversely, given a principal bundle ξ and an effective action of G on a manifold F, we can construct the fibre bundle ξ[F] obtained by replacing G by F. The Borel construction provides a direct construction of ξ[F]. The maps ξ 7→ξ[F] and ξ 7→P(ξ) induce a bijection between equivalence classes of principal G-bundles and fibre bundles (with structure group G). Furthermore, ξ is a trivial bundle iffP(ξ) is a trivial bundle. The equivalence of fibre bundles and principal bundles (over the same base B, and with the same structure group G) is the key to the classification of fibre bundles, but we do not discuss this deep result. Section 10.10 is devoted to principal bundles that arise from proper and free actions of a Lie group. When the base space is a homogenous space, which means that it arises from a transitive action of a Lie group, then the total space is a principal bundle. There are many illustrations of this situation involving SO(n + 1) and SU(n + 1). 10.1 Fibre Bundles We begin by carefully stating the definition of a fibre bundle because we believe that it clarifies the notions of vector bundles and principal fibre bundles, the concepts that are our primary concern. The following definition is not the most general, but it is sufficient for our needs. Definition 10.1. A fibre bundle with (typical) fibre F and structure group G is a tuple ξ = (E, π, B, F, G), where E, B, F are smooth manifolds, π: E →B is a smooth surjective map, G is a Lie group of diffeomorphisms of F, and there is some open cover U = (Uα)α∈I 408 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES of B and a family ϕ = (ϕα)α∈I of diffeomorphisms ϕα : π−1(Uα) →Uα × F. The space B is called the base space, E is called the total space, F is called the (typical) fibre, and each ϕα is called a (local) trivialization. The pair (Uα, ϕα) is called a bundle chart, and the family {(Uα, ϕα)} is a trivializing cover. For each b ∈B, the space π−1(b) is called the fibre above b; it is also denoted by Eb, and π−1(Uα) is also denoted by E ↾Uα; see Figure 10.1. B E F Figure 10.1: The spiky cylinder E is a typical fibre bundle with base B as the purple cylinder and fibre isomorphic to a line segment. The following properties hold: (a) (local triviality) The diagram π−1(Uα) π $ ϕα / Uα × F p1 { Uα commutes for all α ∈I, where p1 : Uα × F →Uα is the first projection. Equivalently, for all (b, y) ∈Uα × F, π ◦ϕ−1 α (b, y) = b. (b) (fibre diffeomorphism) For every (Uα, ϕα) and every b ∈Uα, because p1 ◦ϕα = π, by (a) the restriction of ϕα to Eb = π−1(b) is a diffeomorphism between Eb and {b} × F, so we have the diffeomorphism ϕα,b : Eb →F 10.1. FIBRE BUNDLES 409 given by ϕα,b(Z) = (p2 ◦ϕα)(Z), for all Z ∈Eb; see Figure 10.2. Furthermore, for all Uα, Uβ in U such that Uα ∩Uβ ̸= ∅, for every B Uα Uα E b F b b Z φ φ α α(Z ) = (b,y) y p 2 Figure 10.2: An illustration of ϕα,b : Eb →F over B = S1. b ∈Uα ∩Uβ, there is a relationship between ϕα,b and ϕβ,b which gives the twisting of the bundle. (c) (fibre twisting) The diffeomorphism ϕα,b ◦ϕ−1 β,b : F →F is an element of the group G. (d) (transition maps) The map gαβ : Uα ∩Uβ →G defined by gαβ(b) = ϕα,b ◦ϕ−1 β,b, is smooth. The maps gαβ are called the transition maps of the fibre bundle. A fibre bundle ξ = (E, π, B, F, G) is also referred to, somewhat loosely, as a fibre bundle over B or a G-bundle, and it is customary to use the notation F − →E − →B, or F / E  B, 410 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES even though it is imprecise (the group G is missing!), and it clashes with the notation for short exact sequences. Observe that the bundle charts (Uα, ϕα) are similar to the charts of a manifold. Actually, Definition 10.1 is too restrictive because it does not allow for the addition of compatible bundle charts, for example when considering a refinement of the cover U. This problem can easily be fixed using a notion of equivalence of trivializing covers analogous to the equivalence of atlases for manifolds (see Definition 10.2 below). Also observe that (b), (c), and (d) imply that the isomorphism ϕα ◦ϕ−1 β : (Uα ∩Uβ) × F →(Uα ∩Uβ) × F is related to the smooth map gαβ : Uα ∩Uβ →G by the identity ϕα ◦ϕ−1 β (b, x) = (b, gαβ(b)(x)), (∗) for all b ∈Uα ∩Uβ and all x ∈F. We interpret gαβ(b)(x) as the action of the group element gαβ(b) on x; see Figure 10.4. Intuitively, a fibre bundle over B is a family E = (Eb)b∈B of spaces Eb (fibres) indexed by B and varying smoothly as b moves in B, such that every Eb is diffeomorphic to F. The bundle E = B × F, where π is the first projection, is called the trivial bundle (over B). The trivial bundle B ×F is often denoted ϵF. The local triviality Condition (a) says that locally, that is over every subset Uα from some open cover of the base space B, the bundle ξ ↾Uα is trivial. Note that if G is the trivial one-element group, then the fibre bundle is trivial. In fact, the purpose of the group G is to specify the “twisting” of the bundle; that is, how the fibre Eb gets twisted as b moves in the base space B. A M¨ obius strip is an example of a nontrivial fibre bundle where the base space B is the circle S1, the fibre space F is the closed interval [−1, 1], and the structural group is G = {1, −1}, where −1 is the reflection of the interval [−1, 1] about its midpoint 0. The total space E is the strip obtained by rotating the line segment [−1, 1] around the circle, keeping its midpoint in contact with the circle, and gradually twisting the line segment so that after a full revolution, the segment has been tilted by π. See Figure 10.3. Note that U1 = {−π < x < π 2}, U2 = {0 < x < 3π 2 }, while U1 ∩U2 = V ∪W where V = {0 < x < π 2} and W = {−π < x < −π 2}. The transition map is ϕ1 ◦ϕ−1 2 (b, x) = (b, g12(b)x) where g12(b) = 1 if b ∈V and g12(b) = −1 if b ∈W. A Klein bottle is also a fibre bundle for which both the base space and the fibre are the circle, S1, while G = {−1, 1}. Again, the reader should work out the details for this example. Other examples of fibre bundles are: (1) SO(n + 1), an SO(n)-bundle over the sphere Sn with fibre SO(n). (for n ≥0). (2) SU(n + 1), an SU(n)-bundle over the sphere S2n+1 with fibre SU(n) (for n ≥0). (3) SL(2, R), an SO(2)-bundle over the upper-half space H, with fibre SO(2). 10.1. FIBRE BUNDLES 411 U1 U2 V W U1 U2 0 π π -π π 2 3 2 0 V π 2 π π 3 2 π 2 V (b,x) π π 3 2 0 W W (b,x) π 2 (b,x) W -π -V π 2 0 (b,x) U1 U2 φ φ 1 2 Figure 10.3: The M¨ obius strip as a line bundle over the unit circle. (4) GL(n, R), an O(n)-bundle over the space SPD(n) of symmetric, positive definite matrices, with fibre O(n). (5) GL+(n, R), an SO(n)-bundle over the space, SPD(n) of symmetric, positive definite matrices, with fibre SO(n). (6) SO(n + 1), an O(n)-bundle over the real projective space RPn, with fibre O(n) (for n ≥0). (7) SU(n + 1), an U(n)-bundle over the complex projective space CPn, with fibre U(n) (for n ≥0). (8) O(n), an O(k) × O(n −k)-bundle over the Grassmannian G(k, n), with fibre O(k) × O(n −k). (9) SO(n), an S(O(k) × O(n −k))-bundle over the Grassmannian G(k, n), with fibre S(O(k) × O(n −k)). (10) SO(n), an SO(n −k)-bundle over the Stiefel manifold S(k, n), with 1 ≤k ≤n −1. 412 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES (11) The Lorentz group, SO0(n, 1), is an SO(n)-bundle over the space H+ n (1) consisting of one sheet of the hyperbolic paraboloid Hn(1), with fibre SO(n) (see Gallier and Quaintance ). Observe that in all the examples above, F = G; that is, the typical fibre is identical to the group G. Special bundles of this kind are called principal fibre bundles. The above definition is slightly different (but equivalent) to the definition given in Bott and Tu , page 47-48. Definition 10.1 is closer to the one given in Hirzebruch (Chapter I, Section 3.2(a)). Bott and Tu and Hirzebruch assume that G acts effectively (or faithfully) on the left on the fibre F. This means that there is a smooth action ·: G × F →F. If G is a Lie group and if F is a manifold, an action ϕ: G × F →F is smooth if ϕ is smooth. Also recall that G acts effectively (or faithfully) on F ifffor every g ∈G, if g · x = x for all x ∈F, then g = 1. Every g ∈G induces a diffeomorphism ϕg : F →F, defined by ϕg(x) = g · x, for all x ∈F. The fact that G acts effectively on F means that the map g 7→ϕg is injective. This justifies viewing G as a group of diffeomorphisms of F. If instead of viewing G as a group of diffeomorphisms of F we assume that we have a smooth effective action of G on F we modify Definition 10.1 by deleting Condition (c) and replacing Condition (d) by the following condition: (c’) For all α, β, there is a smooth map gαβ : Uα ∩Uβ →G called a transition function such that ϕα,b ◦ϕ−1 β,b(x) = gαβ(b) · x, b ∈Uα ∩Uβ, x ∈F, and thus ϕα ◦ϕ−1 β (b, x) = (b, gαβ(b) · x), b ∈Uα ∩Uβ, x ∈F. With Condition (c’) replacing Conditions (c) and (d) Definition 10.1 is essentially Defi-nition 4.2 in Davis and Kirk . Since G acts effectively on F, there is a bijection between G and a subgroup of the group of diffeomorphisms of F so we can view the group element gαβ(b) as the corresponding diffeomorphism and denote gαβ(b) · x as gαβ(b)(x). There are situations in which it is desirable to generalize the notion of fibre bundle by dropping the assumption that the action of G on F is effective. If the action of G on F is not effective, two different group elements of G may induce the same diffeomorphism of F, so two distinct group elements gαβ(b) and gα′β′(b) could define the same diffeomorphism ϕα,b ◦ϕ−1 β,b = ϕα′,b ◦ϕ−1 β′,b, but this is not a problem. We observed that Definition 10.1 is too restrictive because it does not allow for the addition of compatible bundle charts. We can fix this problem as follows: 10.1. FIBRE BUNDLES 413 Definition 10.2. Let ξ = (E, π, B, F, G) be fibre bundle defined as in Definition 10.1. Given a trivializing cover {(Uα, ϕα)}, for any open U of B and any diffeomorphism ϕ: π−1(U) →U × F, we say that (U, ϕ) is compatible with the trivializing cover {(Uα, ϕα)} iffwhenever U ∩Uα ̸= ∅, there is some smooth map gα : U ∩Uα →G, so that ϕ ◦ϕ−1 α (b, x) = (b, gα(b) · x), for all b ∈U ∩Uα and all x ∈F, which according to the notational convention introduced above is also written as ϕ ◦ϕ−1 α (b, x) = (b, gα(b)(x)). Two trivializing covers are equivalent iffevery bundle chart of one cover is compatible with the other cover. This is equivalent to saying that the union of two trivializing covers is a trivializing cover. Definition 10.2 implies the following Definition 10.3. Using the conventions of Definition 10.1, a fibre bundle is a tuple (E, π, B, F, G, {(Uα, ϕα)}), where {(Uα, ϕα)} is an equivalence class of trivializing covers. As for manifolds, given a trivializing cover {(Uα, ϕα)}, the set of all bundle charts compatible with {(Uα, ϕα)} is a maximal trivializing cover equivalent to {(Uα, ϕα)}; see Figure 10.4. b (b, x) φ α φ α (b,x) -1 α φ (b,x) -1 α φ (b, g (b)(x)) = φ U U α x F x F Figure 10.4: A schematic illustration of the transition between two elements of a trivializing cover. A special case of the above occurs when we have a trivializing cover {(Uα, ϕα)} with U = {Uα} an open cover of B, and another open cover V = (Vβ)β∈J of B which is a refinement 414 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES of U. This means that there is a map τ : J →I, such that Vβ ⊆Uτ(β) for all β ∈J. Then for every Vβ ∈V, since Vβ ⊆Uτ(β), the restriction of ϕτ(β) to Vβ is a trivialization ϕ′ β : π−1(Vβ) →Vβ × F, and Conditions (b) and (c) are still satisfied, so (Vβ, ϕ′ β) is compatible with {(Uα, ϕα)}. The family of transition functions (gαβ) satisfies the following crucial conditions. Definition 10.4. Given a fibre bundle ξ = (E, π, B, F, G, {(Uα, ϕα)}) with family of transi-tion functions (gαβ), the cocycle condition is the set of equations gαβ(b)gβγ(b) = gαγ(b) (where gαβ(b), gβγ(b), gαγ(b) ∈G), for all α, β, γ such that Uα ∩Uβ ∩Uγ ̸= ∅and all b ∈ Uα ∩Uβ ∩Uγ; see Figure 10.5. U x F U x F U x F (b,x) (b, g (b)x) αγ γ αγ g β (b, g (b)x) βγ gβγ α αβ g (b, g (b)g (b)x) βγ αβ = φ φ φ γ β α Figure 10.5: A schematic illustration of the cocycle condition. The three sheets of the bundle actually glue together into a single sheet. Setting α = β = γ, we get gαα = id, 10.1. FIBRE BUNDLES 415 and setting γ = α, we get gβα = g−1 αβ. Again, beware that this means that gβα(b) = g−1 αβ(b), where g−1 αβ(b) is the inverse of gβα(b) in G. In general, g−1 αβ is not the functional inverse of gβα. Experience shows that most objects of interest in geometry (vector fields, differential forms, etc.) arise as sections of certain bundles. Furthermore, deciding whether or not a bundle is trivial often reduces to the existence of a (global) section. Thus, we define the important concept of a section right away. Definition 10.5. Given a fibre bundle ξ = (E, π, B, F, G), a smooth section of ξ is a smooth map s: B →E, so that π ◦s = idB. Given an open subset U of B, a (smooth) section of ξ over U is a smooth map s: U →E, so that π ◦s(b) = b, for all b ∈U; we say that s is a local section of ξ. The set of all sections over U is denoted Γ(U, ξ), and Γ(B, ξ) (for short, Γ(ξ)) is the set of global sections of ξ; see Figure 10.6. B s(B) Figure 10.6: An illustration of a global section of E ∼ = B × R where B is the unit disk. Here is an observation that proves useful for constructing global sections. Let s: B →E be a global section of a bundle ξ. For every trivialization ϕα : π−1(Uα) →Uα × F, let sα : Uα →E and σα : Uα →F be given by sα = s ↾Uα and σα = pr2 ◦ϕα ◦sα, so that sα(b) = ϕ−1 α (b, σα(b)). Obviously, π ◦sα = id, so sα is a local section of ξ, and σα is a function σα : Uα →F. We claim that on overlaps, we have σα(b) = gαβ(b)σβ(b). See Figure 10.7. 416 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES B U U α β β s α = s(U ) α sβ= s(U ) β Uα x F U x F σ σ φ φ φ α α α s ( ) ( α ) β σ φ ( ) ( sβ ) β β gαβ Figure 10.7: An illustration of the gluing of two local sections sα and sβ. Proof. Indeed, recall that ϕα ◦ϕ−1 β (b, x) = (b, gαβ(b)x), for all b ∈Uα ∩Uβ and all x ∈F, and as sα = s ↾Uα and sβ = s ↾Uβ, sα and sβ agree on Uα ∩Uβ. Consequently, from sα(b) = ϕ−1 α (b, σα(b)) and sβ(b) = ϕ−1 β (b, σβ(b)), we get ϕ−1 α (b, σα(b)) = sα(b) = sβ(b) = ϕ−1 β (b, σβ(b)) = ϕ−1 α (b, gαβ(b)σβ(b)), which implies σα(b) = gαβ(b)σβ(b), as claimed. Conversely, assume that we have a collection of functions σα : Uα →F, satisfying σα(b) = gαβ(b)σβ(b) on overlaps. Let sα : Uα →E be given by sα(b) = ϕ−1 α (b, σα(b)). Each sα is a local section and we claim that these sections agree on overlaps, so they patch and define a global section s. Proof. We need to show that sα(b) = ϕ−1 α (b, σα(b)) = ϕ−1 β (b, σβ(b)) = sβ(b), 10.2. BUNDLE MORPHISMS, EQUIVALENT AND ISOMORPHIC BUNDLES 417 for b ∈Uα ∩Uβ; that is, (b, σα(b)) = ϕα ◦ϕ−1 β (b, σβ(b)), and since ϕα ◦ϕ−1 β (b, σβ(b)) = (b, gαβ(b)σβ(b)), and by hypothesis, σα(b) = gαβ(b)σβ(b), our equation sα(b) = sβ(b) is verified. 10.2 Bundle Morphisms, Equivalent and Isomorphic Bundles Now that we have defined a fibre bundle, it is only natural to analyze mappings between two fibre bundles. The notion of a map between fibre bundles is more subtle than one might think because of the structure group G. Let us begin with the simpler case where G = Diff(F), the group of all smooth diffeomorphisms of F. Definition 10.6. If ξ1 = (E1, π1, B1, F, Diff(F)) and ξ2 = (E2, π2, B2, F, Diff(F)) are two fibre bundles with the same typical fibre F and the same structure group G = Diff(F), a bundle map (or bundle morphism) f : ξ1 →ξ2 is a pair f = (fE, fB) of smooth maps fE : E1 →E2 and fB : B1 →B2, such that (a) The following diagram commutes: E1 π1  fE / E2 π2  B1 fB / B2 (b) For every b ∈B1, the map of fibres fE ↾π−1 1 (b): π−1 1 (b) →π−1 2 (fB(b)) is a diffeomorphism (preservation of the fibre). A bundle map f : ξ1 →ξ2 is an isomorphism if there is some bundle map g: ξ2 →ξ1, called the inverse of f, such that gE ◦fE = id and fE ◦gE = id. The bundles ξ1 and ξ2 are called isomorphic. Given two fibre bundles ξ1 = (E1, π1, B, F, Diff(F)) and ξ2 = (E2, π2, B, F, Diff(F)) over the same base space B, a bundle map (or bundle morphism) f : ξ1 →ξ2 is a pair f = (fE, fB), where fB = id (the identity map). Such a bundle map is an isomorphism if it has an inverse as defined above. In this case, we say that the bundles ξ1 and ξ2 over B are isomorphic. 418 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES Observe that the commutativity of the diagram in Definition 10.6 implies that fB is actually determined by fE. Also, when f is an isomorphism, the surjectivity of π1 and π2 implies that gB ◦fB = id and fB ◦gB = id. Thus when f = (fE, fB) is an isomorphism, both fE and fB are diffeomorphisms. Remark: Some authors do not require the “preservation” of fibres. However, it is automatic for bundle isomorphisms. Let us take a closer look at what it means for a bundle map to preserve fibres. When we have a bundle map f : ξ1 →ξ2 as above, for every b ∈B, for any trivializations ϕα : π−1 1 (Uα) →Uα × F of ξ1 and ϕ′ β : π−1 2 (Vβ) →Vβ × F of ξ2, with b ∈Uα and fB(b) ∈Vβ, we have the map ϕ′ β ◦fE ◦ϕ−1 α : (Uα ∩f −1 B (Vβ)) × F →Vβ × F. Consequently, as ϕα and ϕ′ β are diffeomorphisms and as f is a diffeomorphism on fibres, we have a map ρα,β : Uα ∩f −1 B (Vβ) →Diff(F), such that ϕ′ β ◦fE ◦ϕ−1 α (b, x) = (fB(b), ρα,β(b)(x)), for all b ∈Uα ∩f −1 B (Vβ) and all x ∈F; see Figure 10.8. (b,x) α α B φ 1 B2 ( fB(b) , ρα,β (b)(x)) V x F β φβ ‘ f f E B f B -1 ( V ) β U -1 h ( ) x F b fB(b) Figure 10.8: The construction of the map ϕ′ β ◦fE ◦ϕ−1 α between the M¨ obius strip bundle ξ1 and the cylinder bundle ξ2. 10.2. BUNDLE MORPHISMS, EQUIVALENT AND ISOMORPHIC BUNDLES 419 Since we may always pick Uα and Vβ so that fB(Uα) ⊆Vβ, we may also write ρα instead of ρα,β, with ρα : Uα →G. Then observe that locally, fE is given as the composition π−1 1 (Uα) ϕα / Uα × F e fα / Vβ × F ϕ′ β −1 / π−1 2 (Vβ) z / (b, x) / (fB(b), ρα(b)(x)) / ϕ′ β −1(fB(b), ρα(b)(x)), with e fα(b, x) = (fB(b), ρα(b)(x)), that is, fE(z) = ϕ′ β −1(fB(b), ρα(b)(x)), with z ∈π−1 1 (Uα) and (b, x) = ϕα(z). Conversely, if (fE, fB) is a pair of smooth maps satisfying the commutative diagram of Definition 10.6 and the above conditions hold locally, then as ϕα, ϕ′−1 β , and ρα(b) are diffeo-morphisms on fibres, we see that fE is a diffeomorphism on fibres. In the general case where the structure group G is not the whole group of diffeomorphisms Diff(F), there is no guarantee that ρα(b) ∈G. This is the case if ξ is a vector bundle or a principal bundle, but if ξ is a fibre bundle, following Hirzebruch , we use the local conditions above to define the “right notion” of bundle map, namely Definition 10.7. Another advantage of this definition is that two bundles (with the same fibre, structure group, and base) are isomorphic iffthey are equivalent (see Proposition 10.1 and Proposition 10.2). Definition 10.7. Given two fibre bundles ξ1 = (E1, π1, B1, F, G) and ξ2 = (E2, π2, B2, F, G) with the same fibre and the same structure group, a bundle map f : ξ1 →ξ2 is a pair f = (fE, fB) of smooth maps fE : E1 →E2 and fB : B1 →B2, such that: (a) The diagram E1 π1  fE / E2 π2  B1 fB / B2 commutes. (b) There is an open cover U = (Uα)α∈I for B1, an open cover V = (Vβ)β∈J for B2, a family ϕ = (ϕα)α∈I of trivializations ϕα : π−1 1 (Uα) →Uα × F for ξ1, a family ϕ′ = (ϕ′ β)β∈J of trivializations ϕ′ β : π−1 2 (Vβ) →Vβ × F for ξ2, such that for every b ∈B, there are some trivializations ϕα : π−1 1 (Uα) →Uα × F and ϕ′ β : π−1 2 (Vβ) →Vβ × F, with fB(Uα) ⊆Vβ, b ∈Uα and some smooth map ρα : Uα →G, such that ϕ′ β ◦fE ◦ϕ−1 α : Uα × F →Vα × F is given by ϕ′ β ◦fE ◦ϕ−1 α (b, x) = (fB(b), ρα(b)(x)), for all b ∈Uα and all x ∈F. 420 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES See Figure 10.8. A bundle map is an isomorphism if it has an inverse as in Definition 10.6. If the bundles ξ1 and ξ2 are over the same base B, then we also require fB = id. As we remarked in the discussion before Definition 10.7, Condition (b) insures that the maps of fibres fE ↾π−1 1 (b): π−1 1 (b) →π−1 2 (fB(b)) are diffeomorphisms. In the special case where ξ1 and ξ2 have the same base, B1 = B2 = B, we require fB = id, and we can use the same cover (i.e., U = V), in which case Condition (b) becomes: There is some smooth map ρα : Uα →G, such that ϕ′ α ◦f ◦ϕα −1(b, x) = (b, ρα(b)(x)), for all b ∈Uα and all x ∈F. Definition 10.8. We say that a bundle ξ with base B and structure group G is trivial iff ξ is isomorphic to the product bundle B × F, according to the notion of isomorphism of Definition 10.7. We can also define the notion of equivalence for fibre bundles over the same base space B (see Hirzebruch , Section 3.2, Chern , Section 5, and Husemoller , Chapter 5). We will see shortly that two bundles over the same base are equivalent iffthey are isomorphic. Definition 10.9. Given two fibre bundles ξ1 = (E1, π1, B, F, G) and ξ2 = (E2, π2, B, F, G) over the same base space B, we say that ξ1 and ξ2 are equivalent if there is an open cover U = (Uα)α∈I for B, a family ϕ = (ϕα)α∈I of trivializations ϕα : π−1 1 (Uα) →Uα × F for ξ1, a family ϕ′ = (ϕ′ α)α∈I of trivializations ϕ′ α : π−1 2 (Uα) →Uα × F for ξ2, and a family (ρα)α∈I of smooth maps ρα : Uα →G, such that g′ αβ(b) = ρα(b)gαβ(b)ρβ(b)−1, for all b ∈Uα ∩Uβ; see Figure 10.9. Since the trivializations are bijections, the family (ρα)α∈I is unique. The conditions for two fibre bundles to be equivalent are local. Nevertheless, they are strong enough to imply that equivalent bundles over the same base are isomorphic (see Proposition 10.2). The following proposition shows that isomorphic fibre bundles are equivalent. Proposition 10.1. If two fibre bundles ξ1 = (E1, π1, B, F, G) and ξ2 = (E2, π2, B, F, G) over the same base space B are isomorphic, then they are equivalent. Proof. Let f : ξ1 →ξ2 be a bundle isomorphism. In a slight abuse of notation, we also let f : E1 →E2 be the isomorphism between E1 and E2. Then by Definition 10.7 we know that 10.2. BUNDLE MORPHISMS, EQUIVALENT AND ISOMORPHIC BUNDLES 421 0 1 2 3 z z’ U U U U α α Uα Uα β β Uβ Uβ x F x F x F x F φ φ φ φ α α β β ‘ ‘ ρ ρβ α g (b) (b) (b) αβ -1 b b ξ ξ 1 2 Figure 10.9: An illustration of the mapping g′ αβ(b) = ρα(b)gαβ(b)ρβ(b)−1. Point 0 is ϕ′ β(z′) = (b, x). Point 1 is (b, ρ−1 β (b)(x)). Point 2 is (b, gαβ(b)ρ−1 β (b)(x)), while Point 3 is (b, ρα(b)gαβ(b)ρ−1 β (b)(x)) = (b, g′ αβ(b)(x)). for some suitable open cover of the base B, and some trivializing families (ϕα) for ξ1 and (ϕ′ α) for ξ2, there is a family of maps ρα : Uα →G, so that ϕ′ α ◦f ◦ϕα −1(b, x) = (b, ρα(b)(x)), for all b ∈Uα and all x ∈F. Recall that ϕα ◦ϕ−1 β (b, x) = (b, gαβ(b)(x)), for all b ∈Uα ∩Uβ and all x ∈F. This is equivalent to ϕ−1 β (b, x) = ϕ−1 α (b, gαβ(b)(x)), so it is notationally advantageous to introduce ψα such that ψα = ϕ−1 α . Then we have ψβ(b, x) = ψα(b, gαβ(b)(x)), (∗) and ϕ′ α ◦f ◦ϕ−1 α (b, x) = (b, ρα(b)(x)) becomes ψα(b, x) = f −1 ◦ψ′ α(b, ρα(b)(x)). (∗∗) 422 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES By applying () and () we have ψβ(b, x) = ψα(b, gαβ(b)(x)) = f −1 ◦ψ′ α(b, ρα(b)(gαβ(b)(x))). On the other hand applying () then () gives ψβ(b, x) = f −1 ◦ψ′ β(b, ρβ(b)(x)) = f −1 ◦ψ′ α(b, g′ αβ(b)(ρβ(b)(x))), from which we deduce ρα(b)(gαβ(b)(x)) = g′ αβ(b)(ρβ(b)(x)), that is g′ αβ(b) = ρα(b)gαβ(b)ρβ(b)−1, for all b ∈Uα ∩Uβ, as claimed. Remark: If ξ1 = (E1, π1, B1, F, G) and ξ2 = (E2, π2, B2, F, G) are two bundles over different bases and f : ξ1 →ξ2 is a bundle isomorphism, with f = (fB, fE), then fE and fB are diffeomorphisms, and it is easy to see that we get the conditions g′ αβ(fB(b)) = ρα(b)gαβ(b)ρβ(b)−1, for all b ∈Uα ∩Uβ. The converse of Proposition 10.1 also holds. Proposition 10.2. If two fibre bundles ξ1 = (E1, π1, B, F, G) and ξ2 = (E2, π2, B, F, G) over the same base space B are equivalent, then they are isomorphic. Proof. Assume that ξ1 and ξ2 are equivalent. Then for some suitable open cover of the base B and some trivializing families (ϕα) for ξ1 and (ϕ′ α) for ξ2, there is a family of maps ρα : Uα →G, so that g′ αβ(b) = ρα(b)gαβ(b)ρβ(b)−1, for all b ∈Uα ∩Uβ, which can be written as g′ αβ(b)ρβ(b) = ρα(b)gαβ(b). For every Uα, define fα as the composition π−1 1 (Uα) ϕα / Uα × F e fα / Uα × F ϕ′ α −1 / π−1 2 (Uα) z / (b, x) / (b, ρα(b)(x)) / ϕ′ α −1(b, ρα(b)(x)); that is, fα(z) = ϕ′ α −1(b, ρα(b)(x)), with z ∈π−1 1 (Uα) and (b, x) = ϕα(z). 10.2. BUNDLE MORPHISMS, EQUIVALENT AND ISOMORPHIC BUNDLES 423 Since fα = ϕ′−1 α ◦e fα ◦ϕα, the definition of fα implies that ϕ′ α ◦fα ◦ϕα −1(b, x) = (b, ρα(b)(x)), for all b ∈Uα and all x ∈F, and locally fα is a bundle isomorphism with respect to ρα. If we can prove that any two fα and fβ agree on the overlap Uα ∩Uβ, then the fα’s patch and yield a bundle isomorphism between ξ1 and ξ2. Now, on Uα ∩Uβ, ϕα ◦ϕ−1 β (b, x) = (b, gαβ(b)(x)) yields ϕ−1 β (b, x) = ϕ−1 α (b, gαβ(b)(x)). We need to show that for every z ∈Uα ∩Uβ, fα(z) = ϕ′ α −1(b, ρα(b)(x)) = ϕ′ β −1(b, ρβ(b)(x′)) = fβ(z), where ϕα(z) = (b, x) and ϕβ(z) = (b, x′). From ϕ1 α(b, x) = z = ϕ−1 β (b, x′) = ϕ−1 α (b, gαβ(b)(x′)), we get x = gαβ(b)(x′). We also have ϕ′ β −1(b, ρβ(b)(x′)) = ϕ′ α −1(b, g′ αβ(b)(ρβ(b)(x′))), and since g′ αβ(b)ρβ(b) = ρα(b)gαβ(b) and x = gαβ(b)(x′), we get ϕ′ β −1(b, ρβ(b)(x′)) = ϕ′ α −1(b, g′ αβ(b)(ρβ(b))(x′)) = ϕ′ α −1(b, ρα(b)(gαβ(b))(x′)) = ϕ′ α −1(b, ρα(b)(x)), as desired. Therefore, the fα’s patch to yield a bundle map f, with respect to the family of maps ρα : Uα →G. The map f is bijective because it is an isomorphism on fibres, but it remains to show that it is a diffeomorphism. This is a local matter, and as the ϕα and ϕ′ α are diffeomorphisms, it suffices to show that the map e fα : Uα × F − →Uα × F given by (b, x) 7→(b, ρα(b)(x)) is a diffeomorphism. For this, observe that in local coordinates, the Jacobian matrix of this map is of the form J = I 0 C J(ρα(b))  , where I is the identity matrix and J(ρα(b)) is the Jacobian matrix of ρα(b). Since ρα(b) is a diffeomorphism, det(J) ̸= 0, and by the inverse function theorem, the map e fα is a diffeomorphism, as desired. 424 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES Remark: If in Proposition 10.2, ξ1 = (E1, π1, B1, F, G) and ξ2 = (E2, π2, B2, F, G) are two bundles over different bases and if we have a diffeomorphism fB : B1 →B2, and the conditions g′ αβ(fB(b)) = ρα(b)gαβ(b)ρβ(b)−1, for all b ∈Uα ∩Uβ hold, then there is a bundle isomorphism (fB, fE) between ξ1 and ξ2. It follows from Proposition 10.1 and Proposition 10.2 that two bundles over the same base are equivalent iffthey are isomorphic, a very useful fact. Actually, we can use the proof of Proposition 10.2 to show that any bundle morphism f : ξ1 →ξ2 between two fibre bundles over the same base B is a bundle isomorphism. Because a bundle morphism f as above is fibre preserving, f is bijective, but it is not obvious that its inverse is smooth. Proposition 10.3. Any bundle morphism f : ξ1 →ξ2 between two fibre bundles over the same base B is an isomorphism. Proof. Since f is bijective this is a local matter, and it is enough to prove that each e fα : Uα × F − →Uα × F is a diffeomorphism, since f can be written as f = ϕ′ α −1 ◦e fα ◦ϕα, with e fα(b, x) = (b, ρα(b)(x)). However, the end of the proof of Proposition 10.2 shows that e fα is a diffeomorphism. 10.3 Bundle Constructions Via the Cocycle Condition Given a fibre bundle ξ = (E, π, B, F, G), we observed that the family g = (gαβ) of transition maps gαβ : Uα ∩Uβ →G induced by a trivializing family ϕ = (ϕα)α∈I relative to the open cover U = (Uα)α∈I for B satisfies the cocycle condition gαβ(b)gβγ(b) = gαγ(b), for all α, β, γ such that Uα ∩Uβ ∩Uγ ̸= ∅and all b ∈Uα ∩Uβ ∩Uγ. Without altering anything, we may assume that gαβ is the (unique) function from ∅to G, when Uα ∩Uβ = ∅. Then we call a family g = (gαβ)(α,β)∈I×I as above a U-cocycle, or simply a cocycle. Remarkably, given such a cocycle g relative to U, a fibre bundle ξg over B with fibre F and structure group G having g as family of transition functions can be constructed. In view of Proposition 10.1, we make the following definition. Definition 10.10. We say that two cocycles g = (gαβ)(α,β)∈I×I and g′ = (g′ αβ)(α,β)∈I×I are equivalent if there is a family (ρα)α∈I of smooth maps ρα : Uα →G, such that g′ αβ(b) = ρα(b)gαβ(b)ρβ(b)−1, for all b ∈Uα ∩Uβ. 10.3. BUNDLE CONSTRUCTIONS VIA THE COCYCLE CONDITION 425 Theorem 10.4. Given two smooth manifolds B and F, a Lie group G acting effectively on F, an open cover U = (Uα)α∈I of B, and a cocycle g = (gαβ)(α,β)∈I×I, there is a fibre bundle ξg = (E, π, B, F, G) whose transition maps are the maps in the cocycle g. Furthermore, if g and g′ are equivalent cocycles, then ξg and ξg′ are isomorphic. Proof sketch. First, we define the space Z as the disjoint sum Z = a α∈I (Uα × F). We define the relation ≃on Z × Z as follows: For all (b, x) ∈Uβ × F and (b, y) ∈Uα × F, if Uα ∩Uβ ̸= ∅, (b, x) ≃(b, y) iff y = gαβ(b)(x). We let E = Z/ ≃, and we give E the largest topology such that the injections ηα : Uα × F →Z are smooth. The cocycle condition insures that ≃is indeed an equivalence relation. We define π: E →B by π([b, x]) = b. If p: Z →E is the the quotient map, observe that the maps p ◦ηα : Uα × F →E are injective, and that π ◦p ◦ηα(b, x) = b. Thus, p ◦ηα : Uα × F →π−1(Uα) is a bijection, and we define the trivializing maps by setting ϕα = (p ◦ηα)−1. It is easily verified that the corresponding transition functions are the original gαβ. There are some details to check. A complete proof (the only one we could find!) is given in Steenrod , Part I, Section 3, Theorem 3.2. The fact that ξg and ξg′ are isomorphic when g and g′ are equivalent follows from Proposition 10.2 (see Steenrod , Part I, Section 2, Lemma 2.10). Remark: (For readers familiar with sheaves) Hirzebruch defines the sheaf G∞, where G∞(U) = Γ(U, G∞) is the group of smooth functions g: U →G, where U is some open subset of B and G is a Lie group acting effectively (on the left) on the fibre F. The group operation on Γ(U, G∞) is induced by multiplication in G; that is, given two (smooth) functions g: U →G and h: U →G, gh(b) = g(b)h(b), for all b ∈U.  Beware that gh is not function composition, unless G itself is a group of functions, which is the case for vector bundles. 426 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES Our conditions (b) and (c) are then replaced by the following equivalent condition: For all Uα, Uβ in U such that Uα ∩Uβ ̸= ∅, there is some gαβ ∈Γ(Uα ∩Uβ, G∞) such that ϕα ◦ϕ−1 β (b, x) = (b, gαβ(b)(x)), for all b ∈Uα ∩Uβ and all x ∈F. The classic source on fibre bundles is Steenrod . The most comprehensive treatment of fibre bundles and vector bundles is probably given in Husemoller . A more extensive list of references is given at the end of Section 10.9. Remark: (The following paragraph is intended for readers familiar with ˇ Cech cohomology.) The cocycle condition makes it possible to view a fibre bundle over B as a member of a certain ( ˇ Cech) cohomology set ˇ H1(B, G), where G denotes a certain sheaf of functions from the manifold B into the Lie group G, as explained in Hirzebruch , Section 3.2. However, this requires defining a noncommutative version of ˇ Cech cohomology (at least, for ˇ H1), and clarifying when two open covers and two trivializations define the same fibre bundle over B, or equivalently, defining when two fibre bundles over B are equivalent. If the bundles under considerations are line bundles (see Definition 10.13), then ˇ H1(B, G) is actually a group. In this case, G = GL(1, R) ∼ = R∗in the real case, and G = GL(1, C) ∼ = C∗in the complex case (where R∗= R−{0} and C∗= C−{0}), and the sheaf G is the sheaf of smooth (real-valued or complex-valued) functions vanishing nowhere. The group ˇ H1(B, G) plays an important role, especially when the bundle is a holomorphic line bundle over a complex manifold. In the latter case, it is called the Picard group of B. Remark: (The following paragraph is intended for readers familiar with ˇ Cech cohomology.) Obviously, if we start with a fibre bundle ξ = (E, π, B, F, G) whose transition maps are the cocycle g = (gαβ), and form the fibre bundle ξg, the bundles ξ and ξg are equivalent. This leads to a characterization of the set of equivalence classes of fibre bundles over a base space B as the cohomology set ˇ H1(B, G). In the present case, the sheaf G is defined such that Γ(U, G) is the group of smooth maps from the open subset U of B to the Lie group G. Since G is not abelian, the coboundary maps have to be interpreted multiplicatively. If we define the sets of cochains Ck(U, G), so that C0(U, G) = Y α G(Uα), C1(U, G) = Y α<β G(Uα ∩Uβ), C2(U, G) = Y α<β<γ G(Uα ∩Uβ ∩Uγ), etc., then it is natural to define δ0 : C0(U, G) →C1(U, G) by (δ0g)αβ = g−1 α gβ, 10.3. BUNDLE CONSTRUCTIONS VIA THE COCYCLE CONDITION 427 for any g = (gα), with gα ∈Γ(Uα, G). As to δ1 : C1(U, G) →C2(U, G), since the cocycle condition in the usual case is gαβ + gβγ = gαγ, we set (δ1g)αβγ = gαβgβγg−1 αγ , for any g = (gαβ), with gαβ ∈Γ(Uα ∩Uβ, G). Note that a cocycle g = (gαβ) is indeed an element of Z1(U, G), and the condition for being in the kernel of δ1 : C1(U, G) →C2(U, G) is the cocycle condition gαβ(b)gβγ(b) = gαγ(b), for all b ∈Uα ∩Uβ ∩Uγ. In the commutative case, two cocycles g and g′ are equivalent if their difference is a boundary, which can be stated as g′ αβ + ρβ = gαβ + ρα = ρα + gαβ, where ρα ∈Γ(Uα, G), for all α ∈I. In the present case, two cocycles g and g′ are equivalent iffthere is a family (ρα)α∈I, with ρα ∈Γ(Uα, G), such that g′ αβ(b) = ρα(b)gαβ(b)ρβ(b)−1, for all b ∈Uα ∩Uβ. This is the same condition of equivalence defined earlier. Thus, it is easily seen that if g, h ∈Z1(U, G), then ξg and ξh are equivalent iffg and h correspond to the same element of the cohomology set ˇ H1(U, G). As usual, ˇ H1(B, G) is defined as the direct limit of the directed system of sets ˇ H1(U, G) over the preordered directed family of open covers. For details, see Hirzebruch , Section 3.1. In summary, there is a bijection between the equivalence classes of fibre bundles over B (with fibre F and structure group G) and the cohomology set ˇ H1(B, G). In the case of line bundles, it turns out that ˇ H1(B, G) is in fact a group. As an application of Theorem 10.4, we define the notion of pullback (or induced) bundle. Definition 10.11. Let ξ = (E, π, B, F, G) is a fibre bundle and assume we have a smooth map f : N →B. The pullback bundle f ∗ξ = (f ∗E, π∗, N, F, G) is the bundle over N, induced by the bundle map (f ∗, f): f ∗ξ →ξ f ∗E f∗ / π∗  E π  N f / B, 428 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES where f ∗E is a pullback in the categorical sense. This means that for any other bundle ξ′ over N and any bundle map E′ f′ / π′  E π  N f / B, there is a unique bundle map (e f ′, id): ξ′ →f ∗ξ, so that (f ′, f) = (f ∗, f)◦(e f ′, id) as illustrated by the following diagram. E′ π′ f′ $ e f′ ! f ∗E π∗  f∗ / E π  N f / B Definition 10.11 implies there is an isomorphism (natural) Hom(ξ′, ξ) ∼ = Hom(ξ,′ f ∗ξ). As a consequence, by Proposition 10.3, for any bundle map between ξ′ and ξ, E′ π′  f′ / E π  N f / B, there is an isomorphism, ξ′ ∼ = f ∗ξ. The bundle f ∗ξ can be constructed as follows: Pick any open cover (Uα) of B, then (f −1(Uα)) is an open cover of N, and check that if (gαβ) is a cocycle for ξ, then the maps gαβ ◦f : f −1(Uα) ∩f −1(Uβ) →G satisfy the cocycle conditions. Then, f ∗ξ is the bundle defined by the cocycle (gαβ ◦f). We leave as an exercise to show that the pullback bundle f ∗ξ can be defined explicitly if we set f ∗E = {(n, e) ∈N × E | f(n) = π(e)}, π∗= pr1 and f ∗= pr2. For any trivialization ϕα : π−1(Uα) →Uα × F of ξ, we have (π∗)−1(f −1(Uα)) = {(n, e) ∈N × E | n ∈f −1(Uα), e ∈π−1(f(n))}, and so we have a bijection e ϕα : (π∗)−1(f −1(Uα)) →f −1(Uα) × F, given by e ϕα(n, e) = (n, pr2(ϕα(e))). 10.4. VECTOR BUNDLES 429 By giving f ∗E the smallest topology that makes each e ϕα a diffeomorphism, e ϕα is a trivial-ization of f ∗ξ over f −1(Uα), and f ∗ξ is a smooth bundle. Note that the fibre of f ∗ξ over a point n ∈N is isomorphic to the fibre π−1(f(n)) of ξ over f(n). If g: M →N is another smooth map of manifolds, it is easy to check that (f ◦g)∗ξ = g∗(f ∗ξ). Definition 10.12. Given a bundle ξ = (E, π, B, F, G) and a submanifold N of B, we define the restriction of ξ to N as the bundle ξ ↾N = (π−1(N), π ↾π−1(N), B, F, G). There are two particularly interesting special cases of fibre bundles: (1) Vector bundles, which are fibre bundles for which the typical fibre is a finite-dimensional vector space V , and the structure group is a subgroup of the group of linear isomor-phisms (GL(n, R) or GL(n, C), where n = dim V ). (2) Principal fibre bundles, which are fibre bundles for which the fibre F is equal to the structure group G, with G acting on itself by left translation. First we discuss vector bundles. 10.4 Vector Bundles Given a real vector space V , we denote by GL(V ) (or Aut(V )) the group of linear invertible maps from V to V . If V has dimension n, then GL(V ) has dimension n2. Obviously, GL(V ) is isomorphic to GL(n, R), so we often write GL(n, R) instead of GL(V ), but this may be slightly confusing if V is the dual space W ∗of some other space W. If V is a complex vector space, we also denote by GL(V ) (or Aut(V )) the group of linear invertible maps from V to V , but this time GL(V ) is isomorphic to GL(n, C), so we often write GL(n, C) instead of GL(V ). Definition 10.13. A rank n real smooth vector bundle with fibre V is a tuple ξ = (E, π, B, V ) such that (E, π, B, V, GL(V )) is a smooth fibre bundle, the fibre V is a real vector space of dimension n, the action of the group GL(V ) on V is given by f ·u = f(u) for all f ∈GL(V ) and all u ∈V , and the following conditions hold: (a) For every b ∈B, the fibre π−1(b) is an n-dimensional (real) vector space. (b) For every trivialization ϕα : π−1(Uα) →Uα × V , for every b ∈Uα, the restriction of ϕα to the fibre π−1(b) is a linear isomorphism π−1(b) − →V . More precisely, the maps ϕα,b : π−1(b) →V are linear isomorphisms for all b ∈Uα. 430 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES A rank n complex smooth vector bundle with fibre V is a tuple, ξ = (E, π, B, V ), where (E, π, B, V, GL(V )) is a smooth fibre bundle such that the fibre V is an n-dimensional complex vector space (viewed as a real smooth manifold), the group GL(V ) acts on V as above, and Conditions (a) and (b) above hold (for complex vector spaces). When n = 1, a vector bundle is called a line bundle. Observe that the action of GL(V ) on V is obviously effective. Definition 10.14. A holomorphic vector bundle is a fibre bundle where E, B are complex manifolds, V is a complex vector space of dimension n, the map π is holomorphic, the ϕα are biholomorphic, and the transition functions gαβ are holomorphic. When n = 1, a holomorphic vector bundle is called a holomorphic line bundle. Just as fibre bundles can be generalized by considering a smooth action of a group G on the fibre F that is not necessarily effective, vector bundles can be generalized. This time we assume that we have a representation ρ: G →GL(V ), where the fibre V is a vector space and the diffeomorphism ϕα,b ◦ϕ−1 β,b : V →V of the fibre V is a linear automorphism specified in terms of a smooth map gαβ : Uα ∩Uβ →G such that ϕα,b ◦ϕ−1 β,b(x) = ρ(gαβ(b))(x), b ∈Uα ∩Uβ, x ∈V, so that ϕα ◦ϕ−1 β (b, x) = (b, ρ(gαβ(b))(x)), b ∈Uα ∩Uβ, x ∈V. The trivial vector bundle E = B ×V is often denoted ϵV . When V = Rk, we also use the notation ϵk. Given a (smooth) manifold M of dimension n, the tangent bundle TM and the cotangent bundle T ∗M are rank n vector bundles. Let us compute the transition functions for the tangent bundle TM, where M is a smooth manifold of dimension n. For every p ∈M, the tangent space TpM consists of all equivalence classes of triples (U, ϕ, x), where (U, ϕ) is a chart with p ∈U, x ∈Rn, and the equivalence relation on triples is given by (U, ϕ, x) ≡(V, ψ, y) iff (ψ ◦ϕ−1)′ ϕ(p)(x) = y. We have a natural isomorphism θU,ϕ,p : Rn →TpM between Rn and TpM given by θU,ϕ,p : x 7→[(U, ϕ, x)], x ∈Rn. Observe that for any two overlapping charts (U, ϕ) and (V, ψ), θ−1 V,ψ,p ◦θU,ϕ,p = (ψ ◦ϕ−1)′ z for all p ∈U ∩V , with z = ϕ(p) = ψ(p). We let TM be the disjoint union TM = [ p∈M TpM, 10.4. VECTOR BUNDLES 431 define the projection π: TM →M so that π(v) = p if v ∈TpM, and we give TM the weakest topology that makes the functions e ϕ: π−1(U) →R2n given by e ϕ(v) = (ϕ ◦π(v), θ−1 U,ϕ,π(v)(v)) continuous, where (U, ϕ) is any chart of M. Each function e ϕ: π−1(U) →ϕ(U) × Rn is a homeomorphism, and given any two overlapping charts (U, ϕ) and (V, ψ), since θ−1 V,ψ,p ◦θU,ϕ,p = (ψ ◦ϕ−1)′ z, with z = ϕ(p) = ψ(p), the transition map e ψ ◦e ϕ−1 : ϕ(U ∩V ) × Rn − →ψ(U ∩V ) × Rn is given by e ψ ◦e ϕ−1(z, x) = (ψ ◦ϕ−1(z), (ψ ◦ϕ−1)′ z(x)), (z, x) ∈ϕ(U ∩V ) × Rn. It is clear that e ψ ◦e ϕ−1 is smooth. Moreover, the bijection τU : π−1(U) →U × Rn given by τU(v) = (π(v), θ−1 U,ϕ,π(v)(v)) satisfies pr1 ◦τU = π on π−1(U) and is a linear isomorphism restricted to fibres, so it is a trivialization for TM. For any two overlapping charts (Uα, ϕα) and (Uβ, ϕβ), the transition function, gαβ : Uα ∩Uβ →GL(n, R) is given by gαβ(p) = (ϕα ◦ϕ−1 β )′ ϕ(p). See Figure 10.10. We can also compute trivialization maps for T ∗M. This time, T ∗M is the disjoint union T ∗M = [ p∈M T ∗ p M, and π: T ∗M →M is given by π(ω) = p if ω ∈T ∗ p M, where T ∗ p M is the dual of the tangent space TpM. For each chart (U, ϕ), by dualizing the map θU,ϕ,p : Rn →TpM, we obtain an isomorphism θ⊤ U,ϕ,p : T ∗ p M →(Rn)∗. Composing θ⊤ U,ϕ,p with the isomorphism ι: (Rn)∗→Rn (induced by the canonical basis (e1, . . . , en) of Rn and its dual basis), we get an isomorphism θ∗ U,ϕ,p = ι ◦θ⊤ U,ϕ,p : T ∗ p M →Rn. Then define the bijection e ϕ∗: π−1(U) →ϕ(U) × Rn ⊆R2n by e ϕ∗(ω) = (ϕ ◦π(ω), θ∗ U,ϕ,π(ω)(ω)), 432 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES M U V p φ φ (U) ψ ψ (V) V p U TM v φ(U) R R R (z,x) x R x ψ (V) φ U ~ U x ψ ~ V V x ψ ~ ( φ ~ -1(z,x) ) τ τ Figure 10.10: An illustration of the line bundle T M over the curve M. The diagram in the upper left illustrates U and V , two overlapping charts of M. The middle illustrates the tangent bundle T M, the two trivialization maps, τU and τV , and the chart maps, e ϕ and e ψ. with ω ∈π−1(U). We give T ∗M the weakest topology that makes the functions e ϕ∗contin-uous, and then each function e ϕ∗is a homeomorphism. Given any two overlapping charts (U, ϕ) and (V, ψ), as θ−1 V,ψ,p ◦θU,ϕ,p = (ψ ◦ϕ−1)′ ϕ(p), by dualization we get θ⊤ U,ϕ,p ◦(θ⊤ V,ψ,p)−1 = θ⊤ U,ϕ,p ◦(θ−1 V,ψ,p)⊤= ((ψ ◦ϕ−1)′ ϕ(p))⊤, then θ⊤ V,ψ,p ◦(θ⊤ U,ϕ,p)−1 = (((ψ ◦ϕ−1)′ ϕ(p))⊤)−1, and so ι ◦θ⊤ V,ψ,p ◦(θ⊤ U,ϕ,p)−1 ◦ι−1 = ι ◦(((ψ ◦ϕ−1)′ ϕ(p))⊤)−1 ◦ι−1; that is, θ∗ V,ψ,p ◦(θ∗ U,ϕ,p)−1 = ι ◦(((ψ ◦ϕ−1)′ ϕ(p))⊤)−1 ◦ι−1. 10.4. VECTOR BUNDLES 433 Consequently, the transition map e ψ∗◦(e ϕ∗)−1 : ϕ(U ∩V ) × Rn − →ψ(U ∩V ) × Rn is given by e ψ∗◦(e ϕ∗)−1(z, x) = (ψ ◦ϕ−1(z), ι ◦(((ψ ◦ϕ−1)′ z)⊤)−1 ◦ι−1(x)), (z, x) ∈ϕ(U ∩V ) × Rn. If we view (ψ ◦ϕ−1)′ z as a matrix, then we can forget ι and the second component of e ψ∗◦(e ϕ∗)−1(z, x) is (((ψ ◦ϕ−1)′ z)⊤)−1x. We also have trivialization maps τ ∗ U : π−1(U) →U × (Rn)∗for T ∗M, given by τ ∗ U(ω) = (π(ω), θ⊤ U,ϕ,π(ω)(ω)), for all ω ∈π−1(U). The transition function g∗ αβ : Uα ∩Uβ →GL(n, R) is given by g∗ αβ(p)(η) = θ⊤ Uα,ϕα,π(η) ◦(θ⊤ Uβ,ϕβ,π(η))−1(η) = ((θ−1 Uα,ϕα,π(η) ◦θUβ,ϕβ,π(η))⊤)−1(η) = (((ϕα ◦ϕ−1 β )′ ϕ(p))⊤)−1(η), with η ∈(Rn)∗. Also note that GL(n, R) should really be GL((Rn)∗), but GL((Rn)∗) is isomorphic to GL(n, R). We conclude that g∗ αβ(p) = (gαβ(p)⊤)−1, for every p ∈M. This is a general property of dual bundles; see Property (f) in Section 10.5. Maps of vector bundles are maps of fibre bundles such that the isomorphisms between fibres are linear. Definition 10.15. Given two vector bundles ξ1 = (E1, π1, B1, V ) and ξ2 = (E2, π2, B2, V ) with the same typical fibre V , a bundle map (or bundle morphism) f : ξ1 →ξ2 is a pair f = (fE, fB) of smooth maps fE : E1 →E2 and fB : B1 →B2, such that: (a) The following diagram commutes: E1 π1  fE / E2 π2  B1 fB / B2 (b) For every b ∈B1, the map of fibres fE ↾π−1 1 (b): π−1 1 (b) →π−1 2 (fB(b)) is a bijective linear map. 434 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES A bundle map isomorphism f : ξ1 →ξ2 is defined as in Definition 10.6. Given two vector bundles ξ1 = (E1, π1, B, V ) and ξ2 = (E2, π2, B, V ) over the same base space B, we require fB = id. Remark: Some authors do not require the preservation of fibres; that is, the map fE ↾π−1 1 (b): π−1 1 (b) →π−1 2 (fB(b)) is simply a linear map. It is automatically bijective for bundle isomorphisms. Note that Definition 10.15 does not include Condition (b) of Definition 10.7. However, because the restrictions of the maps ϕα, ϕ′ β, and fE to the fibres are linear isomorphisms, it turns out that Condition (b) (of Definition 10.7) does hold. Indeed, if fB(Uα) ⊆Vβ, then ϕ′ β ◦fE ◦ϕ−1 α : Uα × V − →Vβ × V is a smooth map of the form ϕ′ β ◦fE ◦ϕ−1 α (b, x) = (fB(b), ρα(b)(x)) for all b ∈Uα and all x ∈V , where ρα(b) is some linear isomorphism of V . Because ϕ′ β ◦fE ◦ϕ−1 α is smooth, the map b 7→ρα(b) is smooth, therefore, there is a smooth map ρα : Uα →GL(V ) so that ϕ′ β ◦f ◦ϕ−1 α (b, x) = (fB(b), ρα(b)(x)), and a vector bundle map is a fibre bundle map. Definition 10.9 (equivalence of bundles) also applies to vector bundles (just replace G by GL(n, R) or GL(n, C)) and defines the notion of equivalence of vector bundles over B. Since vector bundle maps are fibre bundle maps, Propositions 10.1 and 10.2 immediately yield Proposition 10.5. Two vector bundles ξ1 = (E1, π1, B, V ) and ξ2 = (E2, π2, B, V ) over the same base space B are equivalent iffthey are isomorphic. Since a vector bundle map is a fibre bundle map, Proposition 10.3 also yields the useful fact: Proposition 10.6. Any vector bundle map f : ξ1 →ξ2 between two vector bundles over the same base B is an isomorphism. Proposition 10.6 is proved in Milnor and Stasheff for continuous vector bundles (see Lemma 2.3), and in Madsen and Tornehave for smooth vector bundles as well as continuous vector bundles (see Lemma 15.10). The definition of a continuous vector bundle is similar to the definition of a smooth vector bundle, except that the manifolds are topological manifolds instead of smooth maninfolds, and the maps involved are continuous rather than smooth. Theorem 10.4 also holds for vector bundles and yields a technique for constructing new vector bundles over some base B. 10.4. VECTOR BUNDLES 435 Theorem 10.7. Given a smooth manifold B, an n-dimensional (real, resp. complex) vector space V , an open cover U = (Uα)α∈I of B, and a cocycle g = (gαβ)(α,β)∈I×I (with gαβ : Uα ∩Uβ →GL(n, R), resp. gαβ : Uα ∩Uβ →GL(n, C)), there is a vector bundle ξg = (E, π, B, V ) whose transition maps are the maps in the cocycle g. Furthermore, if g and g′ are equivalent cocycles, then ξg and ξg′ are equivalent. Observe that a cocycle g = (gαβ)(α,β)∈I×I is given by a family of matrices in GL(n, R) (resp. GL(n, C)). A vector bundle ξ always has a global section, namely the zero section, which assigns the element 0 ∈π−1(b) to every b ∈B. A global section s is a nonzero section iffs(b) ̸= 0 for all b ∈B. It is usually difficult to decide whether a bundle has a nonzero section. This question is related to the nontriviality of the bundle, and there is a useful test for triviality. Assume ξ is a trivial rank n vector bundle. There is a bundle isomorphism f : B×V →ξ. For every b ∈B, we know that f(b, −) is a linear isomorphism, so for any choice of a basis (e1, . . . , en) of V , we get a basis (f(b, e1), . . . , f(b, en)) of the fibre π−1(b). Thus, we have n global sections s1(b) = f(b, e1), . . . , sn(b) = f(b, en) such that (s1(b), . . . , sn(b)) forms a basis of the fibre π−1(b), for every b ∈B. Definition 10.16. Let ξ = (E, π, B, V ) be a rank n vector bundle. For any open subset U ⊆ B, an n-tuple of local sections (s1, . . . , sn) over U is called a frame over U iff(s1(b), . . . , sn(b)) is a basis of the fibre π−1(b), for every b ∈U. See Figure 10.11. If U = B, then the si are global sections and (s1, . . . , sn) is called a frame (of ξ). b b b 0 b0 1 b1 2 b2 e1 e1 e1 e2 e2 e2 b0 b1 b2 U x V s1 s1 s1 s2 s2 s2 b0 ( ( ( ( b1 b2 ( ( ) ) ) ) ) ) B U ξ φ Figure 10.11: A frame of ξ = (E, π, B, R2) over U obtained from a local trivialization. For i ≤0 ≤2, s1(bi) = ϕ−1(bi, e1) and s2(bi) = ϕ−1(bi, e2), where e1 and e2 are the standard basis vectors of R2. The notion of a frame is due to ´ Elie Cartan who (after Darboux) made extensive use of them under the name of moving frame (and the moving frame method). Cartan’s terminology 436 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES is intuitively clear: As a point b moves in U, the frame (s1(b), . . . , sn(b)) moves from fibre to fibre. Physicists refer to a frame as a choice of local gauge. The converse of the property established just before Definition 10.16 is also true. Proposition 10.8. A rank n vector bundle ξ is trivial iffit possesses a frame of global sections. Proof. (Adapted from Milnor and Stasheff, Theorem 2.2.) We only need to prove that if ξ has a frame (s1, . . . , sn), then it is trivial. Pick a basis (e1, . . . , en) of V , and define the map f : B × V →ξ as follows: f(b, v) = n X i=1 visi(b), where v = Pn i=1 viei. Clearly, f is bijective on fibres, smooth, and a map of vector bundles. By Proposition 10.6, the bundle map, f, is an isomorphism. The above considerations show that if ξ is any rank n vector bundle, not necessarily trivial, then for any local trivialization ϕα : π−1(Uα) →Uα ×V , there are always frames over Uα. Indeed, for every choice of a basis (e1, . . . , en) of the typical fibre V , if we set sα i (b) = ϕ−1 α (b, ei), b ∈Uα, 1 ≤i ≤n, then (sα 1, . . . , sα n) is a frame over Uα. See Figure 10.11. Definition 10.17. Given any two vector spaces V and W, both of dimension n, we denote by Iso(V, W) the space of all linear isomorphisms between V and W. The space of n-frames F(V ) is the set of bases of V . Since every basis (v1, . . . , vn) of V is in one-to-one correspondence with the map from Rn to V given by ei 7→vi, where (e1, . . . , en) is the canonical basis of Rn (so, ei = (0, . . . , 1, . . . 0) with the 1 in the ith slot), we have an isomorphism, F(V ) ∼ = Iso(Rn, V ). (The choice of a basis in V also yields an isomorphism Iso(Rn, V ) ∼ = GL(n, R), so F(V ) ∼ = GL(n, R).) Definition 10.18. For any rank n vector bundle ξ, we can form the frame bundle F(ξ), by replacing the fibre π−1(b) over any b ∈B by F(π−1(b)). In fact, F(ξ) can be constructed using Theorem 10.4. Indeed, identifying F(V ) with Iso(Rn, V ), the group GL(n, R) acts on F(V ) effectively on the left via A · v = v ◦A−1. 10.4. VECTOR BUNDLES 437 (The only reason for using A−1 instead of A is that we want a left action.) The resulting bundle has typical fibre F(V ) ∼ = GL(n, R), and turns out to be a principal bundle. We will take a closer look at principal bundles in Section 10.9. We conclude this section with an example of a bundle that plays an important role in algebraic geometry, the canonical line bundle on RPn. Let HR n ⊆RPn × Rn+1 be the subset HR n = {(L, v) ∈RPn × Rn+1 | v ∈L}, where RPn is viewed as the set of lines L in Rn+1 through 0, or more explicitly, HR n = {((x0 : · · · : xn), λ(x0, . . . , xn)) | (x0 : · · · : xn) ∈RPn, λ ∈R}. Geometrically, HR n consists of the set of lines [(x0, . . . , xn)] associated with points (x0 : · · · : xn) of RPn. If we consider the projection π: HR n →RPn of HR n onto RPn, we see that each fibre is isomorphic to R. We claim that HR n is a line bundle. For this, we exhibit trivializations, leaving as an exercise the fact that HR n is a manifold of dimension n + 1. Recall the open cover U0, . . . , Un of RPn, where Ui = {(x0 : · · · : xn) ∈RPn | xi ̸= 0}. Then the maps ϕi : π−1(Ui) →Ui × R given by ϕi((x0 : · · · : xn), λ(x0, . . . , xn)) = ((x0 : · · · : xn), λxi) are trivializations. The transition function gij : Ui ∩Uj →GL(1, R) is given by gij(x0 : · · · : xn)(u) = xi xj u, where we identify GL(1, R) and R∗= R −{0}. Interestingly, the bundle HR n is nontrivial for all n ≥1. For this, by Proposition 10.8 and since HR n is a line bundle, it suffices to prove that every global section vanishes at some point. So, let σ be any section of HR n . Composing the projection, p: Sn − →RPn, with σ, we get a smooth function, s = σ ◦p: Sn − →HR n , and we have s(x) = (p(x), f(x)x), for every x ∈Sn, where f : Sn →R is a smooth function. Moreover, f satisfies f(−x) = −f(x), since s(−x) = (p(−x), −f(−x)x) = (p(x), −f(−x)x) = (p(x), f(x)x) = s(x). As Sn is connected and f is continuous, by the intermediate value theorem, there is some x such that f(x) = 0, and thus, σ vanishes, as desired. 438 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES The reader should look for a geometric representation of HR 1 . It turns out that HR 1 is an open M¨ obius strip; that is, a M¨ obius strip with its boundary deleted (see Milnor and Stasheff, Chapter 2). There is also a complex version of the canonical line bundle on CPn, with Hn = {(L, v) ∈CPn × Cn+1 | v ∈L}, where CPn is viewed as the set of lines L in Cn+1 through 0. These bundles are also nontrivial. Furthermore, unlike the real case, the dual bundle H∗ n is not isomorphic to Hn. Indeed, H∗ n turns out to have nonzero global holomorphic sections! 10.5 Operations on Vector Bundles Because the fibres of a vector bundle are vector spaces all isomorphic to some given space V , we can perform operations on vector bundles that extend familiar operations on vector spaces, such as: direct sum, tensor product, (linear) function space, and dual space. Basically, the same operation is applied on fibres. It is usually more convenient to define operations on vector bundles in terms of operations on cocycles, using Theorem 10.7. (a) (Whitney Sum or Direct Sum) If ξ = (E, π, B, V ) is a rank m vector bundle and ξ′ = (E′, π′, B, W) is a rank n vector bundle, both over the same base B, then their Whitney sum ξ ⊕ξ′ is the rank (m + n) vector bundle whose fibre over any b ∈B is the direct sum Eb ⊕E′ b; that is, the vector bundle with typical fibre V ⊕W (given by Theorem 10.7) specified by the cocycle whose matrices are gαβ(b) 0 0 g′ αβ(b)  , b ∈Uα ∩Uβ. (b) (Tensor Product) If ξ = (E, π, B, V ) is a rank m vector bundle and ξ′ = (E′, π′, B, W) is a rank n vector bundle, both over the same base B, then their tensor product ξ ⊗ξ′ is the rank mn vector bundle whose fibre over any b ∈B is the tensor product Eb ⊗E′ b; that is, the vector bundle with typical fibre V ⊗W (given by Theorem 10.7) specified by the cocycle whose matrices are gαβ(b) ⊗g′ αβ(b), b ∈Uα ∩Uβ. (Here, we identify a matrix with the corresponding linear map.) (c) (Tensor Power) If ξ = (E, π, B, V ) is a rank m vector bundle, then for any k ≥0, we can define the tensor power bundle ξ⊗k, whose fibre over any b ∈B is the tensor power E⊗k b , and with 10.5. OPERATIONS ON VECTOR BUNDLES 439 typical fibre V ⊗k. (When k = 0, the fibre is R or C). The bundle ξ⊗k is determined by the cocycle g⊗k αβ (b), b ∈Uα ∩Uβ. (d) (Exterior Power) If ξ = (E, π, B, V ) is a rank m vector bundle, then for any k ≥0, we can define the exterior power bundle Vk ξ, whose fibre over any b ∈B is the exterior power Vk Eb, and with typical fibre Vk V . The bundle Vk ξ is determined by the cocycle k ^ gαβ(b), b ∈Uα ∩Uβ. Using (a), we also have the exterior algebra bundle V ξ = Lm k=0 Vk ξ. (When k = 0, the fibre is R or C). (e) (Symmetric Power) If ξ = (E, π, B, V ) is a rank m vector bundle, then for any k ≥0, we can define the symmetric power bundle Sk ξ, whose fibre over any b ∈B is the symmetric power Sk Eb, and with typical fibre Sk V . (When k = 0, the fibre is R or C). The bundle Skξ is determined by the cocycle Sk gαβ(b), b ∈Uα ∩Uβ. (f) (Tensor Bundle of type (r, s)) If ξ = (E, π, B, V ) is a rank m vector bundle, then for any r, s ≥0, we can define the bundle T r,s ξ whose fibre over any b ∈ξ is the tensor space T r,s Eb, and with typical fibre T r,s V . The bundle T r,sξ is determined by the cocycle g⊗r αβ(b) ⊗((gαβ(b)⊤)−1)⊗s(b), b ∈Uα ∩Uβ. (g) (Dual Bundle) If ξ = (E, π, B, V ) is a rank m vector bundle, then its dual bundle ξ∗is the rank m vector bundle whose fibre over any b ∈B is the dual space E∗ b ; that is, the vector bundle with typical fibre V ∗(given by Theorem 10.7) specified by the cocycle whose matrices are (gαβ(b)⊤)−1, b ∈Uα ∩Uβ. The reason for this seemingly complicated formula is this: For any trivialization ϕα : π−1(Uα) →Uα × V , for any b ∈B, recall that the restriction ϕα,b : π−1(b) →V of 440 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES ϕα to π−1(b) is a linear isomorphism. By dualization we get a map ϕ⊤ α,b : V ∗→(π−1(b))∗, and thus ϕ∗ α,b for ξ∗is given by ϕ∗ α,b = (ϕ⊤ α,b)−1 : (π−1(b))∗→V ∗. As g∗ αβ(b) = ϕ∗ α,b ◦(ϕ∗ β,b)−1, we get g∗ αβ(b) = (ϕ⊤ α,b)−1 ◦ϕ⊤ β,b = ((ϕ⊤ β,b)−1 ◦ϕ⊤ α,b)−1 = ((ϕ−1 β,b)⊤◦ϕ⊤ α,b)−1 = ((ϕα,b ◦ϕ−1 β,b)⊤)−1 = (gαβ(b)⊤)−1, as claimed. (h) (Hom Bundle) If ξ = (E, π, B, V ) is a rank m vector bundle and ξ′ = (E′, π′, B, W) is a rank n vector bundle, both over the same base B, then their Hom bundle Hom(ξ, ξ′) is the rank mn vector bundle whose fibre over any b ∈B is Hom(Eb, E′ b); that is, the vector bundle with typical fibre Hom(V, W). The transition functions of this bundle are obtained as follows: For any trivializations ϕα : π−1(Uα) →Uα × V and ϕ′ α : (π′)−1(Uα) →Uα ×W, for any b ∈B, recall that the restrictions ϕα,b : π−1(b) →V and ϕ′ α,b : (π′)−1(b) →W are linear isomorphisms. We have a linear isomorphism ϕHom α,b : Hom(π−1(b), (π′)−1(b)) − →Hom(V, W) given by ϕHom α,b (f) = ϕ′ α,b ◦f ◦ϕ−1 α,b, f ∈Hom(π−1(b), (π′)−1(b)). Then, gHom αβ (b) = ϕHom α,b ◦(ϕHom β,b )−1. See Figure 10.12. As an illustration of (d), consider the exterior power Vr T ∗M, where M is a manifold of dimension n. We have trivialization maps τ ∗ U : π−1(U) →U ×Vr(Rn)∗for Vr T ∗M, given by τ ∗ U(ω) = (π(ω), r ^ θ⊤ U,ϕ,π(ω)(ω)), for all ω ∈π−1(U). The transition function g Vr αβ : Uα ∩Uβ →GL(n, R) is given by g Vr αβ (p)(ω) = ( r ^ (((ϕα ◦ϕ−1 β )′ ϕ(p))⊤)−1)(ω), for all ω ∈π−1(U). Consequently, g Vr αβ (p) = r ^ (gαβ(p)⊤)−1, 10.5. OPERATIONS ON VECTOR BUNDLES 441 b b b Eb Eb ‘ f Uα Uα Uα ξ ξ‘ x V x W -1 b b φ φ α , ‘ α , Figure 10.12: A schematic illustration of ϕHom α,b . for every p ∈M, a special case of (f). In view of the canonical isomorphism Hom(V, W) ∼ = V ∗⊗W, it is easy to show the following results. Proposition 10.9. The vector bundle Hom(ξ, ξ′), is isomorphic to ξ∗⊗ξ′. Similarly, ξ∗∗ is isomorphic to ξ. We also have the isomorphism T r,sξ ∼ = ξ⊗r ⊗(ξ∗)⊗s.  Do not confuse the space of bundle morphisms Hom(ξ, ξ′) with the bundle Hom(ξ, ξ′). However, observe that Hom(ξ, ξ′) is the set of global sections of Hom(ξ, ξ′). Remark: For rank 1 vector bundles, namely line bundles, it is easy to show that the set of equivalence classes of line bundles over a base B forms a group, where the group operation is ⊗, the inverse is ∗(dual), and the identity element is the trivial bundle. This is the Picard group of B. In general, the dual ξ∗of a bundle is not isomorphic to the original bundle ξ. This is because V ∗is not canonically isomorphic to V , and to get a bundle isomorphism between ξ and ξ∗, we need canonical isomorphisms between the fibres. However, if ξ is real, then (using a partition of unity), ξ can be given a Euclidean metric and so, ξ and ξ∗are isomorphic.  It is not true in general that a complex vector bundle is isomorphic to its dual because a Hermitian metric only induces a canonical isomorphism between E∗and E, where E is the conjugate of E, with scalar multiplication in E given by (z, w) 7→wz. Remark: Given a real vector bundle, ξ, the complexification ξC of ξ is the complex vector bundle defined by ξC = ξ ⊗R ϵC, 442 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES where ϵC = B × C is the trivial complex line bundle. Given a complex vector bundle ξ, by viewing its fibre as a real vector space we obtain the real vector bundle ξR. Proposition 10.10. The following facts hold. (1) For every real vector bundle ξ, (ξC)R ∼ = ξ ⊕ξ. (2) For every complex vector bundle ξ, (ξR)C ∼ = ξ ⊕ξ∗. 10.6 Properties of Vector Bundle Sections It can be shown (see Madsen and Tornehave , Chapter 15) that for every real smooth vector bundle ξ, there is some integer k such that ξ has a complement η in ϵk, where ϵk = B × Rk is the trivial rank k vector bundle, so that ξ ⊕η = ϵk. This fact can be used to prove an interesting property of the space of global sections Γ(ξ). First, observe that Γ(ξ) is not just a real vector space, but also a C∞(B)-module (see Section 2.14). Indeed, for every smooth function f : B →R and every smooth section s: B →E, the map fs: B →E given by (fs)(b) = f(b)s(b), b ∈B, is a smooth section of ξ. In general, Γ(ξ) is not a free C∞(B)-module unless ξ is trivial. However, the above remark implies that Γ(ξ) ⊕Γ(η) = Γ(ϵk), where Γ(ϵk) is a free C∞(B)-module of dimension dim(ξ) + dim(η). This proves that Γ(ξ) is a finitely generated C∞(B)-module which is a summand of a free C∞(B)-module. Such modules are projective modules; see Definition 2.28 in Section 2.14. Therefore, Γ(ξ) is a finitely generated projective C∞(B)-module. The following isomorphisms can be shown (see Madsen and Tornehave , Chapter 16). Proposition 10.11. The following isomorphisms hold for vector bundles: Γ(Hom(ξ, η)) ∼ = HomC∞(B)(Γ(ξ), Γ(η)) Γ(ξ ⊗η) ∼ = Γ(ξ) ⊗C∞(B) Γ(η) Γ(ξ∗) ∼ = HomC∞(B)(Γ(ξ), C∞(B)) = (Γ(ξ))∗ Γ( k ^ ξ) ∼ = k ^ C∞(B) (Γ(ξ)). 10.6. PROPERTIES OF VECTOR BUNDLE SECTIONS 443 Using the operations on vector bundles described in Section 10.5, we can define the set of vector valued differential forms Ak(M; F) defined in Section 4.5 as the set of smooth sections of the vector bundle Vk T ∗M  ⊗ϵF; that is, as Ak(M; F) = Γ  k ^ T ∗M  ⊗ϵF  , where ϵF is the trivial vector bundle ϵF = M × F. Proposition 10.12. We have the following isomorphisms: Ak(M; F) ∼ = Ak(M) ⊗C∞(M) C∞(M; F) ∼ = Altk C∞(M)(X(M); C∞(M; F)), Proof. By Proposition 10.11 and since Γ(ϵF) ∼ = C∞(M; F) and Ak(M) = Γ Vk T ∗M  , we have Ak(M; F) = Γ  k ^ T ∗M  ⊗ϵF  ∼ = Γ  k ^ T ∗M  ⊗C∞(M) Γ(ϵF) = Ak(M) ⊗C∞(M) C∞(M; F) ∼ = k ^ C∞(M) (Γ(TM))∗⊗C∞(M) C∞(M; F) ∼ = HomC∞(M)  k ^ C∞(M) Γ(TM), C∞(M; F)  ∼ = Altk C∞(M)(X(M); C∞(M; F)), with all of the spaces viewed as C∞(M)-modules, and where we used the fact that X(X) = Γ(TM) is a projective module, and that Proposition 3.5 is still valid for exterior powers over a commutative ring. Therefore, Ak(M; F) ∼ = Ak(M) ⊗C∞(M) C∞(M; F) ∼ = Altk C∞(M)(X(M); C∞(M; F)), which reduces to Proposition 4.15 when F = R. In Section 11.2, we will consider a generalization of the above situation where the trivial vector bundle ϵF is replaced by any vector bundle ξ = (E, π, B, V ), and where M = B. 444 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES 10.7 Duality between Vector Fields and Differential Forms, Covariant Derivatives of Tensor Fields Given a manifold M, the covariant derivative ∇X given by a connection ∇on TM can be extended to a covariant derivative ∇r,s X defined on tensor fields in Γ(M, T r,s(M)) for all r, s ≥0, where T r,s(M) = T ⊗rM ⊗(T ∗M)⊗s. We already have ∇1,0 X = ∇X and it is natural to set ∇0,0 X f = X[f] = d f(X). Recall that there is an isomorphism between the set of tensor fields Γ(M, T r,s(M)), and the set of C∞(M)-multilinear maps Φ: A1(M) × · · · × A1(M) | {z } r × X(M) × · · · × X(M) | {z } s − →C∞(M), where A1(M) and X(M) are C∞(M)-modules. The next proposition is left as an exercise. For help, see O’Neill , Chapter 2, Propo-sition 13 and Theorem 15. Proposition 10.13. for every vector field X ∈X(M), there is a unique family of R-linear map ∇r,s : Γ(M, T r,s(M)) →Γ(M, T r,s(M)), with r, s ≥0, such that (a) ∇0,0 X f = d f(X), for all f ∈C∞(M) and ∇1,0 X = ∇X, for all X ∈X(M). (b) ∇r1+r2,s1+s2 X (S ⊗T) = ∇r1,s1 X (S) ⊗T + S ⊗∇r2,s2 X (T), for all S ∈Γ(M, T r1,s1(M)) and all T ∈Γ(M, T r2,s2(M)). (c) ∇r−1,s−1 X (cij(S)) = cij(∇r,s X (S)), for all S ∈Γ(M, T r,s(M)) and all contractions, cij, of Γ(M, T r,s(M)). Furthermore, (∇0,1 X θ)(Y ) = X[θ(Y )] −θ(∇XY ), for all X, Y ∈X(M) and all one-forms, θ ∈A1(M), and for every S ∈Γ(M, T r,s(M)), with r + s ≥2, the covariant derivative ∇r,s X (S) is given by (∇r,s X S)(θ1, . . . , θr, X1, . . . , Xs) = X[S(θ1, . . . , θr, X1, . . . , Xs)] − r X i=1 S(θ1, . . . , ∇0,1 X θi, . . . , θr, X1, . . . , Xs) − s X j=1 S(θ1, . . . , . . . , θr, X1, . . . , ∇XXj, . . . , Xs), for all X1, . . . , Xs ∈X(M) and all one-forms, θ1, . . . , θr ∈A1(M). 10.7. COVARIANT DERIVATIVES OF TENSOR FIELDS 445 In particular, for S = g, the Riemannian metric on M (a (0, 2) tensor), we get ∇X(g)(Y, Z) = X(g(Y, Z)) −g(∇XY, Z) −g(Y, ∇XZ), for all X, Y, Z ∈X(M). We will see later on that a connection on M is compatible with a metric g iff∇X(g) = 0. Definition 10.19. The covariant differential ∇r,sS of a tensor S ∈Γ(M, T r,s(M)) is the (r, s + 1)-tensor given by (∇r,sS)(θ1, . . . , θr, X, X1, . . . , Xs) = (∇r,s X S)(θ1, . . . , θr, X1, . . . , Xs), for all X, Xj ∈X(M) and all θi ∈A1(M). For simplicity of notation we usually omit the superscripts r and s. In particular, if r = 1 and s = 0, in which case S is a vector field, the covariant derivative ∇S is defined so that (∇S)(X) = ∇XS. If (M, ⟨−, −⟩) is a Riemannian manifold, then the inner product ⟨−, −⟩p on TpM, estab-lishes a canonical duality between TpM and T ∗ p M, as explained in Section 2.2. Namely, we have the isomorphism ♭: TpM →T ∗ p M, defined such that for every u ∈TpM, the linear form u♭∈T ∗ p M is given by u♭(v) = ⟨u, v⟩p v ∈TpM. The inverse isomorphism ♯: T ∗ p M →TpM is defined such that for every ω ∈T ∗ p M, the vector ω♯is the unique vector in TpM so that ⟨ω♯, v⟩p = ω(v), v ∈TpM. The isomorphisms ♭and ♯induce isomorphisms between vector fields X ∈X(M) and one-forms ω ∈A1(M): A vector field X ∈X(M) yields the one-form X♭∈A1(M) given by (X♭)p = (Xp)♭, and a one-form ω ∈A1(M) yields the vector field ω♯∈X(M) given by (ω♯)p = (ωp)♯, so that ωp(v) = ⟨(ωp)♯, v⟩p, v ∈TpM, p ∈M. In particular, for every smooth function f ∈C∞(M), the vector field corresponding to the one-form d f is the gradient grad f, of f. The gradient of f is uniquely determined by the condition ⟨(grad f)p, v⟩p = d fp(v), v ∈TpM, p ∈M. 446 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES Recall from Proposition 10.13 that the covariant derivative ∇Xω of any one-form ω ∈A1(M) is the one-form given by (∇Xω)(Y ) = X(ω(Y )) −ω(∇XY ). If ∇is a metric connection, then the vector field (∇Xω)♯corresponding to ∇Xω is nicely expressed in terms of ω♯. Indeed, we have the following proposition. Proposition 10.14. If ∇is a metric connection on a smooth manifold M, then for every vector field X and every one-form ω we have (∇Xω)♯= ∇Xω♯. Proof. We have (∇Xω)(Y ) = X(ω(Y )) −ω(∇XY ) = X(⟨ω♯, Y ⟩) −⟨ω♯, ∇XY ⟩ = ⟨∇Xω♯, Y ⟩+ ⟨ω♯, ∇XY ⟩−⟨ω♯, ∇XY ⟩ = ⟨∇Xω♯, Y ⟩, where we used the fact that the connection is compatible with the metric in the third line and so, (∇Xω)♯= ∇Xω♯, as claimed. 10.8 Metrics on Vector Bundles, Reduction of Structure Groups, Orientation Because the fibres of a vector bundle are vector spaces, the definition of a Riemannian metric on a manifold can be lifted to vector bundles. Definition 10.20. Given a (real) rank n vector bundle ξ = (E, π, B, V ), we say that ξ is Euclidean iffthere is a family (⟨−, −⟩b)b∈B of inner products on each fibre π−1(b), such that ⟨−, −⟩b depends smoothly on b, which means that for every trivializing map ϕα : π−1(Uα) →Uα × V , for every frame, (s1, . . . , sn), on Uα, the maps b 7→⟨si(b), sj(b)⟩b, b ∈Uα, 1 ≤i, j ≤n are smooth. We say that ⟨−, −⟩is a Euclidean metric (or Riemannian metric) on ξ. If ξ is a complex rank n vector bundle ξ = (E, π, B, V ), we say that ξ is Hermitian iffthere is a family (⟨−, −⟩b)b∈B of Hermitian inner products on each fibre π−1(b), such that ⟨−, −⟩b depends smoothly on b. We say that ⟨−, −⟩is a Hermitian metric on ξ. For any smooth manifold M, if TM is a Euclidean vector bundle, then we say that M is a Riemannian manifold. 10.8. METRICS ON BUNDLES, REDUCTION, ORIENTATION 447 Now, given a real (resp. complex) vector bundle ξ, since B is paracompact because it is a manifold a Euclidean metric (resp. Hermitian metric) exists on ξ. This is a consequence of the existence of partitions of unity (see Warner , Chapter 1, or Gallier and Quaintance ). Theorem 10.15. Every real (resp. complex) vector bundle admits a Euclidean (resp. Her-mitian) metric. In particular, every smooth manifold admits a Riemannian metric. Proof. Let (Uα) be a trivializing open cover for ξ and pick any frame (sα 1, . . . , sα n) over Uα. For every b ∈Uα, the basis (sα 1(b), . . . , sα n(b)) defines a Euclidean (resp. Hermitian) inner product ⟨−, −⟩b on the fibre π−1(b), by declaring (sα 1(b), . . . , sα n(b)) orthonormal w.r.t. this inner product. (For x = Pn i=1 xisα i (b) and y = Pn i=1 yisα i (b), let ⟨x, y⟩b = Pn i=1 xiyi, resp. ⟨x, y⟩b = Pn i=1 xiyi, in the complex case.) The ⟨−, −⟩b (with b ∈Uα) define a metric on π−1(Uα), denote it ⟨−, −⟩α. Now, using a partition of unity, glue these inner products using a partition of unity (fα) subordinate to (Uα), by setting ⟨x, y⟩= X α fα⟨x, y⟩α. We verify immediately that ⟨−, −⟩is a Euclidean (resp. Hermitian) metric on ξ. The existence of metrics on vector bundles allows the so-called reduction of structure group. Recall that the transition maps of a real (resp. complex) vector bundle ξ are functions gαβ : Uα ∩Uβ →GL(n, R) (resp. GL(n, C)). Let GL+(n, R) be the subgroup of GL(n, R) consisting of those matrices of positive determinant (resp. GL+(n, C) be the subgroup of GL(n, C) consisting of those matrices of positive determinant). Definition 10.21. For every real (resp. complex) vector bundle ξ, if it is possible to find a cocycle g = (gαβ) for ξ with values in a subgroup H of GL(n, R) (resp. of GL(n, C)), then we say that the structure group of ξ can be reduced to H. We say that ξ is orientable if its structure group can be reduced to GL+(n, R) (resp. GL+(n, C)). Proposition 10.16. (a) The structure group of a rank n real vector bundle ξ can be re-duced to O(n); it can be reduced to SO(n) iffξ is orientable. (b) The structure group of a rank n complex vector bundle ξ can be reduced to U(n); it can be reduced to SU(n) iffξ is orientable. Proof. We prove (a), the proof of (b) being similar. Using Theorem 10.15, put a metric on ξ. For every Uα in a trivializing cover for ξ and every b ∈B, by Gram-Schmidt, orthonormal bases for π−1(b) exist. Consider the family of trivializing maps e ϕα : π−1(Uα) →Uα × V such that e ϕα,b : π−1(b) − →V maps orthonormal bases of the fibre to orthonormal bases of V . Then, it is easy to check that the corresponding cocycle takes values in O(n) and if ξ is orientable, the determinants being positive, these values are actually in SO(n). 448 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES Remark: If ξ is a Euclidean rank n vector bundle, then by Proposition 10.16, we may assume that ξ is given by some cocycle (gαβ), where gαβ(b) ∈O(n), for all b ∈Uα ∩Uβ. We saw in Section 10.5 (f) that the dual bundle ξ∗is given by the cocycle (gαβ(b)⊤)−1, b ∈Uα ∩Uβ. As gαβ(b) is an orthogonal matrix, (gαβ(b)⊤)−1 = gαβ(b), and thus, any Euclidean bundle is isomorphic to its dual. As we noted earlier, this is false for Hermitian bundles. Definition 10.22. Let ξ = (E, π, B, V ) be a rank n vector bundle and assume ξ is orientable. A family of trivializing maps ϕα : π−1(Uα) →Uα ×V is oriented ifffor all α, β, the transition function gαβ(b) has positive determinant for all b ∈Uα ∩Uβ. Two oriented families of trivializing maps ϕα : π−1(Uα) →Uα × V and ψβ : π−1(Wβ) →Wα × V are equivalent ifffor every b ∈Uα ∩Wβ, the map pr2 ◦ϕα ◦ψ−1 β ↾{b} × V : V − →V has positive determinant. It is easily checked that this is an equivalence relation and that it partitions all the oriented families of trivializations of ξ into two equivalence classes. Either equivalence class is called an orientation of ξ. If M is a manifold and ξ = TM, the tangent bundle of M, we know from Section 10.4 that the transition functions of TM are of the form gαβ(p)(u) = (ϕα ◦ϕ−1 β )′ ϕ(p)(u), where each ϕα : Uα →Rn is a chart of M. Consequently, TM is orientable iffthe Jacobian of (ϕα◦ϕ−1 β )′ ϕ(p) is positive, for every p ∈M. This is equivalent to the condition of Definition 7.4 for M to be orientable. Therefore, we have the following result. Proposition 10.17. The tangent bundle TM of a manifold M is orientable iffM is ori-entable.  The notion of orientability of a vector bundle ξ = (E, π, B, V ) is not equivalent to the orientability of its total space E. Indeed, if we look at the transition functions of the total space of TM given in Section 10.4, we see that TM, as a manifold, is always orientable, even if M is not orientable. Indeed, the transition functions of the tangent bundle TM are of the form e ψ ◦e ϕ−1(z, x) = (ψ ◦ϕ−1(z), (ψ ◦ϕ−1)′ z(x)), (z, x) ∈ϕ(U ∩V ) × Rn. Since (ψ ◦ϕ−1)′ z is a linear map, its derivative at any point is equal to itself, and it follows that the derivative of e ψ ◦e ϕ−1 at (z, x) is given by ( e ψ ◦e ϕ−1)′ (z,x)(u, v) = ((ψ ◦ϕ−1)′ z(u), (ψ ◦ϕ−1)′ z(v)), (u, v) ∈Rn × Rn. 10.8. METRICS ON BUNDLES, REDUCTION, ORIENTATION 449 Then the Jacobian matrix of this map is of the form J = A 0 0 A  where A is an n × n matrix, since (ψ ◦ϕ−1)′ z(u) does not involve the variables in v and (ψ ◦ϕ−1)′ z(v) does not involve the variables in u. Therefore det(J) = det(A)2, which shows that the transition functions have positive Jacobian determinant, and thus that TM is orientable. Yet, as a bundle, TM is orientable iffM is orientable. On the positive side, we have the following result. Proposition 10.18. If ξ = (E, π, B, V ) is an orientable vector bundle and its base B is an orientable manifold, then E is orientable too. Proof. To see this, assume that B is a manifold of dimension m, ξ is a rank n vector bundle with fibre V , let ((Uα, ψα))α be an atlas for B, let ϕα : π−1(Uα) →Uα × V be a collection of trivializing maps for ξ, and pick any isomorphism, ι: V →Rn. Then, we get maps (ψα × ι) ◦ϕα : π−1(Uα) − →Rm × Rn. It is clear that these maps form an atlas for E. Check that the corresponding transition maps for E are of the form (x, y) 7→(ψβ ◦ψ−1 α (x), gαβ(ψ−1 α (x))y). Moreover, if B and ξ are orientable, wer can check that these transition maps have positive Jacobian. The notion of subbundle is defined as follows: Definition 10.23. Given two vector bundles ξ = (E, π, B, V ) and ξ′ = (E′, π′, B, V ′) over the same base B, we say that ξ is a subbundle of ξ′ iffE is a submanifold of E′, V is a subspace of V ′, and for every b ∈B, the fibre π−1(b) is a subspace of the fibre (π′)−1(b). If ξ is a subbundle of ξ′, we can form the quotient bundle ξ′/ξ as the bundle over B whose fibre at b ∈B is the quotient space (π′)−1(b)/π−1(b). We leave it as an exercise to define trivializations for ξ′/ξ. In particular, if N is a submanifold of M, then TN is a subbundle of TM ↾N and the quotient bundle (TM ↾N)/TN is called the normal bundle of N in M. The fact that every bundle admits a metric allows us to define the notion of orthogonal complement of a subbundle. We state the following theorem without proof. The reader is invited to consult Milnor and Stasheff for a proof (Chapter 3). 450 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES Proposition 10.19. Let ξ and η be two vector bundles with ξ a subbundle of η. Then there exists a subbundle ξ⊥of η, such that every fibre of ξ⊥is the orthogonal complement of the fibre of ξ in the fibre of η over every b ∈B, and η ∼ = ξ ⊕ξ⊥. In particular, if N is a submanifold of a Riemannian manifold M, then the orthogonal complement of TN in TM ↾N is isomorphic to the normal bundle (TM ↾N)/TN. 10.9 Principal Fibre Bundles We now consider principal bundles. Such bundles arise in terms of Lie groups acting on manifolds. Let L(G) be the group of left translations of the group G, that is, the set of all homomorphisms Lg : G →G given by Lg(g′) = gg′, for all g, g′ ∈G. The map g 7→Lg is an isomorphism between the groups G and L(G) whose inverse is given by L 7→L(1) (where L ∈L(G)). Definition 10.24. Let G be a Lie group. A principal fibre bundle, for short a principal bundle, is a fibre bundle ξ = (E, π, B, G, L(G)) in which the fibre is G and the structure group is L(G), that is, G viewed as its group of left translations (ie., G acts on itself by multiplication on the left). This means that every transition function gαβ : Uα ∩Uβ →L(G) satisfies gαβ(b)(h) = (gαβ(b)(1))h, for all b ∈Uα ∩Uβ and all h ∈G. A principal G-bundle is denoted ξ = (E, π, B, G). In view of the isomorphism between L(G) and G we allow ourself the (convenient) abuse of notation gαβ(b)(h) = gαβ(b)h, where on the left, gαβ(b) is viewed as a left translation of G, and on the right as an element of G. When we want to emphasize that a principal bundle has structure group G, we use the locution principal G-bundle. It turns out that if ξ = (E, π, B, G) is a principal bundle, then G acts on the total space E, on the right. For the next proposition, recall that a right action ·: X × G →X is free iff for every g ∈G, if g ̸= 1, then x · g ̸= x for all x ∈X. Proposition 10.20. If ξ = (E, π, B, G) is a principal bundle, then there is a right action of G on E. This action takes each fibre to itself and is free. Moreover, E/G is diffeomorphic to B. 10.9. PRINCIPAL FIBRE BUNDLES 451 Proof. We show how to define the right action and leave the rest as an exercise. Let {(Uα, ϕα)} be some trivializing cover defining ξ. For every z ∈E, pick some Uα so that π(z) ∈Uα, and let ϕα(z) = (b, h), where b = π(z) and h ∈G. For any g ∈G, we set z · g = ϕ−1 α (b, hg). If we can show that this action does not depend on the choice of Uα, then it is clear that it is a free action. Suppose that we also have b = π(z) ∈Uβ and that ϕβ(z) = (b, h′). By definition of the transition functions, we have h′ = gβα(b)h and ϕβ(z · g) = ϕβ(ϕ−1 α (b, hg)) = (b, gβα(b)(hg)). However, gβα(b)(hg) = (gβα(b)h)g = h′g, hence z · g = ϕ−1 β (b, h′g), which proves that our action does not depend on the choice of Uα. Observe that the action of Proposition 10.20 is defined by z · g = ϕ−1 α (b, ϕα,b(z)g), with b = π(z), for all z ∈E and all g ∈G. It is clear that this action satisfies the following two properties: For every (Uα, ϕα), (1) π(z · g) = π(z), and (2) ϕα(z · g) = ϕα(z) · g, for all z ∈E and all g ∈G, where we define the right action of G on Uα × G so that (b, h) · g = (b, hg). Definition 10.25. A trivializing map ϕα satisfying Condition (2) above is G-equivariant (or equivariant). The following proposition shows that it is possible to define a principal G-bundle using a suitable right action and equivariant trivializations: Proposition 10.21. Let E be a smooth manifold, G be a Lie group, and let ·: E × G →E be a smooth right action of G on E satisfying the following properties: (a) The right action of G on E is free; (b) The orbit space B = E/G is a smooth manifold under the quotient topology, and the projection π: E →E/G is smooth; 452 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES (c) There is a family of local trivializations {(Uα, ϕα)}, where {Uα} is an open cover for B = E/G, and each ϕα : π−1(Uα) →Uα × G is an equivariant diffeomorphism, which means that ϕα(z · g) = ϕα(z) · g, for all z ∈π−1(Uα) and all g ∈G, where the right action of G on Uα × G is (b, h) · g = (b, hg). If π: E →E/G is the quotient map, then ξ = (E, π, E/G, G) is a principal G-bundle. Proof. Since the action of G on E is free, every orbit b = z · G is isomorphic to G, and so every fibre π−1(b) is isomorphic to G. Thus, given that we have trivializing maps, we just have to prove that G acts by left translation on itself. Pick any (b, h) in Uβ × G and let z ∈π−1(Uβ) be the unique element such that ϕβ(z) = (b, h). Then as ϕβ(z · g) = ϕβ(z) · g, for all g ∈G, we have ϕβ(ϕ−1 β (b, h) · g) = ϕβ(z · g) = ϕβ(z) · g = (b, h) · g, which implies that ϕ−1 β (b, h) · g = ϕ−1 β ((b, h) · g). Consequently, ϕα ◦ϕ−1 β (b, h) = ϕα ◦ϕ−1 β ((b, 1) · h) = ϕα(ϕ−1 β (b, 1) · h) = ϕα ◦ϕ−1 β (b, 1) · h, and since ϕα ◦ϕ−1 β (b, h) = (b, gαβ(b)(h)) and ϕα ◦ϕ−1 β (b, 1) = (b, gαβ(b)(1)) we get gαβ(b)(h) = gαβ(b)(1)h. The above shows that gαβ(b): G →G is the left translation by gαβ(b)(1), and thus the transition functions gαβ(b) constitute the group of left translations of G, and ξ is indeed a principal G-bundle. Br¨ ocker and tom Dieck (Chapter I, Section 4) and Duistermaat and Kolk (Ap-pendix A) define principal bundles using the conditions of Proposition 10.21. Propositions 10.20 and 10.21 show that this alternate definition is equivalent to ours (Definition 10.24). It turns out that when we use the definition of a principal bundle in terms of the conditions of Proposition 10.21, it is convenient to define bundle maps in terms of equivariant maps. As we will see shortly, a map of principal bundles is a fibre bundle map. 10.9. PRINCIPAL FIBRE BUNDLES 453 Definition 10.26. If ξ1 = (E1, π1, B1, G) and ξ2 = (E2, π2, B2, G) are two principal bundles, a bundle map (or bundle morphism) f : ξ1 →ξ2 is a pair, f = (fE, fB) of smooth maps fE : E1 →E2 and fB : B1 →B2, such that: (a) The following diagram commutes: E1 π1  fE / E2 π2  B1 fB / B2 (b) The map fE is G-equivariant; that is, fE(a · g) = fE(a) · g, for all a ∈E1 and all g ∈G. A bundle map is an isomorphism if it has an inverse as in Definition 10.6. If the bundles ξ1 and ξ2 are over the same base B, then we also require fB = id. At first glance, it is not obvious that a map of principal bundles satisfies Condition (b) of Definition 10.7. If we define e fα : Uα × G →Vβ × G by e fα = ϕ′ β ◦fE ◦ϕ−1 α , then locally fE is expressed as fE = ϕ′ β −1 ◦e fα ◦ϕα. Furthermore, it is trivial that if a map is equivariant and invertible, then its inverse is equivariant. Consequently, since e fα = ϕ′ β ◦fE ◦ϕ−1 α , as ϕ−1 α , ϕ′ β and fE are equivariant, e fα is also equivariant, and so e fα is a map of (trivial) principal bundles. Thus, it is enough to prove that for every map of principal bundles ϕ: Uα × G →Vβ × G, there is some smooth map ρα : Uα →G, so that ϕ(b, g) = (fB(b), ρα(b)(g)), for all b ∈Uα and all g ∈G. Indeed, we have the following 454 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES Proposition 10.22. For every map of trivial principal bundles ϕ: Uα × G →Vβ × G, there are smooth maps fB : Uα →Vβ and rα : Uα →G, so that ϕ(b, g) = (fB(b), rα(b)g), for all b ∈Uα and all g ∈G. In particular, ϕ is a diffeomorphism on fibres. Proof. As ϕ is a map of principal bundles ϕ(b, 1) = (fB(b), rα(b)), for all b ∈Uα, for some smooth maps fB : Uα →Vβ and rα : Uα →G. Now, using equivariance, we get ϕ(b, g) = ϕ((b, 1)g) = ϕ(b, 1) · g = (fB(b), rα(b)) · g = (fB(b), rα(b)g), as claimed. Consequently, the map ρα : Uα →G given by ρα(b)(g) = rα(b)g for all b ∈Uα and all g ∈G satisfies ϕ(b, g) = (fB(b), ρα(b)(g)), for all b ∈Uα and all g ∈G, and a map of principal bundles is indeed a fibre bundle map (as in Definition 10.7). Since a principal bundle map is a fibre bundle map, Proposition 10.3 also yields the useful fact: Proposition 10.23. Any map f : ξ1 →ξ2 between two principal bundles over the same base B is an isomorphism. A natural question is to ask whether a fibre bundle ξ is isomorphic to a trivial bundle. If so, we say that ξ is trivial. (By the way, the triviality of bundles comes up in physics, in particular, field theory.) Generally, this is a very difficult question, but a first step can be made by showing that it reduces to the question of triviality for principal bundles. Indeed, if ξ = (E, π, B, F, G) is a fibre bundle with fibre F, using Theorem 10.4, we can construct a principal fibre bundle P(ξ) using the transition functions {gαβ} of ξ, but using G itself as the fibre (acting on itself by left translation) instead of F. Definition 10.27. Let ξ = (E, π, B, F, G) is a fibre bundle with fibre F, and P(ξ) be the bundle obtained by replacing the fibre of ξ with G (as described in the preceding paragraph). We call P(ξ) the principal bundle associated to ξ. 10.9. PRINCIPAL FIBRE BUNDLES 455 For example, the principal bundle associated with a vector bundle is the frame bundle, discussed at the end of Section 10.4. Then given two fibre bundles ξ and ξ′, we see that ξ and ξ′ are isomorphic iffP(ξ) and P(ξ′) are isomorphic (Steenrod , Part I, Section 8, Theorem 8.2). More is true: the fibre bundle ξ is trivial iffthe principal fibre bundle P(ξ) is trivial (see Steenrod , Part I, Section 8, Corollary 8.4). Moreover, there is a test for the triviality of a principal bundle, the existence of a (global) section. The following proposition, although easy to prove, is crucial: Proposition 10.24. If ξ is a principal bundle, then ξ is trivial iffit possesses some global section. Proof. If f : B × G →ξ is an isomorphism of principal bundles over the same base B, then for every g ∈G, the map b 7→f(b, g) is a section of ξ. Conversely, let s: B →E be a section of ξ. Then, observe that the map f : B × G →ξ given by f(b, g) = s(b)g is a map of principal bundles. By Proposition 10.23, it is an isomorphism, so ξ is trivial. Generally, in geometry, many objects of interest arise as global sections of some suitable bundle (or sheaf): vector fields, differential forms, tensor fields, etc. Definition 10.28. Given a principal bundle ξ = (E, π, B, G) and given a manifold F, if G acts effectively on F from the left, using Theorem 10.4, we can construct a fibre bundle ξ[F] from ξ, with F as typical fibre, and such that ξ[F] has the same transitions functions as ξ. The fibre bundle ξ[F] is called the fibre bundle induced by ξ. In the case of a principal bundle, there is another slightly more direct construction that takes us from principal bundles to fibre bundles (see Duistermaat and Kolk , Chapter 2, and Davis and Kirk , Chapter 4, Definition 4.6, where it is called the Borel construction). This construction is of independent interest, so we describe it briefly (for an application of this construction, see Duistermaat and Kolk , Chapter 2). As ξ is a principal bundle, recall that G acts on E from the right, so we have a right action of G on E × F, via (z, f) · g = (z · g, g−1 · f). Consequently, we obtain the orbit set E ×F/ ∼, denoted E ×GF, where ∼is the equivalence relation (z, f) ∼(z′, f ′) iff (∃g ∈G)(z′ = z · g, f ′ = g−1 · f). Note that the composed map E × F pr1 − →E π − →B 456 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES factors through E ×G F as a map p: E ×G F →B given by p([z, f)] = π(pr1(z, f)), as illustrated in the diagram below E × F q & pr1 / E π / B E ×G F, p : since π(pr1(z, f)) = π(z) = π(z · g) = π(pr1(z · g, g−1 · f)), which means that the definition of p does not depend on the choice of representative in the equivalence class [(z, f)]. The following proposition is not hard to show: Proposition 10.25. If ξ = (E, π, B, G) is a principal bundle and F is any manifold such that G acts effectively on F from the left, then, ξ[F] = (E ×G F, p, B, F, G) is a fibre bundle with fibre F and structure group G, and ξ[F] and ξ have the same transition functions. Sketch of proof. Let us verify that the charts of ξ yield charts for ξ[F]. For any Uα in an open cover for B, we have a diffeomorphism ϕα : π−1(Uα) →Uα × G. The first step is to show that that there is an isomorphism (Uα × G) ×G F ∼ = Uα × F, where, as usual, G acts on Uα × G via (z, h) · g = (z, hg), Two pairs ((b1, g1), f1) and ((b2, g2), f2) are equivalent iffthere is some g ∈G such that (b2, g2) = (b1, g1) · g, f2 = g−1 · f1, which implies that (b2, g2) = (b1, g1g), so b1 = b2 and g2 = g1g. It follows that g = g−1 1 g2 and g1 · f1 = g2 · f2, so two pairs ((b1, g1), f1) and ((b2, g2), f2) are equivalent iff b1 = b2, and g1 · f1 = g2 · f2. The map θ: (Uα × G) ×G F →Uα × F given by θ([((b1, g1), f1)]) = (b1, g1 · f1) is well-defined on the equivalence class [((b1, g1), f1)], and it is clear that it is a bijection since (b1, g1 · f1) = (b2, g1 · f2) implies that [((b1, g1), f1)] = [((b2, g2), f2)]. 10.9. PRINCIPAL FIBRE BUNDLES 457 We also have an isomorphism p−1(Uα) ∼ = π−1(Uα) ×G F, and since ϕα : π−1(Uα) →Uα × G induces an isomorphism π−1(Uα) ×G F ξ / (Uα × G) ×G F, and we have an isomorphism θ: (Uα × G) ×G F →Uα × F, so we have an isomorphism p−1(Uα) − →Uα × F and we get the commutative diagram p−1(Uα) p $ θ◦ξ / Uα × F pr1 z Uα, which yields a local trivialization θ ◦ξ for ξ[F]. It is easy to see that the transition functions of ξ[F] are the same as the transition functions of ξ. Now if we start with a fibre bundle ξ with fibre F and structure group G, if we make the associated principal bundle P(ξ), and then the induced fibre bundle P(ξ)[F], what is the relationship between ξ and P(ξ)[F]? The answer is: ξ and P(ξ)[F] are equivalent (this is because the transition functions are the same.) Now, if we start with a principal G-bundle ξ, make the fibre bundle ξ[F] as above, and then the principal bundle P(ξ[F]), we get a principal bundle equivalent to ξ. Therefore, the maps ξ 7→ξ[F] and ξ 7→P(ξ) are mutual inverses, and they set up a bijection between equivalence classes of principal G-bundles over B and equivalence classes of fibre bundles over B (with structure group G). Moreover, this map extends to morphisms, so it is functorial (see Steenrod , Part I, Section 2, Lemma 2.6–Lemma 2.10). As a consequence, in order to “classify” equivalence classes of fibre bundles (assuming B and G fixed), it is enough to know how to classify principal G-bundles over B. Given some reasonable conditions on the coverings of B, Milnor solved this classification problem, but this is taking us way beyond the scope of these notes! Proposition 10.25 can be generalized to the situation where the action of the Lie group G on F is not necessarily effective. In this case we need the generalization of the notion of fibre bundle in which the diffeomorphism ϕα,b ◦ϕ−1 β,b : F →F of the fibre F is not a member of G since G can’t generally be identified with a group of diffeomorphims of F if our action is not effective. This more general definition using Condition (c’) is discussed just before Definition 10.2. Recall that we drop Condition (c) and we modify Condition (d) of Definition 10.1 as follows: 458 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES (c’) For all α, β, there is a smooth map gαβ : Uα ∩Uβ →G such that ϕα,b ◦ϕ−1 β,b(x) = gαβ(b) · x, b ∈Uα ∩Uβ, x ∈F, and thus ϕα ◦ϕ−1 β (b, x) = (b, gαβ(b) · x), b ∈Uα ∩Uβ, x ∈F. The transition functions (gαβ) are still coycles, etc. A new version of Proposition 10.25 can be proven, but this time the transition functions of the fibre bundle ξ[F] are defined in terms of the action of G on F, although formally they look like the transition functions of ξ, that involve the action of G on itself. We leave the details as an exercise. Vector bundles can also be generalized (see just after Definition 10.14) by assuming that we have a representation ρ: G →GL(V ), where V is the typical fibre, and the diffeomor-phism ϕα,b ◦ϕ−1 β,b : V →V of the fibre V is a linear automorphism specified in terms of a smooth map gαβ : Uα ∩Uβ →G such that ϕα,b ◦ϕ−1 β,b(x) = ρ(gαβ(b))(x), b ∈Uα ∩Uβ, x ∈V, so that ϕα ◦ϕ−1 β (b, x) = (b, ρ(gαβ(b))(x)), b ∈Uα ∩Uβ, x ∈V. A version of Proposition 10.25 can also be proven for such vector bundles. The classical reference on fibre bundles, vector bundles and principal bundles, is Steenrod . More recent references include Bott and Tu , Madsen and Tornehave , Morita , Griffith and Harris , Wells , Hirzebruch , Milnor and Stasheff, Davis and Kirk , Atiyah , Chern , Choquet-Bruhat, DeWitt-Morette and Dillard-Bleick , Hirsh , Sato , Narasimham , Sharpe and also Husemoller , which covers more, including characteristic classes. Proposition 10.21 shows that principal bundles are induced by suitable right actions, but we still need sufficient conditions to guarantee Conditions (a), (b) and (c). The special situation of homogeneous spaces is considered in the next section. 10.10 Proper and Free Actions, Homogeneous Spaces Now that we have introduced the notion of principal bundle, we can state various results about homogeneous spaces. These results are stronger than those stated in Gallier and Quaintance which apply to groups and sets without any topology or differentiable struc-ture. We need to review the notion of proper map and proper action. Definition 10.29. If X and Y are two Hausdorfftopological spaces,1 a function a ϕ: X →Y is proper iffit is continuous and for every topological space Z, the map ϕ×id: X×Z →Y ×Z 1It is not necessary to assume that X and Y are Hausdorffbut, if X and/or Y are not Hausdorff, we have to replace “compact” by “quasi-compact.” We have no need for this extra generality. 10.10. PROPER AND FREE ACTIONS, HOMOGENEOUS SPACES 459 is a closed map (recall that f is a closed map iffthe image of any closed set by f is a closed set). If we let Z be a one-point space, we see that a proper map is closed. At first glance, it is not obvious how to check that a map is proper just from Definition 10.29. Proposition 10.27 gives a more palatable criterion. The following proposition is easy to prove (see Bourbaki, General Topology , Chapter 1, Section 10). Proposition 10.26. If ϕ: X →Y is any proper map, then for any closed subset F of X, the restriction of ϕ to F is proper. The following result providing a “good” criterion for checking that a map is proper can be shown (see Bourbaki, General Topology , Chapter 1, Section 10). Proposition 10.27. A continuous map ϕ: X →Y is proper iffϕ is closed and if ϕ−1(y) is compact for every y ∈Y . Proposition 10.27 shows that a homeomorphism (or a diffeomorphism) is proper. If ϕ is proper, it is easy to show that ϕ−1(K) is compact in X whenever K is compact in Y . Moreover, if Y is also locally compact, then we have the following result (see Bourbaki, General Topology , Chapter 1, Section 10). Proposition 10.28. If Y is locally compact, a continuous map ϕ: X →Y is a proper map iffϕ−1(K) is compact in X whenever K is compact in Y In particular, this is true if Y is a manifold since manifolds are locally compact. This explains why Lee (Chapter 9) takes the property stated in Proposition 10.28 as the definition of a proper map (because he only deals with manifolds).2 Proper actions are defined as follows. Definition 10.30. Given a Hausdorfftopological group G and a topological space M, a left action ·: G × M →M is proper if it is continuous and if the map θ: G × M − →M × M, (g, x) 7→(g · x, x) is proper. Proposition 10.29. The action ·: H × G →G of a closed subgroup H of a group G on G (given by (h, g) 7→hg) is proper. The same is true for the right action of H on G. 2However, Duistermaat and Kolk seem to have overlooked the fact that a condition on Y (such as local compactness) is needed in their remark on lines 5-6, page 53, just before Lemma 1.11.3. 460 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES Definition 10.31. An action ·: G × M →M is free if for all g ∈G and all x ∈M, if g ̸= 1 then g · x ̸= x. An equivalent way to state that an action ·: G × M →M is free is as follows. For every g ∈G, let τg : M →M be the diffeomorphism of M given by τg(x) = g · x, x ∈M. Then the action ·: G × M →M is free ifffor all g ∈G, if g ̸= 1 then τg has no fixed point. Consequently, an action ·: G × M →M is free ifffor every x ∈M, the stabilizer Gx of x is reduced to the trivial group {1}. If H is a subgroup of G, obviously H acts freely on G (by multiplication on the left or on the right). This fact together with Proposition 10.29 yields the following corollary which provides a large supply of free and proper actions. Corollary 10.30. The action ·: H × G →G of a closed subgroup H of a group G on G (given by (h, g) 7→hg) is free and proper. The same is true for the right action of H on G. Before stating the main results of this section, observe that in the definition of a fibre bundle (Definition 10.1), the local trivialization maps are of the form ϕα : π−1(Uα) →Uα × F, where the fibre F appears on the right. In particular, for a principal fibre bundle ξ, the fibre F is equal to the structure group G, and this is the reason why G acts on the right on the total space E of ξ (see Proposition 10.20). To be more precise, we call a right bundle a bundle ξ = (E, π, B, F, G) where the group G acts effectively on the left on the fibre F and where the local trivialization maps are of the form ϕα : π−1(Uα) →Uα × F. If ξ is a right principal bundle, the group G acts on E on the right . We call a a left bundle a bundle ξ = (E, π, B, F, G) where the group G acts effectively on the right on the fibre F and the local trivialization maps are of the form ϕα : π−1(Uα) →F × Uα. Then if ξ is a left principal bundle, the group G acts on E on the left. Duistermaat and Kolk address this issue at the end of their Appendix A, and prove the theorem stated below (Chapter 1, Section 11). Beware that in Duistermaat and Kolk , this theorem is stated for right bundles. However, the weaker version that does not mention principal bundles is usually stated for left actions; for instance, see Lee (Chapter 9, Theorem 9.16). We formulate both versions at the same time. 10.10. PROPER AND FREE ACTIONS, HOMOGENEOUS SPACES 461 Theorem 10.31. Let M be a smooth manifold, G be a Lie group, and let ·: M ×G →M be a right smooth action (resp. ·: G × M →M a left smooth action) which is proper and free. Then, M/G is a principal right G-bundle (resp. left G-bundle) of dimension dim M −dim G. Moreover, the canonical projection π: M →M/G is a submersion,3 and there is a unique manifold structure on M/G with this property. Theorem 10.31 has some interesting corollaries. Because a closed subgroup H of a Lie group G is a Lie group, and because the action of a closed subgroup is free and proper, we get the following result (proofs can also be found in Br¨ ocker and tom Dieck (Chapter I, Section 4) and in Duistermaat and Kolk (Chapter 1, Section 11)). Theorem 10.32. If G is a Lie group and H is a closed subgroup of G, then the right action of H on G defines a principal (right) H-bundle ξ = (G, π, G/H, H), where π: G →G/H is the canonical projection. Moreover, π is a submersion, and there is a unique manifold structure on G/H with this property. In the special case where G acts transitively on M, for any x ∈M, if Gx is the stabilizer of x, then with H = Gx, we get Proposition 10.33 below. Recall the definition of a homogeneous space. Definition 10.32. A homogeneous space is a smooth manifold M together with a smooth transitive action ·: G × M →M, of a Lie group G on M. The following result can be shown as a corollary of Theorem 10.32 (see Lee , Chapter 9, Theorem 9.24). It is also mostly proved in Br¨ ocker and tom Dieck , Chapter I, Section 4): Proposition 10.33. Let ·: G × M →M be smooth transitive action of a Lie group G on a manifold M. Then, G/Gx and M are diffeomorphic, and G is the total space of a principal bundle ξ = (G, π, M, Gx), where Gx is the stabilizer of any element x ∈M. Furthermore, the projection π: G →G/Gx is a submersion. Thus, we finally see that homogeneous spaces induce principal bundles. Going back to some of the examples mentioned earlier (also, see Gallier and Quaintance ), we see that (1) SO(n + 1) is a principal SO(n)-bundle over the sphere Sn (for n ≥0). (2) SU(n + 1) is a principal SU(n)-bundle over the sphere S2n+1 (for n ≥0). (3) SL(2, R) is a principal SO(2)-bundle over the upper-half space H. (4) GL(n, R) is a principal O(n)-bundle over the space SPD(n) of symmetric, positive definite matrices. 3Recall that this means that the derivative dπp : TpM →Tπ(p)M/G is surjective for every p ∈M. 462 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES (5) GL+(n, R), is a principal SO(n)-bundle over the space SPD(n) of symmetric, positive definite matrices, with fibre SO(n). (6) SO(n + 1) is a principal O(n)-bundle over the real projective space RPn (for n ≥0). (7) SU(n + 1) is a principal U(n)-bundle over the complex projective space CPn (for n ≥0). (8) O(n) is a principal O(k) × O(n −k)-bundle over the Grassmannian G(k, n). (9) SO(n) is a principal S(O(k) × O(n −k))-bundle over the Grassmannian G(k, n). (10) SO(n) is a principal SO(n −k)-bundle over the Stiefel manifold S(k, n), with 1 ≤k ≤ n −1. (11) The Lorentz group SO0(n, 1) is a principal SO(n)-bundle over the space H+ n (1), con-sisting of one sheet of the hyperbolic paraboloid Hn(1). Thus, we see that both SO(n + 1) and SO0(n, 1) are principal SO(n)-bundles, the dif-ference being that the base space for SO(n + 1) is the sphere Sn, which is compact, whereas the base space for SO0(n, 1) is the (connected) surface H+ n (1), which is not compact. Many more examples can be given, for instance, see Arvanitoyeorgos . 10.11 Problems Problem 10.1. Show that a Klein bottle is a fibre bundle with B = F = S1 and G = {−1, 1}. Problem 10.2. Adjust the proof of Proposition 10.1 to prove the following: If ξ1 = (E1, π1, B1, F, G) and ξ2 = (E2, π2, B2, F, G) are two bundles over different bases and f : ξ1 →ξ2 is a bundle isomorphism, with f = (fB, fE), then fE and fB are diffeomor-phisms, and g′ αβ(fB(b)) = ρα(b)gαβ(b)ρβ(b)−1, for all b ∈Uα ∩Uβ. Problem 10.3. Adjust the proof of Proposition 10.2 to prove the following: If ξ1 = (E1, π1, B1, F, G) and ξ2 = (E2, π2, B2, F, G) are two bundles over different bases and if there is a diffeomorphism fB : B1 →B2, and the conditions g′ αβ(fB(b)) = ρα(b)gαβ(b)ρβ(b)−1, for all b ∈Uα ∩Uβ hold, then there is a bundle isomorphism (fB, fE) between ξ1 and ξ2. Problem 10.4. Complete the proof details of Theorem 10.4. In particular check that the cocycle condition produces an equivalence relation on Z×Z. Also verify that the correspond-ing transition functions are the original gαβ. Finally prove that ξg and ξg′ are isomorphic when g and g′ are equivalent cocycles. 10.11. PROBLEMS 463 Hint. See Steenrod , Part I, Section 3, Theorem 3.2. Also see Steenrod , Part I, Section 2, Lemma 2.10. Problem 10.5. Show that the transitions functions of the pullback bundle f ∗ξ can be constructed as follows: Pick any open cover (Uα) of B, then (f −1(Uα)) is an open cover of N, and check that if (gαβ) is a cocycle for ξ, then the maps gαβ ◦f : f −1(Uα) ∩f −1(Uβ) →G satisfy the cocycle conditions. Problem 10.6. Show that the pullback bundle f ∗ξ of Definition 10.11 can be defined ex-plicitly as follows. Set f ∗E = {(n, e) ∈N × E | f(n) = π(e)}, π∗= pr1 and f ∗= pr2. For any trivialization ϕα : π−1(Uα) →Uα × F of ξ, we have (π∗)−1(f −1(Uα)) = {(n, e) ∈N × E | n ∈f −1(Uα), e ∈π−1(f(n))}, and so we have a bijection e ϕα : (π∗)−1(f −1(Uα)) →f −1(Uα) × F, given by e ϕα(n, e) = (n, pr2(ϕα(e))). By giving f ∗E the smallest topology that makes each e ϕα a diffeomorphism, prove that each e ϕα is a trivialization of f ∗ξ over f −1(Uα). Problem 10.7. If g: M →N is another smooth map of manifolds, show that that (f ◦g)∗ξ = g∗(f ∗ξ). Problem 10.8. Let HR n ⊆RPn × Rn+1 be the subset HR n = {(L, v) ∈RPn × Rn+1 | v ∈L}, where RPn is viewed as the set of lines L in Rn+1 through 0, or more explicitly, HR n = {((x0 : · · · : xn), λ(x0, . . . , xn)) | (x0 : · · · : xn) ∈RPn, λ ∈R}. Show that that HR n is a manifold of dimension n + 1. Problem 10.9. For rank 1 vector bundles, show that the set of equivalence classes of line bundles over a base B forms a group, where the group operation is ⊗, the inverse is ∗(dual), and the identity element is the trivial bundle. Problem 10.10. Given a real vector bundle, ξ, recall that the complexification ξC of ξ is the complex vector bundle defined by ξC = ξ ⊗R ϵC, where ϵC = B ×C is the trivial complex line bundle. Given a complex vector bundle ξ, recall that by viewing its fibre as a real vector space we obtain the real vector bundle ξR. Prove the following. 464 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES (1) For every real vector bundle ξ, (ξC)R ∼ = ξ ⊕ξ. (2) For every complex vector bundle ξ, (ξR)C ∼ = ξ ⊕ξ∗. Problem 10.11. Given ξ, a subbundle of ξ′, we can form the quotient bundle ξ′/ξ as the bundle over B whose fibre at b ∈B is the quotient space (π′)−1(b)/π−1(b). Define the trivializations and transition maps for ξ′/ξ. Problem 10.12. Prove Proposition 10.13. Hint. See O’Neill , Chapter 2, Proposition 13 and Theorem 15. Problem 10.13. Prove the following: If ξ = (E, π, B, V ) is an orientable vector bundle and its base B is an orientable manifold, then E is orientable too. Hint. Assume that B is a manifold of dimension m, ξ is a rank n vector bundle with fibre V , let ((Uα, ψα))α be an atlas for B, let ϕα : π−1(Uα) →Uα ×V be a collection of trivializing maps for ξ, and pick any isomorphism, ι: V →Rn. (a) Show that the maps (ψα × ι) ◦ϕα : π−1(Uα) − →Rm × Rn. form an atlas for E. (b) Check that the corresponding transition maps for E are of the form (x, y) 7→(ψβ ◦ψ−1 α (x), gαβ(ψ−1 α (x))y). (c) Since B and ξ are orientable, check that these transition maps have positive Jacobian. Problem 10.14. Prove Proposition 10.19. Hint. See Milnor and Stasheff, Chapter 3. Problem 10.15. Prove Proposition 10.11. Hint. See Madsen and Tornehave , Chapter 16. Problem 10.16. Complete the proof of Proposition 10.20. Recall that right action is defined as follows: Let {(Uα, ϕα)} be some trivializing cover defining ξ. For every z ∈E, pick some Uα so that π(z) ∈Uα, and let ϕα(z) = (b, h), where b = π(z) and h ∈G. For any g ∈G, we set z · g = ϕ−1 α (b, hg). Show this action takes each fibre to itself. Also show that E/G is diffeomorphic to B. 10.11. PROBLEMS 465 Problem 10.17. Complete the proof details of Proposition 10.25. In particular show that the map θ: (Uα × G) ×G F →Uα × F given by θ([((b1, g2), g−1 2 g1 · f1)]) = (b1, g1 · f1) is well-defined on the equivalence class [((b1, g1), f1)] and that it is a bijection. Also show that the transition functions of ξ[F] are the same as the transition functions of ξ. 466 CHAPTER 10. BUNDLES, METRICS ON BUNDLES, HOMOGENEOUS SPACES Chapter 11 Connections and Curvature in Vector Bundles 11.1 Introduction to Connections in Vector Bundles A connection on a manifold B is a means of relating different tangent spaces. In particular, a connection on B is a R-bilinear map ∇: X(B) × X(B) →X(B), (†) such that the following two conditions hold: ∇fXY = f∇XY ∇X(fY ) = X[f]Y + f∇XY, for all smooth vector fields X, Y ∈X(B) and all f ∈C∞(B); see Gallot, Hulin, Lafontaine, , Do Carmo , or Gallier and Quaintance . Given p ∈B and X, Y ∈X(B), we know that Equation (†) is related to the directional derivative DXY (p) of Y with respect to X, namely DXY (p) = lim t→0 Y (p + tX(p)) −Y (p) t , since DXY (p) = ∇XY (p) + (Dn)XY (p), where its horizontal (or tangential) component is ∇XY (p) ∈TpB, and its normal component is (Dn)XY (p). A natural question is to wonder whether we can generalize this notion of directional derivative to the case of a vector bundle ξ = (E, π, B, V ). The answer is yes if we let Y be a smooth global vector field of V instead of a smooth global vector field of tangent vectors. In other words, since X(B) is the set of smooth sections of the tangent bundle TB, we may rewrite (†) as ∇: X(B) × Γ(TB) →Γ(TB), (††) 467 468 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES replace the two occurrence of Γ(TB) with Γ(ξ) and say a connection on ξ is an R-bilinear map ∇: X(B) × Γ(ξ) →Γ(ξ), such that the following two conditions hold: ∇fXs = f∇Xs ∇X(fs) = X[f]s + f∇Xs, for all s ∈Γ(ξ), all X ∈X(B) and all f ∈C∞(B). We refer to ∇Xs as the covariant derivative of s relative to X. This definition of a connection on a vector bundle has the advantage in that it readily allows us to transfer all the concepts of connections on a manifold to the context of con-nections in vector bundles. In particular, we will show that connections in vector bundles exist and are local operators; see Propositions 11.4 and 11.1 respectively. We will be able to define the notion of parallel transport along a curve of B in terms of the R-linear map D dt where DX dt (t0) = (∇γ′(t0) s)γ(t0), whenever X is induced by a global section s ∈Γ(ξ), i.e. X(t0) = s(γ(t0)) for all t0 ∈[a, b]; see Proposition 11.6 and Definition 11.7. We will also be able to define the notion of a metric connection in a vector bundle as follows. Given any metric ⟨−, −⟩on a vector bundle ξ, a connection ∇on ξ is compatible with the metric if and only if X(⟨s1, s2⟩) = ⟨∇Xs1, s2⟩+ ⟨s1, ∇Xs2⟩, for every vector field X ∈X(B) and sections s1, s2 ∈Γ(ξ); see Definition 11.17. We can also generalize the notion of curvature in a Riemannian manifold to the context of vector bundles if we define the curvature tensor of Γ(ξ) as R(X, Y ) = ∇X∇Y −∇Y ∇X −∇[X,Y ], where X, Y ∈X(B). Note that this definition of curvature implies that R: X(B) × X(B) × Γ(ξ) − →Γ(ξ) is a R-trilinear map where R(X, Y )s = ∇X∇Y s −∇Y ∇Xs −∇[X,Y ]s, whenever X, Y ∈X(B) and s ∈Γ(ξ). The reason we are interested in having a definition of curvature on a vector bundle ξ = (E, π, B, V ) is that it allows us to define global invariants on ξ called the Pontrjagin and Chern classes; see Section 11.11. However in order to define the Pontrjagin and Chern 11.1. INTRODUCTION TO CONNECTIONS IN VECTOR BUNDLES 469 classes in an accessible manner, we will need to associate R(X, Y ) with a vector valued two-form R∇. We are able to make this association if we realize that a connection on ξ = (E, π, B, V ) is actually a vector valued one-form with the vector values taken from Γ(ξ). Therefore, following the lead of Appendix C in Milnor and Stasheff, we will rephrase the notions of connection, metric connection, and curvature in terms of vector valued forms. This vector valued form approach has another advantage in that it allows for elegant proofs of the essential properties of connections on vector bundles. In Section 11.2 we define connections on a vector bundle. This can be done in two equivalent ways. One of the two definitions is more abstract than the other because it involves a tensor product, but it is technically more convenient. This definition states that a connection on a vector bundle ξ, as an R-linear map ∇: Γ(ξ) →A1(B) ⊗C∞(B) Γ(ξ) (∗) that satisfies the “Leibniz rule” ∇(fs) = d f ⊗s + f∇s, with s ∈Γ(ξ) and f ∈C∞(B), where Γ(ξ) and A1(B) are treated as C∞(B)-modules. Here, A1(B) = Γ(T ∗B) is the space of 1-forms on B. Since there is an isomorphism A1(B) ⊗C∞(B) Γ(ξ) ∼ = Γ(T ∗B ⊗ξ), a connection can be defined equivalently as an R-linear map ∇: Γ(ξ) →Γ(T ∗B ⊗ξ) satisfying the Leibniz rule. Milnor and Stasheff (Appendix C) use this second version, and Madsen and Tornehave (Chapter 17) use the equivalent version stated in (∗). We show that a connection is a local operator. In Section 11.3, we show how a connection can be represented in a chart in terms of a certain matrix called a connection matrix. We prove that every vector bundle possesses a connection, and we give a formula describing how a connection matrix changes if we switch from one chart to another. In Section 11.4 we define the notion of covariant derivative along a curve and parallel transport. Section 11.5 is devoted to the very important concept of curvature form R∇of a connec-tion ∇on a vector bundle ξ. We show that the curvature form is a vector-valued two-form with values in Γ(Hom(ξ, ξ)). We also establish the relationhip between R∇and the more familiar definition of the Riemannian curvature in terms of vector fields. In Section 11.6 we show how the curvature form can be expressed in a chart in terms of a matrix of two-forms called a curvature matrix. The connection matrix and the curvature 470 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES matrix are related by the structure equation. We also give a formula describing how a curvature matrix changes if we switch from one chart to another. Bianchi’s identity gives an expression for the exterior derivative of the curvature matrix in terms of the curvature matrix itself and the connection matrix. Section 11.8 deals with connections compatible with a metric, and the Levi-Civita con-nection, which arises in the Riemannian geometry of manifolds. One way of characterizing the Levi-Civita connection involves defining the notion of connection on the dual bundle. This is achieved in Section 11.9. Levi-Civita connections on the tangent bundle of a manifold are investigated in Section 11.10. The purpose of Section 11.11 is to introduce the reader to Pontrjagin Classes and Chern Classes, which are fundamental invariants of real (resp. complex) vector bundles. Here we are dealing with one of the most sophisticated and beautiful parts of differential geometry. Pontrjagin, Stiefel, and Chern (starting from the late 1930’s) discovered that invariants with “good” properties could be defined if we took these invariants to belong to various cohomology groups associated with B. Such invariants are usually called characteristic classes. Roughly, there are two main methods for defining characteristic classes: one using topology, and the other due to Chern and Weil, using differential forms. A masterly exposition of these methods is given in the classic book by Milnor and Stasheff . Amazingly, the method of Chern and Weil using differential forms is quite accessible for someone who has reasonably good knowledge of differential forms and de Rham cohomology, as long as one is willing to gloss over various technical details. We give an introduction to characteristic classes using the method of Chern and Weil. If ξ is a real orientable vector bundle of rank 2m, and if ∇is a metric connection on ξ, then it is possible to define a closed global form eu(R∇), and its cohomology class e(ξ) is called the Euler class of ξ. This is shown in Section 11.13. The Euler class e(ξ) turns out to be a square root of the top Pontrjagin class pm(ξ) of ξ. A complex rank m vector bundle can be viewed as a real vector bundle of rank 2m, which is always orientable. The Euler class e(ξ) of this real vector bundle is equal to the top Chern class cm(ξ) of the complex vector bundle ξ. The global form eu(R∇) is defined in terms of a certain polynomial Pf(A) associated with a real skew-symmetric matrix A, which is a kind of square root of the determinant det(A). The polynomial Pf(A), called the Pfaffian, is defined in Section 11.12. The culmination of this chapter is a statement of the generalization due to Chern of a classical theorem of Gauss and Bonnet. This theorem known as the generalized Gauss– Bonnet formula expresses the Euler characteristic χ(M) of an orientable, compact smooth manifold M of dimension 2m as χ(M) = Z M eu(R∇), 11.2. CONNECTIONS IN VECTOR BUNDLES AND RIEMANNIAN MANIFOLDS 471 where eu(R∇) is the Euler form associated with the curvature form R∇of a metric connection ∇on M. 11.2 Connections and Connection Forms in Vector Bundles The goal of this section is to generalize the notions of a connection to vector bundles. Among other things, this material has applications to theoretical physics. This chapter makes heavy use of differential forms (and tensor products), so the reader may want to brush up on these notions before reading it. Given a manifold M, as X(M) = Γ(M, TM) = Γ(TM), the set of smooth sections of the tangent bundle TM, it is natural that for a vector bundle ξ = (E, π, B, V ), a connection on ξ should be some kind of bilinear map, X(B) × Γ(ξ) − →Γ(ξ), that tells us how to take the covariant derivative of sections. Technically, it turns out that it is cleaner to define a connection on a vector bundle ξ, as an R-linear map ∇: Γ(ξ) →A1(B) ⊗C∞(B) Γ(ξ) (∗) that satisfies the “Leibniz rule” ∇(fs) = d f ⊗s + f∇s, with s ∈Γ(ξ) and f ∈C∞(B), where Γ(ξ) and A1(B) are treated as C∞(B)-modules. Since A1(B) = Γ(B, T ∗B) = Γ(T ∗B) is the space of 1-forms on B, and by Proposition 10.11, A1(B) ⊗C∞(B) Γ(ξ) = Γ(T ∗B) ⊗C∞(B) Γ(ξ) ∼ = Γ(T ∗B ⊗ξ) ∼ = Γ(Hom(TB, ξ)) ∼ = HomC∞(B)(Γ(TB), Γ(ξ)) = HomC∞(B)(X(B), Γ(ξ)), the range of ∇can be viewed as a space of Γ(ξ)-valued differential forms on B. Milnor and Stasheff (Appendix C) use the version where ∇: Γ(ξ) →Γ(T ∗B ⊗ξ), and Madsen and Tornehave (Chapter 17) use the equivalent version stated in (∗). A thorough presentation of connections on vector bundles and the various ways to define them 472 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES can be found in Postnikov which also constitutes one of the most extensive references on differential geometry. If we use the isomorphism A1(B) ⊗C∞(B) Γ(ξ) ∼ = HomC∞(B)(X(B), Γ(ξ)), then a connection is an R-linear map ∇: Γ(ξ) − →HomC∞(B)(X(B), Γ(ξ)) satisfying a Leibniz-type rule, or equivalently, an R-bilinear map ∇: X(B) × Γ(ξ) − →Γ(ξ) such that, for any X ∈X(B) and s ∈Γ(ξ), if we write ∇Xs instead of ∇(X, s), then the following properties hold for all f ∈C∞(B): ∇fXs = f∇Xs ∇X(fs) = X[f]s + f∇Xs. This second version may be considered simpler than the first since it does not involve a tensor product. Since, by Proposition 2.19, A1(B) = Γ(T ∗B) ∼ = HomC∞(B)(X(B), C∞(B)) = (X(B))∗, using Proposition 2.41, the isomorphism α: A1(B) ⊗C∞(B) Γ(ξ) ∼ = HomC∞(B)(X(B), Γ(ξ)) can be described in terms of the evaluation map EvX : A1(B) ⊗C∞(B) Γ(ξ) →Γ(ξ), given by EvX(ω ⊗s) = ω(X)s, X ∈X(B), ω ∈A1(B), s ∈Γ(ξ). Namely, for any θ ∈A1(B) ⊗C∞(B) Γ(ξ), α(θ)(X) = EvX(θ). In particular, we have EvX(d f ⊗s) = d f(X)s = X[f]s. Then it is easy to see that we pass from the first version of ∇, where ∇: Γ(ξ) →A1(B) ⊗C∞(B) Γ(ξ) (∗) 11.2. CONNECTIONS IN VECTOR BUNDLES AND RIEMANNIAN MANIFOLDS 473 with the Leibniz rule ∇(fs) = d f ⊗s + f∇s, to the second version of ∇, denoted ∇′, where ∇′ : X(B) × Γ(ξ) →Γ(ξ) (∗∗) is R-bilinear and where the two conditions ∇′ fXs = f∇′ Xs ∇′ X(fs) = X[f]s + f∇′ Xs hold, via the equation ∇′ X = EvX ◦∇. From now on, we will simply write ∇Xs instead of ∇′ Xs, unless confusion arise. As summary of the above discussion, we make the following definition. Definition 11.1. Let ξ = (E, π, B, V ) be a smooth real vector bundle. A connection on ξ is an R-linear map ∇: Γ(ξ) →A1(B) ⊗C∞(B) Γ(ξ) (∗) such that the Leibniz rule ∇(fs) = d f ⊗s + f∇s holds, for all s ∈Γ(ξ) and all f ∈C∞(B). For every X ∈X(B), we let ∇X = EvX ◦∇ where the evaluation map EvX : A1(B) ⊗C∞(B) Γ(ξ) →Γ(ξ), is given by EvX(ω ⊗s) = ω(X)s, X ∈X(B), ω ∈A1(B), s ∈Γ(ξ), and for every s ∈Γ(ξ), we call ∇Xs the covariant derivative of s relative to X. Then the family (∇X) induces a R-bilinear map also denoted ∇, ∇: X(B) × Γ(ξ) →Γ(ξ), (∗∗) such that the following two conditions hold: ∇fXs = f∇Xs ∇X(fs) = X[f]s + f∇Xs, for all s ∈Γ(ξ), all X ∈X(B) and all f ∈C∞(B). We refer to (∗) as the first version of a connection and to (∗∗) as the second version of a connection. 474 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Every vector bundle admits a connection. We need some technical tools to prove this, so we postpone the proof until Proposition 11.4. Remark: Given two connections, ∇1 and ∇2, we have ∇1(fs) −∇2(fs) = d f ⊗s + f∇1s −d f ⊗s −f∇2s = f(∇1s −∇2s), which shows that ∇1 −∇2 is a C∞(B)-linear map from Γ(ξ) to A1(B)⊗C∞(B) Γ(ξ). However HomC∞(B)(Γ(ξ), Ai(B) ⊗C∞(B) Γ(ξ)) ∼ = (Γ(ξ))∗⊗C∞(B) (Ai(B) ⊗C∞(B) Γ(ξ)) ∼ = Ai(B) ⊗C∞(B) ((Γ(ξ))∗⊗C∞(B) Γ(ξ)) ∼ = Ai(B) ⊗C∞(B) HomC∞(B)(Γ(ξ), Γ(ξ)) ∼ = Ai(B) ⊗C∞(B) Γ(Hom(ξ, ξ)). Therefore, ∇1 −∇2 is a one-form with values in Γ(Hom(ξ, ξ)). But then, the vector space Γ(Hom(ξ, ξ)) acts on the space of connections (by addition) and makes the space of connec-tions into an affine space. Given any connection, ∇and any one-form ω ∈Γ(Hom(ξ, ξ)), the expression ∇+ ω is also a connection. Equivalently, any affine combination of connections is also a connection. If ξ = TM, the tangent bundle of some smooth manifold M, then a connection on TM, also called a connection on M, is a linear map ∇: X(M) − →A1(M) ⊗C∞(M) X(M) ∼ = HomC∞(M)(X(M), X(M)), since Γ(TM) = X(M). Then for fixed Y ∈X(M), the map ∇Y is C∞(M)-linear, which implies that ∇Y is a (1, 1) tensor. In a local chart, (U, ϕ), we have ∇ ∂ ∂xi  ∂ ∂xj  = n X k=1 Γk ij ∂ ∂xk , where the Γk ij are Christoffel symbols. A basic property of ∇is that it is a local operator. Proposition 11.1. Let ξ = (E, π, B, V ) be a smooth real vector bundle and let ∇be a connection on ξ. For every open subset U ⊆B, for every section s ∈Γ(ξ), if s ≡0 on U, then ∇s ≡0 on U; that is, ∇is a local operator. Proof. Using a bump function applied to the constant function with value 1, for every p ∈U, there is some open subset, V ⊆U, containing p and a smooth function, f : B →R, such that supp f ⊆U and f ≡1 on V . Consequently, fs is a smooth section which is identically zero. By applying the Leibniz rule, we get 0 = ∇(fs) = d f ⊗s + f∇s, which, evaluated at p yields (∇s)(p) = 0, since f(p) = 1 and d f ≡0 on V . 11.2. CONNECTIONS IN VECTOR BUNDLES AND RIEMANNIAN MANIFOLDS 475 As an immediate consequence of Proposition 11.1, if s1 and s2 are two sections in Γ(ξ) that agree on U, then s1 −s2 is zero on U, so ∇(s1 −s2) = ∇s1 −∇s2 is zero on U, that is, ∇s1 and ∇s2 agree on U. Proposition 11.1 implies the following fact. Proposition 11.2. A connection ∇on ξ restricts to a connection ∇↾U on the vector bundle ξ ↾U, for every open subset U ⊆B. Proof. Indeed, let s be a section of ξ over U. Pick any b ∈U and define (∇s)(b) as follows: Using a bump function, there is some open subset, V1 ⊆U, containing b and a smooth function, f1 : B →R, such that supp f1 ⊆U and f1 ≡1 on V1 so, let s1 = f1s, a global section of ξ. Clearly, s1 = s on V1, and set (∇s)(b) = (∇s1)(b). This definition does not depend on (V1, f1), because if we had used another pair, (V2, f2), as above, since b ∈V1 ∩V2, we have s1 = f1s = s = f2s = s2 on V1 ∩V2 so, by Proposition 11.1, (∇s1)(b) = (∇s2)(b). It should also be noted that (∇Xs)(b) only depends on X(b). Proposition 11.3. For any two vector fields X, Y ∈X(B), if X(b) = Y (b) for some b ∈B, then (∇Xs)(b) = (∇Y s)(b), for every s ∈Γ(ξ). Proof. As above, by linearity, it it enough to prove that if X(b) = 0, then (∇Xs)(b) = 0 (this argument is due to O’Neill , Chapter 2, Lemma 3). To prove this, pick any local chart, (U, ϕ), with b ∈U. We can write X ↾U = d X i=1 Xi ∂ ∂xi . However, as before, we can find a pair, (V, f), with b ∈V ⊆U, supp f ⊆U and f = 1 on V , so that f ∂ ∂xi is a smooth vector field on B and f ∂ ∂xi agrees with ∂ ∂xi on V , for i = 1, . . . , n. Clearly, fXi ∈C∞(B) and fXi agrees with Xi on V so if we write e X = f 2X, then e X = f 2X = d X i=1 fXi f ∂ ∂xi and we have f 2∇Xs = ∇e Xs = d X i=1 fXi ∇f ∂ ∂xi s. Since Xi(b) = 0 and f(b) = 1, we get (∇Xs)(b) = 0, as claimed. 476 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Using the above property, for any point, p ∈B, we can define the covariant derivative (∇us)(p) of a section s ∈Γ(ξ), with respect to a tangent vector u ∈TpB. Definition 11.2. Pick any vector field X ∈X(B) such that X(p) = u (such a vector field exists locally over the domain of a chart, then extend it using a bump function), and define ∇us by (∇us)(p) = (∇Xs)(p). Proof. By the above property, if X(p) = Y (p), then (∇Xs)(p) = (∇Y s)(p) so (∇us)(p) is well-defined. Since ∇is a local operator, (∇us)(p) is also well defined for any tangent vector u ∈TpB, and any local section s ∈Γ(U, ξ) defined in some open subset U, with p ∈U. From now on, we will use this property without any further justification. Since ξ is locally trivial, it is interesting to see what ∇↾U looks like when (U, ϕ) is a local trivialization of ξ. This can be done in terms of connection matrices. 11.3 Connection Matrices Fix once and for all some basis (v1, . . . , vn) of the typical fibre V (n = dim(V )). To every local trivialization ϕ: π−1(U) →U × V of ξ (for some open subset, U ⊆B), we associate the frame (s1, . . . , sn) over U, given by si(b) = ϕ−1(b, vi), b ∈U. (∗) Then every section s over U can be written uniquely as s = Pn i=1 fisi, for some functions fi ∈C∞(U), and we have ∇s = n X i=1 ∇(fisi) = n X i=1 (d fi ⊗si + fi∇si). On the other hand, each ∇si can be written as ∇si = n X j=1 ωji ⊗sj, for some n × n matrix ω = (ωij) of one-forms ωij ∈A1(U), which we represent in matrix form as ∇s1 · · · ∇sn  = s1 · · · sn     ω11 · · · ω1n . . . ... . . . ωn1 · · · ωnn   . Thus we get ∇s = n X i=1 d fi ⊗si + n X i=1 fi∇si = n X i=1 d fi ⊗si + n X i,j=1 fiωji ⊗sj = n X j=1 (d fj + n X i=1 fiωji) ⊗sj, which we efficiently record as follows: 11.3. CONNECTION MATRICES 477 Definition 11.3. With respect to the frame (s1, . . . , sn) over the open subset U the connec-tion ∇has the matrix form ∇    f1 . . . fn   =    d f1 . . . d fn   + ω    f1 . . . fn   , where the matrix ω = (ωij) of one-forms ωij ∈A1(U) is called the connection form or connection matrix of ∇with respect to ϕ: π−1(U) →U × V . The above computation also shows that on U, any connection is uniquely determined by a matrix of one-forms, ωij ∈A1(U). Example 11.1. Let B = R3 and ξ be the tangent bundle of R3, i.e. ξ = (TR3, π, R3, R3). Following O’Neil, we describe the spherical frame (s1, s2, s3) for the tangent space at each point of R3. Recall that each point of R3 may be parametrized via spherical coordinates as follows: x = r cos ϕ sin θ y = r sin ϕ sin θ z = r cos θ, where r ≥0, 0 ≤θ ≤π, and 0 ≤ϕ < 2π. For each p ∈R3, we define the orthogonal spherical frame for TR3 p as s1 = ∂ ∂r = (cos ϕ sin θ, sin ϕ sin θ, cos θ) s2 = ∂ ∂θ ∂ ∂θ = (cos ϕ cos θ, sin ϕ cos θ, −sin θ) s3 = ∂ ∂ϕ ∂ ∂ϕ = (−sin θ, cos θ, 0). See Figure 11.1. By utilizing an attitude matrix (see O’Neil , Chapter 2, Section 2.7), the connection form for (s1, s2, s3) is given by (∇s1, ∇s2, ∇s3) = (s1, s2, s3)   0 −dθ −sin θdϕ dθ 0 −cos θdϕ sin θdϕ cos θdϕ 0  . Definition 11.4. The connection on U for which ∇s1 = 0, . . . , ∇sn = 0, corresponding to the zero matrix is called the flat connection on U (w.r.t. (s1, . . . , sn)). 478 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES s1 s2 s3 r Θ φ Figure 11.1: The spherical frame (s1, s2, s3) associated with the spherical coordinates (r, θ, ϕ). Note s1 is normal to the sphere, s2 is normal to the teal cone with θ = θ1, while s3 is normal the peach plane ϕ = ϕ1.  We are following the convention in Morita in expressing ∇si as ∇si = Pn j=1 ωji ⊗sj, except that Morita denotes the matrix ω as (ωi j) where i is the row index and j is the column index, that is, ∇si = n X j=1 ωj i ⊗sj. Other authors such as Milnor and Stasheff and Madsen and Tornehave define ∇si as ∇si = Pn j=1 e ωij ⊗sj, in matrix form    ∇s1 . . . ∇sn   =    e ω11 · · · e ω1n . . . ... . . . e ωn1 · · · e ωnn       s1 . . . sn   , so that their matrix e ω is the transpose of our matrix ω. As a consequence, some of the results differ either by a sign (as in ω ∧ω) or by a permutation of matrices (as in the formula for a change of frame). As we will see shortly, the advantage of Morita’s convention is that it is consistent with the representation of a linear map by a matrix. This will show up in Proposition 11.5. Remark: If (σ1, . . . , σn) is a local frame of TB over U, and if (θ1, . . . , θn) is the dual frame 11.3. CONNECTION MATRICES 479 of (σ1, . . . , σn), that is, θi ∈A1(U) is the one-form defined so that θi(b)(σj(b)) = δij, for all b ∈U, 1 ≤i, j ≤n, then we can write ωik = Pn j=1 Γk jiθj and so, ∇si = n X j,k=1 Γk ji(θj ⊗sk), where the Γk ji ∈C∞(U) are the Christoffel symbols. Proposition 11.4. Every vector bundle ξ possesses a connection. Proof. Since ξ is locally trivial, we can find a locally finite open cover (Uα)α of B such that π−1(Uα) is trivial. If (fα) is a partition of unity subordinate to the cover (Uα)α and if ∇α is any flat connection on ξ ↾Uα, then it is immediately verified that ∇= X α fα∇α is a connection on ξ. If ϕα : π−1(Uα) →Uα ×V and ϕβ : π−1(Uβ) →Uβ ×V are two overlapping trivializations, we know that for every b ∈Uα ∩Uβ, we have ϕα ◦ϕ−1 β (b, u) = (b, gαβ(b)u), where gαβ : Uα ∩Uβ →GL(V ) is the transition function. As ϕ−1 β (b, u) = ϕ−1 α (b, gαβ(b)u), if (s1, . . . , sn) is the frame over Uα associated with ϕα and (t1, . . . , tn) is the frame over Uβ associated with ϕβ, since si(b) = ϕ−1 α (b, vi) and ti(b) = ϕ−1 β (b, vi) = ϕ−1 α (b, gαβ(b)vi), if (gij) is the matrix of the linear map gαβ with respect to the basis (v1, . . . , vn), that is gαβ(b)vj = n X i=1 gijvi, (∗∗) which in matrix form is gαβ(b)v1 · · · gαβ(b)vn  = v1 · · · vn     g11 · · · g1n . . . ... . . . gn1 · · · gnn   , 480 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES we obtain ti(b) = ϕ−1 α (b, gαβ(b)vi) = ϕ−1 α  b, n X j=1 gjivj  = n X j=1 gjiϕ−1 α (b, vj) = n X j=1 gjisj(b), that is ti = n X j=1 gjisj on Uα ∩Uβ. Proposition 11.5. With the notations as above, the connection matrices, ωα and ωβ respec-tively over Uα and Uβ obey the transformation rule ωβ = g−1 αβωαgαβ + g−1 αβ(dgαβ), where gαβ is viewed as the matrix function (gij) given by gαβ(b)vj = Pn i=1 gijvi for j = 1, . . . , n and for every b ∈Uα ∩Uβ. Proof. To prove the above proposition, apply ∇to both sides of the equations ti = n X j=1 gjisj on Uα ∩Uβ we obtain ∇ti = n X j=1 dgji ⊗sj + n X j=1 gji∇sj. Since ∇ti = Pn k=1(ωβ)ki ⊗tk, ∇sj = Pn k=1(ωα)kj ⊗sk, and tk = Pn j=1 gjksj, we get ∇ti = n X j,k=1 (ωβ)kigjk ⊗sj = n X j=1 dgji ⊗sj + n X j,k=1 gji(ωα)kj ⊗sk, and since (s1, . . . , sn) is a frame, the coefficients of sj on both sides must be equal, which yields n X k=1 gjk(ωβ)ki = dgji + n X k=1 (ωα)kjgji for all i, j, which in matrix form means that gαβωβ = dgαβ + ωαgαβ. Since gαβ is invertible, we get ωβ = g−1 αβωαgαβ + g−1 αβ(dgαβ), as claimed. 11.4. PARALLEL TRANSPORT 481 Remark: Everything we did in this section applies to complex vector bundles by considering complex vector spaces instead of real vector spaces, C-linear maps instead of R-linear map, and the space of smooth complex-valued functions, C∞(B; C) ∼ = C∞(B) ⊗R C. We also use spaces of complex-valued differentials forms Ai(B; C) = Ai(B) ⊗R C, and we define Ai(ξ) as Ai(ξ) = Ai(B; C) ⊗C∞(B;C) Γ(ξ). A connection is a C-linear map ∇: Γ(ξ) →A1(ξ), that satisfies the same Leibniz-type rule as before. Obviously, every differential form in Ai(B; C) can be written uniquely as ω + iη, with ω, η ∈Ai(B). The exterior differential, d: Ai(B; C) →Ai+1(B; C) is defined by d(ω + iη) = dω + idη. We obtain complex-valued de Rham cohomology groups, Hi DR(M; C) = Hi DR(M) ⊗R C. The complexification of a real vector bundle ξ is the complex vector bundle ξC = ξ ⊗R ϵ1 C, where ϵ1 C is the trivial complex line bundle B × C. 11.4 Parallel Transport The notion of connection yields the notion of parallel transport in a vector bundle. First we need to define the covariant derivative of a section along a curve. Definition 11.5. Let ξ = (E, π, B, V ) be a vector bundle and let γ : [a, b] →B be a smooth curve in B. A smooth section along the curve γ is a smooth map X : [a, b] →E, such that π(X(t)) = γ(t), for all t ∈[a, b]. When ξ = TB, the tangent bundle of the manifold B, we use the terminology smooth vector field along γ. Recall that the curve γ : [a, b] →B is smooth iffγ is the restriction to [a, b] of a smooth curve on some open interval containing [a, b]. Since a section X along a curve γ does not necessarily extend to an open subset of B (for example, if the image of γ is dense in B), the covariant derivative (∇γ′(t0) X)γ(t0) may not be defined, so we need a proposition showing that the covariant derivative of a section along a curve makes sense. Proposition 11.6. Let ξ be a vector bundle, ∇be a connection on ξ, and γ : [a, b] →B be a smooth curve in B. There is a R-linear map D/dt, defined on the vector space of smooth sections X along γ, which satisfies the following conditions: (1) For any smooth function f : [a, b] →R, D(fX) dt = d f dt X + f DX dt 482 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES (2) If X is induced by a global section s ∈Γ(ξ), that is, if X(t0) = s(γ(t0)) for all t0 ∈[a, b], then DX dt (t0) = (∇γ′(t0) s)γ(t0). Proof. Since γ([a, b]) is compact, it can be covered by a finite number of open subsets Uα such that (Uα, ϕα) is a chart for B and (Uα, ˜ ϕα) is a local trivialization. Thus, we may assume that γ : [a, b] →U for some chart, (U, ϕ), and some local trivialization (U, ˜ ϕ). As ϕ ◦γ : [a, b] →Rn, we can write ϕ ◦γ(t) = (u1(t), . . . , un(t)), where each ui = pri ◦ϕ ◦γ is smooth. Now, for every g ∈C∞(B), as dγt0 d dt t0 ! (g) = d dt(g ◦γ) t0 = d dt((g ◦ϕ−1) ◦(ϕ ◦γ)) t0 = n X i=1 dui dt  ∂ ∂xi  γ(t0) g, since by definition of γ′(t0), γ′(t0) = dγt0 d dt t0 ! , γ′(t0) = n X i=1 dui dt  ∂ ∂xi  γ(t0) . If (s1, . . . , sn) is a frame over U determined by (U, ˜ ϕ), we can write X(t) = n X i=1 Xi(t)si(γ(t)), for some smooth functions, Xi. Then Conditions (1) and (2) imply that DX dt = n X j=1 dXj dt sj(γ(t)) + Xj(t)∇γ′(t)(sj(γ(t)))  and since γ′(t) = n X i=1 dui dt  ∂ ∂xi  γ(t) , there exist some smooth functions, Γk ij, so that ∇γ′(t)(sj(γ(t))) = n X i=1 dui dt ∇ ∂ ∂xi (sj(γ(t))) = X i,k dui dt Γk ijsk(γ(t)). It follows that DX dt = n X k=1 dXk dt + X ij Γk ij dui dt Xj ! sk(γ(t)). 11.4. PARALLEL TRANSPORT 483 Conversely, the above expression defines a linear operator, D/dt, and it is easy to check that it satisfies Conditions (1) and (2). Definition 11.6. The operator D/dt is called the covariant derivative along γ and it is also denoted by ∇γ′(t) or simply ∇γ′. Definition 11.7. Let ξ be a vector bundle and let ∇be a connection on ξ. For every curve γ : [a, b] →B in B, a section X along γ is parallel (along γ) iff DX dt (t0) = 0 for all t0 ∈[a, b]. If ξ was the tangent bundle of a smooth manifold M embedded in Rd (for some d), then to say that X is parallel along γ would mean that the directional derivative, (Dγ′X)(γ(t)), is normal to Tγ(t)M. The following proposition can be shown using the existence and uniqueness of solutions of ODE’s (in our case, linear ODE’s). Proposition 11.7. Let ξ be a vector bundle and let ∇be a connection on ξ. For every C1 curve γ : [a, b] →B in B, for every t ∈[a, b] and every v ∈π−1(γ(t)), there is a unique parallel section X along γ such that X(t) = v. Proof. For the proof of Proposition 11.7 it is sufficient to consider the portions of the curve γ contained in some local trivialization. In such a trivialization, (U, ϕ), as in the proof of Proposition 11.6, using a local frame, (s1, . . . , sn), over U, we have DX dt = n X k=1 dXk dt + X ij Γk ij dui dt Xj ! sk(γ(t)), with ui = pri ◦ϕ ◦γ. Consequently, X is parallel along our portion of γ iffthe system of linear ODE’s in the unknowns, Xk, dXk dt + X ij Γk ij dui dt Xj = 0, k = 1, . . . , n, is satisfied. Remark: Proposition 11.7 can be extended to piecewise C1 curves. Definition 11.8. Let ξ be a vector bundle and let ∇be a connection on ξ. For every curve γ : [a, b] →B in B, for every t ∈[a, b], the parallel transport from γ(a) to γ(t) along γ is the linear map from the fibre π−1(γ(a)) to the fibre π−1(γ(t)), which associates to any v ∈π−1(γ(a)) the vector Xv(t) ∈π−1(γ(t)), where Xv is the unique parallel section along γ with Xv(a) = v. 484 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES The following proposition is an immediate consequence of properties of linear ODE’s: Proposition 11.8. Let ξ = (E, π, B, V ) be a vector bundle and let ∇be a connection on ξ. For every C1 curve γ : [a, b] →B in B, the parallel transport along γ defines for every t ∈[a, b] a linear isomorphism Pγ : π−1(γ(a)) →π−1(γ(t)) between the fibres π−1(γ(a)) and π−1(γ(t)). In particular, if γ is a closed curve, that is, if γ(a) = γ(b) = p, we obtain a linear isomorphism Pγ of the fibre Ep = π−1(p), called the holonomy of γ. The holonomy group of ∇based at p, denoted Holp(∇), is the subgroup of GL(V, R) (where V is the fibre of the vector bundle ξ) given by Holp(∇) = {Pγ ∈GL(V, R) | γ is a closed curve based at p}. If B is connected, then Holp(∇) depends on the basepoint p ∈B up to conjugation and so Holp(∇) and Holq(∇) are isomorphic for all p, q ∈B. In this case, it makes sense to talk about the holonomy group of ∇. If ξ = TB, the tangent bundle of a manifold, B, by abuse of language, we call Holp(∇) the holonomy group of B. 11.5 Curvature, Curvature Form and Curvature Matrix If ξ = B × V is the trivial bundle and ∇is a flat connection on ξ, we obviously have ∇X∇Y −∇Y ∇X = ∇[X,Y ], where [X, Y ] is the Lie bracket of the vector fields X and Y . However, for general bundles and arbitrary connections, the above fails. The error term, R(X, Y ) = ∇X∇Y −∇Y ∇X −∇[X,Y ] measures what’s called the curvature of the connection. In order to write R(X, Y ) as a vector valued two-form, we need the following definition. Definition 11.9. Set A1(ξ) = A1(B; ξ) = A1(B) ⊗C∞(B) Γ(ξ), and more generally, for any i ≥0, set Ai(ξ) = Ai(B; ξ) = Ai(B) ⊗C∞(B) Γ(ξ) ∼ = Γ  i ^ T ∗B  ⊗ξ  . Obviously, A0(ξ) = Γ(ξ) (and recall that A0(B) = C∞(B)). 11.5. CURVATURE AND CURVATURE FORM 485 The space of differential forms Ai(B; ξ) with values in Γ(ξ) is a generalization of the space Ai(M, F) of differential forms with values in F encountered in Section 4.5. Observe that in terms of the Ai(ξ)’s, a connection is a linear map, ∇: A0(ξ) →A1(ξ), satisfying the Leibniz rule. When ξ = TB, a connection (second version) is what is known as an affine connection on the manifold B. The curvature of a connection turns up as the failure of a certain sequence involving the spaces Ai(ξ) = Ai(B) ⊗C∞(B) Γ(ξ) to be a cochain complex. Since the connection on ξ is a linear map ∇: A0(ξ) →A1(ξ) satisfying a Leibniz-type rule, it is natural to ask whether ∇can be extended to a family of operators, d∇: Ai(ξ) →Ai+1(ξ), with properties analogous to d on A∗(B). This is indeed the case, and we get a sequence of maps 0 − →A0(ξ) ∇ − →A1(ξ) d∇ − →A2(ξ) − →· · · − →Ai(ξ) d∇ − →Ai+1(ξ) − →· · · , but in general, d∇◦d∇= 0 fails. In particular, d∇◦∇= 0 generally fails. Definition 11.10. The term R∇= d∇◦∇is the curvature form (or curvature tensor) of the connection ∇. As we will see, it yields our previous curvature R, back. Our next goal is to define d∇. We have the notion of wedge defined for A∗(B). But in order to define d∇, we require a notion of wedge that makes sense on A∗(ξ). Definition 11.11. Let ξ and η be two smooth real vector bundles. We define a C∞(B)-bilinear map ⊼: Ai(ξ) × Aj(η) − →Ai+j(ξ ⊗η) as follows: (ω ⊗s) ⊼(τ ⊗t) = (ω ∧τ) ⊗(s ⊗t), where ω ∈Ai(B), τ ∈Aj(B), s ∈Γ(ξ), and t ∈Γ(η), ω ∧τ is the wedge defined over A∗(B), and where we used the fact that Γ(ξ ⊗η) = Γ(ξ) ⊗C∞(B) Γ(η). In order to help with the calculations associated with the propositions of this section, we need to consider the special case of ⊼where ξ = ϵ1 = B × R, the trivial line bundle over B. In this case, Ai(ξ) = Ai(B) and we have a bilinear map ⊼: Ai(B) × Aj(η) − →Ai+j(η) 486 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES given by ω ⊼(τ ⊗t) = (ω ∧τ) ⊗t, τ ∈Aj(B), t ∈Γ(η). (1) For j = 0, we have the bilinear map ⊼: Ai(B) × Γ(η) − →Ai(η) given by ω ⊼t = ω ⊗t. (2) It can be shown that the bilinear map ⊼: Ar(B) × As(η) − →Ar+s(η) has the following properties: (ω ∧τ) ⊼θ = ω ⊼(τ ⊼θ) (3) 1 ⊼θ = θ, for all ω ∈Ai(B), τ ∈Aj(B) with i + j = r, θ ∈As(ξ), and where 1 denotes the constant function in C∞(B) with value 1. Proposition 11.9. For every vector bundle ξ, for all j ≥0, there is a unique R-linear map (resp. C-linear if ξ is a complex VB) d∇: Aj(ξ) →Aj+1(ξ), such that (i) d∇= ∇for j = 0. (ii) d∇(ω ⊼t) = dω ⊼t + (−1)iω ⊼d∇t, for all ω ∈Ai(B) and all t ∈Aj(ξ). Proof. Recall that Aj(ξ) = Aj(B) ⊗C∞(B) Γ(ξ), and define ˆ d∇: Aj(B) × Γ(ξ) →Aj+1(ξ) by ˆ d∇(ω, s) = dω ⊗s + (−1)jω ⊼∇s, for all ω ∈Aj(B) and all s ∈Γ(ξ). We claim that ˆ d∇induces an R-linear map on Aj(ξ), but there is a complication as ˆ d∇is not C∞(B)-bilinear. The way around this problem is to use Proposition 2.42. For this we need to check that ˆ d∇satisfies the condition of Proposition 2.42, where the right action of C∞(B) on Aj(B) is equal to the left action, namely wedging: f ∧ω = ω ∧f f ∈C∞(B) = A0(B), ω ∈Aj(B). As ⊼and ∧are C∞(B)-bilinear, for all ω ∈Ai(B) and all s ∈Γ(ξ), we have ˆ d∇(ωf, s) = d(ωf) ⊗s + (−1)j(ωf) ⊼∇s = d(ωf) ⊼s + (−1)jfω ⊼∇s, by (2) = ((dω)f + (−1)jω ∧d f) ⊼s + (−1)jfω ⊼∇s, by Proposition 4.12 = fdω ⊼s + ((−1)jω ∧d f) ⊼s + (−1)jfω ⊼∇s 11.5. CURVATURE AND CURVATURE FORM 487 and ˆ d∇(ω, fs) = dω ⊗(fs) + (−1)jω ⊼∇(fs) = dω ⊼(fs) + (−1)jω ⊼∇(fs), by (2) = fdω ⊼s + (−1)jω ⊼(d f ⊗s + f∇s), by Definition 11.1 = fdω ⊼s + (−1)jω ⊼(d f ⊼s + f∇s), by (2) = fdω ⊼s + ((−1)jω ∧d f) ⊼s + (−1)jfω ⊼∇s, by (3). Thus, ˆ d∇(ωf, s) = ˆ d∇(ω, fs), and Proposition 2.42 shows that d∇: Aj(ξ) →Aj+1(ξ) given by d∇(ω ⊗s) = ˆ d∇(ω, s) is a well-defined R-linear map for all j ≥0. Furthermore, it is clear that d∇= ∇for j = 0. Now, for ω ∈Ai(B) and t = τ ⊗s ∈Aj(ξ) we have d∇(ω ⊼(τ ⊗s)) = d∇((ω ∧τ) ⊗s)), by (1) = d(ω ∧τ) ⊗s + (−1)i+j(ω ∧τ) ⊼∇s, definition of d∇ = (dω ∧τ) ⊗s + (−1)i(ω ∧dτ) ⊗s + (−1)i+j(ω ∧τ) ⊼∇s, by Proposition 4.12 = dω ⊼(τ ⊗s) + (−1)iω ⊼(dτ ⊗s) + (−1)i+jω ⊼(τ ⊼∇s), by (1) and (3) = dω ⊼(τ ⊗s) + (−1)iω ⊼d∇(τ ⊗s), definition of d∇ which proves (ii). As a consequence, we have the following sequence of linear maps 0 − →A0(ξ) ∇ − →A1(ξ) d∇ − →A2(ξ) − →· · · − →Ai(ξ) d∇ − →Ai+1(ξ) − →· · · . but in general, d∇◦d∇= 0 fails. Although generally d∇◦∇= 0 fails, the map d∇◦∇is C∞(B)-linear. Proposition 11.10. The map d∇◦∇: A0(ξ) →A2(ξ) is C∞(B)-linear. Proof. We have (d∇◦∇)(fs) = d∇(d f ⊗s + f∇s), by Definition 11.1 = d∇(d f ⊼s + f ⊼∇s), by (2) = dd f ⊼s −d f ⊼∇s + d f ⊼∇s + f ⊼d∇(∇s), by Proposition 11.9 = f ⊼d∇(∇s), since dd f = 0 = f((d∇◦∇)(s)). Therefore, d∇◦∇: A0(ξ) →A2(ξ) is a C∞(B)-linear map. 488 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Recall that just before Proposition 11.1 we showed that HomC∞(B)(A0(ξ), Ai(ξ)) ∼ = Ai(Hom(ξ, ξ)), therefore, d∇◦∇∈A2(Hom(ξ, ξ)) ∼ = A2(B) ⊗C∞(B) Γ(Hom(ξ, ξ)). Corollary 11.11. The map d∇◦∇is a two-form with values in Γ(Hom(ξ, ξ)). Recall from Definition 11.10 that R∇= d∇◦∇. Although this is far from obvious the curvature form R∇is related to the curvature R(X, Y ) defined at the beginning of Section 11.5. To discover the relationship between R∇and R(−, −), we need to explain how to define R∇ X,Y (s), for any two vector fields X, Y ∈X(B) and any section s ∈Γ(ξ). For any section s ∈Γ(ξ), the value ∇s can be written as a linear combination of elements of the form ω⊗t, with ω ∈A1(B) and t ∈Γ(ξ). If ∇s = ω⊗t = ω⊼t, as above, we have d∇(∇s) = d∇(ω ⊼t) = dω ⊗t −ω ⊼∇t, by Proposition 11.9. But ∇t itself is a linear combination of the form ∇t = X j ηj ⊗tj for some 1-forms ηj ∈A1(B) and some sections tj ∈Γ(ξ), so (1) implies that d∇(∇s) = dω ⊗t − X j (ω ∧ηj) ⊗tj. Thus it makes sense to define R∇ X,Y (s) by R∇ X,Y (s) = dω(X, Y )t − X j (ω ∧ηj)(X, Y )tj = dω(X, Y )t − X j (ω(X)ηj(Y ) −ω(Y )ηj(X))tj = dω(X, Y )t −  ω(X) X j ηj(Y )tj −ω(Y ) X j ηj(X)tj  = dω(X, Y )t −(ω(X)∇Y t −ω(Y )∇Xt), (4) since ∇Xt = P j ηj(X)tj because ∇t = P j ηj ⊗tj, and similarly for ∇Y t. We extend this formula by linearity when ∇s is a linear combinations of elements of the form ω ⊗t. The preceding discussion implies that clean way to define R∇ X,Y is to define the following evaluation map: 11.5. CURVATURE AND CURVATURE FORM 489 Definition 11.12. Let ξ be a smooth real vector bundle. Define EvX,Y : A2(Hom(ξ, ξ)) →A0(Hom(ξ, ξ)) = Γ(Hom(ξ, ξ)) ∼ = HomC∞(B)(Γ(ξ), Γ(ξ)) as follows: For all X, Y ∈X(B), all θ ⊗h ∈A2(Hom(ξ, ξ)) = A2(B) ⊗C∞(B) Γ(Hom(ξ, ξ)), set EvX,Y (θ ⊗h) = θ(X, Y )h. It is clear that this map is C∞(B)-linear and thus well-defined on A2(Hom(ξ, ξ)). (Recall that A0(Hom(ξ, ξ)) = Γ(Hom(ξ, ξ)) = HomC∞(B)(Γ(ξ), Γ(ξ)).) We write R∇ X,Y = EvX,Y (R∇) ∈HomC∞(B)(Γ(ξ), Γ(ξ)). Since R∇is a linear combination of the form R∇= X j θj ⊗hj for some 2-forms θj ∈A2(B) and some sections hj ∈Γ(Hom(ξ, ξ)), for any section s ∈Γ(ξ), we have R∇ X,Y (s) = X j θj(X, Y )hj(s), where hj(s) is some section in Γ(ξ), and then we use the formula obtained above when ∇s is a linear combination of terms of the form ω ⊗s for some 1-forms A1(B) and some sections s ∈Γ(ξ). Proposition 11.12. For any vector bundle ξ, and any connection ∇on ξ, for all X, Y ∈ X(B), if we let R(X, Y ) = ∇X ◦∇Y −∇Y ◦∇X −∇[X,Y ], then R(X, Y ) = R∇ X,Y . Proof. Since for any section s ∈Γ(ξ), the value ∇s can be written as a linear combination of elements of the form ω ⊗t = ω ⊼t, with ω ∈A1(B) and t ∈Γ(ξ), it is sufficient to compute R∇ X,Y (s) when ∇s = ω ⊗t, and we get R∇ X,Y (s) = dω(X, Y )t −(ω(X)∇Y t −ω(Y )∇Xt), by (4) = (X(ω(Y )) −Y (ω(X)) −ω([X, Y ]))t −(ω(X)∇Y t −ω(Y )∇Xt), by Prop. 4.16 = ∇X(ω(Y )t) −∇Y (ω(X)t) −ω([X, Y ])t, by Definition 11.1 = ∇X(∇Y s) −∇Y (∇Xs) −∇[X,Y ]s, since ∇Xs = ω(X)t because ∇s = ω ⊗t (and similarly for the other terms involving ω). Remark: Proposition 11.12 implies that R(Y, X) = −R(X, Y ) and that R(X, Y )(s) is C∞(B)-linear in X, Y and s. 490 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Definition 11.13. For any vector bundle ξ and any connection ∇on ξ, the vector-valued two-form R∇= d∇◦∇∈A2(Hom(ξ, ξ)) is the curvature form (or curvature tensor) of the connection ∇. We say that ∇is a flat connection iffR∇= 0. Remark: The expression R∇is also denoted F ∇or K∇. 11.6 Structure Equations As in the case of a connection, we can express the two-form R∇locally in any local trivial-ization ϕ: π−1(U) →U ×V of ξ. Since R∇∈A2(Hom(ξ, ξ)) = A2(B)⊗C∞(B) Γ(Hom(ξ, ξ)), if (s1, . . . , sn) is the frame associated with (ϕ, U), then R∇(si) = n X j=1 Ωji ⊗sj, for some matrix Ω= (Ωij) of two forms Ωij ∈A2(U). Definition 11.14. The matrix Ω= (Ωij) of two forms such that R∇(si) = n X j=1 Ωji ⊗sj, is called the curvature matrix (or curvature form) associated with the local trivialization. The relationship between the connection form ω and the curvature form Ωis simple. Proposition 11.13. (Structure Equations) Let ξ be any vector bundle and let ∇be any connection on ξ. For every local trivialization ϕ: π−1(U) →U × V , the connection matrix ω = (ωij) and the curvature matrix Ω= (Ωij) associated with the local trivialization (ϕ, U), are related by the structure equation: Ω= dω + ω ∧ω, where the above formula is interpreted in an entry by entry fashion. Proof. By definition, ∇(si) = n X j=1 ωji ⊗sj, 11.6. STRUCTURE EQUATIONS 491 so if we apply d∇and use Property (ii) of Proposition 11.9 we get R∇(si) = d∇(∇(si)) = n X k=1 Ωki ⊗sk = n X j=1 d∇(ωji ⊗sj) = n X j=1 dωji ⊗sj − n X j=1 ωji ⊼∇sj, by definition of d∇ = n X j=1 dωji ⊗sj − n X j=1 ωji ⊼ n X k=1 ωkj ⊗sk ! = n X k=1 dωki ⊗sk − n X k=1  n X j=1 ωji ∧ωkj  ⊗sk, by (1) and so, Ωki = dωki + n X j=1 ωkj ∧ωji, which, means that Ω= dω + ω ∧ω, as claimed.  Some other texts, including Milnor and Stasheff state the structure equations as Ω= dω −ω ∧ω. Example 11.2. In Example 11.1, we showed that the connection matrix for the spherical frame of TR3 is given by ω =   0 −dθ −sin θdϕ dθ 0 −cos θdϕ sin θdϕ cos θdϕ 0  . 492 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Proposition 11.13 shows that the curvature matrix is Ω= dω −ω ∧ω =   0 0 −cos θdθ ∧dϕ 0 0 sin θdθ ∧dϕ cos θdθ ∧dϕ −sin θdθ ∧dϕ 0   +   0 −dθ −sin θdϕ dθ 0 −cos θdϕ sin θdϕ cos θdϕ 0  ∧   0 −dθ −sin θdϕ dθ 0 −cos θdϕ sin θdϕ cos θdϕ 0   =   0 0 −cos θdθ ∧dϕ 0 0 sin θdθ ∧dϕ cos θdθ ∧dϕ −sin θdθ ∧dϕ 0   +   0 0 cos θdθ ∧dϕ 0 0 sin θdθ ∧dϕ −cos θdθ ∧dϕ −sin θdθ ∧dϕ 0   =   0 0 0 0 0 2 sin θdθ ∧dϕ 0 −2 sin θdθ ∧dϕ 0  . If ϕα : π−1(Uα) →Uα × V and ϕβ : π−1(Uβ) →Uβ × V are two overlapping trivializa-tions, the relationship between the curvature matrices Ωα and Ωβ, is given by the following proposition which is the counterpart of Proposition 11.5 for the curvature matrix: Proposition 11.14. If ϕα : π−1(Uα) →Uα × V and ϕβ : π−1(Uβ) →Uβ × V are two over-lapping trivializations of a vector bundle ξ, then we have the following transformation rule for the curvature matrices Ωα and Ωβ: Ωβ = g−1 αβΩαgαβ, where gαβ is viewed as the matrix function representing the linear map gαβ(b) ∈GL(V ) for every b ∈Uα ∩Uβ. Proof. The idea is to take the exterior derivative of the equation ωβ = g−1 αβωαgαβ + g−1 αβ(dgαβ) from Proposition 11.5. To simplify notation, write g for gαβ. Now, since g, Ωα and Ωβ are all matrices, we apply the exterior derivative in a entry by entry fashion. Since g is a matrix of functions such that g−1g = I, we find that 0 = d(g−1g) = dg−1 g + g−1 dg, which is equivalent to dg−1 = −g−1dgg−1. 11.6. STRUCTURE EQUATIONS 493 By recalling that ddη = 0, d(η ∧β) = dη ∧β + (−1)jη ∧dβ, η ∈Ai(B), β ∈Aj(B), we find that dωβ = d(g−1ωαg) + d(g−1dg) = d(g−1ωαg) + dg−1 ∧dg = dg−1 ∧ωαg + g−1 ∧d(ωαg) + dg−1 ∧dg = −g−1dgg−1 ∧ωαg + g−1 ∧d(ωαg) −g−1dgg−1 ∧dg = −g−1dgg−1 ∧ωαg + g−1 ∧(dωαg −ωα ∧dg) −g−1dgg−1 ∧dg = −g−1dgg−1 ∧ωαg + g−1dωαg −g−1ωα ∧dg −g−1dgg−1 ∧dg, so using the structure equation (Proposition 11.13) we get Ωβ = dωβ + ωβ ∧ωβ = −g−1dgg−1 ∧ωαg + g−1dωαg −g−1ωα ∧dg −g−1dgg−1 ∧dg + (g−1ωαg + g−1dg) ∧(g−1ωαg + g−1dg) = −g−1dgg−1 ∧ωαg + g−1dωαg −g−1ωα ∧dg −g−1dgg−1 ∧dg + g−1ωα ∧ωαg + g−1ωα ∧dg + g−1dg ∧g−1ωαg + g−1dg ∧g−1dg = g−1dωαg + g−1ωα ∧ωαg = g−1Ωαg, establishing the desired formula. Proposition 11.13 also yields a formula for dΩ, know as Bianchi’s identity (in local form). Proposition 11.15. (Bianchi’s Identity) For any vector bundle ξ and any connection ∇on ξ, if ω and Ωare respectively the connection matrix and the curvature matrix, in some local trivialization, then dΩ= Ω∧ω −ω ∧Ω. Proof. If we apply d to the structure equation, Ω= dω + ω ∧ω, we get dΩ= ddω + dω ∧ω −ω ∧dω = (Ω−ω ∧ω) ∧ω −ω ∧(Ω−ω ∧ω) = Ω∧ω −ω ∧ω ∧ω −ω ∧Ω+ ω ∧ω ∧ω = Ω∧ω −ω ∧Ω, as claimed. We conclude this section by giving a formula for d∇◦d∇(t), for any t ∈Ai(ξ). 494 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES 11.7 A Formula for d∇◦d∇ Consider the special case of the bilinear map ⊼: Ai(ξ) × Aj(η) − →Ai+j(ξ ⊗η) given in Definition 11.11 with j = 2 and η = Hom(ξ, ξ). This is the C∞(B)-bilinear map ⊼: Ai(ξ) × A2(Hom(ξ, ξ)) − →Ai+2(ξ ⊗Hom(ξ, ξ)). Two applications of Proposition 10.11 show that Γ(ξ ⊗Hom(ξ, ξ)) ∼ = Γ(ξ) ⊗C∞(B) Γ(Hom(ξ, ξ)) ∼ = Γ(ξ) ⊗C∞(B) HomC∞(B)(Γ(ξ), Γ(ξ)). We then have the evaluation map ev: Aj(ξ ⊗Hom(ξ, ξ)) ∼ = Aj(B) ⊗C∞(B) Γ(ξ) ⊗C∞(B) HomC∞(B)(Γ(ξ), Γ(ξ)) − →Aj(B) ⊗C∞(B) Γ(ξ) = Aj(ξ), given by ev(ω ⊗s ⊗h) = ω ⊗h(s), with ω ∈Aj(B), s ∈Γ(ξ) and h ∈HomC∞(B)(Γ(ξ), Γ(ξ)). Definition 11.15. Let ⩞: Ai(ξ) × A2(Hom(ξ, ξ)) − →Ai+2(ξ) be the composition Ai(ξ) × A2(Hom(ξ, ξ)) ⊼ − →Ai+2(ξ ⊗Hom(ξ, ξ)) ev − →Ai+2(ξ). More explicitly, the above map is given (on generators) by (ω ⊗s) ⩞H = ω ⊼H(s), (5) where ω ∈Ai(B), s ∈Γ(ξ) and H ∈HomC∞(B)(Γ(ξ), A2(ξ)) ∼ = A2(Hom(ξ, ξ)). Proposition 11.16. For any vector bundle ξ and any connection ∇on ξ, the composition d∇◦d∇: Ai(ξ) →Ai+2(ξ) maps t to t ⩞R∇, for any t ∈Ai(ξ). Proof. Any t ∈Ai(ξ) is some linear combination of elements ω ⊗s ∈Ai(B)⊗C∞(B) Γ(ξ) and by Proposition 11.9, we have d∇◦d∇(ω ⊗s) = d∇(dω ⊗s + (−1)iω ⊼∇s), by definition of d∇ = ddω ⊗s + (−1)i+1dω ⊼∇s + (−1)idω ⊼∇s + (−1)i(−1)iω ⊼d∇◦∇s = ω ⊼(d∇◦∇s) = (ω ⊗s) ⩞R∇, by (5) as claimed. 11.8. CONNECTIONS COMPATIBLE WITH A METRIC 495 Proposition 11.16 shows that d∇◦d∇= 0 iffR∇= d∇◦∇= 0, that is, iffthe connection ∇is flat. Thus, the sequence 0 − →A0(ξ) ∇ − →A1(ξ) d∇ − →A2(ξ) − →· · · − →Ai(ξ) d∇ − →Ai+1(ξ) − →· · · , is a cochain complex iff∇is flat. Remark: Again everything we did in this section applies to complex vector bundles. 11.8 Connections Compatible with a Metric; Levi-Civita Connections If a vector bundle (or a Riemannian manifold) ξ has a metric, then it is natural to define when a connection ∇on ξ is compatible with the metric. This will require first defining the following three bilinear pairings. Definition 11.16. Let ξ be a smooth real vector bundle ξ with metric ⟨−, −⟩. We can use this metric to define pairings A1(ξ) × A0(ξ) − →A1(B) and A0(ξ) × A1(ξ) − →A1(B) as follows: Set (on generators) ⟨ω ⊗s1, s2⟩= ⟨s1, ω ⊗s2⟩= ⟨s1, s2⟩ω, for all ω ∈A1(B), s1, s2 ∈Γ(ξ) and where ⟨s1, s2⟩is the function in C∞(B) given by b 7→⟨s1(b), s2(b)⟩, for all b ∈B. More generally, we define a pairing Ai(ξ) × Aj(ξ) − →Ai+j(B), by ⟨ω ⊗s1, η ⊗s2⟩= ⟨s1, s2⟩ω ∧η, for all ω ∈Ai(B), η ∈Aj(B), s1, s2 ∈Γ(ξ). Definition 11.17. Given any metric ⟨−, −⟩on a vector bundle ξ, a connection ∇on ξ is compatible with the metric, for short, a metric connection iff d⟨s1, s2⟩= ⟨∇s1, s2⟩+ ⟨s1, ∇s2⟩, for all s1, s2 ∈Γ(ξ). In terms of version-two of a connection, ∇X is a metric connection iff X(⟨s1, s2⟩) = ⟨∇Xs1, s2⟩+ ⟨s1, ∇Xs2⟩, for every vector field, X ∈X(B). Remark: Definition 11.17 remains unchanged if ξ is a complex vector bundle. It is easy to prove that metric connections exist. 496 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Proposition 11.17. Let ξ be a rank n vector with a metric ⟨−, −⟩. Then ξ possesses metric connections. Proof. We can pick a locally finite cover (Uα)α of B such that (Uα, ϕα) is a local trivialization of ξ. Then for each (Uα, ϕα), we use the Gram-Schmidt procedure to obtain an orthonormal frame (sα 1, . . . , sα n) over Uα, and we let ∇α be the trivial connection on π−1(Uα). By con-struction, ∇α is compatible with the metric. We finish the argument by using a partition of unity, leaving the details to the reader. Remark: If ξ is a complex vector bundle, then we use a Hermitian metric and we call a connection compatible with this metric a Hermitian connection . The existence of Hermitian connections is clear. The condition of compatibility with a metric is nicely expressed in a local trivialization. Indeed, let (U, ϕ) be a local trivialization of the vector bundle ξ (of rank n). Then using the Gram-Schmidt procedure, we obtain an orthonormal frame (s1, . . . , sn), over U. Proposition 11.18. Using the above notations, if ω = (ωij) is the connection matrix of ∇ w.r.t. an orthonormal frame (s1, . . . , sn), then ω is skew-symmetric. Proof. Since ∇si = n X j=1 ωji ⊗sj and since ⟨si, sj⟩= δij (as (s1, . . . , sn) is orthonormal), we have d⟨si, sj⟩= 0 on U. Conse-quently, 0 = d⟨si, sj⟩ = ⟨∇si, sj⟩+ ⟨si, ∇sj⟩ = n X k=1 ωki ⊗sk, sj + si, n X l=1 ωlj ⊗sl = n X k=1 ωki⟨sk, sj⟩+ n X l=1 ωlj⟨si, sl⟩ = ωji + ωij, as claimed. Remark: In Proposition 11.18, if ξ is a complex vector bundle, then ω is skew-Hermitian. This means that ω⊤= −ω, where ω is the conjugate matrix of ω; that is, (ω)ij = ωij. If ∇is a metric connection, then the curvature matrices are also skew-symmetric. 11.8. CONNECTIONS COMPATIBLE WITH A METRIC 497 Proposition 11.19. Let ξ be a rank n vector bundle with a metric ⟨−, −⟩. In any local trivialization of ξ, with respect to a orthonormal frame the curvature matrix Ω= (Ωij) is skew-symmetric. If ξ is a complex vector bundle, then Ω= (Ωij) is skew-Hermitian. Proof. By the structure equation (Proposition 11.13), Ω= dω + ω ∧ω, that is, Ωij = dωij + Pn k=1 ωik ∧ωkj. Using the skew symmetry of ωij and wedge, Ωji = dωji + n X k=1 ωjk ∧ωki = −dωij + n X k=1 ωkj ∧ωik = −dωij − n X k=1 ωik ∧ωkj = −Ωij, as claimed. We now restrict our attention to a Riemannian manifold; that is, to the case where our bundle ξ is the tangent bundle ξ = TM of some Riemannian manifold M. We know from Proposition 11.17 that metric connections on TM exist. However, there are many metric connections on TM, and none of them seems more relevant than the others. If M is a Riemannian manifold, the metric ⟨−, −⟩on M is often denoted g. In this case, for every chart (U, ϕ), we let gij ∈C∞(M) be the function defined by gij(p) =  ∂ ∂xi  p ,  ∂ ∂xj  p + p . (Note the unfortunate clash of notation with the transitions functions!) The notations g = P ij gijdxi ⊗dxj or simply g = P ij gijdxidxj are often used to denote the metric in local coordinates. We observed immediately after stating Proposition 10.13 that the covariant differential ∇g of the Riemannian metric g on M is given by ∇X(g)(Y, Z) = d(g(Y, Z))(X) −g(∇XY, Z) −g(Y, ∇XZ), for all X, Y, Z ∈X(M). Therefore, a connection ∇on a Riemannian manifold (M, g) is compatible with the metric iff ∇g = 0. 498 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES It is remarkable that if we require a certain kind of symmetry on a metric connection, then it is uniquely determined. Such a connection is known as the Levi–Civita connection. The Levi–Civita connection can be characterized in several equivalent ways, a rather simple way involving the notion of torsion of a connection. Recall that one way to introduce the curvature is to view it as the “error term” R(X, Y ) = ∇X∇Y −∇Y ∇X −∇[X,Y ]. Another natural error term is the torsion T(X, Y ), of the connection ∇, given by T(X, Y ) = ∇XY −∇Y X −[X, Y ], which measures the failure of the connection to behave like the Lie bracket. Then the Levi– Civita connection is the unique metric and torsion-free connection (T(X, Y ) = 0) on the Riemannian manifold. The first characterization of the Levi–Civita connection is given by the following Proposition. Proposition 11.20. (Levi-Civita, Version 1) Let M be any Riemannian manifold. There is a unique, metric, torsion-free connection ∇on M; that is, a connection satisfying the conditions: X(⟨Y, Z⟩) = ⟨∇XY, Z⟩+ ⟨Y, ∇XZ⟩ ∇XY −∇Y X = [X, Y ], for all vector fields, X, Y, Z ∈X(M). This connection is called the Levi-Civita connection (or canonical connection) on M. Furthermore, this connection is determined by the Koszul formula 2⟨∇XY, Z⟩ = X(⟨Y, Z⟩) + Y (⟨X, Z⟩) −Z(⟨X, Y ⟩) −⟨Y, [X, Z]⟩−⟨X, [Y, Z]⟩−⟨Z, [Y, X]⟩. The proof of Proposition 11.20 can be found in Gallot, Hulin, Lafontaine , Do Carmo , Morita , or Gallier and Quaintance . Another way to characterize the Levi-Civita connection uses the cotangent bundle T ∗M. It turns out that a connection ∇on a vector bundle (metric or not) ξ naturally induces a connection ∇∗on the dual bundle ξ∗. If ∇is a connection on TM, then ∇∗is a connection on T ∗M, namely, a linear map, ∇∗: Γ(T ∗M) →A1(M) ⊗C∞(B) Γ(T ∗M); that is ∇∗: A1(M) →A1(M) ⊗C∞(B) A1(M) ∼ = Γ(T ∗M ⊗T ∗M), since Γ(T ∗M) = A1(M). With a slight abuse of notation, we denote by ∧the map ∧⊗: A1(M) ⊗C∞(B) A1(M) − →A2(M) induced by the C∞(B)-bilinear map ∧: A1(M) × A1(M) − →A2(M). By composition we get the map A1(M) ∇∗ − →A1(M) ⊗C∞(B) A1(M) ∧ − →A2(M). 11.9. CONNECTIONS ON THE DUAL BUNDLE 499 Then miracle, a metric connection is the Levi-Civita connection iff d = ∧◦∇∗, where d: A1(M) →A2(M) is exterior differentiation. There is also a nice local expression of the above equation. Let us now consider the second approach to torsion-freeness. For this, we have to explain how a connection ∇on a vector bundle ξ = (E, π, B, V ) induces a connection ∇∗on the dual bundle ξ∗. 11.9 Connections on the Dual Bundle Let ξ = (E, π, B, V ) be a vector bundle. First, there is an evaluation map Γ(ξ⊗ξ∗) − →Γ(ϵ1) (where ϵ1 = B × R, the trivial line bundle over B), or equivalently ⟨⟨−, −⟩⟩: Γ(ξ) ⊗C∞(B) HomC∞(B)(Γ(ξ), C∞(B)) − →C∞(B), given by ⟨⟨s1, s∗ 2⟩⟩= s∗ 2(s1), s1 ∈Γ(ξ), s∗ 2 ∈HomC∞(B)(Γ(ξ), C∞(B)), and thus a map Ak(ξ ⊗ξ∗) = Ak(B) ⊗C∞(B) Γ(ξ ⊗ξ∗) id⊗⟨⟨−,−⟩⟩ − → Ak(B) ⊗C∞(B) C∞(B) ∼ = Ak(B). Using this map, we obtain a pairing (−, −): Ai(ξ) ⊗Aj(ξ∗) ⊼ − →Ai+j(ξ ⊗ξ∗) − →Ai+j(B) given by (ω ⊗s1, η ⊗s∗ 2) = (ω ∧η) ⊗⟨⟨s1, s∗ 2⟩⟩, where ω ∈Ai(B), η ∈Aj(B), s1 ∈Γ(ξ), s∗ 2 ∈Γ(ξ∗). It is easy to check that this pairing is non-degenerate. Then given a connection ∇on a rank n vector bundle ξ, we define ∇∗on ξ∗by d⟨⟨s1, s∗ 2⟩⟩= ∇(s1), s∗ 2  + s1, ∇∗(s∗ 2)  , where s1 ∈Γ(ξ) and s∗ 2 ∈Γ(ξ∗). Because the pairing (−, −) is non-degenerate, ∇∗is well-defined, and it is immediately that it is a connection on ξ∗. Let us see how it is expressed locally. If (U, ϕ) is a local trivialization and (s1, . . . , sn) is a frame over U, then let (θ1, . . . , θn) be the dual frame (called a coframe). We have ⟨⟨sj, θi⟩⟩= θi(sj) = δij, 1 ≤i, j ≤n. 500 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Recall that ∇sj = n X k=1 ωkj ⊗sk, and write ∇∗θi = n X k=1 ω∗ ki ⊗θk. Applying d to the equation ⟨⟨sj, θi⟩⟩= δij and using the equation defining ∇∗, we get 0 = d⟨⟨sj, θi⟩⟩ = ∇(sj), θi  + sj, ∇∗(θi)  = n X k=1 ωkj ⊗sk, θi  + sj, n X l=1 ω∗ li ⊗θl  = n X k=1 ωkj⟨⟨sk, θi⟩⟩+ n X l=1 ω∗ li⟨⟨sj, θl⟩⟩ = ωij + ω∗ ji. Proposition 11.21. If we write ω∗= (ω∗ ij), then we have ω∗= −ω⊤. If ∇is a metric connection and (s1, . . . , sn) is an orthonormal frame over U, then ω is skew-symmetric; that is, ω⊤= −ω. Corollary 11.22. If ∇is a metric connection on ξ, then ω∗= −ω⊤= ω. Remark: If ξ is a complex vector bundle, then there is a problem because if (s1, . . . , sn) is a frame over U, then the θj(b)’s defined by ⟨⟨si(b), θj(b)⟩⟩= δij are not linear, but instead conjugate-linear. (Recall that a linear form θ is conjugate linear (or semi-linear) iffθ(λu) = λθ(u), for all λ ∈C.) Instead of ξ∗, we need to consider the bundle ξ ∗, which is the bundle whose fibre over b ∈B consist of all conjugate-linear forms over π−1(b). In this case, the evaluation pairing ⟨⟨s, θ⟩⟩is conjugate-linear in s, and we find that ω∗= −ω⊤, where ω∗is the connection matrix of ξ ∗over U. If ξ is a Hermitian bundle, as ω is skew-Hermitian, we find that ω∗= ω, which makes sense since ξ and ξ ∗are canonically isomorphic. However, this does not give any information 11.10. THE LEVI-CIVITA CONNECTION ON TM REVISITED 501 on ξ∗. For this, we consider the conjugate bundle ξ. This is the bundle obtained from ξ by redefining the vector space structure on each fibre π−1(b), with b ∈B, so that (x + iy)v = (x −iy)v, for every v ∈π−1(b). If ω is the connection matrix of ξ over U, then ω is the connection matrix of ξ over U. If ξ has a Hermitian metric, it is easy to prove that ξ∗and ξ are canonically isomorphic (see Proposition 11.33). In fact, the Hermitian product ⟨−, −⟩establishes a pairing between ξ and ξ∗, and basically as above, we can show that if ω is the connection matrix of ξ over U, then ω∗= −ω⊤is the the connection matrix of ξ∗over U. As ω is skew-Hermitian, ω∗= ω. 11.10 The Levi-Civita Connection on TM Revisited If ∇is the Levi-Civita connection of some Riemannian manifold M, for every chart (U, ϕ), in an orthonormal frame we have ω∗= ω, where ω is the connection matrix of ∇over U and ω∗is the connection matrix of the dual connection ∇∗. This implies that the Christoffel symbols of ∇and ∇∗over U are identical. Furthermore, ∇∗is a linear map ∇∗: A1(M) − →Γ(T ∗M ⊗T ∗M). Thus, locally in a chart (U, ϕ), if (as usual) we let xi = pri ◦ϕ, then we can write ∇∗(dxk) = X ij Γj ikdxi ⊗dxj. Now, if we want ∧◦∇∗= d, we must have ∧∇∗(dxk) = X ij Γj ikdxi ∧dxj = ddxk = 0; that is Γj ik = Γj ki, for all i, k. It is known that this condition on the Christoffel symbols is equivalent to torsion-freeness (see Gallot, Hulin, Lafontaine , or Do Carmo ). We record this as Proposition 11.23. Let M be a manifold with connection ∇. Then ∇is torsion-free (i.e., T(X, Y ) = ∇XY −∇Y X −[X, Y ] = 0, for all X, Y ∈X(M)) iff ∧◦∇∗= d, where d: A1(M) →A2(M) is exterior differentiation. Proposition 11.23 together with Proposition 11.20 yield a second version of the Levi-Civita theorem: 502 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Proposition 11.24. (Levi-Civita, Version 2) Let M be any Riemannian manifold. There is a unique, metric connection ∇on M, such that ∧◦∇∗= d, where d: A1(M) →A2(M) is exterior differentiation. This connection is equal to the Levi-Civita connection in Proposition 11.20. Our third version of the Levi-Civita connection is a local version due to ´ Elie Cartan. Recall that locally with respect to a (orthonormal) frame over a chart (U, ϕ), the connection ∇∗is given by the matrix, ω∗, such that ω∗= −ω⊤, where ω is the connection matrix of TM over U. That is, we have ∇∗θi = n X j=1 −ωij ⊗θj, for some one-forms ωij ∈A1(M). Then, ∧◦∇∗θi = − n X j=1 ωij ∧θj so the requirement that d = ∧◦∇∗is expressed locally by dθi = − n X j=1 ωij ∧θj. In addition, since our connection is metric, ω is skew-symmetric, and so ω∗= ω. Then it is not too surprising that the following proposition holds: Proposition 11.25. Let M be a Riemannian manifold with metric ⟨−, −⟩. For every chart (U, ϕ), if (s1, . . . , sn) is an orthonormal frame over over U and (θ1, . . . , θn) is the correspond-ing coframe (dual frame), then there is a unique matrix ω = (ωij) of one-forms ωij ∈A1(M), so that the following conditions hold: (i) ωji = −ωij. (ii) dθi = − n X j=1 ωij ∧θj, or in matrix form, dθ = −ω ∧θ. Proof. There is a direct proof using a combinatorial trick. For instance, see Morita , Chapter 5, Proposition 5.32, or Milnor and Stasheff, Appendix C, Lemma 8. On the other hand, if we view ω = (ωij) as a connection matrix, then we observed that Condition (i) asserts that the connection is metric and Condition (ii) that it is torsion-free. We conclude by applying Proposition 11.24. 11.10. THE LEVI-CIVITA CONNECTION ON TM REVISITED 503 Example 11.3. In Example 11.1, we introduced the spherical frame for TR3 as s1 = ∂ ∂r = (cos ϕ sin θ, sin ϕ sin θ, cos θ) s2 = ∂ ∂θ ∂ ∂θ = (cos ϕ cos θ, sin ϕ cos θ, −sin θ) s3 = ∂ ∂ϕ ∂ ∂ϕ = (−sin θ, cos θ, 0), and found that the connection matrix is ω =   0 −dθ −sin θdϕ dθ 0 −cos θdϕ sin θdϕ cos θdϕ 0  . The dual coframe is then given by θ1 = dr θ2 = r dθ θ3 = r sin θ dϕ. Observe that dθ1 = d dr = 0 = r dθ ∧dθ + r sin ϕ dϕ ∧r sin θ dϕ = −ω12 ∧θ2 −ω13 ∧θ3 dθ2 = dr ∧dθ = −dθ ∧dr + cos θ dϕ ∧r sin θ dϕ = −ω21 ∧θ1 −ω23 ∧θ3 dθ3 = sin θ dr ∧dϕ + r cos θdθ ∧dϕ = −sin θ dϕ ∧dr −cos θ dϕ ∧r dθ = −ω31 ∧θ1 −ω32 ∧θ2, which shows that the connection form obeys Condition (ii) of Proposition 11.25, and hence is the Levi-Civita connection for R3 with the induced Euclidean metric. For another example of Proposition 11.25 consider an orientable (compact) surface M, with a Riemannian metric. Pick any chart (U, ϕ), and choose an orthonormal coframe of one-forms (θ1, θ2), such that VolM = θ1 ∧θ2 on U. Then we have dθ1 = a1θ1 ∧θ2 dθ2 = a2θ1 ∧θ2 for some functions, a1, a2, and we let ω12 = a1θ1 + a2θ2. Clearly,  0 ω12 −ω12 0  θ1 θ2  =  0 a1θ1 + a2θ2 −(a1θ1 + a2θ2) 0  θ1 θ2  = dθ1 dθ2  504 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES which shows that ω = ω∗=  0 ω12 −ω12 0  corresponds to the Levi-Civita connection on M. Since Ω= dω + ω ∧ω, we see that Ω=  0 dω12 −dω12 0  . As M is oriented and as M has a metric, the transition functions are in SO(2). We easily check that  cos t sin t −sin t cos t   0 dω12 −dω12 0  cos t −sin t sin t cos t  =  0 dω12 −dω12 0  , which shows that Ωis a global two-form called the Gauss-Bonnet 2-form of M. There is a function κ, the Gaussian curvature of M, such that dω12 = −κVolM, where VolM = θ1 ∧θ2 is the oriented volume form on M. It should be noted that Milnor and Stasheffdefine the volume form as VolM = −θ1 ∧θ2 so in their work, the curvature κ should be replaced by −κ. The reason for such a choice is explained on page 304 of Milnor and Stasheff. Many other authors (including Warner and Bott and Chern) use the definition VolM = +θ1 ∧θ2 that has been adopted here. The Gauss-Bonnet theorem for orientable surfaces asserts that Z M dω12 = 2πχ(M), where χ(M) is the Euler characteristic of M. Remark: The Levi-Civita connection induced by a Riemannian metric g can also be defined in terms of the Lie derivative of the metric g. This is the approach followed in Petersen (Chapter 2). If θX is the one-form given by θX = iXg; that is, (iXg)(Y ) = g(X, Y ) for all X, Y ∈X(M), and if LXg is the Lie derivative of the symmetric (0, 2) tensor g, defined so that (LXg)(Y, Z) = X(g(Y, Z)) −g(LXY, Z) −g(Y, LXZ) (see Proposition 5.13), then it is proved in Petersen (Chapter 2, Theorem 1) that the Levi-Civita connection is defined implicitly by the formula 2g(∇XY, Z) = (LY g)(X, Z) + (dθY )(X, Z). 11.11. PONTRJAGIN CLASSES AND CHERN CLASSES, A GLIMPSE 505 11.11 Pontrjagin Classes and Chern Classes, a Glimpse The purpose of this section is to introduce the reader to Pontrjagin Classes and Chern Classes, which are fundamental invariants of real (resp. complex) vector bundles. We focus on motivations and intuitions and omit most proofs, but we give precise pointers to the literature for proofs. Given a real (resp. complex) rank n vector bundle ξ = (E, π, B, V ), we know that locally, ξ “looks like” a trivial bundle U ×V , for some open subset U of the base space B. Globally, ξ can be very twisted, and one of the main issues is to understand and quantify “how twisted” ξ really is. Now we know that every vector bundle admits a connection, say ∇, and the curvature R∇of this connection is some measure of the twisting of ξ. However, R∇depends on ∇, so curvature is not intrinsic to ξ, which is unsatisfactory as we seek invariants that depend only on ξ. Pontrjagin, Stiefel, and Chern (starting from the late 1930’s) discovered that invariants with “good” properties could be defined if we took these invariants to belong to various cohomology groups associated with B. Such invariants are usually called characteristic classes. Roughly, there are two main methods for defining characteristic classes: one using topology, and the other due to Chern and Weil, using differential forms. A masterly exposition of these methods is given in the classic book by Milnor and Stasheff . Amazingly, the method of Chern and Weil using differential forms is quite accessible for someone who has reasonably good knowledge of differential forms and de Rham cohomology, as long as one is willing to gloss over various technical details. As we said earlier, one of the problems with curvature is that is depends on a connection. The way to circumvent this difficulty rests on the simple, yet subtle observation, that locally, given any two overlapping local trivializations (Uα, ϕα) and (Uβ, ϕβ), the transformation rule for the curvature matrices Ωα and Ωβ is Ωβ = g−1 αβΩαgαβ, where gαβ : Uα ∩Uβ →GL(V ) is the transition function. The matrices of two-forms Ωα and Ωα are local, but the stroke of genius is to glue them together to form a global form using invariant polynomials. Indeed, the Ωα are n × n matrices, so consider the algebra of polynomials R[X1, . . . , Xn2] (or C[X1, . . . , Xn2] in the complex case) in n2 variables X1, . . . , Xn2, considered as the entries of an n × n matrix. It is more convenient to use the set of variables {Xij | 1 ≤i, j ≤n}, and to let X be the n × n matrix X = (Xij). Definition 11.18. A polynomial P ∈R[{Xij | 1 ≤i, j ≤n}] (or P ∈C[{Xij | 1 ≤i, j ≤ n}]) is invariant iff P(AXA−1) = P(X), for all A ∈GL(n, R) (resp. A ∈GL(n, C)). The algebra of invariant polynomials over n×n matrices is denoted by In. 506 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Examples of invariant polynomials are the trace tr(X) and the determinant det(X) of the matrix X. We will characterize shortly the algebra In. Now comes the punch line: For any homogeneous invariant polynomial P ∈In of degree k, we can substitute Ωα for X; that is, substitute ωij for Xij, and evaluate P(Ωα). This is because Ωis a matrix of two-forms, and the wedge product is commutative for forms of even degree. Therefore, P(Ωα) ∈A2k(Uα). But the formula for a change of trivialization yields P(Ωα) = P(g−1 αβΩαgαβ) = P(Ωβ), so the forms P(Ωα) and P(Ωβ) agree on overlaps, and thus they define a global form denoted P(R∇) ∈A2k(B). Definition 11.19. For any vector bundle ξ = (E, π, B, V ), for any homogeneous invariant polynomial P ∈In of degree k, the global form P(R∇) ∈A2k(B) defined above is called the global curvature form on the vector bundle ξ. Now we know how to obtain global 2k-forms P(R∇) ∈A2k(B), but they still seem to depend on the connection, and how do they define a cohomology class? Both problems are settled thanks to the following theorems: Theorem 11.26. For every real rank n vector bundle ξ, for every connection ∇on ξ, for every invariant homogeneous polynomial P of degree k, the 2k-form P(R∇) ∈A2k(B) is closed. If ξ is a complex vector bundle, then the 2k-form P(R∇) ∈A2k(B; C) is closed. Theorem 11.26 implies that the 2k-form P(R∇) ∈A2k(B) defines a cohomology class [P(R∇)] ∈H2k DR(B). We will come back to the proof of Theorem 11.26 later. Theorem 11.27. For every real (resp. complex) rank n vector bundle ξ, for every invariant homogeneous polynomial P of degree k, the cohomology class [P(R∇)] ∈H2k DR(B) (resp. [P(R∇)] ∈H2k DR(B; C)) is independent of the choice of the connection ∇. Definition 11.20. The cohomology class [P(R∇)], which does not depend on ∇, is denoted P(Rξ) (or P(Kξ)) and is called the characteristic class of ξ corresponding to P. Remark: Milnor and Stasheff use the notation P(K), Madsen and Tornehave use the notation P(F ∇), and Morita use the notation f(E) (where E is the total space of the vector bundle ξ). The proof of Theorem 11.27 involves a kind of homotopy argument; see Madsen and Tornehave (Lemma 18.2), Morita (Proposition 5.28), or Milnor and Stasheff (Appendix C). The upshot is that Theorems 11.26 and 11.27 give us a method for producing invariants of a vector bundle that somehow reflect how curved (or twisted) the bundle is. However, it appears that we need to consider infinitely many invariants. Fortunately, we can do better because the algebra In of invariant polynomials is finitely generated, and in fact, has very nice sets of generators. For this, we recall the elementary symmetric functions in n variables. 11.11. PONTRJAGIN CLASSES AND CHERN CLASSES, A GLIMPSE 507 Definition 11.21. Given n variables λ1, . . . , λn, we can write n Y i=1 (1 + tλi) = 1 + σ1t + σ2t2 + · · · + σntn, where the σi are symmetric, homogeneous polynomials of degree i in λ1, . . . , λn, called ele-mentary symmetric functions in n variables. For example, σ1 = n X i=1 λi, σ2 = X 1≤i<j≤n λiλj, σn = λ1 · · · λn. To be more precise, we write σi(λ1, . . . , λn) instead of σi. Definition 11.22. Given any n × n matrix X = (Xij), we define σi(X) by the formula det(I + tX) = 1 + σ1(X)t + σ2(X)t2 + · · · + σn(X)tn. Proposition 11.28. Let X be an n × n matrix. Then σi(X) = σi(λ1, . . . , λn), where λ1, . . . , λn are the eigenvalues of X. Proof. Indeed, λ1, . . . , λn are the roots the the polynomial det(λI −X) = 0, and as det(λI −X) = n Y i=1 (λ −λi) = λn n Y i=1  1 −λi λ  = λn + n X i=1 (−1)iσi(λ1, . . . , λn)λn−i, by factoring λn and replacing λ−1 by −λ−1, we get det(I + (−λ−1)X) = 1 + n X i=1 = σi(λ1, . . . , λn)(−λ−1)i, which proves our claim. Observe that σ1(X) = tr(X), σn(X) = det(X). Also, σk(X⊤) = σk(X), since det(I + tX) = det((I + tX)⊤) = det(I + tX⊤). It is not very difficult to prove the following theorem. Theorem 11.29. The algebra In of invariant polynomials in n2 variables is generated by σ1(X), . . . , σn(X); that is, In ∼ = R[σ1(X), . . . , σn(X)] (resp. In ∼ = C[σ1(X), . . . , σn(X)]). 508 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Proof sketch. For a proof of Theorem 11.29, see Milnor and Stasheff (Appendix C, Lemma 6), Madsen and Tornehave (Appendix B), or Morita (Theorem 5.26). The proof uses the fact that for every matrix X, there is an upper-triangular matrix T, and an invertible matrix B, so that X = BTB−1. Then we can replace B by the matrix diag(ϵ, ϵ2, . . . , ϵn)B, where ϵ is very small, and make the offdiagonal entries arbitrarily small. By continuity, it follows that P(X) depends only on the diagonal entries of BTB−1, that is, on the eigenvalues of X. So, P(X) must be a symmetric function of these eigenvalues, and the classical theory of symmetric functions completes the proof. It turns out that there are situations where it is more convenient to use another set of generators instead of σ1, . . . , σn. Define si(X) by si(X) = tr(Xi). Of course, si(X) = λi 1 + · · · + λi n, where λ1, . . . , λn are the eigenvalues of X. Now the σi(X) and si(X) are related to each other by Newton’s formula, namely: si(X) −σ1(X)si−1(X) + σ2(X)si−2(X) + · · · + (−1)i−1σi−1(X)s1(X) + (−1)iiσi(X) = 0, with 1 ≤i ≤n. A “cute” proof of the Newton formulae is obtained by computing the derivative of log(h(t)), where h(t) = n Y i=1 (1 + tλi) = 1 + σ1t + σ2t2 + · · · + σntn, see Madsen and Tornehave (Appendix B) or Morita (Exercise 5.7). Consequently, we can inductively compute si in terms of σ1, . . . , σi, and conversely σi in terms of s1, . . . , si. For example, s1 = σ1, s2 = σ2 1 −2σ2, s3 = σ3 1 −3σ1σ2 + 3σ3. It follows that In ∼ = R[s1(X), . . . , sn(X)] (resp. In ∼ = C[s1(X), . . . , sn(X)]). Using the above, we can give a simple proof of Theorem 11.26, using Theorem 11.29. 11.11. PONTRJAGIN CLASSES AND CHERN CLASSES, A GLIMPSE 509 Proof. (Proof of Theorem 11.26). Since s1, . . . , sn generate In, it is enough to prove that si(R∇) is closed. We need to prove that dsi(R∇) = 0, and for this, it is enough to prove it in every local trivialization (Uα, ϕα). To simplify notation, we write Ωfor Ωα. Now, si(Ω) = tr(Ωi), so dsi(Ω) = dtr(Ωi) = tr(dΩi), and we use Bianchi’s identity (Proposition 11.15), dΩ= Ω∧ω −ω ∧Ω. We have dΩi = dΩ∧Ωi−1 + Ω∧dΩ∧Ωi−2 + · · · + Ωk ∧dΩ∧Ωi−k−1 + · · · + Ωi−1 ∧dΩ = (Ω∧ω −ω ∧Ω) ∧Ωi−1 + Ω∧(Ω∧ω −ω ∧Ω) ∧Ωi−2 + · · · + Ωk ∧(Ω∧ω −ω ∧Ω) ∧Ωi−k−1 + Ωk+1 ∧(Ω∧ω −ω ∧Ω) ∧Ωi−k−2 + · · · + Ωi−1 ∧(Ω∧ω −ω ∧Ω) = −ω ∧Ωi + Ω∧ω ∧Ωi−1 −Ω∧ω ∧Ωi−1 + Ω2 ∧ω ∧Ωi−2 + · · · + −Ωk ∧ω ∧Ωi−k + Ωk+1 ∧ω ∧Ωi−k−1 −Ωk+1 ∧ω ∧Ωi−k−1 + Ωk+2 ∧ω ∧Ωi−k−2 + · · · −Ωi−1 ∧ω ∧Ω+ Ωi ∧ω = Ωi ∧ω −ω ∧Ωi. However, the entries in ω are one-forms, the entries in Ωare two-forms, and since η ∧θ = θ ∧η for all η ∈A1(B) and all θ ∈A2(B) and tr(XY ) = tr(Y X) for all matrices X and Y with commuting entries, we get tr(dΩi) = tr(ω ∧Ωi −Ωi ∧ω) = tr(Ωi ∧ω) −tr(ω ∧Ωi) = 0, as required. A more elegant proof (also using Bianchi’s identity) can be found in Milnor and Stasheff (Appendix C, page 296-298). For real vector bundles, only invariant polynomials of even degrees matter. Proposition 11.30. If ξ is a real vector bundle, then for every homogeneous invariant polynomial P of odd degree k, we have P(Rξ) = 0 ∈H2k DR(B). Proof. As In ∼ = R[s1(X), . . . , sn(X)] and si(X) is homogeneous of degree i, it is enough to prove Proposition 11.30 for si(X) with i odd. By Theorem 11.27, we may assume that we pick a metric connection on ξ, so that Ωα is skew-symmetric in every local trivialization. Then, Ωi α is also skew symmetric and tr(Ωi α) = 0, since the diagonal entries of a real skew-symmetric matrix are all zero. It follows that si(Ωα) = tr(Ωi α) = 0. 510 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Proposition 11.30 implies that for a real vector bundle ξ, non-zero characteristic classes can only live in the cohomology groups H4k DR(B) of dimension 4k. This property is specific to real vector bundles and generally fails for complex vector bundles. Before defining Pontrjagin and Chern classes, we state another important property of the homology classes P(Rξ); see Madsen and Tornehave (Chapter 18, Theorem 18.5). Proposition 11.31. If ξ = (E, π, B, V ) and ξ′ = (E′, π′, B′, V ) are real (resp. complex) vector bundles, for every bundle map E fE / π  E′ π′  B f / B′, for every homogeneous invariant polynomial P of degree k, we have P(Rξ) = f ∗(P(Rξ′)) ∈H2k DR(B) (resp. P(Rξ) = f ∗(P(Rξ′)) ∈H2k DR(B; C)). In particular, for every smooth map g: N →B, we have P(Rg∗ξ) = g∗(P(Rξ)) ∈H2k DR(N) (resp. P(Rg∗ξ) = g∗(P(Rξ)) ∈H2k DR(N; C)), where g∗ξ is the pullback bundle of ξ along g. The above proposition implies that if (fE, f): ξ →ξ′ is an isomorphism of vector bundles, then the pullback map f ∗maps the characteristic classes of ξ′ to the characteristic classes of ξ bijectively. We finally define Pontrjagin classes and Chern classes. Definition 11.23. If ξ be a real rank n vector bundle, then the kth Pontrjagin class of ξ, denoted pk(ξ), where 1 ≤2k ≤n, is the cohomology class pk(ξ) =  1 (2π)2k σ2k(R∇)  ∈H4k DR(B), for any connection ∇on ξ. If ξ be a complex rank n vector bundle, then the kth Chern class of ξ, denoted ck(ξ), where 1 ≤k ≤n, is the cohomology class ck(ξ) = " −1 2πi k σk(R∇) # ∈H2k DR(B), for any connection ∇on ξ. We also set p0(ξ) = 1, and c0(ξ) = 1 in the complex case. 11.11. PONTRJAGIN CLASSES AND CHERN CLASSES, A GLIMPSE 511 The strange coefficient in pk(ξ) is present so that our expression matches the topological definition of Pontrjagin classes. The equally strange coefficient in ck(ξ) is there to insure that ck(ξ) actually belongs to the real cohomology group H2k DR(B), as stated (from the definition, we can only claim that ck(ξ) ∈H2k DR(B; C)). This requires a proof which can be found in Morita (Proposition 5.30), or in Madsen and Tornehave (Chapter 18). One can use the fact that every complex vector bundle admits a Hermitian connection. Locally, the curvature matrices are skew-Hermitian and this easily implies that the Chern classes are real, since if Ωis skew-Hermitian, then iΩis Hermitian. (Actually, the topological version of Chern classes shows that ck(ξ) ∈H2k(B; Z).) If ξ is a real rank n vector bundle and n is odd, say n = 2m+1, then the “top” Pontrjagin class pm(ξ) corresponds to σ2m(R∇), which is not det(R∇). However, if n is even, say n = 2m, then the “top” Pontrjagin class pm(ξ) corresponds to det(R∇). Definition 11.24. The Pontrjagin polynomial p(ξ)(t) ∈H• DR(B)[t], given by p(ξ)(t) =  det  I + t 2π R∇  = 1 + p1(ξ)t + p2(ξ)t2 + · · · + p⌊n 2 ⌋(ξ)t⌊n 2 ⌋ and the Chern polynomial c(ξ)(t) ∈H• DR(B)[t], given by c(ξ)(t) =  det  I − t 2πi R∇  = 1 + c1(ξ)t + c2(ξ)t2 + · · · + cn(ξ)tn. If a vector bundle is trivial, then all its Pontrjagin classes (or Chern classes) are zero for all k ≥1. If ξ is the real tangent bundle ξ = TB of some manifold B of dimension n, then the ⌊n 4⌋Pontrjagin classes of TB are denoted p1(B), . . . , p⌊n 4 ⌋(B). For complex vector bundles, the manifold B is often the real manifold corresponding to a complex manifold. If B has complex dimension n, then B has real dimension 2n. In this case, the tangent bundle TB is a rank n complex vector bundle over the real manifold of dimension 2n, and thus, it has n Chern classes, denoted c1(B), . . . , cn(B). The determination of the Pontrjagin classes (or Chern classes) of a manifold is an impor-tant step for the study of the geometric/topological structure of the manifold. For example, it is possible to compute the Chern classes of complex projective space CPn (as a complex manifold). The Pontrjagin classes of a real vector bundle ξ are related to the Chern classes of its complexification ξC = ξ ⊗R ϵ1 C (where ϵ1 C is the trivial complex line bundle B × C). Proposition 11.32. For every real rank n vector bundle ξ = (E, π, B, V ), if ξC = ξ ⊗R ϵ1 C is the complexification of ξ, then pk(ξ) = (−1)kc2k(ξC) ∈H4k DR(B) k ≥0. 512 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Basically, the reason why Proposition 11.32 holds is that 1 (2π)2k = (−1)k  −1 2πi 2k For details, see Morita (Chapter 5, Section 5, Proposition 5.38). We conclude this section by stating a few more properties of Chern classes. Proposition 11.33. For every complex rank n vector bundle ξ, the following properties hold: (1) If ξ has a Hermitian metric, then we have a canonical isomorphism ξ∗∼ = ξ. (2) The Chern classes of ξ, ξ∗and ξ satisfy: ck(ξ∗) = ck(ξ) = (−1)kck(ξ). (3) For any complex vector bundles ξ and η, ck(ξ ⊕η) = k X i=0 ci(ξ)ck−i(η), or equivalently c(ξ ⊕η)(t) = c(ξ)(t)c(η)(t), and similarly for Pontrjagin classes when ξ and η are real vector bundles. To prove (2), we can use the fact that ξ can be given a Hermitian metric. Then we saw earlier that if ω is the connection matrix of ξ over U then ω = −ω⊤is the connection matrix of ξ over U. However, it is clear that σk(−Ω⊤ α) = (−1)kσk(Ωα), and so ck(ξ) = (−1)kck(ξ). For details, see Morita (Chapter 5, Section 5, Theorem 5.37 and Proposition 5.40). Remark: For a real vector bundle ξ, it is easy to see that (ξC)∗= (ξ∗)C, which implies that ck((ξC)∗) = ck(ξC) (as ξ ∼ = ξ∗) and (2) implies that ck(ξC) = 0 for k odd. This proves again that the Pontrjagin classes exit only in dimension 4k. A complex rank n vector bundle ξ can also be viewed as a rank 2n vector bundle ξR and we have: Proposition 11.34. For every rank n complex vector bundle ξ, if pk = pk(ξR) and ck = ck(ξ), then we have 1 −p1 + p2 + · · · + (−1)npn = (1 + c1 + c2 + · · · + cn)(1 −c1 + c2 + · · · + (−1)ncn). For a proof, see Morita (Chapter 5, Section 5, Proposition 5.41). Besides defining the Chern and Pontrjagin classes, the curvature form R∇also defines an Euler class. But in order to efficiently define the Euler class, we need a technical tool, the Pfaffian polynomial. 11.12. THE PFAFFIAN POLYNOMIAL 513 11.12 The Pfaffian Polynomial The results of this section will be needed to define the Euler class of a real orientable rank 2n vector bundle; see Section 11.13. Let so(2n) denote the vector space (actually, Lie algebra) of 2n×2n real skew-symmetric matrices. It is well-known that every matrix A ∈so(2n) can be written as A = PDP ⊤, where P is an orthogonal matrix and where D is a block diagonal matrix D =      D1 D2 ... Dn      consisting of 2 × 2 blocks of the form Di =  0 −ai ai 0  . For a proof, see Horn and Johnson (Corollary 2.5.14), Gantmacher (Chapter IX), or Gallier (Chapter 11). Since det(Di) = a2 i and det(A) = det(PDP ⊤) = det(D) = det(D1) · · · det(Dn), we get det(A) = (a1 · · · an)2. The Pfaffian is a polynomial function Pf(A) in skew-symmetric 2n × 2n matrices A (a polynomial in (2n −1)n variables) such that Pf(A)2 = det(A), and for every arbitrary matrix B, Pf(BAB⊤) = Pf(A) det(B). The Pfaffian shows up in the definition of the Euler class of a vector bundle. There is a simple alternative way to define the Pfaffian using some exterior algebra. Let (e1, . . . , e2n) be any basis of R2n. For any matrix A ∈so(2n), let ω(A) = X i<j aij ei ∧ej, where A = (aij). Then Vn ω(A) is of the form Ce1 ∧e2 ∧· · · ∧e2n for some constant C ∈R. 514 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Definition 11.25. For every skew symmetric matrix A ∈so(2n), the Pfaffian polynomial or Pfaffian, is the degree n polynomial Pf(A) defined by n ^ ω(A) = n! Pf(A) e1 ∧e2 ∧· · · ∧e2n. Clearly, Pf(A) is independent of the basis chosen. If A is the block diagonal matrix D, a simple calculation shows that ω(D) = −(a1e1 ∧e2 + a2e3 ∧e4 + · · · + ane2n−1 ∧e2n) and that n ^ ω(D) = (−1)nn! a1 · · · an e1 ∧e2 ∧· · · ∧e2n, and so Pf(D) = (−1)na1 · · · an. Since Pf(D)2 = (a1 · · · an)2 = det(A), we seem to be on the right track. Proposition 11.35. For every skew symmetric matrix A ∈so(2n) and every arbitrary matrix B, we have: (i) Pf(A)2 = det(A) (ii) Pf(BAB⊤) = Pf(A) det(B). Proof. If we assume that (ii) is proved then, since we can write A = PDP ⊤for some orthogonal matrix P and some block diagonal matrix D as above, as det(P) = ±1 and Pf(D)2 = det(A), we get Pf(A)2 = Pf(PDP ⊤)2 = Pf(D)2 det(P)2 = det(A), which is (i). Therefore, it remains to prove (ii). Let fi = Bei for i = 1, . . . , 2n, where (e1, . . . , e2n) is any basis of R2n. Since fi = P k bkiek, we have τ = X i,j aij fi ∧fj = X i,j X k,l bkiaijblj ek ∧el = X k,l (BAB⊤)kl ek ∧el, and so, as BAB⊤is skew symmetric and ek ∧el = −el ∧ek, we get τ = 2ω(BAB⊤). Consequently, n ^ τ = 2n n ^ ω(BAB⊤) = 2nn! Pf(BAB⊤) e1 ∧e2 ∧· · · ∧e2n. 11.12. THE PFAFFIAN POLYNOMIAL 515 Now, n ^ τ = C f1 ∧f2 ∧· · · ∧f2n, for some C ∈R. If B is singular, then the fi are linearly dependent, which implies that f1 ∧f2 ∧· · · ∧f2n = 0, in which case Pf(BAB⊤) = 0, as e1 ∧e2 ∧· · · ∧e2n ̸= 0. Therefore, if B is singular, det(B) = 0 and Pf(BAB⊤) = 0 = Pf(A) det(B). If B is invertible, as τ = P i,j aij fi ∧fj = 2 P i<j aij fi ∧fj, we have n ^ τ = 2nn! Pf(A) f1 ∧f2 ∧· · · ∧f2n. However, as fi = Bei, we have f1 ∧f2 ∧· · · ∧f2n = det(B) e1 ∧e2 ∧· · · ∧e2n, so n ^ τ = 2nn! Pf(A) det(B) e1 ∧e2 ∧· · · ∧e2n and as n ^ τ = 2nn! Pf(BAB⊤) e1 ∧e2 ∧· · · ∧e2n, we get Pf(BAB⊤) = Pf(A) det(B), as claimed. Remark: It can be shown that the polynomial Pf(A) is the unique polynomial with integer coefficients such that Pf(A)2 = det(A) and Pf(diag(S, . . . , S)) = +1, where S =  0 1 −1 0  ; see Milnor and Stasheff (Appendix C, Lemma 9). There is also an explicit formula for Pf(A), namely Pf(A) = 1 2nn! X σ∈S2n sgn(σ) n Y i=1 aσ(2i−1) σ(2i). For example, if A = 0 −a a 0  , 516 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES then Pf(A) = −a, and if A =     0 a1 a2 a3 −a1 0 a4 a5 −a2 −a4 0 a6 −a3 −a5 −a6 0    , then Pf(A) = a1a6 −a2a5 + a4a3. It is easily checked that det(A) = (Pf(A))2 = (a1a6 −a2a5 + a4a3)2.  Beware, some authors use a different sign convention and require the Pfaffian to have the value +1 on the matrix diag(S′, . . . , S′), where S′ = 0 −1 1 0  . For example, if R2n is equipped with an inner product ⟨−, −⟩, then some authors define ω(A) as ω(A) = X i<j ⟨Aei, ej⟩ei ∧ej, where A = (aij). But then, ⟨Aei, ej⟩= aji and not aij, and this Pfaffian takes the value +1 on the matrix diag(S′, . . . , S′). This version of the Pfaffian differs from our version by the factor (−1)n. In this respect, Madsen and Tornehave seem to have an incorrect sign in Proposition B6 of Appendix C. We will also need another property of Pfaffians. Recall that the ring Mn(C) of n × n matrices over C is embedded in the ring M2n(R) of 2n × 2n matrices with real coefficients, using the injective homomorphism that maps every entry z = a + ib ∈C to the 2 × 2 matrix a −b b a  . If A ∈Mn(C), let AR ∈M2n(R) denote the real matrix obtained by the above process. Observe that every skew Hermitian matrix A ∈u(n) (i.e., with A∗= A ⊤= −A) yields a matrix AR ∈so(2n). Proposition 11.36. For every skew Hermitian matrix A ∈u(n), we have Pf(AR) = in det(A). 11.13. EULER CLASSES AND THE GENERALIZED GAUSS-BONNET THEOREM517 Proof. It is well-known that a skew Hermitian matrix can be diagonalized with respect to a unitary matrix U and that the eigenvalues are pure imaginary or zero, so we can write A = U diag(ia1, . . . , ian)U ∗, for some reals aj ∈R. Consequently, we get AR = UR diag(D1, . . . , Dn)U ⊤ R , where Dj =  0 −aj aj 0  and Pf(AR) = Pf(diag(D1, . . . , Dn)) = (−1)n a1 · · · an, as we saw before. On the other hand, det(A) = det(diag(ia1, . . . , ian)) = in a1 · · · an, and as (−1)n = inin, we get Pf(AR) = in det(A), as claimed.  Madsen and Tornehave state Proposition 11.36 using the factor (−i)n, which is wrong. 11.13 Euler Classes and The Generalized Gauss-Bonnet Theorem Let ξ = (E, π, B, V ) be a real vector bundle of rank n = 2m and let ∇be any metric connection on ξ. Then if ξ is orientable (as defined in Section 10.8, see Definition 10.21 and the paragraph following it), it is possible to define a global form eu(R∇) ∈A2m(B), which turns out to be closed. Furthermore, the cohomology class [eu(R∇)] ∈H2m DR(B) is independent of the choice of ∇. This cohomology class, denoted e(ξ), is called the Euler class of ξ, and has some very interesting properties. For example, pm(ξ) = e(ξ)2. As ∇is a metric connection, in a trivialization (Uα, ϕα), the curvature matrix Ωα is a skew symmetric 2m × 2m matrix of 2-forms. Therefore, we can substitute the 2-forms in Ωα for the variables of the Pfaffian of degree m (see Section 11.12), and we obtain the 2m-form, Pf(Ωα) ∈A2m(B). Now as ξ is orientable, the transition functions take values in SO(2m), so by Proposition 11.14, since Ωβ = g−1 αβΩαgαβ, 518 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES we conclude from Proposition 11.35 (ii) that Pf(Ωα) = Pf(Ωβ). Therefore, the local 2m-forms Pf(Ωα) patch and define a global form Pf(R∇) ∈A2m(B). The following propositions can be shown. Proposition 11.37. For every real, orientable, rank 2m vector bundle ξ, for every metric connection ∇on ξ, the 2m-form Pf(R∇) ∈A2m(B) is closed. Proposition 11.38. For every real, orientable, rank 2m vector bundle ξ, the cohomology class [Pf(R∇)] ∈H2m DR(B) is independent of the metric connection ∇on ξ. Proofs of Propositions 11.37 and 11.38 can be found in Madsen and Tornehave (Chapter 19) or Milnor and Stasheff (Appendix C) (also see Morita , Chapters 5 and 6). Definition 11.26. Let ξ = (E, π, B, V ) be any real, orientable, rank 2m vector bundle. For any metric connection ∇on ξ, the Euler form associated with ∇is the closed form eu(R∇) = 1 (2π)m Pf(R∇) ∈A2m(B), and the Euler class of ξ is the cohomology class e(ξ) = eu(R∇) ∈H2m DR(B), which does not depend on ∇.  Some authors, including Madsen and Tornehave , have a negative sign in front of R∇ in their definition of the Euler form; that is, they define eu(R∇) by eu(R∇) = 1 (2π)m Pf(−R∇). However these authors use a Pfaffian with the opposite sign convention from ours and this Pfaffian differs from ours by the factor (−1)n (see the warning in Section 11.12). Madsen and Tornehave seem to have overlooked this point and with their definition of the Pfaffian (which is the one we have adopted) Proposition 11.41 is incorrect. Here is the relationship between the Euler class e(ξ), and the top Pontrjagin class pm(ξ): Proposition 11.39. For every real, orientable, rank 2m vector bundle ξ = (E, π, B, V ), we have pm(ξ) = e(ξ)2 ∈H4m DR(B). 11.13. EULER CLASSES AND THE GENERALIZED GAUSS-BONNET THEOREM519 Proof. The top Pontrjagin class pm(ξ) is given by pm(ξ) =  1 (2π)2m det(R∇)  , for any (metric) connection ∇, and e(ξ) = eu(R∇) , with eu(R∇) = 1 (2π)m Pf(R∇). From Proposition 11.35 (i), we have det(R∇) = Pf(R∇)2, which yields the desired result. A rank m complex vector bundle ξ = (E, π, B, V ) can be viewed as a real rank 2m vector bundle ξR, by viewing V as a 2m dimensional real vector space. Proposition 11.40. For any complex vector bundle ξ = (E, π, B, V ), the real vector bundle ξR is naturally orientable. Proof. For any basis, (e1, . . . , em), of V over C, observe that (e1, ie1, . . . , em, iem) is a basis of V over R (since v = Pm i=1(λi +iµi)ei = Pm i=1 λiei +Pm i=1 µiiei). But, any m×m invertible matrix A, over C becomes a real 2m × 2m invertible matrix AR, obtained by replacing the entry ajk + ibjk in A by the real 2 × 2 matrix ajk −bjk bjk ajk.  Indeed, if vk = Pm j=1 ajkej + Pm j=1 bjkiej, then ivk = Pm j=1 −bjkej + Pm j=1 ajkiej and when we express vk and ivk over the basis (e1, ie1, . . . , em, iem), we get a matrix AR consisting of 2 × 2 blocks as above. Clearly, the map r: A 7→AR is a continuous injective homomorphism from GL(m, C) to GL(2m, R). Now, it is known that GL(m, C) is connected, thus Im(r) = r(GL(m, C)) is connected, and as det(I2m) = 1, we conclude that all matrices in Im(r) have positive determinant.1 Therefore, the transition functions of ξR which take values in Im(r) have positive determinant, and ξR is orientable. We can give ξR an orientation by fixing some basis of V over R. We have the following relationship between e(ξR) and the top Chern class, cm(ξ). 1One can also prove directly that every matrix in Im(r) has positive determinant by expressing r(A) as a product of simple matrices whose determinants are easily computed. 520 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Proposition 11.41. For every complex, rank m vector bundle ξ = (E, π, B, V ), we have cm(ξ) = e(ξ) ∈H2m DR(B). Proof. Pick some metric connection ∇on the complex vector bundle ξ . Recall that cm(ξ) =  −1 2πi m det(R∇)  = im  1 2π m det(R∇)  . On the other hand, e(ξ) =  1 (2π)m Pf(R∇ R )  . Here, R∇ R denotes the global 2m-form, which locally, is equal to ΩR, where Ωis the m × m curvature matrix of ξ over some trivialization. By Proposition 11.36, Pf(ΩR) = im det(Ω), so cm(ξ) = e(ξ), as claimed. The Euler class enjoys many other nice properties. For example, if f : ξ1 →ξ2 is an orientation preserving bundle map, then e(f ∗ξ2) = f ∗(e(ξ2)), where f ∗ξ2 is given the orientation induced by ξ2. Also, the Euler class can be defined by topological means and it belongs to the integral cohomology group H2m(B; Z). Although this result lies beyond the scope of these notes, we cannot resist stating one of the most important and most beautiful theorems of differential geometry usually called the Generalized Gauss-Bonnet theorem or Gauss-Bonnet-Chern theorem. For this we need the notion of Euler characteristic. Since we haven’t discussed trian-gulations of manifolds, we will use a definition in terms of cohomology. Although concise, this definition is hard to motivate, and we apologize for this. Given a smooth n-dimensional manifold M, we define its Euler characteristic χ(M), as χ(M) = n X i=0 (−1)i dim(Hi DR(M)). The integers bi = dim(Hi DR(M)) are known as the Betti numbers of M. For example, χ(S2) = 2. It turns out that if M is an odd dimensional manifold, then χ(M) = 0. This explains partially why the Euler class is only defined for even dimensional bundles. The Generalized Gauss-Bonnet theorem (or Gauss-Bonnet-Chern theorem) is a general-ization of the Gauss-Bonnet theorem for surfaces. In the general form stated below it was first proved by Allendoerfer and Weil (1943), and Chern (1944). 11.14. PROBLEMS 521 Theorem 11.42. (Generalized Gauss-Bonnet Formula) Let M be an orientable, smooth, compact manifold of dimension 2m. For every metric connection ∇on TM, (in particular, the Levi-Civita connection for a Riemannian manifold), we have Z M eu(R∇) = χ(M). A proof of Theorem 11.42 can be found in Madsen and Tornehave (Chapter 21), but beware of some sign problems. The proof uses another famous theorem of differential topology, the Poincar´ e-Hopf theorem. A sketch of the proof is also given in Morita , Chapter 5. Theorem 11.42 is remarkable because it establishes a relationship between the geometry of the manifold (its curvature) and the topology of the manifold (the number of “holes”), somehow encoded in its Euler characteristic. Characteristic classes are a rich and important topic and we’ve only scratched the surface. We refer the reader to the texts mentioned earlier in this section as well as to Bott and Tu for comprehensive expositions. 11.14 Problems Problem 11.1. Complete the proof of Proposition 11.4. In particular show that ∇= X α fα∇α is a connection on ξ. Problem 11.2. Prove Proposition 11.8. Problem 11.3. Show that the bilinear map ⊼: Ar(B) × As(η) − →Ar+s(η), as defined in Definition 11.11 (with ξ = B × R), satisfies the following properties: (ω ∧τ) ⊼θ = ω ⊼(τ ⊼θ) (3) 1 ⊼θ = θ, for all ω ∈Ai(B), τ ∈Aj(B) with i + j = r, θ ∈As(ξ), and where 1 denotes the constant function in C∞(B) with value 1. Problem 11.4. Complete the partition of unity argument for Proposition 11.17. 522 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Problem 11.5. Show that the pairing (−, −): Ai(ξ) ⊗Aj(ξ∗) ⊼ − →Ai+j(ξ ⊗ξ∗) − →Ai+j(B) given by (ω ⊗s1, η ⊗s∗ 2) = (ω ∧η) ⊗⟨⟨s1, s∗ 2⟩⟩, where ω ∈Ai(B), η ∈Aj(B), s1 ∈Γ(ξ), s∗ 2 ∈Γ(ξ∗), is non-degenerate. Problem 11.6. Let ξ is a complex vector bundle with connection matrix ω. Consider the bundle ξ ∗, which is the bundle whose fibre over b ∈B consist of all conjugate-linear forms over π−1(b). (i) Show that the evaluation pairing ⟨⟨s, θ⟩⟩is conjugate-linear in s. (ii) Show that ω∗= −ω⊤, where ω∗is the connection matrix of ξ ∗over U. Problem 11.7. Let ξ is a complex vector bundle. Consider the conjugate bundle ξ, which is obtained from ξ by redefining the vector space structure on each fibre π−1(b), with b ∈B, so that (x + iy)v = (x −iy)v, for every v ∈π−1(b). If ω is the connection matrix of ξ over U, prove that ω is the connection matrix of ξ over U. Problem 11.8. Let θX is the one-form given by θX = iXg; that is, (iXg)(Y ) = g(X, Y ) for all X, Y ∈X(M). If LXg is the Lie derivative of the symmetric (0, 2) tensor g, defined so that (LXg)(Y, Z) = X(g(Y, Z)) −g(LXY, Z) −g(Y, LXZ), show that the Levi-Civita connection is defined implicitly by the formula 2g(∇XY, Z) = (LY g)(X, Z) + (dθY )(X, Z). Hint. See Petersen , Chapter 2, Theorem 1. Problem 11.9. Investigate the following sources, namely Madsen and Tornehave (Lemma 18.2), Morita (Proposition 5.28), and Milnor and Stasheff (Appendix C), and prove Theorem 11.27. Problem 11.10. Complete the proof sketch of Theorem 11.29. Hint. See Milnor and Stasheff, Appendix C, Lemma 6, Madsen and Tornehave , Appendix B, or Morita , Theorem 5.26. 11.14. PROBLEMS 523 Problem 11.11. Let X be an n × n matrix. Recall that si(X) = tr(Xi). Prove Newton’s formula for symmetric polynomial namely: si(X) −σ1(X)si−1(X) + σ2(X)si−2(X) + · · · + (−1)i−1σi−1(X)s1(X) + (−1)iiσi(X) = 0, with 1 ≤i ≤n. See Madsen and Tornehave , Appendix B, or Morita , Exercise 5.7. Problem 11.12. Prove Proposition 11.31. Problem 11.13. Complete the proof details of Proposition 11.33. Problem 11.14. (i) Show that Pf(A) is independent of the basis chosen. (ii) If A is the block diagonal matrix D, a show that ω(D) = −(a1e1 ∧e2 + a2e3 ∧e4 + · · · + ane2n−1 ∧e2n) and that n ^ ω(D) = (−1)nn! a1 · · · an e1 ∧e2 ∧· · · ∧e2n, and so Pf(D) = (−1)na1 · · · an. Problem 11.15. Use Definition 11.25 to that Pf(A) is the unique polynomial with integer coefficients such that Pf(A)2 = det(A) and Pf(diag(S, . . . , S)) = +1, where S =  0 1 −1 0  . Hint. See Milnor and Stasheff, Appendix C, Lemma 9. Problem 11.16. Show that Pf(A) is explicitly calculated as Pf(A) = 1 2nn! X σ∈S2n sgn(σ) n Y i=1 aσ(2i−1) σ(2i). Problem 11.17. Prove Propositions 11.37 and 11.38. Hint. See Madsen and Tornehave , Chapter 19 or Milnor and Stasheff, Appendix C. 524 CHAPTER 11. CONNECTIONS AND CURVATURE IN VECTOR BUNDLES Chapter 12 Connections and Curvature in Principal Bundles In this very short chapter we explain, mostly without proofs, how to define the notion of connection and curvature on a principal bundle ξ. Connections on a principal bundle differ in flavor from connections on vector bundles as discussed in Chapter 11 because connections on a vector bundle define the notion of covariant derivative (on sections), whereas connections on a principal bundle reflect the idea of relating two fibres Eb0 and Eb1 using parallel transport, in terms of the horizontal lift of a curve in the base manifold (where b0 and b1 are points in the base manifold). So a connection in a principal bundle does not define a notion of covariant derivative (however, it does if the principal bundle is a frame bundle associated with a vector bundle). But as in the case of vector bundles, a connection is a way of measuring the twisting of the fibres, which is technically achieved by the notion of curvature. Since a vector bundle gives rise to a principal bundle, namely the corresponding frame bundle (obtained by replacing the fibre Rn by the group GL(n, R), see Definition 10.18), it is natural to wonder what is the relationship between connections on a vector bundle and connections on the associated frame bundle. The anwser is that there is a bijection between these two families of connections. A connection on a principal bundle ξ with structure group G and base manifold B induces a notion of curvature. Chern and Weil defined an algebra W(g) called the Weil algebra and a homomorphism w from W(g) to the space of differential forms A∗(ξ) on ξ (here g is the Lie algebra of G). This homomorphism depends on the choice of a connection on ξ. However, one can define a subset I(G) of W(g) consisiting of basic forms, and then it is a remarkable result of Chern and Weil that there is a homomorphism w: I(G) →H• DR(B; R) (where B is the base manifold of the principal bundle ξ) which is independent of the connection chosen on ξ. This homomorphism assigns characteristic classes to the elements in I(G), and these characteristic classes are de Rham cohomology classes (more specifically, if f ∈Ik(G), then w(f) ∈H2k DR(B; R)). Our presentation relies heavily on Morita and Kobayashi and Nomizu . Exposi-525 526 CHAPTER 12. CONNECTIONS AND CURVATURE IN PRINCIPAL BUNDLES tions of this material from the physics point of view can be found in Choquet–Bruhat and DeWitt–Morette (Chapter Vbis) and Nakahara (Chapters 9-11). The point of view of physics sheds some light on the rather obscure motivations behind this material and we contend that it makes it even more beautiful. Some of it is actually used in control theory and robotics. The Chern–Weil theory is very nicely presented in Morita , Chapter 6, in particular Section 6.3-6.6. The reader may finally come to appreciate the notion of a vector-valued differential form. 12.1 Connections and Connection Forms in Principal Bundles The goal of this section is to define the notion of a connection on a principal bundle. It turns out that it is technically more convenient to use the definition of a principal bundle in terms of a free right action as in Proposition 10.21 so for the reader’s convenience we provide this definition. Definition 12.1. A principal G-bundle is a quadruple ξ = (E, π, E/G, G), where E be a smooth manifold, G is Lie group, and ·: E × G →E is a smooth right action of G on E satisfying the following properties: (1) The right action of G on E is free; (2) The orbit space B = E/G is a smooth manifold under the quotient topology, and the projection π: E →E/G is smooth; (3) There is some open cover U = (Uα)α∈I of B = E/G and a family ψ = (ψα)α∈I of diffeomorphisms called (local) trivializations ψα : π−1(Uα) →Uα × G such that (a) (local triviality) the diagram π−1(Uα) π $ ψα / Uα × G pr1 { Uα commutes. 12.1. CONNECTIONS AND CONNECTION FORMS IN PRINCIPAL BUNDES 527 (b) Every map ψα : π−1(Uα) →Uα×G is an equivariant diffeomorphism, which means that ψα(z · h) = ψα(z) · h for all z ∈π−1(Uα) and all h ∈G, where the right action of G on Uα × G is (x, h1) · h = (x, h1h). Observe that if ψα(z) = (x, h1), then since ψα(z) · h = (x, h1h), we have pr1(ψα(z) · h) = pr1(ψα(z)) = x = π(z). Recall that the action ·: E × G →E is free if it acts without fixed points, that is, for every h ∈G, if h ̸= 1, then x · h ̸= x for all x ∈E. By Conditions (a) and (b) and the definition of the right action of G on Uα × G, for all z ∈π−1(Uα) and all h ∈G, we have π(z · h) = pr1(ψα(z · h)) = pr1(ψα(z) · h) = pr1(ψα(z)) = π(z), so for any x ∈B = E/G and any z ∈Ex = π−1(x), we have z ·h ∈Ex. In fact, for any z ∈Ex, it is easily shown that Ex = {z · h | h ∈G}, namely the orbits of the right action of G on E are the fibres Ex, with x ∈B. Since the action of G on E is free, the action of G on Ex is also free. The restriction of ψα to Ex for any x ∈Uα is a diffeomorphism from Ex onto {x} × G given by ψα(z) = (x, ψα,x(z)), where ψα,x : Ex →G is a diffeomorphism between Ex and G. For all α, β such that Uα ∩Uβ ̸= ∅, for every x ∈Uα ∩Uβ, we have a diffeomorphism ψα,x ◦ψ−1 β,x : G − →G, which yields the map gαβ : Uα ∩Uβ →Diff(G) called a transition map given by gαβ(x) = ψα,x ◦ψ−1 β,x, x ∈Uα ∩Uβ. Intuitively, the transition functions express how the fibre Ex twists as x moves in Uα ∩Uβ. From the definition above, the isomorphism ψα ◦ψ−1 β : (Uα ∩Uβ) × G →(Uα ∩Uβ) × G is given by (ψα ◦ψ−1 β )(x, h) = (x, gαβ(x)(h)), x ∈Uα ∩Uβ, h ∈G. A priori, the map gαβ(x) is a diffeomorphism of the Lie group G, but because the transition maps ψα are equivariant, it is shown in Proposition 10.21 that gαβ(x) is the left translation by gαβ(x)(1) ∈G, that is, gαβ(x)(h) = gαβ(x)(1)h, x ∈Uα ∩Uβ, h ∈G. 528 CHAPTER 12. CONNECTIONS AND CURVATURE IN PRINCIPAL BUNDLES Since the group of left translations of G (the maps Lh : G →G given by Lh(h1) = hh1 (h, h1 ∈G)) is isomorphic to G, we usually view the map gαβ(x) as the element gαβ(x)(1) of G, and thus we view the transition function gαβ as a map gαβ : Uα ∩Uβ →G such that (ψα ◦ψ−1 β )(x, h) = (x, gαβ(x)h), x ∈Uα ∩Uβ, h ∈G. Definition 12.2. Given a principal bundle ξ, for any fixed z ∈E, we have the map σz : G → E given by σz(g) = z · g, g ∈G. From the discussion above, the map σz : G →E is a diffeomorphism from G to the fibre Eπ(z) and the derivative of this map defines a linear map d(σz)e : g →TzE. The family of maps σz allows us to associate a vector field A∗on TE to every A ∈g defined as follows: A∗ z = d(σz)e(A), z ∈E, A ∈g. The vector field A∗is called the fundamental vector field corresponding to A by Kobayashi and Nomizu , Chapter I, Section 5. Proposition 12.1. The following properties hold. (1) For any A ∈g, the vector field A∗can also be defined as follows: for every z ∈E, A∗ z = d dt (z · exp(tA)) t=0 . In other words, the vector field A∗generates the 1-parameter group of diffeomorphisms t 7→Rexp(tA). (2) The map d(σz)e : g →TEz is injective. (3) For all A, B ∈g, we have [A, B]∗= [A∗, B∗]. In other words, the map A 7→A∗is Lie algebra homomorphism. Proof. (1) For any A ∈g, consider the curve in G given by c(t) = exp(tA). Observe that c(0) = e and c′(0) = A. Since z · exp(tA) = σz(exp(tA)), using the chain rule, we have d dt (z · exp(tA)) t=0 = (σz(exp(tA)))′(0) = (σz(c(t)))′(0) = d(σz)c(0)(c′(0)) = d(σz)e(A) = A∗ z, 12.1. CONNECTIONS AND CONNECTION FORMS IN PRINCIPAL BUNDES 529 as claimed. (2) It suffices to prove that the kernel of d(σz)e is reduced to the zero vector. The proof uses the chain rule in a tricky way as explained in Duistermaat and Kolk ; see the proof of Theorem 1.11.4, Chapter I. If d(σz)e(A) = 0, then for all t0, s ∈R we have d dt (z · exp(tA)) t=t0 = d ds (z · exp((s + t0)A)) s=0 = d ds ((z · exp(sA)) · exp(t0A)) s=0 = (Rexp(t0A)(σz(exp(sA))))′(0) = d(Rexp(t0A))z(d(σz)e(A)) = 0, where Rg is the map Rg : E →E defined by Rgz = z · g, z ∈E, g ∈G. Since d dt (z · exp(tA)) t=t0 = 0 for all t0 ∈R, the map t 7→z · exp(tA) is constant, and since it has the value z for t = 0, we have z · exp(tA) = z for all t. Since the action of G on E is free, we must have exp tA = e for all t, and by taking the derivative at t = 0 we get A = 0. Property (3) is proven in Kobayashi and Nomizu , Chapter I, Proposition 4.1. Remark: It is also shown in Kobayashi and Nomizu that the map A 7→A∗is injective if the action is effective; see , Chapter I, Proposition 4.1 The following property is the key to the definition of connections in terms of g-valued differential forms. Proposition 12.2. For every z ∈E we have the following exact sequence: 0 / g (dσz)e / TzE dπz / Tπ(z)B / 0. This means that d(σz)e is injective, dπz is surjective, and that Ker dπz = Im d(σz)e. In particular, Ker dπz = Im d(σz)e is isomorphic to g. Proof. We already know that d(σz)e is injective. For any x ∈B = E/G, because Ex = {z ·h | h ∈G} for any z ∈Ex, we have π ◦σz(h) = π(z · h) = x, h ∈G, a constant map, so by taking the derivative we obtain dπz ◦d(σz)e = 0. 530 CHAPTER 12. CONNECTIONS AND CURVATURE IN PRINCIPAL BUNDLES It follows that Im d(σz)e ⊆Ker dπz. By taking derivatives, the local triviality diagram π−1(Uα) π $ ψα / Uα × G pr1 { Uα implies that for any z ∈π−1(Uα) we have the equation dπz = d(pr1)ψα(z) ◦d(ψα)z, and if we write ψα(z) = (b, h) with b = π(z), we have isomorphisms TzE ∼ = T(b,h)(Uα × G) ∼ = TbB × ThG ThG ∼ = g Ker dπz ∼ = ThG ∼ = g Im dπz = TbB. The last equation shows that dπz is surjective. Since d(σz)e is injective, dim g = dim Im d(σz)e, and since dim Ker dπz = dim g and Im d(σz)e ⊆Ker dπz we deduce that Ker dπz = Im d(σz)e. This completes the proof that we have an exact sequence. If we write Vz = Ker dπz = Im d(σz)e, the tangent vectors in Vz ⊆TzE are called vertical vectors. The idea due to Ehresmann is to split each tangent space TzE into a direct sum TzE = Vz ⊕Hz = Ker dπz ⊕Hz = Im d(σz)e ⊕Hz, where the vectors in Hz are called horizontal vectors, and since G acts on E, we require that G also acts nicely on the spaces Hz, namely (dRg)zHz = Hz·g, z ∈E, g ∈G. Of course we also require that Hz varies diffentiably in z. Such a choice of horizontal subspaces is called a connection (or Ehresmann connection). The exact sequence of Proposition 12.2 makes it possible to give an alternative definition of a connection in terms of certain g-valued linear maps. The idea is to define the horizontal subspaces as the kernels of certain linear maps. Indeed, if we have a direct sum decomposition TzE = Vz ⊕Hz = Ker dπz ⊕Hz = Im d(σz)e ⊕Hz, 12.1. CONNECTIONS AND CONNECTION FORMS IN PRINCIPAL BUNDES 531 for any Z ∈TzE, if we write Z = Zv +Zh with Zv ∈Vz and Zh ∈Hz, since d(σz)e is injective, we can define a linear map ωz : TzE →g (a retraction) by ωz(Z) = d(σz)−1 e (Zv), so that ωz ◦d(σz)e = idg Ker ωz = Hz. Conversely, given a linear map ωz : TzE →g such that ωz ◦d(σz)e = idg, we claim that TzE = Im d(σz)e ⊕Ker ωz = Ker dπz ⊕Hz. First, observe that ωz(Z −d(σz)e(ωz(Z))) = ωz(Z) −ωz(d(σz)e(ωz(Z))) = ωz(Z) −ωz(Z) = 0, namely Z −d(σz)e(ωz(Z)) ∈Ker ωz. Since Z = d(σz)e(ωz(Z)) + Z −d(σz)e(ωz(Z)), we have TzE = Im d(σz)e + Ker ωz. Second, if Y ∈Im d(σz)e ∩Ker ωz, then ωz(Y ) = 0 and Y = d(σz)e(A) for some A ∈g. Since ωz ◦d(σz)e = idg, we have 0 = ωz(Y ) = ωz(d(σz)e(A)) = A, so A = 0, and thus Y = 0 since Y = d(σz)e(A). Therefore the choice of a horizontal subspace Hz such that TzE = Ker dπz ⊕Hz is equiv-alent to the choice of a linear map ωz : TzE →g such that ωz ◦d(σz)e = idg, with Hz = Ker ωz. By Definition 12.2, observe that the condition ωz ◦d(σz)e = idg, z ∈E, is equivalent to ωz(A∗) = A, A ∈g, z ∈E. Also note that each vector A∗ z is vertical and that the restriction of dπz to Hz is an isomor-phism between Hz and Tπ(z)B. The above discussion yields a definition of a connection in terms of g-valued one-forms. 532 CHAPTER 12. CONNECTIONS AND CURVATURE IN PRINCIPAL BUNDLES Definition 12.3. (Version I) Let ξ = (E, π, E/G, G) be a principal G-bundle and write B = E/G. A connection form on ξ is a family of linear maps ωz : TzE →g satisfying the following conditions: (1) We have ωz(A∗) = A for all A ∈g and all z ∈E, or equivalently ωz ◦d(σz)e = idg, z ∈E. (2) With Hz = Ker ωz, we have (dRg)zHz = Hz·g, z ∈E, g ∈G. Furthermore, the map z 7→ωz is smooth. In terms of Definition 4.20, ω = (ωz)z∈E ∈A1(E; g), that is, ω is a smooth g-valued one-form on E. As explained before, we have a direct sum decomposition of the tangent space TzE into vertical vectors (in Ker dπz) and horizontal vectors (in Hz = Ker ωz), TzE = Ker dπz ⊕Hz = Im d(σz)e ⊕Hz. It is often technically more convenient to restate Condition (2) directly in terms of ωz. Proposition 12.3. Condition (2) of Definition 12.3 is equivalent to the condition R∗ gω = Ad(g−1)ω, g ∈G, or more explicitly, ωz·g(d(Rg)z(Z)) = Ad(g−1)(ωz(Z)), z ∈E, Z ∈TzE. A proof of Proposition 12.3 is given in Kobayashi and Nomizu ; see Chapter II, Proposition 1.1. In view of Proposition 12.3 we have the following equivalent definition of a connection form which appears to be used most often in the literature. Definition 12.4. (Version II) Let ξ = (E, π, E/G, G) be a principal G-bundle and write B = E/G. A connection form on ξ is a family of linear maps ωz : TzE →g satisfying the following conditions: (1) We have ωz(A∗) = A for all A ∈g and all z ∈E, or equivalently ωz ◦d(σz)e = idg, z ∈E. (2) R∗ gω = Ad(g−1)ω, g ∈G, or more explicitly, ωz·g(d(Rg)z(Z)) = Ad(g−1)(ωz(Z)), z ∈E, Z ∈TzE. 12.1. CONNECTIONS AND CONNECTION FORMS IN PRINCIPAL BUNDES 533 Furthermore, the map z 7→ωz is smooth. In terms of Definition 4.20, ω = (ωz)z∈E ∈A1(E; g), that is, ω is a smooth g-valued one-form on E. Example 12.1. Consider the trivial principal bundle E = B × G, with π: E →B the projection onto B (in fact, π = pr1). A special connection form on E arises from the Maurer– Cartan form on G, but here to conform to the customary notation, instead of denoting it ωMC, we denote it by ω0, that is, we define the g-valued one-form ω0 ∈A1(G; g) by (ω0)g(X) = d(L−1 g )g, (X) g ∈G, X ∈TgG. The form ω0 brings back a vector X ∈TgG to g = TeG; see Definition 4.22. Then we define the g-valued one-form ωE,MC on E by ωE,MC = pr∗ 2 ω0, where pr2 : B × G →G is the second projection. Thus, for every (b, g) ∈B × G, (ωE,MC)(b,g) = d(Lg−1 ◦pr2)(b,g). It is easily verified that ωE,MC is a connection form on E. The connection form ωE,MC is called the Maurer–Cartan connection or flat connection or canonical connection on E = B × G. Using Example 12.1 and a partition of unity argument the following result can be shown. See Morita , Chapter 6, Proposition 6.3. Proposition 12.4. Every principal bundle ξ admits a connection form. For every open subset Uα in the open cover of B we have the Maurer–Cartan connection form on Uα × G ≃π−1(Uα), and we glue these connection forms using a partition of unity. Interestingly the set of connection forms on a trivial principal bundle E = B × G is in bijection with the set of g-valued one-forms in A1(B; g). Let i: B →E be the injection i(b) = (b, e). Proposition 12.5. Given a trivial principal bundle E = B × G, with π: B × G →B, for any one-form ω ∈A1(B; g), if e ω ∈A1(E; g) is given by e ω(b,g) = Ad(g−1) ◦ωb ◦dπ(b,g), then the map ω 7→e ω + ωE,MC is a bijection between the set of one-forms in A1(B; g) and the set of connection forms in A1(E; g). It is easy to see that e ω is the unique form in A1(E; g) such that i∗(e ω) = ω and 534 CHAPTER 12. CONNECTIONS AND CURVATURE IN PRINCIPAL BUNDLES (1) e ωz(Z) = 0 for all vertical Z ∈TzE and all z ∈E. (2) R∗ g e ω = Ad(g−1) e ω for all g ∈G. We leave it as an exercise to prove Proposition 12.5 using the above facts. If ξ is a vector bundle with typical fibre V = Rn, for any open subset Uα in the open cover of the base space B, recall from Section 11.3 that to every local trivialization ϕα : π−1(Uα) → Uα × Rn and any basis (v1, . . . , vn) in Rn, we associate the frame (s1, . . . , sn) over Uα, given by si(b) = ϕ−1 α (b, vi), b ∈Uα. (∗) Then for any connection ∇on this vector bundle, each ∇si can be written as ∇si = n X j=1 ωji ⊗sj, for some n × n matrix ω = (ωij) of one-forms ωij ∈A1(Uα). This matrix ωα = (ωij) defines an Mn(R)-valued one form in A1(Uα; g). If G = GL(n, R), then g(n, R) = Mn(R), so locally on Uα, a connection ∇on the trivial vector bundle ξ is equivalent to a one-form ωα ∈A1(Uα; g(n, R)). Since the frame bundle associated with the trivial vector bundle Uα × Rn is the trivial principal bundle Uα × GL(n, R), Proposition 12.5 and the preceding paragraph implies that locally on Uα, there is a bijection between the space of connections on the trivial vector bundle Uα × Rn and connection forms on the principal bundle Uα × GL(n, R). We can also define connection forms locally. Definition 12.5. Let ξ = (E, π, E/G, G) be a principal G-bundle and write B = E/G. For each α, given the local trivialization ψα : π−1(Uα) →Uα × G, let sα : Uα →E be the section given by sα(b) = ψ−1 α (b, e), b ∈Uα. For any connection form ω ∈A1(E; g), for every α, let ωα ∈A1(Uα; g) be the one-form given by ωα = s∗ αω for all α. For all α, β, recall that we view the transition function gαβ as a map gαβ : Uα ∩Uβ →G so that ψα ◦ψ−1 β (b, h) = (b, gαβ(b)h), b ∈Uα ∩Uβ, h ∈G, as explained after Definition 12.1. Also, ω0 is the Maurer–Cartan form on G. The following result is proven in Kobayashi and Nomizu ; see Chapter II, Proposition 1.4. 12.1. CONNECTIONS AND CONNECTION FORMS IN PRINCIPAL BUNDES 535 Proposition 12.6. Let ξ = (E, π, E/G, G) be a principal G-bundle and write B = E/G. For any connection form ω ∈A1(E; g), the forms ωα satisfy the following conditions for all α, β: ωβ(b) = Ad(g−1 αβ(b))(ωα(b)) + (g∗ αβ ω0)(b), b ∈Uα ∩Uβ. Conversely, for any family (ωα) of one forms ωα ∈A1(Uα; g), if these one-forms satisfy the above condition, then there is a unique connection form ω ∈A1(E; g) such that ωα = s∗ αω for all α. Definition 12.6. A family (ωα) of one forms ωα ∈A1(Uα; g) satisfying the property of Proposition 12.6 is called a family of local connection forms. Remark: In physics (Yang–Mills theories) the one-forms ωα are called gauge potentials, and the local trivialization maps ψα are called local gauges. Proposition 12.6 and the fact that locally there is an equivalence between connections on a trivial vector bundle and connection forms on the corresponding frame bundle can be used to prove that for any vector bundle, there is an equivalence between vector bundle connections and connection forms on the corresponding frame bundle. As in the case of vector bundles, a connection on a principal bundle can be used to define a notion of parallel transport. First we define the horizontal lift of a curve in the base space. Definition 12.7. Let ξ = (E, π, E/G, G) be a principal G-bundle with base space B = E/G and let ω be a connection form on ξ. For every piecewise smooth curve c: [a, b] →B in the base space B a curve e c: [a, b] →E is a horizontal lift of c if c(t) = π(e c(t)) for all t ∈[a, b] and if every tangent vector (e c)′(t) is a horizontal vector in He c(t). The following result is proven in Morita (Chapter 6, Proposition 6.36 and Kobayashi and Nomizu ; Chapter II, Section 3, Proposition 3.1.. Proposition 12.7. Let ξ = (E, π, E/G, G) be a principal G-bundle with base space B = E/G and let ω be a connection form on ξ. For every piecewise smooth curve c: [a, b] →B in the base space B and every point z0 in the fibre Ec(a), there is a unique horizontal lift e c of c such that e c(a) = z0. Proposition 12.7 allows the definition of the parallel transport along the curve c. This is the map from the fibre Ec(a) to the fibre Ec(b) which sends any point z0 ∈Ec(a) to the point e c(b) ∈Ec(b). The notion of parallel transport can be used to define holonomy groups but we will not discuss this here and instead refer the reader to Morita , Chapter 6, Section 3, and Kobayashi and Nomizu ; see Chapter II, Section 4. 536 CHAPTER 12. CONNECTIONS AND CURVATURE IN PRINCIPAL BUNDLES Given a principal bundle ξ = (E, π, E/G, G) with B = E/G and a representation ρ: G → GL(m, R) the construction of Proposition 10.25 (the Borel construction) yields a vector bundle ξ[Rm] = (E ×G Rm, p, B, Rm, G), where G acts on the typical fibre Rm in terms of the representation ρ. This is a generalization of the notion of vector bundle given in Definition 10.13 as explained just after Definition 10.14. In this more general situation the transition functions of the vector bundle ξ[Rm] are linear maps in the subgroup ρ(G) ⊆GL(m, R). In Chapter II, Section 7, of Kobayashi and Nomizu , given a connection form in ξ, it is expained how to define a notion of parallel transport in ξ[Rm], and then in Chapter III, Section 1, to define the notion of covariant derivative ∇Xs of a section s of the vector bundle ξ[Rm] in the direction X ∈TxB, with x ∈B. Then Kobayashi and Nomizu show that this notion of covariant derivative satisfies the axioms of the second version of a connection on a vector bundle given in Definition 11.1. The above constructions are also presented in a more concrete way in Nakahara (Section 10.4) and in Choquet–Bruhat and DeWitt–Morette (Chapter Vbis, Section 3). We are back where we started in Chapter 11. Connections on a principal bundle are the more general notion of connection, but for manifolds and vector bundles, it is preferable to define a notion of connection that does not depend on the notion of parallel transport. The next important concept is the curvature form induced by a connection form. 12.2 Curvature Form Given a connection form ω ∈A1(E; g) on a principal bundle ξ, the exterior derivative dω is a 2-form, dω ∈A2(E; g). If ξ is the trivial principal bundle E = B × G and if ω = ωE,MC is the Maurer–Cartan form, then Proposition 4.32 implies that dω = −1 2[ω, ω]. However, in general, the above equation fails. The quantity dω + 1 2[ω, ω] is 2-form in A2(E; g) that quantifies this failure; it is the curvature form of ω. Definition 12.8. Let ξ = (E, π, E/G, G) be a principal G-bundle and write B = E/G. For any connection form ω ∈A1(E; g), the curvature form Ωω ∈A2(E; g) of ω, for notational simplicity denoted by Ω, is defined by Ωz(Y, Z) = dωz(Y, Z) + 1 2[ωz(Y ), ωz(Z)], z ∈E, Y, Z ∈TzE, which is often abbreviated as the equation dω = −1 2[ω, ω] + Ω and called the structure equation of E. Cartan. 12.2. CURVATURE FORM 537 If we pick a basis X1, . . . , Xm of the Lie algebra g, then the connection form ω ∈A1(E; g) can written as ω = m X i=1 ωiXi, where the ωi are ordinary one-forms ωi ∈A1(E), and the curvature form Ω∈A2(E; g) can written as Ω= m X i=1 ΩiXi, where the Ωi are ordinary 2-forms Ωi ∈A2(E). Recall that the Lie brackets [Xi, Xj] are defined in terms of the basis X1, . . . , Xm in terms of the structure constants ck ij ∈R by the equations [Xi, Xj] = m X k=1 ck ijXk, 1 ≤i, j, k ≤m. Then it is easy to see that the structure equation is equivalent to the system of m equations dωi = −1 2 i X j,k=1 ci jkωj ∧ωk + Ωi, 1 ≤i ≤m; compare Proposition 4.31. The following theorem is proven in Morita , Chapter 6, Proposition 6.39, and Kobayashi and Nomizu , Chapter II, Section 5 (in particular, Theorem 5.2 and Theorem 5.4). Theorem 12.8. Let ξ = (E, π, E/G, G) be a principal G-bundle and write B = E/G. For any connection form ω ∈A1(E; g), the curvature form Ωof ω satisfies the following properties: (1) For all g ∈G, R∗ g Ω= Ad(g−1) Ω. (2) For every z ∈E and all Y, Z ∈TzE, we have Ωz(Y, Z) = dωz(Yh, Zh), where Yh, Zh ∈Ker ωz are the horizontal components of Y and Z. (3) If Y, Z ∈Ker ωz are horizontal tangent vectors, then Ωz(Y, Z) = −1 2ωz([Y, Z]). 538 CHAPTER 12. CONNECTIONS AND CURVATURE IN PRINCIPAL BUNDLES (4) (Bianchi’s identity) dΩ= [Ω, ω]. In particular, dΩz(X, Y, Z) = 0 for all horizontal vectors X, Y, Z ∈TzE. If (ωα) is a family a local connection forms defining a connection form ω, then it can be shown that the curvature form Ωof ω is defined by the family (Ωα) of 2-forms Ωα ∈A2(Uα; g) given by Ωα = dωα + 1 2[ωα, ωα]. Remark: In physics, the 2-form Ωα is called a field strength in the gauge ψα. Definition 12.9. Let ξ = (E, π, E/G, G) be a principal G-bundle. A connection ω on ξ is flat if the curvature form Ωof ω is identically zero, equivalently dω + 1 2[ω, ω] ≡0. For example, the Maurer–Cartan connection on a trivial principal bundle E = B × G is flat. There are various conditions equivalent to flatness; see Kobayashi and Nomizu , Chapter II, Section 9, in particular Theorem 9.1 that we state below. Proposition 12.9. Let ξ = (E, π, E/G, G) be a principal G-bundle and write B = E/G. A connection ω on ξ is flat iffthere is there an open cover (Uα) of B and some trivializing maps ψα : π−1(Uα) →Uα × G such that the restriction of ω to π−1(Uα) is equal to ψ∗ αωπ−1(Uα),MC. The next step would be to define the Weil algebra and to discuss basic forms and the Weil homomorphism, which allows the definition of the characteristic classes of a principal bundle. We refer to Morita , Chapter 6, Sections 4-5, for a complete exposition. When ξ is a principal bundle (G, π, G/H, H) where G is a Lie group and H is a closed subgroup of G, it is interesting to know when the homogeneous space B = G/H is reductive, which means that there is a subspace m of g (where m is usually not a Lie subalgebra) such that g = h ⊕m Adh(m) = m, h ∈H. See Gallier and Quaintance , Chapter 23, Definition 23.8. Theorem 11.1 from Kobayashi and Nomizu (Chapter II) gives an interesting criterion for (G, π, G/H, H) to be reductive in terms of connection forms. Proposition 12.10. Let G be a Lie group, H a closed subgroup of G, and consider the principal bundle (G, π, G/H, H). 12.2. CURVATURE FORM 539 (1) If the homogeneous space G/H is reductive, then the h-component of the Maurer– Cartan form ω0 on G with respect to the direct sum h ⊕m defines a connection form on (G, π, G/H, H) which is invariant by the operations Lg on G/H, with Lg(g1H) = (gg1)H (g, g1 ∈G). (2) Conversely, if there is a connection form ω on (G, π, G/H, H) and if ω is invariant by the operations Lg, then G/H is reductive. (3) The curvature form Ωassociated with a connection form invariant by the operations Lg is given by Ω(Y, Z) = −1 2[Y, Z]h, where Y and Z are left-invariant vector fields on G with values in m and [Y, Z]h is the h-component of [Y, Z], which takes values in g. More results along this vein are discussed in Kobayashi and Nomizu (Chapter II, Section 11). 540 CHAPTER 12. CONNECTIONS AND CURVATURE IN PRINCIPAL BUNDLES Chapter 13 Clifford Algebras, Clifford Groups, and the Groups Pin(n) and Spin(n) 13.1 Introduction: Rotations As Group Actions The main goal of this chapter is to explain how rotations in Rn are induced by the action of a certain group Spin(n) on Rn, in a way that generalizes the action of the unit complex numbers U(1) on R2, and the action of the unit quaternions SU(2) on R3 (i.e., the action is defined in terms of multiplication in a larger algebra containing both the group Spin(n) and Rn). The group Spin(n), called a spinor group, is defined as a certain subgroup of units of an algebra Cln, the Clifford algebra associated with Rn. Furthermore, for n ≥3, we are lucky, because the group Spin(n) is topologically simpler than the group SO(n). Indeed, for n ≥3, the group Spin(n) is simply connected (a fact that it not so easy to prove without some machinery), whereas SO(n) is not simply connected. Intuitively speaking, SO(n) is more twisted than Spin(n). In fact, we will see that Spin(n) is a double cover of SO(n). Since the spinor groups are certain well chosen subgroups of units of Clifford algebras, it is necessary to investigate Clifford algebras to get a firm understanding of spinor groups. This chapter provides a tutorial on Clifford algebra and the groups Spin and Pin, including a study of the structure of the Clifford algebra Clp,q associated with a nondegenerate symmetric bilinear form of signature (p, q) and culminating in the beautiful “8-periodicity theorem” of ´ Elie Cartan and Raoul Bott (with proofs). We also explain when Spin(p, q) is a double-cover of SO(p, q). The reader should be warned that a certain amount of algebraic (and topological) background is expected. This being said, perseverant readers will be rewarded by being exposed to some beautiful and nontrivial concepts and results, including ´ Elie Cartan and Raoul Bott “8-periodicity theorem.” Going back to rotations as transformations induced by group actions, recall that if V is a vector space, a linear action (on the left) of a group G on V is a map α: G × V →V satisfying the following conditions, where, for simplicity of notation, we denote α(g, v) by g · v: 541 542 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN (1) g · (h · v) = (gh) · v, for all g, h ∈G and v ∈V ; (2) 1 · v = v, for all v ∈V , where 1 is the identity of the group G; (3) The map v 7→g · v is a linear isomorphism of V for every g ∈G. For example, the (multiplicative) group U(1) of unit complex numbers acts on R2 (by identifying R2 and C) via complex multiplication: For every z = a + ib (with a2 + b2 = 1), for every (x, y) ∈R2 (viewing (x, y) as the complex number x + iy), z · (x, y) = (ax −by, ay + bx). Now every unit complex number is of the form cos θ + i sin θ, and thus the above action of z = cos θ + i sin θ on R2 corresponds to the rotation of angle θ around the origin. In the case n = 2, the groups U(1) and SO(2) are isomorphic, but this is an exception. To represent rotations in R3 and R4, we need the quaternions. For our purposes, it is convenient to define the quaternions as certain 2 × 2 complex matrices. Let 1, i, j, k be the matrices 1 = 1 0 0 1  , i = i 0 0 −i  , j =  0 1 −1 0  , k = 0 i i 0  , and let H be the set of all matrices of the form X = a1 + bi + cj + dk, a, b, c, d ∈R. Thus, every matrix in H is of the form X =  a + ib c + id −(c −id) a −ib  , a, b, c, d ∈R. The quaternions 1, i, j, k satisfy the famous identities discovered by Hamilton: i2 = j2 = k2 = ijk = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j. As a consequence, it can be verified that H is a skew field (a noncommutative field) called the quaternions. It is also a real vector space of dimension 4 with basis (1, i, j, k); thus as a vector space, H is isomorphic to R4. The unit quaternions are the quaternions such that det(X) = a2 + b2 + c2 + d2 = 1. Given any quaternion X = a1 + bi + cj + dk, the conjugate X of X is given by X = a1 −bi −cj −dk. 13.1. INTRODUCTION: ROTATIONS AS GROUP ACTIONS 543 It is easy to check that the matrices associated with the unit quaternions are exactly the matrices in SU(2). Thus, we call SU(2) the group of unit quaternions. Now we can define an action of the group of unit quaternions SU(2) on R3. For this, we use the fact that R3 can be identified with the pure quaternions in H, namely, the quaternions of the form x1i + x2j + x3k, where (x1, x2, x3) ∈R3. Then we define the action of SU(2) over R3 by Z · X = ZXZ−1 = ZXZ, where Z ∈SU(2) and X is any pure quaternion. Now it turns out that the map ρZ (where ρZ(X) = ZXZ) is indeed a rotation, and that the map ρ: Z 7→ρZ is a surjective homomor-phism ρ: SU(2) →SO(3) whose kernel is {−1, 1}, where 1 denotes the multiplicative unit quaternion. (For details, see Gallier , Chapter 8). We can also define an action of the group SU(2)×SU(2) over R4, by identifying R4 with the quaternions. In this case, (Y, Z) · X = Y XZ, where (Y, Z) ∈SU(2)×SU(2) and X ∈H is any quaternion. Then the map ρY,Z is a rotation (where ρY,Z(X) = Y XZ), and the map ρ: (Y, Z) 7→ρY,Z is a surjective homomorphism ρ: SU(2)×SU(2) →SO(4) whose kernel is {(1, 1), (−1, −1)}. (For details, see Gallier , Chapter 8). Thus, we observe that for n = 2, 3, 4, the rotations in SO(n) can be realized via the linear action of some group (the case n = 1 is trivial, since SO(1) = {1, −1}). It is also the case that the action of each group can be somehow be described in terms of multiplication in some larger algebra “containing” the original vector space Rn (C for n = 2, H for n = 3, 4). However, these groups appear to have been discovered in an ad hoc fashion, and there does not appear to be any universal way to define the action of these groups on Rn. It would certainly be nice if the action was always of the form Z · X = ZXZ−1(= ZXZ). A systematic way of constructing groups realizing rotations in terms of linear action, using a uniform notion of action, does exist. Such groups are the spin groups. We just observed that the rotations in SO(3) can be realized by the linear action of the group of unit quaternions SU(2) on R3, and how the rotations in SO(4) can be realized by the linear action of the group SU(2) × SU(2) on R4. The main reasons why the rotations in SO(3) can be represented by unit quaternions are the following: (1) For every nonzero vector u ∈R3, the reflection su about the hyperplane perpendicular to u is represented by the map v 7→−uvu−1, where u and v are viewed as pure quaternions in H (i.e., if u = (u1, u2, u2), then view u as u1i + u2j + u3k, and similarly for v). 544 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN (2) The group SO(3) is generated by the reflections. As one can imagine, a successful generalization of the quaternions, i.e., the discovery of a group G inducing the rotations in SO(n) via a linear action, depends on the ability to generalize Properties (1) and (2) above. Fortunately, it is true that the group SO(n) is generated by the hyperplane reflections. In fact, this is also true for the orthogonal group O(n), and more generally for the group of isometries O(Φ) of any nondegenerate quadratic form Φ, by the Cartan-Dieudonn´ e theorem (for instance, see Bourbaki , or Gallier , Chapter 7, Theorem 7.2.1). In order to generalize (1), we need to understand how the group G acts on Rn. The case n = 3 is special, because the underlying space R3 on which the rotations act can be embedded as the pure quaternions in H. The case n = 4 is also special, because R4 is the underlying space of H. The generalization to n ≥5 requires more machinery, namely, the notions of Clifford groups and Clifford algebras. As we will see, for every n ≥2, there is a compact, connected (and simply connected when n ≥3) group Spin(n), the “spinor group,” and a surjective homomorphism ρ: Spin(n) → SO(n) whose kernel is {−1, 1}, where 1 denotes the multiplicative unit of Spin(n). This time, Spin(n) acts directly on Rn, because Spin(n) is a certain subgroup of the group of units of the Clifford algebra Cln, and Rn is naturally a subspace of Cln. The group of unit quaternions SU(2) turns out to be isomorphic to the spinor group Spin(3). Because Spin(3) acts directly on R3, the representation of rotations in SO(3) by elements of Spin(3) may be viewed as more natural than the representation by unit quaternions. The group SU(2) × SU(2) turns out to be isomorphic to the spinor group Spin(4), but this isomorphism is less obvious. In summary, we are going to define a group Spin(n) representing the rotations in SO(n), for any n ≥1, in the sense that there is a linear action of Spin(n) on Rn which induces a surjective homomorphism ρ: Spin(n) →SO(n) whose kernel is {−1, 1}. Furthermore, the action of Spin(n) on Rn is given in terms of multiplication in an algebra Cln containing Spin(n), and in which Rn is also embedded. It turns out that as a bonus, for n ≥3, the group Spin(n) is topologically simpler than SO(n), since Spin(n) is simply connected, but SO(n) is not. By being astute, we can also construct a group Pin(n) and a linear action of Pin(n) on Rn that induces a surjective homomorphism ρ: Pin(n) →O(n) whose kernel is {−1, 1}. The difficulty here is the presence of the negative sign in (1). We will see how Atiyah, Bott and Shapiro circumvent this problem by using a “twisted adjoint action,” as opposed to the usual adjoint action (where v 7→uvu−1). Let us now outline in more detail the contents of this chapter. The first step for gener-alizing the quaternions is to define the notion of a Clifford algebra. Let K be any field of characteristic different from 2. Let V be a finite-dimensional vector space over a field K of characteristic ̸= 2, let ϕ: V × V →K be a possibly degenerate symmetric bilinear form, 13.1. INTRODUCTION: ROTATIONS AS GROUP ACTIONS 545 and let Φ(v) = ϕ(v, v) be the corresponding quadratic form. Roughly speaking, a Clifford algebra associated with V and Φ is a K-algebra Cl(V, Φ) satisfying the condition v2 = v · v = Φ(v) · 1 for all v ∈V , where 1 is the multiplicative unit of the algebra. In all rigor, V is not contained in Cl(V, Φ) but there is an injection of V into Cl(V, Φ). The algebra Cl(V, Φ) is the quotient T(V )/A of the tensor algebra T(V ) over V modulo the ideal A of T(V ) generated by all elements of the form v ⊗v −Φ(v) · 1, where v ∈V . If V is finite dimensional and if (e1, . . . , en) is a basis of V , then Cl(V, Φ) has a basis consisting of the 2n −1 products ei1ei2 · · · eik, 1 ≤i1 < i2 < . . . < ik ≤n, and 1. Thus Cl(V, Φ), also denoted by Cl(Φ), has dimension 2n. If (e1, . . . , en) is an orthog-onal basis of V with respect to Φ, then we can view Cl(Φ) as the algebra presented by the generators (e1, . . . , en) and the relations e2 j = Φ(ej) · 1, 1 ≤j ≤n, and ejek = −ekej, 1 ≤j, k ≤n, j ̸= k. If V has finite dimension n and (e1, . . . , en) is a basis of V , we can define two maps t and α as follows. The map t is defined on basis elements by t(ei) = ei t(ei1ei2 · · · eik) = eikeik−1 · · · ei1, where 1 ≤i1 < i2 · · · < ik ≤n, and of course, t(1) = 1. The map α is defined on basis elements by α(ei) = −ei α(ei1ei2 · · · eik) = (−1)kei1ei2 · · · eik where 1 ≤i1 < i2 < · · · < ik ≤n, and of course, α(1) = 1. The even-graded elements (the elements of Cl0(Φ)) are those generated by 1 and the basis elements consisting of an even number of factors ei1ei2 · · · ei2k, and the odd-graded elements (the elements of Cl1(Φ)) are those generated by the basis elements consisting of an odd number of factors ei1ei2 · · · ei2k+1. The second step is to define the Clifford group, which is a subgroup of the group Cl(Φ)∗ of units of Cl(Φ). The Clifford group of Φ is the group Γ(Φ) = {x ∈Cl(Φ)∗| α(x)vx−1 ∈V for all v ∈V }. 546 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN For any x ∈Γ(Φ), let ρx : V →V be the map defined by v 7→α(x)vx−1, v ∈V. The map ρ: Γ(Φ) →GL(V ) given by x 7→ρx is a linear action called the twisted adjoint rep-resentation. It was introduced by Atiyah, Bott and Shapiro and has technical advantages over the earlier adjoint representation ρ0 given by v 7→xvx−1. The group Γ+(Φ), called the special Clifford group, is defined by Γ+(Φ) = Γ(Φ) ∩Cl0(Φ). The key property of the Clifford groups is that if the bilinear form ϕ is nondegenerate, then the map ρ: Γ(Φ) →GL(V ) is actually a surjection ρ: Γ(Φ) →O(Φ) onto the orthog-onal group O(Φ) associated with the quadratic form Φ, and the map ρ: Γ+(Φ) →SO(Φ) is a surjection onto the special orthogonal group SO(Φ) associated with the quadratic form Φ. In both cases, the kernel of ρ is K∗· 1. In order to cut down on the size of the kernel of ρ, we need to define groups smaller than Γ(Φ) and Γ+(Φ). To do so, we introduce a notion of norm on Cl(V, Φ). If ϕ is nondegenerate, then the restriction of the norm N to Γ(Φ) is a map N : Γ(Φ) →K∗· 1. We can now define the groups Pin and Spin as follows. Assume ϕ is a nondegenerate bilinear map on V . We define the pinor group Pin(Φ) as the group Pin(Φ) = {x ∈Γ(Φ) | N(x) = ±1}, and the spinor group Spin(Φ) as Pin(Φ) ∩Γ+(Φ). If the field K is not R or C, it is not obvious that the restriction of ρ to Pin(Φ) is surjective onto O(Φ), and that the restriction of ρ to Spin(Φ) is surjective onto SO(Φ). These maps are surjective if K = R and K = C, but in general it is not surjective. In all cases the kernel of ρ is equal to {−1, 1}. When Φ(x1, . . . , xn) = −(x2 1 + · · · + x2 n), the group Spin(Φ), denoted Spin(n), is exactly the generalization of the unit quaternions (and when n = 3, Spin(n) ∼ = SU(2), the unit quaternions). Some preliminaries on algebras and tensor algebras are reviewed in Section 13.2. In Section 13.3, we define Clifford algebras over the field K = R. The Clifford groups (over K = R) are defined in Section 13.4. In the second half of this section we restrict our attention to the real quadratic form Φ(x1, . . . , xn) = −(x2 1 + · · · + x2 n). The corresponding Clifford algebras are denoted Cln and the corresponding Clifford groups as Γn. In Section 13.5, we define the groups Pin(n) and Spin(n) associated with the real quadratic form Φ(x1, . . . , xn) = −(x2 1+· · ·+x2 n). We prove that the maps ρ: Pin(n) →O(n) 13.1. INTRODUCTION: ROTATIONS AS GROUP ACTIONS 547 and ρ: Spin(n) →SO(n) are surjective with kernel {−1, 1}. We determine the groups Spin(n) for n = 2, 3, 4. Section 13.6 is devoted to the Spin and Pin groups associated with the real nondegenerate quadratic form Φ(x1, . . . , xp+q) = x2 1 + · · · + x2 p −(x2 p+1 + · · · + x2 p+q). We obtain Clifford algebras Clp,q, Clifford groups Γp,q, and groups Pin(p, q) and Spin(p, q). We show that the maps ρ: Pin(p, q) →O(p, q) and ρ: Spin(p, q) →SO(p, q) are surjective with kernel {−1, 1}. In Section 13.7 we show that the Lie groups Pin(p, q) and Spin(p, q) are double covers of O(p, q) and SO(p, q). In Section 13.8 we prove an amazing result due to ´ Elie Cartan and Raoul Bott, namely the 8-periodicity of the Clifford algebras Clp,q. This result says that: for all n ≥0, we have the following isomorphisms: Cl0,n+8 ∼ = Cl0,n ⊗Cl0,8 Cln+8,0 ∼ = Cln,0 ⊗Cl8,0. Furthermore, Cl0,8 = Cl8,0 = R(16), the real algebra of 16 × 16 matrices. Section 13.9 is devoted to the complex Clifford algebras Cl(n, C). In this case, we have a 2-periodicity, Cl(n + 2, C) ∼ = Cl(n, C) ⊗C Cl(2, C), with Cl(2, C) = C(2), the complex algebra of 2 × 2 matrices. Finally, in the last section, Section 13.10, we outline the theory of Clifford groups and of the Pin and Spin groups over any field K of characteristic ̸= 2. Our presentation is heavily influenced by Br¨ ocker and tom Dieck (Chapter 1, Section 6), where most details can be found. This chapter is almost entirely taken from the first 11 pages of the beautiful and seminal paper by Atiyah, Bott and Shapiro , Clifford Modules, and we highly recommend it. Another excellent (but concise) exposition can be found in Kirillov . A very thorough exposition can be found in two places: 1. Lawson and Michelsohn , where the material on Pin(p, q) and Spin(p, q) can be found in Chapter I. 2. Lounesto’s excellent book . One may also want to consult Baker , Curtis , Porteous , Fulton and Harris (Lecture 20) , Choquet-Bruhat , Bourbaki , and Chevalley , a classic. The original source is Elie Cartan’s book (1937) whose translation in English appears in . 548 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN 13.2 Preliminaries We begin by recalling what is an algebra over a field. Let K denote any (commutative) field, although for our purposes we may assume that K = R (and occasionally, K = C). Since we will only be dealing with associative algebras with a multiplicative unit, we only define algebras of this kind. Definition 13.1. Given a field K, a K-algebra is a K-vector space A together with a bilinear operation ⋆: A × A →A, called multiplication, which makes A into a ring with unity 1 (or 1A, when we want to be very precise). This means that ⋆is associative and that there is a multiplicative identity element 1 so that 1 ⋆a = a ⋆1 = a, for all a ∈A. Given two K-algebras A and B, a K-algebra homomorphism h: A →B is a linear map that is also a ring homomorphism, with h(1A) = 1B. For example, the ring Mn(K) of all n × n matrices over a field K is a K-algebra with multiplicative identity element 1 = In. There is an obvious notion of ideal of a K-algebra: Definition 13.2. An ideal A ⊆A is a linear subspace of a K-algebra A that is also a two-sided ideal with respect to multiplication in A. If the field K is understood, we usually simply say an algebra instead of a K-algebra. We will also need tensor products. A rather detailed exposition of tensor products is given in Chapter 2 and the reader may want to review Section 2.2. For the reader’s convenience, we recall the definition of the tensor product of vector spaces. The basic idea is that tensor products allow us to view multilinear maps as linear maps. The maps become simpler, but the spaces (product spaces) become more complicated (tensor products). For more details, see Section 2.2 or Atiyah and Macdonald . Definition 13.3. Given two K-vector spaces E and F, a tensor product of E and F is a pair (E ⊗F, ⊗), where E ⊗F is a K-vector space and ⊗: E ×F →E ⊗F is a bilinear map, so that for every K-vector space G and every bilinear map f : E × F →G, there is a unique linear map f⊗: E ⊗F →G with f(u, v) = f⊗(u ⊗v) for all u ∈E and all v ∈V , as in the diagram below. E × F ⊗ / f % E ⊗F f⊗  G The vector space E ⊗F is defined up to isomorphism. The vectors u ⊗v, where u ∈E and v ∈F, generate E ⊗F. 13.2. PRELIMINARIES 549 Remark: We should really denote the tensor product of E and F by E ⊗K F, since it depends on the field K. Since we usually deal with a fixed field K, we use the simpler notation E ⊗F. As shown in Section 2.4, we have natural isomorphisms (E ⊗F) ⊗G ∼ = E ⊗(F ⊗G) and E ⊗F ∼ = F ⊗E. Given two linear maps f : E →F and g: E′ →F ′, we have a unique bilinear map f × g: E × E′ →F × F ′ so that (f × g)(a, a′) = (f(a), g(a′)) for all a ∈E and all a′ ∈E′. Thus, we have the bilinear map ⊗◦(f × g): E × E′ →F ⊗F ′, and so, there is a unique linear map f ⊗g: E ⊗E′ →F ⊗F ′ so that (f ⊗g)(a ⊗a′) = f(a) ⊗g(a′) for all a ∈E and all a′ ∈E′. Let us now assume that E and F are K-algebras. We want to make E ⊗F into a K-algebra. Since the multiplication operations mE : E × E →E and mF : F × F →F are bilinear, we get linear maps m′ E : E ⊗E →E and m′ F : F ⊗F →F, and thus the linear map m′ E ⊗m′ F : (E ⊗E) ⊗(F ⊗F) →E ⊗F. Using the isomorphism τ : (E ⊗E) ⊗(F ⊗F) →(E ⊗F) ⊗(E ⊗F), we get a linear map mE⊗F : (E ⊗F) ⊗(E ⊗F) →E ⊗F, which defines a multiplication m on E ⊗F (namely, m(α, β) = mE⊗F(α ⊗β) for all α, β ∈ E ⊗F). It is easily checked that E ⊗F is indeed a K-algebra under the multiplication m. Using the simpler notation · for m, we have (a ⊗a′) · (b ⊗b′) = (ab) ⊗(a′b′) (∗) for all a, b ∈E and all a′, b′ ∈F. Given any vector space V over a field K, there is a special K-algebra T(V ) together with a linear map i: V →T(V ), with the following universal mapping property: Given any K-algebra A, for any linear map f : V →A, there is a unique K-algebra homomorphism f : T(V ) →A so that f = f ◦i, as in the diagram below. V i / f " T(V ) f  A 550 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN The algebra T(V ) is the tensor algebra of V ; see Section 2.6. The algebra T(V ) may be constructed as the direct sum T(V ) = M i≥0 V ⊗i, where V 0 = K, and V ⊗i is the i-fold tensor product of V with itself. For every i ≥0, there is a natural injection ιn : V ⊗n →T(V ), and in particular, an injection ι0 : K →T(V ). The multiplicative unit 1 of T(V ) is the image ι0(1) in T(V ) of the unit 1 of the field K. Since every v ∈T(V ) can be expressed as a finite sum v = v1 + · · · + vk, where vi ∈V ⊗ni and the ni are natural numbers with ni ̸= nj if i ̸= j, to define multiplication in T(V ), using bilinearity, it is enough to define the multiplication V ⊗m ×V ⊗n − →V ⊗(m+n). Of course, this is defined by (v1 ⊗· · · ⊗vm) · (w1 ⊗· · · ⊗wn) = v1 ⊗· · · ⊗vm ⊗w1 ⊗· · · ⊗wn. (This has to be made rigorous by using isomorphisms involving the associativity of tensor products; for details, see see Jacobson ) The algebra T(V ) is an example of a graded algebra, where the homogeneous elements of rank n are the elements in V ⊗n. Remark: It is important to note that multiplication in T(V ) is not commutative. Also, in all rigor, the unit 1 of T(V ) is not equal to 1, the unit of the field K. The field K is embedded in T(V ) using the mapping λ 7→λ1. More generally, in view of the injections ιn : V ⊗n →T(V ), we identify elements of V ⊗n with their images in T(V ). Most algebras of interest arise as well-chosen quotients of the tensor algebra T(V ). This is true for the exterior algebra V• V (also called Grassmann algebra), where we take the quotient of T(V ) modulo the ideal generated by all elements of the form v ⊗v, where v ∈V , see Section 3.4. From now on, we assume that K is a field of characteristic different from 2. Given a symmetric bilinear form ϕ: V × V →K, recall that the quadratic form Φ associated with ϕ is given by Φ(v) = ϕ(v, v) for all v ∈V , and that ϕ can be recovered from Φ by the polarization identity ϕ(u, v) = 1 2(Φ(u + v) −Φ(u) −Φ(v)). The symmetric bilinear form ϕ is nondegenerate ifffor every u ∈V , if ϕ(u, v) = 0 for all v ∈V , then u = 0. Definition 13.4. Let (V, ϕ) be a vector space equipped with a nondegenerate symmetric bilinear form ϕ. The the set of linear maps f : V →V such that ϕ(f(u), f(v)) = ϕ(u, v) for all u, v ∈V forms a group denoted O(V, Φ) (or O(V, ϕ)) which is called the group of isometries or orthogonal group of (V, ϕ). 13.3. CLIFFORD ALGEBRAS 551 The condition ϕ(f(u), f(v)) = ϕ(u, v) for all u, v ∈V is equivalent to the condition Φ(f(v)) = Φ(v) for all v ∈V . Definition 13.5. The subgroup of O(V, Φ) denoted SO(V, Φ) (or SO(V, ϕ)) is defined by SO(V, Φ) = {f ∈O(V, Φ) | det(f) = 1} and is called the special orthogonal group or group of rotations of (V, ϕ). We often abbreviate O(V, Φ) as O(Φ) and SO(V, Φ) as SO(Φ). Definition 13.6. If K = R and Φn is the Euclidean quadratic form Φn(x1, . . . , xn) = x2 1 + · · · + x2 n, we write O(n, R) or even O(n) for O(Rn, Φn) and SO(n, R) or even SO(n) for SO(Rn, Φn). Similarly when K = C and Φn(x1, . . . , xn) = x2 1 + · · · + x2 n, we write O(n, C) for O(Cn, Φn) and SO(n, C) for SO(Cn, Φn). If K = R and if Φp,q(x1, . . . , xp+q) = x2 1 + · · · + x2 p −(x2 p+1 + · · · + x2 p+q), with n = p + q we write O(p, q) for O(Rn, Φp,q) and SO(p, q) for SO(Rn, Φp,q). Observe that Φn,0 = Φn. It is not hard to show that O(p, q) and O(q, p) are isomorphic, and similarly SO(p, q) and SO(q, p) are isomorphic. In the special cases where p = 0 or q = 0, we have Φ0,n(x1, . . . , xn) = −(x2 1 + · · · + x2 n) = −Φn(x1, . . . , xn) = −Φn,0(x1, . . . , xn), so for any linear map f we have Φ0,n(f(x1, . . . , xn)) = Φ0,n(x1, . . . , xn) iffΦn(f(x1, . . . , xn)) = Φn(x1, . . . , xn), which shows that O(0, n) = O(n, 0) = O(n) and SO(0, n) = SO(0, n) = SO(n). 13.3 Clifford Algebras A Clifford algebra may be viewed as a refinement of the exterior algebra, in which we take the quotient of T(V ) modulo the ideal generated by all elements of the form v ⊗v −Φ(v) · 1, where Φ is the quadratic form associated with a symmetric bilinear form ϕ: V × V →K, and ·: K × T(V ) →T(V ) denotes the scalar product of the algebra T(V ). For simplicity, let us assume that we are now dealing with real algebras. Definition 13.7. Let V be a real finite-dimensional vector space together with a symmetric bilinear form ϕ: V × V →R and associated quadratic form Φ(v) = ϕ(v, v). A Clifford algebra associated with V and Φ is a real algebra Cl(V, Φ) together with a linear map iΦ : V → 552 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN Cl(V, Φ) satisfying the condition (iΦ(v))2 = Φ(v) · 1 for all v ∈V , and so that for every real algebra A and every linear map f : V →A with (f(v))2 = Φ(v) · 1A for all v ∈V , there is a unique algebra homomorphism f : Cl(V, Φ) →A so that f = f ◦iΦ, as in the diagram below. V iΦ/ f $ Cl(V, Φ) f  A We use the notation λ·u for the product of a scalar λ ∈R and of an element u in the algebra Cl(V, Φ), and juxtaposition uv for the multiplication of two elements u and v in the algebra Cl(V, Φ). By a familiar argument, any two Clifford algebras associated with V and Φ are isomorphic. We often denote iΦ by i. To show the existence of Cl(V, Φ), observe that T(V )/A does the job, where A is the ideal of T(V ) generated by all elements of the form v ⊗v −Φ(v) · 1, where v ∈V . The map iΦ : V →Cl(V, Φ) is the composition V ι1 − →T(V ) π − →T(V )/A, where π is the natural quotient map. We often denote the Clifford algebra Cl(V, Φ) simply by Cl(Φ). Remark: Observe that Definition 13.7 does not assert that iΦ is injective or that there is an injection of R into Cl(V, Φ), but we will prove later that both facts are true when V is finite-dimensional. Also, as in the case of the tensor algebra, the unit 1 of the algebra Cl(V, Φ) and the unit 1 of the field R are not equal. Since Φ(u + v) −Φ(u) −Φ(v) = 2ϕ(u, v) and (i(u + v))2 = (i(u))2 + (i(v))2 + i(u)i(v) + i(v)i(u), using the fact that i(u)2 = Φ(u) · 1, we get i(u)i(v) + i(v)i(u) = 2ϕ(u, v) · 1. (∗) 13.3. CLIFFORD ALGEBRAS 553 As a consequence, if (u1, . . . , un) is an orthogonal basis w.r.t. ϕ (which means that ϕ(uj, uk) = 0 for all j ̸= k), we have i(uj)i(uk) + i(uk)i(uj) = 0 for all j ̸= k. Remark: Certain authors drop the unit 1 of the Clifford algebra Cl(V, Φ) when writing the identities i(u)2 = Φ(u) · 1 and 2ϕ(u, v) · 1 = i(u)i(v) + i(v)i(u), where the second identity is often written as ϕ(u, v) = 1 2(i(u)i(v) + i(v)i(u)). This is very confusing and technically wrong, because we only have an injection of R into Cl(V, Φ), but R is not a subset of Cl(V, Φ).  We warn the readers that Lawson and Michelsohn adopt the opposite of our sign convention in defining Clifford algebras, i.e., they use the condition (f(v))2 = −Φ(v) · 1 for all v ∈V . The most confusing consequence of this is that their Cl(p, q) is our Cl(q, p). Observe that when Φ ≡0 is the quadratic form identically zero everywhere, then the Clifford algebra Cl(V, 0) is just the exterior algebra V• V . Example 13.1. Let V = R, e1 = 1, and assume that Φ(x1e1) = −x2 1. Then Cl(Φ) is spanned by the basis (1, e1). We have e2 1 = −1. Under the bijection e1 7→i, the Clifford algebra Cl(Φ), also denoted by Cl1, is isomorphic to the algebra of complex numbers C. Example 13.2. Now let V = R2, (e1, e2) be the canonical basis, and assume that Φ(x1e1 + x2e2) = −(x2 1 + x2 2). Then Cl(Φ) is spanned by the basis (1, e1, e2, e1e2). Furthermore, we have e2e1 = −e1e2, e2 1 = −1, e2 2 = −1, (e1e2)2 = −1. Under the bijection e1 7→i, e2 7→j, e1e2 7→k 1 7→1, 554 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN it is easily checked that the quaternion identities i2 = j2 = k2 = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j, hold, and thus the Clifford algebra Cl(Φ), also denoted by Cl2, is isomorphic to the algebra of quaternions H. Our prime goal is to define an action of Cl(Φ) on V in such a way that by restricting this action to some suitably chosen multiplicative subgroups of Cl(Φ), we get surjective homomorphisms onto O(Φ) and SO(Φ), respectively. The key point is that a reflection in V about a hyperplane H orthogonal to a vector w can be defined by such an action, but some negative sign shows up. A correct handling of signs is a bit subtle and requires the introduction of a canonical anti-automorphism t, and of a canonical automorphism α, defined as follows: Proposition 13.1. Every Clifford algebra Cl(Φ) possesses a canonical anti-automorphism t: Cl(Φ) →Cl(Φ) satisfying the properties t(xy) = t(y)t(x), t ◦t = id, and t(i(v)) = i(v), for all x, y ∈Cl(Φ) and all v ∈V . Furthermore, such an anti-automorphism is unique. Proof. Consider the opposite algebra Cl(Φ)o, in which the product of x and y is given by yx. It has the universal mapping property. Thus, we get a unique isomorphism t, as in the diagram below. V i / i # Cl(V, Φ) t  Cl(Φ)o We also denote t(x) by xt. When V is finite-dimensional, for a more palatable description of t in terms of a basis of V , see the paragraph following Theorem 13.4. The canonical automorphism α is defined using the proposition. Proposition 13.2. Every Clifford algebra Cl(Φ) has a unique canonical automorphism α: Cl(Φ) →Cl(Φ) satisfying the properties α ◦α = id, and α(i(v)) = −i(v), for all v ∈V . 13.3. CLIFFORD ALGEBRAS 555 Proof. Consider the linear map α0 : V →Cl(Φ) defined by α0(v) = −i(v), for all v ∈V . We get a unique homomorphism α as in the diagram below. V i / α0 # Cl(V, Φ) α  Cl(Φ) Furthermore, every x ∈Cl(Φ) can be written as x = x1 · · · xm, with xj ∈i(V ), and since α(xj) = −xj, we get α ◦α = id. It is clear that α is bijective. When V is finite-dimensional, a more palatable description of α in terms of a basis of V can be given; see the paragraph following Theorem 13.4. If (e1, . . . , en) is a basis of V , then the Clifford algebra Cl(Φ) consists of certain kinds of “polynomials,” linear combinations of monomials of the form P J λJeJ, where J = {i1, i2, . . . , ik} is any subset (possibly empty) of {1, . . . , n}, with 1 ≤i1 < i2 · · · < ik ≤n, and the monomial eJ is the “product” ei1ei2 · · · eik. We now show that if V has dimension n, then i is injective and Cl(Φ) has dimension 2n. A clever way of doing this is to introduce a graded tensor product. First, observe that Cl(Φ) = Cl0(Φ) ⊕Cl1(Φ), where Cli(Φ) = {x ∈Cl(Φ) | α(x) = (−1)ix}, where i = 0, 1. We say that we have a Z/2-grading, which means that if x ∈Cli(Φ) and y ∈Clj(Φ), then xy ∈Cli+j (mod 2)(Φ). When V is finite-dimensional, since every element of Cl(Φ) is a linear combination of the form P J λJeJ as explained earlier, in view of the description of α given above, we see that the elements of Cl0(Φ) are those for which the monomials eJ are products of an even number of factors, and the elements of Cl1(Φ) are those for which the monomials eJ are products of an odd number of factors. Remark: Observe that Cl0(Φ) is a subalgebra of Cl(Φ), whereas Cl1(Φ) is not. Definition 13.8. Given two Z/2-graded algebras A = A0 ⊕A1 and B = B0 ⊕B1, their graded tensor product A b ⊗B is defined by (A b ⊗B)0 = (A0 ⊗B0) ⊕(A1 ⊗B1), (A b ⊗B)1 = (A0 ⊗B1) ⊕(A1 ⊗B0), 556 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN with multiplication (a′ ⊗b)(a ⊗b′) = (−1)ij(a′a) ⊗(bb′), for a ∈Ai and b ∈Bj. The reader should check that A b ⊗B is indeed Z/2-graded. Proposition 13.3. Let V and W be finite dimensional vector spaces with quadratic forms Φ and Ψ. Then there is a quadratic form Φ ⊕Ψ on V ⊕W defined by (Φ + Ψ)(v, w) = Φ(v) + Ψ(w). If we write i: V →Cl(Φ) and j : W →Cl(Ψ), we can define a linear map f : V ⊕W →Cl(Φ) b ⊗Cl(Ψ) by f(v, w) = i(v) ⊗1 + 1 ⊗j(w). Furthermore, the map f induces an isomorphism (also denoted by f) f : Cl(Φ + Ψ) →Cl(Φ) b ⊗Cl(Ψ). Proof. See Br¨ ocker and tom Dieck , Chapter 1, Section 6, page 57. As a corollary, we obtain the following result: Theorem 13.4. For every vector space V of finite dimension n, the map i: V →Cl(Φ) is injective. Given a basis (e1, . . . , en) of V , the 2n −1 products i(ei1)i(ei2) · · · i(eik), 1 ≤i1 < i2 · · · < ik ≤n, and 1 form a basis of Cl(Φ). Thus, Cl(Φ) has dimension 2n. Proof. The proof is by induction on n = dim(V ). For n = 1, the tensor algebra T(V ) is just the polynomial ring R[X], where i(e1) = X. Thus, Cl(Φ) = R[X]/(X2 −Φ(e1)), and the result is obvious ((1, X) is a basis). Since i(ej)i(ek) + i(ek)i(ej) = 2ϕ(ei, ej) · 1, it is clear that the products i(ei1)i(ei2) · · · i(eik), 1 ≤i1 < i2 < · · · < ik ≤n, and 1 generate Cl(Φ). In order to conclude that these vectors form a basis it suffices to show that the dimension of Cl(Φ) is 2n. Now there is always a basis that is orthogonal with respect to ϕ (for example, see Artin , Chapter 7, or Gallier , Chapter 6, Problem 6.14), and thus, we have a splitting (V, Φ) ∼ = n M k=1 (Vk, Φk), where Vk has dimension 1. Choosing a basis so that ek ∈Vk, the theorem follows by induction from Proposition 13.3. 13.4. CLIFFORD GROUPS 557 Since i is injective, for simplicity of notation, from now on we write u for i(u). Theorem 13.4 implies the following result. Proposition 13.5. If (e1, . . . , en) is an orthogonal basis of V with respect to Φ, then Cl(Φ) is the algebra presented by the generators (e1, . . . , en) and the relations e2 j = Φ(ej) · 1, 1 ≤j ≤n, and ejek = −ekej, 1 ≤j, k ≤n, j ̸= k. If V has finite dimension n and (e1, . . . , en) is a basis of V , by Theorem 13.4, the maps t and α are completely determined by their action on the basis elements. Namely, t is defined by t(ei) = ei t(ei1ei2 · · · eik) = eikeik−1 · · · ei1, where 1 ≤i1 < i2 · · · < ik ≤n, and of course, t(1) = 1. The map α is defined by α(ei) = −ei α(ei1ei2 · · · eik) = (−1)kei1ei2 · · · eik where 1 ≤i1 < i2 < · · · < ik ≤n, and of course, α(1) = 1. Furthermore, the even-graded elements (the elements of Cl0(Φ)) are those generated by 1 and the basis elements consisting of an even number of factors ei1ei2 · · · ei2k, and the odd-graded elements (the elements of Cl1(Φ)) are those generated by the basis elements consisting of an odd number of factors ei1ei2 · · · ei2k+1. We are now ready to define the Clifford group and investigate some of its properties. 13.4 Clifford Groups Definition 13.9. Let V be a real finite-dimensional vector space with a quadratic form Φ. Let Cl(Φ) be the Clifford algebra (see Definition 13.7). The multiplicative group of invertible elements of Cl(Φ) is denoted by Cl(Φ)∗. Proposition 13.6. For any x ∈V , Φ(x) ̸= 0 if and only if x is invertible. Proof. This follows from the fact that x2 = Φ(x) (where we abused notation and wrote Φ(x) · 1 = Φ(x)). If Φ(x) ̸= 0, then x−1 = x(Φ(x))−1, and if x is invertible then x ̸= 0 and x = Φ(x)x−1, so Φ(x) ̸= 0. We would like Cl(Φ)∗to act on V via x · v = α(x)vx−1, where x ∈Cl(Φ)∗and v ∈V . In general, there is no reason why α(x)vx−1 should be in V or why this action defines an automorphism of V , so we restrict this map to the subset Γ(Φ) of Cl(Φ)∗as follows. 558 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN Definition 13.10. Given a finite dimensional vector space V and a quadratic form Φ on V , the Clifford group of Φ is the group Γ(Φ) = {x ∈Cl(Φ)∗| α(x)vx−1 ∈V for all v ∈V }. Definition 13.11. For any x ∈Γ(Φ), let ρx : V →V be the map defined by v 7→α(x)vx−1, v ∈V. It is not entirely obvious why the map ρ: Γ(Φ) →GL(V ) given by x 7→ρx is a linear action, and for that matter, why Γ(Φ) is a group. This is because V is finite-dimensional and α is an automorphism. Proposition 13.7. The set Γ(Φ) is a group and ρ is a linear representation. Proof. For any x ∈Γ(Φ), the map ρx from V to V defined by v 7→α(x)vx−1 is clearly linear. If α(x)vx−1 = 0, since by hypothesis x is invertible and since α is an automorphism α(x) is also invertible, so v = 0. Thus our linear map is injective, and since V has finite dimension, it is bijective. This proves that ρ is a linear representation. To prove that x−1 ∈Γ(Φ), pick any v ∈V . Since the linear map ρx is bijective, there is some w ∈V such that ρx(w) = v, which means that α(x)wx−1 = v. Since x is invertible and α is an automorphism, we get α(x−1)vx = w, so α(x−1)vx ∈V ; since this holds for any v ∈V , we have x−1 ∈Γ(Φ). Since α is an automorphism, if x, y ∈Γ(Φ), for any v ∈V we have ρy(ρx(v)) = α(y)α(x)vx−1y−1 = α(yx)v(yx)−1 = ρyx(v), which shows that ρyx is a linear automorphism of V , so yx ∈Γ(Φ) and ρ is a homomorphism. Therefore, Γ(Φ) is a group and ρ is a linear representation. Definition 13.12. Given a finite dimensional vector space V and quadratic form Φ on V , the special Clifford group of Φ is the group Γ+(Φ) = Γ(Φ) ∩Cl0(Φ). Remarks: 13.4. CLIFFORD GROUPS 559 1. The map ρ: Γ(Φ) →GL(V ) given by x 7→ρx is called the twisted adjoint representa-tion. It was introduced by Atiyah, Bott and Shapiro . It has the advantage of not introducing a spurious negative sign, i.e., when v ∈V and Φ(v) ̸= 0, the map ρv is the reflection sv about the hyperplane orthogonal to v (see Theorem 13.9). Further-more, when Φ is nondegenerate, the kernel Ker (ρ) of the representation ρ is given by Ker (ρ) = R∗· 1, where R∗= R −{0}. The earlier adjoint representation ρ0 (used by Chevalley and others) is given by v 7→xvx−1. Unfortunately, in this case ρ0 v represents −sv, where sv is the reflection about the hy-perplane orthogonal to v. Furthermore, the kernel of the representation ρ0 is generally bigger than R∗·1. This is the reason why the twisted adjoint representation is preferred (and must be used for a proper treatment of the Pin group). 2. According to Lounesto (in Riesz ), the Clifford group was actually discoved by Rudolf Lipschitz in 1880 and not by Clifford, two years after Clifford’s discovery of Clifford algebras. Lounesto says (page 219): “Chevalley introduced the exterior expo-nential of bivectors and used it to scrutinize properties of Lipschitz’s covering group of rotations (naming it unjustly a “Clifford group”).” Proposition 13.8. The maps α and t induce an automorphism and an anti-automorphism of the Clifford group, Γ(Φ). Proof. It is not very instructive; see Br¨ ocker and tom Dieck , Chapter 1, Section 6, page 58. The following key result shows why Clifford groups generalize the quaternions. Theorem 13.9. Let V be a finite dimensional vector space and let Φ a quadratic form on V . For every element x of the Clifford group Γ(Φ), if x ∈V then Φ(x) ̸= 0 and the map ρx : V →V given by v 7→α(x)vx−1 for all v ∈V is the reflection about the hyperplane H orthogonal to the non-isotropic vector x. Proof. We already observed that if x ∈V is an invertible element then Φ(x) ̸= 0. Recall that the reflection s about the hyperplane H orthogonal to the vector x is given by s(u) = u −2 ϕ(u, x) Φ(x) · x. However, we have x2 = Φ(x) · 1 and ux + xu = 2ϕ(u, x) · 1. 560 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN Thus, we have s(u) = u −2 ϕ(u, x) Φ(x) · x = u −2ϕ(u, x) ·  1 Φ(x) · x  = u −2ϕ(u, x) · x−1 = u −2ϕ(u, x) · (1x−1) = u −(2ϕ(u, x) · 1)x−1 = u −(ux + xu)x−1 = −xux−1 = α(x)ux−1, since α(x) = −x, for x ∈V . Recall that the linear representation ρ: Γ(Φ) →GL(V ) is given by ρx(v) = α(x)vx−1, for all x ∈Γ(Φ) and all v ∈V . We would like to show that ρ is a surjective homomorphism from Γ(Φ) onto O(Φ), and a surjective homomorphism from Γ+(Φ) onto SO(Φ). For this, we will need to assume that ϕ is nondegenerate, which means that for every v ∈V , if ϕ(v, w) = 0 for all w ∈V , then v = 0. In order to prove that ρx ∈O(Φ) for any x ∈Γ(Φ), we define a notion of norm on Γ(Φ), and for this we need to define a notion of conjugation on Cl(Φ). Definition 13.13. We define conjugation on a Clifford algebra Cl(Φ) as the map x 7→x = t(α(x)) for all x ∈Cl(Φ). Observe that since (t ◦α)(v) = (α ◦t)(v) for all v ∈V and since α is an automorphism and t is an anti-automorphism, we have t ◦α = α ◦t on Cl(Φ). For all x, y ∈Cl(Φ) we also have xy = t(α(xy)) = t(α(x)α(y)) = t(α(y))t(α(x)) = y x. Thus we showed the following fact. 13.4. CLIFFORD GROUPS 561 Proposition 13.10. Conjugation is an anti-automorphism. If V has finite dimension n and (e1, . . . , en) is a basis of V , in view of previous remarks, conjugation is defined by ei = −ei ei1ei2 · · · eik = (−1)keikeik−1 · · · ei1 where 1 ≤i1 < i2 · · · < ik ≤n, and of course, 1 = 1. Definition 13.14. The map N : Cl(Φ) →Cl(Φ) given by N(x) = xx is called the norm of Cl(Φ). Observe that N(v) = vv = −v2 = −Φ(v) · 1 for all v ∈V , that is, N(v) = −Φ(v) · 1 for all v ∈V . Also, if (e1, . . . , en) is a basis of V , since conjugation is an anti-automorphism, we obtain N(ei1ei2 · · · eik) = ei1ei2 · · · eikei1ei2 · · · eik = ei1ei2 · · · eik(−1)keik · · · ei2ei1 = (−1)kΦ(ei1)Φ(ei2) · · · Φ(eik) · 1. In general, for an arbitrary element x ∈Cl(Φ), there is no guarantee that N(x) is a scalar multiple of 1. However, we will show in Proposition 13.12 that if x ∈Γ(Φ), then N(x) ∈R∗·1. For simplicity of exposition, we first assume that Φ is the quadratic form on Rn defined by Φ(x1, . . . , xn) = Φ0,n(x1, . . . , xn) = −(x2 1 + · · · + x2 n). Note that the isometry groups associated with Φ = Φ0,n are O(0, n) and SO(0, n), but we know that O(0, n) = O(n) and SO(0, n) = SO(n). Let Cln denote the Clifford algebra Cl(Φ) and Γn denote the Clifford group Γ(Φ). The following lemma plays a crucial role. Lemma 13.11. The kernel of the map ρ: Γn →GL(n) is R∗· 1, the multiplicative group of nonzero scalar multiples of 1 ∈Cln. Proof. If ρx = id, then α(x)v = vx for all v ∈Rn. (1) Since Cln = Cl0 n ⊕Cl1 n, we can write x = x0 + x1, with xi ∈Cli n for i = 0, 1. Then Equation (1) becomes x0v = vx0 and −x1v = vx1 for all v ∈Rn. (2) 562 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN Using Theorem 13.4, we can express x0 as a linear combination of monomials in the canonical basis (e1, . . . , en), so that x0 = a0 + e1b1, with a0 ∈Cl0 n, b1 ∈Cl1 n, where neither a0 nor b1 contains a summand with a factor e1. Applying the first relation in (2) to v = e1, we get e1a0 + e2 1b1 = a0e1 + e1b1e1. (3) Now the basis (e1, . . . , en) is orthogonal w.r.t. Φ, which implies that ejek = −ekej for all j ̸= k. Since each monomial in a0 is of even degree and contains no factor e1, we get a0e1 = e1a0. Similarly, since b1 is of odd degree and contains no factor e1, we get e1b1e1 = −e2 1b1. But then from (3), we get e1a0 + e2 1b1 = a0e1 + e1b1e1 = e1a0 −e2 1b1, and so, e2 1b1 = 0. However, e2 1 = −1, and so, b1 = 0. Therefore, x0 contains no monomial with a factor e1. We can apply the same argument to the other basis elements e2, . . . , en, and thus, we just proved that x0 ∈R · 1. A similar argument applying to the second equation in (2), with x1 = a1+e1b0 and v = e1 shows that b0 = 0. By repeating the argument for the other basis elements, we ultimately conclude that x1 = 0. Finally, x = x0 ∈(R · 1) ∩Γn = R∗· 1. Remark: If Φ is any nondegenerate quadratic form, we know (for instance, see Artin , Chapter 7, or Gallier , Chapter 6, Problem 6.14) that there is an orthogonal basis (e1, . . . , en) with respect to ϕ (i.e. ϕ(ej, ek) = 0 for all j ̸= k and ϕ(ej, ej) ̸= 0 for all j). Thus, the commutation relations e2 j = Φ(ej) · 1, with Φ(ej) ̸= 0, 1 ≤j ≤n, and ejek = −ekej, 1 ≤j, k ≤n, j ̸= k hold, and since the proof only rests on these facts, Lemma 13.11 holds for the Clifford group Γ(Φ) associated with any nondegenerate quadratic form. 13.4. CLIFFORD GROUPS 563  However, Lemma 13.11 may fail for degenerate quadratic forms. For example, if Φ ≡0, then Cl(V, 0) = V• V . Consider the element x = 1 + e1e2. Clearly, x−1 = 1 −e1e2. But now, for any v ∈V , we have α(1 + e1e2)v(1 + e1e2)−1 = (1 + e1e2)v(1 −e1e2) = v. Yet, 1 + e1e2 is not a scalar multiple of 1. If instead of the twisted adjoint action we had used the action ρ0 : Γn →GL(n) given by ρ0 x(v) = xvx−1, then when n is odd the kernel of ρ0 contains other elements besides scalar multiples of 1. Indeed, if (e1, . . . , en) is an orthogonal basis, we have eiej = −ejei for all j ̸= i and e2 i = −1 for all i, so the element e1 · · · en ∈Cl∗ n commutes with all ei (it belongs to the center of Cln), and thus e1 · · · en ∈Ker ρ0. Thus, we see that another subtle consequence of the “Atiyah– Bott–Shapiro trick” of using the action ρx(v) = α(x)vx where α takes care of the parity of x ∈Γn is to cut down the kernel of ρ to R∗· 1. The following proposition shows that the notion of norm is well-behaved. Proposition 13.12. If x ∈Γn, then N(x) ∈R∗· 1. Proof. The trick is to show that N(x) = xx is in the kernel of ρ. To say that x ∈Γn means that α(x)vx−1 ∈Rn for all v ∈Rn. Applying t, we get t(x)−1vt(α(x)) = α(x)vx−1, since t is the identity on Rn. Thus, we have v = t(x)α(x)v(t(α(x))x)−1 = t(x)α(x)v(xx)−1 = α(α(t(x)))α(x)v(xx)−1, since α ◦α = id = α(t(α(x)))α(x)v(xx)−1, since α ◦t = t ◦α = α(x)α(x)v(xx)−1 = α(xx)v(xx)−1, so xx ∈Ker (ρ). By Proposition 13.8, we have x ∈Γn, and so, xx = x x ∈Ker (ρ). Remark: Again, the proof also holds for the Clifford group Γ(Φ) associated with any non-degenerate quadratic form Φ. When Φ(v) = −∥v∥2, where ∥v∥is the standard Euclidean norm of v, we have N(v) = ∥v∥2 · 1 for all v ∈V . However, for other quadratic forms, it is possible that N(x) = λ · 1 where λ < 0, and this is a difficulty that needs to be overcome. 564 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN Proposition 13.13. The restriction of the norm N to Γn is a homomorphism N : Γn → R∗· 1, and N(α(x)) = N(x) for all x ∈Γn. Proof. We have N(xy) = xyy x = xN(y)x = xxN(y) = N(x)N(y), where the third equality holds because N(x) ∈R∗· 1. Next, observe that since α and t commute we have α(x) = t(α(α(x))) = α(t(α(x))) = α(x), so we get N(α(x)) = α(x)α(x) = α(x)α(x) = α(xx) = α(N(x)) = N(x), since N(x) ∈R∗· 1. Remark: The proof also holds for the Clifford group Γ(Φ) associated with any nondegen-erate quadratic form Φ. Proposition 13.14. We have Rn −{0} ⊆Γn and ρ(Γn) ⊆O(n). Proof. Let x ∈Γn and v ∈Rn, with v ̸= 0. We have N(ρx(v)) = N(α(x)vx−1) = N(α(x))N(v)N(x−1) = N(x)N(v)N(x)−1 = N(v), since N : Γn →R∗· 1. However, for v ∈Rn, we know that N(ρxv) = −Φ(ρxv) · 1, and N(v) = −Φ(v) · 1. Thus, ρx is norm-preserving, and so ρx ∈O(n). Remark: The proof that ρ(Γ(Φ)) ⊆O(Φ) also holds for the Clifford group Γ(Φ) associated with any nondegenerate quadratic form Φ. The first statement needs to be replaced by the fact that every non-isotropic vector in Rn (a vector is non-isotropic if Φ(x) ̸= 0) belongs to Γ(Φ). Indeed, x2 = Φ(x) · 1, which implies that x is invertible. We are finally ready for the introduction of the groups Pin(n) and Spin(n). 13.5. THE GROUPS PIN(n) AND SPIN(n) 565 13.5 The Groups Pin(n) and Spin(n) Definition 13.15. We define the pinor group Pin(n) as the kernel Ker (N) of the homo-morphism N : Γn →R∗· 1, and the spinor group Spin(n) as Pin(n) ∩Γ+ n . Observe that if N(x) = 1, then x is invertible, and x−1 = x since xx = N(x) = 1. Thus, we can write Pin(n) = {x ∈Cln | α(x)vx−1 ∈Rn for all v ∈Rn, N(x) = 1} = {x ∈Cln | α(x)vx ∈Rn for all v ∈Rn, xx = 1}, and Spin(n) = {x ∈Cl0 n | xvx−1 ∈Rn for all v ∈Rn, N(x) = 1} = {x ∈Cl0 n | xvx ∈Rn for all v ∈Rn, xx = 1} Remark: According to Atiyah, Bott and Shapiro, the use of the name Pin(k) is a joke due to Jean-Pierre Serre (Atiyah, Bott and Shapiro , page 1). Theorem 13.15. The restriction of ρ: Γn →O(n) to the pinor group Pin(n) is a surjective homomorphism ρ: Pin(n) →O(n) whose kernel is {−1, 1}, and the restriction of ρ to the spinor group Spin(n) is a surjective homomorphism ρ: Spin(n) →SO(n) whose kernel is {−1, 1}. Proof. By Proposition 13.14, we have a map ρ: Pin(n) →O(n). The reader can easily check that ρ is a homomorphism. By the Cartan-Dieudonn´ e theorem (see Bourbaki , or Gallier , Chapter 7, Theorem 7.2.1), every isometry f ∈O(n) is the composition f = s1 ◦· · ·◦sk of hyperplane reflections sj. If we assume that sj is a reflection about the hyperplane Hj orthogonal to the nonzero vector wj, by Theorem 13.9, ρ(wj) = sj. Since N(wj) = ∥wj∥2 ·1, we can replace wj by wj/ ∥wj∥, so that N(w1 · · · wk) = 1, and then f = ρ(w1 · · · wk), and ρ is surjective. Note that Ker (ρ | Pin(n)) = Ker (ρ) ∩Ker (N) = {t ∈R∗· 1 | N(t) = 1} = {−1, 1}. As to Spin(n), we just need to show that the restriction of ρ to Spin(n) maps Γn into SO(n). If this was not the case, there would be some improper isometry f ∈O(n) so that ρx = f, where x ∈Γn∩Cl0 n. However, we can express f as the composition of an odd number of reflections, say f = ρ(w1 · · · w2k+1). Since ρ(w1 · · · w2k+1) = ρx, 566 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN we have x−1w1 · · · w2k+1 ∈Ker (ρ). By Lemma 13.11, we must have x−1w1 · · · w2k+1 = λ · 1 for some λ ∈R∗, and thus w1 · · · w2k+1 = λ · x, where x has even degree and w1 · · · w2k+1 has odd degree, which is impossible. Let us denote the set of elements v ∈Rn with N(v) = 1 (with norm 1) by Sn−1. We have the following corollary of Theorem 13.15. Corollary 13.16. The group Pin(n) is generated by Sn−1, and every element of Spin(n) can be written as the product of an even number of elements of Sn−1. Example 13.3. In Example 13.1 we showed that Cl1 is isomorphic to C. The reader should verify that Pin(1) ∼ = Z/4Z as follows. By definition Pin(1) = {x ∈Cl1 | α(x)vx−1 ∈R for all v ∈R, N(x) = 1}. A typical element in Pin(1) has the form a1 + be1 where e2 1 = −1. Set e1 7→i and 1 7→1 as in Example 13.1. The condition N(x) = 1 implies that N(x) = xx = (a + bi)(a −bi) = a2 + b2 = 1. Thus x−1 = x a2 + b2 = x. and x ∈Pin(1) implies that α(x)x−1 ∈R where α(x)x−1 = (a −ib)(a −ib) = a2 −b2 −2abi. Thus either a = 0 or b = 0. This constraint, along with a2 + b2 = 1, implies that Pin(1) = {1, i, −1, −i} ∼ = Z/4Z since i generates Pin(1) and i4 = 1. Since Spin(1) = Pin(1) ∩Cl0 n, we conclude that Spin(1) = {−1, 1} ∼ = Z/2Z. 13.5. THE GROUPS PIN(n) AND SPIN(n) 567 Example 13.4. Definition 13.15 also implies Pin(2) ∼ = {ae1 + be2 | a2 + b2 = 1} ∪{c1 + de1e2 | c2 + d2 = 1}, Spin(2) = U(1). We may also write Pin(2) = U(1) + U(1), where U(1) is the group of complex numbers of modulus 1 (the unit circle in R2). Let us take a closer look at Spin(2). The Clifford algebra Cl2 is generated by the four elements 1, e1, e2, e1e2, and they satisfy the relations e2 1 = −1, e2 2 = −1, e1e2 = −e2e1. We saw in Example 13.2 that Cl2 is isomorphic to the algebra of quaternions H. According to Corollary 13.16, the group Spin(2) consists of all products 2k Y i=1 (aie1 + bie2) consisting of an even number of factors and such that a2 i + b2 i = 1. In view of the above relations, every such element can be written as x = a1 + be1e2, where x satisfies the conditions that xvx−1 ∈R2 for all v ∈R2, and N(x) = 1. Since 1 = 1, e1 = −e1, e2 = −e2, e1e2 = −e1e2, the definition of conjugation implies that x = t(α(x)) = t(α(a1 + be1e2)) = at(α(1)) + bt(α(e1e2)) = a1 + be1e2 = a1 −be1e2. Then we get N(x) = xx = (a2 + b2) · 1, and the condition N(x) = 1 is simply a2 + b2 = 1. We claim that if x ∈Cl0 2, then xvx−1 ∈R2. Indeed, since x ∈Cl0 2 and v ∈Cl1 2, we have xvx−1 ∈Cl1 2, which implies that xvx−1 ∈R2, since the only elements of Cl1 2 are those in R2. This observation provides an alternative proof of the fact that Spin(2) consists of those elements x = a1 + be1e2 so that a2 + b2 = 1. If we let i = e1e2, we observe that i2 = −1, e1i = −ie1 = −e2, e2i = −ie2 = e1. 568 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN Thus, Spin(2) is isomorphic to U(1). Also note that e1(a1 + bi) = (a1 −bi)e1. Let us find out explicitly what is the action of Spin(2) on R2. Given X = a1 + bi, with a2 + b2 = 1, then X = a1 −bi, and for any v = v1e1 + v2e2, we have α(X)vX−1 = X(v1e1 + v2e2)X−1 = X(v1e1 + v2e2)(−e1e1)X = X(v1e1 + v2e2)(−e1)(e1X) = X(v11 + v2i)Xe1 = X2(v11 + v2i)e1 = (((a2 −b2)v1 −2abv2)1 + ((a2 −b2)v2 + 2abv1)i)e1 = ((a2 −b2)v1 −2abv2)e1 + ((a2 −b2)v2 + 2abv1)e2. Since a2 +b2 = 1, we can write X = a1+bi = (cos θ)1+(sin θ)i, and the above derivation shows that α(X)vX−1 = (cos 2θv1 −sin 2θv2)e1 + (cos 2θv2 + sin 2θv1)e2. This means that the rotation ρX induced by X ∈Spin(2) is the rotation of angle 2θ around the origin. Observe that the maps v 7→v(−e1), X 7→Xe1 establish bijections between R2 and Spin(2) ∼ = U(1). Also, note that the action of X = cos θ + i sin θ viewed as a complex number yields the rotation of angle θ, whereas the action of X = (cos θ)1 + (sin θ)i viewed as a member of Spin(2) yields the rotation of angle 2θ. There is nothing wrong. In general, Spin(n) is a two–to–one cover of SO(n). Next let us take a closer look at Spin(3). Example 13.5. The Clifford algebra Cl3 is generated by the eight elements 1, e1, e2, e3, e1e2, e2e3, e3e1, e1e2e3, and they satisfy the relations e2 i = −1, eiej = −ejei, 1 ≤i, j ≤3, i ̸= j. It is not hard to show that Cl3 is isomorphic to H ⊕H. By Corollary 13.16, the group Spin(3) consists of all products 2k Y i=1 (aie1 + bie2 + cie3) 13.5. THE GROUPS PIN(n) AND SPIN(n) 569 consisting of an even number of factors and such that a2 i + b2 i + c2 i = 1. In view of the above relations, every such element can be written as x = a1 + be2e3 + ce3e1 + de1e2, where x satisfies the conditions that xvx−1 ∈R3 for all v ∈R3, and N(x) = 1. Since e2e3 = −e2e3, e3e1 = −e3e1, e1e2 = −e1e2, we observe that x = a1 −be2e3 −ce3e1 −de1e2. We then get N(x) = (a2 + b2 + c2 + d2) · 1, and the condition N(x) = 1 is simply a2 + b2 + c2 + d2 = 1. It turns out that the conditions x ∈Cl0 3 and N(x) = 1 imply that xvx−1 ∈R3 for all v ∈R3. To prove this, first observe that N(x) = 1 implies that x−1 = x, and that v = −v for any v ∈R3, and so, xvx−1 = x−1 v x = −xvx−1. Also, since x ∈Cl0 3 and v ∈Cl1 3, we have xvx−1 ∈Cl1 3. Thus, we can write xvx−1 = u + λe1e2e3, for some u ∈R3 and some λ ∈R. But e1e2e3 = −e3e2e1 = e1e2e3, and so, xvx−1 = −u + λe1e2e3 = −xvx−1 = −u −λe1e2e3, which implies that λ = 0. Thus, xvx−1 ∈R3, as claimed. By using this observation, we once again conclude that Spin(3) consists of those elements x = a1 + be2e3 + ce3e1 + de1e2 so that a2 + b2 + c2 + d2 = 1. Under the bijection i 7→e2e3, j 7→e3e1, k 7→e1e2, we can check that we have an isomorphism between the group SU(2) of unit quaternions and Spin(3). If X = a1 + be2e3 + ce3e1 + de1e2 ∈Spin(3), observe that X−1 = X = a1 −be2e3 −ce3e1 −de1e2. Now using the identification i 7→e2e3, j 7→e3e1, k 7→e1e2, 570 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN we can easily check that (e1e2e3)2 = 1, (e1e2e3)i = i(e1e2e3) = −e1, (e1e2e3)j = j(e1e2e3) = −e2, (e1e2e3)k = k(e1e2e3) = −e3, (e1e2e3)e1 = −i, (e1e2e3)e2 = −j, (e1e2e3)e3 = −k. Then if X = a1 + bi + cj + dk ∈Spin(3), for every v = v1e1 + v2e2 + v3e3, we have α(X)vX−1 = X(v1e1 + v2e2 + v3e3)X−1 = X(e1e2e3)2(v1e1 + v2e2 + v3e3)X−1 = (e1e2e3)X(e1e2e3)(v1e1 + v2e2 + v3e3)X−1 = −(e1e2e3)X(v1i + v2j + v3k)X−1. This shows that the rotation ρX ∈SO(3) induced by X ∈Spin(3) can be viewed as the rotation induced by the quaternion a1+bi+cj+dk on the pure quaternions, using the maps v 7→−(e1e2e3)v, X 7→−(e1e2e3)X to go from a vector v = v1e1 + v2e2 + v3e3 to the pure quaternion v1i + v2j + v3k, and back. We close this section by taking a closer look at Spin(4). Example 13.6. We will show in Section 13.8 that Cl4 is isomorphic to M2(H), the algebra of 2 × 2 matrices whose entries are quaternions. According to Corollary 13.16, the group Spin(4) consists of all products 2k Y i=1 (aie1 + bie2 + cie3 + die4) consisting of an even number of factors and such that a2 i + b2 i + c2 i + d2 i = 1. Using the relations e2 i = −1, eiej = −ejei, 1 ≤i, j ≤4, i ̸= j, every element of Spin(4) can be written as x = a11 + a2e1e2 + a3e2e3 + a4e3e1 + a5e4e3 + a6e4e1 + a7e4e2 + a8e1e2e3e4, where x satisfies the conditions that xvx−1 ∈R4 for all v ∈R4, and N(x) = 1. Let i = e1e2, j = e2e3, k = e3e1, i′ = e4e3, j′ = e4e1, k′ = e4e2, 13.5. THE GROUPS PIN(n) AND SPIN(n) 571 and I = e1e2e3e4. The reader will easily verify that ij = k jk = i ki = j i2 = −1, j2 = −1, k2 = −1 iI = Ii = i′ jI = Ij = j′ kI = Ik = k′ I2 = 1, I = I i = −i, j = −j, k = −k i′ = −i′, j′ = −j′, k′ = −k′. Then every x ∈Spin(4) can be written as x = u + Iv, with u = a1 + bi + cj + dk and v = a′1 + b′i + c′j + d′k, with the extra conditions stated above. Using the above identities, we have (u + Iv)(u′ + Iv′) = uu′ + vv′ + I(uv′ + vu′). Furthermore, the identities imply u + Iv = t(α(u + Iv)) = t(α(u)) + t(α(Iv)) = u + t(α(I)α(v)) = u + t(α(v))t(α(I)) = u + vI = u + vI = u + Iv. As a consequence, N(u + Iv) = (u + Iv)(u + Iv) = uu + vv + I(uv + vu), and thus, N(u + Iv) = 1 is equivalent to uu + vv = 1 and uv + vu = 0. As in the case n = 3, it turns out that the conditions x ∈Cl0 4 and N(x) = 1 imply that xvx−1 ∈R4 for all v ∈R4. The only change to the proof is that xvx−1 ∈Cl1 4 can be written as xvx−1 = u + X i,j,k λi,j,keiejek, for some u ∈R4, with {i, j, k} ⊆{1, 2, 3, 4}. 572 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN As in the previous proof, we get λi,j,k = 0. So once again, Spin(4) consists of those elements u + Iv so that uu + vv = 1 and uv + vu = 0, with u and v of the form a1 + bi + cj + dk. Finally, we see that Spin(4) is isomorphic to Spin(3) × Spin(3) under the isomorphism u + vI 7→(u + v, u −v). Indeed, we have N(u + v) = (u + v)(u + v) = 1, and N(u −v) = (u −v)(u −v) = 1, since uu + vv = 1 and uv + vu = 0, and (u + v, u −v)(u′ + v′, u′ −v′) = (uu′ + vv′ + uv′ + vu′, uu′ + vv′ −(uv′ + vu′)). In summary, we have shown that Spin(3) ∼ = SU(2) and Spin(4) ∼ = SU(2) × SU(2). The group Spin(5) is isomorphic to the symplectic group Sp(2), and Spin(6) is isomorphic to SU(4) (see Curtis or Porteous ). Remark: It can be shown that the assertion if x ∈Cl0 n and N(x) = 1, then xvx−1 ∈Rn for all v ∈Rn, is true up to n = 5 (see Porteous , Chapter 13, Proposition 13.58). However, this is already false for n = 6. For example, if X = 1/ √ 2(1 + e1e2e3e4e5e6), it is easy to see that N(X) = 1, and yet, Xe1X−1 / ∈R6. 13.6 The Groups Pin(p, q) and Spin(p, q) For every nondegenerate quadratic form Φ over R, there is an orthogonal basis with respect to which Φ is given by Φ(x1, . . . , xp+q) = x2 1 + · · · + x2 p −(x2 p+1 + · · · + x2 p+q), where p and q only depend on Φ. The quadratic form corresponding to (p, q) is denoted Φp,q and we call (p, q) the signature of Φp,q. Let n = p + q. We define the group O(p, q) as the group of isometries w.r.t. Φp,q, i.e., the group of linear maps f so that Φp,q(f(v)) = Φp,q(v) for all v ∈Rn and the group SO(p, q) as the subgroup of O(p, q) consisting of the isometries f ∈O(p, q) with det(f) = 1. We denote the Clifford algebra Cl(Φp,q) where Φp,q has signature (p, q) by Clp,q, the corresponding Clifford group by Γp,q, and the special Clifford group Γp,q ∩Cl0 p,q by Γ+ p,q. Note that with this new notation, Cln = Cl0,n. 13.6. THE GROUPS PIN(p, q) AND SPIN(p, q) 573  As we mentioned earlier, since Lawson and Michelsohn adopt the opposite of our sign convention in defining Clifford algebras; their Cl(p, q) is our Cl(q, p). As we mentioned in Section 13.4, we have the problem that N(v) = −Φ(v)·1, but −Φ(v) is not necessarily positive (where v ∈Rn). The fix is simple: Allow elements x ∈Γp,q with N(x) = ±1. Definition 13.16. We define the pinor group Pin(p, q) as the group Pin(p, q) = {x ∈Γp,q | N(x) = ±1}, and the spinor group Spin(p, q) as Pin(p, q) ∩Γ+ p,q. Remarks: (1) It is easily checked that the group Spin(p, q) is also given by Spin(p, q) = {x ∈Cl0 p,q | xvx ∈Rn for all v ∈Rn, N(x) = ±1}. This is because Spin(p, q) consists of elements of even degree. (2) One can check that if N(x) = xx ̸= 0, then x−1 = x(N(x))−1. This in turn implies α(x)vx−1 = α(x)vx(N(x))−1 = α(x)vα(t(x))(N(x))−1 = α(x)α(−v)α(t(x))(N(x))−1, since α(v) = −v = α(−xvt(x))(N(x))−1 = xvt(x)(N(x))−1. Thus, we have Pin(p, q) = {x ∈Clp,q | xvt(x) ∈Rn for all v ∈Rn, N(x) = ±1} = {x ∈Clp,q | xvx ∈Rn for all v ∈Rn, N(x) = ±1}. Theorem 13.15 generalizes as follows: Theorem 13.17. The restriction of ρ: Γp,q →GL(n) to the pinor group Pin(p, q) is a surjective homomorphism ρ: Pin(p, q) →O(p, q) whose kernel is {−1, 1}, and the restriction of ρ to the spinor group Spin(p, q) is a surjective homomorphism ρ: Spin(p, q) →SO(p, q) whose kernel is {−1, 1}. Proof. The Cartan-Dieudonn´ e also holds for any nondegenerate quadratic form Φ, in the sense that every isometry in O(Φ) is the composition of reflections defined by hyperplanes 574 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN orthogonal to non-isotropic vectors (see Dieudonn´ e , Chevalley , Bourbaki , or Gal-lier , Chapter 7, Problem 7.14). Thus, Theorem 13.15 also holds for any nondegenerate quadratic form Φ. The only change to the proof is the following: Since N(wj) = −Φ(wj) · 1, we can replace wj by wj/ p |Φ(wj)|, so that N(w1 · · · wk) = ±1, and then f = ρ(w1 · · · wk), and ρ is surjective. If f ∈SO(p, q), then k is even and w1 · · · wk ∈Γ+ p,q and by replacing wj by wj/ p |Φ(wj)| we obtain w1 · · · wk ∈Spin(p, q). If we consider Rn equipped with the quadratic form Φp,q (with n = p + q), we denote the set of elements v ∈Rn with N(v) = ±1 by Sn−1 p,q . We have the following corollary of Theorem 13.17 (generalizing Corollary 13.16). Corollary 13.18. The group Pin(p, q) is generated by Sn−1 p,q , and every element of Spin(p, q) can be written as the product of an even number of elements of Sn−1 p,q . Example 13.7. In Example 13.1 we showed that Cl0,1 ∼ = C. We use a similar argument to calculate Cl1,0. The basis for Cl1,0 is 1, e1 where e2 1 = 1. By using the bijection 1 7→1 + 0, e1 7→0 + 1 we find that Cl1,0 ∼ = R ⊕R, where the multiplication on R ⊕R is given by (a1 + b1)(a2 + b2) = (a1a2 + b1b2) + (a1b2 + a2b1) ∼ = (a11 + b1e1)(a21 + b2e1) = (a1a2 + b1b1)1 + (a1b2 + a2b1)e1. From Example 13.3 we have Pin(0, 1) ∼ = Z/4Z. To calculate Pin(1, 0) = {x ∈Cl1,0 | α(x)vx−1 ∈R for all v ∈R, N(x) = ±1}, we first observe that a typical element of Cl1,0 has the form x = a1 + be1, where e2 1 = 1. Then N(x) = xx = (a1 + be1)(a1 −be1) = (a2 −b2)1 = ±1, 13.6. THE GROUPS PIN(p, q) AND SPIN(p, q) 575 which in turn implies a2 −b2 = ±1, and that x−1 = xN(x)−1 = a1 −be1 a2 −b2 . Since x ∈Pin(1, 0), we have α(x)x−1 ∈R, or equivalently (a1 −be1)a1 −be1 a2 −b2 = 1 a2 −b2[(a2 + b2)1 −2abe1] ∈R. This implies that a = 0 or b = 0. If a = 0, we set a2 −b2 = −1 to obtain b = ±1. If b = 0, we set a2 −b2 = 1 to obtain a = ±1. Thus Pin(1, 0) = {1, e1, −e1, −1} ∼ = Z/2Z × Z/2Z, since 12 = e2 1 = −e2 1. Since Spin(1, 0) = Pin(1, 0) ∩Γ+ 1,0, we deduce that Spin(1, 0) = {1, −1} ∼ = Z/2Z. Example 13.8. We now turn our attention to Clifford algebras over R2. In Example 13.2 we showed that Cl0,2 ∼ = H. To calculate Cl2,0 we first observe that Cl2,0 is spanned by the basis {1, e1, e2, e1e2}, where e2 1 = 1, e2 2 = 1, e1e2 = −e2e1. Define the bijection 1 7→ 1 0 0 1  , e1 7→ 1 0 0 −1  , e2 7→ 0 1 1 0  . Then e1e2 = 1 0 0 −1  0 1 1 0  =  0 1 −1 0  . A few basic computations show that 1 0 0 1  , 1 0 0 −1  , 0 1 1 0  ,  0 1 −1 0  form a basis for M2(R). Furthermore 1 0 0 −1 2 = 1 0 0 1  0 1 1 0 2 = 1 0 0 1  0 1 1 0  1 0 0 −1  = 0 −1 1 0  . 576 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN From this bijection we conclude that Cl2,0 ∼ = M2(R). A similar calculation shows that Cl1,1 ∼ = M2(R). But this time 1 7→ 1 0 0 1  , e1 7→ 1 0 0 −1  , e2 7→  0 1 −1 0  , which implies that e1e2 = 1 0 0 −1   0 1 −1 0  = 0 1 1 0  , and that e2 1 = 1, e2 2 = −1, e1e2 = −e2e1. One can also work out what are Pin(2, 0), Pin(1, 1), and Pin(0, 2). See Choquet-Bruhat , Chapter I, Section 7, page 26, for details As far as Spin(0, 2), we know from Example 13.3 that Spin(0, 2) = Spin(2) ∼ = U(1). By applying the results of the following paragraph regarding the isomorphism between Cl0 p,q and Cl0 q,p, we may deduce that Spin(0, 2) = Spin(2, 0) ∼ = U(1). Finally an application of Corollary 13.18 implies that Spin(1, 1) = {a1 + be1e2 | a2 −b2 = ±1}, and Pin(1, 1) = {a1 + be1e2 | a2 −b2 = ±1} ∪{ae1 + be2 | a2 −b2 = ±1}. Observe that Spin(1, 1) has four connected components and Pin(1, 1) has eight connected components. It is easy to show that SO(1, 1) = a b b a  a2 −b2 = 1  , which has two connected components, and O(1, 1) = a b b a  a2 −b2 = ±1  , which has four connected components. 13.7. THE GROUPS PIN(p, q) AND SPIN(p, q) AS DOUBLE COVERS 577 More generally, it can be shown that Cl0 p,q and Cl0 q,p are isomorphic, from which it follows that Spin(p, q) and Spin(q, p) are isomorphic, but Pin(p, q) and Pin(q, p) are not isomor-phic in general, and in particular, Pin(p, 0) and Pin(0, p) are not isomorphic in general (see Choquet-Bruhat , Chapter I, Section 7). However, due to the “8-periodicity” of the Clif-ford algebras (to be discussed in Section 13.8), it follows that Clp,q and Clq,p are isomorphic when |p −q| = 0 mod 4. Remark: We can also define the group Spin+(p, q) as Spin+(p, q) = {x ∈Cl0 p,q | xvx ∈Rn for all v ∈Rn, N(x) = 1}, and SO0(p, q) as the connected component of SO(p, q) containing the identity. Then it can be shown that the map ρ: Spin+(p, q) →SO0(p, q) is a surjective homomorphism with kernel {−1, 1}; see Lounesto (Chapter 17, Section 17.2). In particular, Spin+(1, 1) = {a1 + be1e2 | a2 −b2 = 1}. This group has two connected components, but it can be shown that for p + q ≥2 and (p, q) ̸= (1, 1) the groups Spin+(p, q) are connected; see Lounesto (Chapter 17, Section 17.2). 13.7 The Groups Pin(p, q) and Spin(p, q) as double cov-ers of O(p, q) and SO(p, q) It turns out that the groups Pin(p, q) and Spin(p, q) have nice topological properties w.r.t. the groups O(p, q) and SO(p, q). To explain this, we review the definition of covering maps and covering spaces (for details, see Fulton , Chapter 11). Another interesting source is Chevalley , where it is proved that Spin(n) is a universal double cover of SO(n) for all n ≥3. Since Clp,q is an algebra of dimension 2p+q, it is a vector space isomorphic to V = R2p+q. Proposition 13.19. The spaces Cl∗ p,q, Pin(p, q), and Spin(p, q) are Lie groups. Proof. The group Cl∗ p,q of units of Clp,q is open in Clp,q, because x ∈Clp,q is a unit if the linear map Lx is a bijection iffdet(Lx) ̸= 0 (where Lx is defined by Lx(y) = xy for all y ∈Clp,q). Thus we have a continuous map L: Clp,q →R given by L(x) = det(Lx) and since Cl∗ p,q = L−1(R−{0}) and R−{0} is open, Cl∗ p,q is open. Thus, Cl∗ p,q is a Lie group, and since Pin(p, q) and Spin(p, q) are clearly closed subgroups of Cl∗ p,q, they are also Lie groups. The definition below is analogous to the definition of a covering map (see Gallot, Hulin, Lafontaine or Gallier and Quaintance ), except that now, we are only dealing with topological spaces and not manifolds. 578 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN Definition 13.17. Given two topological spaces X and Y , a covering map is a continuous surjective map p: Y →X with the property that for every x ∈X, there is some open subset U ⊆X with x ∈U, so that p−1(U) is the disjoint union of open subsets Vα ⊆Y , and the restriction of p to each Vα is a homeomorphism onto U. We say that U is evenly covered by p. We also say that Y is a covering space of X. A covering map p: Y →X is called trivial if X itself is evenly covered by p (i.e., Y is the disjoint union of open subsets Yα each homeomorphic to X), and nontrivial otherwise. When each fiber p−1(x) has the same finite cardinality n for all x ∈X, we say that p is an n-covering (or n-sheeted covering). See Figure 13.1. U x U x p V1 2 V V3 V1 2 V V3 Figure 13.1: Two coverings of S1. The left illustration is p: R →S1 with π(t) = (cos(2πt), sin(2πt)), while the right illustration is the trivial 3-fold covering. Note that a covering map p: Y →X is not always trivial, but always locally trivial (i.e., for every x ∈X, it is trivial in some open neighborhood of x). A covering is trivial iffY is isomorphic to a product space of X × T, where T is any set with the discrete topology. See Figure 13.1. Also, if Y is connected, then the covering map is nontrivial. Definition 13.18. An isomorphism ϕ between covering maps p: Y →X and p′ : Y ′ →X is a homeomorphism ϕ: Y →Y ′ so that p = p′ ◦ϕ. Typically, the space X is connected, in which case it can be shown that all the fibers p−1(x) have the same cardinality. One of the most important properties of covering spaces is the path–lifting property, a property that we will use to show that Spin(n) is path-connected. 13.7. THE GROUPS PIN(p, q) AND SPIN(p, q) AS DOUBLE COVERS 579 Proposition 13.20. (Path lifting) Let p: Y →X be a covering map, and let γ : [a, b] →X be any continuous curve from xa = γ(a) to xb = γ(b) in X. If y ∈Y is any point so that p(y) = xa, then there is a unique curve e γ : [a, b] →Y so that y = e γ(a) and p ◦e γ(t) = γ(t) for all t ∈[a, b]. See Figure 13.2. = ~ = ~ Y a b p X Υ Υ Υ y (a) p(y) Υ (a) = xa Figure 13.2: The lift of a curve γ when π: R →S1 is π(t) = (cos(2πt), sin(2πt)). Proof. See Fulton , Chapter 11, Lemma 11.6. Many important covering maps arise from the action of a group G on a space Y . If Y is a topological space, recall that an action (on the left) of a group G on Y is a map α: G × Y →Y satisfying the following conditions, where for simplicity of notation, we denote α(g, y) by g · y: (1) g · (h · y) = (gh) · y, for all g, h ∈G and y ∈Y ; (2) 1 · y = y, for all ∈Y , where 1 is the identity of the group G; (3) The map y 7→g · y is a homeomorphism of Y for every g ∈G. 580 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN We define an equivalence relation on Y as follows: x ≡y iffy = g · x for some g ∈G (check that this is an equivalence relation). The equivalence class G · x = {g · x | g ∈G} of any x ∈Y is called the orbit of x. We obtain the quotient space Y/G and the projection map p: Y →Y/G sending every y ∈Y to its orbit. The space Y/G is given the quotient topology (a subset U of Y/G is open iffp−1(U) is open in Y ). Given a subset V of Y and any g ∈G, we let g · V = {g · y | y ∈V }. Definition 13.19. We say that G acts evenly on Y if for every y ∈Y , there is an open subset V containing y so that g · V and h · V are disjoint for any two distinct elements g, h ∈G. The importance of the notion a group acting evenly is that such actions induce a covering map. See Figure 13.3. q -1(q) -1 (q) V U 1 2 p V p p Figure 13.3: The 2-fold antipodal covering of RP2 induced by {−1, 1} acting evenly on S2. Proposition 13.21. If G is a group acting evenly on a space Y , then the projection map p: Y →Y/G is a covering map. Proof. See Fulton , Chapter 11, Lemma 11.17. The following proposition shows that Pin(p, q) and Spin(p, q) are nontrivial covering spaces, unless p = q = 1. 13.7. THE GROUPS PIN(p, q) AND SPIN(p, q) AS DOUBLE COVERS 581 Proposition 13.22. For all p, q ≥0, the groups Pin(p, q) and Spin(p, q) are double covers of O(p, q) and SO(p, q), respectively. Furthermore, they are nontrivial covers unless p = q = 1. Proof. We know that kernel of the homomorphism ρ: Pin(p, q) →O(p, q) is Z2 = {−1, 1}. If we let Z2 act on Pin(p, q) in the natural way, then O(p, q) ∼ = Pin(p, q)/Z2, and the reader can easily check that Z2 acts evenly. By Proposition 13.21, we get a double cover. The argument for ρ: Spin(p, q) →SO(p, q) is similar. Since Pin(1, 1) = {a1 + be1e2 | a2 −b2 = ±1} ∪{ae1 + be2 | a2 −b2 = ±1} and O(1, 1) = a b b a  a2 −b2 = ±1  , we see that Pin(1, 1) is the disjoint union of two open subsets each homeomorphic with O(1, 1), and so the covering is trivial. Similarly, since Spin(1, 1) = {a1 + be1e2 | a2 −b2 = ±1}, and SO(1, 1) = a b b a  a2 −b2 = 1  , Spin(1, 1) is the disjoint union of two open subsets each homeomorphic with SO(1, 1), so the covering is also trivial. Let us now assume that p ̸= 1 or q ̸= 1. In order to prove that we have nontrivial covers, it is enough to show that −1 and 1 are connected by a path in Pin(p, q) (If we had Pin(p, q) = U1 ∪U2 with U1 and U2 open, disjoint, and homeomorphic to O(p, q), then −1 and 1 would not be in the same Ui, and so, they would be in disjoint connected components. Thus, −1 and 1 can’t be path–connected, and similarly with Spin(p, q) and SO(p, q).) Since (p, q) ̸= (1, 1), we can find two orthogonal vectors e1 and e2 so that either Φp,q(e1) = Φp,q(e2) = 1 or Φp,q(e1) = Φp,q(e2) = −1. Then, γ(t) = ± cos(2t) 1 + sin(2t) e1e2 = (cos t e1 + sin t e2)(sin t e2 −cos t e1), for 0 ≤t ≤π, defines a path in Spin(p, q), since (± cos(2t) 1 + sin(2t) e1e2)−1 = ± cos(2t) 1 −sin(2t) e1e2, as desired. In particular, if n ≥2, since the group SO(n) is path-connected, the group Spin(n) is also path-connected. Indeed, given any two points xa and xb in Spin(n), there is a path 582 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN γ from ρ(xa) to ρ(xb) in SO(n) (where ρ: Spin(n) →SO(n) is the covering map). By Proposition 13.20, there is a path e γ in Spin(n) with origin xa and some origin e xb so that ρ( e xb) = ρ(xb). However, ρ−1(ρ(xb)) = {−xb, xb}, and so e xb = ±xb. The argument used in the proof of Proposition 13.22 shows that xb and −xb are path-connected, and so, there is a path from xa to xb, and Spin(n) is path-connected. In fact, for n ≥3, it turns out that Spin(n) is simply connected. Such a covering space is called a universal cover (for instance, see Chevalley ). This last fact requires more algebraic topology than we are willing to explain in detail, and we only sketch the proof. The notions of fibre bundle, fibration, and homotopy sequence associated with a fibration are needed in the proof. We refer the perseverant readers to Bott and Tu (Chapter 1 and Chapter 3, Sections 16–17) or Rotman (Chapter 11) for a detailed explanation of these concepts. Recall that a topological space is simply connected if it is path connected and if π1(X) = (0), which means that every closed path in X is homotopic to a point. Since we just proved that Spin(n) is path connected for n ≥2, we just need to prove that π1(Spin(n)) = (0) for all n ≥3. The following facts are needed to prove the above assertion: (1) The sphere Sn−1 is simply connected for all n ≥3. (2) The group Spin(3) ∼ = SU(2) is homeomorphic to S3, and thus, Spin(3) is simply connected. (3) The group Spin(n) acts on Sn−1 in such a way that we have a fibre bundle with fibre Spin(n −1): Spin(n −1) − →Spin(n) − →Sn−1. Fact (1) is a standard proposition of algebraic topology, and a proof can found in many books. A particularly elegant and yet simple argument consists in showing that any closed curve on Sn−1 is homotopic to a curve that omits some point. First, it is easy to see that in Rn, any closed curve is homotopic to a piecewise linear curve (a polygonal curve), and the radial projection of such a curve on Sn−1 provides the desired curve. Then we use the stereographic projection of Sn−1 from any point omitted by that curve to get another closed curve in Rn−1. Since Rn−1 is simply connected, that curve is homotopic to a point, and so is its preimage curve on Sn−1. Another simple proof uses a special version of the Seifert—van Kampen’s theorem (see Gramain ). Fact (2) is easy to establish directly, using (1). To prove (3), we let Spin(n) act on Sn−1 via the standard action: x·v = xvx−1. Because SO(n) acts transitively on Sn−1 and there is a surjection Spin(n) − →SO(n), the group Spin(n) also acts transitively on Sn−1. Now we have to show that the stabilizer of any element of Sn−1 is Spin(n −1). For example, we can do this for e1. This amounts to some simple calculations taking into account the identities among basis elements. Details of this 13.7. THE GROUPS PIN(p, q) AND SPIN(p, q) AS DOUBLE COVERS 583 proof can be found in Mneimn´ e and Testard , Chapter 4. Then by Proposition 10.33, the Lie group Spin(n) is a principal fibre bundle over Sn−1 with fibre Spin(n −1). Now a fibre bundle is a fibration (as defined in Bott and Tu , Chapter 3, Section 16, or in Rotman , Chapter 11). For a proof of this fact, see Rotman , Chapter 11, or Mneimn´ e and Testard , Chapter 4. So, there is a homotopy sequence associated with the fibration (Bott and Tu , Chapter 3, Section 17, or Rotman , Chapter 11, Theorem 11.48), and in particular, we have the exact sequence π1(Spin(n −1)) − →π1(Spin(n)) − →π1(Sn−1). Since π1(Sn−1) = (0) for n ≥3, we get a surjection π1(Spin(n −1)) − →π1(Spin(n)), and so, by induction and (2), we get π1(Spin(n)) ∼ = π1(Spin(3)) = (0), proving that Spin(n) is simply connected for n ≥3. We can also show that π1(SO(n)) = Z/2Z for all n ≥3. For this, we use Theorem 13.15 and Proposition 13.22, which imply that Spin(n) is a fibre bundle over SO(n) with fibre {−1, 1}, for n ≥2: {−1, 1} − →Spin(n) − →SO(n). Again, the homotopy sequence of the fibration exists, and in particular we get the exact sequence π1(Spin(n)) − →π1(SO(n)) − →π0({−1, +1}) − →π0(Spin(n)). Since π0({−1, +1}) = Z/2Z, π0(Spin(n)) = (0), and π1(Spin(n)) = (0), when n ≥3, we get the exact sequence (0) − →π1(SO(n)) − →Z/2Z − →(0), and so, π1(SO(n)) = Z/2Z. Therefore, SO(n) is not simply connected for n ≥3. Remark: Of course, we have been rather cavalier in our presentation. Given a topological space X, the group π1(X) is the fundamental group of X, i.e. the group of homotopy classes of closed paths in X (under composition of loops). But π0(X) is generally not a group! Instead, π0(X) is the set of path-connected components of X. However, when X is a Lie group, π0(X) is indeed a group. Also, we have to make sense of what it means for the sequence to be exact. All this can be made rigorous (see Bott and Tu , Chapter 3, Section 17, or Rotman , Chapter 11). 584 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN 13.8 Periodicity of the Clifford Algebras Clp,q It turns out that the real algebras Clp,q can be build up as tensor products of the basic algebras R, C, and H. As pointed out by Lounesto (Section 23.16 ), the description of the real algebras Clp,q as matrix algebras and the 8-periodicity was first observed by Elie Cartan in 1908; see Cartan’s article, Nombres Complexes, based on the original article in German by E. Study, in Molk , article I-5 (fasc. 3), pages 329-468. These algebras are defined in Section 36 under the name “‘Systems of Clifford and Lipschitz,” pages 463-466. Of course, Cartan used a very different notation; see page 464 in the article cited above. These facts were rediscovered independently by Raoul Bott in the 1960’s (see Raoul Bott’s comments in Volume 2 of his collected papers.). We adopt the notation K(n) for the algebra of n × n matrices over a ring or a field K; here K = R, C, H. This is the notation used in most of the literature on Clifford algebras, for instance Atiyah, Bott and Shapiro , and it is a departure from the notation Mn(K) that we have been using all along. As mentioned in Examples 13.3 and 13.7, it is not hard to show that Cl0,1 = C Cl1,0 = R ⊕R Cl0,2 = H Cl2,0 = R(2), and Cl1,1 = R(2). The key to the classification is the following lemma: Lemma 13.23. We have the isomorphisms Cl0,n+2 ∼ = Cln,0 ⊗Cl0,2 Cln+2,0 ∼ = Cl0,n ⊗Cl2,0 Clp+1,q+1 ∼ = Clp,q ⊗Cl1,1, for all n, p, q ≥0. Proof. Let Φ0,n+2(x) = −∥x∥2, where ∥x∥is the standard Euclidean norm on Rn+2, and let (e1, . . . , en+2) be an orthonormal basis for Rn+2 under the standard Euclidean inner product. We also let (e′ 1, . . . , e′ n) be a set of generators for Cln,0 and (e′′ 1, e′′ 2) be a set of generators for Cl0,2. We can define a linear map f : Rn+2 →Cln,0 ⊗Cl0,2 by its action on the basis (e1, . . . , en+2) as follows: f(ei) =  e′ i ⊗e′′ 1e′′ 2 for 1 ≤i ≤n 1 ⊗e′′ i−n for n + 1 ≤i ≤n + 2. Observe that for 1 ≤i, j ≤n, we have f(ei)f(ej) + f(ej)f(ei) = (e′ ie′ j + e′ je′ i) ⊗(e′′ 1e′′ 2)2 = −2δij1 ⊗1, 13.8. PERIODICITY OF THE CLIFFORD ALGEBRAS CLp,q 585 since e′′ 1e′′ 2 = −e′′ 2e′′ 1, (e′′ 1)2 = −1, and (e′′ 2)2 = −1, and e′ ie′ j = −e′ je′ i, for all i ̸= j, and (e′ i)2 = 1, for all i with 1 ≤i ≤n. Also, for n + 1 ≤i, j ≤n + 2, we have f(ei)f(ej) + f(ej)f(ei) = 1 ⊗(e′′ i−ne′′ j−n + e′′ j−ne′′ i−n) = −2δij1 ⊗1, and f(ei)f(ek) + f(ek)f(ei) = 2e′ i ⊗(e′′ 1e′′ 2e′′ k−n + e′′ k−ne′′ 1e′′ 2) = 0, for 1 ≤i, j ≤n and n + 1 ≤k ≤n + 2 (since e′′ k−n = e′′ 1 or e′′ k−n = e′′ 2). Thus, we have (f(x))2 = −∥x∥2 · 1 ⊗1 for all x ∈Rn+2, and by the universal mapping property of Cl0,n+2, we get an algebra map e f : Cl0,n+2 →Cln,0 ⊗Cl0,2. Since e f maps onto a set of generators, it is surjective. However dim(Cl0,n+2) = 2n+2 = 2n · 2 = dim(Cln,0)dim(Cl0,2) = dim(Cln,0 ⊗Cl0,2), and e f is an isomorphism. The proof of the second identity is analogous. For the third identity, we have Φp,q(x1, . . . , xp+q) = x2 1 + · · · + x2 p −(x2 p+1 + · · · + x2 p+q), and let (e1, . . . , ep+1, ϵ1, . . . , ϵq+1) be an orthogonal basis for Rp+q+2 so that Φp+1,q+1(ei) = +1 and Φp+1,q+1(ϵj) = −1 for i = 1, . . . , p+1 and j = 1, . . . , q+1. Also, let (e′ 1, . . . , e′ p, ϵ′ 1, . . . , ϵ′ q) be a set of generators for Clp,q and (e′′ 1, ϵ′′ 1) be a set of generators for Cl1,1. We define a linear map f : Rp+q+2 →Clp,q ⊗Cl1,1 by its action on the basis as follows: f(ei) =  e′ i ⊗e′′ 1ϵ′′ 1 for 1 ≤i ≤p 1 ⊗e′′ 1 for i = p + 1, and f(ϵj) =  ϵ′ j ⊗e′′ 1ϵ′′ 1 for 1 ≤j ≤q 1 ⊗ϵ′′ 1 for j = q + 1. We can check that (f(x))2 = Φp+1,q+1(x) · 1 ⊗1 for all x ∈Rp+q+2, and we finish the proof as in the first case. To apply this lemma, we need some further isomorphisms among various matrix algebras. 586 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN Proposition 13.24. The following isomorphisms hold: R(m) ⊗R(n) ∼ = R(mn) for all m, n ≥0 R(n) ⊗R K ∼ = K(n) for K = C or K = H and all n ≥0 C ⊗R C ∼ = C ⊕C C ⊗R H ∼ = C(2) H ⊗R H ∼ = R(4). Proof. Details can be found in Lawson and Michelsohn . The first two isomorphisms are quite obvious. The third isomorphism C ⊕C →C ⊗R C is obtained by sending (1, 0) 7→1 2(1 ⊗1 + i ⊗i), (0, 1) 7→1 2(1 ⊗1 −i ⊗i). The field C is isomorphic to the subring of H generated by i. Thus, we can view H as a C-vector space under left scalar multiplication. Consider the R-bilinear map π: C × H →HomC(H, H) given by πy,z(x) = yxz, where y ∈C and x, z ∈H. Thus, we get an R-linear map π: C ⊗R H →HomC(H, H). However, we have HomC(H, H) ∼ = C(2). Furthermore, since πy,z ◦πy′,z′ = πyy′,zz′, the map π is an algebra homomorphism π: C × H →C(2). We can check on a basis that π is injective, and since dimR(C × H) = dimR(C(2)) = 8, the map π is an isomorphism. The last isomorphism is proved in a similar fashion. We now have the main periodicity theorem. Theorem 13.25. (Cartan/Bott) For all n ≥0, we have the following isomorphisms: Cl0,n+8 ∼ = Cl0,n ⊗Cl0,8 Cln+8,0 ∼ = Cln,0 ⊗Cl8,0. Furthermore, Cl0,8 = Cl8,0 = R(16). 13.8. PERIODICITY OF THE CLIFFORD ALGEBRAS CLp,q 587 Proof. By Lemma 13.23 we have the isomorphisms Cl0,n+2 ∼ = Cln,0 ⊗Cl0,2 Cln+2,0 ∼ = Cl0,n ⊗Cl2,0, and thus, Cl0,n+8 ∼ = Cln+6,0 ⊗Cl0,2 ∼ = Cl0,n+4 ⊗Cl2,0 ⊗Cl0,2 ∼ = · · · ∼ = Cl0,n ⊗Cl2,0 ⊗Cl0,2 ⊗Cl2,0 ⊗Cl0,2. Since Cl0,2 = H and Cl2,0 = R(2), by Proposition 13.24, we get Cl2,0 ⊗Cl0,2 ⊗Cl2,0 ⊗Cl0,2 ∼ = H ⊗H ⊗R(2) ⊗R(2) ∼ = R(4) ⊗R(4) ∼ = R(16). The second isomorphism is proved in a similar fashion. From all this, we can deduce the following table. n 0 1 2 3 4 5 6 7 8 Cl0,n R C H H ⊕H H(2) C(4) R(8) R(8) ⊕R(8) R(16) Cln,0 R R ⊕R R(2) C(2) H(2) H(2) ⊕H(2) H(4) C(8) R(16) A table of the Clifford groups Clp,q for 0 ≤p, q ≤7 can be found in Kirillov , and for 0 ≤p, q ≤8, in Lawson and Michelsohn (but beware that their Clp,q is our Clq,p). It can also be shown that Clp+1,q ∼ = Clq+1,p Clp,q ∼ = Clp−4,q+4 with p ≥4 in the second identity (see Lounesto , Chapter 16, Sections 16.3 and 16.4). Using the second identity, if |p−q| = 4k, it is easily shown by induction on k that Clp,q ∼ = Clq,p, as claimed in the previous section. We also have the isomorphisms Clp,q ∼ = Cl0 p,q+1, frow which it follows that Spin(p, q) ∼ = Spin(q, p) (see Choquet-Bruhat , Chapter I, Sections 4 and 7). However, in general, Pin(p, q) and Pin(q, p) are not isomorphic. In fact, Pin(0, n) and Pin(n, 0) are not isomorphic if n ̸= 4k, with k ∈N (see Choquet-Bruhat , Chapter I, Section 7, page 27). 588 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN 13.9 The Complex Clifford Algebras Cl(n, C) One can also consider Clifford algebras over the complex field C. In this case, it is well-known that every nondegenerate quadratic form can be expressed by ΦC n(x1, . . . , xn) = x2 1 + · · · + x2 n in some orthonormal basis. Also, it is easily shown that the complexification C ⊗R Clp,q of the real Clifford algebra Clp,q is isomorphic to Cl(ΦC n). Thus, all these complex algebras are isomorphic for p + q = n, and we denote them by Cl(n, C). Theorem 13.23 yields the following periodicity theorem: Theorem 13.26. The following isomorphisms hold: Cl(n + 2, C) ∼ = Cl(n, C) ⊗C Cl(2, C), with Cl(2, C) = C(2). Proof. Since Cl(n, C) = C ⊗R Cl0,n = C ⊗R Cln,0, by Lemma 13.23, we have Cl(n + 2, C) = C ⊗R Cl0,n+2 ∼ = C ⊗R (Cln,0 ⊗R Cl0,2) ∼ = (C ⊗R Cln,0) ⊗C (C ⊗R Cl0,2). However, Cl0,2 = H, Cl(n, C) = C ⊗R Cln,0, and C ⊗R H ∼ = C(2), so we get Cl(2, C) = C(2) and Cl(n + 2, C) ∼ = Cl(n, C) ⊗C C(2), and the theorem is proven. As a corollary of Theorem 13.26, we obtain the fact that Cl(2k, C) ∼ = C(2k) and Cl(2k + 1, C) ∼ = C(2k) ⊕C(2k). The table of the previous section can also be completed as follows n 0 1 2 3 4 5 6 7 8 Cl0,n R C H H ⊕H H(2) C(4) R(8) R(8) ⊕R(8) R(16) Cln,0 R R ⊕R R(2) C(2) H(2) H(2) ⊕H(2) H(4) C(8) R(16) Cl(n, C) C 2C C(2) 2C(2) C(4) 2C(4) C(8) 2C(8) C(16), where 2C(k) is an abbreviation for C(k) ⊕C(k). 13.10. CLIFFORD GROUPS OVER A FIELD K 589 13.10 Clifford Groups Over a Field K In this final section we quickly indicate which of the results about Clifford algebras, Clifford groups, and the Pin and Spin groups obtained in Sections 13.3–13.6 for vector spaces over the fields R and C generalize to nondegenerate bilinear forms on vector spaces over an arbitrary field K of characteristic different from 2. As we will see, most results generalize, except for some of the surjectivity results such as Theorem 13.17. Let V be a finite-dimensional vector space over a field K of characteristic ̸= 2, let ϕ: V × V →K be a possibly degenerate symmetric bilinear form, and let Φ(v) = ϕ(v, v) be the corresponding quadratic form. Definition 13.20. A Clifford algebra associated with V and Φ is a K-algebra Cl(V, Φ) together with a linear map iΦ : V →Cl(V, Φ) satisfying the condition (iΦ(v))2 = Φ(v) · 1 for all v ∈V , and so that for every K-algebra A and every linear map f : V →A with (f(v))2 = Φ(v) · 1A for all v ∈V , there is a unique algebra homomorphism f : Cl(V, Φ) →A so that f = f ◦iΦ, as in the diagram below. V iΦ/ f $ Cl(V, Φ) f  A We use the notation λ · u for the product of a scalar λ ∈K and of an element u in the algebra Cl(V, Φ), and juxtaposition uv for the multiplication of two elements u and v in the algebra Cl(V, Φ). By a familiar argument, any two Clifford algebras associated with V and Φ are isomorphic. We often denote iΦ by i. To show the existence of Cl(V, Φ), since the tensor algebra T(V ) makes sense for a vector space V over any field K, observe that T(V )/A does the job, where A is the ideal of T(V ) generated by all elements of the form v⊗v−Φ(v)·1, where v ∈V . The map iΦ : V →Cl(V, Φ) is the composition V ι1 − →T(V ) π − →T(V )/A, where π is the natural quotient map. We often denote the Clifford algebra Cl(V, Φ) simply by Cl(Φ). Proposition 13.27. Every Clifford algebra Cl(Φ) possesses a canonical anti-automorphism t: Cl(Φ) →Cl(Φ) satisfying the properties t(xy) = t(y)t(x), t ◦t = id, and t(i(v)) = i(v), for all x, y ∈Cl(Φ) and all v ∈V . Furthermore, such an anti-automorphism is unique. 590 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN Proposition 13.28. Every Clifford algebra Cl(Φ) has a unique canonical automorphism α: Cl(Φ) →Cl(Φ) satisfying the properties α ◦α = id, and α(i(v)) = −i(v), for all v ∈V . Write Cl(Φ) = Cl0(Φ) ⊕Cl1(Φ), where Cli(Φ) = {x ∈Cl(Φ) | α(x) = (−1)ix}, where i = 0, 1. We say that we have a Z/2-grading. The theorem about the existence of a nice basis of Cl(Φ) only depends on the fact that there is always a basis of V that is orthogonal with respect to ϕ (even if ϕ is degenerate) so we have Theorem 13.29. For every vector space V of finite dimension n, the map i: V →Cl(Φ) is injective. Given a basis (e1, . . . , en) of V , the 2n −1 products i(ei1)i(ei2) · · · i(eik), 1 ≤i1 < i2 · · · < ik ≤n, and 1 form a basis of Cl(Φ). Thus, Cl(Φ) has dimension 2n. Since i is injective, for simplicity of notation, from now on we write u for i(u). Theorem 13.29 implies that if (e1, . . . , en) is an orthogonal basis of V with respect to Φ, then Cl(Φ) is the algebra presented by the generators (e1, . . . , en) and the relations e2 j = Φ(ej) · 1, 1 ≤j ≤n, and ejek = −ekej, 1 ≤j, k ≤n, j ̸= k. If V has finite dimension n and (e1, . . . , en) is a basis of V , by Theorem 13.29, the maps t and α are completely determined by their action on the basis elements. Namely, t is defined by t(ei) = ei t(ei1ei2 · · · eik) = eikeik−1 · · · ei1, where 1 ≤i1 < i2 · · · < ik ≤n, and of course, t(1) = 1. The map α is defined by α(ei) = −ei α(ei1ei2 · · · eik) = (−1)kei1ei2 · · · eik 13.10. CLIFFORD GROUPS OVER A FIELD K 591 where 1 ≤i1 < i2 < · · · < ik ≤n, and of course, α(1) = 1. Furthermore, the even-graded elements (the elements of Cl0(Φ)) are those generated by 1 and the basis elements consisting of an even number of factors ei1ei2 · · · ei2k, and the odd-graded elements (the elements of Cl1(Φ)) are those generated by the basis elements consisting of an odd number of factors ei1ei2 · · · ei2k+1. The definition of the Clifford group given in Section 13.4 does not depend on the field K or on the fact that the symmetric bilinear form ϕ is nondegenerate. Definition 13.21. Given a finite dimensional vector space V over a field K and a quadratic form Φ on V , the Clifford group of Φ is the group Γ(Φ) = {x ∈Cl(Φ)∗| α(x)vx−1 ∈V for all v ∈V }. For any x ∈Γ(Φ), let ρx : V →V be the map defined by v 7→α(x)vx−1, v ∈V. As in Section 13.4, the map ρ: Γ(Φ) →GL(V ) given by x 7→ρx is a linear action, and Γ(Φ) is a group. This is because V is finite-dimensional and α is an automorphism. We also define the group Γ+(Φ), called the special Clifford group, by Γ+(Φ) = Γ(Φ) ∩Cl0(Φ). Proposition 13.30. The maps α and t induce an automorphism and an anti-automorphism of the Clifford group, Γ(Φ). The following key result obtained in Section 13.4 still holds because its proof does not depend on the field K. Theorem 13.31. Let V be a finite dimensional vector space over a field K and let Φ a quadratic form on V . For every element x of the Clifford group Γ(Φ), if x ∈V then Φ(x) ̸= 0 and the map ρx : V →V given by v 7→α(x)vx−1 for all v ∈V is the reflection about the hyperplane H orthogonal to the non-isotropic vector x. We would like to show that ρ is a surjective homomorphism from Γ(Φ) onto O(Φ), and a surjective homomorphism from Γ+(Φ) onto SO(Φ). In order to prove that ρx ∈O(Φ) for any x ∈Γ(Φ), we define a notion of norm on Γ(Φ), and for this we need to define a notion of conjugation on Cl(Φ). Definition 13.22. We define conjugation on a Clifford algebra Cl(Φ) as the map x 7→x = t(α(x)) for all x ∈Cl(Φ). 592 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN Conjugation is an anti-automorphism. If V has finite dimension n and (e1, . . . , en) is a basis of V , in view of previous remarks, conjugation is defined by ei = −ei ei1ei2 · · · eik = (−1)keikeik−1 · · · ei1 where 1 ≤i1 < i2 · · · < ik ≤n, and of course, 1 = 1. Definition 13.23. The map N : Cl(Φ) →Cl(Φ) given by N(x) = xx is called the norm of Cl(Φ). Observe that N(v) = vv = −v2 = −Φ(v) · 1 for all v ∈V . Up to this point, there is no assumption regarding the degeneracy of ϕ. Now we will need to assume that ϕ is nondegenerate. We observed that the proof of Lemma 13.11 goes through as long as ϕ is nondegenerate. Thus we have Lemma 13.32. Assume ϕ is a nondegenerate bilinear map on V . The kernel of the map ρ: Γ(Φ) →GL(V ) is K∗·1, the multiplicative group of nonzero scalar multiples of 1 ∈Cl(Φ). We also observed that the proof of Proposition 13.12 goes through as long as ϕ is non-degenerate. Proposition 13.33. Assume ϕ is a nondegenerate bilinear map on V . If x ∈Γ(Φ), then N(x) ∈K∗· 1. Similarly the following holds. Proposition 13.34. Assume ϕ is a nondegenerate bilinear map on V . The restriction of the norm N to Γ(Φ) is a homomorphism N : Γ(Φ) →K∗· 1, and N(α(x)) = N(x) for all x ∈Γ(Φ). Finally we obtain the following result. Proposition 13.35. Assume ϕ is a nondegenerate bilinear map on V . The set of non-isotropic vectors in V (those x ∈V such that Φ(x) ̸= 0) is a subset of Γ(Φ), and ρ(Γ(Φ)) ⊆ O(Φ). We have the following theorem. Theorem 13.36. Assume ϕ is a nondegenerate bilinear map on V . The map ρ: Γ(Φ) → O(Φ) is a surjective homomorphism whose kernel is K∗·1, and the map ρ: Γ+(Φ) →SO(Φ) is a surjective homomorphism whose kernel is K∗· 1. 13.10. CLIFFORD GROUPS OVER A FIELD K 593 Proof. The Cartan-Dieudonn´ e holds for any nondegenerate quadratic form Φ over a field of characteristic ̸= 2, in the sense that every isometry f ∈O(Φ) is the composition f = s1◦· · ·◦sk of reflections sj defined by hyperplanes orthogonal to non-isotropic vectors wj ∈V . (see Dieudonn´ e , Chevalley , Bourbaki , or Gallier , Chapter 7, Problem 7.14). Then we have f = ρ(w1 · · · wk), and since the wj are non-isotrotpic Φ(wj) ̸= 0, so wj ∈Γ(Φ) and we have w1 · · · wk ∈Γ(Φ). For the second statement, we need to show that ρ maps Γ+(Φ) into SO(Φ). If this was not the case, there would be some improper isometry f ∈O(Φ) so that ρx = f, where x ∈Γ(Φ) ∩Cl0(Φ). However, we can express f as the composition of an odd number of reflections, say f = ρ(w1 · · · w2k+1). Since ρ(w1 · · · w2k+1) = ρx, we have x−1w1 · · · w2k+1 ∈Ker (ρ). By Lemma 13.32, we must have x−1w1 · · · w2k+1 = λ · 1 for some λ ∈K∗, and thus w1 · · · w2k+1 = λ · x, where x has even degree and w1 · · · w2k+1 has odd degree, which is impossible. The groups Pin and Spin are defined as follows. Definition 13.24. Assume ϕ is a nondegenerate bilinear map on V . We define the pinor group Pin(Φ) as the group Pin(Φ) = {x ∈Γ(Φ) | N(x) = ±1}, equivalently Pin(Φ) = {x ∈Cl(Φ) | xvx ∈V for all v ∈V , N(x) = ±1}, and the spinor group Spin(Φ) as Pin(Φ) ∩Γ+(Φ). Equivalently, Spin(Φ) = {x ∈Cl0(Φ) | xvx ∈V for all v ∈V , N(x) = ±1}. This time, if the field K is not R or C, it is not obvious that the restriction of ρ to Pin(Φ) is surjective onto O(Φ) and that the restriction of ρ to Spin(Φ) is surjective onto SO(Φ). To understand this better, assume that ρx = ρ(y) = f 594 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN for some x, y ∈Γ(Φ) and some f ∈O(Φ). Then ρ(yx−1) = id, which by Lemma 13.32 implies that yx−1 = λ1 for some λ ∈K∗, that is, y = λx. Then we obtain N(y) = yy = λx λx = λ2xx = λ2N(x). This suggests defining a map σ from O(Φ) to the group K∗/(±(K∗)2) by σ(f) = [N(x)] for any x ∈Γ(Φ) with ρx = f, where ±(K∗)2 denotes the multiplicative subgroup of K∗consisting of all elements of the form ±λ2, with λ ∈K∗, and [N(x)] denotes the equivalence class of N(x) in K∗/(±(K∗)2). Then we have the following result. Proposition 13.37. Assume ϕ is a nondegenerate bilinear map on V . We have the exact sequences (1) / {−1, 1} / Pin(Φ) ρ / O(Φ) σ / Im σ / (1) and (1) / {−1, 1} / Spin(Φ) ρ / SO(Φ) σ / Im σ / (1). Proof. Since by Lemma 13.32 the kernel of the map ρ: Γ(Φ) →GL(V ) is K∗· 1, and since N(x) = ±1 if x ∈Pin(Φ), the sequence is exact at Pin(Φ). For any x ∈Pin(Φ), since N(x) = ±1, we have σ(ρx) = 1, which means that Im ρ ⊆Ker σ. Assume that f ∈Ker σ, which means that ρx = f some x ∈Γ(Φ) such that N(x) = ±λ2 for some λ ∈K∗. Then N(λ−1x) = ±1 so λ−1x ∈Pin(Φ) and since ρ is a homomorphism, ρ(λ−1x) = ρ(λ−1)ρx = id ◦f = f, which shows that Ker σ ⊆Im ρ. The fact that the second sequence is exact follows from the fact that the first sequence is exact and by definition of Spin(Φ). If K = R or K = C, every element of K is of the form ±λ2, so ρ is surjective, which gives another proof of Theorem 13.15. Remarks: (1) Our definition of σ is inspired by the definition of Mnemn´ e and Testard (Chapter 4, Section 4.7) who define σ from SO(Φ) to the group K∗/(K∗)2 by σ(f) = [N(x)] for any x ∈Γ(Φ) with ρx = f. Allowing negative squares as well as positive squares yields the surjectivity of ρ when K = R or C. 13.10. CLIFFORD GROUPS OVER A FIELD K 595 (2) We define the subgroup Spin+(Φ) of Spin(Φ) by Spin+(Φ) = {x ∈Cl0(Φ) | xvx ∈V for all v ∈V , N(x) = 1}. The image of Spin+(Φ) by ρ is a subgroup of SO(Φ) denoted by SO+(Φ). For example, when K = R and Φ = Φp,q, we have SO+(Φp,q) = SO0(p, q), the connected component of the identity. The group Spin+(1, 1) has two connected components but Spin+(p, q) is connected for p + q ≥2 and (p, q) ̸= (1, 1). The groups Spin+(n, 1) ∼ = Spin+(1, n) are simply connected for n ≥3, but in general Spin+(p, q) is not simply connected; for example, Spin+(3, 3) is not simply connected; see Lounesto (Chapter 17). If K is a an arbitrary field, we can’t expect that the periodicity results of Section 13.8 and Section 13.9 hold for the Clifford algebra Cl(Φ). Still some interesting facts about the structure of the algebras Cl(Φ) can be established. For this, we need to define the notion of a central simple K-algebra. If A is an associative K-algebra over a field K with identity element 1, then there is an injection of the field K into A given by λ 7→λ · 1, so that we can view K · 1 as a copy of K in A. Observe that every element λ · 1 ∈K · 1 commutes with every element u ∈A, since by K-bilinearity of the multiplication operation (u, v) 7→uv on A, we have (λ · 1)u = λ · (1u) = λ · u and u(λ · 1) = λ · (u1) = λ · u. This shows that K · 1 is a contained in the center of A, which is defined as follows. Definition 13.25. Given any associative K-algebra with identity element 1 (where K is a field), the center Z(A) of A is the subalgebra of A given by Z(A) = {u ∈A | uv = vu for all v ∈A}. The K-algebra A is called a central algebra if Z(A) = K · 1. As we just observed, K · 1 ⊆Z(A). A typical example of a central K-algebra is the algebra Mn(K) of n × n matrices over a field K. Definition 13.26. An associative K-algebra with identity element 1 is simple if for any two-sided ideal A in A, either A = (0) or A = A. In other words A contains no nonzero proper two-sided ideals. Again, a typical example of a simple K-algebra is the algebra Mn(K) of n × n matrices over a field K. By a theorem of Wedderburn, any finite-dimensional central simple K-algebra is isomorphic to the algebra Mn(∆) of n × n matrices over some division ring (also called a skew field) ∆whose center is K, for some n ≥1; see Dummit and Foote , Chapter 17, Section 4, and Chapter 18, Section 2, Theorem 4. Based on results of Chevalley , the following results are proved in Bourbaki (§9, no 4, Theorem 2, its Corollary, and Theorem 3). 596 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN Theorem 13.38. If m = 2r is even, for any nondegenerate quadratic form Φ over a K-vector space E of dimension m, the Clifford algebra Cl(Φ) is a central simple algebra of dimension 2m. If m > 0, the Clifford algebra Cl0(Φ) has a center Z of dimension 2, and either Cl0(Φ) is simple if Z is a field, or Cl0(Φ) is the direct sum of two simple subalgebras of dimension 2m−2. Remark: More is true when Φ is a neutral form (which means that E is the direct sum of two totally isotropic subspaces). In this case, Cl(Φ) is isomorphic to the algebra of endomorphisms End(V N) of the exterior product V N of some totally isotropic subspace N of E of dimension r. Theorem 13.39. If m = 2r + 1 is odd, for any nondegenerate quadratic form Φ over a K-vector space E of dimension m (with char(K) ̸= 2), the Clifford algebra Cl0(Φ) is a central simple algebra of dimension 22r. The Clifford algebra Cl(Φ) has a center Z of dimension 2, and Cl(Φ) is isomorphic to Z ⊗K Cl0(Φ); as a consequence, Cl(Φ) is either simple or the direct sum of two simple subalgebras. A related result due to Chevalley asserts that Cl(Φ) is isomorphic to a subalgebra of the algebra of endomorphisms End(V E). To prove this, Chevalley introduced a product operation on V E called the Clifford product. The reader is referred to Lounesto (Chapter 22) for details on this construction, as well as a simpler construction due to Riesz (who introduces an exterior product in Cl(Φ)). The above results have some interesting applications to representation theory. Indeed, they lead to certain irreducible representations known as spin representations or half-spin representations first discovered by ´ Elie Cartan. The spaces that they act on are called spinors or half-spinors. Such representations play an important role in theoretical physics. The interested reader is referred to Fulton and Harris (Lecture 20) or Jost (Section 2.4). 13.11 Problems Problem 13.1. The “right way” (meaning convenient and rigorous) to define the unit quaternions is to define them as the elements of the unitary group SU(2), namely the group of 2 × 2 complex matrices of the form  α β −β α  α, β ∈C, αα + ββ = 1. Then, the quaternions are the elements of the real vector space H = R SU(2). Let 1, i, j, k be the matrices 1 = 1 0 0 1  , i = i 0 0 −i  , j =  0 1 −1 0  , k = 0 i i 0  , 13.11. PROBLEMS 597 then H is the set of all matrices of the form X = a1 + bi + cj + dk, a, b, c, d ∈R. Indeed, every matrix in H is of the form X =  a + ib c + id −(c −id) a −ib  , a, b, c, d ∈R. (1) Prove that the quaternions 1, i, j, k satisfy the famous identities discovered by Hamil-ton: i2 = j2 = k2 = ijk = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j. Prove that H is a skew field (a noncommutative field) called the quaternions, and a real vector space of dimension 4 with basis (1, i, j, k); thus as a vector space, H is isomorphic to R4. A concise notation for the quaternion X defined by α = a + ib and β = c + id is X = [a, (b, c, d)]. We call a the scalar part of X and (b, c, d) the vector part of X. With this notation, X∗= [a, −(b, c, d)], which is often denoted by X. The quaternion X is called the conjugate of q. If q is a unit quaternion, then q is the multiplicative inverse of q. A pure quaternion is a quaternion whose scalar part is equal to zero. (2) Given a unit quaternion q =  α β −β α  ∈SU(2), the usual way to define the rotation ρq (of R3) induced by q is to embed R3 into H as the pure quaternions, by ψ(x, y, z) =  ix y + iz −y + iz −ix  , (x, y, z) ∈R3. Observe that the above matrix is skew-Hermitian (ψ(x, y, z)∗= −ψ(x, y, z)). But, the space of skew-Hermitian matrices is the Lie algebra su(2) of SU(2), so ψ(x, y, z) ∈su(2). Then, q defines the map ρq (on R3) given by ρq(x, y, z) = ψ−1(qψ(x, y, z)q∗), 598 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN where q∗is the inverse of q (since SU(2) is a unitary group) and is given by q∗= α −β β α  . Actually, the adjoint representation of the group SU(2) is the group homomorphism Ad: SU(2) →GL(su(2)) defined such that for every q ∈SU(2), Adq(A) = qAq∗, A ∈su(2). Therefore, modulo the isomorphism ψ, the linear map ρq is the linear isomorphism Adq. In fact, ρq is a rotation (and so is Adq), which you will prove shortly. Since the matrix ψ(x, y, z) is skew-Hermitian, the matrix −iψ(x, y, z) is Hermitian, and we have −iψ(x, y, z) =  x z −iy z + iy −x  = xσ3 + yσ2 + zσ1, where σ1, σ2, σ3 are the Pauli spin matrices σ1 = 0 1 1 0  , σ2 = 0 −i i 0  , σ3 = 1 0 0 −1  . Check that i = iσ3, j = iσ2, k = iσ1. Prove that matrices of the form xσ3 + yσ2 + zσ1 (with x, y, x ∈R) are exactly the 2 × 2 Hermitian matrix with zero trace. (3) Prove that for every q ∈SU(2), if A is any 2 × 2 Hermitian matrix with zero trace as above, then qAq∗is also a Hermitian matrix with zero trace. Prove that det(xσ3 + yσ2 + zσ1) = det(qAq∗) = −(x2 + y2 + z2). We can embed R3 into the space of Hermitian matrices with zero trace by ϕ(x, y, z) = xσ3 + yσ2 + zσ1. Check that ϕ = −iψ and ϕ−1 = iψ−1. Prove that every quaternion q induces a map rq on R3 by rq(x, y, z) = ϕ−1(qϕ(x, y, z)q∗) = ϕ−1(q(xσ3 + yσ2 + zσ1)q∗) which is clearly linear, and an isometry. Thus, rq ∈O(3). 13.11. PROBLEMS 599 (4) Find the fixed points of rq, where q = (a, (b, c, d)). If (b, c, d) ̸= (0, 0, 0), then show that the fixed points (x, y, z) of rq are solutions of the equations −dy + cz = 0 cx −by = 0 dx −bz = 0. This linear system has the nontrivial solution (b, c, d) and the matrix of this system is   0 −d c c −b 0 d 0 −b  . Prove that the above matrix has rank 2, so the fixed points of rq form the one-dimensional space spanned by (b, c, d). Deduce from this that rq must be a rotation. Prove that r: SU(2) →SO(3) given by r(q) = rq is a group homomorphism whose kernel is {I, −I}. (5) Find the matrix Rq representing rq explicitly by computing q(xσ3 + yσ2 + zσ1)q∗=  α β −β α   x z −iy z + iy −x  α −β β α  . You should find Rq =   a2 + b2 −c2 −d2 2bc −2ad 2ac + 2bd 2bc + 2ad a2 −b2 + c2 −d2 −2ab + 2cd −2ac + 2bd 2ab + 2cd a2 −b2 −c2 + d2  . Since a2 + b2 + c2 + d2 = 1, this matrix can also be written as Rq =   2a2 + 2b2 −1 2bc −2ad 2ac + 2bd 2bc + 2ad 2a2 + 2c2 −1 −2ab + 2cd −2ac + 2bd 2ab + 2cd 2a2 + 2d2 −1  . Prove that rq = ρq. (6) To prove the surjectivity of r algorithmically, proceed as follows. First, prove that tr(Rq) = 4a2 −1, so a2 = tr(Rq) + 1 4 . If R ∈SO(3) is any rotation matrix and if we write R =   r11 r12 r13 r21 r22 r23 r31 r32 r33,   600 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN we are looking for a unit quaternion q ∈SU(2) such that rq = R. Therefore, we must have a2 = tr(R) + 1 4 . We also know that tr(R) = 1 + 2 cos θ, where θ ∈[0, π] is the angle of the rotation R. Deduce that |a| = cos θ 2  (0 ≤θ ≤π). There are two cases. Case 1. tr(R) ̸= −1, or equivalently θ ̸= π. In this case a ̸= 0. Pick a = p tr(R) + 1 2 . Then, show that b = r32 −r23 4a , c = r13 −r31 4a , d = r21 −r12 4a . Case 2. tr(R) = −1, or equivalently θ = π. In this case a = 0. Prove that 4bc = r21 + r12 4bd = r13 + r31 4cd = r32 + r23 and b2 = 1 + r11 2 c2 = 1 + r22 2 d2 = 1 + r33 2 . Since q ̸= 0 and a = 0, at least one of b, c, d is nonzero. If b ̸= 0, let b = √1 + r11 √ 2 , and determine c, d using 4bc = r21 + r12 4bd = r13 + r31. 13.11. PROBLEMS 601 If c ̸= 0, let c = √1 + r22 √ 2 , and determine b, d using 4bc = r21 + r12 4cd = r32 + r23. If d ̸= 0, let d = √1 + r33 √ 2 , and determine b, c using 4bd = r13 + r31 4cd = r32 + r23. (7) Given any matrix A ∈su(2), with A =  iu1 u2 + iu3 −u2 + iu3 −iu1  , write θ = p u2 1 + u2 2 + u2 3 and prove that eA = cos θI + sin θ θ A, θ ̸= 0, with e0 = I. Therefore, eA is a unit quaternion representing the rotation of angle 2θ and axis (u1, u2, u3) (or I when θ = kπ, k ∈Z). The above formula shows that we may assume that 0 ≤θ ≤π. An equivalent but often more convenient formula is obtained by assuming that u = (u1, u2, u3) is a unit vector, equivalently det(A) = −1, in which case A2 = −I, so we have eθA = cos θI + sin θA. Using the quaternion notation, this read as eθA = [cos θ, sin θ u]. Prove that the logarithm A ∈su(2) of a unit quaternion q =  α β −β α  with α = a + bi and β = c + id can be determined as follows: 602 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN If q = I (i.e. a = 1) then A = 0. If q = −I (i.e. a = −1), then A = ±π i 0 0 −i  . Otherwise, a ̸= ±1 and (b, c, d) ̸= (0, 0, 0), and we are seeking some A = θB ∈su(2) with det(B) = 1 and 0 < θ < π, such that q = eθB = cos θI + sin θB. Then, cos θ = a (0 < θ < π) (u1, u2, u3) = 1 sin θ(b, c, d). Since a2+b2+c2+d2 = 1 and a = cos θ, the vector (b, c, d)/ sin θ is a unit vector. Furthermore if the quaternion q is of the form q = [cos θ, sin θu] where u = (u1, u2, u3) is a unit vector (with 0 < θ < π), then A = θ  iu1 u2 + iu3 −u2 + iu3 −iu1  is a logarithm of q. Show that the exponential map exp: su(2) →SU(2) is surjective, and injective on the open ball {θB ∈su(2) | det(B) = 1, 0 ≤θ < π}. (8) You are now going to derive a formula for interpolating between two quaternions. This formula is due to Ken Shoemake, once a Penn student and my TA! Since rotations in SO(3) can be defined by quaternions, this has applications to computer graphics, robotics, and computer vision. First, we observe that multiplication of quaternions can be expressed in terms of the inner product and the cross-product in R3. Indeed, if q1 = [a, u1] and q2 = [a2, u2], then check that q1q2 = [a1, u1][a2, u2] = [a1a2 −u1 · u2, a1u2 + a2u1 + u1 × u2]. We will also need the identity u × (u × v) = (u · v)u −(u · u)v. Given a quaternion q expressed as q = [cos θ, sin θ u], where u is a unit vector, we can interpolate between I and q by finding the logs of I and q, interpolating in su(2), and then exponentiating. We have A = log(I) = 0 0 0 0  , B = log(q) = θ  iu1 u2 + iu3 −u2 + iu3 −iu1  . 13.11. PROBLEMS 603 Since SU(2) is a compact Lie group and since the inner product on su(2) given by ⟨X, Y ⟩= tr(X⊤Y ) is Ad(SU(2))-invariant, it induces a biinvariant Riemannian metric on SU(2), and the curve λ 7→eλB, λ ∈[0, 1] is a geodesic from I to q in SU(2). We write qλ = eλB. Given two quaternions q1 and q2, because the metric is left invariant, the curve λ 7→Z(λ) = q1(q−1 1 q2)λ, λ ∈[0, 1] is a geodesic from q1 to q2. Remarkably, there is a closed-form formula for the interpolant Z(λ). Say q1 = [cos θ, sin θ u] and q2 = [cos ϕ, sin ϕ v], and assume that q1 ̸= q2 and q1 ̸= −q2. Define Ωby cos Ω= cos θ cos ϕ + sin θ sin ϕ(u · v). Since q1 ̸= q2 and q1 ̸= −q2, we have 0 < Ω< π. Prove that Z(λ) = q1(q−1 1 q2)λ = sin(1 −λ)Ω sin Ω q1 + sin λΩ sin Ωq2. (9) We conclude by discussing the problem of a consistent choice of sign for the quaternion q representing a rotation R = ρq ∈SO(3). We are looking for a “nice” section s: SO(3) → SU(2), that is, a function s satisfying the condition ρ ◦s = id, where ρ is the surjective homomorphism ρ: SU(2) →SO(3). I claim that any section s: SO(3) →SU(2) of ρ is neither a homomorphism nor contin-uous. Intuitively, this means that there is no “nice and simple ” way to pick the sign of the quaternion representing a rotation. To prove the above claims, let Γ be the subgroup of SU(2) consisting of all quaternions of the form q = [a, (b, 0, 0)]. Then, using the formula for the rotation matrix Rq corresponding to q (and the fact that a2 + b2 = 1), show that Rq =   1 0 0 0 2a2 −1 −2ab 0 2ab 2a2 −1  . Since a2 + b2 = 1, we may write a = cos θ, b = sin θ, and we see that Rq =   1 0 0 0 cos 2θ −sin 2θ 0 sin 2θ cos 2θ  , 604 CHAPTER 13. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN a rotation of angle 2θ around the x-axis. Thus, both Γ and its image are isomorphic to SO(2), which is also isomorphic to U(1) = {w ∈C | |w| = 1}. By identifying i and i and identifying Γ and its image to U(1), if we write w = cos θ + i sin θ ∈Γ, show that the restriction of the map ρ to Γ is given by ρ(w) = w2. Prove that any section s of ρ is not a homomorphism. (Consider the restriction of s to the image ρ(Γ)). Prove that any section s of ρ is not continuous. Problem 13.2. Let A = A0 ⊕A1 and B = B0 ⊕B1 be two Z/2-graded algebras. (i) Show that A b ⊗B is Z/2-graded by (A b ⊗B)0 = (A0 ⊗B0) ⊕(A1 ⊗B1), (A b ⊗B)1 = (A0 ⊗B1) ⊕(A1 ⊗B0). (ii) Prove Proposition 13.3. Hint. See See Br¨ ocker and tom Dieck , Chapter 1, Section 6, page 57. Problem 13.3. Prove Proposition 13.8. Hint. 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Symbol Index ((−, −)), 395 ((−, −)): Hk DR(M) × Hn−k DR (M) − →R, 394 ((A, B)), 396 ((ui)i∈Σ, ≤), 99 (−, −): Ai(ξ) ⊗Aj(ξ∗) ⊼ − →Ai+j(ξ ⊗ξ∗), 499 (K(I), ι), 36 (Uα, ϕα), 408 (X, Y )X, 396 (uk)k∈K, 287 (ω, η), 385 (dω)x, 145 (fi)i∈I, 35 (gij), 27 (gij), 27 (u∅), 101 ∗α, 111 1M, 271 A, 56 A b ⊗B, 555 AR ∈M2n(R), 516 B, 407 B(x, y), 346 B∗(U), 151 Bk(M), 166 Bp(U), 151 C(Sn), 324 Ck(U, G), 426 CC(Sn), 324 D(p), 218 Dp, 218 E, 407 E ↾Uα, 408 E ×G F, 455 E∗, 25 E1 ⊗· · · ⊗En, 37 Eb, 408 Ek, 325 F, 407 F(V ), 436 F(ξ), 436 F ∇, 490 Fk(σ, τ), 333 G, 407 G∞, 425 G∞(U), 425 Hk DR(M), 166 Hm l (r, θ, ϕ), 309 Hp(U), 151 Hp DR(U), 151 H•(U), 151 H• DR(M), 166 H• DR(U), 151 Hi, 269 HR n , 437, 463 In, 506 Iσ(f), 262 K ⋆f, 341 K(n), 584 K(I), 35 K∇, 490 L(G), 450 L2(S1), 298 L2(S2), 301, 308 L2(Sn), 323 L2 C(Sn), 324 613 614 SYMBOL INDEX LXY , 173 LXω, 174 Lv∗ 1,...,v∗ n : E1 ⊗· · · ⊗En − →K, 49 M(m, J, k1, . . . , kn), 77 N(x), 242, 265 N m l , 307 P(Kξ), 506 P(R∇), 506 P(Rξ), 506 P(Ωα), 506 P(ξ), 454 P m k (t), 305 Pk(t), 304 P λ k (t), 341 Pγ, 484 Pk,n(t), 330, 332 Qk(t), 304 R∇, 485, 488, 490 R∇ X,Y , 488 Sn−1 p,q , 574 TM, 430 T, 267 T(V ) = L m≥0 V ⊗m, 56, 58 T ∗M, 430 T •,•(V ), 60 T •(V ), 56 T r,s(M), 160, 444 T r,s(V ), 60 T r,s ξ, 439 T r s (V ), 62 V ⊗0, 56 V ⊗m, 56 X[f], 173 X♭, 371, 445 Y mk k , 326 Z(A), 595 Z∗(U), 151 Z1(U, G), 427 Zk(M), 166 Zp(U), 151 Zτ k, 329, 332 [P(R∇)], 506 [Pf(R∇)], 518 ∆f, 299, 301 ∆(g), 274 ∆C, 387 Γ, 339 Γ(B, ξ), 415 Γ(U, G∞), 425 Γ(U, ξ), 415 Γ(Φ), 558, 591 Γ(ξ), 415 Γ+(Φ), 558, 591 Γn, 561 Γp,q, 572 Γ+ p,q, 572 Ω= (Ωij), 490 ΦC n, 588 Φn, 551 Φp,q, 572 Φp,q, 551 ⊼: Ai(ξ) × Aj(η) − →Ai+j(ξ ⊗η), 485 N V , 56 Nm V , 56 V T ∗M, 159 V ξ, 439 V(E; F), 131 V(V ), 59 Vk ξ, 439 Vn E, 97 Vn(E), 97 Vn(V ), 106 V•(V ), 106 ˇ H1(B, G), 426, 427 χ(M), 504, 520 C[X1, . . . , Xn2], 505 ` α∈I, 425 δi j, 49 dom(M), 74, 75 ⩞: Ai(ξ) × A2(Hom(ξ, ξ)) − →Ai+2(ξ), 494 ϵF, 410 ϵV , 430 ϵk, 430 ϵi,J,H, 127 SYMBOL INDEX 615 ∂(ψi1,ψi2,...,ψin) ∂(u1,u2,...,un) , 242 L2(T), 297 L2([a, b]), 286 R G f, 273 R G f(g)dg, 275 R G fω, 273 R M ω, 246 R M f, 271 R M f(t)dt, 271 R M f VolM, 271 R M : An c (M) − →R, 246 R M ω, 247 R U ω, 245 R S1 f(θ)g(θ) dθ, 298 R Rn ω, 244 R Rn f(x)dx1 · · · dxn, 244 ι⊙, 73 ι⊗, 38 ι∧, 98 ⟨−, −⟩E∗, 29 ⟨f, g⟩Sn, 323 ⟨f, h⟩M, 318 ⟨⟨−, −⟩⟩, 499 ⟨⟨A, B⟩⟩, 396 ai1...ir j1...js  , 63 ⌟: Vp E × Vp+q E∗− →Vq E∗, 114 Cl(V, Φ), 551, 589 Cl(Φ), 552, 589 Cl(Φ)∗, 557 Cl(Φ)o, 554 Cl(n, C), 588 Cli(Φ), 555, 590 Cl1, 553 Cl2, 554 Cln, 561 Clp,q, 572 Pin(n), 565 Pin(p, q), 573 Spin(n), 565 Spin(p, q), 573 (f ∗, f): f ∗ξ →ξ, 427 ∗: V(V ) →V(V ), 112 ∗: Vk V →Vn−k V , 111 D: M →TM, 218 Lv∗ 1,...,v∗ n : Vn(E) →K, 102 Lv∗ 1,...,v∗ n : Sn(E) →K, 79 N : Cl(Φ) →Cl(Φ), 561, 592 ∆: Ak(M) →Ak(M), 382 α⊗: E∗⊗F →Hom(E, F), 53 α: Cl(Φ) →Cl(Φ), 554, 590 Vp f : Vp E →Vp F, 104 ·: Altn(E; R) × F →Altn(E; F), 133 δ: Vp E∗→Vn−p E, 122 δ: Ak(M) →Ak−1(M), 381 η: En →E⊗n, 81, 105 ♭: E →E∗, 27 γ : Vp E →Vn−p E∗, 122 ˆ d∇: Aj(B) × Γ(ξ) →Aj+1(ξ), 486 ι0 : K →T(V ), 57 ιn : V ⊗n →T(V ), 57 ι: (Rn)∗→Rn, 431 ι: I →K(I), 35 ⟨−, −⟩: E × F →K, 25 Evx : P ∗⊗R Q →Q, 87 EvX,Y : A2(Hom(ξ, ξ)) →A0(Hom(ξ, ξ)), 489 µ: V n →R, 249 µ: Vn V →R, 249 ∇∗: Γ(T ∗M) →A1(M) ⊗C∞(B) Γ(T ∗M), 498 ∇: Γ(ξ) →A1(B) ⊗C∞(B) Γ(ξ), 473 ∇: X(B) × Γ(ξ) →Γ(ξ), 473 ∇: X(M) × X(M) →X(M), 214 ∇: Ak(M) →HomC∞(M)(X(M), Ak(M)), 396 ∆: G →R, 274 f : V(V ) →A, 108 π: E →B, 407 ψα,x : Ex →G, 527 ρx : V →V , 546, 558, 591 ρα : Uα →G, 419 ρ: Γ(Φ) →GL(V ), 558 ♯: E∗→E, 27 σ: {1, . . . , n} →{1, . . . , n}, 71 τU : π−1(U) →U × Rn, 431 τg : M →M, 460 τ ∗ U : π−1(U) →U × (Rn)∗, 433 616 SYMBOL INDEX θ∗ U,ϕ,p = ι ◦θ⊤ U,ϕ,p : T ∗ p M →Rn, 431 θ⊤ U,ϕ,p : T ∗ p M →(Rn)∗, 431 θU,ϕ,p : Rn →TpM, 430 ϕ∗: H•(N) →H•(M), 166 ϕ∗: Ak(N) →Ak(M), 161 ϕ∗: Ap(V ) →Ap(U), 153 ϕg : F →F, 412 ϕα,b : Eb →F, 408 e ϕ∗: π−1(U) →ϕ(U) × Rn, 431 e ϕ: π−1(U) →ϕ(U) × Rn, 431 ci,j : T r,s(V ) →T r−1,s−1(V ), 68, 69 d∇: Ai(ξ) →Ai+1(ξ), 485, 486 dp : Ap(U) →Ap+1(U), 145 d: A∗(M) →A∗(M), 164 d: Ap(U) →Ap+1(U), 145 f ⊗g: E ⊗F →E′ ⊗F ′, 40 f ♮: E →E, 30 gαβ : Uα ∩Uβ →G, 409 gαβ : Uα ∩Uβ →Diff(G), 527 i(X): Ak(M) →Ak−1(M), 172 iG : G(p, n) →RP(n p)−1, 128 iΦ : V →Cl(V, Φ), 552, 589 o: V n →R, 251 p1 : Uα × F →Uα, 408 pri : Rn →R, 146 rσ : En →E⊗n, 81 s: B →E, 415 s: U →E, 415 t: Cl(Φ) →Cl(Φ), 554, 589 vg : C0(M) →R, 271 xi : U →R, 147 H, 542 Hk(M), 393 Hm, 254 I, 571 1, 542 GL(V ), 429 GL(n, C), 447 GL+(n, R), 447 O(V, Φ), 550 O(V, ϕ), 550 O(Φ), 551 Pin(Φ), 593 SO(V, Φ), 551 SO(V, ϕ), 551 SO(Φ), 551 Spin(Φ), 593 Spin+(Φ), 595 Spin+(p, q), 577 i, 542 j, 542 k, 542 Alt(E), 109 Altk C∞(M)(X(M)), 170 Altn(E; F), 94 Aut(V ), 429 Diff(F), 417 EvX, 473 Hess(f), 372 Holp(∇), 484 Hom(E, E), 30 Hom(E, E; K), 30 Hom(E, K), 25 Hom(E1, . . . , En; F), 32 Hom(ξ, ξ′), 441 HomR(M, R), 86 HomC∞(M)(X(M), C∞(M)), 171 Homalg(A, B), 57 Homsymlin(En, F), 71 Int(M), 255 Int(Hm), 254 Iso(V, W), 436 L(E1, . . . , En; F), 32 Mn(K), 57 Or(V ), 251 Pf(A), 513, 514 Pf(R∇), 518 Pf(Ωα), 517, 518 Sym, 73 Sym(V ), 59 Symn(E; F), 71 S(V ), 82 Sk ξ, 439 Sn(E), 72 SYMBOL INDEX 617 S•(V ), 82 VolM, 240 VolSn, 241 VolRPn, 243, 279 curlF, 266 den(M), 251 den(V ), 249 divF, 265 div X, 310 eu(R∇), 518 ev, 494 grad f, 310, 371 shuffle(m, n), 84 sign(λ), 251 supp(ω), 246 vol(Sn), 324 I(D), 226 Ik(D), 226 Ia, 106 Sn, 71 µ′, 104 µ: E∗ 1 ⊗· · · ⊗E∗ n ∼ = Hom(E1, . . . , En; K), 50 µ: Vn(E∗) ∼ = Altn(E; K), 104 µ: Sn(E∗) − →Symn(E; K), 83 µF : (Vn(E∗)) ⊗F − →Altn(E; F), 132 µF : (Vp(Rn)∗) ⊗F − →Altp(Rn; F), 179 ∇∇f, 373 ∇ω, 396 ∇d f, 373 ∇(X, Y ), 214 ∇∗, 397 ∇2 X,Y ω, 398 ∇r,sS, 445 ∇XY , 214 ∇Xθ, 373 ∇Xs, 468, 473 ∇r,s X , 444 ∥f∥, 324 ∥f∥2, 324 ∥f∥∞, 324 ω = (ωij), 477 ω(A), 513 ω(p), 159 ω(x), 143 ω ∈A∗ c(M), 246 ω ∈An c (M), 246 ω ∈An c (U), 245 ω ∧Φ η, 131 ω∗= (ω∗ ij), 500 ω♯, 27, 309, 445 ωx, 143 ωSn, 236 ωRn, 235 Rm +, 269 ξ, 501 ξ ∗, 500 x, 560 ⟨E, ∥∥⟩, 285 ⟨E, ϕ⟩, 285 ∂M, 255 ∂Rm +, 269 ∂(P), 321 ∂Hm, 254 ±(K∗)2, 594 ψ = (ψα)α∈I, 526 ψ(v) = ⟨−, v⟩, 26 1, 57 RXp, 217 R[X1, . . . , Xn2] , 505 ⌞: Vp+q E∗× Vp E − →Vq E∗, 114 ρH,L, 117 σ1, . . . , σak,n+1, 337 σi, 507 ⋆, 548 A∗(M), 159 A∗(U), 141 A∗(U; F), 179 A0(U), 141 A1(B; ξ), 484 A1(U), 141 A1(ξ), 484 Ai(B; ξ), 484 Ai(ξ), 484 Ak(M), 159 618 SYMBOL INDEX Ak(M; F), 186, 443 Ap(U), 141 Ap(U; F), 179 A•(U), 142 C(T), 297 C∞[a, b], 286 F = {Fα}α, 229 Hom(ξ, ξ′), 440 HC k (Sn), 320 Hk(S1), 298 Hk(S2), 301 Hk(Sn), 320 Hk(n + 1), 320 HC k (n + 1), 320 Hl(S2), 308 Pk(Sn), 320 Pk(n + 1), 320 PC k (Sn), 320 PC k (n + 1), 320 ˜ ωRn, 241 ϕ = (ϕα)α∈I, 408 ϕ(u) = ⟨u, −⟩, 26 ϕ∗ω, 153 ϕ∗(ω), 153 ϕ∗(ω)x, 153 ϕHom α,b , 440 u1 ⊙· · · ⊙un, 72 u1 ⊗· · · ⊗un, 37 u1 ∧· · · ∧un, 97 u⊙k, 77 ∧: A1(M) × A1(M) − →A2(M), 498 ∧R(u): Vq E − →Vp+q E, 114 ∧Φ, 131, 181 b ⊗, 110 ξ = (E, π, E/G, G), 526 ξ, 407 ξ′/ξ, 449 ξ[F], 455 ξ ⊕ξ′, 438 ξ ⊗ξ′, 438 ξ∗, 439 ξ⊗k, 438 ξg, 425 ξC, 441 ξR, 442 ak,n, 323 c(ξ)(t), 511 ck, 287 ck,mk, 326 ck(ξ), 510 dS, 265 dr, 267 dI, 227 dF, 181 ds, 267 e(ξ), 518 e∗ I, 142 f = (fE, fB), 417 f ∗ξ, 427 f⊙, 71 f⊗, 32 f∧, 96 gHom αβ , 440 i(u)z∗, 122 iXg, 371 l2, 286 l2(K), 292 lv∗ 1,...,v∗ n : (u1, . . . , un) 7→det(v∗ j(ui)), 102 lv∗ 1,...,v∗ n : (u1, . . . , un) 7→v∗ 1(u1) · · · v∗ n(un), 49 lv∗ 1,...,v∗ n : (u1, . . . , un), 79 ni = ni(x), 242 ni1,i2,··· ,in(x), 242 p(ξ)(t), 511 pk(ξ), 510 si, 508 u ⌟z∗, 115 u♭, 27, 309 uI, 99 xt, 554 ym l , 307 z ⌞u∗, 120 z∗⌞u, 119 Sn−1, 566 divC, 387 Index (m, n)-shuffle, 84 (r, s)-tensor fields, 195 Ck-manifold with boundary boundary, 255 definition, 255 interior, 255 K-algebra Z/2-graded graded tensor product, 555 center, 595 central, 595 definition, 57, 548 graded, 59, 550 homogeneous elements rank n, 59, 550 homomorphism, 57, 548 ideal, 57, 548 multiplication, 548 simple, 595 L2-norm, 281 d: Ak(M) →Ak+1(M), 381 adjoint, 385 Hom bundle, 440 k-dimensional foliation of smooth manifold, 229 flat chart, 229 involutive distribution, 229 leaves, 229 r-dimensional tangential distribution completely integrable, 218 definition, 218 flat chart, 221 integral manifold, 218 involutive, 219 differential ideal criterion, 227 integrability conditions, 228 lannihilating k-form, 226 locally defining one-form, 226 locally spanned, 218 vector space I(D), 226 ˇ Cech cohomology set ˇ H1(B, G), 426 Picard group, 426 algebra of differential forms A∗(U), 141 wedge product, 143 alternating multilinear map, 94 antiderivation on differential forms, 167 degree, 167 Bessel inequality, 289 beta function, 346 Betti numbers, 520 bi-invariant of integral on Lie group, 275 Bochner’s formula, 399 Bochner(connection) Laplacian definition, 397 harmonic form, 397 relationship to Laplacian, 400 via second covariant derivative, 399 Borel construction for fibre bundles, 455 bundle map(morphism) definition, 417 isomorphism, 417 isomorphism(local definition), 420 isomorphsim, 417 local definition, 420 preservation of fibre, 417 619 620 INDEX cannonical isomorphism ♯: E∗→E, 27 canonical isomorphism ♭: E →E∗, 27 ♭: TpM →T ∗ p M, 309 ♯: T ∗ p M →TpM, 310 definition, 26 Cartan’s formula for Lie derivative, 176 Cartan’s moving frame method, 435 Cartan-Dieudonn´ e Theorem, 544 central K-algebra, 595 characteristic class, 470, 505 Chern class, 510 Chern polynomial, 511 de Rham cohomology class [P(R∇)], 506 definition, 506 Euler class, 518 Euler form, 518 global form Pf(Ωα), 517 global form P(R∇), 506 Pontrjagin class, 510 Pontrjagin polynomial, 511 chart with corners, 269 Clifford algebra canonical anti-automorphsim, 554, 590 canonical automorphism, 554, 590 Z/2-grading, 555, 590 Clifford product, 596 conjugation, 560, 591 definition(for real vector space), 552 definition(over field K), 545, 589 existence, 552 generators and relations representation, 545, 557, 590 group of invertible elements, 557 monomial basis, 556, 590 norm, 561, 592 periodicity theorem, 586–588 relationship to exterior algebra, 553 Clifford group definition(arbitrary vector space), 545, 591 definition(for real vector space), 558 representation ρx : V →V , 546, 591 representation ρ: Γ(Φ) →GL(V ), 558 geometric interpretation, 559 special Clifford group, 546, 558, 591 twisted adjoint representation, 559 closed convex cone Rm +, 269 boundary, 269 corner point, 269 closed upper half space Hm, 254 boundary ∂Hm, 254 interior, 254 cocycle definition, 424 equivalent, 425 reduction, 447 cocycle condition cocycle, 424 complete metric space, 285 complex-valued differential forms, 481 conjugate linear form, 500 connection form(matrix) definition, 477 transformation rule, 480 connection Laplacian of a function, 388 connection on manifold definition, 214, 467 connection on vector bundle connection matrix(form), 477 connection on dual bundle connection matrix, 500 connection on dual bundle, 499 covariant derivative, 473 curvature form R∇= d∇◦∇, 488 definition first version, 473 definition second version, 473 dual basis representation, 479 Christoffel symbols, 479 evaluation map, 473 existence, 479 flat, 477 Hermitian, 496 INDEX 621 Leibniz rule, 473 local section representation, 476 metric connection, 495 existence, 496 parallel transport, 484 convex function definiton via Hessian, 376 cotangent bundle definition, 432 transition map, 433 trivialization map, 433 covariant derivative k-form, 396 one-form, 446 Riemannian metric, 445 tensor fields, 444 covariant derivative ∇XY , 214 covariant derivative of vector bundle, 473 covariant differential ∇r,sS , 445 covering map topological, 578 n-covering, 578 covering space, 578 isomorphism, 578 path lifting, 579 covering space, 578 curl of F : R3 →R3, 266 curvature form on vector bundle R∇= d∇◦∇, 490 R∇= d∇◦∇, 488 curvature matrix, 490 Bianchi’s Identity, 493 global form P(R∇), 506 structure equation, 490 transformation rule, 492 definition, 490 flat, 490 local section representation, 490 Darboux derivative, 192 de Rham complex de Rham cohomology algebra, 151, 166 de Rham cohomology group Hp(U), 151 for manifold, 140, 166 Hodge Theorem, 393 Poincar´ e duality, 394 de-Rham complex, 138, 151 density of vector space, 249 n-form definition, 249, 279 properites, 250 vector space den(V ), 250 derivation on differential forms, 167 differential p-form Rn basis representation, 143 closed, 151 compact support, 244 definition, 141 exact, 151 integration, 245 pull-back, 153 pull-back properties, 155 vector space Ap(U), 141 on manifold, 139, 159 Altk C∞(M)(X(M)), 170 compact support, 244, 246 coordinate representation, 160 integration, 247 pull-back, 161 vector space ω ∈Ap c(M), 246 vector space A∗ c(M), 244 vector space Ak(M), 139, 159 vector-valued basis representation, 180, 184 definition, 179 exterior derivative, 181, 183 exterior derivative basis representation, 184 manifolds, 186 pull-back, 183 vector space Ap(U; F), 179 wedge product, 180, 181 wedge product basis representation, 184 differential ideal of A•(M), 227 622 INDEX directional derivative DXY (p), 467 divergence divF : R3 →R , 265 div X : M →R, 310 chart representation, 311, 314, 366 connection, 388 chart representaton, 388 Hodge, 384 Divergence Theorem, 265 Green’s Formula, 318 domain with smooth boundary, 257 outward directed tangent vector, 258 dual bundle construction, 439 dual space, 25 dual basis, 27 effective action, 412 Einstein summation convention, 27, 63 elementary symmetric functions in n variables definition, 507 Newton’s formula, 508 Euler characteristic, 520 Betti number, 520 evaluation map Evx : P ∗⊗R Q →Q, 87 exterior algebra, 59, 97 construction, 107 inner product, 110 interior products, 108 insertion operator, 122 left hook, 115 right hook, 119 universal mapping property, 108 wedge product, 106 exterior differential d: Ap(U) →Ap+1(U) anti-derivation degree −1, 148 exterior differential d: Ap(U) →Ap+1(U) basis representation, 145 calculating, 147 curl, 150 definition, 145 divergence, 150 gradient, 149 exterior differential on manifold calculation, 163 definition, 162 vector field interpretation, 171 exterior differentiation on manifold, 164 exterior tensor power linear maps, 98 simple or decomposable Pl¨ ucker’s criteria, 128 criteria, 125 vector space alternating n-forms, 98 alternating n-tensors, 98 basis, 100 compound, 98 construction, 97 definition, 96 duality, 104 simple or decomposable, 98 universal mapping property, 96 faithfull action, 412 fibre bundle associated principal bundle, 455 frame bundle, 455 base space, 408 bundle chart, 408 bundle map, 417, 420 cocycle condition, 414, 424 definition, 408 equivalent, 420 fibre, 408 global section, 415 construction, 416 isomorphic, 417 left bundle, 460 local section, 415 local trivialization, 408 principal fibre bundle, 412, 429 pullback or induced, 427 construction, 428 INDEX 623 restriction, 429 right bundle, 460 smooth section, 415 structure group, 408 total space, 408 transition maps, 409 trivial bundle, 420 trivializing cover, 408 compatible with, 413 equivalent, 413 vector bundle, 429 Fourier coefficient, 282 Bessel inequality, 289 properties, 289 Fourier series Hilbert space, 287 Fourier coefficient, 287 partial sum, 287 on S1, 298 Parseval identity, 293 frame bundle construction, 437 free vector space generated by I, 34 construction, 35 universal mapping property, 35 Frobenius theorem, 221, 228, 229 fundamental group, 583 Funk-Hecke Formula, 344 Gamma function integral, 339 recurrence relation, 339 Gauss-Bonnet 2-form, 504 Gauss-Bonnet theorem, 504 Gaussian curvature, 504 Gegenbauer (ultraspherical) polynomial, 332 addition formula, 336 differential equation, 340 fundamental system on Sn, 337 generating function, 341 generating spherical harmonics, 338 relationship to Legendre polynomials, 333 reproducing kernel, 336 Rodrigues’ formula, 340 Gelfand pair character of L2 C(K\G/K), 362 definition, 360 Fourier transform, 363 Pontrjagin dual, 363 zonal spherical function, 360 general Legendre equation, 304 Generalized Gauss-Bonnet theorem, 521 gradient f ∈C∞(M), 310, 371 chart representation, 311, 372 f ∈C∞(M), 446 Grassmann algebra, see exterior algebra, see exterior algebra Grassmannian complex embedding in projective space, 129 Klein quadratic, 129 real embedding in projective space, 128 gradient, 379 Hessian, 379 Green’s Formula, 391 Green’s formula, 318 group action acting evenly, 580 diffeomorphism τg : M →M, 460 faithful or effective, 412 free, 450, 460, 527 left action, 580 linear action, 542 orbit, 580 proper, 459 smooth, 412 Haar measure compact Lie group, 353 left, 278 modular function, 278 right, 278 half-spin representation, 596 624 INDEX half-spinor, 596 harmonic form, 382 space of Hk(M), 393 harmonic function, 382 harmonic functions on R2, 300 harmonic polynomial complex coefficients, 320 real coefficients, 320 restricted to S1, 301 restricted to S2, 309 restriction to Sn, 320 eigenfunction of ∆Sn, 325 reproducing kernel, 333 Hermitian form, 285 positive definite, 285 Hermitian/unitary vector space, 285 Hessian, 372 as covariant derivative, 373 computed via geodesic, 374 defined via gradient, 373 in local coordinates, 375 Hilbert space, 286 L2(S1), 298 Hilbert basis, 298 L2(S2), 301 Hilbert basis, 301, 308 L2(Sn), 326 eigenspace decomposition, 326 Fourier coefficients, 326 Hilbert basis, 326 Hilbert basis via Gegenbauer polynomi-als, 336 psuedo-convolution, 342 L2 C(G), 353 central function, 355 convolution, 353 Hilbert basis, 355 left regular representation, 357 master decomposition, 355 subspace L2 C(K\G), 359 subspace L2 C(G/K), 358 subspace L2 C(K\G/K), 359 l2(K), 292 Cauchy family, 288 Hilbert basis, 287 properties, 295 orthogonal family, 287 Fourier coefficient, 287 Fourier series, 287 partial sum, 287 orthonormal family, 287 real, 286 representation of Lie group, 353 subrepresentation, 353 Riesz-Fischer theorem, 296 separable, 296 summable family, 288 total orthogonal family, see Hilbert basis Hilbert sum decomposition, 13, 282 Hodge ∗-operator, 112 basis application, 112 properties, 112 Riemannian manifold, 380 Hodge Decomposition Theorem, 393 Hodge Laplacian, see Laplace-Beltrami oper-ator holonomy of closed curve, 484 holonomy group, 484 homogeneous function of degree k on R2, 300 on Rn, 320 homogeneous polynomial of degree k complex coefficients, 319 Laplacian, 319 real coefficients, 319 restriction to Sn, 320 homogenous space definition, 461 immersed submanifold integral manifold, 218, 228 maximal, 228 insertion operator, 392 definition, 122 INDEX 625 integral complex valued function on Riemannian manifold, 271 differential form in Rn, 245 change of variable, 245 differential form on smooth oriented man-ifold, 247 real-valued function on Lie group, 273 bi-invariant, 275 left-invariant, 273 right-invariant, 273 real-valued function on Riemannian man-ifold, 271 integral manifold, 218, 228 Frobenius theorem, 221 maximal, 228 interior multiplication, see insertion operator interior product, see insertion operator invariant polynomial algebra In, 506 symmetric polynomial generators , 507 definition, 506 Laplace equation, 300 Laplace-Beltrami operator (Laplacian) chart representation, 311, 314, 387 definition, 310 for differential forms, 382 self-adjoint, 318, 386 Laplacian Euclidean, 12, 282, 299 polar coordinates, 299, 315 restricted to S1, 299 eigenfunctions, 301 restricted to S2, 303 eigenfunctions, 307 restricted to Sn, 315, 318 spherical coordinates, 303 left hook connection to right hook, 120 criteria for decomposability, 125 definition, 115 duality version, 116 properties, 118 left-invariant differential forms, 187 isomorphism with g∗, 187 Maurer-Cartan form, 187 left-invariant of integral on Lie group, 273 left-invariant volume form, 275 Legendre equation, 304 Legendre function (associated), 305 band index, 305 Condon Shortley phase factor, 305 recurrence relation, 306 Legendre functions of the first and second kinds, 304 Legendre polynomials, 304 recursion formula, 305 Rodrigues’ formula, 304 Levi-Civita connection, 498 ´ Elie Cartan’s criteria, 502 Christoffel symbol condition, 501 dual connection local chart representation, 501 dual connection criteria, 501 Koszul formula, 498 via dual connection, 499 via Lie derivative, 373, 504 Lie bracket structure constants, 189 Lie derivative Ck-function, 173 k-form, 174 Cartan’s formula, 177 properties, 174, 178 tensor field, 215 vector field, 173 Lie group function of positive type, 361 integral of smooth function, 273 modular function ∆, 274 left-invariant differential form, 187 volume form, 244 linear action of G on vector space V , 542 SU(2) on R3, 543 626 INDEX SU(2) × SU(2) on R4, 543 U(1) on R2, 542 local operator definition, 166 on differential forms, 166 manifold domain with smooth boundary, 257 smooth orientable, 234 orientation, 239 orientation atlas, 234 volume form, 235 smooth with boundary, 255 manifold with corners, 256, 269 Maurer-Cartan equations, 189 Maurer-Cartan form, 187, 189 flat connection, 192 linear Lie group, 191 properties, 190 metric connection as covariant derivative, 497 metric space complete, 285 modular function of Lie group ∆, 274 unimodular, 274 module over commutative ring, 85 free, 86 projective, 86 torsion element, 85 morphism of representations of Lie group, 351 multilinear map Hom(E1, . . . , En; F), 32 L(E1, . . . , En; F), 32 definition, 32 multiset definition, 74 multiplicity, 74 size, 74 nondegenerate pairing induced linear maps, 26 of vector spaces, 26 nowhere-vanishing n-form, see volume form open cover refinement, 414 orbifold, 580 orbit of group action, 580 ordered basis, 99 orientation of vector space, 233, 251 top form, 233 orientation of vector space properties, 252 orientation preserving diffeomorphism, 239 orientation preserving linear map, 234 oriented manifold, 239 VolM embedded manifold, 242 canonical volume form VolM, 240 positive atlas, 239 positively oriented basis, 239 oriented vector space, 111 negatively oriented, 111 positively oriented, 111 parallel transport vector bundle, 484 Parseval identity, 293, 295 path lifting lemma, 579 permanent, 79 Peter–Weyl Theorem, 348 Peter-Weyl Theorem, 354 trivial ideal, 354 Pfaffian polynomial explicit definition, 515 exterior algebra definition, 514 intrinsic definition, 513 skew Hermitian matrix, 516 Picard group, 441 pinor group Pin(Φ), 593 pinor group Pin(n) Pin(1), 566 Pin(2), 567 relationship to U(1) , 567 INDEX 627 defintion, 565 pinor group Pin(p, q) Pin(1, 0), 574, 575 definition, 573 Pl¨ ucker’s equations, 128 Pl¨ ucker coordinates, 128 Poincar´ e Duality Theorem, 394 Poincar´ e’s Lemma, 158 polar coordinates, 315 principal G-bundle, 526 local trivializations, 526 transition map, 527 principal fibre bundle G-equivariant trivializing map, 451 Borel construction, 455 bundle map(morphism), 453 construction, 452 definition, 450 frame bundle, 437 induced bundle, 455 isomorphism, 453 triviality criteria, 455 proper map criteria, 459 for manifolds, 459 definition, 458 pullback categorical definition, 428 quaternion algebra H, 542 conjugate, 543 defintion, 542 pure, 543 unit, 542 SU(2), 543 quaternions, 596 Radon measure Riemannian manifold, 272 representation of Lie group G-map, 351 (left) regular in L2 C(G), 357 character, 355 equivalent, 351 Hilbert space, 353 Hilbert sum theorem, 357 multiplicity, 357 invariant subspace, 351 irreducible, 351 linear of dimension n, 349 representation space, 349 G-module, 349 special functions, 350 subrepresentation, 351 trivial representation, 349 unitary representation, 349 reprodcing kernel Fk(σ, τ) relationship to zonal function Zτ k(σ), 335 reproducing kernel Fk(σ, τ) in terms of Gegenbauer polynomials, 336 reproducing kernel Fk(σ, τ), 333 Riemannian manifold, 446 Riemannian metric Gram matrix, 27 Riesz-Fischer theorem, 296 right hook connection to left hook, 120 definition, 119 duality version, 120 insertion operator, 122 properties, 121 right-invariant of integral on Lie group, 273 right-invariant volume form, 275 Schur’s Lemma for irreducible representations, 351 second covariant derivative k-form, 398 trace, 398 one-form, 398 Bochner’s formula, 399 semilinear function, 284 sequence Cauchy sequence metric space, 285 628 INDEX sesquilinear form, 285 signed integral differential form in Rn, 245 simply connected, 582 skew-symmetric multilinear map, 94 space of n-frames F(V ) definition, 436 spherical harmonic polynomials, 12, 281 spherical harmonics, 301 spin representation, 596 spinor, 596 spinor group Spin+(p, q), 577, 604 spinor group Spin(Φ), 546, 593 spinor group Spin+(Φ), 595 spinor group Pin(p, q) Pin(1, 1), 576 spinor group Spin(n) Spin(1), 566 Spin(2), 567 relationship to U(1), 568 Spin(3), 569 relationship to unit quaternions, 569 Spin(4), 570 relationship to Spin(4) , 572 definition, 565 spinor group Spin(p, q) Spin(1, 1), 576, 577 Spin(2, 0), 575, 576 connection to U(1), 576 definition, 573 star-shaped, 158 Stiefel manifold real gradient, 378 Stokes’ Theorem classical, 267 Stokes’ Theoreom for domain with smooth boundary, 262 symmetric bilinear form associated quadratic form, 550 signature, 572 group of isometries(orthogonal group), 551 nondegenerate, 550 polarization identity, 550 special orthogonal group, 551 symmetric multilinear map, 71 symmetric tensor algebra, 59 construction, 83 definition, 82 universal mapping property, 83 symmetric tensor power linear maps, 75 vector space basis, 77 compound, 75 construction, 72 definition, 71 duality, 80 simple or decomposable, 75 universal mapping property, 71 tangent bundle definition, 431 orientable criterion, 448 smooth vector field along a curve, 481 transition map, 431 trivializing map, 431 tensor algebra of vector space, 56, 58, 550 exterior algebra, 59 symmetric algebra, 59 universal mapping property, 58 tensor product R-module, 85 linear maps, 40, 549 vector space, 32, 548 m-th tensor power, 56 n-tensor, 40 n-th symmetric power, 71 antisymmetrized tensors, 105 basis, 44 compound, 40 construction, 37 Currying property, 48 INDEX 629 duality, 50 properties, 45 simple or decomposable, 40 symmetrized tensors, 81 tensor space T r,s(V ), 60 universal mapping property, 32 tensor space T r,s(V ), 60 contraction, 68 contravariant, 62 covariant, 62 duality, 61 homogeneous of degree (r, s), 60 trace bilinear form, 30 trivial bundle, 410 vector bundle Hom bundle, 440 bundle map(morphism), 433 canonical line bundle on RPn, 437 conjugate bundle, 501 connection matrix, 501 covariant derivative along curve, 483 definition(complex), 430 definition(real), 429 complexification, 442 dual bundle, 439 Euclidean, 446 Euclidean(Riemannian) metric, 446 existence, 447 exterior algebra, 439 exterior power, 439 frame bundle, 436 frame over U, 435 construction, 436 global frame, 435 global nonzero section, 435 Hermitian, 446 Hermitian metric, 446 holomorphic, 430 holomorphic line bundle, 430 isomorphism, 434 line bundle, 430 orientable, 447 oriented equivalent family trivializing maps, 448 orientation, 448 oriented family trivializing maps, 448 parallel section along curve, 483 quotient bundle, 449 normal bundle, 449, 450 smooth section along a curve, 481 subbundle, 449 orthogonal complement, 449 tensor bundle of type (r, s), 439 tensor power, 439 tensor product, 438 Whitney(direct) sum, 438 zero section, 435 vector field dual, 171 mutually commutative, 219 vector valued differential forms section of vector bundle, 443 vector-valued alternating form basis representation, 131, 133 definition, 131 multiplication, 131 vector-valued differential p-form, see vector-valued alternating form vector-valued form Ai(ξ), 485 d∇: Aj(ξ) →Aj+1(ξ), 486 wedge product, 485 volume form, 235 canonical, 240 chart representation, 240 equivalent, 239 Lie group, 244 left-invariant, 275 right-invariant, 275 relationship to density, 253 relationship to orientation, 253 weak integral, 356 630 INDEX wedge product ⊼: Ai(ξ) × Aj(η) − →Ai+j(ξ ⊗η), 485 definition, 106 Hodge ∗-operator, 112 on A∗(U), 143 skew-symmetry, 107 Weitzenb¨ ock–Bochner Formula, 400, 402 zonal function Zτ k, 332 relationship to reproducing kernel, 335 zonal spherical, 332 geometric interpretation, 338 on S2, 333 zonal harmonics, 329 zonal spherical function Gelfand pair, 360
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https://www.dictionary.com/browse/tolerant
Advertisement Skip to tolerant Advertisement tolerant [tol-er-uhnt] adjective inclined or disposed to tolerate; showing tolerance; forbearing. tolerant of errors. favoring toleration. a tolerant church. Medicine/Medical, Immunology. able to endure or resist the action of a drug, poison, etc. lacking or exhibiting low levels of immune response to a normally immunogenic substance. tolerant / ˈtɒlərənt / adjective able to tolerate the beliefs, actions, opinions, etc, of others permissive able to withstand extremes, as of heat and cold med (of a patient) exhibiting tolerance to a drug Other Word Forms Word History and Origins Origin of tolerant1 Example Sentences Mahmood will argue that "fair migration" and secure borders are integral parts of an "open, generous, tolerant" country. Scotland is becoming less tolerant towards immigration, its former first minister claims. County, at its best, comes across as urbane, lively, tolerant, expansive, with a sense of humor about itself. She really believed that she was seeking consensus, tolerant of all perspectives as long as they didn’t impinge on her beliefs, the origins of which are poignantly related later in the play. “The U.K. is a fair, tolerant and decent country, so the last thing that British people want is dangerous and inflammatory language which threatens violence and intimidation on our streets,” Pares said. Advertisement Related Words Advertisement Advertisement Advertisement Browse Follow us Get the Word of the Day every day! By clicking "Sign Up", you are accepting Dictionary.com Terms & Conditions and Privacy Policies.
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https://www.icliniq.com/articles/ear-health/lateral-sinus-thrombosis
What Is Lateral Sinus Thrombosis? Free Consultation [x] Ask a Doctor Online Now HomeHealth articleslateral sinus thrombosis Ear healthVerified data Lateral Sinus Thrombosis - Symptoms, Diagnosis, and Treatment 4 min read 0 Share Verified data 0 4 min read Share Outline Lateral sinus thrombosis is a rare compilation of infections of the middle ear and the mastoid forming a blood clot in the venous sinus of the brain. Written by Dr. Monisha. G Medically reviewed by Dr. Akshay. B. K. May 19, 2023 Published on March 13, 2024 Last reviewed on Contents Share What Is Sinus Thrombosis? What Is Lateral Sinus Thrombosis? What Are the Signs and Symptoms of Lateral Sinus Thrombosis? What Is the Diagnosis of Lateral Sinus Thrombosis? What Is the Treatment for Lateral Sinus Thrombosis? What Are the Complications of Lateral Sinus Thrombosis? Contents What Is Sinus Thrombosis? What Is Lateral Sinus Thrombosis? What Are the Signs and Symptoms of Lateral Sinus Thrombosis? What Is the Diagnosis of Lateral Sinus Thrombosis? What Is the Treatment for Lateral Sinus Thrombosis? What Are the Complications of Lateral Sinus Thrombosis? Introduction The network of veins between the two layers of the dura mater that help drain the venous blood from the brain is called the dural venous sinus. It collects the deoxygenated blood from the skull and returns it to the heart for oxygenation. There are seven main dural venous sinuses between the dural layers: the periosteal or endosteal layer and the meningeal layer. They are listed below: Superior sagittal sinus. Inferior sagittal sinus. Cavernous sinus. Straight sinus. Transverse sinus. Sigmoid sinus. Superior petrosal sinus. The most important sinus among the seven sinuses is the cavernous sinus. There are no valves in the dural venous sinuses, meaning the direction of blood flow can either be forward or backward. The backward flow can spread microbes and cancer cells to the other adjacent parts of the brain through which the venous drains. A dangerous triangular area on the face is marked by the inner corner of the eyes as the apex, the sides of the nose, and the upper lip as the base. It is where any infectious pathogen can travel from the face to the brain and cause a cavernous sinus infection. The infection will form a blood clot in the cavernous sinus called cavernous sinus thrombosis. The thrombus causes the sinus to swell and press on adjacent nerve structures. What Is Sinus Thrombosis? The formation of a blood clot is known as thrombosis. Hence, the formation of a thrombus in the venous sinuses of the brain cavity is called dural venous sinus thrombosis. Venous sinus thrombosis is of three types which are given as follows. Cavernous Sinus Thrombosis (CST): The most important and clinically significant sinus thrombosis occurring in the cavernous sinus is called cavernous sinus thrombosis. They can get infected through the dangerous area of the face, and the irregularity in their shape and location on the skull base makes them more vulnerable to infection. The infection can be from the ear or the upper teeth or due to trauma. Lateral Sinus Thrombosis (LST): The formation of a thrombus in the mastoid sinus is called lateral sinus thrombosis. Superior Sagittal Sinus Thrombosis: Superior sagittal sinus is the largest venous channel in the skull because many veins drain into the sinus. So, clot formation is rare because of its big size. Subscribe By subscribing, I agree to iCliniq's Terms&Privacy Policy. What Is Lateral Sinus Thrombosis? Lateral sinus thrombosis is the blood clot formation in the sigmoid and transverse sinuses. It usually occurs as a complication of a middle ear infection called otitis media. The occurrence of complications is rare in recent times with the use of antibiotics for the treatment of otitis media. When antibiotics were not used for treating otitis media, lateral sinus thrombosis was the second most common fatal complication, with meningitis being the first. It is relatively rare but requires immediate treatment. The death percentage ranges from 5 % to 10 %. The anatomical location of mastoid air cells and the middle ear in the proximity of the dural venous sinuses makes it more susceptible to secondary infection during otitis media. What Are the Signs and Symptoms of Lateral Sinus Thrombosis? There are variations in the signs and symptoms because of different locations of clot formation. They include the following: Chronic otitis media. Severe headache. Ear and neck pain. Fever (picket fence fever - temperature spikes in the afternoon or early evening and then comes down). Vomiting. Loss of sharpness in vision. Otorrhea (drainage of liquid from the ear). Papilledema (swelling of the optic nerve). Increased intracranial pressure. The typical clinical feature of lateral sinus thrombosis and reflex thrombosis in the mastoid emissary vein is Griesinger’s sign - edema and tenderness over the mastoid bone behind the ear. Neck swelling and tenderness if the thrombosis spreads to the internal jugular vein. Paralysis of the 9th, 10th, and 11th cranial nerve if thrombosis spreads to the jugular bulb. What Is the Diagnosis of Lateral Sinus Thrombosis? The diagnosis is made based on laboratory tests and radiology. Blood Tests: The number of red blood cells getting reduced or diminished hemoglobin concentration, called anemia, is present, and an elevated number of white blood cells called leukocytes are common laboratory findings. Blood Culture: The bacteria causing the infection was mainly beta-hemolytic streptococcus. But now, due to antibiotics, cultures show a mixed flora of microbes, including Bacteroides, Staphylococcus, Proteus, Pseudomonas, Enterobacteriaceae, and other species. The blood culture is often negative since ear infections are pre-treated by antibiotics. Computed Tomography (CT) Scan: This imaging technique is used to demonstrate the specific ‘delta sign’-dural enhancement around the sinus and abnormality in the opacification of the sinus cavity. The scan also helps in detecting the compilation that occurs in the brain. Magnetic Resonance Imaging (MRI) Scan: It is the choice of investigation for lateral venous thrombosis. The imaging method is more sensitive than CT and shows the location of sinus occlusion and the sequel of backflow of blood. In addition, Gadolinium-enhanced MRI gives the thrombus a vascular and bright appearance for the dural walls. Magnetic Resonance Venography:This type of MRI can detect the signals lost in the sinus and the absence of blood flow. Tobey-Ayer Test: It monitors cerebrospinal fluid pressure during a lumbar puncture. When the internal jugular vein is compressed, if there is no increase in CSF pressure on the affected side and an exaggerated response on the unaffected side suggests lateral sinus thrombosis. What Is the Treatment for Lateral Sinus Thrombosis? The different treatment options available are: Antibiotics: Broad-spectrum antibiotics are administered intravenously at the earliest. The use of antibiotics has brought down the percentage of complications. Myringotomy:It is a cut made on the tympanic membrane to drain pus. Mastoidectomy:It is an incision made on the lateral sinus of the mastoid to remove the clot. Local packing of the cavity is done. Anticoagulation Therapy: The effectiveness of anticoagulants (drugs that prevent the clotting of blood) is poorly understood. But they can be used to slow down the thrombus propagation and improve the neurological condition in patients with raised intracranial pressure. Internal Jugular Vein Ligation:This procedure involves removing the affected part of the event from the body. But it was done in older days to prevent septic emboli formation. But now, the surgery is performed infrequently. What Are the Complications of Lateral Sinus Thrombosis? The incidence of complications has reduced but is still present, which are: Meningitis (inflammation of membranes that cover the brain and spinal cord). An intracranial abscess (swelling filled with pus in the brain). Otitic hydrocephalus. Internal jugular vein thrombosis. Conclusion Ear infection leading to lateral sinus thrombosis has now become rare due to the implementation of antibiotics to treat infections, and it can prove to be fatal and dangerous if not treated appropriately. No improvement in symptoms, even with antibiotics, should raise suspicion and be investigated with MRI and venography to diagnose lateral sinus thrombosis. The treatment should be provided at the airline to prevent further complications. Listen to related tracks in our music library Sources Neuroanatomy, Dural Venous Sinuses Otogenic Lateral Sinus Thrombosis Lateral Sinus Thrombosis in Otology Lateral Sinus Thrombophlebitis Dr. Akshay. B. K. Otolaryngology (E.N.T) Consult this doctor Rate this article Tags: lateral sinus thrombosis Related topics Anal Pathology - Types, Symptoms, and TreatmentForehead Reconstruction - An OverviewSurgical Management of Cushing Syndrome - An OverviewTolosa-Hunt Syndrome- Causes, Symptoms, Diagnosis, and Management Most popular articles Ovulation and Safe Period: What is the Safe Period to Have Sex? Dr. Sabita Laskar Unwanted 72 - Uses, Dosage, Side Effects, Drug Warnings, and Precautions Dr. Divya Banu M Jelqing - Penis Enlargement Exercise Dr. Ramchandra Lamba Dolo 650 MG Tablet Dr. Anshul Varshney I-Pill - Uses, Dosage, Side Effects, Drug Warnings, and Precautions Dr. Vasantha. K. S Ask your health query to a doctor online Otolaryngology (E.N.T) guaranteed answer within 4 hours Disclaimer: No content published on this website is intended to be a substitute for professional medical diagnosis, advice or treatment by a trained physician. Seek advice from your physician or other qualified healthcare providers with questions you may have regarding your symptoms and medical condition for a complete medical diagnosis. Do not delay or disregard seeking professional medical advice because of something you have read on this website. Read our Editorial Process to know how we create content for health articles and queries. Your health is our #1 priority and it never changes Ask a doctor online Explore all specialties Company About usOur Medical Review TeamBlogsCareersPressSuccess Stories Customer Care ContactFAQSupportGet iCliniq Disclaimer: No content published on this website is intended to be a substitute for professional medical diagnosis, advice or treatment by a trained physician. Seek advice from your physician or other qualified healthcare providers with questions you may have regarding your symptoms and medical condition for a complete medical diagnosis. Do not delay or disregard seeking professional medical advice because of something you have read on this website. Read ourEditorial Processto know how we create content for health articles and queries. Terms-Privacy-Ads Policy-Sitemap Copyright © 2025, iCliniq - All Rights Reserved Visit other versions in US,UK,Australia,India,PhilippinesandIntl. Home Screen Reader Access -+
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https://www.quora.com/If-a-number-n-is-not-divisible-from-2-to-square-root-of-n-then-it-is-said-to-be-prime-Why
If a number n is not divisible from 2 to square root of n, then it is said to be prime. Why? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Divisibility Tests Square Root Symbol Large Prime Numbers Arithmetic Theory of Numbers Square Roots (numbers) Prime Integers. Divisibility 5 If a number n is not divisible from 2 to square root of n, then it is said to be prime. Why? All related (36) Sort Recommended Alexandre Ribeiro Miquilino Number Theory Addict · Author has 729 answers and 3.1M answer views ·7y Suppose a given positive integer n n can be written as the product of two positive integers k k and r r. Assume that k≥√n k≥n. We know that k∗r=n k∗r=n; since k k and r r are both positive, we can multiply our inequality by r r to obtain k∗r≥r√n k∗r≥r n, and thus n≥r√n n≥r n. Dividing both sides by √n n, we arrive at r≤√n r≤n. The same argument can be applied for the case k≤√n k≤n; we have that k∗r≤r√n k∗r≤r n, and thus n≤r√n n≤r n, which yields r≥√n r≥n. Therefore, if one factor is greater than or equal to √n n, the other must necessarily be less than or equa Continue Reading Suppose a given positive integer n n can be written as the product of two positive integers k k and r r. Assume that k≥√n k≥n. We know that k∗r=n k∗r=n; since k k and r r are both positive, we can multiply our inequality by r r to obtain k∗r≥r√n k∗r≥r n, and thus n≥r√n n≥r n. Dividing both sides by √n n, we arrive at r≤√n r≤n. The same argument can be applied for the case k≤√n k≤n; we have that k∗r≤r√n k∗r≤r n, and thus n≤r√n n≤r n, which yields r≥√n r≥n. Therefore, if one factor is greater than or equal to √n n, the other must necessarily be less than or equal to √n n; what this means is that if we check all cases for factors between 2 2 and √n n, we’re also checking the other values as well. However, if there are no factors between 2 2 and √n n, there cannot be any other factors between √n n and n n, because of the relationship we established earlier - one factor must be either greater or equal to √n n, and the other less than or equal this value.. Upvote · 99 13 9 1 Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Work faster with AI, while ensuring your writing always makes the right impression. Download 999 210 Related questions More answers below The only primes less than the square root of 400=20 are 2, 3, 5, 7, 11, 13, 17, and 19. Suppose that n is a number less than 400 which is not divisible by any of these primes. What can you conclude about n? What is the square root of a number that is divisible by all prime numbers? If n is a prime number and n/2 is also a prime number then what does it say about n+n/2? Why is 2 n−1 2 n−1 always prime and 2 n+1 2 n+1 always divisible by 3 3? If n is a perfect square, then 2 n can never be a perfect squared, but what if n is equal to 2? Satish Patel Algoricursive · Author has 133 answers and 411.9K answer views ·11y Originally Answered: If a number n is not divisible from 2 to square root of n then it is said to be prime, why? · To check whether a number n is prime or not we proceed from 2 2 to s q r t(n)s q r t(n). suppose n n is not a prime then n=a∗b n=a∗b if both a>s q r t(n),b>s q r t(n)a>s q r t(n),b>s q r t(n) then a∗b>n a∗b>n this implies at least one of these factors must be less than or equal to s q r t(n)s q r t(n). And hence we only check from 2 2 to s q r t(n)s q r t(n). Upvote · 99 17 9 4 David Schulman Amateur mathematician · Author has 3K answers and 2.8M answer views ·5y Because any composite integer n n must have at least one factor ≤√n≤n. Put another way, if an integer does not have a prime factor less than or equal to its own square root, that fact alone constitutes proof of its primality. Consider, for example, 283. Its square root is about 16.82, so we can perform a primality test by checking its divisibility by the first six primes {2, 3, 5, 7, 11, 13}. It turns out that none of them divide 283, therefore 283 is prime. To see why, let’s consider 437. Its square root is about 20.9, so the largest prime we have to check is 19 — which turns out to be a f Continue Reading Because any composite integer n n must have at least one factor ≤√n≤n. Put another way, if an integer does not have a prime factor less than or equal to its own square root, that fact alone constitutes proof of its primality. Consider, for example, 283. Its square root is about 16.82, so we can perform a primality test by checking its divisibility by the first six primes {2, 3, 5, 7, 11, 13}. It turns out that none of them divide 283, therefore 283 is prime. To see why, let’s consider 437. Its square root is about 20.9, so the largest prime we have to check is 19 — which turns out to be a factor, as 437 = 19 × 23. In fact, 437 is a member of a class of composite integers known as semiprimes, which have only two prime factors “close” to each other (and to √n n), meaning that they represent the “worst case” for this trial-division method of factoring or primality testing. But notice how the factors of composite numbers must occur in pairs {p,q}{p,q}: those factors are in an inverse relationship with each other, meaning that p∝q−1 p∝q−1 and both are therefore maximized when p=q p=q (i.e. n=p 2=q 2 n=p 2=q 2, a perfect square). The trial-division factoring method is fully deterministic and guaranteed to work, but it doesn’t scale well because it’s also very slow, especially once n n gets large (specifically, when its smallest prime factor exceeds 10 9 10 9 or so). If we return to the example of 437, we could factor it a lot faster than having to check its divisibility by the first eight primes, by recognizing it as a difference of squares a 2−b 2 a 2−b 2, specifically 21 2−2 2 21 2−2 2. Such numbers can always be represented as (a−b)(a+b)(a−b)(a+b) (this method is due to the French mathematician Pierre de Fermat), so 437=(21−2)(21+2)=19×23.437=(21−2)(21+2)=19×23. Upvote · 9 1 Nandeesh H N (ನಂದೀಶ್ ಎಚ್ ಎನ್) An engineer who likes Maths, English, Kannada, linguistics · Author has 2.3K answers and 3.4M answer views ·7y Suppose I want to check whether n = 83 is a prime. There are three ranges of numbers between 2 and n-1=82. First range of numbers is between n/2 =41 and n-1=82. These numbers will always give a quotient of 1 and some remainder. This is true even whether the given number n is prime or not. Second range of numbers is between √n and n/2= 9 (integer portion). These numbers give a quotient between 2 and √n. It means testing with numbers between 2 and √n (this is the third range of numbers) will suffice to check if the given number n is prime or not. Even within the third range of numbers, we can avoid t Continue Reading Suppose I want to check whether n = 83 is a prime. There are three ranges of numbers between 2 and n-1=82. First range of numbers is between n/2 =41 and n-1=82. These numbers will always give a quotient of 1 and some remainder. This is true even whether the given number n is prime or not. Second range of numbers is between √n and n/2= 9 (integer portion). These numbers give a quotient between 2 and √n. It means testing with numbers between 2 and √n (this is the third range of numbers) will suffice to check if the given number n is prime or not. Even within the third range of numbers, we can avoid the composite numbers to check divisibility of n. We need to check divisibility of n only with the prime numbers between 2 and √n. For example check if 83 is prime. Divide by 2, quotient 41. Remainder 1. Divide by 3 , quotient 27. Remainder 2. No need to divide by 4. If n was divisible by 4, it would have been divisible by 2. Divide by 5, quotient 16. Remainder 3. No need to check with 6 or 8. Divide by 7, quotient 11. Remainder 6. No need to check with 9. So 83 is a prime number. So, checking divisibility with prime numbers between 2 and square root is enough to check whether the given number is prime. To check whether 87 is a prime. To check with 2,3,5 and 7. It fails at 3. No need to proceed with 5 and 7. 87 is not a prime. Upvote · 9 1 Promoted by HP HP Tech Takes Tech Enthusiast | Insights, Tips & Guidance ·Updated Sep 18 Which is a good affordable wireless laser printer that prints both sides of paper automatically? Finding an affordable wireless laser printer that supports automatic duplex printing can be a smart investment if you want to save paper and streamline your workflow. Duplex printing allows the printer to print on both sides of the page without manual intervention, which is especially useful for producing professional documents, reports, or booklets. Wireless connectivity adds convenience by enabling printing from multiple devices, including smartphones, tablets, and laptops. HP offers several models that combine these features with reliable performance and cost-efficiency, making them ideal f Continue Reading Finding an affordable wireless laser printer that supports automatic duplex printing can be a smart investment if you want to save paper and streamline your workflow. Duplex printing allows the printer to print on both sides of the page without manual intervention, which is especially useful for producing professional documents, reports, or booklets. Wireless connectivity adds convenience by enabling printing from multiple devices, including smartphones, tablets, and laptops. HP offers several models that combine these features with reliable performance and cost-efficiency, making them ideal for home offices or small businesses. One strong recommendation is the HP LaserJet Pro MFP 3102fdw. This model is priced around £229 to £329 and includes automatic duplex printing, wireless connectivity, and multifunction capabilities such as scanning and copying. It’s designed for users who need consistent, high-speed printing with minimal maintenance. The compact design makes it suitable for smaller workspaces, while the efficient toner system helps keep running costs low. Its compatibility with the HP Smart app allows for easy mobile printing, adding flexibility to your workflow. For a more budget-conscious alternative, the HP LaserJet M234sdw is another excellent option. Priced from £136 to £210, it offers automatic duplex printing and wireless functionality, along with fast print speeds of up to 29 pages per minute. This model is ideal for users who primarily need high-volume monochrome printing without the added features of scanning or copying. It supports mobile printing through platforms like Apple AirPrint and the HP Smart app, making it easy to print from various devices. It uses toner really efficiently, so you won’t be spending loads on refills, thus great for everyday printing. LaserJet Printers - Black & White or Color Document Printers In conclusion, if you need a multifunction printer with duplex printing and wireless capabilities, the HP LaserJet Pro MFP 3102fdw offers excellent value and performance. If your focus is on fast, reliable black and white printing with duplex support at a lower price point, the HP LaserJet M234sdw is a dependable and cost-effective solution. Both models can give you strong results without exceeding your budget, and the choice depends on whether you require additional features beyond printing. For more duplex printer model recommendations, check out this blog post: Top Multifunction Printers for Small Businesses Features And Recommendations | HP® Tech Takes By Lizzie - HP Tech Expert Upvote · 99 16 9 6 Related questions More answers below If n is a prime number greater than 3, what is the reminder when the square of those prime is divided by twelve? Why is 2^n -1 a prime number if n is a prime number? How do you find a square root of a number if it is not prime and not divisible by four or three? Is it possible to generate more primes from n numbers than a given value of n? Is a number n always co-prime with n-1? Assistant Bot · 1y A number n n is defined as prime if it has exactly two distinct positive divisors: 1 and itself. To understand why it suffices to check for divisibility from 2 to √n n, consider the following: Divisor Pairs: If n n can be expressed as a product of two integers a a and b b (i.e., n=a×b n=a×b), then at least one of those integers must be less than or equal to √n n. This is because if both a a and b b were greater than √n n, their product a×b a×b would be greater than n n. Checking Up to √n n: Therefore, when checking for factors of n n, it is enough to test divisibility using numbers from 2 Continue Reading A number n n is defined as prime if it has exactly two distinct positive divisors: 1 and itself. To understand why it suffices to check for divisibility from 2 to √n n, consider the following: Divisor Pairs: If n n can be expressed as a product of two integers a a and b b (i.e., n=a×b n=a×b), then at least one of those integers must be less than or equal to √n n. This is because if both a a and b b were greater than √n n, their product a×b a×b would be greater than n n. Checking Up to √n n: Therefore, when checking for factors of n n, it is enough to test divisibility using numbers from 2 up to √n n. If n n is not divisible by any of these numbers, then it cannot have any factors other than 1 and n n itself, confirming that n n is prime. Examples: For n=29 n=29: √29≈5.39 29≈5.39. We check divisibility by 2, 3, and 5. None divide 29, so it is prime. For n=36 n=36: √36=6 36=6. We check divisibility by 2, 3, and 4. Since 36 is divisible by 2 (and others), it is not prime. In summary, checking for divisibility only up to √n n is sufficient to determine if n n is prime because any factor larger than √n n would necessarily require a corresponding factor smaller than √n n. Upvote · Michael Mark Ross Autodidactic number empiricist · Author has 2.6K answers and 11.2M answer views ·Updated 9y Originally Answered: If a number n is not divisible from 2 to square root of n, then it is said to be prime. What is the scientifical and mathematical proof to that? · Here is a commonsense explanation - no proof required: If the number is not prime it must have two or more prime factors. Suppose all the prime factors were greater than the square root, the product of the factors would be greater than the number (which needless to say it cannot be). Therefore, for the product of the prime factors to equal the number, one of them must be smaller than the square root. And so, one needs to test up to the square root only. (Exactly the same principle applies when considering composite factors.) Upvote · 9 3 Eleftherios Argyropoulos B.S. in Mathematics&Physics, Northeastern University (Graduated 2002) · Upvoted by Jeremy Collins , M.A. Mathematics, Trinity College, Cambridge and David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 2K answers and 2.5M answer views ·Updated 5y Related Why is 2 n−1 2 n−1 always prime and 2 n+1 2 n+1 always divisible by 3 3? The number 2 n−1 2 n−1 is prime only for n n prime, but not for every prime n n. Actually, 2 n−1 2 n−1 is prime for very few primes n n and when this is true, the number 2 n−1 2 n−1 is called Mersenne prime. Therefore, we have to note that in order the number 2 n−1 2 n−1 to be prime, n n has to be prime, but this is not an if and only if relation. When 2 n−1 2 n−1 is Mersenne prime, then the product 2 n−12 n−1 is perfect number. As of today’s date 14 March 2020, only 51 Mersenne primes are known. These are: 2 2−1 2 2−1, 2 3−1 2 3−1, 2 5−1 2 5−1, 2 7−1 2 7−1, 2 13−1 2 13−1, 2 17−1 2 17−1, 2 19−1 2 19−1, 2 31−1,2 61−1 2 31−1,2 61−1, 2 89−1 2 89−1, 2 107−1 2 107−1, Continue Reading The number 2 n−1 2 n−1 is prime only for n n prime, but not for every prime n n. Actually, 2 n−1 2 n−1 is prime for very few primes n n and when this is true, the number 2 n−1 2 n−1 is called Mersenne prime. Therefore, we have to note that in order the number 2 n−1 2 n−1 to be prime, n n has to be prime, but this is not an if and only if relation. When 2 n−1 2 n−1 is Mersenne prime, then the product 2 n−12 n−1 is perfect number. As of today’s date 14 March 2020, only 51 Mersenne primes are known. These are: 2 2−1 2 2−1, 2 3−1 2 3−1, 2 5−1 2 5−1, 2 7−1 2 7−1, 2 13−1 2 13−1, 2 17−1 2 17−1, 2 19−1 2 19−1, 2 31−1,2 61−1 2 31−1,2 61−1, 2 89−1 2 89−1, 2 107−1 2 107−1, 2 127−1 2 127−1, 2 521−1 2 521−1, 2 607−1 2 607−1, 2 1279−1 2 1279−1, 2 2203−1 2 2203−1,2 2281−1 2 2281−1, 2 3217−1 2 3217−1, 2 4253−1 2 4253−1, 2 4423−1 4423−1, 2 9689−1 2 9689−1, 2 9941−1 9941−1, 2 11213−1 2 11213−1, 2 19937−1 2 19937−1, 2 21701−1 2 21701−1, 2 23209−1 2 23209−1, 2 44497−1 2 44497−1, 2 86243−1 2 86243−1, 2 110503−1 2 110503−1, 2 132049−1 2 132049−1,2 216091−1 2 216091−1, 2 756839−1 2 756839−1, 2 859433−1 2 859433−1, 2 1257787−1 2 1257787−1, 2 1398269−1 2 1398269−1, 2 2976221−1 2 2976221−1, 2 3021377−1 2 3021377−1, 2 6972593−1 2 6972593−1, 2 13466917−1 2 13466917−1, 2 20996011−1 2 20996011−1, 2 24036583−1 2 24036583−1, 2 25964951−1 2 25964951−1, 2 30402457−1 2 30402457−1, 2 32582657−1 2 32582657−1, 2 37156667−1 2 37156667−1, 2 42643801−1 2 42643801−1, 2 43112609−1 2 43112609−1, 2 57885161−1 2 57885161−1, 2 74207281−1 2 74207281−1, 2 77232917−1 2 77232917−1, 2 82589933−1 2 82589933−1. Now, concerning the divisibility of the number 2 n+1 2 n+1 with the number 3 3, we have: For n odd⟹2 n+1≡0(mod 3)n odd⟹2 n+1≡0(mod 3) For n even⟹2 n+1≡2(mod 3)n even⟹2 n+1≡2(mod 3) Therefore, neither of the two claims of the question is correct. Please before asking any questions here, be certain that you know well what you write. Upvote · 99 47 9 6 9 1 Sponsored by Visit Dubai Plan your dream Dubai getaway. Explore traditional souks, luxury malls and desert adventures in one epic trip! Learn More 99 41 Vishal Chandratreya Former Senior Engineer at Samsung Semiconductor India (2019–2021) · Author has 931 answers and 2.9M answer views ·7y Related Why do we run a loop only from 2 till square root(n) to check if a number is prime or not? If a number n∈N n∈N(n≠1)(n≠1) is composite (i.e. not prime), it has at least one factor less than or equal to √n n. Assume that n n is the product of two numbers a a and b b, where a,b∈N a,b∈N. n=a b n=a b a a and b b cannot both be greater than √n n. Let us prove this by contradiction. Let a>√n a>n and b>√n b>n. Using this, we arrive at n=a b>√n√n=n n=a b>n n=n, which is a contraction. Hence, one of them must be less than or equal to√n n. (Of course, the ‘equal to’ part applies only if n n is a perfect square.) Therefore, you iterate a loop only beginning from 2 until the squ Continue Reading If a number n∈N n∈N(n≠1)(n≠1) is composite (i.e. not prime), it has at least one factor less than or equal to √n n. Assume that n n is the product of two numbers a a and b b, where a,b∈N a,b∈N. n=a b n=a b a a and b b cannot both be greater than √n n. Let us prove this by contradiction. Let a>√n a>n and b>√n b>n. Using this, we arrive at n=a b>√n√n=n n=a b>n n=n, which is a contraction. Hence, one of them must be less than or equal to√n n. (Of course, the ‘equal to’ part applies only if n n is a perfect square.) Therefore, you iterate a loop only beginning from 2 until the square root of the number. (a=1 a=1 or b=1 b=1 will not make n n composite!) If at all n n is composite, it definitely has at least one factor in the range [2,√n][2,n]. Upvote · 99 14 A Das JRF in Mathematics&Abstract Algebra, University of Göttingen (Graduated 2017) ·Updated 7y Related Why does a positive integer n is prime if it is not divisible by any prime less than or equal to √n? Let's say m=√n m=√n , then m 2=n m 2=n . Now, if n n is not a prime then n n can be written as n=a×b n=a×b, so m×m=a×b m×m=a×b . Notice that m m is a real number , since the square root of a number is not necessarily natural , whereas n,a&b n,a&b are natural numbers. Now there can be 3 cases: a>m⇒bm⇒b<m a=m⇒b=m a=m⇒b=m am am In all 3 cases, m i n(a,b)≤m m i n(a,b)≤m. Hence if we search till m m, we are bound to find at least one factor of n n, which is enough to show that n n is not prime. Since , this happens only in cases where n is not a prime , therefore the reverse happens in cases where n is a prime . Here , the calculati Continue Reading Let's say m=√n m=√n , then m 2=n m 2=n . Now, if n n is not a prime then n n can be written as n=a×b n=a×b, so m×m=a×b m×m=a×b . Notice that m m is a real number , since the square root of a number is not necessarily natural , whereas n,a&b n,a&b are natural numbers. Now there can be 3 cases: a>m⇒bm⇒b<m a=m⇒b=m a=m⇒b=m am am In all 3 cases, m i n(a,b)≤m m i n(a,b)≤m. Hence if we search till m m, we are bound to find at least one factor of n n, which is enough to show that n n is not prime. Since , this happens only in cases where n is not a prime , therefore the reverse happens in cases where n is a prime . Here , the calculation will be much faster if we skip all other numbers to be tested for the factor of n n , other than the prime numbers (which are to be tested for prime factors) . This explains why a positive integer n is prime if it is not divisible by any other primes equal to or less than √n . To cite an example , The square root of 100 is 10. Let's say a x b = 100, for various pairs of a and b. If a = b, then they are equal, and are the square root of 100, exactly. Which is 10. If one of them is less than 10, the other has to be greater. For example, 5 x 20 = 100. One is greater than 10, the other is less than 10. Thinking about a x b, if one of them goes down, the other must get bigger to compensate, so the product stays at 100. They pivot around the square root. The square root of 101 is about 10.049875621. So if you're testing the number 101 for primality, you only need to try the integers upto 10, including 10. But 8, 9, and 10 are not themselves prime, so you only have to test upto 7, which is prime . (Even better if we test the numbers 2,3,5 and 7 only , as all other numbers till 10 are their multiples ). Because if there's a pair of factors with one of the numbers bigger than 10, the other of the pair has to be less than 10. If the smaller one doesn't exist, there is no matching larger factor of 100 . If you're testing 121, the square root is 11. You have to test the prime integers 1 through 11 (inclusive) to see if it goes in evenly. 11 goes in 11 times, so 121 is not prime. If you had stopped at 10, and not tested 11, you would have missed 11. You have to test every prime integer greater than 2, but less than or equal to the square root, assuming you are only testing odd numbers . Hope this helps . Upvote · 9 3 Promoted by Bata India Dhruti Shah Visualiser | Graphic Designer (2018–present) ·Sep 12 What are the best professional affordable and comfortable shoes for women? I usually look at three things when I’m buying work shoes: comfort, cushioning and arch support; how sturdy the sole is; and whether I can actually afford to get more than one pair if I want them in different colours. Ballerinas by Bata though, are what I wear the most. I didn’t know about them until recently, when a coworker recommended them to me, also spotted my favorite creator Siddhi Karwa styling them across Europe and I have been absolutely loving them. They’re professional enough for work wear but don’t feel heavy and keep me comfortable throughout the day, even when I’m commuting to the Continue Reading I usually look at three things when I’m buying work shoes: comfort, cushioning and arch support; how sturdy the sole is; and whether I can actually afford to get more than one pair if I want them in different colours. Ballerinas by Bata though, are what I wear the most. I didn’t know about them until recently, when a coworker recommended them to me, also spotted my favorite creator Siddhi Karwa styling them across Europe and I have been absolutely loving them. They’re professional enough for work wear but don’t feel heavy and keep me comfortable throughout the day, even when I’m commuting to the office. I got mine for around ₹999 from Bata, which felt like a steal compared to some other brands I looked at. They’ve held up really well, and I can easily pair them with trousers, skirts for my work outfits. If you’re on a budget but still want something that is comfortable and follows fashion trends, Ballerinas by Bata are the perfect choice. I picked up mine from a Bata store near me, you can grab yours too. Upvote · 1.1K 1.1K 99 91 99 13 Amitabha Tripathi have more than a working knowledge of Z · Author has 4.7K answers and 13.9M answer views ·7y Related Why does a positive integer n is prime if it is not divisible by any prime less than or equal to √n? Every integer n>1 n>1 can be uniquely expressed as a product of primes. So if n n is composite, with least prime divisor p 0 p 0, then n=m p 0 n=m p 0, with m>1 m>1 and no prime divisor of m m less than p 0 p 0. Hence, m≥p 0 m≥p 0 and so n≥p 2 0 n≥p 0 2. It follows that n must have at least one prime divisor ≤√n≤n whenever n n is composite. Therefore, if n n has no prime divisor ≤√n≤n and n>1 n>1, then n n must be prime. ■◼ Upvote · 99 12 9 1 9 1 Nandeesh H N (ನಂದೀಶ್ ಎಚ್ ಎನ್) An engineer who likes Maths, English, Kannada, linguistics · Author has 2.3K answers and 3.4M answer views ·Updated 5y Related Why does a positive integer n is prime if it is not divisible by any prime less than or equal to √n? Suppose a number N is divisible by a number a. It means N has another factor b equal to N/a. If a and b are equal, then N= aa and hence a is the square root of N. If a is less than the square root, b is obviously more than the square root. So in order to find whether N is divisible by any number, it is enough if we try to divide N by numbers ranging from 2 to the square root of N. Moreover, we need to try with only prime numbers below the square root of N. This is because, if N is not divisible by a number m, it is not divisible by any multiple of m. So if you cross out all such multiples, you ar Continue Reading Suppose a number N is divisible by a number a. It means N has another factor b equal to N/a. If a and b are equal, then N= aa and hence a is the square root of N. If a is less than the square root, b is obviously more than the square root. So in order to find whether N is divisible by any number, it is enough if we try to divide N by numbers ranging from 2 to the square root of N. Moreover, we need to try with only prime numbers below the square root of N. This is because, if N is not divisible by a number m, it is not divisible by any multiple of m. So if you cross out all such multiples, you are left ith only prime numbers to check. It cannot happen that N is not divisible by any prime number less than the square root and is divisible by a number greater than the square root. For any number N, the number of prime numbers less than the square root are much less compared to the number of prime numbers greater than the square root. So to check whether a number N is prime or not, it is enough to check divisibility by a few numbers (prime numbers less than the square root). Upvote · 9 4 Utsav Dey Information technology in Bachelor of Technology Degrees, Jadavpur University (Expected 2026) · Author has 726 answers and 1.8M answer views ·7y Related Why does a positive integer n is prime if it is not divisible by any prime less than or equal to √n? I assume a number N N So we all know a number is a product form of some primes . N=(m∏k=1 p k)2 N=(∏k=1 m p k)2 Since we have taken a perfect square so √N=m∏k=1 p k N=∏k=1 m p k So if we consider a M=(m∏k=1 p k−1)2 p k+1.M=(∏k=1 m p k−1)2 p k+1. So obviously M M∤N∤N Or if we take a prime p k+1<N p k+1<N p k+1∤N p k+1∤N Hence that's for p k+j∀p k+j∀j>0 j>0 So we can say on Continue Reading I assume a number N N So we all know a number is a product form of some primes . N=(m∏k=1 p k)2 N=(∏k=1 m p k)2 Since we have taken a perfect square so √N=m∏k=1 p k N=∏k=1 m p k So if we consider a M=(m∏k=1 p k−1)2 p k+1.M=(∏k=1 m p k−1)2 p k+1. So obviously M M∤N∤N Or if we take a prime p k+1<N p k+1<N p k+1∤N p k+1∤N Hence that's for p k+j∀p k+j∀j>0 j>0 So we can say only Some primes less than its root divides N But since the question asks if N N is prime . S... Upvote · 9 2 Aman Arora Been there, done that. Not doing it anymore! ·7y Related Why does a positive integer n is prime if it is not divisible by any prime less than or equal to √n? Suppose it is not true. Let Square Root of n = X. i.e. n is composite despite the fact that none of the primes less than X divide it. => n has 2 factors between X and n. — Statement A Product of any two numbers which are greater than X will be greater than n. Suppose they are X X + k and X X+ l. Product = n + (l+k)X X + lk which is >n. This is a contradiction to Statement A. Upvote · 9 2 Related questions The only primes less than the square root of 400=20 are 2, 3, 5, 7, 11, 13, 17, and 19. Suppose that n is a number less than 400 which is not divisible by any of these primes. What can you conclude about n? What is the square root of a number that is divisible by all prime numbers? If n is a prime number and n/2 is also a prime number then what does it say about n+n/2? Why is 2 n−1 2 n−1 always prime and 2 n+1 2 n+1 always divisible by 3 3? If n is a perfect square, then 2 n can never be a perfect squared, but what if n is equal to 2? If n is a prime number greater than 3, what is the reminder when the square of those prime is divided by twelve? Why is 2^n -1 a prime number if n is a prime number? How do you find a square root of a number if it is not prime and not divisible by four or three? Is it possible to generate more primes from n numbers than a given value of n? Is a number n always co-prime with n-1? When is the nth prime number n? Are the square roots of all prime numbers irrational? Can we separate if the number n is prime or not by calculating (n-1)! (mod n)? If n square + 2 n - 8 is a prime number where n is a natural number, then is “n” a legitimate number? What is the largest possible square root of a non-prime number? Related questions The only primes less than the square root of 400=20 are 2, 3, 5, 7, 11, 13, 17, and 19. Suppose that n is a number less than 400 which is not divisible by any of these primes. What can you conclude about n? What is the square root of a number that is divisible by all prime numbers? If n is a prime number and n/2 is also a prime number then what does it say about n+n/2? Why is 2 n−1 2 n−1 always prime and 2 n+1 2 n+1 always divisible by 3 3? If n is a perfect square, then 2 n can never be a perfect squared, but what if n is equal to 2? If n is a prime number greater than 3, what is the reminder when the square of those prime is divided by twelve? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025 Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. 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https://stackoverflow.com/questions/50613928/how-to-solve-a-linear-system-for-only-one-component-in-matlab
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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to solve a linear system for only one component in MATLAB Ask Question Asked 7 years, 4 months ago Modified7 years, 4 months ago Viewed 212 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I need to solve the linear system matlab A x = b which can be done efficiently by matlab x = A \ b But now A is very large and I actually only need one component, say x(1). Is there a way to solve this more efficiently than to compute all components of x? A is not sparse. Here, efficiency is actually an issue because this is done for many b. Also, storing the inverse of K and multiplying only its first row to b is not possible because K is badly conditioned. Using the \ operator employs the LDL solver in this case, and accuracy is lost when the inverse is explicitly used. matlab linear-algebra equation-solving Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited May 30, 2018 at 22:09 rehctawratsrehctawrats asked May 30, 2018 at 22:00 rehctawratsrehctawrats 231 5 5 silver badges 16 16 bronze badges Add a comment| 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. I don't think you'd technically get a speed-up over the very optimized Matlab routine however if you understand how it is solved then you can just solve for one part of x. E.g the following. in traditional solver you use backsub for QR solve for instance. In LU solve you use both back sub and front sub. I could get LU. Unfortunately, it actually starts at the end due to how it solves it. The same is true for LDL which would employ both. That doesn't preclude that fact there may be more efficient ways of solving whatever you have. ```matlab function [Q,R] = qrcgs(A) %Classical Gram Schmidt for an m x n matrix [m,n] = size(A); % Generates the Q, R matrices Q = zeros(m,n); R = zeros(n,n); for k = 1:n % Assign the vector for normalization w = A(:,k); for j=1:k-1 % Gets R entries R(j,k) = Q(:,j)'w; end for j = 1:k-1 % Subtracts off orthogonal projections w = w-R(j,k)Q(:,j); end % Normalize R(k,k) = norm(w); Q(:,k) = w./R(k,k); end end function x = backsub(R,b) % Backsub for upper triangular matrix. [m,n] = size(R); p = min(m,n); x = zeros(n,1); for i=p:-1:1 % Look from bottom, assign to vector r = b(i); for j=(i+1):p % Subtract off the difference r = r-R(i,j)x(j); end x(i) = r/R(i,i); end end ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited May 31, 2018 at 0:47 answered May 31, 2018 at 0:12 user1911226 user1911226 1 Comment Add a comment rehctawrats rehctawratsOver a year ago That's a nice and basic answer. However I will not try it out because as you say it will most likely still be slower than the built-in methods. 2018-06-01T10:44:04.387Z+00:00 0 Reply Copy link This answer is useful 1 Save this answer. Show activity on this post. The method mldivide, generally represented as \ accepts solving many systems with the same A at once. matlab x = A\[b1 b2 b3 b4] # where bi are vectors with n rows Solves the system for each b, and will return an nx4 matrix, where each column is the solution of each b. Calling mldivide like this should improve efficiency becaus the descomposition is only done once. As in many decompositions like LU od LDL' (and in the one you are interested in particular) the matrix multiplying x is upper diagonal, the first value to be solved is x(n). However, having to do the LDL' decomposition, a simple backwards substitution algorithm won't be the bottleneck of the code. Therefore, the decomposition can be saved in order to avoid repeating the calculation for every bi. Thus, the code would look similar to this: matlab [LA,DA] = ldl(A); DA = sparse(DA); % LA = sparse(LA); %LA can also be converted to sparse matrix % loop over bi xi = LA'\(DA\(LA\bi)); % end loop As you can see in the documentation of mldivide (Algorithms section), it performs some checks on the input matrixes, and having defined LA as full and DA as sparse, it should directly go for a triangular solver and a tridiagonal solver. If LA was converted to sparse, it would use a triangular solver too, and I don't know if the conversion to sparse would represent any improvement. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Jun 1, 2018 at 11:06 answered May 31, 2018 at 1:44 OriolAbrilOriolAbril 8,963 4 4 gold badges 33 33 silver badges 41 41 bronze badges 2 Comments Add a comment rehctawrats rehctawratsOver a year ago Storing the factorization makes a lot of sense. I've actually done it before in another context but couldn't think of it now. Thanks for the reminder. 2018-06-01T10:41:02.193Z+00:00 0 Reply Copy link OriolAbril OriolAbrilOver a year ago Happy to help, the key is that `` uses many different solvers depending on the input matrix, as explained in the Algorithms section, so applying mldivide to the alredy decomposed matrixes should be much faster than on the original system. 2018-06-01T11:04:08.107Z+00:00 0 Reply Copy link Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. 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https://www.quora.com/Is-there-a-proof-why-numbers-are-divisible-by-3-if-the-sum-of-their-digits-is-divisible-by-3
Something went wrong. Wait a moment and try again. Divisibility Tests Proofs of Mathematics For... Arithmetic Number Theory Studying Number Theory Proofs (mathematics) Theory of Numbers 5 Is there a proof why numbers are divisible by 3 if the sum of their digits is divisible by 3? Felix Mcpeake might do Maths one day · Upvoted by Horst H. von Brand , PhD Computer Science & Mathematics, Louisiana State University (1987) and Yair Livne , Master's Mathematics, Hebrew University of Jerusalem (2007) · 12y Consider a number, x with the digits c,b and a in that order. So; x=100c+10b+a x=99c+9b+c+b+a x=3(33c+3b)+c+b+a Since 3(33a+3b) is divisible by 3, then we can say that x is divisible by 3 if, and only if the sum of it's digits is divisible by 3. Notice I've only shown this for 3 digits, but it's clear that this approach follows for more digits. It's also worth noting that; x=9(11c+b)+c+b+a Hence this "trick" works for divisions by 9 also. Related questions A number is divisible by 3 if the sum of its digits is divisible by 3. It is true? Why is it true that if the sum of its digits is divisible by 3 then the integer is also divisible by 3? How can it be demonstrated that every number which has the sum of its digits divisible by 3, is itself divisible by 3? If the sum of the digits is divisible by 3 (or 9), then the number itself is divisible by 3 (or 9). Why does such a divisibility rule hold good? If the sum of digits of a number is divisible by 9, it's divisible by 9 (a new rule like 3)? Can you prove by example if not. Wes Browning Studied Mathematics at University of Washington · Author has 8.9K answers and 6.1M answer views · 6y Yes. I won’t give the entire proof here, just a big hint. 1–1 is 0, which is divisble by 3. 10–1 is 9. That’s divisble by 3. 100–1 is 99, that’s divisble by 3, 1000–1 is 999, that’s divisible by 3. Consider the number 4578. Is it divisible by 3? Well, it equals, by definition, 41000 + 5100 + 710 + 18. How much more than 4 + 5 +7 +8 = 24 is this number? It’s this much more: 4(1000–1) + 5(100–1) + 7(10–1) + 8(1–1). All that is divisible by 3. So the number 4578 is a number divisible by 3 plus 24 (the sum of the digits) and 24 is also divisible by 3. The sum of two numbers each divisible by 3 Yes. I won’t give the entire proof here, just a big hint. 1–1 is 0, which is divisble by 3. 10–1 is 9. That’s divisble by 3. 100–1 is 99, that’s divisble by 3, 1000–1 is 999, that’s divisible by 3. Consider the number 4578. Is it divisible by 3? Well, it equals, by definition, 41000 + 5100 + 710 + 18. How much more than 4 + 5 +7 +8 = 24 is this number? It’s this much more: 4(1000–1) + 5(100–1) + 7(10–1) + 8(1–1). All that is divisible by 3. So the number 4578 is a number divisible by 3 plus 24 (the sum of the digits) and 24 is also divisible by 3. The sum of two numbers each divisible by 3 is always divisible by 3. Extend that to longer numbers and all possible positive integers in decimal notation. Assistant Bot · 1y Yes, there is a proof for why a number is divisible by 3 if the sum of its digits is divisible by 3. This property relies on the concept of congruences in modular arithmetic. Proof: Understanding the Number Representation: Any number n in decimal form can be expressed as: n=dk⋅10k+dk−1⋅10k−1+…+d1⋅101+d0⋅100 where di are the digits of the number n. Modulo 3 Property of Powers of 10: We need to consider how powers of 10 behave under modulo 3: 10≡1mod3 Therefore, for any integer i, 10i≡1i≡1mod3 Reduc Yes, there is a proof for why a number is divisible by 3 if the sum of its digits is divisible by 3. This property relies on the concept of congruences in modular arithmetic. Proof: Understanding the Number Representation: Any number n in decimal form can be expressed as: n=dk⋅10k+dk−1⋅10k−1+…+d1⋅101+d0⋅100 where di are the digits of the number n. Modulo 3 Property of Powers of 10: We need to consider how powers of 10 behave under modulo 3: 10≡1mod3 Therefore, for any integer i, 10i≡1i≡1mod3 Reducing the Number Modulo 3: Using the property from step 2, we can reduce n modulo 3: n≡dk⋅1+dk−1⋅1+…+d1⋅1+d0⋅1mod3 This simplifies to: n≡dk+dk−1+…+d1+d0mod3 This means that n is congruent to the sum of its digits modulo 3. Conclusion: If the sum of the digits S is divisible by 3, then: S≡0mod3 Consequently, since n≡Smod3, it follows that: n≡0mod3 Therefore, n is divisible by 3. Summary: The property that a number is divisible by 3 if the sum of its digits is divisible by 3 can be proven through the use of modular arithmetic, specifically showing that each digit contributes equally under modulo 3 due to the behavior of powers of 10. Shreshth Mehrotra Author has 137 answers and 418.6K answer views · 7y Originally Answered: For any integer, as long as the sum of all digits is a multiple of 3, the integer is divisible by 3. Is there a formal proof for this? · I know of a beautiful number theory proof. Let p denote the polynomial function: p(x)=a0xn+a1xn−1+a2xn−2+...+an =n∑i=0aixn−i Then, x≡y(modk)⟹p(x)≡p(y)(modk) The proof of this is rather elementary, considering the fact that congruences can be added and multiplied. Now, Any number can be written in terms of increasing powers of I know of a beautiful number theory proof. Let p denote the polynomial function: p(x)=a0xn+a1xn−1+a2xn−2+...+an =n∑i=0aixn−i Then, x≡y(modk)⟹p(x)≡p(y)(modk) The proof of this is rather elementary, considering the fact that congruences can be added and multiplied. Now, Any number can be written in terms of increasing powers of 10 (by the definition of base 10, or the decimal system). For instance, 39=3∗101+9∗100 486=4∗102+8∗101+6∗100 6942=6∗103+9∗102+4∗101+2∗100 This seems to match with the definition of the polynomial function. As a result, we can conclude that any number can be written as a polynomial function of 10. (notice that 10 behaves exactly like x does in the definition). so, if N=p(x)=axn+bxn−1+cxn−2+...+k, p(10)=abc...k Also, observe that p(1)=a+b+c+...+k, that is, the digit sum. Let us denote this by S. Now, we come to the meat of the proof. We need to establish divisilibity by 3, hence we need to use mod 3. We have 10≡1(mod3) From the theorem, it follows, p(10)≡p(1)(mod3) We have already established that [math]... Related questions What is the proof that if the sum of the digits of a number is divisible by 3, then that number is a multiple of 3? Why is a number whose sum is divisible by 3 always divisible by 3? If the sum of the digits of a number is divisible by 9, then it is divisible by 9 (a new rule). Can you prove by example if not? This also means that it is divisible by 3. Can you give an example of a number that is divisible by 9 but not by 3. What is proof that not all the sum of a 2-digit number and the number formed by reversing its digits will always be divisible by 11? What is the sum of the numbers of three digits divisible by 7? Mathivanan Palraj Author of GMAT:Problem Solving Techniques for Top Score · Author has 1.7K answers and 3.9M answer views · 6y Let ab is a two digit number. This can be written as 10a+b = 9a+a+b. Since 9a is divisible by 3, we have to check only for a+b, that is sum of digits. Let abc be a three digit number. This can be written as 100a+10b+c or 99a+9b+a+b+c. Since 99a+9b is divisible by 3, we have to check for only a+b+c, that is the sum of digits. Similarly, we can prove for any number. This proof also works for 9. Alan Bustany Trinity Wrangler, Diophantine equations are hard · Upvoted by Jeremy Collins , M.A. Mathematics, Trinity College, Cambridge and Horst H. von Brand , PhD Computer Science & Mathematics, Louisiana State University (1987) · Author has 9.8K answers and 58.5M answer views · Updated 2y Originally Answered: Why is it true that if the sum of its digits is divisible by 3 then the integer is also divisible by 3? · Let's not restrict ourselves to just decimals being divisible by three. Let's look at divisibility by any number m in any base b. In general a number in Positional notation is represented as a (reverse) sequence of digits, di, interpreted as the sum of those digits multiplied by a power of the base, b. That is the number N represented as dn…d2d1d0 has the numeric value given by N=n∑i=0dibi=dnbn+⋯+d2b2+d1b+d0 So the number we represent in decimal as 573 actually means 5×102+7×101+3×100=500+70+3 Tests for divisibility depend on a n Let's not restrict ourselves to just decimals being divisible by three. Let's look at divisibility by any number m in any base b. In general a number in Positional notation is represented as a (reverse) sequence of digits, di, interpreted as the sum of those digits multiplied by a power of the base, b. That is the number N represented as dn…d2d1d0 has the numeric value given by N=n∑i=0dibi=dnbn+⋯+d2b2+d1b+d0 So the number we represent in decimal as 573 actually means 5×102+7×101+3×100=500+70+3 Tests for divisibility depend on a number leaving a remainder of zero when divided by a number. The general way to say that in Modular arithmetic is: N≡r(modm) N is equivalent to r modulo m, or N leaves a remainder of r when divided by m. One of the great things about this form of arithmetic is the following identity: bi≡(bmodm)i(modm) As a result we can calculate the remainder just using the digits because N≡∑di(bmodm)i(modm) This allows us to do great things when b≡1(modm), because one raised to any power is just one. In particular 10≡1(mod3) so N≡∑di(mod3). That is a number is divisible by three if and only if the sum of its decimal digits is divisible by three. Note also that 10≡1(mod9) so N≡∑di(mod9). That is a number is divisible by nine if and only if the sum of its decimal digits is divisible by nine. This trick works for divisibility by: Seven in octal; Fifteen, five, or three in hexadecimal; and so on. Other tricks follow when b≡0(modm), because zero raised to any power in just zero. Thus you can ignore all the digits except the last one when checking for divisibility by a factor of the base. In decimal this means it is easy to see if a number is divisible by five or two because 10≡0(mod5) and 10≡0(mod2). Some would argue that base sixty is better because it includes factors of 2,3,4,5,6 making divisibility easy to see in all of those numbers. This allows us to easily have a quarter past the hour which would get a bit messy in decimal. It also makes it easier for Lawyers and other professional services to charge you in 6-minute intervals... Finally b≡−1(modm) allows us to add and subtract alternate digits because powers of −1 alternate between plus and minus one. Hence the test for divisibility by eleven in decimal is to add and subtract alternate digits in the number. This trick also works for divisibility by: Nine in octal; Seventeen in hexadecimal; and so on. Kurt Mager Enjoys solving math problems · Author has 17.6K answers and 7.4M answer views · 5y Originally Answered: Why is it true that if the sum of its digits is divisible by 3 then the integer is also divisible by 3? · Why is it true that if the sum of its digits is divisible by 3 then the integer is also divisible by 3? It's easily demonstrated. Consider the number 123. Right now focus less on that it is divisible by 3, but how it is constructed. 123=(1×100)+(2×10)+3 Seems obvious enough, right? Well let's construct a similar number but instead with the digits a,b,and c. 100a+10b+c Now subtract one instance of each of the digits. (100a+10b+c)−(a+b+c)=99a+9b Now, for argument sake lets assume that a+b+c=5, and add that back to what we have so far, and then try to divide by 3. \dfrac{99a+9b+5}{3}=33a+3b+1\dfrac{2}{3 Why is it true that if the sum of its digits is divisible by 3 then the integer is also divisible by 3? It's easily demonstrated. Consider the number 123. Right now focus less on that it is divisible by 3, but how it is constructed. 123=(1×100)+(2×10)+3 Seems obvious enough, right? Well let's construct a similar number but instead with the digits a,b,and c. 100a+10b+c Now subtract one instance of each of the digits. (100a+10b+c)−(a+b+c)=99a+9b Now, for argument sake lets assume that a+b+c=5, and add that back to what we have so far, and then try to divide by 3. 99a+9b+53=33a+3b+123 The sum of the digits did not add to a multiple of 3, and so integer created cannot be evenly divided by 3. Now assume that the sum of the digits is a multiple of 3 instead. 99a+9b+3n3=33a+3b+n Thus if the sum of an integer’s digits are divisible by 3, then the integer itself will be as well. Kenneth Chen does lots and lots of math · Author has 455 answers and 1.1M answer views · 7y Originally Answered: Why is it true that if the sum of its digits is divisible by 3 then the integer is also divisible by 3? · User-10592860167709128945’s answer does it best, but it might be hard to follow for some. I think that’s why Quora asked me to answer this. I’ll make it easy to understand. If we have a four-digit number with digits A, B, C, and D for example, our number could be written as: 1000A+100B+10C+D =(999+1)A+(99+1)B+(9+1)C+D =(999A+99B+9C)+A+B+C+D You can see that the first set of parentheses is divisible by 3 (and 9). The problem is A+B+C+D. In order for the whole number to be divisible by 3, that sum also needs to be divisible by 3. This works for a number of any size because EVERY power of 10 is 1 more User-10592860167709128945’s answer does it best, but it might be hard to follow for some. I think that’s why Quora asked me to answer this. I’ll make it easy to understand. If we have a four-digit number with digits A, B, C, and D for example, our number could be written as: 1000A+100B+10C+D =(999+1)A+(99+1)B+(9+1)C+D =(999A+99B+9C)+A+B+C+D You can see that the first set of parentheses is divisible by 3 (and 9). The problem is A+B+C+D. In order for the whole number to be divisible by 3, that sum also needs to be divisible by 3. This works for a number of any size because EVERY power of 10 is 1 more than a multiple of 3 (and 9). As I’ve hinted throughout my answer, this can also be applied to check divisibility by 9. Dan That's Numberwang. · Author has 8.9K answers and 6.6M answer views · 4y Originally Answered: What is the mathematics behind "a number whose digit sum becomes three is divisible by three"? · Let’s do this in reverse and also find an even easier test… This is a long answer. But what I’ve aimed to do is start with Kindergarten arithmetic, hand-hold through some very basic algebra and show the most basic Mathematics behind this idea. I’ll assume you know division and remainder? Well in my head it tells us that any non-negative whole number can be written as a=3q+r and r is 0,1 or 2. a, q and r are placeholders for numbers. Algebra isn’t Kindergarten, but you should realise that division and remainder means ‘so many 3s’ and ‘so many’ left-over and the left-over is less than 3. Well that’s w Let’s do this in reverse and also find an even easier test… This is a long answer. But what I’ve aimed to do is start with Kindergarten arithmetic, hand-hold through some very basic algebra and show the most basic Mathematics behind this idea. I’ll assume you know division and remainder? Well in my head it tells us that any non-negative whole number can be written as a=3q+r and r is 0,1 or 2. a, q and r are placeholders for numbers. Algebra isn’t Kindergarten, but you should realise that division and remainder means ‘so many 3s’ and ‘so many’ left-over and the left-over is less than 3. Well that’s wordy. So we call the ‘so many 3s’ q and ‘so many left over’ r. At this level, that’s algebra. Letters standing in for numbers. If you can live with that and multiplying out brackets we’re probably going to get you through! Feel free to hold my hand. That means ask questions in the comments! We can’t start however without remember that ‘a is divisible by 3 if (and only if) r=0’. That’s what divisible means. You’ll have been taught long division probably but I’m being ignorant today. Just count how many times you can subtract 3 from a until you’ve got less than 3 left. That count is q (quotient) and what’s left is r (remainder). That’s a terrible way to do division but we’re not going to do much and then only easy examples. OK, what you may have noticed or not is that when we’re faced with a sum such as 700+77 we can do the division and remainder thing by adding them and then dividing or we could divide them individually and then add the separate quotients (q) and the remainders together separately. It doesn’t quite work because the sum of the remainders might be ‘oversized’ (more than 3) to be the real remainder. It is here. 700=3233+1 and 77=325+2. So adding the quotients (233 and 25) we get 258 and adding the remainders we get 3. Snag! 3 isn’t a remainder. That has to be 0,1 or 2. But the fix is easy and hopefully fairly obvious. We then further divide the remainder (which is nice and small) by 3 then add any 3s that ‘come out’ to the quotient and find the the ‘real’ remainder and the ‘real’ quotient. in the 700+77 case that means divide 3 by 3 get 1 and no remainder, add 1 to 258 and get the final quotient of 259 and remainder 0. The correct answer. 777 is divisible by 3 and the quotient is 259 (as if I cared about quotient). So instead of adding it all up and doing a big divide, we opted for separate divides, add up then ‘fix-up; as needed if the ‘rough’ remainder is invalid. I got a ‘rough’ result for the quotient then found the remainder was ‘oversized’, reduced it and tidied up to an exact result. That’s why I wanted to see division as subtract until done. It’s easier to see this is tidying up repeated subtraction. No the monster of long division! That do and fix up can simply arithmetic problems. But the idea is going to crack this challenge in two taps! What is even more important to notice here is that when I did the fix up I needed to know the remainders to determine any fix-up to the ‘rough’ quotient but I didn’t needed need the quotient to fix-up the ‘rough’ remainder. If you tell me the ‘rough’ remainder I can tell you the ‘right’ remainder. This method extends to a sum of any number of numbers. What’s that? You had a sum of 7 numbers, divided them individually and got remainders of 2,1,0,2,0,2 and 1? Well those remainders sum to 8 (oversized) and so the ‘right’ remainder is 2 and you need to add 2 onto the ‘rough’ quotient. That’s because 8=32+2. I still don’t know what the quotient is but I know the remainder. Quotients are boring anyway! At least today… This is where we start to get smart. Don’t worry you don’t need to get smart. Some unknown person was smart thousands of years ago. And the smart thing they noticed is very simple. Smart but simple should be everybody’s favourite kind of smart and we’re in its presence. If I’m asking about divisibility I told you I don’t even care about the quotient! We need that to complete a division. We don’t need it to determine divisibility. We’ve just spotted that because the remainder didn’t depend on the quotient. Getting it? So murdering the point, if I happen to know the remainders for the terms of a sum I don’t even need to think about each division. I can determine the remainder of the sum from the remainders (possibly with a final remainder division to reduce it to 0,1 or 3). I sum the remainders and then reduce (on a low heat) to a number between 0 and 2 with a quick division by 3. We’re getting close. Back to Kindergarten! 777=700+70+7. That is definitely kindergarten. We called it Hundreds, Tens and Units and fancy types call it place-value but whatever it is it’s what 777 even means. 7 hundreds, 7 tens and a 7. If only! If only!! (wink wink) I knew the remainders of 700, 70 and 7 I only want the remainder so I could cut out a bunch of division work and do our sum and fix trick straight of. Surprise! I do. And maybe not how you think… All the remainders are 1. So while we’re here we can notice that 1+1+1 is 3 which leaves remainder 0 divided by three so 777 is divisible by 3. If you want to check, the divisions are 700=3233+1, 70=323+1 and 7=32+1. Quotient in italics, remainder in bold. The key result to this whole idea of adding the digits relies on the fact that: There is no coincidence that 700,70 and 7 leave the same remainder (1) when divided by 3. It arises from an important structural fact about the relationship between 3 and 10. Key Fact: The remainder of 7 followed by any number of zeroes divided by 3 is the same as the remainder of 7 on its own (no zeroes) divided by 3. Super Key Fact: Any number ending in any number of zeroes has the same remainder when divided by 3 as that number (without the zeroes). 7 is just our example. It’s 3 and 10 that have a special relationship here makes this trick work for 3 but not any number! Let’s do times 10 (one zero). We have our form above a=3q+r and r=0,1 or 2 as before. q is ‘so many 3s’ and r is ‘so many left over’ as before. So if a=3q+r then 10a=10(3q+r). I’m going to labour the algebra for a moment. Hopefully its clear that if a=3q+r then 10a=10(3q+r) because 10 times the same thing (even if expressed differently). So expanding brackets we get 10a=103q+10r=310q+10r. OK so that’s also 30q+ 10r but stop precisely where I have it above. Remember our fix-up trick? It’s back. r is 0,1 or 2. So 10r is 0,10 or 20. If r is 0 the ‘fixed’ remainder is 0 (easy). If r is 1 the ‘fixed’ remainder is also 1 (because 10=33+1) and if r is 2 the fixed remainder is (drum-roll) also 2 because 20=36+2). There are only 3 values for r and we’ve done them and the remainder for 10a divided by 3 is the same a divided by 3 not matter what it was! So look at 5. 5 is remainder 2 divided by 3. So is 50 (316+2). Try some more. What is the remainder of 80 divided by 3 by only thinking about 8 divided by 3. It doesn’t work with all divisors (3 is the divisor here). For example 15 divided by 6 is 2 and 3 remainder but 1015=150 and 150 is 256 and no remainder. Not the same. There’s so advanced maths about all this. But stick to remainder divided by 3 and multiplying by 10 and we’ll get there today. But back to our breakthrough. Multiply a number by 10 and the result will have the same remainder divided by 3 as the number itself. Be it 0,1 or 2. We’ve just shown that by brute-force of examining the cases for 0,1 and 2. What happens to the remainder if we times by 10 again? Same answer - nothing - it’s the same. Mathematicians have this fancy thing called ‘Mathematical Induction’ but today I’m appealing to ‘obviously’. Multiplying by 10 didn’t change the remainder so doing it again won’t and doing it any number of times will leave the remainder unchanged. The quotient (q)? Well that’s been going up and up (times 10 plus ‘extras’ of 0, 3 or 6 from the fix up from ‘rough’ remainders). But we don’t care about the quotient (q). Quotients are dead to us today. We’re all about remainders for divisibility remember. So pulling that together for any number defg where d,e,f,g are digits we know that’s defg=d1000+e100+f10+g Because that’s what place-value means! And if we want its remainder divided by 3 we only have to look at the remainders of d1000, e100, f100 and g divided by 3. But by our interesting result about multiplying by 10 (any number of times) we only really need to look at the remainders of d, e, f and g divided by 3. How do we do that? We add the the digits together and look at their remainder divided by 3 (fix-up stylie) and if that is 0 we know the remainder of defg is zero. We know multiplying by 10 changes nothing (so we didn’t do it) and we know how to fix-up the remainder for sums which is precisely adding them all together and seeing if that divides by 3. Job done. That’s the result the question asked for! Something is divisible by 3 if and only if the sum of its digits is divisible by 3. We need to wink at this mysterious ‘Mathematical Induction’ thing. I’ve only really shown that for 4 digit numbers. But because of the multiply by ten makes no odds thing it’s hopefully clear this extends to numbers of any length. ‘Mathematical Induction’ actually needs another wink because I hope it’s also clear we can just keep adding the digits of the sum of the digits until we get something easy or inevitably reach a single digit number and see if it’s divisible by 3. OK, so the proof took a while but we just bank that and take the result away. My problem is that while adding the digits is easy it could be easier! Here’s a case 396817. 3+9+6+8+1+7 = 34 and 3+4 is 7 and 7 is remainder 1 divided by 3 so 369817 is not divisible by 3. Well first up forget about 0, 3, 6 and 9. They’re multiples of 3 and not invited to the remainder party! They add nothing (to the remainder). They’re shunned whole to the quotient (we don’t care about). So you could skip them and calculate 8+1+7 which gives 16 and comes back to 7 which is still remainder 1 divided by 3. But when the numbers get proper long we can simplify further. Take 363798363805734756 Write something like this: 363798363805734756 1 2 2 21 112 I told you 0, 3, 6 and 9 aren’t invited. I’ve blanked them! Your sort should go and play with your la-di-dah quotient friends. I also don’t care about 3s hiding in numbers like the two inside 8. So I’ve said that’s a remainder of 2 snuggled up with some 3s. That’s reduce this to summing some 1s and 2s. So all we want is to know the remainder of 1+2+2+2+1+1+1+2 divided by 3. Sums are still to hard. Let’s make it even easier. What jumps out is that the first two terms sum to 3. So scrub them. I also see a run of 3 1s. So scrub them. Then we’ve got [1+2]+2+2+[1+1+1]+2. Quora won’t do strikethrough. So think of the things in square brackets as struck-out. So now all we’ve got is 3 twos. So scrub the lot. It turns out 363798363805734756 is divisible by 3 and we hardly did anything. Going through the rigmarole of 3+6+3+7+9+8+3+6+3+8+0+5+7+3+4+7+5+6 =93 then 9+3 = 12 and 1+3 = 3 so the remainder is 0 and that big fella is divisible by 3 (I don’t even care what the quotient is). Is too many steps. With skipping 0,3,6 and 9 its still 7+8+8+5+7+4+7+5 = 51 giving 5+1 = 6 which is divisible by 3. Done. I’m not suggesting you’d do all this in a practical task to test divisibility! I’m pointing out that this trick of ignoring mutliples of 3s goes further than the common trick of “It divides by 3 if the digit sum divides by 3”. Alex Moon BS in Pure Mathematics, Michigan State University · Author has 3.4K answers and 2.1M answer views · Updated 4y This holds because we’re in base 10, and 3 divides 9 cuz (9=10–1). Sound like a weird reason? Well let me prove it: A number x can be expressed as just ∑n−1k=0bkak (for n digit number) where a is digit values in your baseb, and if you sum the digits themselves it’s clearly just ∑n−1k=0ak we’ll call this y. x−y=∑n−1k=0bkak−∑n−1k=0ak=∑n−1k=0(bk−1)ak b−1|bk−1 for any natural number k, (b−1)(1+b+…+bk−2+bk−1)=bk−1 This means x−y is always divisible by b−1 and anything that divid This holds because we’re in base 10, and 3 divides 9 cuz (9=10–1). Sound like a weird reason? Well let me prove it: A number x can be expressed as just ∑n−1k=0bkak (for n digit number) where a is digit values in your baseb, and if you sum the digits themselves it’s clearly just ∑n−1k=0ak we’ll call this y. x−y=∑n−1k=0bkak−∑n−1k=0ak=∑n−1k=0(bk−1)ak b−1|bk−1 for any natural number k, (b−1)(1+b+…+bk−2+bk−1)=bk−1 This means x−y is always divisible by b−1 and anything that divides b−1 also divides bk−1 and x−y too! This means x and y have the same remainder when divided by the factors of b−1, which means “3 either divides the number AND the sum of its digits, or neither, in base10” to give the example that you asked for specifically. But it’s actually deeper than just 3 in base10, which comes from 9 in base10. Scott I write code · Author has 11.7K answers and 12M answer views · 4y Originally Answered: What is the mathematics behind "a number whose digit sum becomes three is divisible by three"? · Imagine you have four piles of matchsticks, and that you take seven matchsticks out of one pile, throw three away, and put the remaining four back in a different pile. You can see that the remainder when you divide the sum of the matchsticks in all the piles by 3 would remain unchanged, since we threw away a number that’s a multiple of 3. That’s how it works. No, really. We have a pile over here that we call the “ones place”, and another that we call the “tens place.” By convention, we don’t allow there to be more than 9 things in any of these piles. But imagine that that convention weren’t ther Imagine you have four piles of matchsticks, and that you take seven matchsticks out of one pile, throw three away, and put the remaining four back in a different pile. You can see that the remainder when you divide the sum of the matchsticks in all the piles by 3 would remain unchanged, since we threw away a number that’s a multiple of 3. That’s how it works. No, really. We have a pile over here that we call the “ones place”, and another that we call the “tens place.” By convention, we don’t allow there to be more than 9 things in any of these piles. But imagine that that convention weren’t there, and picture what we’re doing. Let’s say you have 27 matchsticks, all in the ones pile. We take 10 of them, throw 9 away, and move the remaining 1 into the tens pile. Since the number we threw away is divisible by 3, the divisibility of the digit sum didn’t change. We do the same thing if we take 10 out of the tens pile, throw 9 away, and put the remaining 1 in the hundreds pile. The same kind of logic can also apply in reverse: we could take 1 out of the tens pile, add 9 new matchsticks, and place all 10 in the ones pile without changing the divisibility by 3. Lucas Curtis lover of numbers · Author has 7.4K answers and 22.1M answer views · 3y I’ll start with a three-digit number, and you can generalize the proof to any number of digits you wish. Let N=100a+10b+c, where a, b, and c are the digits of the number. N=(99+1)a+(9+1)b+c N=99a+9b+a+b+c Now clearly 99a+9b is divisible by 3. 99a+9b3=33a+3b Which means that if a+b+c is divisible by 3, then 100a+10b+c is also divisible by 3. You can write this in proper mathematical language that generalizes the proof for any number of digits, but hopefully you can deduce from this short exercise that the divisibility rule works for any integer. Gary Russell Former Professor at University of Iowa (1996–2025) · Author has 6K answers and 3.1M answer views · Updated 11mo Yes. To see this, we need the concept of the digital root (DR). This is the remainder when you divide by 9. Formally, for a number y, you have y = DR mod 9 = DR + 9k where k is an integer. You can prove that DR can be obtained by repeatedly summing the digits of a number until you get a number in the range {1,2, …, 9}. For example, suppose that your number is 307. Then, 3+0+7 = 10. Now, compute the sum 1+0 =1. So, 1 is the DR of 307. Note that 307 = 9k + 1, where, in this case, k =34. Suppose that DR = 9. Then, the number is divisible by 9. Hence, it is divisible by 3. Suppose that DR = 3. Then, x Yes. To see this, we need the concept of the digital root (DR). This is the remainder when you divide by 9. Formally, for a number y, you have y = DR mod 9 = DR + 9k where k is an integer. You can prove that DR can be obtained by repeatedly summing the digits of a number until you get a number in the range {1,2, …, 9}. For example, suppose that your number is 307. Then, 3+0+7 = 10. Now, compute the sum 1+0 =1. So, 1 is the DR of 307. Note that 307 = 9k + 1, where, in this case, k =34. Suppose that DR = 9. Then, the number is divisible by 9. Hence, it is divisible by 3. Suppose that DR = 3. Then, x = 3 + 9k, which is divisible by 3. So, we realize that if the DR is divisible by 3, the number is divisible by 3. OK, do we have to do all the repeated summations? No, if you just do it once, you will find that [Summation Once] = DR + 9k for some k. Again, if this is divisible by 3, then DR is either 3 or 9 and x is divisible by 3. So, if the sum of the digits of a number is divisible by 3, the number is divisible by 3. For more information on the digital root, click here. Digital root - Wikipedia Repeated sum of a number's digits The digital root (also repeated digital sum ) of a natural number in a given radix is the (single digit) value obtained by an iterative process of summing digits , on each iteration using the result from the previous iteration to compute a digit sum. The process continues until a single-digit number is reached. For example, in base 10, the digital root of the number 12345 is 6 because the sum of the digits in the number is 1 + 2 + 3 + 4 + 5 = 15, then the addition process is repeated again for the resulting number 15, so that the sum of 1 + 5 equals 6, which is the digital root of that number. In base 10, this is equivalent to taking the remainder upon division by 9 (except when the digital root is 9, where the remainder upon division by 9 will be 0), which allows it to be used as a divisibility rule . Let n {\displaystyle n} be a natural number. For base b > 1 {\displaystyle b>1} , we define the digit sum F b : N → N {\displaystyle F_{b}:\mathbb {N} \rightarrow \mathbb {N} } to be the following: F b ( n ) = ∑ i = 0 k − 1 d i {\displaystyle F_{b}(n)=\sum _{i=0}^{k-1}d_{i}} where k = ⌊ log b ⁡ n ⌋ + 1 {\displaystyle k=\lfloor \log _{b}{n}\rfloor +1} is the number of digits in the number in base b {\displaystyle b} , and d i = n mod b i + 1 − n mod b i b i {\displaystyle d_{i}={\frac {n{\bmod {b^{i+1}}}-n{\bmod {b}}^{i}}{b^{i}}}} is the value of each digit of the number. A natural number n {\displaystyle n} is a digital root if it is a fixed point for F b {\displaystyle F_{b}} , which occurs if F b ( n ) = n {\displaystyle F_{b}(n)=n} . All natural numbers n {\displaystyle n} are preperiodic points for F b {\displaystyle F_{b}} , regardless of the base. This is because if n ≥ b {\displaystyle n\geq b} , then n = ∑ i = 0 k − 1 d i b i {\displaystyle n=\sum _{i=0}^{k-1}d_{i}b^{i}} and therefore F b ( n ) = ∑ i = 0 k − 1 d i < ∑ i = 0 k − 1 d i b i = n {\displaystyle F_{b}(n)=\sum _{i=0}^{k-1}d_{i}<\sum _{i=0}^{k-1}d_{i}b^{i}=n} because b > 1 {\displaystyle b>1} . If n < b {\displaystyle n<b} , then trivially F b ( n ) = n {\displaystyle F_{b}(n)=n} Therefore, the only possible digital roots are the natural numbers 0 ≤ n < b {\displaystyle 0\leq n<b} , and there are no cycles other than the fixed points of 0 ≤ n < b {\displaystyle 0\leq n<b} . In base 12 , 8 is the additive digital root of the base 10 number 3110, as for n = 3110 {\displaystyle n=3110} d 0 = 3110 mod 12 0 + 1 − 3110 mod 1 2 0 12 0 = 3110 mod 12 − 3110 mod 1 1 = 2 − 0 1 = 2 1 = 2 {\displaystyle d_{0}={\frac {3110{\bmod {12^{0+1}}}-3110{\bmod {1}}2^{0}}{12^{0}}}={\frac {3110{\bmod {12}}-3110{\bmod {1}}}{1}}={\frac {2-0}{1}}={\frac {2}{1}}=2} d 1 = 3110 mod 12 1 + 1 − 3110 mod 1 2 1 12 1 = 3110 mod 144 − 3110 mod 1 2 12 = 86 − 2 12 = 84 12 = 7 {\displaystyle d_{1}={\frac {3110{\bmod {12^{1+1}}}-3110{\bmod {1}}2^{1}}{12^{1}}}={\frac {3110{\bmod {144}}-3110{\bmod {1}}2}{12}}={\frac {86-2}{12}}={\frac {84}{12}}=7} d 2 = 3110 mod 12 2 + 1 − 3110 mod 1 2 2 12 Here’s another proof. Suppose that we have a 4 digit number x = “abcd”. Then, x = a(10^3) + b(10^2) + c(10) + d x mod 3 = a(1^3) + b(1^2) + c(1) + d mod 3 = a+b+c+d mod 3 If a+b+c+d is divisible by 3, then a+b+c+d mod 3 = 0 mod 3 x = 0 mod 3 = 3k for some integer k. You can repeat this proof for any number of digits in x. You will get the same result. Thus, if the sum of the digits of a number is divisible by 3, the number is divisible by 3. Related questions A number is divisible by 3 if the sum of its digits is divisible by 3. It is true? Why is it true that if the sum of its digits is divisible by 3 then the integer is also divisible by 3? How can it be demonstrated that every number which has the sum of its digits divisible by 3, is itself divisible by 3? If the sum of the digits is divisible by 3 (or 9), then the number itself is divisible by 3 (or 9). Why does such a divisibility rule hold good? If the sum of digits of a number is divisible by 9, it's divisible by 9 (a new rule like 3)? Can you prove by example if not. What is the proof that if the sum of the digits of a number is divisible by 3, then that number is a multiple of 3? Why is a number whose sum is divisible by 3 always divisible by 3? If the sum of the digits of a number is divisible by 9, then it is divisible by 9 (a new rule). Can you prove by example if not? This also means that it is divisible by 3. Can you give an example of a number that is divisible by 9 but not by 3. What is proof that not all the sum of a 2-digit number and the number formed by reversing its digits will always be divisible by 11? What is the sum of the numbers of three digits divisible by 7? How many 3-digit numbers are divisible by 99? What are 3 digit numbers divisible by 3 and 5? What is the proof of divisibility test of 3? What is the greatest 3 digit number which is exactly divisible by 3, 4, and 5? What is the small three-digit number divisible by 3? Related questions A number is divisible by 3 if the sum of its digits is divisible by 3. It is true? Why is it true that if the sum of its digits is divisible by 3 then the integer is also divisible by 3? How can it be demonstrated that every number which has the sum of its digits divisible by 3, is itself divisible by 3? If the sum of the digits is divisible by 3 (or 9), then the number itself is divisible by 3 (or 9). Why does such a divisibility rule hold good? If the sum of digits of a number is divisible by 9, it's divisible by 9 (a new rule like 3)? Can you prove by example if not. What is the proof that if the sum of the digits of a number is divisible by 3, then that number is a multiple of 3? Why is a number whose sum is divisible by 3 always divisible by 3? If the sum of the digits of a number is divisible by 9, then it is divisible by 9 (a new rule). Can you prove by example if not? This also means that it is divisible by 3. Can you give an example of a number that is divisible by 9 but not by 3. What is proof that not all the sum of a 2-digit number and the number formed by reversing its digits will always be divisible by 11? 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https://wiki.sei.cmu.edu/confluence/display/c/FLP32-C.+Prevent+or+detect+domain+and+range+errors+in+math+functions
Skip to main content assistive.skiplink.to.breadcrumbs assistive.skiplink.to.header.menu assistive.skiplink.to.action.menu assistive.skiplink.to.quick.search Skip to sidebarSkip to main contentSkip to breadcrumbsSkip to search Log in SEI CERT C Coding Standard 1 Front Matter 2 Rules Rule 01. Preprocessor (PRE) Rule 02. Declarations and Initialization (DCL) Rule 03. Expressions (EXP) Rule 04. Integers (INT) Rule 05. Floating Point (FLP) FLP30-C. Do not use floating-point variables as loop counters FLP32-C. Prevent or detect domain and range errors in math functions FLP34-C. Ensure that floating-point conversions are within range of the new type FLP36-C. Preserve precision when converting integral values to floating-point type FLP37-C. Do not use object representations to compare floating-point values Rule 06. Arrays (ARR) Rule 07. Characters and Strings (STR) Rule 08. Memory Management (MEM) Rule 09. Input Output (FIO) Rule 10. Environment (ENV) Rule 11. Signals (SIG) Rule 12. Error Handling (ERR) Rule 13. Application Programming Interfaces (API) Rule 14. Concurrency (CON) Rule 48. Miscellaneous (MSC) Rule 50. POSIX (POS) Rule 51. Microsoft Windows (WIN) 3 Recommendations 4 Back Matter 5 Admin Wiki Contents CERT manifest files Created by Admin, last modified by David Svoboda on Aug 06, 2025 The C Standard, 7.12.1 [ISO/IEC 9899:2024], defines three types of errors that relate specifically to math functions in <math.h>. Paragraph 2 states A domain error occurs if an input argument is outside the domain over which the mathematical function is defined. Paragraph 3 states A pole error (also known as a singularity or infinitary) occurs if the mathematical function has an exact infinite result as the finite input argument(s) are approached in the limit. Paragraph 4 states arange error occurs if and only if the result overflows or underflows An example of a domain error is the square root of a negative number, such as sqrt(-1.0), which has no meaning in real arithmetic. Contrastingly, 10 raised to the 1-millionth power, pow(10., 1e6), cannot be represented in many floating-point implementations because of the limited range of the type double and consequently constitutes a range error. In both cases, the function will return some value, but the value returned is not the correct result of the computation. An example of a pole error is log(0.0), which results in negative infinity. Programmers can prevent domain and pole errors by carefully bounds-checking the arguments before calling mathematical functions and taking alternative action if the bounds are violated. Range errors usually cannot be prevented because they are dependent on the implementation of floating-point numbers as well as on the function being applied. Instead of preventing range errors, programmers should attempt to detect them and take alternative action if a range error occurs. The following table lists the double forms of standard mathematical functions, along with checks that should be performed to ensure a proper input domain, and indicates whether they can also result in range or pole errors, as reported by the C Standard. Both float and long double forms of these functions also exist but are omitted from the table for brevity. If a function has a specific domain over which it is defined, the programmer must check its input values. The programmer must also check for range errors where they might occur. The standard math functions not listed in this table, such as fabs(), have no domain restrictions and cannot result in range or pole errors. | Function | Domain | Range | Pole | | acos(x) | -1 <= x && x <= 1 | No | No | | asin(x) | -1 <= x && x <= 1 | Yes | No | | atan(x) | None | Yes | No | | atan2(y, x) | None | No | No | | acosh(x) | x >= 1 | Yes | No | | asinh(x) | None | Yes | No | | atanh(x) | -1 < x && x < 1 | Yes | Yes | | cosh(x), sinh(x) | None | Yes | No | | exp(x), exp2(x), expm1(x) | None | Yes | No | | ldexp(x, exp) | None | Yes | No | | log(x), log10(x), log2(x) | x >= 0 | No | Yes | | log1p(x) | x >= -1 | No | Yes | | ilogb(x) | x != 0 && !isinf(x) && !isnan(x) | Yes | No | | logb(x) | x != 0 | Yes | Yes | | scalbn(x, n), scalbln(x, n) | None | Yes | No | | hypot(x, y) | None | Yes | No | | pow(x,y) | x > 0 || (x == 0 && y > 0) || (x < 0 && y is an integer) | Yes | Yes | | sqrt(x) | x >= 0 | No | No | | erf(x) | None | Yes | No | | erfc(x) | None | Yes | No | | lgamma(x), tgamma(x) | x != 0 && ! (x < 0 && x is an integer) | Yes | Yes | | lrint(x), lround(x) | None | Yes | No | | fmod(x, y), remainder(x, y), remquo(x, y, quo) | y != 0 | Yes | No | | nextafter(x, y), nexttoward(x, y) | None | Yes | No | | fdim(x,y) | None | Yes | No | | fma(x,y,z) | None | Yes | No | Domain and Pole Checking The most reliable way to handle domain and pole errors is to prevent them by checking arguments beforehand, as in the following exemplar: ``` double safe_sqrt(double x) { if (x < 0) { fprintf(stderr, "sqrt requires a nonnegative argument"); / Handle domain / pole error / } return sqrt (x); } ``` Range Checking Programmers usually cannot prevent range errors, so the most reliable way to handle them is to detect when they have occurred and act accordingly. The exact treatment of error conditions from math functions is tedious. The C Standard, 7.12.1 paragraph 5 [ISO/IEC 9899:2024], defines the following behavior for floating-point overflow: A floating result overflows if a finite result value with ordinary accuracy would have magnitude (absolute value) too large for the representation with full precision in the specified type. A result that is exactly an infinity does not overflow. If a floating result overflows and default rounding is in effect, then the function returns the value of the macro HUGE_VAL, HUGE_VALF, or HUGE_VALL according to the return type, with the same sign as the correct value of the function; however, for the types with reduced-precision representations of numbers beyond the overflow threshold, the function may return a representation of the result with less than full precision for the type. If a floating resultoverflowsanddefaultroundingisineffectandtheintegerexpressionmath_errhandling & MATH_ERRNO is nonzero, then the integer expression errno acquires the value ERANGE. If a floating result overflows, and the integer expression math_errhandling & MATH_ERREXCEPT is nonzero, the "overflow" floating-point exception is raised (regardless of whether default rounding is in effect). It is preferable not to check for errors by comparing the returned value against HUGE_VAL or 0 for several reasons: These are, in general, valid (albeit unlikely) data values. Making such tests requires detailed knowledge of the various error returns for each math function. Multiple results aside from HUGE_VAL and 0 are possible, and programmers must know which are possible in each case. Different versions of the library have varied in their error-return behavior. It can be unreliable to check for math errors using errno because an implementation might not set errno. For real functions, the programmer determines if the implementation sets errno by checking whether math_errhandling & MATH_ERRNO is nonzero. The C Standard, 7.3.2, paragraph 1 [ISO/IEC 9899:2024], states: an implementation may set errno but is not required to. The obsolete System V Interface Definition (SVID3) [UNIX 1992] provides more control over the treatment of errors in the math library. The programmer can define a function named matherr() that is invoked if errors occur in a math function. This function can print diagnostics, terminate the execution, or specify the desired return value. The matherr() function has not been adopted by C or POSIX, so it is not generally portable. The following error-handing template uses C Standard functions for floating-point errors when the C macro math_errhandling is defined and indicates that they should be used; otherwise, it examines errno: ``` include include include / ... / / Use to call a math function and check errors / { #pragma STDC FENV_ACCESS ON if (math_errhandling & MATH_ERREXCEPT) { feclearexcept(FE_ALL_EXCEPT); } errno = 0; / Call the math function / if ((math_errhandling & MATH_ERRNO) && errno != 0) { / Handle range error / } else if ((math_errhandling & MATH_ERREXCEPT) && fetestexcept(FE_INVALID | FE_DIVBYZERO | FE_OVERFLOW | FE_UNDERFLOW) != 0) { / Handle range error / } } ``` See FLP03-C. Detect and handle floating-point errors for more details on how to detect floating-point errors. Subnormal Numbers A subnormal number is a nonzero number that does not use all of its precision bits [IEEE 754 2006]. These numbers can be used to represent values that are closer to 0 than the smallest normal number (one that uses all of its precision bits). However, the asin(), asinh(), atan(), atanh(), and erf() functions may produce range errors, specifically when passed a subnormal number. When evaluated with a subnormal number, these functions can produce an inexact, subnormal value, which is an underflow error. The C Standard, 7.12.1, paragraph 6 [ISO/IEC 9899:2024], defines the following behavior for floating-point underflow: The result underflows if a nonzero result value with ordinary accuracy would have magnitude (absolute value) less than the minimum normalized number in the type; however a zero result that is specified to be an exact zero does not underflow. Also, a result with ordinary accuracy and the magnitude of the minimum normalized number may underflow.269) If the result underflows, the function returns an implementation-defined value whose magnitude is no greater than the smallest normalized positive number in the specified type; if the integer expression math_errhandling & MATH_ERRNO is nonzero, whether errno acquires the value ERANGE is implementation-defined; if the integer expression math_errhandling & MATH_ERREXCEPT s nonzero, whether the"underflow" floating-point exception is raised is implementation-defined. Implementations that support floating-point arithmetic but do not support subnormal numbers, such as IBM S/360 hex floating-point or nonconforming IEEE-754 implementations that skip subnormals (or support them by flushing them to zero), can return a range error when calling one of the following families of functions with the following arguments: fmod((min+subnorm), min) remainder((min+subnorm), min) remquo((min+subnorm), min, quo) where min is the minimum value for the corresponding floating point type and subnorm is a subnormal value. If Annex F is supported and subnormal results are supported, the returned value is exact and a range error cannot occur. The C Standard, F.10.7.1 paragraph 2 [ISO/IEC 9899:2024], specifies the following for the fmod(), remainder(), and remquo() functions: When subnormal results are supported, the returned value is exact and is independent of the current rounding direction mode. Annex F, subclause F.10.7.2, paragraph 2, and subclause F.10.7.3, paragraph 2, of the C Standard identify when subnormal results are supported. Noncompliant Code Example (sqrt()) This noncompliant code example determines the square root of x: ``` include void func(double x) { double result; result = sqrt(x); } ``` However, this code may produce a domain error if x is negative. Compliant Solution (sqrt()) Because this function has domain errors but no range errors, bounds checking can be used to prevent domain errors: ``` include void func(double x) { double result; if (isless(x, 0.0)) { / Handle domain error / } result = sqrt(x); } ``` Noncompliant Code Example (sinh(), Range Errors) This noncompliant code example determines the hyperbolic sine of x: ``` include void func(double x) { double result; result = sinh(x); } ``` This code may produce a range error if x has a very large magnitude. Compliant Solution (sinh(), Range Errors) Because this function has no domain errors but may have range errors, the programmer must detect a range error and act accordingly: ``` include include include void func(double x) { double result; { #pragma STDC FENV_ACCESS ON if (math_errhandling & MATH_ERREXCEPT) { feclearexcept(FE_ALL_EXCEPT); } errno = 0; result = sinh(x); if ((math_errhandling & MATH_ERRNO) && errno != 0) { / Handle range error / } else if ((math_errhandling & MATH_ERREXCEPT) && fetestexcept(FE_INVALID | FE_DIVBYZERO | FE_OVERFLOW | FE_UNDERFLOW) != 0) { / Handle range error / } } / Use result... / } ``` Noncompliant Code Example (pow()) This noncompliant code example raises x to the power of y: ``` include void func(double x, double y) { double result; result = pow(x, y); } ``` This code may produce a domain error if x is negative and y is not an integer value or if x is 0 and y is 0. A domain error or pole error may occur if x is 0 and y is negative, and a range error may occur if the result cannot be represented as a double. Compliant Solution (pow()) Because the pow() function can produce domain errors, pole errors, and range errors, the programmer must first check that x and y lie within the proper domain and do not generate a pole error and then detect whether a range error occurs and act accordingly: ``` include include include void func(double x, double y) { double result; if (((x == 0.0f) && islessequal(y, 0.0)) || isless(x, 0.0)) { / Handle domain or pole error / } { #pragma STDC FENV_ACCESS ON if (math_errhandling & MATH_ERREXCEPT) { feclearexcept(FE_ALL_EXCEPT); } errno = 0; result = pow(x, y); if ((math_errhandling & MATH_ERRNO) && errno != 0) { / Handle range error / } else if ((math_errhandling & MATH_ERREXCEPT) && fetestexcept(FE_INVALID | FE_DIVBYZERO | FE_OVERFLOW | FE_UNDERFLOW) != 0) { / Handle range error / } } / Use result... / } ``` Noncompliant Code Example (asin(), Subnormal Number) This noncompliant code example determines the inverse sine of x: ``` include void func(float x) { float result = asin(x); / ... / } ``` Compliant Solution (asin(), Subnormal Number) Because this function has no domain errors but may have range errors, the programmer must detect a range error and act accordingly: ``` include include include void func(float x) { float result; { #pragma STDC FENV_ACCESS ON if (math_errhandling & MATH_ERREXCEPT) { feclearexcept(FE_ALL_EXCEPT); } errno = 0; result = asin(x); if ((math_errhandling & MATH_ERRNO) && errno != 0) { / Handle range error / } else if ((math_errhandling & MATH_ERREXCEPT) && fetestexcept(FE_INVALID | FE_DIVBYZERO | FE_OVERFLOW | FE_UNDERFLOW) != 0) { / Handle range error / } } / Use result... / } ``` Risk Assessment Failure to prevent or detect domain and range errors in math functions may cause unexpected results. | Rule | Severity | Likelihood | Detectable | Repairable | Priority | Level | | FLP32-C | Medium | Probable | Yes | Yes | P12 | L1 | Automated Detection | Tool | Version | Checker | Description | | Astrée | 24.04 | stdlib-limits | Partially checked | | Axivion Bauhaus Suite | 7.2.0 | CertC-FLP32 | Partially implemented | | CodeSonar | 9.1p0 | MATH.DOMAIN.ATAN MATH.DOMAIN.TOOHIGH MATH.DOMAIN.TOOLOW MATH.DOMAIN MATH.RANGE MATH.RANGE.GAMMA MATH.DOMAIN.LOG MATH.RANGE.LOG MATH.DOMAIN.FE_INVALID MATH.DOMAIN.POW MATH.RANGE.COSH.TOOHIGH MATH.RANGE.COSH.TOOLOW MATH.DOMAIN.SQRT | Arctangent Domain Error Argument Too High Argument Too Low Floating Point Domain Error Floating Point Range Error Gamma on Zero Logarithm on Negative Value Logarithm on Zero Raises FE_INVALID Undefined Power of Zero cosh on High Number cosh on Low Number sqrt on Negative Value | | Helix QAC | 2025.2 | C5025 C++5033 | | | Parasoft C/C++test | 2024.2 | CERT_C-FLP32-a | Validate values passed to library functions | | PC-lint Plus | 1.4 | 2423 | Partially supported: reports domain errors for functions with the Semantics dom_1, dom_lt0, or dom_lt1, including standard library math functions | | Polyspace Bug Finder | R2024b | CERT-C: Rule FLP32-C | Checks for invalid use of standard library floating point routine (rule fully covered) | | RuleChecker | 24.04 | stdlib-limits | Partially checked | | TrustInSoft Analyzer | 1.38 | out-of-range argument | Partially verified. | Related Vulnerabilities Search for vulnerabilities resulting from the violation of this rule on the CERT website. Related Guidelines Key here (explains table format and definitions) | Taxonomy | Taxonomy item | Relationship | | CERT C Secure Coding Standard | FLP03-C. Detect and handle floating-point errors | Prior to 2018-01-12: CERT: Unspecified Relationship | | CWE 2.11 | CWE-682, Incorrect Calculation | 2017-07-07: CERT: Rule subset of CWE | CERT-CWE Mapping Notes Key here for mapping notes CWE-391 and FLP32-C Intersection( CWE-391, FLP32-C) = Failure to detect range errors in floating-point calculations CWE-391 - FLP32-C Failure to detect errors in functions besides floating-point calculations FLP32-C – CWE-391 = Failure to detect domain errors in floating-point calculations CWE-682 and FLP32-C Independent( INT34-C, FLP32-C, INT33-C) CWE-682 = Union( FLP32-C, list) where list = Incorrect calculations that do not involve floating-point range errors Bibliography | | | --- | | [ISO/IEC 9899:2024] | 7.3.2, "Conventions" 7.12.1, "Treatment of Error Conditions" F.10.7, "Remainder Functions" | | [IEEE 754 2006 ] | | | [Plum 1985] | Rule 2-2 | | [Plum 1989] | Topic 2.10, "conv—Conversions and Overflow" | | [UNIX 1992] | System V Interface Definition (SVID3) | fortify nptc rose-na-macros error-handling rule flp cwe-682 nptc-complexity in-cpp 25 Comments Geoff Clare The treatment of pow() in this rule is somewhat lacking. The specified bounds check (x != 0 || y > 0) is insufficient: C99 says a domain error also occurs "if x is finite and negative and y is finite and not an integer value." The CCE for pow() says "This code tests x and y to ensure that there will be no range or domain errors" but it does not detect the range error given in the introductory text at the top of the page: pow(10., 1e6) Given that the page begins with "Prevent or detect domain errors and range errors ...", and given the complexity involved in preventing pow() errors, I think that the document should recommend detection for this function instead of prevention. On the subject of detection, the subsection entitled "Non-Compliant Coding Example (Error Checking)" only talks about return values and errno. There is no mention of error checking by examining the exception flags. C90 required the maths functions to set errno on error. C99 requires them (the non-complex ones, that is) either to set errno or to set the exception flags, or both. So I think the recommendation (for non-complex maths functions) should be: If there is a simple bounds check that can be done to prevent domain and range errors, then do it. (This applies to all the current examples except pow().) Otherwise, detect errors as follows: include if defined(math_errhandling) && (math_errhandling & MATH_ERREXCEPT) include endif [...] if defined(math_errhandling) && (math_errhandling & MATH_ERREXCEPT) feclearexcept(FE_ALL_EXCEPT); endif errno = 0; / call the function / if !defined(math_errhandling) || (math_errhandling & MATH_ERRNO) if (errno != 0){ / handle error / } endif if defined(math_errhandling) && (math_errhandling & MATH_ERREXCEPT) if (fetestexcept(FE_INVALID | FE_DIVBYZERO | FE_OVERFLOW) != 0){ / handle error / } endif Other functions besides pow() where detection should be used because prevention would be too complicated include erfc(), lgamma() and tgamma(). Permalink Apr 04, 2008 Douglas A. Gwyn There is essentially no reason for a program to invoke pow() with a negative base. Permalink Apr 16, 2008 David Svoboda I've addressed these comments. I included Geoff's code sample under 'Compliant Example: Error Checking' Permalink May 02, 2008 Geoff Clare Your changes helped a lot, but there were still some problems relating to pow(). I have attempted to fix them. Permalink May 06, 2008 Alex Volkovitsky Shaun, regarding the second NCCE under pow(), what does "result cannot be represented as a double" mean? It means the result is either a NaN or Infty... we can check for those two after computing the pow() to ensure no range errors happened, ahh... the beauty of floating point Permalink Jun 04, 2008 Geoff Clare "result cannot be represented as a double" means the true (mathematical) result is outside the range of values that can be represented by double. It does not mean the result is NaN or Infinity. E.g. for pow(10.,1e6) the true result is ten to the power of one million, which is larger than DBL_MAX and therefore cannot be represented as a double. The true result is not infinity. Also note that some implementations do not support Inifinity and/or NaN, and so applications cannot reply on them being returned. Permalink Jun 05, 2008 David Svoboda This should be checkable by Rose. But there is a snag. The isLess() etc. functions, which are being used to do range checking, are not defined. If we could simply check for "x > 0.0", then we can do it. Is that what isGreater(x, 0) really means? Permalink Jun 18, 2008 Geoff Clare The difference between x > 0.0 and isgreater(x,0) is that isgreater(x,0) will not raise a floating-point exception if x is a NaN. If you want to switch to using the operators, you would have to add explicit isnan() checks (in the cases where there isn't one already). Permalink Jun 20, 2008 Geoff Clare For consistency with the way pow() was treated, shouldn't the new tgamma() examples have a NCCE that does no checking, then a NCCE that does only the domain checks, and then just give a reference to the Error Checking and Detection section instead of having a CS that duplicates code from it Permalink Jul 24, 2008 Alex Volkovitsky In that case, we should add the FE_UNDERFLOW flag to the Error Checking section Permalink Jul 24, 2008 David Svoboda agreed; the pow() section and the tgamma() section both follow the same outline. Permalink Jul 24, 2008 Jonathan Paulson Java universally deals with this issue by returning NaN; it might be worth a guideline to check if the result a math operation is Nan? Removing the exportable-java guideline. Permalink Mar 28, 2011 Geoff Clare The new safe_sqrt() exemplar seems to me to be not very exemplary, as it handles the domain error in a particularly unhelpful way. Is there a reason not to use the usual convention of a / Handle error / comment here? Permalink Apr 15, 2014 David Svoboda Agreed...fixed. Permalink Apr 15, 2014 Jeremy Hall I think row 6 in the table should be asinh() and not asin() Permalink Nov 08, 2015 Aaron Ballman Agreed. I've fixed it, thanks! Permalink Nov 09, 2015 Jeremy Hall For log(x), log10(x), log2(x) I think the domain should be x > 0 rather than x >= 0 because they produce a pole error for x == 0. Log1p looks correct, the domain is given as x > -1.0 Permalink Nov 08, 2015 Geoff Clare No, you have it backwards. If x == 0 was outside the domain of the function, log(0) would produce a domain error, not a pole error. The mistake is for log1p(x) which should give the domain as x >= -1.0. Permalink Nov 08, 2015 Aaron Ballman I looked at the wording in the standard, and I agree that log(), log10(), and log2() seem to be correct, while log1p() was incorrect with its domain. I've corrected. Permalink Nov 09, 2015 Eddie O'Hagan I noticed that in the documentation for cos(), sin(), and tan(): Errors are reported as specified in math_errhandling If the implementation supports IEEE floating-point arithmetic (IEC 60559), if the argument is ±0, the result is 1.0 if the argument is ±∞, NaN is returned and FE_INVALID is raised if the argument is NaN, NaN is returned I think it makes sense to add cos(), sin(), and tan() to the list as well. Permalink Oct 13, 2016 Aaron Ballman tan() could certainly be in that list with a pole check, since tan() returns an infinity as the function approaches pi / 2. However, I don't believe any IEEE floating point representation can exactly represent pi / 2, so I don't believe a pole error can technically occur in practice. I'm uncertain of how we might want to represent that in the list. We would have to do a lot more research before adding cos() and sin() to that list, because that's a difference in specification between POSIX and C. The C standard does not define any domain error when given infinity or NaN, while POSIX does. I suspect similar distinctions occur for other functions. We would need to make mention of where POSIX and C differ. Permalink Oct 13, 2016 Wolfgang Stanglmeier The domain of atan2( y, x ) should be x != 0 || y != 0. x or y may be zero, only the case x=0 ∧ y=0 is undefined. Permalink Jan 31, 2022 Joseph C. Sible Wait, isn't even that case defined too? In the C17 standard, under F.10.1.4, it says "atan2(±0, −0) returns ±π" and "atan2(±0, +0) returns ±0". Permalink Jan 31, 2022 David Svoboda Correct. The C23 draft standard (sF.10.1.4) says: atan2(±0,−0)returns ±π.404 Footnote 404: atan2(0, 0) does not raise the "invalid" floating-point exception, nor does atan2(y, 0) raise the "divide-by-zero" floating- point exception. Whether passing zeroes in constitutes a domain error depends on how you define atan2(y,x). Every definition I've seen references the "intermediate value" of y/x, which is undefined for x=y=0. But ISO C is clear on atan2(0,0) for both positive and negative zero :) So I took out the domain error. Permalink Jan 31, 2022 Wolfgang Stanglmeier Thanks for the clarifications and the reference to the standards. I noticed that the domain was obviously too narrow, but didn't realize that there is no domain restriction at all. Permalink Jan 31, 2022 Content Tools {"serverDuration": 188, "requestCorrelationId": "786dbc99291ddfce"} You are not logged in. Any changes you make will be marked as anonymous. You may want to Log In if you already have an account. 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Skip to lesson content AP®︎/College Chemistry Course: AP®︎/College Chemistry>Unit 1 Lesson 6: Periodic trends Periodic trends and Coulomb's law Atomic and ionic radii Ionization energy: group trend Ionization energy: period trend First and second ionization energy Worked example: Identifying an element from successive ionization energies Electron affinity: period trend Electronegativity Periodic trends Science> AP®︎/College Chemistry> Atomic structure and properties> Periodic trends © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Atomic and ionic radii AP.Chem: SAP‑2 (EU), SAP‑2.A (LO), SAP‑2.A.2 (EK), SAP‑2.A.3 (EK) Google Classroom Microsoft Teams About About this video Transcript Atomic and ionic radii are found by measuring the distances between atoms and ions in chemical compounds. On the periodic table, atomic radius generally decreases as you move from left to right across a period (due to increasing nuclear charge) and increases as you move down a group (due to the increasing number of electron shells). Similar trends are observed for ionic radius, although cations and anions need to be considered separately.Created by Jay. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Maryam Syeda 4 years ago Posted 4 years ago. Direct link to Maryam Syeda's post “How do you even measure t...” more How do you even measure that small a distance? ( Atomic radius ) Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Richard 4 years ago Posted 4 years ago. Direct link to Richard's post “Probably the most common ...” more Probably the most common way to determine these distances is using a method called X-ray crystallography. Along with being able to measure distances between atoms it allows us to determine the structure of a molecule. In effect being able take a picture of the molecule. This method involves first creating a crystal composed of the molecule with sufficient size, high purity, and a rectangular prism shape. Which in my experience is the most time consuming. Afterwards we place the crystal in an instrument where X-rays are directed toward it and the molecules in the crystals scatter those X-rays. Those scatterings essentially give us pictures of the molecule which we can combine to yield the structure of the molecule and allows us to determine the distances between the atoms. Hope that helps. 2 comments Comment on Richard's post “Probably the most common ...” (19 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... PAULLLLLLLLLL 2 years ago Posted 2 years ago. Direct link to PAULLLLLLLLLL's post “I do not quite understand...” more I do not quite understand why the added electron to a neutral chlorine ino would increase the size of the ion. Since the electron configuration for the chloride anion is the same as the noble gas Ar, and the atomic radius generally dicreases from left of the periodic table to the right of the periodic table, so why won't the added electron makes the chiride anion smaller in terms of size? Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Richard 2 years ago Posted 2 years ago. Direct link to Richard's post “If you have a neutral chl...” more If you have a neutral chlorine atom and all you do is add an electron, then you’ve added to the repulsive force felt by the electrons. You’ve decreased the effective nuclear charge felt by the electrons towards the nucleus and so they feel less attractive force towards the nucleus and the valence electrons orbit farther from the nucleus resulting in a larger atomic radius. This is different from the trend of decreasing atomic radii as you move left to right along a period. As you move left to right, you’re changing the type of element the atom is which means you’re adding an extra proton each step to the right. While you are also adding an extra electron, the extra proton results in a net increase in the effective nuclear charge because the attractive pull of a proton is greater than the shielding of an extra electron in the same shell. If the effective nuclear charge for elements increases as you move to the right, the electrons feel a greater force of attraction for the nucleus and the valence electrons orbit closer resulting in a smaller atomic radius. The chlorine ion example is keeping the same number of protons but adding an electron. So, while a chloride ion has the same electron configuration as a neutral argon atom, they have different radii because of the different number of protons in the nuclei. Hope that helps. Comment Button navigates to signup page (14 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Marvyn Greco 2 years ago Posted 2 years ago. Direct link to Marvyn Greco's post “what does pm mean?” more what does pm mean? Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Richard 2 years ago Posted 2 years ago. Direct link to Richard's post “Post meridiem, or after m...” more Post meridiem, or after midday. Na just kidding. In a chemistry context pm stands for picometers. It’s a unit of length equal to 10^(-12) m, or a trillionth of a meter. We have length this small to measure the sizes of atoms. Hope that helps. 1 comment Comment on Richard's post “Post meridiem, or after m...” (14 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more A.T.8301032100 3 years ago Posted 3 years ago. Direct link to A.T.8301032100's post “Will this help in grindin...” more Will this help in grinding for IChO in Canada? Answer Button navigates to signup page •1 comment Comment on A.T.8301032100's post “Will this help in grindin...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Richard 3 years ago Posted 3 years ago. Direct link to Richard's post “KA would most likely help...” more KA would most likely help. At the very least it would serve as an introduction to many of the topics on the IChO. It would be best to seek out additional chemistry teaching materials too. Studying from a variety of sources and becoming proficient in them would certainly help your odds. I have to warn you though that the IChO exam is formidable for secondary school students. The difficulty of the material they test on is what you would expect at a college or university chemistry course. Here on KA they can give you a good start in general chemistry with their AP chemistry material, but the IChO also tests organic chemistry (also found on KA), physical chemistry, and spectroscopic techniques like proton NMR, C-13 NMR, and IR. Hope that helps. Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Lauren Kong 2 years ago Posted 2 years ago. Direct link to Lauren Kong's post “Why exactly does electron...” more Why exactly does electron repulsion increase in anions once there is an extra electron that is added? I did some research and many sources say that this is because once an electron is added, there is an increase in shielding. But when you add an electron, you're adding a valence electron, not a core electron. So why exactly does shielding increase? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Richard 2 years ago Posted 2 years ago. Direct link to Richard's post “Shielding happens not onl...” more Shielding happens not only from core electrons, but also from electrons in the same shell. So valence electrons shield other valence electrons from the nucleus. This should make sense since all electrons have negative charges, so an electron in feels a repulsive force from a nearby electron in the same shell. Shielding for effective nuclear charge is often first introduced as just originating from core electron electrons. But in reality shielding is more complex. Using a simpler definition of shielding works fine as an introduction, but eventually becomes problematic for more complex problems. We use a set of guidelines called Slater’s rules to give a more accurate shielding value which takes into account the shielding from electrons in the same shell. So if you have a neutral atom, and all you do is add an extra valence electron, you’re adding an extra repulsive force to the existing valence electrons without any increase in the attractive force from the nucleus. The net change is that the valence electrons have less effective nuclear charge be applied to them and so feel less attracted to the nucleus and orbit at a greater distance from the nucleus resulting in a larger atomic radius as compared to the neutral atom. Hope that helps. Comment Button navigates to signup page (7 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Jerry Weng a year ago Posted a year ago. Direct link to Jerry Weng's post “perhaps an alternative ex...” more perhaps an alternative explanation for why an anion is bigger is because less force is exerted on the valence electrons on average due to the addition of an extra electron, hence making the electrons a bit farther away from the nucleus? Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Richard a year ago Posted a year ago. Direct link to Richard's post “Yeah, that’s pretty much ...” more Yeah, that’s pretty much it. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more sairoshanpatra 2 years ago Posted 2 years ago. Direct link to sairoshanpatra's post “'On the periodic table, a...” more 'On the periodic table, atomic radius generally decreases as you move from left to right across a period (due to increasing nuclear charge) and increases as you move down a group (due to the increasing number of electron shells).' but the nuumber of protons also increases down the group, Isnt it? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Richard 2 years ago Posted 2 years ago. Direct link to Richard's post “Yes, but the higher shell...” more Yes, but the higher shell numbers cause a greater increase in the atomic radii than is countered by the contraction of the additional protons. The net change is an increase in atomic radius. Hope that helps. 2 comments Comment on Richard's post “Yes, but the higher shell...” (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Junxi Wu 2 years ago Posted 2 years ago. Direct link to Junxi Wu's post “When going left to right ...” more When going left to right in the d or f block, isn't that adding electrons to a non-valence shell? So why does the shielding effect stay the same? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Richard 2 years ago Posted 2 years ago. Direct link to Richard's post “Well first elements in th...” more Well first elements in the d and f block do have d and f electrons as a part of their valence electrons. In a simplified version here, they only consider the shielding due to core electrons in lower shells. However, in reality shielding is caused by valence electrons in the same shell as well. Which should make sense since all electrons are negatively charged and they all repel each other. This also means the shield would change as we add electrons. Hope that helps. Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more mymoajoy 3 years ago Posted 3 years ago. Direct link to mymoajoy's post “If two ions have the same...” more If two ions have the same number of electrons but one has more protons which will be larger One with more protons or less protons? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Richard 3 years ago Posted 3 years ago. Direct link to Richard's post “The other replier is corr...” more The other replier is correct, I would just like to expand on it. The ion with more protons will have a smaller atomic radius because the effective nuclear charge felt by the electrons is greater. The effective nuclear charge being the force of attraction felt between the protons of the nucleus and the electrons in orbit around them. If the protons and electrons have a stronger force of attraction between them, they will be able to move closer together resulting in a smaller electron cloud. Hope that helps. Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Rafa 2 years ago Posted 2 years ago. Direct link to Rafa's post “Why does the electrons ou...” more Why does the electrons outside of our interested electron does not show shielding effect? Don't they repulse the electron also? Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Amaranthine a year ago Posted a year ago. Direct link to Amaranthine's post “Those are far enough for ...” more Those are far enough for us to think their repulsion to be neglible. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Video transcript In this video, we're going to look at atomic and ionic radii. And first, we'll start with the atomic radius. So if you think about an atom as a sphere, the idea of atomic radius is simple. You would just take this as a sphere here, and then a sphere of course would have fixed and defined radius. And so that would be one way of thinking about it. The problem is that an atom doesn't really have a fixed, defined radius like this sphere example, because there's a nucleus and then there's this electron cloud, or this probability of finding your electron. So there's no real, clear defined boundary there, and so it's difficult to have a fixed and defined radius. So what chemists do is they take two identical atoms. So let's say these are two atoms bonded together, the same element. And if you find their nuclei-- so let's say that that's their nuclei here-- and you measure the distance between those two nuclei, so this would be our distance d between our two nuclei. If you take half of that distance, that would be a good approximation of the atomic radius of one of those atoms. And so that's the idea behind the definition of atomic radius. Let's look at the trends for atomic radius, and first we'll start with group trends. And so here we have two elements found in group one, so hydrogen and lithium. And let's go ahead and sketch out the atoms first. And so we start with hydrogen, which has atomic number of 1, which means that it has one proton in the nucleus. So here's our nucleus for hydrogen, so one proton. In a neutral atom, the number of protons equals the number of electrons, and so therefore there must be one electron. So go ahead and sketch in our electron here. And we'll make things really simple and just show this simple version of the atom, even though we know it doesn't really exactly look like this. And when we do lithium, atomic number of 3, so that means three protons in the nucleus of lithium. So this is representative of lithium's nucleus with three protons and three electrons. Two of those electrons are in the inner shell. So let me go ahead and show two of lithium's electrons in the inner shell, so that would be in the first energy level. And then we would need to account for one more, so lithium's third electron is in the second energy level or at the outer shell in this example. And so here we have our two atoms. And you can see as you go down a group, you're going to get an increase in the atomic radius. And that's because as you go down a group, you're adding electrons in higher energy levels that are farther away. So in this case, we added this electron to a higher energy level which is farther away from the nucleus, which means that the atoms of course would get larger and larger. So you're adding more stuff to it, so it's kind of a simple idea. Let's look at period trends next. As you're going across a period this way, so as you're going this way, you're actually going to get a decrease in the atomic radius. And let's see if we can figure out why by once again drawing some simple pictures of our atoms. And so lithium with atomic number of 3, so we've already talked about that. So there are three protons in the nucleus of lithium. So I'm going to go ahead and write that in here. So 3 positive charge for the nucleus of lithium. And we have to account for the three electrons. So once again two of those electrons were in an inner shell, so there we go, and then we had one electron in an outer shell, so the picture is something like this. Now, let's think about what's going to happen to that outer electron as a result of where it is. So this outer electron, this one right here in magenta, would be pulled closer to the nucleus. The nucleus is positively charged, that electron is negatively charged, and so the positively charged nucleus is going to pull that electron in closer to it. At the same time, those negatively charged inner shell electrons are going to repel it. So let me go ahead and highlight these guys right here. These are our inner shell electrons. Like charges repel. And so you could think about this electron right here wanting to push this outer electron that way, and this electron wanting to push this electron that way. And so the nucleus attracts a negative charge, and the inner shell electrons repel the outer electron. And then we call this shielding, because the inner shell electrons are shielding that magenta electron from the pole of the nucleus. So this is called electronic shielding or electron screening. Now, it's going to be important concepts. So now let's go ahead and draw the atom for beryllium, so atomic number 4. And so here's our nucleus for beryllium. With an atomic number of 4, that means there are four protons in the nucleus, so a charge of four plus in our nucleus. And we have four electrons to worry about this time, so I'll go ahead and put in the two electrons in my inner orbital in our first energy level. And then we have two electrons in our outer orbital, or our second energy level. And so again, this is just a rough approximation for an idea of what beryllium might look like. And so when we think about what's happening, we're moving from a charge of 3 plus with lithium to a charge of 4 plus with beryllium. And the more positive your charges, the more it's going to attract those outer electrons. And when you think about the idea of electron screening, so once again we have these electrons in green here shielding our outer shell electrons from the effect of that positively charged nucleus. Now, you might think that outer shell electrons could shield, too. So you might think that oh, this electron right here in magenta could shield the other electron in magenta. But the problem is they're both at pretty much the same distance from the nucleus, so outer shell electrons don't really shield each other. It's more of these inner shell electrons. And because you have the same number of inner shell electrons shielding as in the lithium example-- so let me go ahead and highlight those again. So we have two inner shell electrons shielding a beryllium. We also have two inner shell electrons shielding in lithium. Because you have the same number of shielding but you have a higher positive charge, those outer electrons are going to feel more of a pull from the nucleus. And they're going to be pulled in even tighter than you might imagine, or at least tighter than our previous example. So these electrons are pulled in even more. And because of that, you're going to get the beryllium atom as being smaller than the lithium atom, hence the trend. Hence as you go across the period, you're always going to increase in the number of protons and that increased whole is going to pull those outer electrons in closer, therefore decreasing the size of the atom. All right. Let's look at ionic radius now. And ionic radius can be kind of complicated depending on what chemistry you are involved in. So this is going to be just a real simple version. If I took a neutral lithium atom again, so lithium-- so we've drawn this several times. Let me go ahead and draw it once more. So we have our lithium nucleus, which we have three electrons. So once again I'll go ahead and sketch in our three electrons real fast. Two electrons in the inner shell, and one electron in the outer shell like that. And let's say you were going to form a cation, so we are going to take away an electron from our neutral atom. So we have-- let me go ahead and draw this in here-- we had a three protons in the nucleus and three electrons those cancel each other out to be a neutral atom. And if we were to take away one of those electrons, so let's go ahead and show lithium losing an electron. So if lithium loses an electron, it's going to lose that outer electron. So the nucleus still has a plus 3 charge, because it has three protons in it. And we still have our two inner shell electrons like that, but we took away that outer shell electron. So we took away this electron in magenta, so let me go ahead and label this. So we lost an electron, so that's this electron right here, and so you could just show it over here like that. And by doing so, now we have three positive charges in our nucleus and only two electrons. And so therefore our lithium gets a plus 1 charge. So it's Li plus, it's a cation. And so we formed a cation, which is smaller than the neutral atom itself. And that just makes intuitive sense. If you take away this outer electron, now you have three positive charges in the nucleus and only two electrons here. So it's pulling those electrons in, you lost that outer electron, it's getting smaller. And so the cation is smaller than the neutralize atom. And so we've seen that neutral atoms will shrink when you convert them to cations, so it kind of makes sense that if you take a neutral atom and add an electron, it's going to get larger. And so that's our next concept here. So if we took something like chlorine, so a neutral chlorine atom, and we added an electron to chlorine, that would give it a negative charge. So we would get chlorine with a negative charge, or the chloride anion, I should say. And so in terms of sizes, let's go ahead and draw a representative atom here. So if this is our neutral chlorine atom and we add an electron to it, it actually gets a lot bigger. So the anion is bigger than the neutral atom. And let's see if we can think about why here. So if we were to draw an electron configuration, or to write a noble gas electron configuration for the neutral chlorine-- so you should already know how to do this-- you would just write your noble gas in brackets. So neon and then 3s2, 3p5, so seven electrons in the outer shell for the neutral chlorine atom. For the chloride anion, you would start off the same way. You would say neon in brackets, 3s2. And you'd be adding an electron to it. So it wouldn't be 3p5, it would be 3p6 like that. And so now we would have so this would give us eight electrons in our outer shell, and this would give us only seven electrons in our outer shell. Now, the explanation for the larger size of the chloride anion in most textbooks is, you'll see people say that the addition of this extra electron here, so that means that those electrons are going to repel each other more. You have eight of them instead of seven, and so because they repel each other more, it gets a little bit bigger. And that makes sense, but you'll see some people disagree with that explanation, and I haven't really seen a great alternative offered. And so however you want to think about it, generally the anion is larger than the neutral atom. But in terms of the explanation for that, you could think about it as electrons are repelling each other if you wanted to, despite the fact that people disagree with that. You could think about just more stuff as a really simple way of thinking about it. But again, in general for exams, think about the anion being larger. 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https://www.texmacs.org/joris/chinese-macis/chinese-macis.pdf
Fast Chinese remaindering in practice by Joris van der Hoeven Laboratoire d'informatique, UMR 7161 CNRS Campus de l'École polytechnique 1, rue Honoré d'Estienne d'Orves Bâtiment Alan Turing, CS35003 91120 Palaiseau October 24, 2017 Abstract The Chinese remainder theorem is a key tool for the design of efficient multi-modular algorithms. In this paper, we study the case when the moduli m1; :::; mare fixed and can even be chosen by the user. If is small or moderately large, then we show how to choose gentle moduli that allow for speedier Chinese remaindering. The multiplication of integer matrices is one typical application where we expect practical gains for various common matrix dimensions and bitsizes of the coeffi-cients. Keywords: Chinese remainder theorem, algorithm, complexity, integer matrix multiplication 1 Introduction Modular reduction is an important tool for speeding up computations in com-puter arithmetic, symbolic computation, and elsewhere. The technique allows to reduce a problem that involves large integer or polynomial coefficients to one or more similar problems that only involve small modular coefficients. Depending on the application, the solution to the initial problem is reconstructed via the Chinese remainder theorem or Hensel's lemma. We refer to [9, chapter 5] for a gentle introduction to this topic. In this paper, we will mainly be concerned with multi-modular algorithms over the integers that rely on the Chinese remainder theorem. Given a; m 2 Z with m > 1, we will denote by a rem m 2 Rm := f0; :::; m ¡ 1g the remainder of the Euclidean division of a by m. Given an r × r matrix A 2 Zr×r with integer coefficients, we will also denote A rem m 2 Zr×r for the matrix with coefficients (A rem m)i;j = Ai;j rem m. One typical application of Chinese remaindering is the multiplication of r × r integer matrices A; B 2 Zr×r. Assuming that we have a bound M with 2 j(A B)i;jj < M for all i; j, we proceed as follows: 0. Select moduli m1; :::; mwith m1 ··· m > M that are mutually coprime. 1. Compute A rem mk and B rem mk for k = 1; :::; . 2. Multiply C rem mk := (A rem mk) (B rem mk) rem mk for k = 1; :::;. 1 3. Reconstruct C rem M from the C rem mk with k = 1; :::; . The simultaneous computation of Ai;j rem mk from Ai;j for all k = 1; :::; is called the problem of multi-modular reduction. In step 1, we need to perform 2 r2 multi-modular reductions for the coefficients of A and B. The inverse problem of reconstructing Ci;j rem M from the Ci;j rem mk with k = 1; :::; is called the problem of multi-modular reconstruction. We need to perform r2 such reconstructions in step 3. Our hypothesis on M allows us to recover C from C rem M. Let us quickly examine when and why the above strategy pays off. In this paper, the number should be small or moderately large, say 664. The moduli m1; :::; m typically fit into a machine word. Denoting by µ the bitsize of a machine word (say µ=32 or µ=64), the coefficients of A and B should therefore be of bitsize ≈µ/2. For small, integer multiplications of bitsize µ /2 are usually carried out using a naive algorithm, of complexity Θ(2). If we directly compute the product A B using r3 naive integer multiplications, the computation time is therefore of order Θ(r3 2). In comparison, as we will see, one naive multi-modular reduc-tion or reconstruction for moduli roughly requires Θ(2) machine operations, whereas an r × r matrix multiplication modulo any of the mk can be done in time Θ(r3). Altogether, this means that the above multi-modular algorithm for integer matrix multiplication has running time Θ(2 r2 + r3 ), which is Θ(min(; r)) times faster than the naive algorithm. If ≪r, then the cost Θ(2 r2) of steps 1 and 3 is negligible with respect to the cost Θ(r3 ) of step 2. However, if and r are of the same order of magnitude, then Chinese remaindering may take an important part of the computation time; the main purpose of this paper is to reduce this cost. If ≫r, then we notice that other algorithms for matrix multiplication usually become faster, such as naive multiplication for small, Karatsuba multiplication for larger , or FFT-based techniques for very large. Two observations are crucial for reducing the cost of Chinese remaindering. First of all, the moduli m1; :::; mare the same for all 2 r2 multi-modular reductions and r2 multi-modular reconstructions in steps 1 and 3. If r is large, then this means that we can essentially assume that m1; :::; m were fixed once and for all. Secondly, we are free to choose m1;:::;min any way that suits us. We will exploit these observations by precomputing gentle moduli for which Chinese remaindering can be performed more efficiently than for ordinary moduli. The first idea behind gentle moduli is to consider moduli mi of the form 2sw ¡ "i 2, where w is somewhat smaller than µ, where s is even, and "i 2 < 2w. In section 3.1, we will show that multi-modular reduction and reconstruction both become a lot simpler for such moduli. Secondly, each mi can be factored as mi = (2sw/2 ¡ "i) (2sw/2 + "i) and, if we are lucky, then both 2sw/2 ¡ "i and 2sw/2 + "i can be factored into s/2 moduli of bitsize <µ. If we are very lucky, then this allows us to obtain w moduli mi;j of bitsize ≈w that are mutually coprime and for which Chinese remaindering can be implemented efficiently. 2 Section 1 Let us briefly outline the structure of this paper. In section 2, we rapidly recall basic facts about Chinese remaindering and naive algorithms for this task. In section 3, we introduce gentle moduli and describe how to speed up Chinese remaindering with respect to such moduli. The last section 4 is dedicated to the brute force search of gentle moduli for specific values of s and w. We imple-mented a sieving method in Mathemagix which allowed us to compute tables with gentle moduli. For practical purposes, it turns out that gentle moduli exist in sufficient number for s 6 8. We expect our technique to be efficient for . s2, but this still needs to be confirmed via an actual implementation. The application to integer matrix multiplication in section 4.3 also has not been implemented yet. Let us finally discuss a few related results. In this paper, we have chosen to highlight integer matrix multiplication as one typical application in com-puter algebra. Multi-modular methods are used in many other areas and the operations of multi-modular reduction and reconstruction are also known as con-versions between the positional number system (PNS) and the residue number system (RNS). Asymptotically fast algorithms are based on remainder trees [8, 14, 3], with recent improvements in [4, 2, 10]; we expect such algorithms to become more efficient when exceeds s2. Special moduli of the form 2n¡" are also known as pseudo-Mersenne moduli. They have been exploited before in cryptography in a similar way as in section 3.1, but with a slightly different focus: whereas the authors of are keen on reducing the number of additions (good for circuit complexity), we rather optimize the number of machine instructions on recent general purpose CPUs (good for software implementations). Our idea to choose moduli 2n ¡" that can be factored into smaller moduli is new. Other algorithms for speeding up multiple multi-modular reductions and reconstructions for the same moduli (while allowing for additional pre-compu-tations) have recently been proposed in . These algorithms essentially replace all divisions by simple multiplications and can be used in conjunction with our new algorithms for conversions between residues modulo mi = mi;1 ··· mi;s and residues modulo mi;1; :::; mi;s. 2 Chinese remaindering 2.1 The Chinese remainder theorem For any integer m > 1, we recall that Rm = f0; :::; m ¡ 1g. For machine computations, it is convenient to use the following effective version of the Chinese remainder theorem: Chinese Remainder Theorem. Let m1; :::; mbe positive integers that are mutually coprime and denote M =m1 ··· mm. There exist c1;:::;c2RM such that for any a1 2 Rm1; :::; a2 Rm, the number x = (c1 a1 + ··· + ca) rem M Chinese remaindering 3 satisfies x rem mi = ai for i = 1; :::; . Proof. For each i = 1; :::;, let πi = M /mi. Since πi and mi are coprime, πi admits an inverse ui modulo mi in Rmi. We claim that we may take ci = πi ui. Indeed, for x = (c1 a1 + ··· + ca) rem M and any i 2 f1; :::; g, we have x ≡a1 π1 u1 + ··· + a πu (mod mi) Since πj is divisible by mi for all j = / i, this congruence relation simplifies into x ≡ai πi ui ≡ai (mod mi): This proves our claim and the theorem. □ Notation. We call c1; :::; cthe cofactors for m1; :::; m in M and also denote these numbers by cm1;M = c1; :::; cm;M = c. 2.2 Modular arithmetic For practical computations, the moduli mi are usually chosen such that they fit into single machine words. Let µ denote the bitsize of a machine word, so that we typically have µ = 32 or µ = 64. It depends on specifics of the processor how basic arithmetic operations modulo mi can be implemented most efficiently. For instance, some processors have instructions for multiplying two µ-bit integers and return the exact (2 µ)-bit product. If not, then we rather have to assume that the moduli mi fit into µ/2 instead of µ bits, or replace µ by µ/2. Some processors do not provide efficient integer arithmetic at all. In that case, one might rather wish to rely on floating point arithmetic and take µ = 52 (assuming that we have hardware support for double precision). For floating point arithmetic it also matters whether the processor offers a “fused-multiply-add” (FMA) instruction; this essentially provides us with an efficient way to multiply two µ-bit integers exactly using floating point arithmetic. It is also recommended to choose moduli mi that fit into slightly less than µ bits whenever possible. Such extra bits can be used to significantly accelerate implementations of modular arithmetic. For a more detailed survey of practically efficient algorithms for modular arithmetic, we refer to . 2.3 Naive multi-modular reduction and reconstruction Let m1; :::; m, M = m1 ··· m, a1 2 Rm1; :::; a2 Rm and x 2 RM be as in the Chinese remainder theorem. We will refer to the computation of a1; :::; aas a function of x as the problem of multi-modular reduction. The inverse problem is called multi-modular reconstruction. In what follows, we assume that m1; :::; m have been fixed once and for all. 4 Section 2 The simplest way to perform multi-modular reduction is to simply take ai := x rem mi (i = 1; :::; ): (1) Inversely, the Chinese remainder theorem provides us with a formula for multi-modular reconstruction: x := (cm1;M a1 + ··· + cm;M a) rem M: (2) Since m1; :::; m are fixed, the computation of the cofactors cm1;M can be regarded as a precomputation. Assume that our hardware provides an instruction for the exact multipli-cation of two integers that fit into a machine word. If mi fits into a machine word, then so does the remainder ai = x rem mi. Cutting cmi;M into machine words, it follows that the product cmi;M ai can be computed using hardware products and hardware additions. Inversely, the Euclidean division of an-word integer x by mi can be done using 2 +O(1) multiplications and 2+O(1) additions/subtractions: we essentially have to multiply the quotient by mi and subtract the result from x; each next word of the quotient is obtained through a one word multiplication with an approximate inverse of mi. The above analysis shows that the naive algorithm for multi-modular reduc-tion of x modulo m1; :::; mrequires 22 + O() hardware multiplications and 22 + O() additions. The multi-modular reconstruction of x rem M can be done using only2 + O() multiplications and2 + O() additions. Depending on the hardware, the moduli mi, and the way we implement things, O(2) more operations may be required for the carry handling—but it is beyond the scope of this paper to go into this level of detail. 3 Gentle moduli 3.1 The naive algorithms revisited for special moduli Let us now reconsider the naive algorithms from section 2.3, but in the case when the moduli m1; :::;mare all close to a specific power of two. More precisely, we assume that mi = 2sw + δi (i = 1; :::;); (3) where jδij 6 2w¡1 and s > 2 a small number. As usual, we assume that the mi are pairwise coprime and we let M = m1 ··· m. We also assume that w is slightly smaller than µ and that we have a hardware instruction for the exact multiplication of µ-bit integers. Gentle moduli 5 For moduli mi as in (3), the naive algorithm for the Euclidean division of a number x 2 R2sw by mi becomes particularly simple and essentially boils down to the multiplication of δi with the quotient of this division. In other words, the remainder can be computed using ∼s hardware multiplications. In comparison, the algorithm from section 2.3 requires ∼2 s2 multiplication when applied to (s w)-bit (instead of w-bit) integers. More generally, the computation of remainders a1 = x rem m1; :::; a = x rem mcan be done using ∼2 s instead of ∼2 2 s2 multiplications. This leads to a potential gain of a factor 2 s, although the remainders are (s w)-bit integers instead of w-bit integers, for the time being. Multi-modular reconstruction can also be done faster, as follows, using sim-ilar techniques as in [1, 5]. Let x2RM. Besides the usual binary representation of x and the multi-modular representation (a1; :::; a) = (x rem m1; :::; x rem m), it is also possible to use the mixed radix representation (or Newton representation) x = b1 + b2 m1 + b3 m1 m2 + ··· + b m1 ··· m¡1; where bi 2 Rmi. Let us now show how to obtain (b1; :::; b) efficiently from (a1; :::; a). Since x rem m1 = b1 = a1, we must take b1 = a1. Assume that b1; :::; bi¡1 have been computed. For j = i ¡ 1; :::; 1 we next compute uj;i = (bj + bj+1 mj + ··· + bi¡1 mj ··· mi¡2) rem mi using ui¡1;i = bi¡1 and uj;i = (bj + uj+1;i mj) rem mi = (bj + uj+1;i · (δj ¡ δi)) rem mi (j = i ¡ 2; :::; 1): Notice that ui¡1;i; :::; u1;i can be computed using (i ¡ 1) (s + 1) hardware multiplications. We have x rem mi = (u1;i + bi m1 ··· mi¡1) rem mi = ai: Now the inverse vi of m1 ··· mi¡1 modulo mi can be precomputed. We finally compute bi = vi (ai ¡ u1;i) rem mi; which requires s2 + O(s) multiplications. For small values of i, we notice that it may be faster to divide successively by m1; :::; mi¡1 modulo mi instead of multiplying with vi. In total, the computation of the mixed radix representation (b1; :::; b) can be done using 2  (s + 1) + s2 + O(s) multiplications. Having computed the mixed radix representation, we next compute xi = bi + bi+1 mi + ··· + b mi ··· m¡1 6 Section 3 for i =; :::; 1, using the recurrence relation xi = bi + xi+1 mi: Since xi+1 2 R2(¡i)sw, the computation of xi requires ( ¡ i) s multiplications. Altogether, the computation of x = x1 from (a1; :::; a) can therefore be done using  2 (2 s + 1) + s2 ≈ s (+ s) hardware multiplications. 3.2 Combining special moduli into gentle moduli For practical applications, we usually wish to work with moduli that fit into one word (instead of s words). With the mi as in the previous subsection, this means that we need to impose the further requirement that each modulus mi can be factored mi = mi;1 ··· mi;s; with mi;1;:::;mi;s <2µ. If this is possible, then the mi are called s-gentle moduli. For given bitsizes w and s>2, the main questions are now: do such moduli indeed exist? If so, then how to find them? If s = 2, then it is easy to construct s-gentle moduli mi = 22w + δi by taking δi = ¡"i 2, where 0 6 "i < 2(w¡1)/2 is odd. Indeed, 22w ¡ "i 2 = (2w + "i) (2w ¡ "i) and gcd(2w + "i; 2w ¡ "i) = gcd(2w + "i; 2 "i) = gcd(2w + "i; "i) = gcd(2w; "i) = 1. Unfortunately, this trick does not generalize to higher values s > 3. Indeed, consider a product (2w + η1) ··· (2w + ηs) = 2sw + (η1 + ··· + ηs) 2(s¡1)w + ((η1 + ··· + ηs)2 ¡ (η1 2 + ··· + ηs 2)) 2(s¡2)w¡1 + ···; where η1;:::; ηs are small compared to 2w. If the coefficient η1+···+ ηs of 2(s¡1)w vanishes, then the coefficient of 2(s¡2)w¡1 becomes the opposite ¡(η1 2 +···+ ηs 2) of a sum of squares. In particular, both coefficients cannot vanish simultaneously, unless η1 = ··· = ηs = 0. If s > 2, then we are left with the option to search s-gentle moduli by brute force. As long as s is “reasonably small” (say s 6 8), the probability to hit an s-gentle modulus for a randomly chosen δi often remains significantly larger than 2¡w. We may then use sieving to find such moduli. By what precedes, it is also desirable to systematically take δi = ¡"i 2 for 0 6 "i < 2(w¡1)/2. This has the additional benefit that we “only” have to consider 2(w¡1)/2 possibilities for "i. Gentle moduli 7 We will discuss sieving in more detail in the next section. Assuming that we indeed have found s-gentle moduli m1; :::; m, we may use the naive algorithms from section 2.3 to compute (x rem mi;1; :::; x rem mi;s) from x rem mi and vice versa for i = 1; :::; . Given x rem mi for all i = 1; :::;, this allows us to compute all remainders xremmi;j using 2 s2+O( s) hardware multiplications, whereas the opposite conversion only requires s2 + O( s) multiplications. Altogether, we may thus obtain the remainders xremmi;j from xremM and vice versa using ∼s ( + 2 s) multiplications. 4 The gentle modulus hunt 4.1 The sieving procedure We implemented a sieving procedure in Mathemagix that uses the Mpari package with an interface to Pari-GP . Given parameters s; w; w0 and µ, the goal of our procedure is to find s-gentle moduli of the form M = (2sw/2 + ") (2sw/2 ¡ ") = m1 ··· ms with the constraints that mi < 2w0 gcd(mi; 2µ!) = 1; for i=1;:::;s, and m16···6ms. The parameter s is small and even. One should interpret w and w0 as the intended and maximal bitsize of the small moduli mi. The parameter µ stands for the minimal bitsize of a prime factor of mi. The parameter " should be such that 4 "2 fits into a machine word. In Table 1 below we have shown some experimental results for this sieving procedure in the case when s = 6, w = 22, w0 = 25 and µ = 4. For " < 1000000, the table provides us with ", the moduli m1; :::; ms, as well as the smallest prime power factors of the product M. Many hits admit small prime factors, which increases the risk that different hits are not coprime. For instance, the number 17 divides both 2132¡3113852 and 2132¡3765632, whence these 6-gentle moduli cannot be selected simultaneously (except if one is ready to sacrifice a few bits by working modulo lcm(2132 ¡ 3113852; 2132 ¡ 3765632) instead of (2132 ¡ 3113852) · (2132 ¡ 3765632)). In the case when we use multi-modular arithmetic for computations with rational numbers instead of integers (see [9, section 5 and, more particularly, section 5.10]), then small prime factors should completely be prohibited, since they increase the probability of divisions by zero. For such applications, it is therefore desirable that m1; :::; ms are all prime. In our table, this occurs for " = 57267 (we indicated this by highlighting the list of prime factors of M). 8 Section 4 In order to make multi-modular reduction and reconstruction as efficient as possible, a desirable property of the moduli mi is that they either divide 2sw/2¡" or 2sw/2+". In our table, we highlighted the " for which this happens. We notice that this is automatically the case if m1; :::; ms are all prime. If only a small number of mi (say a single one) do not divide either 2sw/2 ¡" or 2sw/2+", then we remark that it should still be possible to design reasonably efficient ad hoc algorithms for multi-modular reduction and reconstruction. Another desirable property of the moduli m1 6 ··· 6 ms is that ms is as small as possible: the spare bits can for instance be used to speed up matrix multiplication modulo ms. Notice however that one “occasional” large modulus ms only impacts on one out of s modular matrix products; this alleviates the negative impact of such moduli. We refer to section 4.3 below for more details. For actual applications, one should select gentle moduli that combine all desirable properties mentioned above. If not enough such moduli can be found, then it it depends on the application which criteria are most important and which ones can be released. " m1 m2 m3 m4 m5 m6 p1 ν1; p2 ν2; ::: 27657 28867 4365919 6343559 13248371 20526577 25042063 29; 41; 43; 547; ::: 57267 416459 1278617 2041469 6879443 25754563 28268089 416459; ::: 77565 7759 8077463 8261833 18751793 19509473 28741799 59; 641; ::: 95253 724567 965411 3993107 4382527 19140643 23236813 43; 724567; ::: 294537 190297 283729 8804561 19522819 19861189 29537129 232; 151; 1879; ::: 311385 145991 4440391 4888427 6812881 7796203 32346631 17; 79; 131; ::: 348597 114299 643619 6190673 11389121 32355397 32442427 31; 277; ::: 376563 175897 1785527 2715133 7047419 30030061 30168739 17; 127; 1471; ::: 462165 39841 3746641 7550339 13195943 18119681 20203643 67; 641; 907; ::: 559713 353201 873023 2595031 11217163 18624077 32569529 19; 59; 14797; ::: 649485 21727 1186571 14199517 15248119 31033397 31430173 19; 109; 227; ::: 656997 233341 1523807 5654437 8563679 17566069 18001723 79; 89; 63533; ::: 735753 115151 923207 3040187 23655187 26289379 27088541 53; 17419; ::: 801687 873767 1136111 3245041 7357871 8826871 26023391 23; 383777; ::: 826863 187177 943099 6839467 11439319 12923753 30502721 73; 157; 6007; ::: 862143 15373 3115219 11890829 18563267 19622017 26248351 31; 83; 157; ::: 877623 514649 654749 4034687 4276583 27931549 33525223 41; 98407; ::: 892455 91453 2660297 3448999 12237457 21065299 25169783 29; 397; 2141; ::: Table 1. List of 6-gentle moduli for w = 22, w0 = 25, µ = 4 and " < 1000000. 4.2 Influence of the parameters s, w and w0 Ideally speaking, we want s to be as large as possible. Furthermore, in order to waste as few bits as possible, w0 should be close to the word size (or half of it) and w0 ¡ w should be minimized. When using double precision floating point arithmetic, this means that we wish to take w02f24;25;26;50;51;52g. Whenever we have efficient hardware support for integer arithmetic, then we might prefer w 2 f30; 31; 32; 62; 63; 64g. The gentle modulus hunt 9 Let us start by studying the influence of w0 ¡ w on the number of hits. In Table 2, we have increased w by one with respect to Table 1. This results in an approximate 5% increase of the “capacity” s w of the modulus M. On the one hand, we observe that the hit rate of the sieve procedure roughly decreases by a factor of thirty. On the other hand, we notice that the rare gentle moduli that we do find are often of high quality (on four occasions the moduli m1; :::;ms are all prime in Table 2). " m1 m2 m3 m4 m5 m6 p1 ν1; p2 ν2; ::: 936465 543889 4920329 12408421 15115957 24645539 28167253 19; 59; 417721; ::: 2475879 867689 4051001 11023091 13219163 24046943 28290833 867689; ::: 3205689 110161 12290741 16762897 22976783 25740731 25958183 59; 79; 509; ::: 3932205 4244431 5180213 5474789 8058377 14140817 25402873 4244431; ::: 5665359 241739 5084221 18693097 21474613 23893447 29558531 31; 41; 137; ::: 5998191 30971 21307063 21919111 22953967 31415123 33407281 101; 911; 941; ::: 6762459 3905819 5996041 7513223 7911173 8584189 29160587 43; 137; 90833; ::: 9245919 2749717 4002833 8274689 9800633 15046937 25943587 2749717; ::: 9655335 119809 9512309 20179259 21664469 22954369 30468101 17; 89; 149; ::: 12356475 1842887 2720359 7216357 13607779 23538769 30069449 1842887; ::: 15257781 1012619 5408467 9547273 11431841 20472121 28474807 31; 660391; ::: Table 2. List of 6-gentle moduli for w = 23, w0 = 25, µ = 4 and " < 16000000. Without surprise, the hit rate also sharply decreases if we attempt to increase s. The results for s = 8 and w = 22 are shown in Table 3. A further infortunate side effect is that the quality of the gentle moduli that we do find also decreases. Indeed, on the one hand, M tends to systematically admit at least one small prime factor. On the other hand, it is rarely the case that each mi divides either 2sw/2 ¡ " or 2sw/2 + " (this might nevertheless happen for other recombinations of the prime factors of M, but only modulo a further increase of ms). " m1 m2 m3 m4 m5 m6 m7 m8 p1 ν1; p2 ν2; ::: 889305 50551 1146547 4312709 5888899 14533283 16044143 16257529 17164793 17; 31; 31; 59; ::: 2447427 53407 689303 3666613 4837253 7944481 21607589 25976179 32897273 31; 61; 103; ::: 2674557 109841 1843447 2624971 5653049 7030883 8334373 18557837 29313433 103; 223; 659; ::: 3964365 10501 2464403 6335801 9625841 10329269 13186219 17436197 25553771 23; 163; 607; ::: 4237383 10859 3248809 5940709 6557599 9566959 11249039 22707323 28518509 23; 163; 1709; ::: 5312763 517877 616529 879169 4689089 9034687 11849077 24539909 27699229 43; 616529; ::: 6785367 22013 1408219 4466089 7867589 9176941 12150997 26724877 29507689 23; 41; 197; ::: 7929033 30781 730859 4756351 9404807 13807231 15433939 19766077 22596193 31; 307; 503; ::: 8168565 10667 3133103 3245621 6663029 15270019 18957559 20791819 22018021 43; 409; 467; ::: 8186205 41047 2122039 2410867 6611533 9515951 14582849 16507739 30115277 23; 167; 251; ::: Table 3. List of 8-gentle moduli for w = 22, w0 = 25, µ = 4 and " < 10000000. An increase of w0 while maintaining s and w0 ¡ w fixed also results in a decrease of the hit rate. Nevertheless, when going from w0 = 25 (floating point arithmetic) to w0 = 31 (integer arithmetic), this is counterbalanced by the fact that " can also be taken larger (namely " < 2w0); see Table 4 for a concrete example. When doubling w and w0 while keeping the same upper bound for ", the hit rate remains more or less unchanged, but the rate of high quality hits tends to decrease somewhat: see Table 5. 10 Section 4 It should be possible to analyze the hit rate as a function of the parameters s, w, w0 and µ from a probabilistic point of view, using the idea that a random number n is prime with probability (logn)¡1. However, from a practical perspec-tive, the priority is to focus on the case when w0 6 64. For the most significant choices of parameters µ < w < w0 6 64 and s, it should be possible to compile full tables of s-gentle moduli. Unfortunately, our current implementation is still somewhat inefficient for w0 >32. A helpful feature for upcoming versions of Pari would be a function to find all prime factors of an integer below a specified maximum 2w0 (the current version only does this for prime factors that can be tabulated). " m1 m2 m3 m4 m5 m6 p1 ν1; p2 ν2; ::: 303513 42947057 53568313 331496959 382981453 1089261409 1176003149 292; 1480933; ::: 851463 10195123 213437143 470595299 522887483 692654273 1008798563 17; 41; 67; ::: 1001373 307261 611187931 936166801 1137875633 1196117147 1563634747 47; 151; ::: 1422507 3950603 349507391 490215667 684876553 693342113 1164052193 29; 211; 349; ::: 1446963 7068563 94667021 313871791 877885639 1009764377 2009551553 23; 71; 241; ::: 1551267 303551 383417351 610444753 1178193077 2101890797 2126487631 29; 43; 2293; ::: 1555365 16360997 65165071 369550981 507979403 1067200639 1751653069 17; 23; 67; ::: 4003545 20601941 144707873 203956547 624375041 655374931 1503716491 47; 67; ::: 4325475 11677753 139113383 210843443 659463289 936654347 1768402001 19; 41; ::: 4702665 8221903 131321017 296701997 496437899 1485084431 1584149417 8221903; ::: 5231445 25265791 49122743 433700843 474825677 907918279 1612324823 17; 1486223; ::: 5425527 37197571 145692101 250849363 291039937 456174539 2072965393 37197571; ::: 6883797 97798097 124868683 180349291 234776683 842430863 858917923 97798097; ::: 7989543 4833137 50181011 604045619 638131951 1986024421 2015143349 23; 367; ::: Table 4. List of 6-gentle moduli for w =28, w0=31, µ=4 and "<1600000. Followed by some of the next gentle moduli for which each mi divides either 2sw/2¡ or 2sw/2+ . " m1 m2 ··· m5 m6 p1 ν1; p2 ν2; ::: 15123 380344780931 774267432193 ··· 463904018985637 591951338196847 37; 47; 239; ::: 34023 9053503517 13181369695139 ··· 680835893479031 723236090375863 29; 35617; ::: 40617 3500059133 510738813367 ··· 824394263006533 1039946916817703 23; 61; 347; ::: 87363 745270007 55797244348441 ··· 224580313861483 886387548974947 71; 9209; ::: 95007 40134716987 2565724842229 ··· 130760921456911 393701833767607 19; 67; ::: 101307 72633113401 12070694419543 ··· 95036720090209 183377870340761 41; 401; ::: 140313 13370367761 202513228811 ··· 397041457462499 897476961701171 379; 1187; ::: 193533 35210831 15416115621749 ··· 727365428298107 770048329509499 59; 79; ::: 519747 34123521053 685883716741 ··· 705516472454581 836861326275781 127; 587; ::: 637863 554285276371 1345202287357 ··· 344203886091451 463103013579761 79; 1979; ::: 775173 322131291353 379775454593 ··· 194236314135719 1026557288284007 322131291353; ::: 913113 704777248393 1413212491811 ··· 217740328855369 261977228819083 37; 163; 677; ::: 1400583 21426322331 42328735049 ··· 411780268096919 626448556280293 21426322331; ::: T able 5. List of 6-gentle moduli for w =44, w0=50, µ=4 and " <200000. Followed by some of the next gentle moduli for which each mi divides either 2sw/2¡ or 2sw/2+ . 4.3 Application to matrix multiplication Let us finally return to our favourite application of multi-modular arithmetic to the multiplication of integer matrices A; B 2 Zr×r. From a practical point of view, the second step of the algorithm from the introduction can be implemented very efficiently if r mi 2 fits into the size of a word. The gentle modulus hunt 11 When using floating point arithmetic, this means that we should have r mi 2< 252 for all i. For large values of r, this is unrealistic; in that case, we subdivide the r ×r matrices into smaller ri ×ri matrices with ri mi 2<252. The fact that ri may depend on i is very significant. First of all, the larger we can take ri, the faster we can multiply matrices modulo mi. Secondly, the mi in the tables from the previous sections often vary in bitsize. It frequently happens that we may take all ri large except for the last modulus m. The fact that matrix multiplications modulo the worst modulus m are somewhat slower is compensated by the fact that they only account for one out of every modular matrix products. Several of the tables in the previous subsections were made with the applica-tion to integer matrix multiplication in mind. Consider for instance the modulus M = m1 ··· m6 = 2132 ¡ 6569972 from Table 1. When using floating point arithmetic, we obtain r1 6 82713, r2 6 1939, r3 6 140, r4 6 61, r5 6 14 and r6 6 13. Clearly, there is a trade-offbetween the efficiency of the modular matrix multiplications (high values of ri are better) and the bitsize ≈ w of M (larger capacities are better). Bibliography J.-C. Bajard, M. E. Kaihara, and T. Plantard. Selected rns bases for modular multiplica-tion. In Proc. of the 19th IEEE Symposium on Computer Arithmetic, pages 25–32, 2009. D. Bernstein. Scaled remainder trees. Available from scaledmod-20040820.pdf, 2004. A. Borodin and R. T. Moenck. Fast modular transforms. Journal of Computer and System Sciences, 8:366–386, 1974. A. Bostan, G. Lecerf, and É. Schost. Tellegen's principle into practice. In Proceedings of ISSAC 2003, pages 37–44. ACM Press, 2003. A. Bostan and É. Schost. Polynomial evaluation and interpolation on special sets of points. Journal of Complexity, 21(4):420–446, August 2005. Festschrift for the 70th Birthday of Arnold Schönhage. J. W. Cooley and J. W. Tukey. An algorithm for the machine calculation of complex Fourier series. Math. Computat., 19:297–301, 1965. J. Doliskani, P. Giorgi, R. Lebreton, and É. Schost. Simultaneous conversions with the Residue Number System using linear algebra. Technical Report HAL, 2016. C. M. Fiduccia. Polynomial evaluation via the division algorithm: the fast fourier trans-form revisited. In A. L. Rosenberg, editor, Fourth annual ACM symposium on theory of computing, pages 88–93, 1972. J. von zur Gathen and J. Gerhard. Modern Computer Algebra. Cambridge University Press, New York, NY, USA, 3rd edition, 2013. J. van der Hoeven. Faster Chinese remaindering. Technical report, HAL, 2016. http:// hal.archives-ouvertes.fr/hal-01403810. J. van der Hoeven, G. Lecerf, B. Mourrain, et al. Mathemagix, 2002. http:// www.mathemagix.org. J. van der Hoeven, G. Lecerf, and G. Quintin. Modular SIMD arithmetic in Mathemagix. ACM Trans. Math. Softw., 43(1):5:1–5:37, 2016. A. Karatsuba and J. Ofman. Multiplication of multidigit numbers on automata. Soviet Physics Doklady, 7:595–596, 1963. 12 Section R. T. Moenck and A. Borodin. Fast modular transforms via division. In Thirteenth annual IEEE symposium on switching and automata theory, pages 90–96, Univ. Mary-land, College Park, Md., 1972. The PARI Group, Bordeaux. PARI/GP, 2012. Software available from http:// pari.math.u-bordeaux.fr. Bibliography 13
10419
https://courses.lumenlearning.com/hccs-macroeconomics-3/chapter/the-expenditure-output-model/
Reading: The Expenditure-Output Model | Macroeconomics – Haci Skip to main content Macroeconomics – Haci Module: Keynesian and Neoclassical Economics Search for: Reading: The Expenditure-Output Model The Axes of the Expenditure-Output Diagram The fundamental ideas of Keynesian economics were developed before the AD–AS model was popularized. From the 1930s until the 1970s, Keynesian economics was usually explained with a different model, known as the expenditure-output approach. This approach is strongly rooted in the fundamental assumptions of Keynesian economics: it focuses on the total amount of spending in the economy, with no explicit mention of aggregate supply or of the price level (although as you will see, it is possible to draw some inferences about aggregate supply and price levels based on the diagram). The expenditure-output model, sometimes also called the Keynesian cross diagram, determines the equilibrium level of real GDP by the point where the total or aggregate expenditures in the economy are equal to the amount of output produced. The axes of the Keynesian cross diagram presented in Figure B.1 show real GDP on the horizontal axis as a measure of output and aggregate expenditures on the vertical axis as a measure of spending. Figure B.1. The Expenditure-Output Diagram The aggregate expenditure-output model shows aggregate expenditures on the vertical axis and real GDP on the horizontal axis. A vertical line shows potential GDP where full employment occurs. The 45-degree line shows all points where aggregate expenditures and output are equal. The aggregate expenditure schedule shows how total spending or aggregate expenditure increases as output or real GDP rises. The intersection of the aggregate expenditure schedule and the 45-degree line will be the equilibrium. Equilibrium occurs at E0, where aggregate expenditure AE0 is equal to the output level Y0. Remember that GDP can be thought of in several equivalent ways: it measures both the value of spending on final goods and also the value of the production of final goods. All sales of the final goods and services that make up GDP will eventually end up as income for workers, for managers, and for investors and owners of firms. The sum of all the income received for contributing resources to GDP is called national income (Y). At some points in the discussion that follows, it will be useful to refer to real GDP as “national income.” Both axes are measured in real (inflation-adjusted) terms. THE POTENTIAL GDP LINE AND THE 45-DEGREE LINE The Keynesian cross diagram contains two lines that serve as conceptual guideposts to orient the discussion. The first is a vertical line showing the level of potential GDP. Potential GDP means the same thing here that it means in the AD/AS diagrams: it refers to the quantity of output that the economy can produce with full employment of its labor and physical capital. The second conceptual line on the Keynesian cross diagram is the 45-degree line, which starts at the origin and reaches up and to the right. A line that stretches up at a 45-degree angle represents the set of points (1, 1), (2, 2), (3, 3) and so on, where the measurement on the vertical axis is equal to the measurement on the horizontal axis. In this diagram, the 45-degree line shows the set of points where the level of aggregate expenditure in the economy, measured on the vertical axis, is equal to the level of output or national income in the economy, measured by GDP on the horizontal axis. When the macroeconomy is in equilibrium, it must be true that the aggregate expenditures in the economy are equal to the real GDP—because by definition, GDP is the measure of what is spent on final sales of goods and services in the economy. Thus, the equilibrium calculated with a Keynesian cross diagram will always end up where aggregate expenditure and output are equal—which will only occur along the 45-degree line. THE AGGREGATE EXPENDITURE SCHEDULE The final ingredient of the Keynesian cross or expenditure-output diagram is the aggregate expenditure schedule, which will show the total expenditures in the economy for each level of real GDP. The intersection of the aggregate expenditure line with the 45-degree line—at point E 0 in Figure B.1—will show the equilibrium for the economy, because it is the point where aggregate expenditure is equal to output or real GDP. After developing an understanding of what the aggregate expenditures schedule means, we will return to this equilibrium and how to interpret it. Building the Aggregate Expenditure Schedule Aggregate expenditure is the key to the expenditure-income model. The aggregate expenditure schedule shows, either in the form of a table or a graph, how aggregate expenditures in the economy rise as real GDP or national income rises. Thus, in thinking about the components of the aggregate expenditure line—consumption, investment, government spending, exports and imports—the key question is how expenditures in each category will adjust as national income rises. CONSUMPTION AS A FUNCTION OF NATIONAL INCOME How do consumption expenditures increase as national income rises? People can do two things with their income: consume it or save it (for the moment, let’s ignore the need to pay taxes with some of it). Each person who receives an additional dollar faces this choice. The marginal propensity to consume (MPC), is the share of the additional dollar of income a person decides to devote to consumption expenditures. The marginal propensity to save (MPS) is the share of the additional dollar a person decides to save. It must always hold true that: MPC+MPS=1 For example, if the marginal propensity to consume out of the marginal amount of income earned is 0.9, then the marginal propensity to save is 0.1. With this relationship in mind, consider the relationship among income, consumption, and savings shown in Figure B.2. (Note that we use “Aggregate Expenditure” on the vertical axis in this and the following figures, because all consumption expenditures are parts of aggregate expenditures.) An assumption commonly made in this model is that even if income were zero, people would have to consume something. In this example, consumption would be $600 even if income were zero. Then, the MPC is 0.8 and the MPS is 0.2. Thus, when income increases by $1,000, consumption rises by $800 and savings rises by $200. At an income of $4,000, total consumption will be the $600 that would be consumed even without any income, plus $4,000 multiplied by the marginal propensity to consume of 0.8, or $ 3,200, for a total of $ 3,800. The total amount of consumption and saving must always add up to the total amount of income. (Exactly how a situation of zero income and negative savings would work in practice is not important, because even low-income societies are not literally at zero income, so the point is hypothetical.) This relationship between income and consumption, illustrated in Figure B.2 and Table B.1, is called the consumption function. Figure B.2. The Consumption Function. In the expenditure-output model, how does consumption increase with the level of national income? Output on the horizontal axis is conceptually the same as national income, since the value of all final output that is produced and sold must be income to someone, somewhere in the economy. At a national income level of zero, $600 is consumed. Then, each time income rises by $1,000, consumption rises by $800, because in this example, the marginal propensity to consume is 0.8. The pattern of consumption shown in Table B.1 is plotted in Figure B.2. To calculate consumption, multiply the income level by 0.8, for the marginal propensity to consume, and add $600, for the amount that would be consumed even if income was zero. Consumption plus savings must be equal to income. Table B.1. The Consumption Function| Income | Consumption | Savings | --- | $0 | $600 | –$600 | | $1,000 | $1,400 | –$400 | | $2,000 | $2,200 | –$200 | | $3,000 | $3,000 | $0 | | $4,000 | $3,800 | $200 | | $5,000 | $4,600 | $400 | | $6,000 | $5,400 | $600 | | $7,000 | $6,200 | $800 | | $8,000 | $7,000 | $1,000 | | $9,000 | $7,800 | $1,200 | However, a number of factors other than income can also cause the entire consumption function to shift. These factors were summarized in the earlier discussion of consumption, and listed in Table B.1. When the consumption function moves, it can shift in two ways: either the entire consumption function can move up or down in a parallel manner, or the slope of the consumption function can shift so that it becomes steeper or flatter. For example, if a tax cut leads consumers to spend more, but does not affect their marginal propensity to consume, it would cause an upward shift to a new consumption function that is parallel to the original one. However, a change in household preferences for saving that reduced the marginal propensity to save would cause the slope of the consumption function to become steeper: that is, if the savings rate is lower, then every increase in income leads to a larger rise in consumption. INVESTMENT AS A FUNCTION OF NATIONAL INCOME Investment decisions are forward-looking, based on expected rates of return. Precisely because investment decisions depend primarily on perceptions about future economic conditions, they do not depend primarily on the level of GDP in the current year. Thus, on a Keynesian cross diagram, the investment function can be drawn as a horizontal line, at a fixed level of expenditure. Figure B.3 shows an investment function where the level of investment is, for the sake of concreteness, set at the specific level of 500. Just as a consumption function shows the relationship between consumption levels and real GDP (or national income), the investment function shows the relationship between investment levels and real GDP. Figure B.3. The Investment Function. The investment function is drawn as a flat line because investment is based on interest rates and expectations about the future, and so it does not change with the level of current national income. In this example, investment expenditures are at a level of 500. However, changes in factors like technological opportunities, expectations about near-term economic growth, and interest rates would all cause the investment function to shift up or down. The appearance of the investment function as a horizontal line does not mean that the level of investment never moves. It means only that in the context of this two-dimensional diagram, the level of investment on the vertical aggregate expenditure axis does not vary according to the current level of real GDP on the horizontal axis. However, all the other factors that vary investment—new technological opportunities, expectations about near-term economic growth, interest rates, the price of key inputs, and tax incentives for investment—can cause the horizontal investment function to shift up or down. GOVERNMENT SPENDING AND TAXES AS A FUNCTION OF NATIONAL INCOME In the Keynesian cross diagram, government spending appears as a horizontal line, as in Figure B.4, where government spending is set at a level of 1,300. As in the case of investment spending, this horizontal line does not mean that government spending is unchanging. It means only that government spending changes when Congress decides on a change in the budget, rather than shifting in a predictable way with the current size of the real GDP shown on the horizontal axis. Figure B.4. The Government Spending Function. The level of government spending is determined by political factors, not by the level of real GDP in a given year. Thus, government spending is drawn as a horizontal line. In this example, government spending is at a level of 1,300. Congressional decisions to increase government spending will cause this horizontal line to shift up, while decisions to reduce spending would cause it to shift down. The situation of taxes is different because taxes often rise or fall with the volume of economic activity. For example, income taxes are based on the level of income earned and sales taxes are based on the amount of sales made, and both income and sales tend to be higher when the economy is growing and lower when the economy is in a recession. For the purposes of constructing the basic Keynesian cross diagram, it is helpful to view taxes as a proportionate share of GDP. In the United States, for example, taking federal, state, and local taxes together, government typically collects about 30–35 % of income as taxes. Table B.2 revises the earlier table on the consumption function so that it takes taxes into account. The first column shows national income. The second column calculates taxes, which in this example are set at a rate of 30%, or 0.3. The third column shows after-tax income; that is, total income minus taxes. The fourth column then calculates consumption in the same manner as before: multiply after-tax income by 0.8, representing the marginal propensity to consume, and then add $600, for the amount that would be consumed even if income was zero. When taxes are included, the marginal propensity to consume is reduced by the amount of the tax rate, so each additional dollar of income results in a smaller increase in consumption than before taxes. For this reason, the consumption function, with taxes included, is flatter than the consumption function without taxes, as Figure B.5 shows. Figure B.5. The Consumption Function. Before and After Taxes The upper line repeats the consumption function from Figure B.2. The lower line shows the consumption function if taxes must first be paid on income, and then consumption is based on after-tax income. Table B.2. The Consumption Function Before and After Taxes| Income | Taxes | After-Tax Income | Consumption | Savings | --- --- | $0 | $0 | $0 | $600 | –$600 | | $1,000 | $300 | $700 | $1,160 | –$460 | | $2,000 | $600 | $1,400 | $1,720 | –$320 | | $3,000 | $900 | $2,100 | $2,280 | –$180 | | $4,000 | $1,200 | $2,800 | $2,840 | –$40 | | $5,000 | $1,500 | $3,500 | $3,400 | $100 | | $6,000 | $1,800 | $4,200 | $3,960 | $240 | | $7,000 | $2,100 | $4,900 | $4,520 | $380 | | $8,000 | $2,400 | $5,600 | $5,080 | $520 | | $9,000 | $2,700 | $6,300 | $5,640 | $660 | EXPORTS AND IMPORTS AS A FUNCTION OF NATIONAL INCOME The export function, which shows how exports change with the level of a country’s own real GDP, is drawn as a horizontal line, as in the example in Figure B.6(a) where exports are drawn at a level of $840. Again, as in the case of investment spending and government spending, drawing the export function as horizontal does not imply that exports never change. It just means that they do not change because of what is on the horizontal axis—that is, a country’s own level of domestic production—and instead are shaped by the level of aggregate demand in other countries. More demand for exports from other countries would cause the export function to shift up; less demand for exports from other countries would cause it to shift down. Figure B.6. The Export and Import Functions. (a) The export function is drawn as a horizontal line because exports are determined by the buying power of other countries and thus do not change with the size of the domestic economy. In this example, exports are set at 840. However, exports can shift up or down, depending on buying patterns in other countries. (b) The import function is drawn in negative territory because expenditures on imported products are a subtraction from expenditures in the domestic economy. In this example, the marginal propensity to import is 0.1, so imports are calculated by multiplying the level of income by –0.1. Imports are drawn in the Keynesian cross diagram as a downward-sloping line, with the downward slope determined by the marginal propensity to import (MPI), out of national income. In Figure B.6(b), the marginal propensity to import is 0.1. Thus, if real GDP is $5,000, imports are $500; if national income is $6,000, imports are $600, and so on. The import function is drawn as downward sloping and negative, because it represents a subtraction from the aggregate expenditures in the domestic economy. A change in the marginal propensity to import, perhaps as a result of changes in preferences, would alter the slope of the import function. Candela Citations CC licensed content, Shared previously Appendix B. Authored by: OpenStax College. Provided by: Rice University. Located at: License: CC BY: Attribution. License Terms: Download for free at Licenses and Attributions CC licensed content, Shared previously Appendix B. Authored by: OpenStax College. Provided by: Rice University. Located at: License: CC BY: Attribution. License Terms: Download for free at PreviousNext Privacy Policy
10420
https://www.facebook.com/80sruled/videos/a-negative-times-a-negative-equals-a-positive-stand-and-deliver/449003689196228/
A Negative Times A Negative Equals A Positive - Powerful Scene from Stand and Deliver | The 80s Ruled | Facebook Log In Log In Forgot Account? Video ===== Home Live Reels Explore More Home Live Reels Explore 0:00 / 3:45 A Negative Times A Negative Equals A Positive - Stand and Deliver Like Comment Share 841 · 77 comments · 219K views The 80s Ruled June 5, 2019 · Follow A Negative Times A Negative Equals A Positive - Powerful Scene from Stand and Deliver Comments Most relevant Jonny Mase Why don’t gang bangers write short paragraphs? Because that ain’t a proper essay! 5y 7 View 1 reply Chris Martindale Ok, I have been through 6 college math courses. I still don’t know why a negative times a negative equals a positive. 5y View more comments 2 of 53 Related Reels The 80s Ruled 13K Related Videos 0:37 I think we're gonna change our Christmas dinner up this year! The 80s Ruled 10.3K views · December 7, 2024 0:15 Hello, I Must Be Going! is the second solo studio album by Phil Collins. It was released 42 years ago today on November 5, 1982 and featured the top ten hit “You Can’t Hurry Love”! The 80s Ruled 5.1K views · November 6, 2024 0:15 Loved this show!!! The 80s Ruled 7.7K views · September 18, 2024 0:51 Lita Ford - Kiss Me Deadly The 80s Ruled 120K views · October 5, 2023 0:19 Just let your Soul Glo and have a great Labor Day weekend! The 80s Ruled 4.2K views · September 1, 2023 0:22 80s Celebs were so natural on camera. The 80s Ruled 8.7K views · September 1, 2023 0:42 80s Roller Derby was insane! The 80s Ruled 14.5K views · August 31, 2023 0:24 This blew our minds in the '80s The 80s Ruled 5.4K views · August 31, 2023 0:44 Sounds of the 80s! The 80s Ruled 918.3K views · August 31, 2023 Pages Other Brand Website Entertainment website The 80s Ruled Videos A Negative Times A Negative Equals A Positive - Stand and Deliver Home Live Explore Related Pages Flashback to the 80's Interest 690K Followers Golden 80s Music Chart 458K Followers Remembering the 70s 80s and 90s Entertainment website 180K Followers Growing up in the 80's and 90's I remember? Video Game 20K Followers Totally 80's Entertainment website 43K Followers Classic Rock in Pics Music Chart 757K Followers Privacy · Consumer Health Privacy · Terms · Advertising · Ad Choices · Cookies · More See more on Facebook See more on Facebook Email or phone number Password Log In Forgot password? or Create new account
10421
https://www.gauthmath.com/solution/Question-10-The-proof-of-the-identity-4xy-x-y-2-x-y-2-is-shown-Drag-each-explana-1715376945571862
Solved: The proof of the identity 4xy=(x+y)^2-(x-y)^2 is shown. Drag each explanation to the step [Math] Search Drag Image or Click Here to upload Command+to paste Upgrade Sign in Homework Homework Assignment Solver Assignment Calculator Calculator Resources Resources Blog Blog App App Gauth Unlimited answers Gauth AI Pro Start Free Trial Homework Helper Study Resources Math Expression Questions Question The proof of the identity 4xy=(x+y)^2-(x-y)^2 is shown. Drag each explanation to the step of the proof that it justifies. + Instructions 4xy=(x+y)^2 Original equation -(x-y)^2 4xy=x^2+2xy Commutative Property +y^2-(x-y)^2 4xy=x^2+2xy +y^2-x^2+2xy Square of a difference -y^2 4xy=x^2-x^2 +2xy+2xy+y^2 Square of a sum -y^2 4xy=4xy Combine like terms Table 1: [] Show transcript Expert Verified Solution 93%(697 rated) Answer $$4xy=(x+y)^{2}-(x-y)^{2}$$4 x y=(x+y)2−(x−y)2 Explanation The original equation is $$4xy=(x+y)^{2}-(x-y)^{2}$$4 x y=(x+y)2−(x−y)2 Expand the squares in the equation to get $$4xy=x^{2}+2xy+y^{2}-(x-y)^{2}$$4 x y=x 2+2 x y+y 2−(x−y)2 Recognize that the expanded form of the square of a sum is $$x^{2}+2xy+y^{2}$$x 2+2 x y+y 2 and the square of a difference is $$x^{2}-2xy+y^{2}$$x 2−2 x y+y 2 Combine like terms to get $$4xy=x^{2}-x^{2}+2xy+2xy+y^{2}-y^{2}$$4 x y=x 2−x 2+2 x y+2 x y+y 2−y 2 Simplify the equation to $$4xy=4xy$$4 x y=4 x y, which proves the identity. Hence, the identity $$4xy=(x+y)^{2}-(x-y)^{2}$$4 x y=(x+y)2−(x−y)2 is proven. Helpful Not Helpful Explain Simplify this solution Related The proof of the identity 4xy=x+y2-x-y2 is shown. Drag each explanation to the step of the proof that it justifies. ① Instructions 4xy=x+y2-x-y2 Onginal equation 4xy=x2+2xy+y2 -x-y2 4xy=x2+2xy+y2-x2 +2xy-y2 100% (2 rated) The proof of the identity 4xy=x+y2-x-y2 is shown. Drag each explanation to the step of the pro ① Instructions 4xy=x+y2-x-y2 4xy=x2+2xy+y2-x-y2 4xy=x2+2xy+y2-x2+2xy -y2 4xy=x2-x2+2xy+2xy+y2 -y2 4xy=4xy 100% (3 rated) Cân The proof of the identity 4 4xy=x+y2-x-y2 is shown. Drag each explanation to the step of the proof that it justifies. Instructions 4xy=x+y2 -x-y2 square 4xy=x2+2xy square +y2 -x-y2 4xy=x2+2xy +y2-x2 +2xy-y2 square 4xy=x2-x2 +2xy+2xy +y2-y2 square 4xy=4xy square Activate Wind 100% (4 rated) ① Instructions 4xy=x+y2-x-y2 Square of a sum 4xy=x2+2xy+y2 -x-y2 Square of a difference 4xy=x2+2xy+y2-x2 +2xy-y2 4xy=x2-x2+2xy+2xy +y2-y2 4xy=4xy Combine like terms. Commutative Property Original equation 86% (11 rated) In assess prod mheducation.c0II ① Instructions 4xy=x+y2-x-y2 4xy=x2+2xy+y2-x-y2 4xy=x2+2xy+y2-x2+2xy -y2 4xy=x2-x2+2xy+2xy+y2 -9 4xy=4xy Commutative Property Original equation Square of a sum Combine like terms. Square of a difference 100% (5 rated) Ch. 3 Module Review 1 - 15 only Question 10 of 20 v Question 10 The proof of the identity 4xy=x+y2-x-y2 is shown. Drag each explanation to the step of the proof that it justifies. Step 1 4xy=x+y2-x-y2 Step 2 4xy=x2+2xy+y2-x-y2 Step 3 4xy=x2+2xy+y2-x2+2xy-y2 Step 4 4xy=x2-x2+2xy+2xy+y2-y2 Step 5 4xy=4xy Instructions 4xy=x+y2-x-y2 100% (668 rated) ALEKS - Erman Gunyil Question 10 - Module Desmos | Graphing Ca X 7w/3w- 12 - Google S= cdn.assess.prod.mheducation.com/dle-assess-delivery-ui/v3.28.6/assessment/page/10?token=f568987567b7f5b26d43ed12f32f615d oclomaris Clever | Log in Final Project Writin... Israel Adesanya Chil... Quickwrite 4/24 Make a Balloon Pow.... 4xy=x+y2-x-y2 4xy=x2+2xy+y2-x-y2 4xy=x2+2xy+y2-x2+2xy -y3 4xy=x2-x2+2xy+2xy+y2 -y3 4xy=4xy Commutative Property Original equation Square of a sum Combine like terms Square of a difference 100% (4 rated) a i How many even numbers are there from 100 to 1000 inclusive? ii I write down m consecutive odd numbers, the smallest of which is л. Find in terms of m and n, the value of the largest odd number I wrote down. b y is a positive whole number such that y/11 lies between 81/13 and 98/15 . Find all possible values of y. 80% (5 rated) A company is using linear programming to decide how many units of each of its two products to make each week. Weekly production will be x units of Product X and y units of Product Y. At least 50 units of X must be produced each week, and at least twice as many units of Y as of X must be produced each week. Each urt of X requires 30 minutes of labour, and each unit of Y requires two hours of labour. There are 5,000 hours of labour available each week. Which of the following is the correct set of constraints? Submit your answer to view the feedback. 0.5x+2y ≤ 5,000 0.5x+2y ≤ 5,000 x ≥ 50 x ≥ 50 y ≤ 2x y ≥ 100 x+4y ≤ 5,000 0.5x+2y ≤ 5,000 x ≥ 50 x ≥ 50 y ≥ 2x y ≥ 2x 100% (4 rated) ① Instructions 4xy=x+y2-x-y2 Original equation 4xy=x2+2xy+y2-x-y2 4xy=x2+2xy+y2-x2+2xy -y2 4xy=x2-x2+2xy+2xy+y2 -y2 Combine like terms. 4xy=4xy 100% (4 rated) Gauth it, Ace it! contact@gauthmath.com Company About UsExpertsWriting Examples Legal Honor CodePrivacy PolicyTerms of Service Download App
10422
https://www.grason-stadler.com/education/guides/eustachian-tube-function
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Settings Accept all powered by: Cookie Information Functional Statistical Marketing Login EN ES DE FR CN RU Products Products view all Audiometers Audiometers view all AudioStar Pro Pello GSI 39 GSI 18 AMTAS Pro AMTAS Flex Tympanometers Tympanometers view all TympStar Pro Allegro GSI 39 OAE OAE view all Corti Audera Pro Novus Evoked Potentials Evoked Potentials view all Audera Pro Novus Otoscope Software Software view all GSI Suite Education Education view all Video Library Courses Courses view all 60 Minute Courses 30 Minute Courses Testing Guides Interviews All Content About About view all Events History Support Blog Locator REQUEST QUOTE Select Language EUSTACHIAN TUBE FUNCTION TYMPANOMETRY GUIDES Eustachian Tube Function (ETF) Testing Intact One purpose of the Eustachian tube is to equalize pressure between the middle ear space and ambient pressure. Normally, the Eustachian tube temporarily opens during a swallow or yawn; thereby, allowing an exchange of air between the middle ear and the nasopharynx. Between swallows, slight fluctuations may occur in the pressure level within the middle-ear since the cells which line the middle-ear absorb air from the cavity. If the Eustachian tube should remain closed for an extended period of time, a negative pressure (relative to atmospheric pressure) may develop within the middle-ear. This causes the tympanic membrane to retract inward, thus stiffening the eardrum. (Air pressure is decreased at the rate of 50 mm H2O/hour if the tube remains closed.) In time, fluid may develop within the middle-ear space further stiffening the middle-ear system and reducing the ability of the ossicular chain to conduct sound to the cochlea. Since a malfunctioning Eustachian tube can lead to middle-ear disease and hearing loss, it is helpful to be able to determine the patency of the Eustachian tube in patients who are susceptible to middle-ear problems. Pressure-swallow test The purpose of this test is to try to force the Eustachian tube to open through use of pressure gradients and swallowing on the part of the patient. A series of three tympanograms are obtained under the three different conditions described below. _NOTE In each case, to avoid the effects of hysteresis, the tympanogram is recorded in the SAME sweep direction._ Condition #1: A tympanogram sweep is performed with the normal pressure range selected (+ 200 to - 400 daPa). Condition #2: A positive pressure of 400 daPa is established within the ear canal, and the eardrum is pushed inward. As this membrane moves inward, the volume of the middle-ear space is reduced. This, in turn, causes air that is present within the middle-ear space to be more compressed, so that as the patient swallows and the Eustachian tube opens, more air than normal flows out of the middle ear. When the Eustachian tube closes and the ear canal is no longer subjected to induced positive pressure, less air than normal is present within the middle ear space. In other words, there is negative pressure within the middle ear. Therefore, when the second tympanogram is recorded, the point of peak mobility will be shifted in the negative direction (approximately 15 to 20 daPa) relative to the point of peak mobility recorded during the first tympanogram. Condition# 3 A negative pressure of 400 daPa is established within the ear canal. As the pressure is reduced within the ear canal, the eardrum moves outward. Stressing the membrane outward leads to a temporary increase in middle-ear cavity volume. When the same amount of air is present within a larger volume, the air pressure within the cavity is reduced. Thus, the middle-ear pressure is negative relative to atmospheric pressure. As the patient drinks some water and the Eustachian tube opens, more air than normal will flow into the middle-ear space. When the Eustachian tube closes and the ear canal is vented, a positive pressure condition will exist within the middle ear relative to atmospheric pressure. When the third tympanogram is recorded, the point of peak mobility will be shifted in the positive direction (approximately 15 to 20 daPa) relative to the first tympanogram. _NOTE If the Eustachian tube is functioning properly, a shift in the pressure peak of approximately 15 to 20 daPa in either direction will be observed. However, if the Eustachian tube is malfunctioning, there will be little, if any, observable difference in the pressure peak recorded from Condition #1, to Condition #2, to Condition #3._ Perforated The operator may follow a protocol similar to that outlined by Holmquist for determining patency of the Eustachian tube in a patient with pressure equalization tubes in place or with a perforated eardrum. During the test, positive (or negative pressure is presented to the ear canal/middle-ear space until a pre-selected pressure limit is reached. The purpose of this test is to determine if the Eustachian tube will open as a direct result of this pressure. Specific opening pressure provides some information about the status of the tube (i.e., properly functioning vs. malfunctioning). It is suggested that the operator perform the ETF test with the maximum pressure (+400 daPa) pre-selected. This eliminates the possibility of the Eustachian tube not opening at a lower pressure which might otherwise be forced open closer to the maximum pressure value. If the tube opens as a direct result of the pressure within the ear canal/middle ear space, only a portion of the positive pressure will escape before the tube closes again. _NOTE If negative pressure is used during the test and the tube opens, some air will enter the ear canal/middle-ear, thereby, reducing the amount of negative pressure. To further check the Eustachian tube function while a pressure gradient exists between the ear canal/middle-ear space and the nasopharynx, the patient is asked to swallow some water. If the tube is functioning properly, some air pressure will be released as the patient swallows the water. If the tube is malfunctioning, very little (if any) pressure will be released. Thus, it is possible to record changes in pressure when the Eustachian tube opens and closes in response to swallowing as a function of time._ Home > Education > Testing Guides > Eustachian Tube Function > GRASON-STADLER Products Education Our Approach History Events Cookie Policy Disclaimer Support PRODUCTS AudioStar Pro Pello GSI 18 AMTAS Pro AMTAS Flex TympStar Pro Allegro GSI 39 Corti Audera Pro Novus GSI Suite Contact Corporate Headquarters 10395 West 70th St. Eden Prairie, MN 55344 General Inquiries +1 800-700-2282 +1 952-278-4402 info@grason-stadler.com(US) international@grason-stadler.com(International) Technical Support Hardware: +1 877-722-4490 Software: +1 952-278-4456 DISTRIBUTOR LOGIN GSI Extranet Request an Account Forgot Password? Get Connected Next Steps Distributor Locator Request Quote © Copyright 2025
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https://www.bbc.co.uk/bitesize/guides/z264jxs/revision/6
National 4 Comparing data setsFinding the median from a frequency table Data sets can be compared by looking at their similarities and differences. This can be done by first calculating an average and a measure of spread for reach. Part of Application of MathsStatistics Save to My Bitesize Finding the median from a frequency table Here are Kieran’s results in a frequency table | | | --- | | Number of tracks on album | Frequency | | 9 | 1 | | 10 | 4 | | 11 | 3 | | 12 | 3 | | 13 | 0 | | 14 | 1 | | | | --- | | Number of tracks on album | 9 | | Frequency | 1 | | | | --- | | Number of tracks on album | 10 | | Frequency | 4 | | | | --- | | Number of tracks on album | 11 | | Frequency | 3 | | | | --- | | Number of tracks on album | 12 | | Frequency | 3 | | | | --- | | Number of tracks on album | 13 | | Frequency | 0 | | | | --- | | Number of tracks on album | 14 | | Frequency | 1 | From the table, we can find the total number of albums (1 + 4 + 3 + 3 + 0 + 1 = 12\, albums) As the results are in a table, they are already ordered for us. In this case, as there are (12) results, the median is between the (6th) and (7th) result. We look at the frequency column and determine when the cumulative frequency passes the (6th) and (7th) results. (There is an explanation of Cumulative Frequency on a previous page in this section) The first row has cumulative frequency (1) The second row has cumulative frequency (5) The third row has cumulative frequency (8) The third row passes the (6th) and (7th) results, so the median must be in there. Median is (11) tracks In general, if there are (n) results, the median will be result ((n+1) \div 2) For example: for five numbers, the median is the ((5+1) \div 2 = 3rd) result. For six numbers, the median is the result between (3rd) and (4th) due to the fact that ((6 + 1) ÷ 2 = 3.5). Question Find the median number of tracks on Suzie's albums. | | | --- | | Number of tracks on album | Frequency | | 6 | 3 | | 7 | 0 | | 8 | 2 | | 9 | 1 | | 10 | 3 | | 11 | 4 | | 12 | 2 | | | | --- | | Number of tracks on album | 6 | | Frequency | 3 | | | | --- | | Number of tracks on album | 7 | | Frequency | 0 | | | | --- | | Number of tracks on album | 8 | | Frequency | 2 | | | | --- | | Number of tracks on album | 9 | | Frequency | 1 | | | | --- | | Number of tracks on album | 10 | | Frequency | 3 | | | | --- | | Number of tracks on album | 11 | | Frequency | 4 | | | | --- | | Number of tracks on album | 12 | | Frequency | 2 | Suzie has (15) albums, so the median is the (8th) result (Remember we can use ((15 + 1) \div 2 )) The cumulative frequency passes the eighth album at the fifth row. The median is (10). Remember, when you are working out the median: Put the results in numerical order (in a frequency table this will already be done) Count the total amount of results and add one Divide this by 2 to find the the position of the middle result Find the middle result in the numerically ordered list or frequency table You will then have the median of the set of results Next page Mode Previous page Median More guides on this topic Statistical diagrams Scatter graphs Video playlist Related links BBC Podcasts: Maths BBC Radio 4: Maths collection BBC Skillswise SQA Lifeskills Maths GOV.UK Maths is Fun NRICH Maths Club
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http://myslu.stlawu.edu/~nkomarov/450/LIP-subgames.pdf
144 Chapter 7. Impartial Games 7.5 Taking-and-Breaking Games There are many natural variations on nim obtained by modifying the legal moves. For example, sometimes a player, in addition to taking counters, might also be permitted to split the remaining heap into two (or sometimes more) heaps. These rule variants yield a rich collection of Taking-and-Breaking games that are discussed in WW [BCG01]. After nim, Taking-and-Breaking games are among the earliest and most studied impartial games; however, by no means is everything known. To the contrary, much of the field remains wide open. Values in games such as grundy’s game (choose a heap and split it into two different sized heaps) and Conway’s couples are forever (choose a heap and split it but heaps of size 2 are not allowed to be split) have been computed up to heaps of size 11 × 109 and 5 × 107, respectively, yet there is no complete analysis for these games. In Taking-and-Breaking variants, the legal moves may vary with the size of the heap and the history of the game. For example, the legal moves might be, “a player must take at least one-quarter of the heap and no more than one-half,” or “a player must take between n/2 and n + 3, where n is the number taken on the last move.” For an example, see Problem 11. Games whose allowed moves depend on the history of the game are typically more difficult to analyze, but when the legal moves are independent of the history (and of moves in other heaps), then the game is a disjunctive sum and we only need analyze games that have a single heap! Definition 7.28. For a given Taking-and-Breaking game G, let G(n) be the nim-value of the game played with a heap of size n. The nim-sequence for the game is G(0), G(1), G(2), . . .. In order to automate the process for finding (and proving) the nim-sequences of selected Taking-and-Breaking games, we need to address two main questions: 1. What types of regularities occur in nim-sequences? 2. When do we know that some regularity observed in a nim-sequence will repeat for eternity? There are three types of regularities that have been observed in many nim-sequences to which we can answer the second question. These are listed in the next definition but we only consider two of the three, those that are periodic and arithmetic periodic, in this book. The reader interested in split-periodicity should read [HN03, HN04]. Definition 7.29. A nim-sequence is • periodic if there is some l ≥0 and p > 0 so that G(n + p) = G(n) for all n ≥l; 7.6. Subtraction Games 145 • arithmetic periodic if there is some l ≥0, p > 0, and s > 0 so that G(n + p) = G(n) + s for all n ≥l;3 and • sapp regular (or split arithmetic periodic/periodic) if there exist integers l ≥0, s > 0, p > 0, and a set S ⊆{0, 1, 2, . . ., p −1} such that for all n ≥l, G(n + p) = ( G(n) if (n mod p) ∈S, G(n) + s if (n mod p) ̸∈S. The subsequence G(0), G(1), . . . , G(l−1) is called the pre-period and its elements are the exceptional values. When l and p are chosen to be as small as possible, subject to meeting the conditions of the definition, we say that l is the pre-period length and p is the period length, while s is the saltus. If there is no pre-period the nim-sequence is called purely periodic, purely arithmetic periodic, purely sapp regular, respectively. Exercise 7.30. Match each sequence on the left one-to-one to a category on the right: 1231451671 . . . periodic 1123123123 . . . purely periodic 1122334455 . . . sapp regular 0123252729 . . . arithmetic periodic 0120120120 . . . purely sapp regular 1112233445 . . . purely arithmetic periodic In each case, identify the period and (when non-zero) the saltus and pre-period. 7.6 Subtraction Games Definition 7.31. A subtraction game denoted subtraction(S), is played with heaps of counters and a set S of positive integers. A move is to choose a heap and remove any number of counters provided that number is in S. • If S = {a1, a2, . . . , ak} is finite, we have a finite subtraction game, which we denote subtraction(a1, a2, . . . , ak). • If, on the other hand, S = {1, 2, 3, . . .}{a1, a2, . . . , ak} consists of all the positive integers except a finite set, we have an all-but subtraction game, denoted allbut(a1, a2, . . . , ak). In Example 7.25, subtraction(1, 2, 4) was shown to be periodic. On the other hand, allbut() is another name for nim and is arithmetic periodic with saltus 1. 3Reminder: + means normal, not nimber, addition! 146 Chapter 7. Impartial Games Finite subtraction games The next table gives the first 15 values of the nim-sequence for several subtrac-tion games subtraction(S). S 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 {1,2,3} 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 {2,3,4} 0 0 1 1 2 2 0 0 1 1 2 2 0 0 1 {3,4,5} 0 0 0 1 1 1 2 2 0 0 0 1 1 1 2 {3,4,6,10} 0 0 0 1 1 1 2 2 2 0 3 3 1 4 0 It is not surprising that the nim-sequence for subtraction(1, 2, 3) is purely periodic with values ˙ 012˙ 3. (Note: we use dots to indicate the first and last values in the period.) The nim-sequence for subtraction(2, 3, 4) is ˙ 00112˙ 2 and ˙ 0001112˙ 2 for subtraction(3, 4, 5). The pattern is not yet evident for subtraction(3, 4, 6, 10), but if we pushed on we would eventually find that the nim-sequence is 00011122203314˙ 0020131˙ 2. Calculating nim-sequences with a Grundy scale The calculation of nim-sequences for subtraction games can be done by a com-puter (CGSuite, Maple or Mathematica, for example, or any spreadsheet pro-gram) but there is an easy, hand technique as well. Use two pieces of graph paper to construct what is known as a Grundy scale. Here, we will work with the example subtraction(3, 4, 6, 10). To begin, take two sheets of lined or graph paper. On each sheet make a scale, with the numbers marked in opposite directions. One of the scales is marked with △s to indicate the positions of numbers in the subtraction set; you can put a ▲to indicate 0. The other sheet will be used to record the nim-values: 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 8 9 10 11 ▲ △ △ △ △ As you fill in nim-values on the top scale, slide the bottom scale to the right. Shown below is the calculation G(9); take the mex of the positions marked by arrows. In this case, mex(1, 1, 2) = 0, so we will fill in a 0 for G(9), now indicated by the ▲: 7.6. Subtraction Games 147 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 0 0 1 1 1 2 2 2 0 0 1 2 3 4 5 6 7 8 9 10 11 ▲ △ △ △ △ Similarly, G(13) = mex(1, 2, 0, 3) = 4: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 0 0 0 1 1 1 2 2 2 0 3 3 1 4 0 1 2 3 4 5 6 7 8 9 10 11 ▲ △ △ △ △ Exercise 7.32. Make a photocopy of the last Grundy scale, make a cut to sep-arate the bottom and top portions, and use it to calculate more of the nim-sequence of subtraction(3, 4, 6, 10). For more practice, try using Grundy scales for the examples in problem 2 until you are comfortable. Periodicity of finite subtraction games After working out a few finite subtraction games, it will come as no surprise that their nim-sequences are always periodic. Theorem 7.33. The nim-sequences of finite subtraction games are periodic. Proof: Consider the finite subtraction game subtraction(a1, a2, . . . , ak) and its nim-sequence. From any position there are at most k legal moves. So, using Observation 7.22, G(n) ≤k for all n. Define a = max{ai}. Since G(n) ≤k for all n there are only finitely many possible blocks of a consecutive values that can arise in the nim-sequence. So we can find positive integers q and r with a ≤q < r such that the a values in the nim-sequence immediately preceding q are the same as those immediately preceding r. Then G(q) = G(r) since G(q) = mex{G(q −ai) | 1 ≤i ≤k} = mex{G(r −ai) | 1 ≤i ≤k} = G(r). In fact, for such q and r and all t ≥0, G(q + t) = G(r + t). This is easily shown by induction — we have just seen the base case, and the inductive step is really just an instance of the base case translated t steps forwards. Now set l = q and p = r −q and we see that the above says that for all n ≥l, G(n + p) = G(n); that is, that the nim-sequence is periodic. 148 Chapter 7. Impartial Games This proof shows that the pre-period and period lengths are at most (k+1)a. However, this is generally a wild overestimate, and using the following corollary the values of the period and pre-period lengths can usually be determined by computer: Corollary 7.34. Let G = subtraction(a1, a2, . . . , ak) and let a = max{ai}. If l and p are positive integers such that G(n) = G(n + p) for l ≤n < l + a, then the nim-sequence for G is periodic with period length p and pre-period length l. Proof: See Problem 5. That is, given conjectured values of l and p, it suffices to inspect the values of G(n) for n ∈{l, l + 1, . . . , l + p + a −1} to confirm the periodicity! It is then rote for a computer to identify the smallest pre-period and period given by the corollary. Applying the corollary to the games in Table 7.6 we see that for: subtraction(1, 2, 3) we have l = 0, p = 4 and a = 3, and these values can be confirmed by inspection of G(n) for n ∈{0, . . ., 6}; subtraction(2, 3, 4) l = 0, p = 6 and a = 4, inspect G(n) for n ∈{0, . . . , 9}; subtraction(3, 4, 5) l = 0, p = 8 and a = 5, inspect G(n), n ∈{0, . . ., 12}; and subtraction(3, 4, 6, 10) l = 14, p = 7 and a = 10, inspect G(n), n ∈ {14, . . ., 30}. Exercise 7.35. Use a Grundy scale to compute values of subtraction(2, 4, 7) until you have enough to apply Corollary 7.34. For more practice, see Problem 2. Two of the main questions, which still attract researchers, are: 1. As a function of the ai, how long can the period of subtraction(a1, a2, . . . , ak) be? 2. Find general forms for the nim-sequence for subtraction(a1, a2, a3). Many games with different subtraction sets are actually the same in the sense that they have the same nim-sequence. For example, in the game subtraction(1), the first player wins precisely when there is an odd number of counters in the heap. The odd-sized heaps only have moves to even-sized heaps; even-sized heaps only have moves to odd-sized heaps and the end posi-tion consist of heaps of size 0; that is, even. Therefore, a heap with n-counters is in N if n is odd, otherwise it is in P. Adjoining any odd numbers to the subtraction set does not change this argument and it is easy to show that the nim-sequence of any game subtraction(1, odds) is ˙ 0˙ 10101. A more general version of this analysis is sufficient to prove: 7.6. Subtraction Games 149 Theorem 7.36. Let G = subtraction(a1, a2, . . . , ak) be purely periodic with period p. Let H = subtraction(a1, a2, . . . , ak, a1 + mp) for m ≥0, then G and H have the same nim-sequence. Proof: See Problem 6. All-but-finite subtraction games The next table gives the first 15 values of the nim-sequence for allbut(S) for several sets S: S 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 {1,2,3} 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 {2,3,4} 0 1 0 1 0 1 2 3 2 3 2 3 4 5 4 {1,2,8,9,10} 0 0 0 1 1 1 2 2 2 3 0 3 4 1 4 Grundy scales can still be used for the all-but subtraction games. This time, the small arrows mark the heaps that are not options. Exercise 7.37. Use a Grundy scale to find the first 20 terms in the nim-sequence of allbut(1, 3, 4). These values certainly do not look periodic, for G(n) appears to steadily increase with n, as confirmed by the following lemma: Lemma 7.38. Let G = allbut(a1, a2, . . . , ak) and a = max{ai}. Then G(n + t) > G(n) for all t > a. Proof: Since any option from n is also an option from n + t, G(n + t) ≥G(n). Additionally, n is an option from n + t, so G(n) occurs as the nim-value of an option from n + t and thus we do not have equality. However, as we will see, all-but subtraction games are arithmetic periodic. From the table above, we might conjecture the following (we denote the saltus information in parentheses with a + sign): S nim-sequence {1,2,3} ˙ 000˙ 0(+1) {2,3,4} ˙ 01010˙ 1(+2) {1,2,8,9,10} 00011122˙ 23˙ 0(+1) Proving a nim-sequence is arithmetic periodic is typically more difficult than proving periodicity. Nonetheless, we can parallel the work done for finite subtraction games, proving first that allbut subtraction games are arithmetic periodic, and then identifying how one (a person or a computer) can automat-ically confirm the arithmetic periodicity. 150 Chapter 7. Impartial Games Theorem 7.39. Let G = allbut(a1, a2, . . . , ak). Then the nim-sequence for G is arithmetic periodic. Proof: Lemmas 7.41 and 7.43, motivated and then proved below, yield the theorem. The proof of this theorem is more technical than that of Theorem 7.33, but in broad strokes is similar. In the proof of Theorem 7.33, we first argued that nim-values cannot get too large, and that therefore some sufficiently long sequence of nimbers must repeat. Once a sufficiently long sequence appears identically twice, an induction argument establishes that those two sequences remain in lock-step ad infinitum. With arithmetic periodicity, the repetition we seek is in the shape of a sequence rather than its values as shown in Figure 7.1 on page 152. As you read through the proofs, keep in mind that two subsequences of nim-values, call them (G(n0), . . . , G(n0 + c)) and (G(n′ 0), . . . , G(n′ 0) + c) have the same shape if 1. the two subsequences differ by a constant: G(n′ 0) −G(n0) = G(n′ 0 + 1) −G(n0 + 1) = · · · = G(n′ 0 + c) −G(n0 + c), 2. or equivalently, both subsequences move up and down the same way: for all 0 ≤i < c, G(n0 + i + 1) −G(n0 + i) = G(n′ 0 + i + 1) −G(n′ 0 + i). It will turn out that the base case for our inductive proof will require a repetition of length about 2a, where a = max{ai}. We show that such a repetition exists in the next two lemmas. Lemma 7.40. Let G = allbut(a1, a2, . . . , ak), and define a = max{ai}. For all n ≥a, k −a ≤G(n + 1) −G(n) ≤a −k + 1. Proof: Fix n > a and let X ⊆{G(0), G(1), G(2), . . . , G(n −1)} be the nim-values of the options of n.4 Now, since G(n) is the mex of X, we know that {0, 1, 2, . . ., G(n) −1} ⊆X. Further, play in G = allbut(a1, a2, . . . , ak) pro-hibits moves to k of the top a heap sizes. Hence, one of {G(0), . . . , G(n−a−1)}, say G(m), must be at least G(n) −1 −(a −k), for only a −k of the terms from X can appear among {G(n −a), . . . , G(n) −1}. Further, m and all the options from m are also options from n + 1. So we have G(n + 1) > G(m), and so G(n + 1) ≥ G(n) −(a −k). 4In other words, X = {G(n −α) | α ̸∈{a1, . . . , ak}}. 7.6. Subtraction Games 151 Similarly, for the second inequality, one of {G(0), . . . , G(n −a −1)} is at least G(n + 1) −2 −(a −k), and it and its options are options of n. So, G(n) ≥ G(n + 1) −(a −k) −1. Lemma 7.41. Let G = allbut(a1, a2, . . . , ak) and a = max{ai}. There exist n0, n′ 0, s, and p = n′ 0 −n0 > 0 such that G(n + p) −G(n) = s for n0 ≤n ≤ n0 + 2a. Proof: By Lemma 7.40, for all n, G(n + 1) −G(n) must lie between k −a and a −k + 1. But there are only 2(a −k) + 2 values in that range. Hence, setting c = 2(a −k) + 2, there are at most c2a possible sequences of the form ( G(n + 1) −G(n), G(n + 2) −G(n + 1), . . . , G(n + 2a) −G(n + 2a −1) ) (7.1) and so eventually, for two values n = n0 and n = n′ 0, the two corresponding sequences are identical. The lemma follows. Exercise 7.42. Complete the algebra to confirm, “The lemma follows.” In par-ticular, given the two identical sequences satisfying (7.1), one with n′ 0, and one with n0, you need to define p and explain why G(n + p) −G(n) is a constant (call it s = G(n′ 0) −G(n0)) for n0 ≤n ≤n0 + 2a. (The matching shapes in Figure 7.1 provide some intuition.) The next lemma completes the inductive step of the proof of the theorem showing that once two sufficiently long sequences have the same shape, they are fated to continue in lock step. Lemma 7.43. Let G = allbut(a1, a2, . . . , ak) and a = max{ai} and suppose G(n + p) −G(n) = s for n0 ≤n ≤n0 + 2a. Then G(n + p) −G(n) = s for all n ≥n0. In particular, G(n) has pre-period length l = n0, period p, and saltus s, and one can identify and confirm the period and saltus by only inspecting the first l + 2a + p + 1 values; i.e., G(n) for 0 ≤n ≤l + 2a + p. Proof: Using induction, it suffices to prove G(n + p) −G(n) = s for n = n0 + 2a + 1. Define the following quantities as shown in Figure 7.1: n′ 0 = n0 + p, n1 = n0 + a, n2 = n0 + 2a, n′ 1 = n′ 0 + a, and n′ 2 = n′ 0 + 2a. As we compute G(n′) = G(n+p) as the mex of the nim-values of its options, by Lemma 7.38, G(n′) exceeds all G(m) for m < n′ −a −1 = n′ 1. And we 152 Chapter 7. Impartial Games b b b b b b b b b b b b b b b b b b n′ n n′ 0 n′ 1 n′ 2 n0 n1 n2 n G(n) a a a a p s Figure 7.1. A diagram sketching the proof of arithmetic periodicity in allbut subtraction games. For the game allbut(a1, . . . , ak), a = max{ai}, s is the proposed saltus, p is the proposed period. We first find two sufficiently long sequences of nim-values with the same shape (but translated upward), and inductively prove that those sequences remain in lock-step. know G(n′) > G(n′ 1) in any event, so we can safely ignore G(m) for m < n′ 0. In other words, G(n′) is the minimum excluded value from {G(n′ 0), . . . , G(n′ 2)} that exceeds G(n′ 1). Since the assignment of G(n′) is unaffected by linear translation of those nim-values, G(n′) −G(n) = s. This last lemma gives an automated method for testing when an all-but subtraction game nim-sequence has become arithmetic periodic. Although Fig-ure 7.1 shows the two subsequences non-overlapping (suggesting that p > 2a), the proof is unaffected by overlap. Exercise 7.44. Apply Lemma 7.43 to find the period length p and the saltus s of the game from Exercise 7.37. In particular, how many values of G(n) need com-puting to confirm the period and saltus? (Hint: The game is purely arithmetic periodic with period between 10 and 15.) 7.6. Subtraction Games 153 Exercise 7.45. We asserted at the start of this section that the nim-sequence for allbut(1, 2, 8, 9, 10) is given by 00011122˙ 23˙ 0(+1). To apply Lemma 7.43, which values of G(n) need be confirmed to be confident of the nim-sequence? As was seen in the table on page 149, some allbut subtraction games have pre-periods. If you do Problems 12, 13, and 14. then you will have shown that the nim-values of allbut games where s has cardinality 1 and 2 are purely arithmetic-periodic. Frequently, the allbut subtraction set can be reduced. While most such reductions remain specific to individual games, we do have one general reduc-tion theorem. Theorem 7.46. Let a1 < a2 < · · · < ak be positive integers, and let b > 2ak. Then, the nim-sequences of allbut(a1, a2, . . . , ak) and allbut(a1, a2, . . . , ak, b) are equal. Proof: Let G(n) denote the nim-sequence of allbut(a1, a2, . . . , ak) and G′(n) that of allbut(a1, a2, . . . , ak, b). Certainly, G(n) = G′(n) for n < b since the options in the two games are identical to that point. Suppose induc-tively that the two nim-sequences agree through n −1, and consider G(n) and G′(n). Since the options of allbut(a1, a2, . . . , ak, b) are a subset of those of allbut(a1, a2, . . . , ak), the only possible way to have G(n) ̸= G′(n) would be if G′(n) = G(n −b), since the latter is the only possible value that does not occur among the options of allbut(a1, a2, . . . , ak, b) but does occur among the options of allbut(a1, a2, . . . , ak). In order for this to be true it would also be the case that no option of n in allbut(a1, a2, . . . , ak, b) had value G(n −b). However, consider m = n −b + ak. By the inductive hypothesis G(m) = G′(m). Moreover, all values smaller than n −b are options from a heap of size m in allbut(a1, a2, . . . , ak), so G(m) ≥G(n −b). As m is an option of n in allbut(a1, a2, . . . , ak, b), in order to avoid a contradiction we would require that G(m) > G(n −b). But then, m would have an option m′ in allbut(a1, a2, . . . , ak) with G(m′) = G(n −b). Since m′ ̸= n −b and m′ < n−ak, m′ is also an option of n in allbut(a1, a2, . . . , ak, b). This contradiction establishes the desired result. For more on all-but-finite subtraction games see [Sieg06]. Kayles and kin kayles is played with a row of pins standing in a row. The players throw balls at the pins. The balls are only wide enough to knock down one or two adjacent pins. † † † † † † † † † † Above is a game in progress. Would you like to take over for the next player? 154 Chapter 7. Impartial Games kayles can also be thought of as a game played with heaps of counters and a player is allowed to take one or two counters and possibly split the remaining heap into two. kayles is clearly a Taking-and-Breaking game but maybe it is too hard right now. We will work up to it with a couple of simpler variants. First, suppose we play a game where a player is only allowed to split-the-heap into two non-empty heaps; no taking. We saw in Example 1.10 of Chapter 1, that this is she loves me she loves me not disguised. That is, G(n) =      0 if n = 0, 0 if n > 0 is odd, 1 if n > 0 is even. In particular, moves from an even heap are to even + even = ∗+ ∗= 0 or odd + odd = 0, while moves from an odd heap are to even + odd = ∗+ 0 = ∗. The version where you are allowed to split-or-take-1-and-must-split (into non-empty heaps) needs a bit more care. For a heap of n, the splitting move gives the options n −i and i for i = 1, 2, . . . , n −1 and the take-1-and-split move leaves the positions n −1 −j and j, j = 1, 2, . . ., n −2. To find the value of the game, we need to find the mex of {G(i) ⊕G(n −i) | i = 1, 2, . . ., n −1} ∪ {G(j) ⊕G(n −1 −j) | j = 1, 2, . . ., n −2}. The first few values are G(0) = 0, G(1) = 0, G(2) = mex{G(1) + G(1)} = 1, G(3) = mex{G(2) ⊕G(1)} ∪{G(1) ⊕G(1)} = 2. This becomes tedious very quickly and a machine will be very useful. If one is not available then a Grundy scale can help. This time both pieces of paper have the nim-values, one from left-to-right as before, the other has them from right-to-left. However, we have to use them twice and record the intermediate values. The next diagram shows the calculation of G(9). First, line up the scales for the split move, 8 + 1, 7 + 2, . . ., 1 + 8. Clearly we only need go to 4 + 5 since we repeat the calculations in reverse. Nim-sum the pairs of numbers and record: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 0 1 2 3 1 4 3 2 0 1 2 3 4 5 6 7 8 9 10 11 0 0 1 2 3 1 4 3 2 This gives {2, 2, 6, 2, 2, 6, 2, 2} as the set of values for these options. Now line them up for the take-one-and-split move. Problems 155 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 0 1 2 3 1 4 3 2 0 1 2 3 4 5 6 7 8 9 10 11 0 0 1 2 3 1 4 3 2 This gives the set {3, 5, 3, 0, 3, 5, 3} for these options. The least non-negative number that does not appear in either set is 1, so record G(9) = 1 on both papers: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 0 1 2 3 1 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 0 0 1 2 3 1 4 3 2 1 Exercise 7.47. Continue using the Grundy scale to compute five more nim-values of the last game where you are allowed to split-or-split-and-take-one-but-always-leave-two-heaps. For a fixed (finite) set S of positive integers, we can define the variant splittles(S) where a player is allowed to take away s, for some s ∈S, from a heap of size at least s and possibly split the remaining heap. In particular, kayles is just splittles(1, 2). What sort of regularities could the nim-sequences of splittles(S) have? In the few examples so far, the nim-values do not grow very quickly at all suggesting that the nim-sequences are periodic but Nobody Knows. All we know is that the nim-sequences are not arithmetic periodic [Aus76, BCG01]. It is believed that splittles(S) is periodic when S is finite, and for periodicity we do have an automatic check. Theorem 7.48. Fix splittles(S) with m = max S. If there exists integers, l ≥0 and p > 0 such that G(n + p) = G(n) for l ≤n ≤2l + 2p + s, then G(n + p) = G(n) for all n ≥l. That is, the period persists forever. Proof: See Problem 15. Problems 1. Find the nim-sequences for subtraction(S), where |S| = 2 and S ⊆ {1, 2, 3, 4}. 156 Chapter 7. Impartial Games 2. Use a Grundy scale and Corollary 7.34 to compute the nim-sequences of (a) subtraction(2, 3, 5); (b) subtraction(3, 5, 8); (c) subtraction(1, 3, 4, 7, 8). 3. Find the period for (a) allbut(1, 2, 3); (b) allbut(5, 6, 7); (c) allbut(3, 4, 6, 10). 4. A game is played like kayles, only you cannot bowl the end of a row of pins. In nim language, you can take one or two counters from a heap and you must split that heap into two non-empty heaps. Using a Grundy scale, calculate the first 15 nim-values for this game. 5. Prove Corollary 7.34 on page 148. 6. Prove Theorem 7.36 on page 149. 7. (This is a generalization of nim.) polynim is played on polynomials with non-negative coefficients. A move is to choose a single polynomial and reduce one coefficient and arbitrarily change or leave alone the coefficients on the smaller powers in this polynomial — 3x2 +15x+3 can be reduced to 0x2 + 19156x + 2345678 = 19156x + 2345678. Analyze polynim. In particular, identify a strategy for determining when a position is a P-position analogous to Theorem 7.12. 8. Find the nim-sequences for subtraction(1, 2q) for q = 1, 2, 3. Find the form of the nim-sequence of subtraction(1, 2q) for arbitrary q. 9. Show that the nim-sequence for subtraction(q, q+1, q+2) is ˙ 00q−11q2˙ 2 if q > 1. (As usual, xb is x repeated b times.) 10. Analyze subtraction(1, 2, 4, 8, 16, . . ., 2i, . . .). 11. Analyze this variant of nim: On any move a player must remove at least half the number of counters from the heap.5 12. Find the periods for allbut(1), allbut(2), and allbut(3). Conjecture and prove your conjecture for the period of allbut(q). 5The nim-sequence for the game in which no more than half can be removed has a re-markable self-similarity property: If you remove the first occurrence of each number in the nim-sequence, then the resulting sequence is the same as the original! See [Lev06]. Problems 157 13. Find the periods for allbut(1, 2), allbut(2, 3), and allbut(3, 4). Con-jecture and prove your conjecture for the period of allbut(q, q + 1). 14. Find the nim-sequence for allbut(q, r), q < r. (Hint: There are two cases r = 2q and r ̸= 2q.) 15. Prove Theorem 7.48 on page 155. 16. The rules of the game turn-a-block are at the textbook website, www. lessonsinplay.com, and you can play the game against the computer. De-termine a winning strategy for turn-a-block. You should be able to consistently beat the computer on the hard setting at 3 × 3 and 5 × 3 turn-a-block (and even bigger boards!). You should be able to determine who should win from any position up to 5 × 5.
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CCSS.Math.Content.HSA.APR.A.1 Understand That Polynomials Form A Discover more Mathematics Math Assessment Educational assessment Educational Education School supplies assessments Sign Up For Our Newsletter Email: Subscribe | Newsletter Index | Daily Dose | About | Site Map Home| SAT/ACT| Common Core| Online Practice| Printables| Grade Level Help| Links PreK-12| Tech| Assessment I4C CCSS.Math.Content.HSA.APR.A.1 Understand That... Home>Common Core>Mathematics>High School Algebra>Arithmetic with Polynomials & Rational Expressions>Understand That Polynomials Form A System Analogous To The Integers,... CCSS.Math.Content.HSA.APR.A.1 - Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials. Authors: National Governors Association Center for Best Practices, Council of Chief State School Officers Title: CCSS.Math.Content.HSA.APR.A.1 Understand That Polynomials Form A System Analogous To The Integers,... Arithmetic with Polynomials & Rational Expressions - High School Algebra Mathematics Common Core State Standards Publisher: National Governors Association Center for Best Practices, Council of Chief State School Officers, Washington D.C. Copyright Date: 2010 (Page last edited 10/08/2017) Add and Subtract Polynomials - Add and subtract polynomials Adding and Subtracting Polynomials - Adding and Subtracting Polynomials 3 Adding and Subtracting Polynomials - Adding and Subtracting Polynomials Adding and Subtracting Polynomials 2 - Adding and Subtracting Polynomials 2 Adding Polynomials - Polynomial Basics: Adding Polynomials Adding Polynomials - Add polynomials to find perimeter Adding polynomials with multiple variables - Basic example of simplifying a polynomial expression with multiple variables. Addition and Subtraction of Polynomials - Sixteen minute video lesson Addition and Subtraction of Polynomials - Addition and Subtraction of Polynomials Arithmetic with Polynomials - Worksheet 1 - 10 problems are presented in a PDF worksheet to test the student's knowledge of understanding that polynomials form a system. Arithmetic with Polynomials - Worksheet 1 - Answer Key - An answer key is provided in PDF form showing answers for the PDF worksheet that tests the student's knowledge of understanding that polynomials form a system. Dividing polynomials with remainders - Dividing polynomials with remainders Factoring Polynomials - Factor polynomials that are not perfect squares, where A=1 Factoring Polynomials: 2 - Factoring Polynomials lesson 2 Factoring trinomials with a common-factor - Example of factoring trinomials with a common-factor FOIL method for multiplying binomials - FOIL method for multiplying binomials example 2 More multiplying polynomials - More multiplying polynomials Multiply Polynomials - Multiply two polynomials using algebra tiles Multiplying binomials - Multiplying binomials to get difference of squares Multiplying binomials 1 - Multiplying binomials example 1 Multiplying binomials example problem - Multiplying binomials word problem Multiplying Monomials - Multiplying Monomials Multiplying Monomials by Polynomials - Multiplying Monomials by Polynomials Multiplying Polynomials - Multiplying Polynomials Multiplying Polynomials 3 - Multiplying Polynomials 3 Polynomial divided by monomial - Polynomial divided by monomial Polynomial Division - Polynomial Division Polynomials - Multiply polynomials Polynomials - Multiply polynomials Polynomials - Multiply two binomials Polynomials - Multiply a polynomial by a monomial Polynomials - Add and subtract polynomials Polynomials - Model polynomials with algebra tiles Polynomials 1 - Polynomials 1 Polynomials with algebra tiles - Add and subtract polynomials using algebra tiles Simplifying Polynomials - simplifying polynomials and algebraic expressions Special Cases of Polynomials - Multiply two binomials: special cases Subtracting Polynomials - Short video lesson Subtracting Polynomials with Multiple Variables - Short video lesson Subtracting polynomials with multiple variables - Subtracting polynomials with multiple variables Understand that Polynomials Form a System - Students should understand that polynomials, like integers, are "closed" when it comes to addition, subtraction, and multiplication. 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To learn more, view the following link: Privacy Policy Open navigation menu Upload 0 ratings0% found this document useful (0 votes) 385 views5 pages 农历干支 本文深入解读干支纪时系统的编算规则，构建了干支纪年、纪月、纪日的计算系统，以提高计算速度和简化推算方法。通过对公式的构建和解析，确保干支纪时的科学性和准确性，使其更好地服务于民众的生产和生活。文中还探讨了干支纪年与公历年之间的转换关系及其在历史研究中的应用。 Uploaded by azure.aoi2 You are on page 1/ 5 2023 年 第 12 期 农历是我国正式行使的传统历法。六十干支周是农历的组成部分，它在生产、生活、科研等领域具有广泛的实用价值和不可替代的重要作用。然而，当前一些日历产品对于干支纪时的历法属性表述各异，农历月与干支月、农历年与干支年编排不一致，干支月推算方法繁琐，给出的一些公式互不相同且缺少对公式意义的解析，引起了公众的困惑和质疑。因此，为了保证干支纪时编算的科学性和准确性，有必要以国家标准《农历的编算和颁行》为基准，深入解读干支纪时的原理和编排规则，并构建完善的纪年纪月纪日计算系统，从而使干支纪时系统更好地服务于民众的生产和生活。 六十干支周是把甲、乙……癸等十个天干和子、丑……亥等十二个地支各循序取一字相配成一对干支，形成由甲子、乙丑……癸亥共六十对干支组成的计序符号系统，循环使用，通常用于纪年、纪月和纪日。如表1所示，如果今年是甲子年，明年就是乙丑年，60年以后又是甲子年。纪月（或纪日）也是如此，以60月（或60日）为周期，循环使用。六十干支周长度稳定，且行用几千年也未中断和改变，因此在计序、纪时、修订历史文献纪日误差、整理古代天文纪录等方面都有着不可替代的作用。下面就依次讨论干支纪年、纪日和纪月三种纪时系统。 陈 亮 （1.华东师范大学 地理信息科学教育部重点实验室, 上海 200241; 2.华东师范大学 地理科学学院, 上海 200241; 3.自然资源部超大城市自然资源时空大数据分析应用重点实验室, 上海 200063; 4.唐山市开滦第一中学, 河北 唐山 063000） 本文基于国家标准《农历的编算和颁行》，深入解读干支纪时系统的编算规则并构建了干支纪年纪月纪日公式。本文通过对公式的构建和对证明过程的解析，全面阐释干支纪时系统的编历原理，建立了完整的干支纪年纪月纪日的计算系统，加快计算速度，简化推算方法，使干支纪时系统能更好地服务于民众的生产和生活。       表1 六十干支周 序号 干支 序号 干支 序号 干支 序号 干支 序号 干支 序号 干支1 甲子 11 甲戌 21 甲申 31 甲午 41 甲辰 51 甲寅2 乙丑 12 乙亥 22 乙酉 32 乙未 42 乙巳 52 乙卯3 丙寅 13 丙子 23 丙戌 33 丙申 43 丙午 53 丙辰4 丁卯 14 丁丑 24 丁亥 34 丁酉 44 丁未 54 丁巳5 戊辰 15 戊寅 25 戊子 35 戊戌 45 戊申 55 戊午6 己巳 16 己卯 26 己丑 36 己亥 46 己酉 56 己未7 庚午 17 庚辰 27 庚寅 37 庚子 47 庚戌 57 庚申8 辛未 18 辛巳 28 辛卯 38 辛丑 48 辛亥 58 辛酉9 壬申 19 壬午 29 壬辰 39 壬寅 49 壬子 59 壬戌10 癸酉 20 癸未 30 癸巳 40 癸卯 50 癸丑 60 癸亥 Download to read ad-free 高中地理 32 2023 年 第 12 期 1.干支纪年系统 纪年是干支纪时系统最常见的一种应用。研究我国历史需要解读史书，并将史书中的干支纪年换算为公历年，这就需要构建公式进行计算。《农历的编算和颁行》中规定了干支纪年的循环参考时间：对应于公历1984年2月2日（春节）0时起到1985年2月19日（除夕）24时截止的农历年为甲子年。本文以此甲子年为计算起点，构建干支纪年与公历年之间的转换关系。1984年为甲子年，定其年干支序号A=1；1985年为乙丑年，则A=2；余类推。年干支序号和公元年都是以“1”为公差的等差数列。为了更清楚地展示公元年与干支序号之间的关系，以公元年Y为横坐标，年干支序号A为纵坐标，画出A与Y函数关系的图像，如图1所示。该图像由许多排列均匀而孤立的点组成，这些点均在斜率为1的相互平行的直线段（图中细点虚线段）上；直线段上满足整数对(Y,A)点的坐标就是某公元年份及其对应的农历年干支序号。 图1 公元年份与其对应农历年干支序号的关系 在图1中，点P 1 (1984,1)、P 2 (1985,2)在直线段L上，易得直线段L的方程为A=Y-1983。因为年干支以60年为周期循环纪年，将上式用同余式表示， 两边对模60同余，即A≡(Y-1983)(mod 60)≡(Y-3)(mod 60)，得某一公元年Y A.D. 的农历干支纪年序号A(Y A.D. )公式： A(Y A.D. )=(Y A.D. -3)(mod 60) （2-1）即某一公元年的干支序号为其年份减去3的差对60的余数。根据公式（2-1），可以快速准确地求算任意公元年的干支序号。例如，公元2023年的干支序号A(2023)=(2023-3)(mod 60)=40，该年为癸卯年；公元1911年的干支序号A(1911)=(1911-3)(mod 60)=48，该年为辛亥年；公元1894年的干支序号A(1894)=(1894-3)(mod 60)=31，该年为甲午年。 需要注意的是，公元纪年没有“0”年。为了使正负运算统一，公元前某年Y B.C. 的干支公式可表示为（Y B.C. 用负数来表示，即公元前一年记为“-1”，前二年记为“-2”，余类推）： A(Y B.C. )=(Y B.C. -2)(mod 60) （2-2）即公元前某年的干支序号为其年份（用负数表示）减去2的差对60的余数。例如，孔子生于公元前551年，其年干支序号A(-551)=(-551-2)(mod 60)=47，即孔子出生于庚戌年。根据上述公式计算：Y=2，A=59；Y=1，A=58；Y=-1，A=57；Y=-2，A=56；余类推。公元前和公元的干支年连续衔接。根据《农历的编算和颁行》规定，干支纪年从每年的农历正月初一北京时间0时开始。 2.干支纪日系统 干支纪日是按顺序用六十干支命名，从甲子日、乙丑日……到癸亥日，六十日一个循环，周而复始。即公元日期每增加1日，农历日干支也增加1日。干支纪日不仅是迄今世界上已知历史最长的一种纪日和计序系统，还是不同历法之间相互转换的重要工具。 1)公式的构建 (1)公式的构建为了方便计算，下面构建干支纪日与公历日换算的关系式。如图2所示，图中展示了时间轴（横轴）上某一日期的干支纪日序号的推算过程。对于需要进行换算的某一日期，Y为其所在年（用四位数表示），m为其所在月，d为其在此月中的日期，O为计算起点，即公元前1年（根据上文公元前年份的规定，其年份记为“-0001”）的岁末，A为所求年Y的前一年岁末，B为所求月m的前一月月末，E为所求年Y的岁末，[x]为下取整函数。 图2 年月日各时段天数及其相互关系 对起算点O，按扁平公历推算其日干支序号D O =15，即日干支序号计算的起点。自O起计量到点d截止的间隔天数为|Od|。因为干支纪日以60日为循环周期，所以用60除|Od|，得整数商q和余数r，即r=|Od|-60q。令|OP|=60q，即|OP|能被60整除，则点P的日干支序号也是D O 。因为公历日每增加1日，干支日也相应增加1日，所以余数r就是点d离开点P的日数，则点d的日干支序号D(d)=D O +r=D O +(|Od|-60q)=D O +|Od|(mod 60)。由图2可以看出，|Od|=|OA|+|AB|+|Bd|=(|OE|-365)+(AB)+d，其中|OE|=365Y+[Y/4]-[Y/100]+[Y/400]，|AB|=30m+[0.6(m+1)]-33。代入上式得D(d)=D O +(365Y+[Y/4]-[Y/100]+[Y/400]+30m+[0.6(m+1)]-398+d)(mod 60)。设年份Y的前两位数为c，后两位数为y，则c=[Y/100]，y=Y-100c，则 Download to read ad-free 高中地理 33 2023 年 第 12 期 Y=100c+y。代入上式并化简，得公历日期的农历日干支序号公式（适用于公元1582年格里历开始应用之后）：D G (d)=([c/4]-16c+5y+[y/4]+30m+[0.6(m+1)]+d-23)(mod 60) （2-3）儒略历是公历的前身，根据其置闰规定和日期编排规则，仿照公式（2-3）的推导思路，可推导出儒略历的干支纪日序号公式： D J (d)=(-15c+5y+[y/4]+30m+[0.6(m+1)]+d-25)(mod 60) （2-4）上述公式中，1、2月要分别看作上一年的13、14月来计算。用（2-4）式减去（2-3）式，就得出上述两种历法的日期差公式： ⊿ d=c-[c/4]-2 （2-5） 2)公式的应用 (2)公式的应用利用公式（2-3）和（2-4）可以准确且方便地将公历或儒略历中的日期换算为干支纪日。例如，开国大典日（公历1949年10月1日）的干支序号D G (d)=([19/4]-16×19+5×49+[49/4]+30×10+[0.6(10+1)]+1-23)-60×4=1，即干支序号为1，该日为农历甲子日，与《农历的编算和颁行》中规定的干支纪日的循环参考时间吻合。 又如，牛顿的出生日期为公历1643年1月4日，根据公式（2-3）可算得该日为农历庚戌日；其儒略历日期为1642年12月25日，根据公式（2-4）亦可算得该日为农历庚戌日。根据公式（2-5）算得 ⊿ d=10，即由于公历和儒略历的不同，同一天在这两种历法中相差10日。 类似地，俄国十月革命发生在公历1917年11月7日，儒略历1917年10月25日，根据公式（2-3）和（2-4）都可算得该日为癸丑日，根据公式（2-5）算得 ⊿ d=13，即将儒略历日期转换为公历日期时，需要将其加上13日得到公历日期。由此可见，干支纪日系统可以将儒略历和公历联系起来。以日干支作为参考，可以计算两种历法之间的日期差，更方便地进行二者之间的转换。 3.干支纪月系统 干支系统也可以用于纪月。学术界对于干支纪月系统的历法属性有两种不同的解读，分别是阳历版本和阴阳历版本。下面以《农历的编算和颁行》为参照标准，解释两种版本的制历原理，拟定制历规则，并构建通项公式简化干支纪月的计算。 1)阳历版本的干支纪月系统 (1)阳历版本的干支纪月系统①原理及编算规则阳历版干支纪月系统的原理相对简单。其以二十四节气即回归年长度定年长和年中的月数，用二十四节气划分出12个干支月，以干支定月序，每月包含2个节气，不设闰月。拟制定的干支月编算规则如下。（A）立春日为干支年首日。（B）包含节气立春在内的干支月为干支年首月。（C）12个节气日(非中气，下同)分别是12个干支月首日。（D）干支月长度是从某个公历月的节气日（含）起计量到次月的节气日（不含）截止的时间间隔。（E）十二地支与12个公历月一一对应，其中子、丑……戌、亥分别对应于12、1……10、11月，子月分别对应农历十一月和公历12月，因为三者均包含中气冬至在内，余类推。（F）十天干与十二地支相配，60月（即5年）一周期，循环使用。②公式的构建和计算仿照公式（2-3）的推导方法，得所求m月的干支序号公式为：M(m)=(M O +((Y-1800)×12+(m-1)))(mod 60)。上式中，计算起点O是公元1800年1月干支丁丑月（M O =14），即所求月干支序号等于自计算起点到所求m月截止的间隔月数加上计算起点的月干支序号的和除以60的余数。将上式化简得M(m)=(12Y+m+13)(mod 60)=(12(100c+y)+m+13)(mod 60)，即：M(m)=(12y+m+13)(mod 60) （2-6）例如，2023年立秋日8月8日的月干支序号M=(12×23+8+13)(mod 60)=57，即该月为庚申月。8月8日立秋日（即当月上半月的节气日）是庚申月首日，庚申月从8月8日起到9月7日截止，9月8日白露日（即次月上半月的节气日）则是辛酉月首日。当月上半月8月8日立秋日（不含）之前的公历日，属于庚申月的前一个干支月范围，因此月干支序号应减去“1”，即M=57-1=56，为己未月。进一步还可分解求得月天干和月地支的序号。天干以10为循环周期，则年份Y（以四位数表示）、Y的前两位数c、后两位数y、个位数u之间的函数关系用同余式可表示为Y≡100c+y≡y≡u(mod 10)。10和12都是60的因数，由公式（2-6）得月天干序号M C (m)和月地支序号M T (m)分别为：M C (m)=(12y+m+13)(mod 10)=(2u+m+3)(mod 10) （2-7）M T (m)=(m+1)(mod 12) （2-8）需要指出的是，干支纪月系统是名副其实的阳历，其编历核心以二十四节气为基准，它标记和度量视太阳在黄道上的位置和时刻以及周年视回归运动，能够 Download to read ad-free 高中地理 34 2023 年 第 12 期 反映季节、气候、农时和物候特征，因而具有较高的科学性和实用价值。公历虽然也是阳历，但是它的公历月却是人为主观划分，且大月与小月日数相差1~3日，大小月排列也较紊乱，从这方面来比较，干支月优于公历月。由于干支纪月系统的上述优越性，学术界对于阳历版干支纪月的支持率比较高。近期比较盛行的《日梭万年历》和《寿星万年历》等日历就是根据上述编算规则制作的。 然而，干支纪月最终并没有被纳入国家标准，其原因是多方面的。首先，我国至迟从殷商时代起的传统历法就是阴阳历，而农历属于阴阳历。干支纪月虽然与公历同属于阳历范畴，但公历是世界通用历法，而干支月的普及率和影响范围比公历小，因此我国的法定历法是公历和农历。其次，和农历数序纪月相比，干支纪月的纪月方式并不具备优越性。数序纪月属于阴阳历，根据朔望月定月长，并设置闰月，拥有阴历和阳历的双重优点，科学性、实用性强。数序纪月具有深厚的华夏历史文化渊源，它所承载的传统文化及重要的传统节日如春节、元宵节、端午节、中秋节、除夕等与亿万民众的生活息息相关，具有广泛的群众基础。再次，干支纪月系统是纯粹的阳历，干支纪月与农历数序纪月、干支月首日（公历月上半月的节气日）与农历月首日（朔日）、干支年首日（立春日）与农历年首日（正月初一）均不一致。最后，如果干支纪月和农历数序纪月并用，则每一农历月必将跨两个干支月，两者不能兼顾。这也是干支纪月没有被纳入国家标准的重要原因之一。虽然阳历版本的干支纪月系统未被纳入国家标准，但它仍属于我国众多传统历法中一种较典型的阳历，在古代天文学史和历法研究、历史、民俗领域中占有重要位置。 2)阴阳历版本的干支纪月系统 (2)阴阳历版本的干支纪月系统 ①原理及编算规则干支系统是农历的一部分，干支纪年、纪日既已被纳入农历的国家标准，干支纪月系统顺其自然应与国家标准保持一致：地支与农历数序纪月相对应，天干与地支相配，以60月为周期循环纪月；以朔望月定干支月长，以二十四节气即回归年定年长；1年分12个月，并通过设置闰月使平均干支年近似等于回归年。仿照《农历的编算和颁行》中农历月的编排规则，拟制定的干支月编算规则如下。（A）以北京时间（即120°E标准时）为标准时间。（B）干支日的计量是从北京时间0时起到北京时间24时截止的时间间隔。 （C）朔日为干支月的第一个干支日。（D）从某个朔日起到下一个朔日（不含）截止的时间间隔为一个干支月。（E）包含节气冬至在内的干支月为子月（对应农历十一月）。（F）寅月为干支年首月（对应农历年正月）；寅月第一个干支日为干支年首日（对应农历正月初一）。（G）十二地支纪月与农历的十二数序纪月一一对应，即子月对应十一月，丑月对应十二月。（H）如果从某个子月开始到下一个子月（不含）之间有13个干支月，则取其中最先出现的一个不包含中气的干支月为闰月；闰月采用在其前一个干支月的名称前加“闰”字的方法命名。（I）十天干与十二地支相配，则各月的干支互不相同，以60月为周期循环纪月。年天干与月干支固定对应关系如表2所示。 表2 年天干（或公元年份的个位数）与月干支对照表 农历月 月地支 甲或己年（4或9） 乙或庚年（5或0） 丙或辛年（6或1） 丁或壬年（7或2） 戊或癸年（8或3）正月 寅 丙寅月 戊寅月 庚寅月 壬寅月 甲寅月二月 卯 丁卯月 己卯月 辛卯月 癸卯月 乙卯月三月 辰 戊辰月 庚辰月 壬辰月 甲辰月 丙辰月四月 巳 己巳月 辛巳月 癸巳月 乙巳月 丁巳月五月 午 庚午月 壬午月 甲午月 丙午月 戊午月六月 未 辛未月 癸未月 乙未月 丁未月 己未月七月 申 壬申月 甲申月 丙申月 戊申月 庚申月八月 酉 癸酉月 乙酉月 丁酉月 己酉月 辛酉月九月 戌 甲戌月 丙戌月 戊戌月 庚戌月 壬戌月十月 亥 乙亥月 丁亥月 己亥月 辛亥月 癸亥月十一月 子 丙子月 戊子月 庚子月 壬子月 甲子月十二月 丑 丁丑月 己丑月 辛丑月 癸丑月 乙丑月注：表头括号中数字指农历年首日对应的公历年的个位数。 ②公式的构建和计算首先，绘制年天干与月干支对照表，再根据年天干与月干支的对应关系构建计算公式。因为干支月以60个月（5年）为周期循环使用，所以本文仅推算自北京 Download to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like 01.王相山 格局决定命运《子平真诠透解》上册 新版248页1 No ratings yet 01.王相山 格局决定命运《子平真诠透解》上册 新版248页1 259 pages 周易基础知识 大六壬入门 No ratings yet 周易基础知识 大六壬入门 45 pages 高级班格局篇 No ratings yet 高级班格局篇 77 pages 正宗阴阳五行八字预知学 (初级教材) No ratings yet 正宗阴阳五行八字预知学 (初级教材) 175 pages 毕法赋纯享版 (凌福之) (Z-Library) No ratings yet 毕法赋纯享版 (凌福之) (Z-Library) 69 pages 八字奥义 100% (1) 八字奥义 55 pages 杨春义大六壬基础、提高班讲义 100% (1) 杨春义大六壬基础、提高班讲义 45 pages 子平真诠 沈孝瞻原著 No ratings yet 子平真诠 沈孝瞻原著 27 pages 段建业盲派命理干支解密 No ratings yet 段建业盲派命理干支解密 45 pages 周易基础知识 大六壬入门 No ratings yet 周易基础知识 大六壬入门 45 pages 不用万年历快速推算四柱干支揭秘 No ratings yet 不用万年历快速推算四柱干支揭秘 6 pages 汉字起源新解 No ratings yet 汉字起源新解 69 pages 六爻精彩卦例集 刘虹言 No ratings yet 六爻精彩卦例集 刘虹言 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10427
https://cdnsciencepub.com/doi/10.1139/z94-090
A review of the classification and distribution of five opalinids from Africa and North America Create a new account Email Returning user Request Username Can't sign in? Forgot your username? Enter your email address below and we will send you your username Email Close If the address matches an existing account you will receive an email with instructions to retrieve your username close LOGIN TO YOUR ACCOUNT INDIVIDUAL LOGIN | REGISTERorINSTITUTIONAL LOGIN Login Register Email Password Forgot password? Reset it here [x] Keep me logged in All fields with are mandatory Email All fields with are mandatory Already have an account? Login here Change Password Old Password New Password Too Short Weak Medium Strong Very Strong Too Long Your password must have 8 characters or more and contain 3 of the following: a lower case character, an upper case character, a special character or a digit Too Short Password Changed Successfully Your password has been changed Can't sign in? 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To learn about our use of cookies and how you can manage your cookie settings, please see our Cookie Policy. × HomeCanadian Journal of ZoologyVolume 72, Number 4, April 1994A review of the classification and distribution of five opalinids from Africa and North America Article Share on A review of the classification and distribution of five opalinids from Africa and North America Authors: Felix-Marie Affa'a and Denis H.LynnAuthors Info & Affiliations Publication: Canadian Journal of Zoology April 1994 6 16 Metrics Total Citations 6 Last 6 Months 0 Last 12 Months 0 Total Downloads 16 Last 6 Months 0 Last 12 Months 1 Get Access Contents Canadian Journal of Zoology Volume 72, Number 4 April 1994 ###### PREVIOUS ARTICLE Helminth infracommunities of Podarcis pityusensis and Podarcis lilfordi (Sauria: Lacertidae) from the Balearic Islands (western Mediterranean basin) Previous###### NEXT ARTICLE Habitat selection by muskrats in experimental marshes undergoing succession Next Abstract Résumé Information & Authors Metrics & Citations Get Access References Figures Tables Media Share Abstract Five multinucleate opalinids from Canada and Cameroon, Africa, commensal in anuran amphibians, are redescribed. The relative size of the falx and the presence or absence of a posterior zone of convergence of kineties in silver-impregnated specimens support three new combinations: Cepedea sudafricana (Fantham, 1923), Cepedea virguloidea (Metcalf, 1923), and Cepedea obtrigonoidea (Metcalf, 1923). These new combinations suggest that Cepedea may be a much more common genus in the Nearctic region than was previously realized. Résumé Les auteurs redécrivent cinq opalines multinucléées commensales de batraciens anoures du Canada et du Cameroun (Afrique). Sur la base de la taille relative de la zone falculaire et de la présence ou absence d'une zone de sécance postérieure sur des spécimens imprégnés à l'argent, trois nouvelles combinaisons sont proposées : Cepedea sudafricana (Fantham, 1923), Cepedea virguloidea (Metcalf, 1923) et Cepedea obtrigonoidea (Metcalf, 1923). Ces nouvelles combinaisons amènent à penser que le genre Cepedea est probablement plus commun dans la région Néarctique qu'on ne le pensait à ce jour. Get full access to this article View all available purchase options and get full access to this article. Get Access Already a Subscriber? Sign in as an individual or via your institution Information & Authors Information Authors Information Published In Canadian Journal of Zoology Volume 72 • Number 4 • April 1994 Pages: 665 - 674 History Version of record online: 15 February 2011 Permissions Request permissions for this article. Request permissions Authors Affiliations Expand All Felix-Marie Affa'a View all articles by this author Denis H.Lynn View all articles by this author Metrics & Citations Metrics Citations 6 Metrics Article Metrics Download Citation No data available. 16 6 Total 6 Months 12 Months Total number of download and citation Other Metrics Citations Cite As Felix-Marie Affa'a and Denis H. Lynn. 1994. A review of the classification and distribution of five opalinids from Africa and North America. Canadian Journal of Zoology. 72(4): 665-674. Export Citations If you have the appropriate software installed, you can download article citation data to the citation manager of your choice. Simply select your manager software from the list below and click Download. Format - [x] Direct import Cited by 1.A revised taxonomy and phylogeny of opalinids (Stramenopiles: Opalinata) inferred from the analysis of complete nuclear ribosomal DNA genes Go to CitationCrossrefGoogle Scholar 2.Cellularisation: les opalines illustrent-elles une tentative de constitution d'un être pluricellulaire ? Go to CitationCrossrefGoogle Scholar 3.Morphological and cytological observations on two opalinid endocommensals of Acanthixalus spinosus (Amphibia, Anura) Félix-Marie Affa'a, Jean-Pierre Mignot, and Jean-Louis Amiet Go to CitationCrossrefGoogle Scholar 4.Opalinidae (Sarcomastigophora) in North American Amphibia. Genus Opalina Purkinje & Valentin, 1835 Go to CitationCrossrefGoogle Scholar 5.Opalinidae (Sarcomastigophora) in North American Amphibia. GenusOpalina Purkinje & Valentin, 1835 Go to CitationCrossrefGoogle Scholar 6.Patterning in Opalinids.III.The Cytoskeleton of Cepedea sudafricana (FANTHAM, 1923) AFFA'A & LYNN 1994, an Intermediate Type between Opalina ranarum and Protoopalina pseudonutti Go to CitationCrossrefGoogle Scholar Figures Tables Media Share Options Share Share the article link Copy Link Copied! Copying failed. Share on social media BlueskyX (formerly Twitter)LinkedInFacebookemail Get Access Get Access Login options Check if you access through your login credentials or your institution to get full access on this article. Personal login Email Password Forgot password? 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10428
https://pmc.ncbi.nlm.nih.gov/articles/PMC11098544/
Hearing preservation surgery for vestibular schwannoma: a systematic review and meta-analysis - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer | PMC Copyright Notice Acta Otorhinolaryngol Ital . 2024 May 13;44(2 Suppl 1):S86–S93. doi: 10.14639/0392-100X-suppl.1-44-2024-N2900 Search in PMC Search in PubMed View in NLM Catalog Add to search Hearing preservation surgery for vestibular schwannoma: a systematic review and meta-analysis Vito Pontillo Vito Pontillo 1 Otolaryngology Unit, Department of Translational Biomedicine and Neurosciences (DiBraiN), University “Aldo Moro” of Bari, Bari, Italy Find articles by Vito Pontillo 1,,✉, Valentina Foscolo Valentina Foscolo 1 Otolaryngology Unit, Department of Translational Biomedicine and Neurosciences (DiBraiN), University “Aldo Moro” of Bari, Bari, Italy Find articles by Valentina Foscolo 1,, Francesco Salonna Francesco Salonna 1 Otolaryngology Unit, Department of Translational Biomedicine and Neurosciences (DiBraiN), University “Aldo Moro” of Bari, Bari, Italy Find articles by Francesco Salonna 1, Francesco Barbara Francesco Barbara 1 Otolaryngology Unit, Department of Translational Biomedicine and Neurosciences (DiBraiN), University “Aldo Moro” of Bari, Bari, Italy Find articles by Francesco Barbara 1, Maria Teresa Bozzi Maria Teresa Bozzi 2 Division of Neurosurgery, Department of Translational Biomedicine and Neurosciences (DiBraiN), University “Aldo Moro” of Bari, Bari, Italy Find articles by Maria Teresa Bozzi 2, Raffaella Messina Raffaella Messina 2 Division of Neurosurgery, Department of Translational Biomedicine and Neurosciences (DiBraiN), University “Aldo Moro” of Bari, Bari, Italy Find articles by Raffaella Messina 2, Francesco Signorelli Francesco Signorelli 2 Division of Neurosurgery, Department of Translational Biomedicine and Neurosciences (DiBraiN), University “Aldo Moro” of Bari, Bari, Italy Find articles by Francesco Signorelli 2, Nicola Antonio Adolfo Quaranta Nicola Antonio Adolfo Quaranta 1 Otolaryngology Unit, Department of Translational Biomedicine and Neurosciences (DiBraiN), University “Aldo Moro” of Bari, Bari, Italy Find articles by Nicola Antonio Adolfo Quaranta 1 Author information Article notes Copyright and License information 1 Otolaryngology Unit, Department of Translational Biomedicine and Neurosciences (DiBraiN), University “Aldo Moro” of Bari, Bari, Italy 2 Division of Neurosurgery, Department of Translational Biomedicine and Neurosciences (DiBraiN), University “Aldo Moro” of Bari, Bari, Italy ✉ Correspondence Vito Pontillo E-mail: pontillovito@gmail.com Vito Pontillo and Valentina Foscolo contributed equally to this work. Received 2024 Jan 15; Accepted 2024 Jan 31; Issue date 2024 May. Società Italiana di Otorinolaringoiatria e Chirurgia Cervico-Facciale, Rome, Italy This is an open access article distributed in accordance with the CC-BY-NC-ND (Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International) license. The article can be used by giving appropriate credit and mentioning the license, but only for non-commercial purposes and only in the original version. For further information: PMC Copyright notice PMCID: PMC11098544 PMID: 38745520 SUMMARY The aim of this systematic review is to analyse the role of hearing preservation surgery for vestibular schwannoma. The complications and hearing outcomes of the single surgical techniques were investigated and compared with those of less invasive strategies, such as stereotactic radiotherapy and wait and scan policy. This systematic review and meta-analysis was performed according to the PRISMA guidelines. All included studies were published in English between 2000 and 2022. Literature data show that hearing preservation is achieved in less than 25% of patients after surgery and in approximately half of cases after stereotactic radiotherapy, even if data on long-term preservation are currently not available. KEY WORDS: vestibular schwannoma, acoustic neuroma, hearing preservation, microsurgery, retrosigmoid approach, middle fossa approach, stereotactic radiosurgery Introduction Vestibular schwannomas (VSs) or “acoustic neuromas” are benign, slowly growing tumours arising from the vestibulo-cochlear nerve and account for 6-7% of all intracranial neoplasms and 80% of the cerebellopontine angle lesions 1. They arise from the inferior vestibular nerve in 73% of cases and from the superior vestibular nerve in 27% 2. The neural area of origin is mainly localised at the Obersteiner-Redlich junction, which is the transition point from central (glial cells) to peripheral (Schwann cells) myelin sheath. Unilateral progressive or sudden sensorineural hearing loss and tinnitus are the main symptoms of presentation 3. Less frequent symptoms include vertigo or dizziness, headache, ataxia, and cranial nerve palsy. The gold standard for VS diagnosis is gadolinium enhanced magnetic resonance (MR) of the internal auditory canal (IAC) and cerebello-pontine angle (CPA), whose easier access in the last decades has allowed earlier diagnosis 4. At present, treatment options include a conservative approach (wait and scan [W&S] policy), stereotactic radiosurgery (SRS), and microsurgery with or without hearing preservation 5-11. The choice depends on different factors, such as patient’s age and comorbidities, size and location of the tumour, and hearing status. W&S is based on systematic follow-up by serial gadolinium enhanced MR and hinges on the often negligible growth of VSs and slow progression of symptoms 9, 12. The goal of conservative treatment is to minimise the risks and complications and to preserve an optimal quality of life in selected patients, such as the elderly, those with minimal symptoms, or with a small or middle-sized tumour. Borsetto et al. have proposed a surveillance protocol consisting in a 10-year minimum follow-up by MR 12. SRS provides high doses of ionising radiation precisely delivered to a target, while limiting irradiation of healthy tissues 13. It is administered in a single session (SRT) or fractionated over several days (FSRT) 14, and performed with Gamma Knife, linear accelerators (LINAC) or Cyber Knife. The choice among the different techniques and modalities is based on tumour size, hearing function, and performance status. Growth control rates after SRS have been reported to be from 90% to 98% at 10 years 14-16. SRS does not represent an option in young patients due to the risk, though minimal (1:1000), of developing radiation-induced cancers 17-20. Surgical treatment of VS can be performed through different approaches, depending on tumour size, location, age, and hearing status. The trans-labyrinthine (TL) approach allows a large exposure of the IAC and CPA with minimal cerebellar retraction and complete control of the facial nerve. However, it does not preserve residual hearing 5 and will not be discussed further in the present review. The retro-sigmoid (RS) approach is a potentially hearing preserving technique that offers a large view of the CPA. However, it requires relevant cerebellar retraction, especially in large and anterior tumours. It is usually indicated in patients with good preoperative hearing and small VS (< 1.5 cm in the CPA), not reaching the midline and fundus of the IAC 21. The retro-labyrinthine approach (RLA) is a trans-mastoid surgical avenue that allows hearing preservation; however, it offers a reduced exposure of the CPA compared to the TL and RS approaches and is nowadays rarely used 22. The middle fossa (MF) or sub-temporal approach allows hearing preservation by reaching the IAC from above. It is indicated in young patients with good preoperative hearing function affected by VSs limited to the IAC or with only minimal (< 0.5 cm) extra-meatal extension 11. The aim of the present systematic review was to evaluate early- and long-term hearing preservation rates after VS hearing preservation surgery (HPS) by comparing them with more conservative approaches such as W&S and SRS. Materials and methods This systematic review was performed in agreement with the PRISMA 2020 Statement Guidelines 23. A specific PICOS question (Population: individuals with unilateral VS and with serviceable hearing function; Intervention: microsurgical approaches with hearing preservation techniques; Comparator: W&S or SRS strategies; Outcomes: hearing preservation; Study design: prospective studies) was constructed. Focused PICOS questions of this review are: ‘is HPS an effective strategy to preserve hearing in specific and tailored cases of VS?’ and ‘is there any difference in terms of early and late hearing preservation rates between HPS, W&S, and SRS?’. Search strategy An electronic literature search was independently conducted by two authors using the PubMed/MEDLINE database as follows: (“vestibular schwannoma” OR “acoustic neuroma”) AND “hearing preservation” AND (randomized OR randomized OR random OR randomly OR randomization OR RCT OR RCTs OR “clinical trial” [Publication Type] or “clinical trials as topic” [MeSH Terms]). The request was done on September 15, 2023 with no time limitations. Study selection Initially, titles and abstracts were independently screened by three authors (VP, VF and FSa) for eligible papers. Next, full-text papers were independently screened and those fulfilling eligibility criteria were included. Reference lists of original studies were hand-searched to identify articles that could have been missed during the electronic search. Any disagreement was resolved by consensus. Articles were included in this systematic review if they met the following inclusion criteria: prospective randomised clinical trial or clinical study; article in English; patients with VS undergoing W&S, SRS or MF, RS or RLA approaches; evaluation of serviceable hearing preservation defined through pure tone and speech audiometry or by the American Academy of Otolaryngology – Head and Neck Surgery (AAO-HNS) 24 or Gardner-Robertson (GR) 25 classification systems. In vitro studies, case series, case reports, animal studies, letters to the editor, opinion articles, abstracts, review papers, book chapters, pre-print and unpublished articles were excluded, together with studies reporting on patients with bilateral VS or type 2 neurofibromatosis (NF2), cases treated by a TL approach or without the specific aim of hearing preservation, patients undergoing dual treatment (gross or near total resection followed by SRS), and patients with less than 3 months of follow-up. Data extraction and comparison The authors performed data extraction individually. Information from the included studies were tabulated according to the study designs, study period, demographics, tumour size, type of treatment, complications, and hearing preservation. Collected data were primarily based on the focused questions outlined above. Series with heterogeneous procedures and methods were carefully screened in order to consider only those cases that met the inclusion criteria. The authors cross-checked all extracted data. Any disagreement was resolved by discussion until consensus was reached. When possible, data were eventually aggregated in subgroups in order to estimate and compare the hearing preservation rates between the different strategies and techniques. Statistical analysis Statistical analysis of the different rates of hearing preservation in the two subgroups (surgery vs. SRS) was performed using the Chi-square test. Results were considered significant for p values < 0.05. The IBM software SPSS Statistics version 26 was used for the analysis. Results Study selection Our initial search yielded 40 records. After initial screening of titles and abstracts, 23 full-text articles were selected for reading. Of these, 13 were further excluded since they did not meet the inclusion criteria. After the final selection stage, 10 studies were included in the present review, of which 3 reporting on surgical approaches, 6 on SRS, and one of both surgical and SRS. No studies reporting data on W&S strategies fulfilled the inclusion criteria. The selection process is shown in Figure 1. Figure 1. Open in a new tab Flow-chart showing selection strategy. General characteristics of studies included Table I shows general characteristics of the included studies 14-15, 26-33. All were unicentric prospective studies published in English between 2000 and 2022, and involved a total of 869 cases, of which 513 treated by hearing preservation microsurgical techniques and 356 by SRS. Mean age of patients ranged between 35 and 66 years. The male to female ratio was not reported in all studies; for this reason, these data were not included in the analysis. The individual studies used different hearing classifications: some authors 14,26-27,29 used the AAO-HNS classification, while others 15,28,30-33 used the GR classification. Similarly, when ambiguity over the concept of ‘serviceable hearing preservation’ was found, these data were standardised by including under this definition only Class A and B according to the AAO-HNS classification, and Class I and II according to the GR classification (pure tone audiometry average threshold lower or equal to 50 dB and word recognition at speech audiometry greater or equal to 50%). Thus, only patients with serviceable pre-treatment hearing and for whom a hearing preserving procedure was attempted were included in the analysis. Table I. General characteristics of the studies included. | # | Author (year) | Study design | No. of patients | Mean age (years) | Treatment | Surgical technique/SRS modality | Hearing classification | :---: :---: :---: :---: | | 1 | Bento et al. (2022) 26 | Unicentric prospective | 22 | 35 | Surgery | RLA | AAO-HNS | | 2 | Colletti et al. (2005) 27 | Unicentric prospective | 70 | 53 | Surgery | 35 RS | AAO-HNS | | 35 MF | | 3 | Tonn et al. (2000) 28 | Unicentric prospective | 399 | 51 | Surgery | RS | GR | | 4 | Pollock et al. (2006) 29 | Unicentric prospective | 52 | 51 | 22 Surgery | RS/MF | AAO-HNS | | 30 SRS | GKRS | | 5 | Saraf et al. (2022) 30 | Unicentric prospective | 20 | 64 | SRS | FSRS (proton) | GR | | 6 | Putz et al. (2020) 14 | Unicentric prospective | 34 | 66 | SRS | 6 SSRS | AAO-HNS | | 28 FSRS | | 7 | Niranjan et al. (2008) 31 | Unicentric prospective | 51 | 54 | SRS | GKRS | GR | | 8 | Chopra et al. (2007) 15 | Unicentric prospective | 106 | 56 | SRS | GKRS | GR | | 9 | Tamura et al. (2009) 32 | Unicentric prospective | 74 | 47 | SRS | GKRS | GR | | 10 | Ikonomidis et al. (2015) 33 | Unicentric prospective | 41 | 55 | SRS | LINAC | GR | | AAO-HNS: American Academy of Otolaryngology – Head and Neck Surgery classification; GR: Gardner-Robertson classification; RS: retro-sigmoid approach; RLA: retro-labyrinthine approach; MF: middle cranial fossa approach; SRS: stereotactic radiosurgery; GKRS: Gamma-knife radiosurgery; FSRS: fractioned stereotactic radiosurgery; SSRS: single-session stereotactic radiosurgery; LINAC: linear accelerator. | Open in a new tab Results of hearing preservation surgery Surgical results of the studies included in this review are summarised in Table II. Table II. Results of hearing preservation surgery. | Author (year) | Surgical technique | No. of patients | Koos stage | Complications | Postoperative FP | Short-term hearing preservation (≤ 12 months) | Long-term hearing preservation (> 12 months) | Mean follow-up (months) | :---: :---: :---: :---: | Bento et al. (2022) 26 | RLA | 22 | I: 36% | 5.5% | 5.5% | 31.8% | NA | 3 | | II: 64% | | Colletti et al. (2005) 27 | RS | 35 | I: 100% | 35% | 20% | 40% | NA | 12 | | MF | 35 | I: 100% | 23% | 23% | 51.4% | NA | 12 | | Tonn et al. (2000) 28 | RS | 399 | II: 54.5% | NA | NA | 19% | NA | 6 | | III: 44.2% | | NA 1.3% | | Pollock et al. (2006) 29 | RS/MF | 22 | NA | NA | 15% | 5% | 5% | 42 | | RS: retro-sigmoid approach; RLA: retro-labyrinthine approach; MF: middle cranial fossa approach; FP: facial palsy; NA: not available. | Open in a new tab Bento et al. 26 included 22 patients in their analysis operated by a RLA for small VSs (36% Koos I, 64% Koos II) with serviceable preoperative hearing. All patients were young (mean age, 35 years) and preoperative hearing levels were AAO-HNS Class A in 2 cases (9%) and Class B in 20 cases (91%). Complete macroscopic tumour removal was obtained in all cases with low rate of complications (1 patient with House-Brackmann grade II facial paralysis). Postoperatively, a serviceable hearing level (Class A and B) was maintained in 31.8% of patients at 3 months. An RS approach was used by Colletti et al. 27 on 35 middle-aged patients (mean age, 52 years) with small intracanalicular (Koos I) VSs and preserved hearing (46% AAO-HNS Class A and 54% Class B), while Tonn et al. 28 used the same approach on 399 middle-aged patients (mean age, 52 years) with larger VSs (Koos II 54% and Koos III 44%) and serviceable hearing. The two authors obtained divergent results in terms of hearing preservation (40% vs. 19%), confirming the likely predictive role of tumour size in cochlear nerve integrity. However, in the series by Tonn et al. 28 a possible bias must be discussed. In fact, only 229 of 399 patients were operated on after the introduction of intraoperative cochlear function monitoring. Thus, when considering only the latter subgroup, better hearing preservation rates were registered (26.8%), but still not remotely comparable to those of the series by Colletti et al. 27. Furthermore, when extracting Tonn’s 28 data in relationship with tumour size, those with extra-meatal diameter < 15 mm were correlated with better hearing preservation rates (22%) compared with larger tumours (15% in tumours with an extra-meatal diameter between 16 and 30 mm; 0% in tumours with an extra-meatal diameter larger than 30 mm). The MF approach was analysed by Colletti et al. 27 on a series of 35 middle-aged patients (mean age, 54 years) with small intra-canalicular (Koos I) VSs and preserved hearing (43% AAO-HNS Class A and 57% Class B). They obtained even better hearing preservation rates (51.4%) compared to the RS subgroup, with a similar rate of complications (23% vs 20% of facial paralysis). Finally, Pollock et al. 29 included 22 patients in their series operated on by RS or MF, obtaining very low rates of hearing preservation (5%). However, no categorisation in terms of technique or tumour size were specified in the manuscript, thus preventing any possible interpretation and analysis of results. Results of stereotactic radiosurgery Results of the different types of SRS of the studies included are summarised in Table III. Table III. Results of stereotactic radiosurgery. | Author (year) | SRS modality | No. of patients | Mean volume (cm 3) | Total dose (Gy) | Max cochlear dose (Gy) | Tumour growth control | Early-term hearing preservation (≤ 12 months) | Long-term hearing preservation (> 12 months) | Mean follow-up (months) | :---: :---: :---: :---: :---: | | Pollock et al. (2006) 29 | GKRS | 30 | 1.5 | 26.4 | NA | 96% | 77% | 63% | 42 | | Saraf et al. (2022) 30 | FSRS (proton) | 20 | 0.81 | 50.4-54 | 50.7 | 100% at 4 years | 53% | 57% | 36 | | Putz et al. (2020) 14 | SSRS | 6 | 13.9 | 12-13 | 13.7 | 100% at 10 years | NA | 53% | 36 | | FSRS | 28 | 13.4 | 50.4-55.8 | 51.1 | 93.8% at 10 years | | Niranjan et al. (2008) 31 | GKRS | 51 | 0.000112 | 18.7-36 | NA | 99% at 3 years | NA | 64.5% | 42 | | Chopra et al. (2007) 15 | GKRS | 106 | 1.3 | 20-26 | NA | 98.3% at 10 years | NA | 56.6% | 68 | | Tamura et al. (2009) 32 | GKRS | 74 | 1.3 | NA | NA | 93.2% at 5 years | NA | 78.4% | 56 | | Ikonomidis et al. (2015) 33 | LINAC | 41 | 2.1 | 15.23 | 11.4 | 75% at 2 years | 51.2% | 36.6% | 39 | | Gy: Gray; NA: not available; SSRS: single-session stereotactic radiosurgery; FSRS: fractioned stereotactic radiosurgery; GKRS: Gamma-knife radiosurgery; LINAC: linear accelerator. | Open in a new tab The traditional delivery modality of SRS is by a single session (SSRS). However, this technique is routinely used in patients with small tumours and non-serviceable hearing level (AAO-HNS Classes C or D) 14. The only 6 cases in which this technique was surprisingly used for patients with serviceable hearing were described by Putz et al. 14, administering on large VSs (mean volume, 13.9 cm 3) a total dose of 12-13 Gy. Unluckily, hearing outcomes were presented by the authors in combination with those of 28 patients treated by fractioned SRS (FSRS), with an excellent but not-interpretable long-term hearing preservation rate of 53%. Fractioned proton radiosurgery was administered by Saraf et al. 30 to 20 patients with smaller VSs (median volume, 0.81 cm 3) with a good hearing preservation at the short- (53% at 1 year) and long-term (57% at 3 years). Gamma-knife radiosurgery (GKRS) was the most widely used technique in the included studies. Pollock et al. 29 administered a mean dose of 26.4 Gy to 30 patients affected by small to medium-sized VSs, with a satisfying tumour control and a serviceable hearing preservation of 77% at 3 months and 63% at last follow-up (mean, 42 months). Similar results were obtained with GKRS by Niranjan et al. 31 on 51 intra-canalicular VSs (64.5% of hearing preservation at 42 months), by Chopra et al. 15 on 106 patients (56.6% at 68 months), and by Tamura et al. 32 on 74 patients (78.4% at 56 months). Ikonomidis et al. 33 performed LINAC SRS on 41 patients with preserved serviceable hearing and with Koos I to III VSs (median volume, 2.1 cm 3). Hearing preservation was registered in 51.2% of cases at 6 months and in 36.6% of cases at the last observation (mean, 39 months). Hearing preservation surgery vs stereotactic radiosurgery When aggregating the results of all the studies (Tab. IV), SRS showed significant better overall hearing preservation rates in comparison to the surgical approaches considered (57.8% vs 23.4%, p value < 0.0001). When comparing single surgical and radiosurgical techniques, the best results were found in the MF and GKRS groups. However, no statistical analysis was performed to confirm this finding, due to the high heterogeneity between the different subgroups. Table IV. Data aggregation and meta-analysis. | Treatment | Hearing preservation (%) | P value | Technique | Hearing preservation (%) | :---: :---: | Surgery | 23.4 | < 0.0001 | MF27 | 51.4 | | RLA26 | 31.8 | | RS27,28 | 20.7 | | Stereotactic radiosurgery | 57.8 | GKRS15,29,31,32 | 62 | | Proton FSRS30 | 57 | | LINAC33 | 36.6 | | MF: middle cranial fossa approach; RLA: retro-labyrinthine approach; RS: retro-sigmoid approach; FSRS: fractioned stereotactic radiosurgery; GKRS: Gamma-knife radiosurgery; LINAC: linear accelerator. | Open in a new tab Discussion VSs are benign tumours, but their progressive growth can lead to severe and life-threatening sequelae. In recent decades, the easier access to MR has allowed VSs to be diagnosed more frequently at a smaller and scarcely symptomatic stage. Furthermore, it has been demonstrated that only one-third of all VSs have the tendency to grow, while approximately 50% of patients maintain their hearing during an observation period of 5 years 10. In this panorama, hearing sparing surgical approaches have gained increasing interest, while the evolution of SRS and the development of W&S strategies have added further options to the current therapeutic armamentarium for management of small and middle-sized VSs. The aim of this review was to analyse the role of HPS for VS by investigating the outcomes of single techniques and comparing them with those of less invasive strategies, such as SRS and W&S. To the best of our knowledge, no other systematic reviews with the same purposes has been published to date. Surgery has demonstrated to provide an adequate rate of hearing preservation (23.4%) when used with the correct principles and indications. The most frequently used hearing sparing approaches are MF and RS, among which the choice depends on the surgeon’s familiarity, preference, and tumour size. The only study comparing these two techniques in terms of hearing preservation is that by Colletti et al. 27. The authors found that the RS approach offers better chances of keeping serviceable hearing in case of adverse anatomic conditions and IAC enlargement greater than 7 mm, while the MF approach provides better preservation rates when the tumour fills the IAC fundus (distance less than 3 mm). Unfortunately, a direct comparison between these three techniques in the included studies was not feasible, since each author used different indication criteria and parameters. However, when looking for a potential predictive factor for hearing preservation in selected series, it is quite easy to speculate over the fact that smaller and intra-canalicular VSs 27 may be correlated with better outcomes (Tab. II). This hypothesis was also confirmed by other authors who found significantly better preservation rates in patients with smaller tumours in both surgical 26, 28 and SRS series 14,31-32. SRS was shown in our review to provide excellent hearing preservation rates (overall 57.8% after a mean follow-up of 53 months) with high rates of growth control (93.2% to 100%). LINAC 33 reported worse results of hearing preservation compared to the other techniques (Tab. IV). However, the large variability in tumour size, and total, marginal and cochlear doses makes it impossible to compare the different techniques routinely in use. Hearing impairment after microsurgery with cochlear nerve integrity preservation is thought to be ascribable to mechanical or thermal neural microvascular damage and is assumed to occur immediately after surgery. On the contrary, hearing deterioration after SRS tends to be progressive over 6 to 24 months and seems to be caused by ischaemic neural damage secondary to tumour swelling, or progressive radiation-induced neural oedema and demyelination 18,38. For this reason, the best results of SRS when compared to microsurgery should be verified over a longer follow-up period. Moreover, the risk of radiation-induced malignant transformation should be always considered in the decision-making process and patient counselling 19. We believe that a comparison of hearing preservation rates among microsurgery and W&S strategies is not reasonable, since they have very different indication criteria. However, literature data show that approximately 50% of patients maintain their hearing over a period of 5 years with a W&S policy 10. One of the main limitations of our review was the high heterogeneity of the different series, which made impossible to compare in detail the data of individual studies. Further randomised clinical trials may be therefore necessary to develop a decisional algorithm based on different patient- and disease-related patterns. Conclusions In the past, surgery was the only possible treatment for VS; today, the development of SRS and other non-surgical conservative strategies has considerably expanded the range. In addition, hearing preservation has become a major challenge. The present review found satisfying preservation rates after both microsurgery and SRS, especially when dealing with intra-canalicular and small-sized VSs. In particular, SRS showed slightly better results, but the observational period of the reviewed series was not long enough to arguably claim the superiority of this approach. Hence, further randomised controlled trials are needed to compare long-term hearing outcomes of the different treatment options. Conflict of interest statement The authors declare no conflict of interest. Funding This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors. Author contributions VP, NQ: conceived, designed, wrote and revised the work; VF, FSa, VP: independently screened, read and selected the articles; VP: realised meta-analysis; VF, FSA, VP, MTB, RM, FSI, NQ: took part to the writing of the present paper. Ethical consideration Not applicable. Figures and tables References 1.Zamani AA. Cerebellopontine angle tumors: role of magnetic resonance imaging. Top Magn Reson Imaging 2000;11:98-107. 10.1097/00002142-200004000-00005 [DOI] [PubMed] [Google Scholar] 2.Okada Y, Takahashi M, Saito A, et al. Electronystagmographic findings in 147 patients with acoustic neuroma. Acta Otolaryngol 1991;487(Suppl):150-155. 10.3109/00016489109130461 [DOI] [PubMed] [Google Scholar] 3.Kentala E, Pyykkö I. Clinical picture of vestibular schwannoma. 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http://www.emersonstatistics.com/GeneralMaterials/Ratios%20and%20Logarithms.pdf
Use of Ratios and Logarithms in Statistical Regression Models Scott S. Emerson, M.D., Ph.D. Department of Biostatistics, University of Washington, Seattle, WA 98195, USA January 22, 2014 Abstract In many regression models, we use logarithmic transformations of either the regression summary measure (a log link), the regression response variable (e.g., when analyzing geometric means), or one or more of the predictors. In this manuscript, I discuss the rationale for using logarithmic transformations, the interpretation of ratios, and the general properties of logarithms. 1 Use of Logarithmic Transformations Logarithmic transformations of data and/or parameters are used extensively in statistics. The fundamental reason for this stems from the following logic: 1. We are most often interested in using statistics to detect associations between two variables. 2. By an association, we mean that the distribution of one variable (we call this the “response variable”) is different in some way between groups that are homogeneous for the other variable (we call this the “predictor of interest” or POI). 3. If the two variables are associated, then that means that some aspect of the distribution (some “sum-mary measure” like the mean, geometric mean, etc.) is unequal between two groups that differ in their value of the POI. In describing the association, we will want to describe (a) How the POI differs between the two (conceptual) groups being compared, and (b) How the summary measure of response compares across two groups. 4. There are two simple arithmetic ways to tell if two numbers (e.g., the level of POI in two separate groups, or the mean of the “response variable” in two separate groups) are unequal: (a) their difference is not 0, or (b) their ratio is not 1. 5. We choose between differences and ratios as methods for comparisons based on a variety of criteria that are sometimes competing: (a) People understand differences more than they understand ratios. i. Part of this is because natural language (e.g., English) is not very precise when describing ratios. 1 Ratios and Logarithms SS Emerson ii. But I may have the cause and effect relationship reversed: Natural language is bad at de-scribing ratios, because the average speaker (i.e., human) does not understand them, and thus humans did not invent natural language sufficiently precise to describe ratios. (b) Differences are better at describing the scientific importance of most comparisons. i. For instance, you probably care less about having 10 times as much money as me if I only have one cent (a difference of $0.09). ii. You probably care more about having $1,000,000 more than me, even if I have $10,000,000 (a ratio of only 1.1). (c) When working with very small numbers, however, a ratio will accentuate an effect better than a difference will. i. Being diagnosed with lung cancer in any given year is a very rare event in nearly every population, including smokers. (In the numbers that follow, I am combining data from many different sources. The order of magnitude is probably correct, but the exact numbers may be a bit off.) ii. In the US, 60-64 year old current or former smokers have a probability of 0.00296 to be diagnosed with lung cancer during the next year. iii. In the US, 60-64 year old never smokers have a probability of 0.000148 to be diagnosed with lung cancer during the next year. iv. The difference in cancer incidence rates is thus a very small 0.002812. v. However, smokers have about a 20-fold higher rate of cancer diagnosis than non-smokers of the same age and sex. (It actually seems to vary by sex, age, race, but the point still obtains.) (d) Sometimes scientific mechanisms dictate that ratios are more generalizable for the summary mea-sure of the response distribution and/or for effects due to predictors. i. Interventions and risk factors often affect the rate that something happens over time. ii. Cellular enzymes affect the rate at which biochemical reactions proceed. Risk factors that affect enzymatic activity will change the amount of some chemical that accumulates. A. Many physiologic actions occur when some agonist A (e.g., a chemical or drug) interacts with a receptor R (e.g., a particular region of an enzyme or portion of a cell membrane). The activity occurs when a receptor-agonist complex RA is formed. B. A simple (simplistic) model for this interaction is given by Michaelis-Menten kinetics, where the relative abundance of the receptor-agonist complex is governed by some reaction constant Keq that relates the concentration [RA] of the receptor-agonist to the product of the concentrations of the free receptor and agonist ([R] and [A]): R + A ⇌RA = ⇒ Keq = [RA] [R] [A] = ⇒ [RA] [R] = Keq[A] C. Clinical outcome measures are often most directly related to the relative abundance of the receptor-agonist complex, and from the above equation we see that drugs or risk factors that affect the rate of reaction through the Keq act multiplicatively, rather than additively, on the concentration of the agonist. iii. The actions of many biochemical pathways are influenced heavily by the rates of absorption and excretion. Quite often these rates are proportional to the concentration of the drug. Hence the biochemical concentration Ct at any point in time t follows an exponential decay model Ct = C0e−Kt and drugs that change the rate parameter K act multiplicatively, rather than additively, on the initial concentration C0. Version: January 22, 2014 2 Ratios and Logarithms SS Emerson iv. When dealing with money (e.g., health care costs), we most often apply taxes and interest as a rate (percentage) and we (therefore) measure inflation on a multiplicative scale. A. Hence, the fact that women’s starting salaries are often lower than men’s starting salaries by some amount leads to even greater differences after a number of years, though the ratio will tend to be constant, because ratios are given as a percentage increase each year. B. Similarly, compared to the prices when I graduated from high school, costs in 2013 are approximately 5-fold higher across a wide variety of products, though the difference in the cost of movies from 1973 to 2013 (about $8.00) and the difference in the cost of cars (that I would buy) from 1973 to 2013 (about $18,000) are not at all the same. (e) Taking differences is generally easiest, and it is generally most stable statistically, because denom-inators that tend toward zero cause wild fluctuations in ratios. (f) And there are some highly technical reasons: In certain distributions, the logarithmic transforma-tion of some common distributional summary measure can be shown to be efficiently estimated using unweighted combinations of the observations (so every subject can be treated equally). That is, for those specific distributions models based on the log link function will in some sense be “nicer”. i. For a Bernoulli random variable (a variable that is binary), the log odds (logit mean) is the “canonical parameter”. ii. For a Poisson random variable, the log rate (log mean) is the “canonical parameter”. iii. For a log normal random variable (with known σ2), the log geometric mean is the “canonical parameter”. 6. When ratios are scientifically or statistically preferred, we gain stability by considering the logarithm of the ratios, because (as will be demonstrated in later sections of this document) the logarithm of a ratio is the difference between the logarithm of the numerator and the logarithm of the denominator. Hence, by using logarithms, we are back on an additive scale. Note: In the above motivation for the use of logarithms, noticeably absent is any reference to transfor-mations to obtain normal distributions. This is not a reason to transform data. I note that it is easy to demonstrate times that non normally distributed data (either as response or predictors) are more efficiently analyzed in their untransformed state than when transformed to a normal distribution. Instead, we do like to work on scales where effects of covariates are additive, rather than multiplicative. But it is true that skewed data often arise through mechanisms described above, so many data analysts (and authors) have gotten “the cart before the horse” and regarded that skewness the reason for log transformation. (For instance, we can have a truly linear relationship between untransformed variable Y and untransformed variable X, but I may have sample X with some large outliers, and thus the distribution of Y is similarly skewed. In this setting, there may be some very influential points, but under the conditions I laid out, there is no justification for transforming either variable.) 2 Examples of Variables that are Often Logarithmically Trans-formed A number of commonly encountered scientific quantities are so typically used after logarithmic transfor-mations, that the measurements themselves are almost always expressed on a logarithmic scale. Examples include: • Acidity / alkalinity of an aqueous solution is measured as the hydrogen ion concentration. However, it is most common to report the pH, which is the negative log base 10 transformation of the hydrogen Version: January 22, 2014 3 Ratios and Logarithms SS Emerson ion concentration. A 1 unit change in the pH thus corresponds to a 10-fold increase or decrease in the hydrogen ion concentration. • In acoustics, the sound pressure is typically measured on a multiplicative scale. Hence, we consider the sound pressure relative to a standard pressure. That standard pressure is typically based on human hearing in the medium (a different standard is used for air versus water). We refer to that ratio on the logarithmic (base 10) scale using the unit “bel”, or more commonly as a value that is 10 times the log10 scale as a “decibel” of ”dB”. A 3 dB increase in sound thus represents an approximate doubling of the sound pressure, because log10 2 = 0.3010 and 10 × .3010 = 3. • In seismology, the strength of earthquakes is quantified by the “moment magnitude scale (MMS)”, which is a successor to the Richter scale. It is two-thirds the logarithmic (base 10) transformation of the seismic moment, minus 10.7. Because the base 10 logarithm is multiplied by 2/3, a 1,000 fold increase in the strength of an earthquake is measured as a 2 unit difference on the MMS. In biomedicine, it is very common to use logarithmic transformations for measurements of antibody concentration and mRNA concentration (gene expression), because these concentrations differ by orders of magnitude across individuals (and sometimes within individuals over time). Similarly owing to the exponen-tial growth associated with viral replication, viral load in hepatitis C or HIV research is generally analyzed after logarithmic transformation. For other measures of concentration, our habits will differ according to the populations considered. Physiologic homeostasis (regulation to maintain balance or equilibrium in a state conducive to healthy life) tends to maintain important components of the blood in relatively tight control in healthy people. Hence, even though we measure concentrations (and the Keq of various chemical reactions would argue that multiplicative effects would still be relevant), the levels are so constant that the log function is relatively constant over the range of observed data. In health, then, it is relatively unimportant that we log transform the measures. For instance, over the “normal” range of serum bilirubin (0.3 - 1.1 mg/dL, or so depending on the laboratory), the following graph displays the natural log (loge) of bilirubin versus bilirubin. A straight line, while not perfect, is still a good approximation to the curve. However, homeostasis is deranged in disease, and levels of specific blood components are uncontrolled. In that setting, the log transformation will be more important to capture the truly important differ-ences in levels. Continuing the example using serum bilirubin, the following graph displays the log of bilirubin versus bilirubin in a population of 418 Mayo Clinic patients who have primary biliary cirrhosis (www.emersonstatistics.com/Datasets/liver.doc). When the pathologic, extremely high measurements of bilirubin are included (the maximum bilirubin in this dataset was 28 mg/dL), there is marked departure from a straight line. The issue then is whether it is likely that the serum bilirubin is linear in its ability to predict severity of disease. That is, if an elevated bilirubin of, say, 2.0 mg/dL is suggestive of more advanced disease and therefore increased risk of death relative to a PBC patient with a “normal” bilirubin of 1.0 mg/dL, would we really expect any such increased risk to be linear over a range that extends to 28 mg/dL? I would argue no for two reasons, one pathophysiologic and one empirical. First, as noted above, many physiologic mechanisms act on a multiplicative scale by virtue of the chemical reactions associated with absorption and excretion kinetics. Primary biliary cirrhosis is a disease of unknown etiology that affects the ability of the body to excrete bilirubin. Hence, we expect the diseased state to accumulate bilirubin on a multiplicative scale: Each “step” in disease progression should result in a multiplicative increase in bilirubin. But it is not the bilirubin that is harmful per se (at least not in adults–in children the worst effects of kernicterus are directly related to high levels of serum bilirubin being deposited in the developing brain). Instead, serum bilirubin is just a marker of more advanced disease, and we would want to use a measure that is more closely aligned with stage of disease. Version: January 22, 2014 4 Ratios and Logarithms SS Emerson 0.4 0.6 0.8 1.0 -1.2 -1.0 -0.8 -0.6 -0.4 -0.2 0.0 Log Bilirubin vs Bilirubin in Normal Range Bilirubin (mg/dL) log Bilirubin (log mg/dL) Figure 2.1 A plot of the logarithmic transformation (loge) of serum bilirubin versus serum biliru-bin for the 177 patients in the Mayo primary biliary cirrhosis data set who have serum bilirubin within a normal range of 0.3 to 1.1 mg/dL. Empirically, we can consider the likelihood that the association between disease outcome of interest (in this instance, death over an observation period that extended up to 13 years) and serum bilirubin (as a marker of disease stage) would be additive or multiplicative over the range of observed values. It is known from clinical experience and many prior studies that treated PBC patients with bilirubin of 2 - 3 mg/dL are at increased risk of death relative to those whose treated bilirubin levels are in the normal range. This increased risk of death has been estimated to be about a 2-fold increase in the rate (hazard) of death between subjects having a bilirubin of about 2 mg/dL and subjects having a bilirubin of about 1 mg/dL. If an additive scale obtains, then someone with a serum bilirubin of 28 would be expected to have a 2-fold increase in risk of death for each 1 mg/dL difference in serum bilirubin. A difference of 27 mg/dL would thus be associated with a 227 = 134, 217, 728-fold higher risk of death at any given time. Now, again from previous clinical experience and scientific studies, treated PBC patients with normal serum bilirubin levels have a death rate of about 2.5 deaths per 100 person-years. If that risk were increased by a factor of 1.34 × 108, the patient with a bilirubin of 28 mg/dL should die in front of our eyes. Hence, empirically, our prior evidence about increased risk of death for patients with mild elevations of serum bilirubin and our observation that some patients have extremely elevated serum bilirubin levels tells us that it is highly unlikely that serum bilirubin is a marker of increased risk of death that is accurate on a linear scale. On a multiplicative scale, our empiric evidence is much more believable. The approximate 2-fold increase in risk of death associated with a treated serum bilirubin of 2 mg/dL compared to 1 mg/dL could also be viewed as a 2-fold increased risk for a doubling of serum creatinine. An observed serum bilirubin of 32 mg/dL (close enough to 28 mg/dL for the purposes of this simple exposition) represents five doublings. So on a multiplicative scale, we might expect the risk to be 25 = 32-fold higher, or a death rate of about 78.7 deaths per 100 person-years. (Technical note: In my “back of the envelope” calculations, I am using analyses appropriate for survival times that follow an exponential distribution. This assumption is unrealistic Version: January 22, 2014 5 Ratios and Logarithms SS Emerson 0 5 10 15 20 25 -1 0 1 2 3 4 5 Log Bilirubin vs Bilirubin in Primary Biliary Cirrhosis Bilirubin (mg/dL) log Bilirubin (log mg/dL) Figure 2.2 A plot of the logarithmic transformation of serum bilirubin versus serum bilirubin for all 418 patients in the Mayo primary biliary cirrhosis data set (range 0.3 - 28 mg/dL). for human survival over a long period of time, but over shorter periods of time and for diseased populations, it sometimes does not do so badly.) Other measurements that are commonly log transformed in diseased populations include • Serum creatinine in kidney disease • C-reactive protein in the presence of a population with inflammatory components to their disease • Prothrombin time in patients with clotting abnormalities • Prostate specific antigen in patients with prostate cancer • Alanine aminotransferase (ALT), Aspartate aminotransferase (AST), alkaline phosphatase in liver dis-ease • Antibody titers in autoimmune diseases 3 Examples of Summary Measures that are Often Logarithmically Modeled In our general regression model, contrasts of some summary measure θ are made across groups defined by covariates ⃗ X = (X0 = 1, X1, X2, . . . Xp) using the expression g(θ) = ⃗ XT ⃗ β = β0 + β1X1 + · · · + βpXp. Version: January 22, 2014 6 Ratios and Logarithms SS Emerson In this expression, we call η = ⃗ XT ⃗ β the “linear predictor” that represents the combined “effect” of all the covariates on the value of θ. For a population having ⃗ X = ⃗ xi we can define the linear predictor ηi = β0 + β1x1i + β2x2i + · · · + βpxpi. Note that the for the ith population having ⃗ X = ⃗ xi and the jth population having ⃗ X = ⃗ xj we find that the difference in linear predictors across the two populations is ηi −ηj = p X ℓ=1 βℓ(xℓi −xℓj) = β1(x1i −x1j) + β2(x2i −x2j) + · · · βp(xpi −xpj). We term g( ) the “link function” that links the linear predictor back to the distributional summary measure θ. A regression model that uses the identity function g(θ) = θ is called “additive” on the linear predictor, because differences (ηi −ηj) in the linear predictors for two populations relate to the difference between the values of θ for the two populations: θi −θj = (ηi −ηj). The commonly used regression model with an identity link is: • Linear regression: θ is the mean of some response variable Y , and g(θ) = θ yielding a regression model θ⃗ x = E[Y | ⃗ X = ⃗ x] = β0 + β1X1 + · · · + βpXp. A regression model that uses the logarithmic function g(θ) = log(θ) is called “multiplicative” on the linear predictor, because differences (ηi −ηj) in the linear predictors for two populations relate to the ratio between the values of θ for the two populations: θi/θj = e(ηi−ηj). Invariably, the log link is defined using the natural logarithm loge. The following commonly used regression models use a log link: • Logistic regression: θ = p/(1 −p) is the odds of some Bernoulli (binary) response variable Y ∼B(1, p), and g(θ) = log(θ) yielding a regression model log(θ⃗ x) = log  p⃗ x (1 −p⃗ x)  = log " E[Y | ⃗ X = ⃗ x] (1 −E[Y | ⃗ X = ⃗ x]) # = β0 + β1X1 + · · · + βpXp. • Poisson regression: θ is the mean of some positive response variable Y , and g(θ) = log(θ) yielding a regression model log(θ⃗ x) = log h E[Y | ⃗ X = ⃗ x] i = β0 + β1X1 + · · · + βpXp. • Proportional hazards regression: θ = λ(t) is the hazard function of some response variable Y , and g(θ) = log(θ) yielding a regression model log(θ⃗ x) = log h λY (t| ⃗ X = ⃗ x) i = λ0(t) + β1X1 + · · · + βpXp. Note: The definition of θ and g( ) in the above regression models is my preferred interpretation. But it should be noted that other interpretations are possible. For instance, in Poisson regression, we could say that θ⃗ x = log(E[Y | ⃗ X = ⃗ x]) and that we use the identity link g(θ) = θ. When the “generalized linear model” was defined, it was always considered that θ was the mean of the response variable. In that setting then, logistic regression is interpreted as a model of θ⃗ x = E[Y | ⃗ X = ⃗ x] with logit link g(θ) = log[θ/(1 −θ)] yielding the exact same regression model: logit(θ⃗ x) = log  p⃗ x (1 −p⃗ x)  = log " E[Y | ⃗ X = ⃗ x] (1 −E[Y | ⃗ X = ⃗ x]) # = β0 + β1X1 + · · · + βpXp. The only thing that changes is the interpretation of the regression parameters in terms of θ and vice versa. Version: January 22, 2014 7 Ratios and Logarithms SS Emerson 4 Review of Logarithms 1. Recall from basic algebra that when you multiply two numbers, you add exponents. That is, if you want to multiply 103 time 107, the answer is 1010. 2. This only works when you express the numbers as exponents of the same base. Hence, we can not so easily multiply 23 times 45. Instead, we would want to convert each number to be a power of the same base. In the problem I have given here, this is easy, because 4 = 22. Hence, 45 = (22)5 = 210. 3. It is possible to raise a base number to a fractional power. For instance, 40.5 is just the square root of 4, or 2. Similarly, 810.25 is the fourth root of 81 (the square root of the square root), or 3. 4. Before calculators were in widespread use (I can remember back that far), logarithms were used to make multiplication problems easier. That is, every number was converted to an exponential form, the exponents were added, and then the answer was converted back. 5. In this process, some common base for the exponential form would have to be chosen. Commonly that base was 10 for tables of logarithms. The logarithm base 10 of a number was just the exponent of the number expressed as a power of 10. For instance, because 102 = 100, the logarithm base 10 of 100 is 2. Similarly, the logarithm base 10 of 1000 is 3, because 103 = 1000. 6. Every positive number can be expressed as a power of 10. For instance, 100.3010 = 2. Finding the appropriate exponent for such a representation (that exponent is termed the logarithm base 10, so the logarithm base 10 of 2 is 0.3010) involves a complicated formula, and in the old days tables were used. Now most calculators have a button you can push to find the logarithm base 10 of a number. 7. More generally, we can talk about the logarithm base k of a number x, which we will write as logk(x). k can be any positive number; it does not need to be an integer. If logk(x) = y, then ky = x. We sometimes speak of the antilog base k of y as being x. 8. In earlier math courses, you probably learned a convention that writing ‘log’ was understood to mean the base 10 logarithm and writing ‘ln’ was the natural logarithm (base e = 2.7182818 . . . ). Like all simple rules, however, this is violated regularly. In fact, it is common in science to use ‘log’ (with no subscript) to mean the natural logarithm. Many statistical software packages use this convention. We will see that this need not be so much of a problem, however. 9. Using different bases for logarithms is just like measuring length in different units (inches, feet, cen-timeters, miles, light years). No matter what base you use log(1) = 0 This is because k0 = 1 for all numbers k. 10. There is a constant of conversion between loge(x) and logk(x) for any base k. For instance, in the following table of selected base 2, base 10, and base e logarithms x log2(x) log10(x) loge(x) = ln(x) 1 0.000000 0.0000000 0.0000000 2 1.000000 0.3010300 0.6931472 3 1.584963 0.4771213 1.0986123 5 2.321928 0.6989700 1.6094379 10 3.321928 1.0000000 2.3025851 20 4.321928 1.3010300 2.9957323 Version: January 22, 2014 8 Ratios and Logarithms SS Emerson You can get every number in one column by multiplying the number in another column by some constant. For instance, every number in the log10(x) column is just .3010300 times the number in the log2(x) column. Similarly, every number in the loge(x) column is just 2.3025851 times the number in the log10(x) column. In general, then, we can find the base k logarithm of any number by either of the following formulas logk(x) = log10(x)/ log10(k) logk(x) = loge(x)/ loge(k) I know of no statistical packages that do not provide loge(x), and most provide log10(x) as well. 11. Important properties of the logarithm come from the properties of exponents: (a) logk(xy) = logk(x) + logk(y) (b) logk x −logk(y) = logk(x/y) (c) logk(xy) = y ∗logk(x) 12. In this class, as in most of science, we will use log(x) = y to mean loge(x) = y. This agrees with the nomenclature of both Stata and R. This also is the base that is used as the link function in regression models, so the antilog of y will be take ey = x. 5 Log Transformations in One- and Two-Sample Problems In one and two sample problems, there is no reason to transform the predictor of interest (POI): • In one-sample problems, the POI is constant and generally does not enter into the analysis in any way. • In two-sample problems, the POI is a binary variable. I usually encourage that POI to be coded as 0 or 1, though it does not truly affect the analysis results returned by standard statistical software for two-sample problems. (It does, however, change the results when two-sample problems are implemented in a regression model.) Hence, the only issue of concern is how to interpret a statistical analysis when the response variable is transformed. Suppose we have random variables Xi and Yi. If we take logarithmic transformations Wi = loge(Xi) and Zi = loge(Yi), then W is the natural log of the geometric mean of X, and Z is the natural log of the geometric mean of Y . It follows, then, that eW and eZ, are respectively the geometric means of X and Y . Furthermore, W −Z is the natural log of the ratio of geometric means. (The log of a ratio is the difference of the logs.) Thus, when we do inference using W and Z, we can easily back transform the data to get the geometric means and ratios of geometric means. Such back transformation works for point estimates and confidence intervals. For instance, eW −Z = eW /eZ is the ratio of the geometric mean for X to the geometric mean for Y . I note that if the log transformed data are symmetric, then the geometric mean and the median are the same number. In that case, we could refer to the ratio of medians. As a general rule, however, a larger sample size is required to be sure that a distribution is symmetric than is required to estimate the geometric means. Hence, I do not really recommend that you presume symmetry. It is safer to just talk about the geometric means. Version: January 22, 2014 9 Ratios and Logarithms SS Emerson 6 Log Transformations in Linear Regression Models We will first consider the linear regression model, because in that model we can consider transformations of both the response and predictor variables. 6.1 Untransformed Predictors Suppose we model E[Y |X = x] = β0 + β1 × x 1. From our standard interpretation of regression slope parameters, we know that every 1 unit difference in X is associated with a β1 unit difference in the expected value of Y : E[Y |X = a + 1] −E[Y |X = a] = (β0 + β1 × (a + 1)) −(β0 + β1 × a) = β1. If a straight line relationship holds, this is exactly true for every choice of a. If the true relationship is nonlinear, then β1 represents some sort of average difference. 2. Similarly, we know that every c unit difference in X is associated with a cβ1 unit difference in the expected value of Y : E[Y |X = a + c] −E[Y |X = a] = (β0 + β1 × (a + c)) −(β0 + β1 × a) = cβ1. 6.2 Transformations of Predictors Suppose we model E[Y |X = x] = β0 + β1 × logk(x) 1. From our standard interpretation of regression slope parameters, we know that every 1 unit difference in logk(X) is associated with a β1 unit difference in the expected value of Y . 2. Similarly, we know that every c unit difference in logk(X) is associated with a cβ1 unit difference in the expected value of Y . 3. Now, a 1 unit difference in logk(X) corresponds to a k-fold increase in X, and a c unit difference in logk(X) corresponds to a kc-fold increase in X. • Ex: A 1 unit change in log10(CHOLEST) corresponds to a 10 fold increase in CHOLEST. A 3 unit change in log2(CHOLEST) corresponds to a 23 = 8 fold increase in cholesterol. 4. If we want to talk about a 10% increase in X, then that would correspond to a c = logk(1.1) unit increase in logk(X). • Ex: Suppose we model predictor HEIGHT on a log base 10 scale. Because we never see a 10 fold increase in height, when interpreting our model parameters it might be better to consider comparisons between populations which differ in height by, say, 10%. We would then estimate the difference in the expected response as log10(1.1)ˆ β1, where ˆ β1 was the least squares estimate for the slope parameter in the regression. Note that we would find a confidence interval for the effect associated with that 10% change in height by multiplying the CI for β1 by log10(1.1) as well. (If you wanted to get a statistical package to do all this for you, just use the base 1.1 logarithm for height in the regression model: htlog = loge(ht)/ loge(1.1). Then a 1 unit change in your predictor corresponds to a 10% change in height.) Version: January 22, 2014 10 Ratios and Logarithms SS Emerson 6.3 Transformation of Response Suppose we model (for arbitrary base j) E[logj(Y )|X = x] = β0 + β1 × x 1. Using the standard interpretation of regression slope parameters, we know that every 1 unit difference in X is associated with a β1 unit difference in the expected value of logj(Y ), and every c unit difference in X is associated with a cβ1 unit difference in the expected value of logj(Y ). 2. Unfortunately, a β1 unit difference in the expected value of logj(Y ) does not have an easy interpretation in the expected value of Y . However, statements made about the distribution of logj(Y ) are generally not well understood by the general population, so we need to find another way. 3. The expected value of logj(Y ) is the log of the geometric mean of Y . Thus, we can make statements about the geometric mean of Y considering our model to be E[logj(Y )|X = x] = logj(GeomMn[Y |X = x]) = β0 + β1 × x 4. Under this modification, a β1 unit difference in the base j logarithm of the geometric mean of Y corresponds to a jβ1-fold change in the geometric mean of Y . Similarly, a cβ1 unit difference in the base j logarithm of the geometric mean of Y corresponds to a jcβ1-fold change in the geometric mean of Y . We can say that jcβ1 is the ratio of geometric means for two populations which differ by c units in their values for X. 5. It is probably easiest to use j = 10 or j = e, because most calculators have a button that will compute the antilogs for those bases. 6. (A very special case in which we can talk about medians. I truly recommend talking about geometric means, instead.) I note that under standard classical assumptions of linear regression (which classical assumptions assume normality of residuals), the expected value of logj(Y ) is also the median of logj(Y ). (Actually, we do not need normality, but we do need the error distribution to be symmetric about its mean. If you do assume normality, then we can state our assumption as being that Y has the lognormal distribution in each subpopulation.) Thus, we can make statements about the median of Y considering our model to be mdn[logj(Y )|X = x] = logj(mdn[Y |X = x]) = β0 + β1 × x Under this modification, a β1 unit difference in the base j logarithm of the median of Y corresponds to a jβ1-fold change in the median of Y . Similarly, a cβ1 unit difference in the base j logarithm of the median of Y corresponds to a jcβ1-fold change in the median of Y . We can say that jcβ1 is the ratio of medians for two populations which differ by c units in their values for X. 6.4 Transformations of the Response and Predictor This is just a combination of the above settings. That is, we talk about the ratio of geometric means of Y associated with a several-fold increase in X. Suppose we model (for arbitrary bases j and k) E[logj(Y )|X = x] = β0 + β1 × logk(x) 1. An r-fold change in X (so a c = logk(r) unit difference in logk(X)) will be associated with an rβ1/ logj k-fold change in the geometric mean of Y . That is, the geometric mean ratio of Y is rβ1/ logj k when comparing two populations, one of which has X r times higher than the other. 2. The above formula becomes much easier if the same base is used for both predictor and response. In this case, j = k, and the geometric mean ratio is simply rβ1 when comparing two populations, one of which has X r times higher than the other. Version: January 22, 2014 11 Ratios and Logarithms SS Emerson 7 Log Transformations in Regression Models using Log Links In logistic, Poisson, and proportional hazards regression we use a log link, and those logarithms are invariably on the natural log scale (loge). Hence we have to consider a different interpretation of the parameters, and in the remaining parts of this document I adopt the standard notation that log(x) = loge(x) and any other base would be explicitly specified (e.g., the base 10 logarithmic function would be written log10(x). 7.1 Untransformed Predictors Suppose we model log(θx) = β0 + β1 × x 1. From our standard interpretation of regression slope parameters, we know that every 1 unit difference in X is associated with a β1 unit difference in log(θx): log(θa+1) −log(θa) = (β0 + β1 × (a + 1)) −(β0 + β1 × a) = β1. If a straight line relationship holds, this is exactly true for every choice of a. If the true relationship is nonlinear, then β1 represents some sort of average difference. 2. Similarly, we know that every c unit difference in X is associated with a cβ1 unit difference in log(θx): log(θa+c) −log(θa) = (β0 + β1 × (a + c)) −(β0 + β1 × a) = cβ1. 3. We do not find it very convenient to talk about log(θ), however. We would rather talk about θ (i.e, the odds, the mean, or the hazard). Hence we back transform to obtain statements about the ratio of θ across groups. So we find that every 1 unit difference in X is associated with a eβ1-fold change in θ: θa+1 θa = eβ1 θa+c θa = ecβ1 = eβ1c . We can similarly say that the odds ratio (in logistic regression), the mean ratio (in Poisson regression), or the hazard ratio (in proportional hazards regression) is eβ1 for each 1 unit difference in the value of X, and the ratio is ecβ1 for each c unit difference in the value of X. 7.2 Transformations of Predictors Suppose we model log(θx) = β0 + β1 × logk(x) 1. From our standard interpretation of regression slope parameters, we know that every 1 unit difference in logk(X) is associated with a eβ1-fold change in the value of θ, and every c unit difference in logk(X) is associated with a ecβ1-fold change in the value of θ. 2. In units of X, we know that a 1 unit difference in logk(X) is a k-fold increase in X, and similarly a c unit difference in logk(X) is a kc-fold increase in X . If k is some convenient multiple (i.e., a doubling when k = 2) we just say that for every k-fold increase in X the value of θ increases eβ1 fold. 3. Note that in Stata and R, the output for logistic, Poisson, or proportional hazards regression can be provided on either the scale of the β’s or as eβ. You have to keep track of which is which. Version: January 22, 2014 12 Ratios and Logarithms SS Emerson • In Stata, logit returns the β’s (and includes the intercept β0), and logistic returns the eβ’s (and suppresses the intercept). • In Stata, poisson returns the β’s (and includes the intercept β0) by default. If you specify option ire, Stata returns the eβ’s (and suppresses the intercept). • In Stata, stcox returns the eβ’s by default, and if you specify option nohr Stata returns the β’s. (Note that the baseline hazard function takes on the role of an intercept in proportional hazards regression, and is never returned as part of the standard regression output.) • In R, summaries of the output of glm() and coxph() will tend to give both. 8 Communicating Ratios in Natural Language It is often quite difficult for people to interpret the many ways that we might talk about ratios. Below I present several examples of how you might describe the output when using a log link with an untransformed predictor. I will use the example of looking at the odds of “response” as a function of dose measured in grams. 1. For an estimated odds ratio of 1.31 I might say any of (a) “the odds of response is 1.31-fold higher in the experimental group taking 1 g of drug than it is in the control group taking placebo” (I tend to use this phrasing when there are only two groups. In this case I would also tend to explicitly state the odds (and/or probability) of response in each group, unless there were other covariates in the model.) (b) “the odds of response is 1.31-fold higher for every 1 g difference in dose” (c) “the odds of response is 31% higher for every 1 g difference in dose” (I tend to prefer this one for 1 < OR < 2) 2. For an estimated odds ratio of 2.31 I might say any of (a) “the odds of response is 2.31-fold higher in the experimental group taking 1 g of drug than it is in the control group taking placebo” it (I tend to use this phrasing when there are only two groups. In this case I would also tend to explicitly state the odds (and/or probability) of response in each group, unless there were other covariates in the model.) (b) “the odds of response is 2.31-fold higher for every 1 g difference in dose” (I tend to prefer this one for OR > 2) (c) “the odds of response is 131% higher for every 1 g difference in dose” 3. For an estimated odds ratio of 0.91 I might say any of (a) “the odds of response in the experimental group taking 1g of drug is 9% lower than the odds in the control group taking placebo” (I tend to use this phrasing when there are only two groups. In this case I would also tend to explicitly state the odds (and/or probability) of response in each group, unless there were other covariates in the model.) (b) “the odds of response is only 0.91 times as high for every 1 g difference in dose” (I tend to prefer this one for OR < 1 when there are more than two groups) (c) “the odds of response is 0.91 times as high for every 1 g difference in dose” (d) “the odds of response is 9% lower for every 1 g difference in dose” (I tend to think this is a little more confusing, because when you are going to consider a difference in dose of c grams, you have to take 0.91c, rather than use 0.09) Note the asymmetry of ratios: If the experimental to control odds ratio is 1.25 (so 25% higher), the control to experimental odds ratio is 0.80 (so 20% lower). Version: January 22, 2014 13
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https://www.youtube.com/watch?v=OPSCKXXvWiM
Writing Equations of Ellipses In Standard Form and Graphing Ellipses - Conic Sections The Organic Chemistry Tutor 9750000 subscribers 16863 likes Description 1309335 views Posted: 29 Apr 2021 This algebra video tutorial explains how to write the equation of an ellipse in standard form as well as how to graph the ellipse when in standard form. It explains how to find the coordinates of the foci, vertices, and co-vertices. This video contains plenty of examples and practice problems. Get The Full 1 Hour 55 Minute Video: Direct Link to The Full Video: Conic Sections - Free Formula Sheet: Full 1 Hour 55 Minute Video: Conic Sections - Video Lessons: 371 comments Transcript: Introduction in this video we're going to focus on ellipses we're going to talk about how to graph an ellipse and also how to find the coordinates of the vertices and the foci of the ellipse so let's begin on the left we have an ellipse with a horizontal major axis on the right the major axis is vertical the left of the major axis is always equal to 2a and that's the case for both types of ellipses this is the length of the minor axis which is vertical on the left but it's horizontal on the right the left of the minor axis it's equal to 2b for this one the minor axis is horizontal but it's still equal to 2b now whenever you have an ellipse where it's centered at the origin here's the equations that you're going to be dealing with for horizontal ellipse it's x squared over a squared plus y squared over b squared for the ellipse on the right it's x squared over b squared plus y squared over a squared equals one now to determine which type of ellipse that you have whether it's elongated in the x direction or elongated in the y direction look at your a value a is always bigger than b when dealing with ellipses if the larger number is under x squared you're dealing with an ellipse with a horizontal major axis if the larger number a is under y squared then the ellipse is vertical it has a major vertical axis so let's say this is the center this is going to be positive a or just a and this is negative a this is b and this is negative b so here are the major vertices and these two points represent the minor vertices for the graph on the right here are the major vertices and here are the minor vertices for the ellipse on the left the coordinates of the major vertices are plus or minus a comma zero for the minor vertices it's zero plus or minus b for the ellipse on the right it's the reverse the coordinates of the major vertices are zero comma plus or minus b and for the minor vertices it's plus or minus a comma zero now let's talk about the coordinates of the foci for the ellipse on the left the foci will be along the major horizontal axis and it's c units away from the origin or the center so once you have the center if you add c you'll get the focus on the right if you subtract it by c you'll get the focus on the left so the coordinates for the foci is going to be plus or minus c comma 0 for the ellipse on the left side for the ellipse on the right side it's 0 plus or minus c so starting from the center need to go up c units and down c units and that will give you the coordinates of the foci now what happens if we have an ellipse that is not centered at the origin if it's not centered at the origin then the coordinates of the center become h comma k this equation changes to x minus h squared over a squared plus y minus k squared over b squared is equal to one and the same is true for this equation everything will be the same but we're going to replace x with x minus h squared and this will be over b squared and then we're going to replace y with y minus k squared and this will be over a squared now let's focus our attention on the horizontal ellipse Horizontal ellipse when the center is the origin i mentioned that the coordinates of the foci is plus or minus c comma zero but now when the center is shifted to some point h comma k the coordinates of the foci or rather the foci will be shifted so all you need to do is add the coordinates on the center to the coordinates of the foci so we're going to add plus or minus c to h so it's going to be h plus or minus c and then we're going to add k to 0 which will be k so this will be the new coordinates of the foci for horizontal ellipse now the coordinates of the vertices are plus or minus a comma zero and zero plus or minus b so when it's shifted all you got to do is add the coordinates to the center to the vertices to get the new one so to get the major vertices is going to be h plus or minus a comma k and for the minor vertices it's going to be h comma k plus or minus b now let's consider the situation if we have a vertical ellipse as a center shifts from the origin to some point h comma k let's see what's going to happen to the foci so the coordinates of the foci if the center is at the origin is 0 plus or minus c when the graph has shifted from the origin it's going to be h comma k plus or minus c the major vertices will be 0 plus or minus a the minor vertices is plus or minus b comma zero so when it's been shifted this becomes h comma k plus or minus a and this becomes h plus or minus b comma k now let's work on some practice problems Practice problems go ahead and graph the ellipse and identify the coordinates of the vertices and the foci by the way in order to find the foci you need to know how to calculate c and here is the equation to calculate c it's a squared minus b squared well c squared is a squared minus b squared not h squared plus b squared as in the case of the pythagorean theorem for hyperbolas is c squared equals a squared plus b squared but for ellipses c squared is equal to a squared minus b squared so that's how you can calculate c using that formula so feel free to try this problem in this problem there's no h or k value so the coordinates of the center is zero comma zero a squared is going to be the larger of the two numbers so nine is greater than four that means a squared is nine which means a is the square root of nine so a is three b squared is the other number four which means b is the square root of four or two so now that we know the values of a and b and the center we can go ahead and graph the ellipse so first let's plot the center a is under the x variable so we're going to travel three units to the right from the center along the x axis and three units to the left so this gives us the coordinates of the major vertices which is going to be plus or minus a comma 0 or plus or minus 3 comma 0. now b is 2 and b or b squared is under y squared so we're going to go up two units from the center along the y axis and down two units so this gives us the coordinates of the minor vertices which is zero plus or minus b or zero plus or minus two so those are the coordinates of the vertices of the ellipse now we can go ahead and graph the ellipse by connecting these four vertices together so that's how you can graph an ellipse now how can we find the coordinates of the foci so the coordinates of the foci is going to be plus or minus c comma 0. it's going to be along the major horizontal axis for this problem so since a is along the x-axis the foci will be along the x-axis as well so we need to determine c we know that c squared is a squared minus b squared a squared is nine b squared is four nine minus four is five taking the square root of both sides we get that c is equal to plus or minus the square root of five so the coordinates of the foci will be plus or minus root five comma zero if you have your calculator and you type in the square root of 5 the square root of 5 is 2.236 approximately so along the x axis that's going to be somewhere in this region so that's how we can plot the coordinates of the foci for this problem now let's try a similar problem to the last one so for the sake of practice feel free to Example problem pause the video and work on this example problem go ahead and graph the ellipse and then find the coordinates of the vertices and the foci so the first thing we should do is identify the coordinates of the center there's no h or k for this problem so the center is going to be the origin it's zero comma zero next we need to determine our a squared value is it nine or is it sixteen a squared will be the larger of the two numbers 16 is greater than 9 so a squared is 16. so once we have a squared we need to calculate a a is going to be the square root of 16 which is four b squared has to be the other number nine thus b is going to be the square root of nine which is three so now that we have a and b we can go ahead and graph the ellipse so b is 3 and b squared is associated with the x variable it's under x squared so what we're going to do is we're going to travel three units to the right from the center and three units to the left from the center so this is going to give us not the major vertices but the minor vertices the minor vertices is always associated with b so the coordinates of the minor vertices for this problem is plus or minus b comma zero so that's plus or minus three comma zero now a is four and it's associated with y a squared is under y squared so a is going to be along the y axis so we're gonna travel four units from the center and four units below the center so the coordinates of the major vertices will be zero comma four and zero negative four and here's the center so it's zero plus or minus a or zero plus or minus zero comma plus or minus four so now let's go ahead and graph the ellipse so that's how we can plot it now the last thing we need to do is find the coordinates of the foci so because the major horizontal axis is the y-axis in this case the foci will be along that major vertical axis but first we need to calculate c let's use this formula c squared is equal to a squared minus b squared a squared is 16 b squared is nine 16 minus nine is seven so c is going to be plus or minus square root seven now the square root of seven it's about two point six so the coordinates of the foci will be in this region it's in the same direction as plus or minus a which is along the y axis now the coordinates of the fill sinus problem is going to be zero comma plus or minus c or zero comma plus or minus the square root of seven so that's how you could find the coordinates of the foci as well as the coordinates of the vertices number three identify the coordinates of the center vertices and foci and then graph the ellipse so the center is going to be h comma k here we have x minus 3 h is simply going to be 3. you just need to change the sign here we have y plus 2 so h is going to be i mean k is going to be negative 2. so that's h and k in this problem so now that we have the coordinates of the center let's determine our a and b values so which one is going to be a squared 16 or 25 well 25 is larger than 16 so a squared will be 25 a is going to be the square root of 25 which is 5 b squared is the other number 16 so b is going to be 4. now let's go ahead and graph it so first let's plot the center which is three negative two so we're going to travel three units to the right and we're going to travel down two units so there is the center now a is five and a is associated with the y variable so from the center we're going to go up five units right now we're at a y value of negative two so we're going to stop at three and then we're gonna go down 5 units and so the coordinates of the vertices rather than being 0 comma plus or minus a it's h comma k plus or minus a h is 3 k is negative 2 a is 5. negative 2 plus 5 is 3 and negative 2 minus 5 is negative 7. so here this point is at three comma three and this point here is three and negative seven so those are the coordinates of the major vertices now let's focus on b b is 4 and it's associated with the x variable so we're going to go four units to the left and four units to the right starting from the center so let's go four units to the right so we get this point here and then four units to left will take us to this point so the coordinates of the minor vertices rather than being plus or minus b comma zero it's going to be h plus or minus b comma k h is three b is four k is negative two three plus four is seven and three minus four is negative one so this point here that's seven negative two and then this other one is negative one negative two so now let's go ahead and graph the ellipse so that's how we can graph this particular ellipse and we have the coordinates of the major and minor vertices now let's focus on finding the coordinates of the foci so the major axis left or the major axis is vertical thus the coordinates of the foci will be along that line so rather than being zero comma plus or minus c it's going to be h comma k plus or minus c so let's begin by finding c c squared is going to be a squared minus b squared a squared is 25 b squared is 16. 25 minus 16 is nine the square root of nine is three so c is going to be plus or minus three by the way technically a is plus or minus five and b is plus or minus four because to get the coordinates of the vertices we would have to like add five and subtract five from the center so it's technically more accurate to say plus or minus 4 and plus minus 5 for a and b so now that we have the value of c we said that c is plus or minus 3. let's use this to find the coordinates of the foci so h is 3 k is negative 2 and we're going to add plus or minus 3 to that negative 2 plus 3 is 1 and negative two minus three is negative five so we have a focus at three one which is here and we have another one at three negative five which is here so technically what we need to do is starting from the center we would go up c units or up 3 units to get this point 3 comma 1 and then we would travel down 3 units to get the other focus which is three negative five number four identify the coordinates of the center vertices and foci and then we'll graph the ellipse so let's begin with this center so here we have a negative one in front of x we're going to change it to positive one and here we have a positive one in front of y so this will become negative one so the center is at one negative one a squared is going to be the larger of the two numbers so a squared is nine a is going to be the square root of nine which is three if you want to write plus or minus 3 that's up to you but if you just put 3 you can still graph it correctly and you'll still get the correct coordinates b squared that's going to be 4. the square root of 4 is 2. now let's go ahead and calculate c so c squared is a squared minus b squared a squared is nine b squared is four so c squared is five the c is going to be the square root of five and you could write that as plus or minus root five if you want to now let's go ahead and graph the ellipse let's first put the appropriate markings on the x and y axis so the center is at one negative one which is here a is 3 and a is associated with the x variable so we're going to travel 3 units to the right and 3 units to the left b is 2 and it's b squared is under y squared so we're going to go up 2 parallel to the y axis and then down 2. so that's how we can graph the ellipse in this problem now let's identify the coordinates of the vertices if you ever forget the formula you can just look at your graph so here we can see that it's four negative one and here the x value is negative two with the y value is negative one so it's negative two comma one and if you remember the formula it's h plus or minus a comma k for a horizontal ellipse h is one a is three k is negative one so you get one plus three which is four and one minus three which is negative two and this should be negative two negative one so those are the ways in which you can find the vertices of the ellipse now let's find the minor vertices of the ellipse so this point here has an x and y value of one and this point here has an x value of one but a y value of negative three or you could use this formula it's h plus or minus b actually take that back it's h comma k plus or minus b so h is one k is negative one b is two so it's negative one plus two which is one and then it's negative one minus two which is negative three giving us those two minor vertices i want you to be familiar with this formula because sometimes a or b could be a radical value and to get the right answer you need to use that formula as will be the case of the foci in this example so the coordinates of the foci for horizontal ellipse is going to be h plus or minus c comma k h is 1 c is the square root of 5 so it's going to be 1 plus or minus root 5 comma k in this case we can't simplify the coordinates of the foci i mean foci so we have to leave it like this so that's our answer for the coordinates of the foci so starting from the center we would have to travel c units to the right the square root of 5 is like 2.2 something 2.236 so going a little bit more than two to the right would take us somewhere to this point and a little bit more than two to the left will take us to that point so that's approximately where the two focal points will be but this is what you would report as your answer for the coordinates of the foci so anytime you have a radical it's going to be hard to identify the points using the graph you need to use this expression so that's why i want you to learn how to use it because you're going to need it for some problems number five write the standard form of the equation for the ellipse shown below so first we need to realize that the center is at the origin and this is a horizontal ellipse where we have a horizontal major axis so a squared is going to be under x squared so this is the general formula of the equation that we need to use x squared over a squared plus y squared over b squared is equal to 1. so if we could find the values of a and b we could write the equation in standard form a is 4 because this point here the major vertices is 4 units away from the center we can see that b is two the minor vertices are two units away from the center so if a is four that means that a squared is four squared which is 16. and if b is 2 b squared or 2 squared is 4. so the formula that we have is x squared over 16 squared plus y squared over 4 squared is equal to one so this is the answer for this problem number six write the standard form of the equation for the ellipse shown below so we're given the center of the ellipse which is negative three comma two and we could find a and b so the distance between the center and one of the vertices we could see it's one two three four five units so it's five units along the x direction and along the y direction it's one two three let me use a different color to count so this is one two three four so we went down four units to get one of the minor vertices so a is going to be the bigger number that means a is five b is 4. if a is 5 a squared is going to be 25 b squared is 16. so now that we have the values of a squared and b squared and we have c our center we can write the standard form of the equation so a is horizontal so what we have is a horizontal ellipse on a list with a horizontal major axis so the formula we're going to use is x minus h squared over a squared since a is parallel to the x axis plus y minus k squared over b squared is equal to one h is negative three k is two so this is going to be x minus h let's use brackets for now h is negative 3 a squared is 25 and then y minus k is positive 2 b squared is 16. so this becomes x plus three so we change negative three to positive three squared over 25 plus y minus two squared over 16 is equal to 1. so that's how we can convert from the graph to the equation in standard form you
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Kelvin - Examples, Definition, Formula, Scales, Effects, Differences Loading [MathJax]/jax/output/HTML-CSS/config.js Subjects English Maths Chemistry Physics Biology Docs Design Business Education Resources English Maths Chemistry Physics Biology Docs Design Business Education AP Practice Tests AP English AP English Language AP English Literature AP Sciences AP Chemistry AP Biology AP Environmental Science AP Physics 1 AP® Physics 2: Algebra-Based AP® Physics C: Electricity and Magnetism AP® Physics C: Mechanics AP® Psychology AP Math and Computer Science AP Calculus AB AP Calculus BC AP® Computer Science A AP® Computer Science Principles AP® Precalculus AP® Statistics AP History and Social Sciences AP® European History AP® Human Geography AP® Macroeconomics AP® Microeconomics AP® United States Government and Politics AP® United States History AP® World History: Modern Accounting CPA CMA CIA Digital Sat Finance CFA CMT More Grad School MCAT Legal MBE Other ACT Digital Sat HomeKelvin Physics Kelvin Planck Statement – Examples, Definition, Uses, FAQ’s Schematic Diagram – 10+ Examples, Symbols, Types, Differences System of Particles and Rotational Dynamics – 10+ Examples, Applications Waves – 20+ Examples, Types, Differences, Properties Units and Measurements – 20+ Examples, Types, Uses Thermal Properties of Matter – 20+ Examples, Types Sound – 40+ Examples, Types, Characteristics, Applications Safety Measures Technology – 15+ Examples, Types, Applications Solid Deformation – 15+ Examples, Types, Properties Scalars and Vectors – 25+ Examples, Differences, Properties Quantum Physics – 10+ Examples, Formulas, Laws, Differences Optics – 10+ Examples, Types, Applications Electrostatics – 15+ Examples, Formulas, Importance Communication System – 15+ Examples, Types, Principles Circuits – 25+ Examples, Types, Rules, Differences Diode – 20+ Examples, Types, Characteristics, Applications Electromagnetic Waves – 20+ Examples, Equation, Types, Applications Oscillation – 35+ Examples, Formula, Types, Differences Nuclear Physics – 10+ Examples, Types, Differences, Applications Constants – List of Constants In Physics with Examples Home Physics Kelvin – Examples, Definition, Formula, Scales, Effects, Differences Last Updated:August 27, 2024 Notes 5 save Save Notes Kelvin – Examples, Definition, Formula, Scales, Effects, Differences The Kelvin scale is a fundamental temperature scale in physics, named after Lord Kelvin. It starts at absolute zero, the theoretical point where all molecular motion ceases, marked as 0 Kelvin (K). Unlike Celsius and Fahrenheit, Kelvin does not use degrees and directly correlates with Celsius for scientific accuracy: 0°C equals 273.15 K. This scale is essential in fields ranging from theoretical physics to practical applications in various technologies, providing a precise measurement system that is crucial for scientific calculations across the globe. What is Kelvin? Kelvin is a unit of temperature in the International System of Units (SI), symbolized as K. It is named after the British physicist Lord Kelvin. This scale measures temperature beginning at absolute zero, where all thermal motion ceases, defined as 0 Kelvin. Unlike Fahrenheit or Celsius, it does not use degrees, making it crucial for scientific and engineering calculations. The Kelvin scale is pivotal in fields such as physics, chemistry, and space science, where precise temperature measurements are essential. Kelvin Scale Formula The Kelvin scale is directly related to the Celsius scale with a simple formula to convert temperatures between them. Here’s the formula for converting Celsius to Kelvin: K = C+273.15 Where: K is the temperature in Kelvin, C is the temperature in Celsius. This formula shows that Kelvin and Celsius scales are offset by 273.15 degrees. This means that 0 degrees Celsius, which is the freezing point of water, corresponds to 273.15 Kelvin. Absolute zero, the theoretical minimum temperature, is 0 Kelvin, which equals -273.15 degrees Celsius. How to Convert a Temperature from Celsius to Kelvin Converting a temperature from Celsius to Kelvin is straightforward. Here’s the simple step-by-step process: Start with your Celsius temperature. This is the temperature you want to convert to Kelvin. Add 273.15 to your Celsius temperature. This conversion factor adjusts for the difference in starting points of the two scales—Celsius starts at the freezing point of water (0°C), whereas Kelvin starts at absolute zero. The result is the temperature in Kelvin. Here’s the formula you use for the conversion: 𝐾 = 𝐶+273.15 Example: If you have a temperature of 25°C and want to convert it to Kelvin: 𝐾=25+273.15=298.15 K Now, you have converted 25°C to Kelvin, which is 298.15 K. This method is used in all scientific contexts where Kelvin is the preferred unit of temperature measurement. Relation Between Celsius and Kelvin | Aspect | Celsius (°C) | Kelvin (K) | --- | Unit Name | Named after Anders Celsius | Named after Lord Kelvin | | Zero Point | Represents the freezing point of water | Represents absolute zero | | Scale Type | Relative temperature scale | Absolute temperature scale | | Unit Increment | Each increment is equal to 1°C | Each increment is equal to 1K | | Common Usage | Widely used in daily life and weather reporting globally | Predominantly used in scientific and engineering contexts | | Base Reference for Zero | 0°C is the freezing point of water | 0 K is the lowest possible temperature, where molecular motion stops | | Conversion Formula | 𝐶 = 𝐾−273.15 | 𝐾 = 𝐶+273.15 | Absolute zero in Kelvin Absolute zero is the theoretical lowest temperature possible, where all molecular motion ceases. In the Kelvin temperature scale, absolute zero is defined as exactly 0 Kelvin (0 K). This temperature corresponds to -273.15 degrees Celsius and -459.67 degrees Fahrenheit. Absolute zero is a cornerstone in the fields of physics and chemistry, particularly in understanding the behavior of gases and the laws of thermodynamics. Scale of the Kelvin The Kelvin scale is an absolute temperature scale used primarily in the scientific community worldwide. Here are some key characteristics and the scale points associated with it: Absolute Zero: 0 Kelvin (K) – This is the starting point of the Kelvin scale, representing the lowest possible temperature where all thermal motion of atoms theoretically ceases. Triple Point of Water: 273.16 K – This is a unique temperature where water can exist simultaneously in gas, liquid, and solid states. This specific point is used to define the Kelvin scale. Increment: The scale increments by 1 Kelvin, which is equivalent to 1 degree Celsius in magnitude. This means that the temperature interval or thermal unit size is the same as that of the Celsius scale, but the Kelvin scale starts at absolute zero. Usage: Kelvin is used extensively in scientific measurements because it allows for precise calculations and descriptions of thermodynamic processes, especially in physics and chemistry. Common Benchmarks: Some common reference temperatures in the Kelvin scale include: Boiling point of water: approximately 373.15 K at standard atmospheric pressure. Freezing point of water: 273.15 K. Room temperature: approximately 293 K to 298 K. Relative Comparison to Other Scales: Kelvin directly relates to the Celsius scale by a simple addition or subtraction of 273.15, making it easy to convert between the two. Unlike Fahrenheit, which requires a more complex conversion formula, Kelvin and Celsius share the same incremental scale, differing only in their starting point (zero value). The Kelvin scale is critical for scientific experiments and calculations, as it provides a thermodynamically absolute frame of reference that simplifies many formulae and theoretical models. Effects of Kelvin Scientific Research and Experimentation Precision in Measurements: The Kelvin scale allows for precise and absolute measurements of temperature, which are crucial in fields like physics, chemistry, and engineering. Thermodynamic Calculations: Kelvin is essential for calculations involving the laws of thermodynamics, as it starts at absolute zero, simplifying many thermodynamic equations by eliminating negative temperature values. Standardization Across Disciplines Universal Scientific Standard: Kelvin provides a consistent temperature measurement across all scientific disciplines, facilitating clearer communication and data comparison. International Standards: The Kelvin scale is part of the International System of Units (SI), making it the standard unit of thermodynamic temperature measurement worldwide. Technological and Industrial Applications Material Science: Understanding how materials behave at various temperatures, especially near absolute zero, is crucial for developing superconductors and other advanced materials. Cryogenics: Kelvin is the standard unit in cryogenics where extremely low temperatures are necessary, for example, in the storage of biological materials or in superconducting magnet applications. Astrophysical Studies Stellar Temperatures: The temperatures of stars and other celestial bodies are measured in Kelvin, which helps astronomers in classifying stars and understanding stellar evolution. Cosmology: Kelvin is used to measure the Cosmic Microwave Background radiation, providing evidence of the Big Bang and aiding in the understanding of the universe’s origins and its temperature over time. Educational Implications Learning and Teaching: The introduction of Kelvin in education facilitates the understanding of absolute temperatures and basic physical concepts, such as entropy and the zeroth law of thermodynamics. FAQs What is the triple point of water and why is it important to the Kelvin scale? The triple point of water is the temperature and pressure at which water can coexist in three states: solid, liquid, and gas. It is precisely defined as 273.16 K. This point is used to calibrate temperature measurements and define the Kelvin scale with high precision. Why does the Kelvin scale not use degrees like Celsius or Fahrenheit? Kelvin is used as an absolute scale for scientific accuracy, so it measures temperature in Kelvins, not degrees. This emphasizes that the scale starts from an absolute zero, which is different from relative temperature scales like Celsius and Fahrenheit. What is the significance of absolute zero on the Kelvin scale? Absolute zero (0 K) is the theoretical lowest possible temperature, where all thermal motion ceases. This concept is crucial for understanding the fundamental principles of thermodynamics and quantum mechanics. Share : Set content Set content Set content Set content Set content Save Download Kelvin – Examples, Definition, Formula, Scales, Effects, Differences The Kelvin scale is a fundamental temperature scale in physics, named after Lord Kelvin. It starts at absolute zero, the theoretical point where all molecular motion ceases, marked as 0 Kelvin (K). Unlike Celsius and Fahrenheit, Kelvin does not use degrees and directly correlates with Celsius for scientific accuracy: 0°C equals 273.15 K. This scale is essential in fields ranging from theoretical physics to practical applications in various technologies, providing a precise measurement system that is crucial for scientific calculations across the globe. What is Kelvin? Kelvin is a unit of temperature in the International System of Units (SI), symbolized as K. It is named after the British physicist Lord Kelvin. This scale measures temperature beginning at absolute zero, where all thermal motion ceases, defined as 0 Kelvin. Unlike Fahrenheit or Celsius, it does not use degrees, making it crucial for scientific and engineering calculations. The Kelvin scale is pivotal in fields such as physics, chemistry, and space science, where precise temperature measurements are essential. Kelvin Scale Formula The Kelvin scale is directly related to the Celsius scale with a simple formula to convert temperatures between them. Here’s the formula for converting Celsius to Kelvin: K = C+273.15 Where: K is the temperature in Kelvin, C is the temperature in Celsius. This formula shows that Kelvin and Celsius scales are offset by 273.15 degrees. This means that 0 degrees Celsius, which is the freezing point of water, corresponds to 273.15 Kelvin. Absolute zero, the theoretical minimum temperature, is 0 Kelvin, which equals -273.15 degrees Celsius. How to Convert a Temperature from Celsius to Kelvin Converting a temperature from Celsius to Kelvin is straightforward. Here’s the simple step-by-step process: Start with your Celsius temperature. This is the temperature you want to convert to Kelvin. Add 273.15 to your Celsius temperature. This conversion factor adjusts for the difference in starting points of the two scales—Celsius starts at the freezing point of water (0°C), whereas Kelvin starts at absolute zero. The result is the temperature in Kelvin. Here’s the formula you use for the conversion: 𝐾 = 𝐶+273.15 Example: If you have a temperature of 25°C and want to convert it to Kelvin: 𝐾=25+273.15=298.15 K Now, you have converted 25°C to Kelvin, which is 298.15 K. This method is used in all scientific contexts where Kelvin is the preferred unit of temperature measurement. Relation Between Celsius and Kelvin | Aspect | Celsius (°C) | Kelvin (K) | --- | Unit Name | Named after Anders Celsius | Named after Lord Kelvin | | Zero Point | Represents the freezing point of water | Represents absolute zero | | Scale Type | Relative temperature scale | Absolute temperature scale | | Unit Increment | Each increment is equal to 1°C | Each increment is equal to 1K | | Common Usage | Widely used in daily life and weather reporting globally | Predominantly used in scientific and engineering contexts | | Base Reference for Zero | 0°C is the freezing point of water | 0 K is the lowest possible temperature, where molecular motion stops | | Conversion Formula | 𝐶 = 𝐾−273.15 | 𝐾 = 𝐶+273.15 | Absolute zero in Kelvin Absolute zero is the theoretical lowest temperature possible, where all molecular motion ceases. In the Kelvin temperature scale, absolute zero is defined as exactly 0 Kelvin (0 K). This temperature corresponds to -273.15 degrees Celsius and -459.67 degrees Fahrenheit. Absolute zero is a cornerstone in the fields of physics and chemistry, particularly in understanding the behavior of gases and the laws of thermodynamics. Scale of the Kelvin The Kelvin scale is an absolute temperature scale used primarily in the scientific community worldwide. Here are some key characteristics and the scale points associated with it: Absolute Zero: 0 Kelvin (K) – This is the starting point of the Kelvin scale, representing the lowest possible temperature where all thermal motion of atoms theoretically ceases. Triple Point of Water: 273.16 K – This is a unique temperature where water can exist simultaneously in gas, liquid, and solid states. This specific point is used to define the Kelvin scale. Increment: The scale increments by 1 Kelvin, which is equivalent to 1 degree Celsius in magnitude. This means that the temperature interval or thermal unit size is the same as that of the Celsius scale, but the Kelvin scale starts at absolute zero. Usage: Kelvin is used extensively in scientific measurements because it allows for precise calculations and descriptions of thermodynamic processes, especially in physics and chemistry. Common Benchmarks: Some common reference temperatures in the Kelvin scale include: Boiling point of water: approximately 373.15 K at standard atmospheric pressure. Freezing point of water: 273.15 K. Room temperature: approximately 293 K to 298 K. Relative Comparison to Other Scales: Kelvin directly relates to the Celsius scale by a simple addition or subtraction of 273.15, making it easy to convert between the two. Unlike Fahrenheit, which requires a more complex conversion formula, Kelvin and Celsius share the same incremental scale, differing only in their starting point (zero value). The Kelvin scale is critical for scientific experiments and calculations, as it provides a thermodynamically absolute frame of reference that simplifies many formulae and theoretical models. Effects of Kelvin Scientific Research and Experimentation Precision in Measurements: The Kelvin scale allows for precise and absolute measurements of temperature, which are crucial in fields like physics, chemistry, and engineering. Thermodynamic Calculations: Kelvin is essential for calculations involving the laws of thermodynamics, as it starts at absolute zero, simplifying many thermodynamic equations by eliminating negative temperature values. Standardization Across Disciplines Universal Scientific Standard: Kelvin provides a consistent temperature measurement across all scientific disciplines, facilitating clearer communication and data comparison. International Standards: The Kelvin scale is part of the International System of Units (SI), making it the standard unit of thermodynamic temperature measurement worldwide. Technological and Industrial Applications Material Science: Understanding how materials behave at various temperatures, especially near absolute zero, is crucial for developing superconductors and other advanced materials. Cryogenics: Kelvin is the standard unit in cryogenics where extremely low temperatures are necessary, for example, in the storage of biological materials or in superconducting magnet applications. Astrophysical Studies Stellar Temperatures: The temperatures of stars and other celestial bodies are measured in Kelvin, which helps astronomers in classifying stars and understanding stellar evolution. Cosmology: Kelvin is used to measure the Cosmic Microwave Background radiation, providing evidence of the Big Bang and aiding in the understanding of the universe’s origins and its temperature over time. Educational Implications Learning and Teaching: The introduction of Kelvin in education facilitates the understanding of absolute temperatures and basic physical concepts, such as entropy and the zeroth law of thermodynamics. FAQs What is the triple point of water and why is it important to the Kelvin scale? The triple point of water is the temperature and pressure at which water can coexist in three states: solid, liquid, and gas. It is precisely defined as 273.16 K. This point is used to calibrate temperature measurements and define the Kelvin scale with high precision. Why does the Kelvin scale not use degrees like Celsius or Fahrenheit? Kelvin is used as an absolute scale for scientific accuracy, so it measures temperature in Kelvins, not degrees. This emphasizes that the scale starts from an absolute zero, which is different from relative temperature scales like Celsius and Fahrenheit. What is the significance of absolute zero on the Kelvin scale? Absolute zero (0 K) is the theoretical lowest possible temperature, where all thermal motion ceases. This concept is crucial for understanding the fundamental principles of thermodynamics and quantum mechanics. AI Generator Text prompt Add Tone Select a Tone Friendly Formal Casual Instructive Professional Empathetic Humorous Serious Optimistic Neutral Generate 10 Examples of Public speaking 20 Examples of Gas lighting Practice Test What is the freezing point of water in Kelvin? Choose the correct answer 0 K 100 K 273 K 373 K of 10 What is the boiling point of water in Kelvin? Choose the correct answer 100 K 273 K 373 K 473 K of 10 If the temperature is 25°C, what is the temperature in Kelvin? Choose the correct answer 248 K 273 K 298 K 323 K of 10 Which of the following is the absolute zero temperature? Choose the correct answer 0°C 0°F 0 K -273 K of 10 A temperature of 300 K is equivalent to how many degrees Celsius? Choose the correct answer 27°C 37°C 47°C 57°C of 10 What is the typical room temperature in Kelvin? Choose the correct answer 273 K 293 K 298 K 310 K of 10 If the temperature of an object increases by 10°C, by how much does the temperature increase in Kelvin? Choose the correct answer 10 K 20 K 30 K 40 K of 10 Which of the following temperatures is the highest? Choose the correct answer 0°C 0°F 0 K 100 K of 10 Which of the following is NOT a valid Kelvin temperature? Choose the correct answer -1 K 0 K 273 K 373 K of 10 If a scientist measures a temperature of 77 K, what is this temperature in Celsius? Choose the correct answer -196°C -173°C -123°C -96°C of 10 Submit Test Free Download school Ready to Test Your Knowledge? close Before you leave, take our quick quiz to enhance your learning! assessment Assess Your Mastery emoji_events Boost Your Confidence speed Instant Results memory Enhance Retention event_available Prepare for Exams repeat Reinforce Learning 👉 Start the Quiz Now Free Interactive Resources ©2025 Examples.com. All rights reserved. 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https://blog.csdn.net/aerchi/article/details/53031790
数学闭区间和开区间的区别是什么_开区间和闭区间的区别-CSDN博客 博客 下载 学习 社区 GitCode InsCodeAI 会议 搜索 AI 搜索 登录 登录后您可以： 复制代码和一键运行 与博主大V深度互动 解锁海量精选资源 获取前沿技术资讯 立即登录 会员·新人礼包 消息 历史 创作中心 创作 数学闭区间和开区间的区别是什么 最新推荐文章于 2025-08-23 11:09:26 发布 转载 于 2016-11-04 09:32:09 发布·1.3w 阅读 · 4 · 4 社区：2048 AI社区加入 自我修练 专栏收录该内容 87 篇文章 订阅专栏 本文详细介绍了数学中区间的基本概念，包括闭区间[a,b]、开区间(a,b)及半开区间[a,b)和(a,b]的定义及其表示方法，并解释了区间端点的意义。 设 a,b 是两个实数,且 a ≤ b. 1）满足 a ≤ x ≤ b 的实数 x 的集合, 表示为 [ a,b ],叫做闭区间; 2）满足 a ＜ x ＜b 的实数 x 的集合, 表示为 ( a,b ),叫做开区间; 3）满足 a ≤ x ＜b,a ＜x ≤ b 的实数 x 的集合, 分别表示为 [ a,b ),( a,b ],叫做半开区间. 这里实数 a,b 叫做区间的端点. 从上边的三个定义你就可以看出来,闭区间是有a,b两个端点的. 确定要放弃本次机会？ 福利倒计时 : : 立减 ¥ 普通VIP年卡可用 立即使用 打杂人 关注关注 4点赞 踩 4 收藏 觉得还不错? 一键收藏 1评论 分享复制链接 分享到 QQ 分享到新浪微博 扫一扫 举报 举报 专栏目录 高等 数学 1.10 闭区间 上连续函数的性质 MowenPan1995的博客 09-12 2422 的任何部分，只要两个自变量的数值接近到一定程度，就可使对应的函数值达到所指定的接近程度。上连续的函数在该区间上有界且一定能取得它的最大值 和 最小值。上连续，且在这取件的端点取不同的函数值。如果对于任意给定的正数。但反过来不一定成立。，那么函数在该区间上不一定有界。根据零点定理，在 开区间。上是连续的，但不是一致连续的。一致连续性表示，不论在区间。不符合一致连续的定义，所以。由上述定义可知，如果函数。内连续，或函数在 闭区间 上。 一致连续性定义是初等函数，它在区间。上的最大值与最小值。 【知识点】二分查找的区间到底是开还是闭？ Macw07的博客 11-27 1852 本文探讨二分查找算法中的区间开闭性问题，从基本原理到实际应用，系统地分析了左闭右 闭区间 和 左闭右 开区间 的特点及差异。 1 条评论 您还未登录，请先 登录 后发表或查看评论 二分法 闭区间 开区间 _说说集合、区间、邻域的那些事 9-28 区间有按照功能分为「开区间」和「闭区间」,以及「半开半 闭区间」,按照长度(所占地盘大小)分为有限区间 和 无限区间.而我们 数学,研究的对象就是实数,因此我们就把具有共同特征的实数构成的集合取专用名字叫区间.所以区间是特殊的集合,特殊在集合中元素都是实数.我们现在学 数学 也只跟实数打交道了,再多的我也用不着啊! 特别说明! 高等 数学 _区间_区间数 9-24 本文介绍了 数学 中区间的基本概念,包括 开区间(不包含端点)与 闭区间(包含端点),以及有限区间 和 无限区间的 区别,同时还提及了邻域的概念,这些都是实数集合理论中的重要组成部分。 1. 区间基本概念 在 数学 里,区间通常是指这样的一类实数集合:如果a 和 b是两个在集合里的数,那么,任何a 和 b之间的数也属于该集合。例如,由... 数学-开区间 和 闭区间 MasterFT的专栏 08-02 2万+ 开区间 用(a,b)来表示，闭区间 用[a,b]来表示。闭区间 包括了两个端点a 和 b，而 开区间 不包含两个端点a 和 b。 下面这个游戏是我以前的老师告诉我的。游戏的规则是：两个人分别在 开区间（a,b）内取一个数，两个人取的数不许相同，谁取的数大，那么谁赢。这个 游戏显然是谁先取谁输！具体一点，假设是在(0,1)这个区间来玩这个游戏。第一个人取的是0.9另一个只要取0.91就可以 区间的概念 mftang的博客 07-25 5921 在高等 数学 中，区间是指由实数构成的一段连续的数值范围。一个区间可以被表示为一个不等式形式的集合，其中包含了所有满足该不等式的实数。本文主要介绍这些概念的定义 和 使用方法。 区间调度问题详解 9-25 在不同的问题中,区间的开闭往往不同,有时是 闭区间,有时是半开半 闭区间。时间区间往往是 闭区间,但是音符中的开始结束区间则是半开半 闭区间,所以在重叠的定义上大家需要具体问题具体分析。稍后你会发现,开闭的 区别 其实只是差一个等号而已。 图1 时间区间示例... 理解Solr中日期区间查询的 闭区间 与 开区间 应用 9-17 [一般情况下使用[]设置范围,而[]表示的是 闭区间,也就是包含端点值。 如果不希望包含端点值,就需要用到 开区间,用大括号来限定{}。 但是对于半 开区间,就无法设置了,在 数学 中常见的例子: (0,1]或者0,1) 如果都使用 闭区间,端点值会重复计算,而都使用 开区间,端点值又会被忽略。 开区间 与开集的 区别 最新发布 Angelaboy的博客 08-23 780 开区间 与开集的 区别 数学:开区间(open)和 闭区间(closed)的含义 ComputerInBook的专栏 03-31 5351 开区间 和 闭区间 的含义 高等 数学:第一章 函数与极限(10)闭区间 上连续函数的性质 9-23 在 开区间 内连续,在右端点 左连续,在左端点 右连续,那未函数 就在 闭区间 上连续。 一、最大值与最小值定理 先介绍最大值与最小值概念: 对于区间 上有定义的函数 ,如果有 ,使得对于任一 都有 则称 是函数 在区间 上的最大值(最小值)。 C++ Range库精要 9-17 区间都是左闭右开的,可以用成员函数begin()/end()或者自由函数begin()/end()获得其两个端点。 range库的特征元函数 range_iterator :返回区间的迭代器类型 range_value :返回区间的值类型 range_reference :返回区间的引用类型 range_pointer :返回区间的指针类型 ... 函数极限＜3＞——连续函数 qq_45491237的博客 01-10 2003 若函数fx在邻域U˚x0δf\left ( x \right )在邻域\mathring{U} \left ( x_{0},\delta \right )fx在邻域U˚x0​δ上有定义,且∀ε0∀ε0∃δ0∃δ0∀x∣x−x0∣δ∀x∣x−x0​∣δ∣fx−fx0∣ε∣fx−fx0​∣ε,也就是lim⁡x→x0fxfx0limx→x0​​。 连续函数（五） 咔咔响 01-21 3375 1.一致连续是比连续要求更严格的概念，核心是ε-δ语言中的δ适用函数所定义的区间上所有x,即δ与x无关2.这样的δ是取每个x最大的(或上确界)δ集合的下确界3.一致连续是连续的充分非必要条... 闭区间、左闭右 开区间 和 开区间 02-07 闭区间 是指包含端点在内的数值范围，在 数学 上表示为 [a,b][a,b]，意味着 a a 和 b b 都属于该区间内的值。任何位于此范围内（含边界）的数都是合法成员[^1]。 #### 使用方法 - 编程: 在某些编程场景下，当... 二分查找总结——左闭右 开区间 和 左闭右 闭区间（C++语言） blue_coffeei的博客 06-14 3766 二分查找： 1.左闭右 开区间，如有相同元素返回查找到的第一个元素。 PS：主循环判断条件都是一样的（left < right），注意这里不能取等号！有相同元素时，如果要返回第一个查找到的元素，则区间包含相同元素时应该从右向左收缩，这时判断条件应该加上等号，并且此时找到的就是第一个元素的秩；如果要返回最后一个查找到的元素，则区间包含相同元素时应该从左向右收缩，这时判断条件没有等号，并且此时... 高等 数学 教学 闭区间 上连续函数的性质PPT课件.pptx 10-25 需要注意的是，这一性质仅适用于 闭区间 和 连续函数，如果区间不是 闭区间，比如是 开区间(a, b)，或者函数在区间内有间断点，那么定理就不一定成立了。 紧接着，有界性定理指出，在 闭区间[a, b]上连续的函数必然是有界... 二分法，太容易写出bug了！ 算法channel 10-26 857 如果你想脚踏实地的学好算法，那么就从这篇文章开始，从最基础的二分查找开始，真正的掌握好基础算法，为未来的转型打下坚实的基础。这篇文章来自承志兄，对二分查找写的入木三分，强烈推荐！欢迎加我... 拉格朗日中值定理求极限什么时候适用。 Cbelieveyouself的博客 05-05 5389 形如：lim ( f[r(x)] - f[g(x)] ) / g = lim f ’ (ζ) (r(x) - g(x) )/g①若r(x) - g(x) 与 g 同阶，则可以用。②若r(x) - g(x) 是低阶 ， 且 r(x) ~ g(x) 则 可用。③若r(x) - g(x) 是低阶 ， 且 r(x) 不等价与 g(x) 则不可以用。 一文彻底搞懂拉格朗日中值定理秒杀复杂极限问题（内含高级秒杀结论） 热门推荐 yuxuezhang的博客 09-12 6万+ 懂不懂拉式中值，在解极限的时候，是两重境界。用一个例题来体会，自行感受。 题目： lim⁡x→0cos(sinx)−cosxx4\lim_{x \rightarrow 0}{\frac{cos(sinx)-cosx}{x^{4}}}limx→0​x4cos(sinx)−cosx​ 当你不会拉式中值的时候，你的解答如下： 可见不用拉式中值的解答，又臭又长！ 但是当你会拉式中值时，你的解答是这样： 运用拉式中值解决极限 三步解决问题它不香么？别人还在写的时候，我的结果都已经口算出来了！ 那么这种化劲是怎么练 漫谈高数 Marvenaaaa的专栏 12-20 5165 漫谈高数 (一) 泰勒级数的物理意义 高等 数学 干吗要研究级数问题? 是为了把简单的问题弄复杂来表明自己的高深? No，是为了把各种简单的问题/复杂的问题，他们的求解过程用一种通用的方法来表示。 提一个问题，9999等于多少? 相信我们不会傻到列式子去算，口算也太难了而是会做一个迂回的方法，99(100-1)，这样更好算。那么995998 【算法基础】第六章：贪心 m0_65787507的博客 05-07 1161 给定 N个 闭区间 [a,b]，请你在数轴上。输出选择的点的最小数量。位于区间端点上的点也算作区间内。① 所有区间按右端点从小到大排序② 遍历每一个区间，如果当前区间的左与前一个区间的右有交集，则只需要一个点就可以覆盖掉两个区间。 11.1每日一题（拉格朗日中值定理求极限） 今天会营业的博客 12-07 172 11.1每日一题（拉格朗日中值定理求极限） 【高数】用拉格朗日中值定理解决极限问题 csq0327的博客 12-26 9758 拉格朗日定理解决复杂极限问题 几个求极限的方法 boring_fish的博客 04-10 8456 1.使用拉格朗日中值定理求极限 有时候当看到所求得题目中出现了 的项，可以考虑使用拉格朗日中值定理，或者说是常规思路写起来很复杂的时候可以使用拉格朗日中值定理简化运算 首先给出定里理： 拉格朗日(Lagrange)中值定理 若函数f满足如下条件： （i）f在 闭区间[a,b]上连续； （ii）f在 开区间（a,b）上可导 则在（a,b）上至少存在一点使得 解题思路 1、首先确定要使用中值定理的项 2、用 进行替换 3.用夹逼定理求出的值，得出答案 例题1、求极限 分析：很快可.. 关于我们 招贤纳士 商务合作 寻求报道 400-660-0108 kefu@csdn.net 在线客服 工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号 京网文〔2020〕1039-165号 经营性网站备案信息 北京互联网违法和不良信息举报中心 家长监护 网络110报警服务 中国互联网举报中心 Chrome商店下载 账号管理规范 版权与免责声明 版权申诉 出版物许可证 营业执照 ©1999-2025北京创新乐知网络技术有限公司 打杂人 博客等级 码龄19年 博客专家认证 441 原创1347 点赞 3698 收藏 1万+粉丝 关注 私信 热门文章 1寸2寸3寸5寸6寸8寸10寸照片的具体尺寸（附常用照片尺寸对照表） 2039505 ip段/数字,如192.168.0.1/24是什么意思? 325845 Flash Helper Service 这个流氓，动不动弹出广告！！ 219256 [乐意黎原创]向上取整⌈⌉和向下取整⌊⌋符号含义及应用 170408 JAVA构造函数(方法) 159294 分类专栏 网页自动化测试19篇 服务器及编程语言配置117篇 Git+SVN2篇 上一篇： [乐意黎原创]向上取整⌈⌉和向下取整⌊⌋符号含义及应用 下一篇： KMP算法Next数组计算 最新评论 计算机网络c类网络划分子网 2301_77980870:第一问bcd最大适用地址是不是写错了，广播地址不是不能作为主机的ip地址吗 photo,image,picture三个词区别以及cinema、film 和movie的用法与差别 Shuuc:"乱画，那就谈不上形象，而只是picture"和 picture⊆image 矛盾了。 中国护照上两行88个字符的含义 2401_86407130:大神 ，最后那个X的校验码看不懂 [乐意黎]Windows 里的环境变量以及%USERPROFILE%等变量设置 weixin_38432413:请教下，HomePath变量建议改吗？改成非系统盘，不想每个软件的数据都默认写到系统盘 软考学习之关于McCabe环路复杂度的计算 li天下:public static void main(String[] args) { 2 int a = 0 , b = 0; 3 Scanner sc = new Scanner(System.in); 4 int i = sc.nextInt(); 5 int j = sc.nextInt(); 6 while (i < 10) { 7 if (j == 0) { 8 a++; 9 } else if(j == 1){ 0 b++; 11 } 12 i++; 13 } 14 System.out.println("a = " + a + ", b = " + b); 15 } 大佬这个代码的控制流图怎么画，不太懂 大家在看 【2026计算机毕业设计】基于Springboot的学院食堂送餐系统 基于重复警情的个人风险预警系统 小杰深度学习（three）——softmax与交叉熵、优化器与优化方法 【2026计算机毕业设计】基于Springboot的网盘系统的设计与实现 用了18个月，跳出运维困局重新启程，我的转行经历或许能给你帮助你！ 最新文章 TASK ERROR: EFI base image ‘/usr/share/pve-edk2-firmware//AAVMF_CODE.fd‘ not found VSCode 正则替换节点增加属性 How to change Google chrome windows order in the Windows task panel 2024年 1篇 2021年 5篇 2020年 22篇 2019年 95篇 2018年 176篇 2017年 340篇 2016年 181篇 2015年 142篇 2014年 238篇 2013年 55篇 2012年 774篇 2011年 121篇 2010年 36篇 2009年 4篇 上一篇： [乐意黎原创]向上取整⌈⌉和向下取整⌊⌋符号含义及应用 下一篇： KMP算法Next数组计算 分类专栏 网页自动化测试19篇 服务器及编程语言配置117篇 Git+SVN2篇 展开全部收起 登录后您可以享受以下权益： 免费复制代码 和博主大V互动 下载海量资源 发动态/写文章/加入社区 ×立即登录 评论 1 被折叠的 条评论 为什么被折叠?到【灌水乐园】发言 查看更多评论 添加红包 祝福语 请填写红包祝福语或标题 红包数量 个 红包个数最小为10个 红包总金额 元 红包金额最低5元 余额支付 当前余额 3.43 元 前往充值 > 需支付：10.00 元 取消 确定 成就一亿技术人! 领取后你会自动成为博主和红包主的粉丝 规则 hope_wisdom 发出的红包 实付 元 使用余额支付 点击重新获取 扫码支付 钱包余额 0 抵扣说明： 1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。 余额充值 确定 取消 举报 选择你想要举报的内容（必选） 内容涉黄 政治相关 内容抄袭 涉嫌广告 内容侵权 侮辱谩骂 样式问题 其他 原文链接（必填） 请选择具体原因（必选） 包含不实信息 涉及个人隐私 请选择具体原因（必选） 侮辱谩骂 诽谤 请选择具体原因（必选） 搬家样式 博文样式 补充说明（选填） 取消 确定 AI助手 AI 搜索 智能体 AI 编程 AI 作业助手 下载APP 程序员都在用的中文IT技术交流社区 公众号 专业的中文 IT 技术社区，与千万技术人共成长 视频号 关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部
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https://baike.baidu.com/item/%E5%8F%98%E5%BC%82%E7%B3%BB%E6%95%B0/6463621
网页 新闻 贴吧 知道 网盘 图片 视频 地图 文库 资讯 采购 百科 帮助 近期有不法分子冒充百度百科官方人员，以删除词条为由威胁并敲诈相关企业。在此严正声明：百度百科是免费编辑平台，绝不存在收费代编服务，请勿上当受骗！详情>> 首页 秒懂百科 特色百科 知识专题 加入百科 : 新人成长 进阶成长 任务广场 百科团队 : 校园团 分类达人团 热词团 繁星团 蝌蚪团 权威合作 : 合作模式 常见问题 联系方式 个人中心 变异系数 播报 锁定 讨论2 上传视频 概率分布离散程度的归一化量度 一分钟了解变异系数 00:47 变异系数：正态分布下的离散程度解析 06:08 变异系数解析：心理统计学必考知识点 06:08 变异系数有单位吗 01:39 变异系数越大说明什么 01:22 标准差和变异系数 01:47 变异系数是什么? 04:15 收藏 查看我的收藏 1012 有用+1 179 本词条由《中国科技信息》杂志社 参与编辑并审核，经科普中国·科学百科认证 。 变异系数（Coefficient of Variation）：当需要比较两组数据 离散程度 大小的时候，如果两组数据的测量尺度相差太大，或者数据 量纲 的不同，直接使用 标准差 来进行比较不合适，此时就应当消除测量尺度和量纲的影响，而变异系数可以做到这一点，它是原始数据标准差与原始数据 平均数 的比。CV没有量纲，这样就可以进行客观比较了。事实上，可以认为变异系数和极差、标准差和 方差 一样，都是反映数据离散程度的绝对值。其数据大小不仅受变量值离散程度的影响，而且还受变量值平均水平大小的影响。 中文名 : 变异系数/变差系数 外文名 : Coefficient of Variation 别 名 : 离散系数、差异系数 应用领域 : 数学 类 别 : 计数方法 定 义 : 概率分布 离散程度的 归一化 量度 目录 1 定义 2. 2 基本含义 3. 3 举例 4 公式 2. 5 优缺点 3. ▪ 优点 ▪ 缺陷 2. 6 应用 定义 播报 在 概率论 和 统计学 中， 变异系数 ，又称“离散系数”（英文：coefficient of variation），是 概率分布 离散程度的一个 归一化 量度，其定义为 标准差 与 平均值 之比： 变异系数（coefficient of variation）只在平均值不为零时有定义，而且一般适用于平均值大于零的情况。变异系数也被称为 标准离差率 或 单位风险 。 变异系数只对由比率标量计算出来的数值有意义。举例来说，对于一个气温的分布，使用 开尔文 或摄氏度来计算的话并不会改变标准差的值，但是温度的平均值会改变，因此使用不同的温标的话得出的变异系数是不同的。也就是说，使用区间标量得到的变异系数是没有意义的。 变异系数分为总体变异系数和样本变异系数。 基本含义 播报 变异系数 (6张) 一般来说，变量值平均水平高，其离散程度的 测度值 越大，反之越小。 变异系数是衡量资料中各观测值变异程度的另一个统计量。当进行两个或多个资料变异程度的比较时，如果度量单位与平均数相同，可以直接利用标准差来比较。如果单位和（或）平均数不同时，比较其变异程度就不能采用标准差，而需采用标准差与平均数的比值（ 相对值 ）来比较。标准差与平均数的比值称为变异系数，记为 C·V 。变异系数可以消除单位和（或）平均数不同对两个或多个资料变异程度比较的影响。 变异系数的计算公式为：变异系数 C·V =（ 标准偏差 SD / 平均值Mean ）× 100% 在进行数据统计分析时，如果变异系数大于15%，则要考虑该数据可能不正常，应该剔除。 举例 播报 已知某良种猪场长白成年母猪平均体重为190 kg ，标准差为10.5 kg ，而大约克成年母猪平均体重为196 kg ，标准差为8.5 kg ，试问两个品种的成年母猪，哪一个体重变异程度大。 此例观测值虽然都是体重，单位相同，但它们的平均数不相同，只能用变异系数来比较其变异程度的大小。 由于，长白成年母猪体重的变异系数：C.V = 10.5 / 190 100% = 5.53% 大约克成年母猪体重的变异系数：C.V = 8.5 / 196 100% = 4.34% 所以，长白成年母猪体重的变异程度大于大约克成年母猪。 注意，变异系数的大小，同时受平均数和标准差两个 统计量 的影响，因而在利用变异系数表示资料的 变异程度 时，最好将平均数和标准差也列出。 公式 播报 (标准偏差SD、平均值MN) 优缺点 播报 优点 比起标准差来，变异系数的好处是不需要参照数据的平均值。变异系数是一个 无量纲量 ，因此在比较两组 量纲 不同或均值不同的数据时，应该用变异系数而不是标准差来作为比较的参考。 缺陷 1. 当平均值接近于0的时候，微小的扰动也会对变异系数产生巨大影响，因此造成精确度不足。 2. 2. 变异系数无法发展出类似于均值的 置信区间 的工具。 应用 播报 变异系数在概率论的许多分支中都有应用，比如说在更新理论、 排队理论 和可靠性理论中。在这些理论中，指数分布通常比 正态分布 更为常见。 由于指数分布的标准差等于其平均值，所以它的变异系数等于一。变异系数小于一的分布，比如爱尔朗分布称为低差别的，而变异系数大于一的分布，如 超指数分布 则被称为高差别的。 词条图册 更多图册 概述图册 (1张) 变异系数 (6张) 1/1 参考资料 1 财务理念 互联网档案馆的存档，暨南大学管理学院会计系 2 罗良清、魏和清 ．统计学 ．中国财政经济出版社 ．2011 ．49 3 吴媚,顾赛赛.变异系数的统计推断及其应用[J].铜仁学院学报,2010,4(1):139-141144 变异系数的概述图（1张） 科普中国 致力于权威的科学传播 本词条认证专家为 尚轶伦 副教授 审核 同济大学数学科学学院 权威合作编辑 : 《中国科技信息》杂志社 《中国科技信息》创刊于1989年10月，是... : 什么是权威编辑 词条统计 浏览次数：3232759次 编辑次数：54次历史版本 最近更新： vera菊花香 （2023-09-14） 突出贡献榜 jasonye163 1 定义 2 基本含义 3 举例 4 公式 5 优缺点 优点 缺陷 6 应用 相关搜索 变异系数大小的意义 变异系数名词解释 变异系数的计算公式cv 变异系数越大说明什么 基因重组 遗传基因 基因分型 基因 北面是什么档次的牌子 xtep 变异系数 选择朗读音色 成熟女声 成熟男声 磁性男声 年轻女声 0 成长任务 编辑入门 编辑规则 本人编辑 内容质疑 在线客服 官方贴吧 意见反馈 举报不良信息 未通过词条申诉 投诉侵权信息 封禁查询与解封 ©2025 Baidu 使用百度前必读 | 百科协议 | 隐私政策 | 百度百科合作平台 | 京ICP证030173号
10435
https://flexbooks.ck12.org/cbook/ck-12-probability-and-statistics-concepts/section/8.2/primary/lesson/calculating-permutations-pst/
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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 8.2 Calculating Permutations Written by:CK-12 Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 If the Olympic High Jump has 24 semifinalists, how many possible ways can the competitors be arranged into Gold, Silver, and Bronze winners? This is a permutation calculation, by the end of the lesson you will have no problem calculating the answer. Calculating Permutations Calculating the number of permutations possible from a group of objects is a rather simple calculation, but understanding the concept can be a little tricky. One way to understand the concept is to think of a group of friends seated around a table. If there are three friends and three seats, then the number of ways the friends can sit around the table is a question of permutations. Suppose the friends are Amber, Brian, and Kelli, and the chairs are one each green, blue, and red. Let’s consider the options, starting with Amber in the green chair: 1. Green: Amber,blue: Brian,red: Kelli2. Green: Amber,red: Brian,blue: Kelli3. Blue: Amber,green: Brian,red: Kelli4. Blue: Amber,red: Brian,green: Kelli5. Red: Amber,blue: Brian,green: Kelli6. Red: Amber,green: Brian,blue: Kelli There are six permutations. We can see that each time the first color is chosen, there are two possibilities left for the next person to choose from, and the last person does not get to choose. That’s why there are two of each color in each column. If Amber chooses green, then Brian can either choose blue, leaving red for Kelli, or Brian can choose red, leaving blue for Kelli. The same goes for Amber choosing blue or red. Each time, Brian has two colors left to choose from, and Kelli gets the remaining color. The chart would look the same regardless of who chooses first, because no matter who the first person is, there are only two ways the others could sit, meaning there should be (and are) two entries for each color under each person’s name. Fortunately, we don’t need to draw out a diagram every time we want to find the number of permutations possible in a situation like this. As long as there are no duplicates or items so similar they can’t be told apart (two green chairs, for instance), all we need to know is how to use factorials. Factorials are notated with an exclamation point “!”, and they indicate that you should start with the number before the exclamation point, and count down to 1, multiplying each number by the next. For instance, the first few factorials are: 0!=1(by definition) 1!=1 2!=2×1=2 3!=3×2×1=6 4!=4×3×2×1=245!=5×4×3×2×1=120 To calculate the number of permutations possible given n non-repeating values, calculate n!. To calculate the number of permutations of r values from set n, calculate the first r numbers of n! Alternatively, the formula for counting permutations is: n!(n−r)! Where n is the number of available values, and r is the number of values to be selected. You may also see this information in the form: nPr Real-World Application: Line How many ways can six people line up as they wait in line to buy tickets? Since this is a basic permutation question, with no duplicates and no indistinguishable members, all we need to do is find six factorial: 6!=6×5×4×3×2×1=720 There are 720 ways that six people could line up. Rearranging Words How many ways can the letters in the word “factory” be arranged? “factory” has seven letters, all unique, and we want the possible number of arrangements using all seven, so the number of permutations is seven factorial: 7!=7×6×5×4×3×2×1=5040 There are 5,040 ways to arrange the letter in the word “factory”. Permutations from Words How many different three-letter permutations are possible using the letters in the word “bread”? This is a slightly different problem than the first two, since we aren’t looking for permutations using all five letters. Since we are only selecting three of the five letters for each arrangement, there will be fewer possible arrangements. As I mentioned at the end of the “Guidance” section, there are a couple of ways to view this type of problem: Conceptually, we need to calculate only the first three numbers of five factorial: 5×4×3=120 Using the formula: n!(n−r)! 5!(5−3)! 5!2!5×4×3×2×12×15×4×3=60 There are 60 three-letter permutations of the letters in the word “bread”. Earlier Problem Revisited If the Olympic High Jump has 24 semifinalists, how many possible ways can the competitors be arranged into Gold, Silver, and Bronze winners? For this question, we need to calculate the number of permutations of three competitors out of the set of 24 semifinalists. Since we only want arrangements of three, we need to calculate the first three numbers in 24! 24×23×22=12,144 There are 12,144 possible gold, silver, bronze rankings of the 24 semifinalists. Examples Example 1 How many ways can the letters in "education" be arranged? There are nine letters in “education”, so we need to calculate 9!: 9×8×7×6×5×4×3×2×1=362,880 Example 2 How many different four letter arrangements can be made from the word "document"? There are eight letters in “document”, but we only want arrangements of four, so we calculate the first four numbers of 8!: 8×7×6×5=1,680 Example 3 How many permutations are represented by 8P5? 8P5 is read as “pick five items from the eight available”. The formula for permutations is n!(n−r)!: 8!(8−5)! 8!3!8×7×6×5×4×3×2×13×2×18×7×6×5×4=6720 Example 4 How many different ways can the basic colors of a rainbow: red, orange, yellow, green, blue, indigo, and violet, be arranged? There are seven colors in a basic rainbow, and we are looking for the number of unique permutations of all seven. 7!=7×6×5×4×3×2×1=5,040 Review For all questions 1-13, assume no duplicated units are allowed. All the answers for a 5 question multiple - choice test are A, B, C, or D. Find the number of possible answer keys a teacher could have. Calculate 6! Calculate 4! You are distributing from a group of 5 items one to each of 2 people. How many different ways can you do this? Calculate 6P3 Find the number of permutations of 7 distinct items. To unlock a school locker, you need a locker combination consisting of 3 unique numbers from 1 to 9. How many possible locker combinations are there? Calculate 6P2 Calculate 8P8 If a bank account number consists of seven unique digits 0-9, how many possible accounts are there? For a dinner party, you need to make a seating arrangement of 9 people. Find the number of different ways of arranging the party. Why can’t you calculate 4.5! ? How many five-letter arrangements can be made from the word “number”? Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. | Image | Reference | Attributions | --- | | | Credit: Steve Fair Source: License: CC BY-NC 3.0 | | | | License: CC BY-NC-SA | Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Save this section to your Library in order to add a Practice or Quiz to it. (Edit Title)24/ 100 This lesson has been added to your library. |Searching in: | | | Looks like this FlexBook 2.0 has changed since you visited it last time. We found the following sections in the book that match the one you are looking for: Go to the Table of Contents No Results Found Your search did not match anything in .
10436
https://www.teacherspayteachers.com/browse/free?search=solving%20linear%20equation%20puzzle
Solving Linear Equation Puzzle | TPT Log InSign Up Cart is empty Total: $0.00 View Wish ListView Cart Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Adult education Resource type Student practice Independent work packet Worksheets Assessment Graphic organizers Task cards Flash cards Teacher tools Classroom management Teacher manuals Outlines Rubrics Syllabi Unit plans Lessons Activities Games Centers Projects Laboratory Songs Clip art Classroom decor Bulletin board ideas Posters Word walls Printables Seasonal Holiday Black History Month Christmas-Chanukah-Kwanzaa Earth Day Easter Halloween Hispanic Heritage Month Martin Luther King Day Presidents' Day St. Patrick's Day Thanksgiving New Year Valentine's Day Women's History Month Seasonal Autumn Winter Spring Summer Back to school End of year ELA ELA by grade PreK ELA Kindergarten ELA 1st grade ELA 2nd grade ELA 3rd grade ELA 4th grade ELA 5th grade ELA 6th grade ELA 7th grade ELA 8th grade ELA High school ELA Elementary ELA Reading Writing Phonics Vocabulary Grammar Spelling Poetry ELA test prep Middle school ELA Literature Informational text Writing Creative writing Writing-essays ELA test prep High school ELA Literature Informational text Writing Creative writing Writing-essays ELA test prep Math Math by grade PreK math Kindergarten math 1st grade math 2nd grade math 3rd grade math 4th grade math 5th grade math 6th grade math 7th grade math 8th grade math High school math Elementary math Basic operations Numbers Geometry Measurement Mental math Place value Arithmetic Fractions Decimals Math test prep Middle school math Algebra Basic operations Decimals Fractions Geometry Math test prep High school math Algebra Algebra 2 Geometry Math test prep Statistics Precalculus Calculus Science Science by grade PreK science Kindergarten science 1st grade science 2nd grade science 3rd grade science 4th grade science 5th grade science 6th grade science 7th grade science 8th grade science High school science By topic Astronomy Biology Chemistry Earth sciences Physics Physical science Social studies Social studies by grade PreK social studies Kindergarten social studies 1st grade social studies 2nd grade social studies 3rd grade social studies 4th grade social studies 5th grade social studies 6th grade social studies 7th grade social studies 8th grade social studies High school social studies Social studies by topic Ancient history Economics European history Government Geography Native Americans Middle ages Psychology U.S. History World history Languages Languages American sign language Arabic Chinese French German Italian Japanese Latin Portuguese Spanish Arts Arts Art history Graphic arts Visual arts Other (arts) Performing arts Dance Drama Instrumental music Music Music composition Vocal music Special education Speech therapy Social emotional Social emotional Character education Classroom community School counseling School psychology Social emotional learning Specialty Specialty Career and technical education Child care Coaching Cooking Health Life skills Occupational therapy Physical education Physical therapy Professional development Service learning Vocational education Other (specialty) Solving Linear Equation Puzzle 30+results Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Search Grade Subject Supports Free Format All filters (1) Filters Free Clear all Grade Elementary 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Higher education Adult education More Subject Math Algebra Algebra 2 Applied math Arithmetic Basic operations Geometry Graphing Math test prep Measurement Precalculus Other (math) For all subjects More Price Free Under $5 $5-10 $10 and up Format Easel Easel Activities Google Apps Microsoft Microsoft PowerPoint Microsoft Word PDF More Resource type Teacher tools Homeschool curricula Printables Hands-on activities Activities Centers Internet activities Games Instruction Handouts Student assessment Assessment Critical thinking and problem solving More Student practice Worksheets Homework Task cards Standard Theme Holiday Halloween More Audience Homeschool Staff & administrators More Supports Special education More Solving Linear Equations Unicorn Puzzle Activity (Free Version) Created by Dr Steve Warner This activity will have students solving linear equations. Students will be given a 9 by 9 grid and worksheets with 45 problems, divided uniformly into 5 levels.. Each problem comes with a puzzle piece. The solutions to the problems tell students where in the grid to copy the figure in that puzzle piece. The end result will be a picture of a unicorn. The levels are as follows: Level 1: One-step equations Level 2: Two-step equations Level 3: Distributive property and combining like terms Level 8 th - 11 th Algebra, Math FREE Rated 4.67 out of 5, based on 6 reviews 4.7(6) Log in to Download Wish List Solving Linear Equations Puzzle Created by Robyn Holloway This puzzle is designed to be used in an algebra or geometry class. It consists of a page of puzzle details, the student puzzle page, and a key. The student puzzle page is a 4 by 4 grid containing linear equations and solutions. It includes equations requiring one-step, two-steps, distributive property, and dealing with variables on both sides. The goal is to put the sixteen squares together to form a block letter I by matching the equations and solutions. I usually have students complete the p 7 th - 10 th Algebra, Geometry FREE Rated 4.92 out of 5, based on 12 reviews 4.9(12) Log in to Download Wish List Get more with resources under $5 See all Solving Linear Equations with Variables on Both Sides Sudoku Puzzle Amusing Algebra $2.00 Original Price $2.00 Rated 4.83 out of 5, based on 6 reviews 4.8(6) Solving Systems of Linear Equations by Elimination and Substitution Puzzles Math And Science Sarah $4.00 Original Price $4.00 Rated 5 out of 5, based on 1 reviews 5.0(1) Solving Linear Equations Digital Google Self-Checking Puzzle Pixel Task Cards MathYouLove $4.00 Original Price $4.00 Solving Linear Equations - Special Cases Digital Puzzle Reveal Math Sister 1 $3.00 Original Price $3.00 Math Puzzle Worksheet Solving One Step Linear Equation (English Fact Series) Created by littleworld Math puzzle worksheet is a great way to make math practice become more fun. This is one of the English-Fact-Series Math-Puzzle-Worksheet about Solving Linear Equation. On each worksheet, students need to solve linear equation and trying to solve the puzzle about English fact at the same time. Something about English that they might not know yet. We can vary the strategies, like give a reward for the first students who solve the problems correctly or having a class discussion once all student 6 th - 10 th Algebra, Algebra 2, Math Test Prep FREE Rated 5 out of 5, based on 3 reviews 5.0(3) Log in to Download Wish List Jigsaw Puzzle #37 Frog - Solving Linear Equations (10 Qs) Created by Jigsaw Math Jigsaw Puzzle #37 Frog - Solving Linear Equations (10 Qs)Solving multi-step equationsOne solution, no solution, & IMSHow to JigsawCut out the puzzle pieces on pg 3Solve the math problems on pg 1 and 2Match each problem with the correct letter from the answer bankUse the problem #s and corresponding letter choice to assemble the puzzleGlue the completed puzzle on a separate sheet of paperGet creative by decorating your completed puzzle Google Doc AccessAccess to a copy of a google doc can be foun 8 th Math, Other (Math) CCSS 8.EE.C.7 , 8.EE.C.7a , 8.EE.C.7b Also included in:Jigsaw Puzzles - 8th Grade Bundle FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List Solving Circle and Linear System of Equations Puzzle Activity Created by Newton's Solutions Looking for a fun and engaging puzzle activity to help students practice solving a circle and linear system of equations? This easy to check activity is perfect for students to gain confidence when finding the solutions to a system of equations by providing repeated exposure of this skill. Just cut out the squares and have students match up a system with a solution. When correct, a 3x3 square will be formed. This puzzle is a great way for students to self-check their work! This activity pro 8 th - 11 th Algebra 2, Math CCSS HSA-REI.C.7 FREE Log in to Download Wish List Algebra 2 - Solving Linear Equations Matching Puzzle with Biblical Integration Created by Lambkins199 In this Matching Puzzle, students will learn about Peter's vision of clean and unclean animals and how God led him to witness to a Roman soldier and his family who were saved and baptized as they solve linear equations. Students will match the problem with the correct corresponding answer. If the line goes through both a number and a letter, the student will record the letter in the appropriate box at the bottom of the worksheet which will reveal a phrase that completes the narrative provided. 8 th - 12 th Algebra, Algebra 2, Math FREE Log in to Download Wish List Magic Square - Solving Multi-step Linear Equations Created by MathSheetz The puzzle is a 4x4 magic square where each row and column add to the same MAGIC number! Each box contains a multi-step linear equation, using inverse operations, combining like terms, and the distributive property. This magic square allows students to self-check their answers - if they are not getting the same number in each row and each column, then they have made an error! If you find this resource useful, please leave a comment and rate it below! Don't forget to follow me! 7 th - 10 th Algebra, Arithmetic, Math CCSS HSA-REI.A.1 , HSA-REI.B.3 FREE Rated 4.8 out of 5, based on 30 reviews 4.8(30) Log in to Download Wish List Solving Systems of Linear Equations using Elimination Digital Activity Created by Hello Algebra In this self-checking activity, students will work to solve systems of linear equations using elimination. As questions are answered correctly, pieces of the mystery puzzle will begin to appear, just like a pixel art activity! This resources comes with two activities. In one activity, students will solve question only focusing on addition and subtraction. In the second activity, students will solve questions where you must multiply one of the equations before using addition/subtraction to so 8 th - 11 th Algebra, Basic Operations, Math CCSS HSA-REI.C.6 Also included in:Algebra 1 Digital Mystery Puzzle Pixel Activities FREE Rated 4.94 out of 5, based on 17 reviews 4.9(17) Add to Google Drive Wish List Solving Multi-Step Equations DIGITAL Sudoku Puzzle Activity Created by Amusing Algebra With this engaging digital activity, your students will enjoy solving math problems to solve the Sudoku puzzle! No prep and self checking, this activity will help your students practice solving multi-step linear equations. Great to use for practice, homework, review, or sub plans. When students fill in the answer to each problem, this answer is automatically filled into a corresponding Sudoku box. When finished solving all of the problems, students must solve the rest of the Sudoku puzzle. St 8 th - 10 th Algebra, Algebra 2, Math CCSS HSA-REI.B.3 Also included in:Solving Linear Equations DIGITAL Activity Bundle - Distance Learning FREE Rated 4.54 out of 5, based on 24 reviews 4.5(24) Add to Google Drive Wish List Solving Two Variable Systems Sudoku Puzzle Created by Amusing Algebra With this engaging activity, your students will enjoy solving math problems to solve the Sudoku puzzle! No prep and ready to print, this activity will help your students practice solving linear systems of equations. Great to use for practice, homework, review, or sub plans. What's included:✅ Printable PDF activity containing 12 problems ✅ 3 problems are special cases ✅ Sudoku puzzle easy in difficulty ✅ Teacher answer key Looking for a digital self-checking version to use? Click here !⭐ You m 8 th - 10 th Algebra, Algebra 2, Math CCSS HSA-REI.C.5 , HSA-REI.C.6 Also included in:Solving Two Variable Systems Activity Bundle FREE Rated 4.71 out of 5, based on 14 reviews 4.7(14) Log in to Download Wish List Finding Missing Angels Puzzle #1 - vertical, supplementary, corresponding..... Created by Joshua Peretti All too often mathematical skills are taught in isolation. This of course, is the source of one of our greatest laments as teachers: “They forget things so quickly! I just taught them _____ last month!” To counter this natural tendency, the activities I create are designed to prompt students to draw from skills learned in multiple lessons over an extended period of time. In this activity, students will need to find the measures of multiple angles by drawing on their equation solving fluency and 7 th - 12 th Algebra, Geometry, Measurement FREE Rated 4.67 out of 5, based on 3 reviews 4.7(3) Log in to Download Wish List Scavenger Hunt: Solving Systems of Linear Equations (Graphing with Calculator) Created by Prachi Parihar This scavenger hunt is a good way to get kids up and moving around the room while they are learning to use their calculators for solving systems of equations. There are 14 problems that they need to solve to complete the puzzle. For each system that they solve, the answer will lead them to the next system and also give them a letter. At the end, they will uncover a secret phrase. The first 15 pages are part of the scavenger hunt (and are in order so you have the solution). The last few pages are 9 th - 10 th Algebra, Algebra 2, Graphing FREE Rated 5 out of 5, based on 2 reviews 5.0(2) Log in to Download Wish List Solving Multi-Step Equations (Variable on both Sides) Algebra 1 Puzzle Activity Created by Ms. P's Math Classroom Help your students master solving multi-step equations with this versatile resource! Perfect for small-group collaboration, this activity challenges students to solve 11 equations featuring variables on one or both sides. A few problems also incorporate fractions and special solutions to keep students engaged and thinking critically. This resource is ready to meet your classroom's needs, with both print and digital versions included for maximum flexibility. Whether you're in-person or online, th 8 th - 12 th Algebra, Algebra 2, Math FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List Solving Trig Equations Puzzle Created by I Teach High School Math A puzzle that is truly upper level high school mathematics. Use this in your Algebra 2, Trigonometry, or Pre Calculus class! The puzzle includes equations of varying levels: multistep linear style, square rooting, factoring with GCF, factoring quadratic style, and using pythagorean identities to solve. Answers to the equations include one, two, three, four, and no solutions. They also have a mixture of degrees and radians. 11 th - 12 th Algebra 2, Other (Math), PreCalculus FREE Log in to Download Wish List LINEAR EQUATIONS IN ONE VARIABLE (math puzzles) - EASY Created by NEO SIMANA The student must complete five different tasks (printable PDF). Tasks are designed to repeat an entire chapter, and require logic, insight, and knowledge. Students will practice solving multi-step equations, including those with distribution and variables on both sides, equivalent equations, tasks of creating equations. Total assignments are designed to engage the student and stimulate the thought process. This file is level C – Easy. Also Included in this resource: Instructional assessmentAnswe 7 th - 10 th Algebra, Math CCSS 7.EE.B.3 , 7.EE.B.4 , 7.EE.B.4a +4 FREE Rated 4.5 out of 5, based on 2 reviews 4.5(2) Log in to Download Wish List Solving Equations Tarsia Puzzle Created by Jennifer Merker Solving linear equations that require 2 or more steps. 6 th - 12 th Algebra, Math FREE Rated 4.6 out of 5, based on 5 reviews 4.6(5) Log in to Download Wish List SLOPE and Y-INTERCEPT EQUATION Lines Practice Activity Puzzle LINEAR EQUATIONS Created by iLoveToTeachKids Allow your students to GET OUT OF THE BOX while practicing finding the SLOPE and Y-intercept of lines. Students will solve a puzzle while working through the problems. This high-interest engaging activity will provide students with opportunities to practice their Algebra skills. In this activity eight lines are graphed on a coordinate plane. Your students must identify the slope and y-intercept of each line. These differentiated activities are NO PREP and READY TO GO! Great for sub (sub 7 th - 10 th Algebra, Math, Other (Math) FREE Rated 5 out of 5, based on 3 reviews 5.0(3) Log in to Download Wish List Solving Linear Equations: Mystery Lesson Created by Coach Roark's Math Corner This is a mystery lesson from the SILVER AND STRONG style where students receive equations in strips like a puzzle that they must put together. This works good as a group activity after students have had exposure to solving multi-step equations. 7 th - 12 th Algebra, Algebra 2, Applied Math FREE Rated 4.5 out of 5, based on 2 reviews 4.5(2) Log in to Download Wish List Solving Equations Tarsia Puzzle-Answer Key Created by Jennifer Merker Answer Key for Linear Equations Tarsia Puzzle 5 th - 12 th Algebra, Math FREE Rated 5 out of 5, based on 3 reviews 5.0(3) Log in to Download Wish List Finding X-INTERCEPT and Y-INTERCEPT EQUATIONS Graphing Linear Equations Created by iLoveToTeachKids Allow your students to GET OUT OF THE BOX while practicing FINDING X- and Y-INTERCEPTS. Students will solve a puzzle while working through the problems. This high-interest engaging activity will provide students with opportunities to practice their Algebra skills. The focus of this activity is to find the x and y intercepts, given the equation of a line expressed in various forms. To complete this activity successfully, your students should have a basic understanding of graphing linear eq 7 th - 10 th Algebra, Math, Other (Math) FREE Rated 4.56 out of 5, based on 15 reviews 4.6(15) Log in to Download Wish List Systems of Linear Equations Task Cards and Review Activity - Sample Created by Teach With Fergy This 16 System of Linear Equations Task Card resource is unique. It can be utilized in a clue gathering, puzzle solving, highly active and engaging activity and/or as an alternative to worksheet review for Linear Systems. If you'd like to receive 50% off your next TeachWithFergy purchase, (includes your entire shopping cart) please CLICK HERE. A new page will open but don't worry, you won't lose this one. This is a portion of my Systems of Linear Equations Task Card Activity, please click HER 7 th - 12 th, Adult Education, Higher Education Algebra, Math, Other (Math) FREE Rated 4.77 out of 5, based on 14 reviews 4.8(14) Log in to Download Wish List Practicing Linear Equations Riddle Created by Turners Teachable Resources In this activity, students can practicing analyzing the slope and y-intercepts of linear functions. Students select the correct answer and input letters into the puzzle at the bottom in order to solve the riddle. I have used this as both a homework and a post-unit assessment in my Algebra class. Students enjoyed it as it keeps them active while still practicing old skills. I think it is important to provide students with a variety of different assessment formats so that the stress of a typical 7 th - 10 th Algebra, Graphing, Math CCSS 8.F.A.3 , 8.F.B.4 , HSF-IF.C.7a +1 Also included in:Algebraic Functions Activities Bundle FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List GRAPHING LINES X-INTERCEPT and Y-INTERCEPT LINEAR EQUATIONS Created by iLoveToTeachKids Allow your students to GET OUT OF THE BOX while practicing USING X- and Y-INTERCEPTS TO GRAPH A LINE. Students will solve a puzzle while working through the problems. This high-interest engaging activity will provide students with opportunities to practice their Algebra skills. For this activity students are to graph the equation of a line given the x- and y-intercepts. Your students will need a ruler. Instruct students to extend the lines to the end of the graph. They will then identif 6 th - 12 th Algebra, Math, Other (Math) FREE Rated 5 out of 5, based on 5 reviews 5.0(5) Log in to Download Wish List EQUATION OF A LINE when Given Two Points Coordinates Slope Linear EQUATIONS Created by iLoveToTeachKids You will receive both the PDF and GOOGLE SLIDES versions of this resource to help meet your specific classroom needs (hybrid, in-person, virtual, etc.). Access the GOOGLE version by clicking the link on the first page of the PDF file. Allow your students to GET OUT OF THE BOX while practicing WRITING THE EQUATION OF A LINE WHEN GIVEN TWO POINTS. Students will solve a puzzle while working through the problems. This high-interest engaging activity will provide students with opportunities to pra 6 th - 12 th Algebra, Math, Other (Math) FREE Rated 5 out of 5, based on 3 reviews 5.0(3) Log in to Download Wish List 1 2 Showing 1-24 of 30+results TPT is the largest marketplace for PreK-12 resources, powered by a community of educators. 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https://www.physio-pedia.com/CREST_Syndrome
Contents Editors Categories Cite Contents loading... Editors loading... Categories loading... When refering to evidence in academic writing, you should always try to reference the primary (original) source. That is usually the journal article where the information was first stated. In most cases Physiopedia articles are a secondary source and so should not be used as references. Physiopedia articles are best used to find the original sources of information (see the references list at the bottom of the article). If you believe that this Physiopedia article is the primary source for the information you are refering to, you can use the button below to access a related citation statement. Cite article CREST Syndrome Jump to:navigation, search Original Editor - Aya Alhindi Top Contributors - Aya Alhindi, Kim Jackson and Khloud Shreif Contents 1 Description 2 Epidemiology 3 Pathophysiology 4 Diagnosis 5 Etiology 5.1 Calcinosis 5.2 Raynaud's phenomenon 5.3 Esophageal dysmotility 5.4 Sclerodactyly 5.5 Telangiectasia 6 Medical Management 6.1 Raynaud’s phenomenon management 6.2 Calcinosis management 6.3 Esophageal dysmotility management 6.4 Cutaneous manifestations management 7 Physical Therapy Management 7.1 Clinical evaluation/ assessment 7.2 Exercise and splinting 7.3 Modalities 7.4 Manual lymph drainage (MLD) 8 Occupational Therapy (OT) 9 References Description[edit | edit source] CREST syndrome (calcinosis and sclerodactyly) CREST syndrome (also known as Cutaneous systemic sclerosis or limited scleroderma ) is an autoimmune disease that has been defined as a subtype of progressive systemic sclerosis (SSc) with limited skin involvement. The word "CREST " is an acronym for the clinical features that are seen in a patient with this disease: Calcinosis- when calcium salts are deposited into the skin and subcutaneous tissue. Raynaud's phenomenon. Esophageal dysmotility-which can cause difficulty in swallowing. Sclerodactyly- scleroderma in which the fingers become thin and shiny with sclerotic skin at the tip due to subcutaneous and intracutaneous calcinosis and diffused fibrosis of the collagen. Telangiectasia-small widened blood vessels on the skin. Epidemiology[edit | edit source] Wide variation in prevalence of systemic sclerosis was observed, with slightly higher estimates reported in North America (13.5–44.3 per 100,000 individuals) compared to Europe (7.2–33.9 per 100,000 individuals), which may be a true reflection of epidemiological variation or an artifact of clinical data analyses. The apparent increase in both incidence and prevalence over the last 50 years is most likely due to improved classification, earlier diagnosis, and survival. CREST syndrome may account for 22-25% of all occurrences of systemic sclerosis, according to serum antibody investigations; however, epidemiologic research specifically looking at CREST syndrome are missing. SSc diagnosis was reported to occur at the ages of 33.5-59.8 years in Europe and 46.1-49.1 years in North America, and to occur more commonly in women (female:male ratio of 3.8-11.5:1 in Europe and 4.6-15:1 in North America). Women have continuously greater prevalence and incidence rates, indicating a clinically significant difference in the occurrence of SSc across genders. Pathophysiology[edit | edit source] The pathogenesis of SSc involves a classical triad of key mechanisms: Endothelial dysfunction and apoptosis of endothelial cells. Uncontrolled activation of adaptive and innate immunity (notably including M1 inflammatory and M2 pro-fibrotic macrophages). Over-production of extracellular matrix (ECM) components by chronically activated myofibroblasts, resulting in the formation of a stiff and fibrotic extracellular matrix in numerous organs, interfering with their function. Myofibroblasts are the primary contributors to ECM formation and fibrosis.In SSc, myofibroblasts are derived from a range of tissue-resident mesenchymal progenitor cell types, such as fibroblasts, pericytes, microvascular endothelial cells, and vascular preadipocytes]. In SSc, myofibroblasts undergo substantial epigenetic remodelling as well as metabolic changes such as increased glycolysis and altered NAD+ homeostasis.Furthermore, myofibroblasts in SSc exhibit apoptosis resistance as well as unregulated production of extracellular matrix (ECM) components such as collagens, tenascin C, and fibronectin. In turn, these released extracellular components can activate myofibroblasts either directly via innate immunological sensors such as TLR-4, or indirectly by mechano-sensing of increased matrix stiffness by integrins in an FAK dependent way. Regarding limited scleroderma, although the primary cause is unknown, it is reasonable to speculate that vascular endothelial cell abnormalities induce mononuclear infiltration, and that the resulting changes in TH1 and/or TH2 cell and cytokine balance result in abnormal fibroblast activity and increased collagen deposition. Diagnosis[edit | edit source] In the absence of a diagnostic test proving the absence or presence of SSc, the diagnosis is based on a combination of clinical and laboratory findings. According to the latest classification scheme from 2013, SSc is confirmed by: Major Criteria: skin thickening of the fingers of both hands extending proximal to the metacarpophalangeal joints (MCP) is sufficient to classify a subject as having SSc. skin thickening sparing the fingers’ are classified as not having SSc. Minor criteria: Skin thickening of the fingers -Puffy fingers and Sclerodactyly of the fingers (distal to the metacarpophalangeal joints but proximal to the proximal interphalangeal joints) Fingertip lesions -Digital tip ulcers and Fingertip pitting scars Telangiectasia Abnormal nailfold capillaries Pulmonary arterial hypertension and/or interstitial lung disease Raynaud’s phenomenon SSc-related autoantibodies: Anticentromere. Anti–topoisomerase I [anti–Scl-70]. Anti–RNA polymerase III. Etiology[edit | edit source] While the exact cause of CREST syndrome is unknown, it is thought to be the result of a complex interplay of genetic, immunological, and environmental variables. Calcinosis[edit | edit source] Approximately 40% of patients with limited cutaneous SSc complicating from calcinosis (or dystrophic calcification) which is the accumulation of insoluble calcified material in the soft tissues, occurring in the presence of normal calcium and phosphate metabolism. The cause of how or why these crystals form in patients with SSc is not well understood.There is some poorly understood factors have all been proposed to contribute to calcinosis including: Chronic hypoxia characterised by: Digital ulcers Loss of digital tip Abnormal capillary drop-outs seen by nailfold Capillaroscopy Repetitive trauma Based on common locations of these deposits such as the fingertips and extensor surfaces of extremities. Localised structural damage Raynaud's phenomenon[edit | edit source] Raynaud’s phenomenon secondary to SSc occurs in 90% of patients and is often the earliest clinical manifestation to occur.It is vasospastic disorder that characterized by frequent and sudden drops in blood flow to the fingertips, often in response to cold temperatures. Raynaud phenomenon is a symptom complex caused by impaired digital perfusion and can occur as a primary phenomenon or secondary to a wide range of underlying causes. The etiology of SSc-associated Raynaud phenomenon includes factors such as endothelial cell injury (possibly autoantibody mediated); an imbalance between vasoconstrictor and vasodilator molecules (such as endothelin 1 and nitric oxide, respectively); structural microvascular changes from progressive microangiopathy; and intravascular events that lead to luminal occlusive disease. Esophageal dysmotility[edit | edit source] SSc and gastrointestinal manifestations have been proven to be associated , with more than 90% of SSc patients also having various types of gastrointestinal dysfunction, which is considered the third greatest cause of death in SSc patients . Esophageal micro-reflux, manifested as dysphagia and reflux heartburn, is the most common indication of gastrointestinal dysfunction involvement and, to some extent, exacerbates the existing interstitial lung diseases. Furthermore, Barrett's oesophagus, esophageal stenosis, and esophageal cancer induced by esophageal motility problems may worsen the prognosis of SSc patients.The etiology of esophageal motility disorders in patients with SSc is still uncertain.Many functional tests show the presence of vascular damage, fibrosis, and inflammatory illness, however these three factors may not be a complete picture and may lead to undetected etiology. Previously, some studies reporting the related pathological observations revealed esophageal muscle atrophy without evidence of vascular damage, fibrosis, or inflammatory infiltration. However, the exact mechanism of esophageal muscle atrophy is not known. Sclerodactyly[edit | edit source] Sclerodactyly develops from a perivascular inflammatory infiltration in the dermis.Although the cause of this inflammatory process is still unknown,it is believed that mucopolysaccharide, glycoprotein, and collagen (types I and III) deposition in the dermis causes the edematous phase of skin involvement.As collagen deposition progresses, the dermis becomes sclerotic rather than edematous. Meanwhile, in small arteries, a similar mechanism happens and in the intima, mucinous deposition occurs. The adventitia is initially invaded by inflammatory cells before becoming fibrotic. This process causes artery narrowing, followed by arterial collapse or thrombosis.As a result the tissue becomes ischemic.Fibrosis typically disappears years after the start of skin changes, leaving atrophic skin behind. Telangiectasia[edit | edit source] Prominent and numerous telangiectasia are a common clinical symptoms of scleroderma.The etiology of telangiectasia in general, and CREST in particular, is unknown. Models have not explained the preference for hands, face, and mucosa, as well as their proclivity to enlarge in diameter with time. Venous hypertension as a cause appears implausible given the rarity of high venous pressures in the foot. While both Raynaud's phenomenon and telangiectasia are present in CREST, telangiectasia is not present in primary Raynaud's disease, implying that telangiectasia is not caused by recurrent vasospasm or vasoconstriction. Telangiectasia is also widespread on the face, which is unaffected by Raynaud's phenomenon. Medical Management[edit | edit source] Because a scleroderma diagnosis will affect a patient's physical and psychological well-being, a comprehensive approach to care should be taken. An assessment of organ involvement is required, as well as patient education about the clinical course, patient and family support, and treatment based on disease severity and organ involvement. A rheumatologist should be consulted. Organ-specific treatment can have a direct effect on outcomes such as mortality, disease progression, and quality of life issues. Organ-specific treatment | Problem | Treatment | | Arthralgias | NSAIDs, methotrexate, Cox-2 inhibitors | | End-stage lung disease | Lung transplantation | | Esophageal reflux | Proton pump inhibitors, metoclopramide | | Intestinal dysmotility | Antibiotics, if malabsorption present; prokinetics | | Pulmonary hypertension | Calcium channel blockers, epoprostenol | | Lung inflammation | Cyclophosphamide | | Inflammatory myositis | Methotrexate, prednisone | | Raynaud’s phenomenon | Calcium channel blockers, cold avoidance, angiotensin receptor blockers, nitroglycerin, digital sympathectomy | | Renal crisis | Aggressive blood pressure control, including angiotensin 1- converting enzyme inhibitors; dialysis | Raynaud’s phenomenon management[edit | edit source] In approximately 50% of the patients Raynaud’s phenomenon is often very severe and can progresses to digital ulceration (DU).First, Raynaud's phenomenon is treated, and then digital ulceration is treated. The two are purposely combined since optimising Raynaud's phenomenon treatment is a critical initial step in the prevention and treatment of SSc-related digital ulcers. Lifestyle changes (including patient education) and vasoactive medication treatments are recommended for the best management of SSc-related Raynaud's phenomenon. Pharmacological management of digital ulceration (DU) can include: Vasoactive therapies. Other pharmacological therapies (excluding procedural treatments) including antibiotic therapy and analgesia. Procedural pharmacological therapies. Treatment of the acute DU (this can be a medical emergency). Calcinosis management[edit | edit source] Calcinosis cutis is difficult to treat pharmacologically, and a range of medications, including bisphosphonates, intralesional corticosteroids, aluminium hydroxide, warfarin, and diltiazem, have been attempted with poor effectiveness. The current available therapeutic option is local excision of uncomfortable or ulcerated nodules, but local recurrence is prevalent. Esophageal dysmotility management[edit | edit source] In those with systemic/localized scleroderma (SSc) or limited scleroderma, the gastrointestinal tract (GI) is the second most affected organ system. SSc can impact any portion of the GI tract, from the mouth cavity to the anorectum. GI involvement and its management in SSc | GI part | Treatment | | Oral cavity | rehabilitation via orofacial exercises and the administration of cevimeline, pilocarpine, muscarinic agonists, and artificial saliva. | | Esophagus | Lifestyle management including head elevation at night, excluding triggering foods/substance abuse, and consuming small/frequent meals during the day. Proton pump inhibitors (PPIs). Endoscopic dilatationEndoscopic ablation or resection of dysplastic epithelium using photochemical, thermal, or radio ablation energy is recommended in Barrett’s esophagus. | | Stomach | Dietary modifications (low-fat/fiber-based diet and vitamin supplementation) are the first line for gastroparesis | | Small intestine | Antibiotics such as ciprofloxacin, norfloxacin, amoxicillin, tetracyclines (doxycycline), metronidazole, and trimethoprim-sulfamethoxazole are effective against small intestinal bacterial overgrowth (SIBO). | | Colon and anorectal | stimulant laxatives and stool softeners for constipation management. | Cutaneous manifestations management[edit | edit source] Current medicines are restricted and inadequate in treating scleroderma's cutaneous symptoms. Autologous fat transfer (AFT) is a surgical procedure for face rejuvenation that has been used for many decades. Adipose stem cells (ASCs) found in fat grafts have also showed promise in terms of anti-inflammatory and regenerative characteristics. AFT has recently been repurposed to treat systemic sclerosis and localised scleroderma skin symptoms. AFT appears to enhance mouth and hand functions, Raynaud's symptoms, and digital ulcerations in scleroderma patients, according to research. AFT is a safe operation with little postoperative problems, making it a prospective option for scleroderma treatment. More research is needed to properly characterise the impact of fat grafts on the recipient site and to define fat transfer criteria in fibrotic skin diseases. Physical Therapy Management[edit | edit source] Physiotherapists are core members of a multi-disciplinary rheumatology team and should be appropriately trained to work with patients and to manage the many different diseases presenting to the service. Clinical evaluation/ assessment[edit | edit source] Before beginning rehabilitation, the physiotherapist must do a thorough assessment of the patient in order to develop a suitable therapeutic programme especially upper limb functional and ROM assessment as the hands are the most commonly involved part of the body in scleroderma, with symptoms such as edema, Raynaud's phenomenon, sclerodactyly producing pain, and joint ROM reduction. Symptoms that are commonly experienced by patients with SSc and have a major impact on carrying out everyday activities, like fatigue, pain, limitations in hand function, and decreased mobility. Assessments specific to scleroderma include: Health Assessment Questionnaire-Disability Index Scleroderma Health Assessment Questionnaire (SHAQ) Hand mobility in scleroderma (HAMIS) test Exercise and splinting[edit | edit source] Exercise is regarded as a non‐pharmaceutical intervention for people with SSc.Regular exercise training may provide anti-inflammatory effects in chronic conditions characterised by systemic low grade inflammation (for example, type 2 diabetes) by lowering inflammatory markers. Given the probable role of inflammation in the aetiology and clinical symptoms of SSc, if exercise training can reduce inflammation, it may be a useful intervention in controlling some problematic SSc symptoms, such as pain.In general the exercises should emphasize flexion of the metacarpophalangeal joint,extension of the proximal interphalangeal joints, and flexion and abduction of the thumb. Stretching exercises for the hand were shown to increase motion and subsequent function in daily tasks.Hand splints to increase joint motion must be used very carefully as dynamic splints were shown to exacerbate Raynaud’s. Static splints may be useful if patients have inflammation but should only be worn at night until the inflammation decreases. Exercise interventions can include: Hand exercises. Mouth exercises. Aerobic exercise. Resistance exercise- should be started before there is any observed loss of motion. Range of motion exercise. Kinesiotherapy. Hydro Kinesiology . Recreational exercise (i.e. lawn bowls). Modalities[edit | edit source] Pulsed dye laser: The concept of selective photothermolysis supports the treatment of telangiectasia with pulsed dye laser . Previous research using equivalent treatment parameters to those utilized herein discovered that telangiectasia responds effectively to pulsed dye laser, with the majority of cases clearing within one to two treatments. Heat therapy: Heat modalities, such as paraffin, in conjunction with exercise programmes, have been shown to be effective in increasing or maintaining joint motion and hand function.Using of paraffin in order to warm up the hands of patients with SSc has been introduced back in 1983 by Askew et al. Few studies have examined the effect of paraffin on the hands of patients with SSc and they all found positive results such as an immediate improvement in patients’ range of motion of hands and fingers after warming up the hands in paraffin. Manual lymph drainage (MLD)[edit | edit source] Manual lymph drainage (MLD) is the use of mild massage to the skin that promotes smooth muscle contraction around lymphatic vessels, increasing lymphatic flow and eliminating edema and extra interstitial fluid.Microvascular alterations, increased sympathetic activity, and inflammatory changes are all part of the complicated pathophysiology of edema in SSc. In addition to the recognized microvascular alterations in SSc, it has been observed that lymphatic circulation in the upper extremities is disrupted. The presence of finger and hand edema in SSc could support the MLD therapy approach.Difficulties in daily living activities occur because of conditions such as hardness in the hand, the presence of edema pain, reduced grip strength and slower hand movements that develop in connection with systemic sclerosis. MLD was found to improve hand function and quality of life in SSc patients when added to the rehabilitation programme.There is also a reduction in edema, skin involvement, discomfort, and respiratory and sleep issues. Occupational Therapy (OT)[edit | edit source] Occupational therapy may also be required to address independence in daily life activities. A home evaluation may be recommended to ensure that the environment is suitable for the individual's needs. To assist with common everyday tasks, assistive gadgets may be prescribed. An ergonomic study of the workstation may allow people with scleroderma to keep their jobs. References[edit | edit source] ↑ Meyer O. CREST syndrome. Ann Med Interne (Paris) [Internet]. 2002;153(3):183–8. ↑ Le C, Bedocs PM. Calcinosis Cutis. StatPearls Publishing; 2023. ↑ Nelson FRT, Blauvelt CT. The Hand and wrist. In: Nelson FRT, Blauvelt CT, editors. A Manual of Orthopaedic Terminology. Elsevier; 2015. p. 307–41. ↑ Telangiectasia (Spider Veins) [Internet]. Pennmedicine.org. [cited 2023 Sep 8]. Available from: ↑ 5.0 5.1 Bergamasco A, Hartmann N, Wallace L, Verpillat P. Epidemiology of systemic sclerosis and systemic sclerosis-associated interstitial lung disease. Clin Epidemiol [Internet]. 2019;11:257–73. ↑ Wangkaew S, Euathrongchit J, Wattanawittawas P, Kasitanon N, Louthrenoo W. Incidence and predictors of interstitial lung disease (ILD) in Thai patients with early systemic sclerosis: Inception cohort study. Mod Rheumatol [Internet]. 2016;26(4):588–93. ↑ 7.0 7.1 Lescoat A, Varga J, Matucci-Cerinic M, Khanna D. New promising drugs for the treatment of systemic sclerosis: pathogenic considerations, enhanced classifications, and personalized medicine. Expert Opin Investig Drugs [Internet]. 2021;30(6):635–52. ↑ De Martinis M, Ciccarelli F, Sirufo MM, Ginaldi L. An overview of environmental risk factors in systemic sclerosis. Expert Rev Clin Immunol [Internet]. 2016;12(4):465–78. ↑ van den Hoogen F, Khanna D, Fransen J, Johnson SR, Baron M, Tyndall A, et al. 2013 classification criteria for systemic sclerosis: an American College of Rheumatology/European League against Rheumatism collaborative initiative. Arthritis and rheumatism [Internet]. 2013 [cited 2023 Sep 8];65(11). ↑ Hsu V, Varga J, Schlesinger N. Calcinosis in scleroderma made crystal clear. Curr Opin Rheumatol [Internet]. 2019;31(6):589–94. ↑ Chang SH, Jun JB, Lee YJ, Kang TY, Moon KW, Ju JH, et al. A clinical comparison of an endothelin receptor antagonist and phosphodiesterase type 5 inhibitors for treating digital ulcers of systemic sclerosis. Rheumatology (Oxford) [Internet]. 2021;60(12):5814–9. ↑ 12.0 12.1 1. Hughes M, Allanore Y, Chung L, Pauling JD, Denton CP, Matucci-Cerinic M. Raynaud phenomenon and digital ulcers in systemic sclerosis. Nat Rev Rheumatol [Internet]. 2020;16(4):208–21. ↑ 13.0 13.1 Li B, Yan J, Pu J, Tang J, Xu S, Wang X. Esophageal dysfunction in systemic sclerosis: An update. Rheumatol Ther [Internet]. 2021;8(4):1535–49. ↑ Adigun R, Goyal A, Hariz A. Systemic Sclerosis. StatPearls Publishing; 2022. ↑ 15.0 15.1 Foti R, De Pasquale R, Dal Bosco Y, Visalli E, Amato G, Gangemi P, et al. Clinical and histopathological features of Scleroderma-like disorders: An update. Medicina (Kaunas) [Internet]. 2021;57(11):1275. ↑ Halachmi S, Gabari O, Cohen S, Koren R, Amitai DB, Lapidoth M. Telangiectasis in CREST syndrome and systemic sclerosis: correlation of clinical and pathological features with response to pulsed dye laser treatment. Lasers Med Sci [Internet]. 2014;29(1):137–40. ↑ Joslin N. Early identification key to scleroderma treatment. Nurse Pract [Internet]. 2004;29(7):24–39; quiz 40–1. ↑ Gelber AC, Wigley FM. Disease severity as a predictor of outcome in scleroderma. Lancet [Internet]. 2002;359(9303):277–9. ↑ Hughes M, Allanore Y, Chung L, Pauling JD, Denton CP, Matucci-Cerinic M. Raynaud phenomenon and digital ulcers in systemic sclerosis. Nat Rev Rheumatol [Internet]. 2020;16(4):208–21. ↑ 20.0 20.1 Herrick AL, Philobos M. Pharmacological management of digital ulcers in systemic sclerosis - what is new? Expert Opin Pharmacother [Internet]. 2023;24(10):1159–70. ↑ Kempanna Y. P28 Recalcitrant exuberant digital calcinosis cutis in a patient of CREST syndrome - A case report. Rheumatol Adv Pract [Internet]. 2022;6(Supplement_1). ↑ Nassar M, Ghernautan V, Nso N, Nyabera A, Castillo FC, Tu W, et al. Gastrointestinal involvement in systemic sclerosis: An updated review. Medicine (Baltimore) [Internet]. 2022;101(45):e31780. ↑ Kawakibi AR, Khouri AN, Cederna PS, Strong AL. Novel indications for autologous fat grafting in reconstruction: scleroderma. Plast Aesthet Res [Internet]. 2023;10(0):48. ↑ Cohen A, Morrow H, Cleary G. Physiotherapy and rheumatological disorders. Paediatr Child Health (Oxford) [Internet]. 2014;24(2):83–8. ↑ Willems LM, Vriezekolk JE, Schouffoer AA, Poole JL, Stamm TA, Boström C, et al. Effectiveness of nonpharmacologic interventions in systemic sclerosis: A systematic review: Nonpharmacologic care in SSc. Arthritis Care Res (Hoboken) [Internet]. 2015;67(10):1426–39. ↑ Johnson SR, Hawker GA, Davis AM. The health assessment questionnaire disability index and scleroderma health assessment questionnaire in scleroderma trials: An evaluation of their measurement properties. Arthritis Rheum [Internet]. 2005;53(2):256–62. ↑ Sandqvist G, Eklund M. Hand mobility in scleroderma (HAMIS) test: The reliability of a novel hand function test. Arthritis Rheum [Internet]. 2000;13(6):369–74. ↑ 28.0 28.1 Frade S, Cameron M, Espinosa-Cuervo G, Suarez-Almazor ME, Lopez-Olivo MA. Exercise and physical therapy for systemic sclerosis. Cochrane Libr [Internet]. 2022;2022(3). ↑ 29.0 29.1 29.2 Krysia Dziedzic, Hammond A. Rheumatology : evidence-based practice for physiotherapists and occupational therapists. Edinburgh: Churchill Livingstone; 2010. ‌ ↑ Halachmi S, Gabari O, Cohen S, Koren R, Amitai DB, Lapidoth M. Telangiectasis in CREST syndrome and systemic sclerosis: correlation of clinical and pathological features with response to pulsed dye laser treatment. Lasers Med Sci [Internet]. 2014;29(1):137–40. ↑ Askew LJ, Backett VL, An K-N, Chao EYS. Objective evaluation of hand function in scleroderman patients to assess effectiveness of physical therapy. Rheumatology (Oxford) [Internet]. 1983;22(4):224–32. ↑ Kristensen LQ, Oestergaard LG, Bovbjerg K, Rolving N, Søndergaard K. Use of paraffin instead of lukewarm water prior to hand exercises had no additional effect on hand mobility in patients with systemic sclerosis: A randomized clinical trial. Hand Ther [Internet]. 2019;24(1):13–21. ↑ 33.0 33.1 Yılmaz A, Çalık BB, Kabul EG, Tasçı M, Çobankara V. Efficacy of manual lymph drainage in systemic sclerosis: A case report. Ann Clin Anal Med [Internet]. 2021;12(Suppl_03):3): S354. ↑ 34.0 34.1 Barange J. Case Study of Physiotherapy Treatment of a Patient with the Diagnosis Scleroderma. 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Read more pPhysiopedia +Physiopedia Plus Physiopedia About News Donations Shop Contact Content Articles Categories Resources Projects Contribute Courses PAI Legal QA Framework Disclaimer Terms Privacy Cookies Report content AI Licensing Physiopedia available in: French German Italian Spanish Ukrainian Back to top Related articles [Scleroderma - Physiopedia Search Search Search Toggle navigation pPhysiopedia pPhysiopedia About News Contribute Courses PAI Resources Shop Contact Login pPhysiopedia About News Contribute Courses PAI Resources Shop Contact p + Contents Editors Categories Cite Contents loading... Editors loading... Categories loading... When refering to evidence in academic writing, you should always try to reference the primary (original) source. That is usually the journal article where the information was first stated. In most cases Physiopedia articles are a secondary source and so should not be used as references. Physiopedia articles are best used to find the original sources of information (see the references list at the bottom of the article). If you believe that this Physiopedia article is the primary source for the information you are refering to, you can use the button below to access a related citation statement. Cite article Scleroderma Jump to:navigation, search Original Editors - Amanda Schoenfelder and Katie Williams from Bellarmine University's Pathophysiology of Complex Patient Problems project. Top Contributors - Bridget Ward , Katie Williams , Amanda Schoenfelder , Lucinda hampton , Admin , Elaine Lonnemann , Khloud Shreif , WikiSysop , Sehriban Ozmen , Wendy Walker and Kim Jackson Contents 1 Introduction 2 Classification 3 Epidemiology 4 Diagnostic Tests/Lab Tests/Lab Values 5 Etiology/Causes 6 Systemic Involvement 7 Medical Management 8 Physical Therapy Management 9 Other Health Professionals 10 Prognosis SSC 11 Differential Diagnosis 12 Case Report/ Case Study 13 References Introduction[edit | edit source] Scleroderma, a rare connective tissue disorder with a complex unknown pathogenesis, that involves changes to the skin, and can also involve changes to the blood vessels and internal organs. It is a chronic, progressive autoimmune disorder where inflammation and the overproduction of collagen accumulate in the body. Classification[edit | edit source] There are two main types of Scleroderma- Systemic and Localized. Systemic Scleroderma (SSc) may affect the skin and multiple systems such as the integumentary, cardiopulmonary, vascular, gastrointestinal, genitourinary, or musculoskeletal. There are three forms of Systemic Scleroderma. With Diffuse Scleroderma skin thickening occurs more rapidly and involves more skin areas than in limited. In addition, people with diffuse scleroderma have a higher risk of developing “sclerosis” or fibrous hardening of the internal organs. Sine Scleroderma involves organ fibrosis with no skin involvement. This is rare. With Limited Scleroderma (includes CREST syndrome) skin thickening is less widespread, typically confined to the fingers, hands and face, and develops slowly over years. Internal problems occur, but they are less frequent and tend to be less severe than in diffuse scleroderma. CREST is an acronym using the letters of the conditions involved; Calcinosis, Raynaud’s, Esophageal dysmotility, Sclerodactyly and Telangiectasia. The video below gives some good examples of these. Localized Scleroderma usually affects the skin and related tissues in part of the body. Two of the main types of Localized Scleroderma are below Morphea is characterized by waxy patches on the skin of varying sizes, shapes and color. The skin under the patches may thicken. The patches may enlarge or shrink, and often may disappear spontaneously within three to five years. In some, rare, cases muscle weakness may be associated. Linear Scleroderma starts as a streak or line of hardened, waxy skin on an arm or leg or on the head and neck. Linear scleroderma tends to involve deeper layers of the skin as well as the surface layers, and sometimes affects the motion of the joints, which lie underneath. Mixed Connective Tissue Disease or Scleroderma Overlap syndrome is when Scleroderma is diagnosed when another disease, such as systemic lupus erythematosus and polymyositis, has been previously diagnosed. Epidemiology[edit | edit source] Scleroderma is a rare disease, with differing prevalence due to ethnicity, gender, and geographic area. Systemic sclerosis SSc is more common in European, North, and South American patients than in East Asian patients, with the Highest prevalence (47 in 100,000) is among indigenous peoples in Canada. In line with other autoimmune diseases, women are at greater risk than men (ratio 4.6:1)but males tend to have more severe SSc. Thirty percent of those with scleroderma have the systemic form of scleroderma and sine scleroderma accounted for nearly 10% of patients with systemic sclerosis. This type appears to be more common in adults. Localised scleroderma is more common in children. Estimates vary. In France, the cutaneous form predominates, with the incidence of systemic sclerosis estimated at 158.3 per million in 2002. Some estimates predict that about 300,000 Americans have scleroderma. Some predict 75,000 to 100,000 It is more common in females than males, and between the ages of 20 to 50 Those with a family member who has scleroderma, or another autoimmune disease such as lupus, may have a slightly higher risk of developing scleroderma. Race and ethnic background, may influence the risk of getting scleroderma, the age of onset, and the pattern or severity of internal organ involvement. Choctaw Native Americans and African-Americans are more likely than Americans of European descent to develop the type of scleroderma that affects internal organs. Diagnostic Tests/Lab Tests/Lab Values[edit | edit source] A diagnosis of scleroderma can be difficult because symptoms may be similar to that’s of other disease. Diagnosis will be based on: Medical history Physical examination. See below for symptoms of ‘Systemic Involvement’ X-rays and CT for bony investigations MRI and US for soft tissue examiantion Lab tests can also confirm a suspected diagnosis:Topoisomerase-1 antibody or Scl-70 antibodies is a marker of systemic sclerosis.; Anticentromere antibodies, or ACA, is a marker of CREST; ESR is found to be increased in those with disease activity or relapse; AMF is positive in those with Mixed Scleroderma Not all people with scleroderma have these antibodies and because not all people with the antibodies have scleroderma, lab test results alone cannot confirm the diagnosis. A skin biopsy (the surgical removal of a small sample of skin for microscopic examination) can aid in or help confirm a diagnosis. Diagnosing scleroderma is easiest when a person has typical symptoms and rapid skin thickening. In other cases, a diagnosis may take months, or even years, as the disease unfolds and reveals itself and as the doctor is able to rule out some other potential causes of the symptoms. Etiology/Causes[edit | edit source] The cause of scleroderma is unknown, but it cannot be transmitted from person to person. Several factors that may increase a person's risk of getting scleroderma include: Abnormal immune or inflammatory activity: In scleroderma, the immune system is thought to stimulate cells called fibroblasts so they produce too much collagen. People who already have rheumatic or autoimmune disorders are at increased risk due to previous abnormal autoimmune activity Genetic makeup: Although genes seem to put certain people at risk for scleroderma and play a role in its course, the disease is not passed from parent to child like some genetic diseases. Research indicates that variations in genes relating to the the body’s immune system, eg.those in the HLA-complex, IRF5 and STAT4, may increase the risk of developing scleroderma. Environmental triggers: Research suggests that exposure to some environmental factors may trigger scleroderma-like disease in people who are genetically predisposed to it. Suspected triggers may include viral or bacterial infections Hormones: Women develop scleroderma more often than men. Scientists suspect that hormonal differences between women and men play a part in the disease. However, the role of estrogen or other female hormones has not been proven. Systemic Involvement[edit | edit source] Medical Management[edit | edit source] Currently, there is no treatment that controls or stops the underlying problem—the overproduction of collagen—in all forms of scleroderma. Treatment and management focus on relieving symptoms and limiting damage. Importantly treatment goals need be holistic and individualised for the client, aiming to optimise their quality of life, as well as preventing further organ damage. Patient Education regarding the disease and advocating involvement in regular exercises, healthy diet and lifestyle is important. Several different specialists may be involved in the care of one person since scleroderma can affect many different organs and organ systems. Typically, care will be managed by a rheumatologist, who may refer the patient to other specialists, depending on the specific problems they are having. These specialists can include: • Dermatologist for the treatment of skin symptoms • Nephrologist for kidney complications • Cardiologist for heart complications • Gastroenterologist for problems of the digestive tract • Pulmonary specialist for lung involvement Medications used to treat scleroderma may include: Anti-inflammatory medicines such as corticosteroids Immune-suppressing medications such as methotrexate and cytoxan. Nonsteroidal anti-inflammatory drugs (NSAIDs) Scleroderma can affect many different organs and organ systems. Additional medications are based on the symptoms that the individual experiences. Examples of other treatments for specific symptoms include: Medicines for heartburn or swallowing problems - antacid drugs, especially proton- pump inhibitors Blood pressure medications (particularly ACE inhibitors) for high blood pressure or kidney problems Light therapy to relieve skin thickening Cyclophosphamide and mycophenolate can be effective in treating interstitial lung disease Medications to treat pulmonary hypertension include prostacyclin-like drugs, endothelin receptor antagonists and PDE-5 inhibitors. These work to open up the blood vessels in the lungs. Medications to treat Raynaud's phenomenon- calcium channel blockers, PDE-5 inhibitors to open up narrowed blood vessels and improve circulation. Intestinal dysfunction Medicines to increase saliva secretion in the mouth to reduce effects of Sjogren’s Syndrome (a chronic autoimmune disease in which a person’s white blood cells attack their moisture-producing glands) Physical Therapy Management[edit | edit source] Physical therapy can be very beneficial in the treatment of scleroderma. Due to the musculoskeletal damage often seen in scleroderma patients, physiotherapy is often needed to prevent lasting damages. Physical therapists/ Physiotherapists, together with other health professionals such as Occupational Therapists, can design a program of regular stretching and gentle exercise to help: • Manage pain • Improve strength • Improve/maintain mobility • Minimize joint contractures • Improve circulation. Education re protecting vulnerable skin areas and keeping them warm • Enhance/maintain performance of activities of daily living in order to encourage independence In the context of physical therapy, various techniques and exercises can be used to aid individuals in regaining or enhancing their physical capabilities including; massage, hydrotherapy, electrical stimulation therapy, exercise movement methods, or physiotherapy techniques. Although randomized controlled trials have evaluated these interventions for their suitability in the treatment of systemic sclerosis (SSc), the results have demonstrated some level of inconsistency. Aerobic exercise such as walking, jogging, cycling, and dancing to enhance cardiovascular efficiency and encompasses a wide range of physical activities, resistance training contributes to the development of muscle strength, anaerobic endurance, and muscle size it may be like lifting groceries, climbing stairs, or transitioning from sitting to standing, or range of motion exercise like stretching, yoga help to enhance the range of motion of specific joints. These sample of exercises can be carried individually or in group even indoors or outdoors. Other Health Professionals[edit | edit source] Other health professionals such as psychologists, and social workers may play a role in patient care. Dentists, orthodontists, and speech therapists may be involved in the management of complications that arise from thickening of mouth and face tissues. Research about the most effective management and treatment is ongoing. Prognosis SSC[edit | edit source] SSc has with high mortality. The prognosis in SSc has improved since the 1990's with the 5-year survival rates now up to 80%. Unfortunately patients with advanced pulmonary arterial hypertension have a poorer outlook, with a 50% 2-year survival rate. Additionally patients with SSc-related pulmonary arterial hypertension are worse off when compared to those with idiopathic pulmonary arterial hypertension. Differential Diagnosis[edit | edit source] A number of other diseases have symptoms similar to those seen in scleroderma. These include: Eosinophilic fasciitis (EF): This rare disease involves the fascia, the thin connective tissue around the muscles. The fascia becomes swollen, inflamed and thick. The skin on the arms, legs, neck, abdomen or feet can be involved. Unlike scleroderma, the fingers are not involved. A skin biopsy distinguishes between the two diseases. Nephrogenic Systemic Fibrosis: An uncommon disease of fibrosis of the skin and organs, caused by gadolinium exposure ( used in imaging) in those patients with renal insufficiency. Generalized scleroderma-like skin thickening: This may occur with scleromyxedema, graft-versus-host disease, porphyria cutanea tarda, and human adjuvant disease Raynauds phenomena: This may occur in isolation, as well as part of the scleroderma disease process. Case Report/ Case Study[edit | edit source] Scleroderma Case Study Scleroderma can present with pulmonary nodules References[edit | edit source] ↑ 1.0 1.1 1.2 Adigun R, Goyal A, Hariz A. Systemic sclerosis. Available: (last accessed 31.3.2024) ↑ 2.0 2.1 Rosendahl AH, Schönborn K, Krieg T. Pathophysiology of systemic sclerosis (scleroderma). The Kaohsiung journal of medical sciences. 2022 Mar;38(3):187-95. ↑ 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 Rosendahl AH, Schönborn K, Krieg T. Pathophysiology of systemic sclerosis (scleroderma). The Kaohsiung journal of medical sciences. 2022 Mar;38(3):187-95. ↑ 4.0 4.1 4.2 Arthritis Foundation. Scleroderma information sheet. ( accessed 15.3.2018) ↑ 5.0 5.1 5.2 5.3 Careta MF, Romiti R. Localized scleroderma: clinical spectrum and therapeutic update. Anais brasileiros de dermatologia. 2015 Jan;90:62-73. ↑ Sato F, Sato M, Yamano T, Yamaguchi K, Miyake T. A Case of Mixed Connective Tissue Disease That Transformed Into Systemic Lupus Erythematosus After a Long Clinical Course. Cureus. 2023 Apr 27;15(4). ↑ 7.0 7.1 Calderon LM, Pope JE. Scleroderma epidemiology update. Current Opinion in Rheumatology. 2021 Mar 1;33(2):122-7. ↑ 8.0 8.1 8.2 Odonwodo A, Badri T, Hariz A. Scleroderma. InStatPearls [Internet] 2022 Aug 1. StatPearls Publishing.Available: (accessed 2.9.2023) ↑ Lescoat A, Huang S, Carreira PE, Siegert E, de Vries-Bouwstra J, Distler JH, Smith V, Del Galdo F, Anic B, Damjanov N, Rednic S. Cutaneous Manifestations, Clinical Characteristics, and Prognosis of Patients With Systemic Sclerosis Sine Scleroderma: Data From the International EUSTAR Database. JAMA dermatology. 2023 Aug 1;159(8):837-47. BibTeXEndNoteRefManRefWorks ↑ 10.0 10.1 10.2 10.3 10.4 American Academy of Dermatology Association. scleroderma: diagnosis and treatment. ↑ 11.0 11.1 11.2 11.3 Mayo Foundation for Medical Education and Research. Patient Care and Health Information. Scleroderma ( accessed 8 April 2018) ↑ Marzano AV, Menni S, Parodi A, Borghi A, Fuligni A, Fabbri P, Caputo R. Localized scleroderma in adults and children. Clinical and laboratory investigations on 239 cases. European Journal of Dermatology. 2003 Apr 15;13(2):171-6. ↑ Careta MF, Romiti R. Localized scleroderma: clinical spectrum and therapeutic update. Anais brasileiros de dermatologia. 2015 Jan;90:62-73. ↑ 14.0 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8 National Institute of Arthritis and Musculoskeletal and Skin Diseases. Health Topics. Scleroderma. ↑ 15.0 15.1 US National Library of Medicine. Genetics Home Reference. Systemic Scleroderma. / (accessed 8 April 2018) ↑ 16.0 16.1 Shah AA, Wigley FM. My approach to the treatment of scleroderma. In Mayo Clinic Proceedings 2013 Apr 1 (Vol. 88, No. 4, pp. 377-393). ↑ Poole JL. Musculoskeletal rehabilitation in the person with scleroderma. Current opinion in rheumatology. 2010 Mar 1;22(2):205-12. ↑ Frade S, Cameron M, Espinosa-Cuervo G, Suarez-Almazor ME, Lopez-Olivo MA. Exercise and physical therapy for systemic sclerosis. The Cochrane Database of Systematic Reviews. 2022;2022(3). ↑ Medscape. Drugs and Diseases. Rheumatology. Scleroderma. Differential Diagnosis. ( accessed 8 April 2018) Retrieved from " Categories: Bellarmine Student Project Conditions Non Communicable Diseases Rheumatology Autoimmune Disorders Get Top Tips Tuesday and The Latest Physiopedia updates Email Address I give my consent to Physiopedia to be in touch with me via email using the information I have provided in this form for the purpose of news, updates and marketing. HP Yes please It's free, and you can unsubscribe any time. Privacy policy. Our Partners The content on or accessible through Physiopedia is for informational purposes only. Physiopedia is not a substitute for professional advice or expert medical services from a qualified healthcare provider. Read more pPhysiopedia +Physiopedia Plus Physiopedia About News Donations Shop Contact Content Articles Categories Resources Projects Contribute Courses PAI Legal QA Framework Disclaimer Terms Privacy Cookies Report content AI Licensing Physiopedia available in: French German Italian Spanish Ukrainian © Physiopedia 2025 | Physiopedia is a registered charity in the UK, no. 1173185 Back to top]( [Raynaud's Phenomenon - Physiopedia Search Search Search Toggle navigation pPhysiopedia pPhysiopedia About News Contribute Courses PAI Resources Shop Contact Login pPhysiopedia About News Contribute Courses PAI Resources Shop Contact p + Contents Editors Categories Cite Contents loading... Editors loading... Categories loading... When refering to evidence in academic writing, you should always try to reference the primary (original) source. That is usually the journal article where the information was first stated. In most cases Physiopedia articles are a secondary source and so should not be used as references. Physiopedia articles are best used to find the original sources of information (see the references list at the bottom of the article). If you believe that this Physiopedia article is the primary source for the information you are refering to, you can use the button below to access a related citation statement. Cite article Raynaud's Phenomenon Jump to:navigation, search Original Editor - Kirenga Bamurange Liliane Top Contributors - Kirenga Bamurange Liliane , Kim Jackson and Lucinda hampton Contents 1 Definition 2 Types of RP 3 Causes of RP 4 Signs and Symptoms 5 Diagnosis of RP 6 Management/Treatment of RP 7 Prevention of RP 8 Risk factors of RP 9 Complications of RP 10 References Definition[edit | edit source] Raynaud's Phenomenon (RP) is a disorder characterized by an episodic color change of the extremities in response to cold exposure or emotional stress. It is represented by fingers that have turned white, then blue and finally become red which represent vasospasm, deoxygenation, and reperfusion hyperemia. It affects primarily the blood vessels in the fingers and toes, but the nose, lips, or ear lobes vessels may also be affected. This phenomenon is named after ' Maurice Raynaud, a medical student who defined the first case in 1862 as " episodic, symmetric, acral vasospasm characterized pallor, cyanosis, and a sense of fullness or tautness, which may be painful." Types of RP[edit | edit source] RP occurs as a primary form on its own or as a secondary form due to an underlying disease . It is referred to as "Raynaud's disease" or " primary Raynaud's phenomenon" when it occurs alone and isn't associated with an underlying condition. It is referred to as Secondary Raynaud's when it's the result of associated diseases and tends to be more serious than the primary form. Causes of RP[edit | edit source] The causes of both primary and secondary Raynaud's phenomenon are unknown. It's maybe linked to some blood disorders by increasing blood thickness due to excess platelets or red blood cells. Both abnormal nerve control of the blood-vessel diameter and nerve sensitivity to cold exposure may also be contributing factors. The following causes can be found: An attack can be triggered by cold temperature in cases such as putting hands in cold water, taking something from the freezer, or exposure to cold air. Attacks can also be triggered by emotional stress Other causes may include: Blood disorders. The Sensitivity of special receptors that control the narrowing of blood vessels. Secondary causes of Raynaud’s include lupus, scleroderma, and other diseases. As well as: The disease of the arteries. Carpal Tunnel Syndrome. Repetitive action or vibration. Smoking. Injuries to the hands and feet such as wrist fracture, surgery, or frostbite. Certain medications such as medications for high blood pressure, migraine, hyperactivity, and chemotherapy. Signs and Symptoms[edit | edit source] The symptoms of RP depend on the severity, frequency, and duration of the blood vessel spasm. These may include: Skin decoloration. Mild tingling and digits' numbness. Pain due to irritation of sensory nerves. Swollen and painful hands Ulceration of the tips of the digits because of poor oxygen supply. Gangrene is caused by a prolonged lack of oxygen. Raynaud's Phenomenon Diagnosis of RP[edit | edit source] RP itself is not a diagnosis and there is no blood test to diagnose it. However, a thorough assessment should be done and should be comprised by: A detailed history including occupation and smoking history. A physical examination of hands to specifically look for signs of connective tissue disease. A full blood count. Nailfold capillaroscopy to look for deformities or swelling of the tiny blood vessels. Management/Treatment of RP[edit | edit source] There is no cure for RP but it can be managed with a proper and well-planed treatment based on the patient's symptoms, age, severity, and overall health. The treatment may include: Avoiding exposure to cold. Keeping warm with gloves, socks, a scarf, and a hat. Stopping smoking. Wearing finger guards over fingers with sores. Avoiding trauma or vibrations to the hand (such as with vibrating tools). Taking blood pressure medicines during the winter months to help reduce constriction of the blood vessels. Patient's education. The treatment includes medical therapy as well in form of antihypertensives, such as calcium-channel blockers (CCB) or sartanes, and vasodilating agents, such as intravenous iloprost. RP patients with no other symptoms than the skin discoloration extremities may require only home-remedy measures to prevent complications. Prevention of RP[edit | edit source] The prevention of RP can be done by: Avoiding all the precipitating factors such as exposure to heat or cold and smoking. Avoiding medications that aggravate symptoms of RP. Risk factors of RP[edit | edit source] The following factors can increase the risk of developing RP : Underlying diseases such as Scleroderma, Lupus, Rheumatoid Arthritis, or Sjögren's syndrome. Cigarette smoking Injury or trauma Side effects of certain medicines Chemical exposure Complications of RP[edit | edit source] People with RP can develop critical ischemia and the development of digital ulcers (DU), that potentially cause superinfection, necrosis, and amputation . References[edit | edit source] ↑ 1.0 1.1 1.2 Herrick, AL. Raynaud’s phenomenon. Journal of Scleroderma and Related Disorders. 2019; 4(2): 89–101. ↑ Raynaud's Phenomenon. Available from: (Accessed, 16th January 2021). ↑ Raynaud Phenomenon. Available from: (Accessed, 16 January 2021). ↑ 4.0 4.1 4.2 4.3 Raynaud's Phenomenon. Available from: Accessed, 16 January 2021). ↑ 5.0 5.1 5.2 5.3 Raynaud's disease. Available from: (Accessed, 17 January 2021). ↑ 6.0 6.1 6.2 6.3 Raynaud's Phenomenon and Disease. Available from: (Accessed, 16 January 2021) ↑ 7.0 7.1 Korsten P, Müller GA, Rademacher JG, Zeisberg M, Tampe B. Rheopheresis for digital ulcers and Raynaud's phenomenon in systemic sclerosis refractory to conventional treatments. Frontiers in medicine. 2019 Sep 18;6:208. ↑ Johns Hopkins Rheumatology. Raynaud’s Phenomenon: What You Should Know | Johns Hopkins Medicine. Available from: [last accessed 27/1/2021] ↑ Brigham And Women's Hospital. Raynaud’s Phenomenon Video – Brigham and Women’s Hospital. Available from: [last accessed 27/1/2021] Retrieved from " Categories: Conditions Medical Blood Disorders Get Top Tips Tuesday and The Latest Physiopedia updates Email Address I give my consent to Physiopedia to be in touch with me via email using the information I have provided in this form for the purpose of news, updates and marketing. HP Yes please It's free, and you can unsubscribe any time. Privacy policy. Our Partners The content on or accessible through Physiopedia is for informational purposes only. Physiopedia is not a substitute for professional advice or expert medical services from a qualified healthcare provider. Read more pPhysiopedia +Physiopedia Plus Physiopedia About News Donations Shop Contact Content Articles Categories Resources Projects Contribute Courses PAI Legal QA Framework Disclaimer Terms Privacy Cookies Report content AI Licensing Physiopedia available in: French German Italian Spanish Ukrainian © Physiopedia 2025 | Physiopedia is a registered charity in the UK, no. 1173185 Back to top]( [Mixed Connective Tissue Disease - Physiopedia Search Search Search Toggle navigation pPhysiopedia pPhysiopedia About News Contribute Courses PAI Resources Shop Contact Login pPhysiopedia About News Contribute Courses PAI Resources Shop Contact p + Contents Editors Categories Cite Contents loading... Editors loading... Categories loading... When refering to evidence in academic writing, you should always try to reference the primary (original) source. That is usually the journal article where the information was first stated. In most cases Physiopedia articles are a secondary source and so should not be used as references. Physiopedia articles are best used to find the original sources of information (see the references list at the bottom of the article). If you believe that this Physiopedia article is the primary source for the information you are refering to, you can use the button below to access a related citation statement. Cite article Mixed Connective Tissue Disease Jump to:navigation, search Original Editors - Students from Bellarmine University's Pathophysiology of Complex Patient Problems project. Top Contributors - Erica Hunter , Laci Sattler , Bridget Ward , Vidya Acharya , Kim Jackson , Elaine Lonnemann , WikiSysop and Sehriban Ozmen Contents 1 Definition/Description 2 Prevalence 3 Aetiology/Causes 4 Characteristics/Clinical Presentation 5 Associated Co-morbidities 6 Medical Management 7 Diagnostic Tests/Lab Tests/Lab Values 8 Physical Therapy Management 9 Differential Diagnosis 10 Case Reports/ Case Studies 11 Resources 12 References Definition/Description[edit | edit source] Mixed connective tissue disease (MCTD) is a systemic disease which consists of clinical symptoms observed in the following three disorders: systemic lupus erythematosus, polymyositis, and systemic sclerosis (also known as systemic scleroderma). MCTD is considered an "overlapping disease" as it contains features of these three disorders. The features can be categorized broadly as arthritic changes, cardiopulmonary dysfunctions, skin changes, muscle weakness, kidney disease, and dysfunctions of the oesophagus. The symptoms associated with the three underlying disorders do not generally present simultaneously. It usually takes several years before the symptoms of each individual disorder present, which ultimately complicates the diagnosis of MCTD. Typically, the first symptom to present is swelling of the fingers or the presentation of “sausage fingers”. As the disease progresses, it can often affect multiple organs such as the lungs, heart, and/or kidneys. There is no cure for MCTD, however side effects can be managed through the use of medications. Prevalence[edit | edit source] It has been reported that 80% of individuals diagnosed with MCTD are women, with the highest prevalence in the age group under thirty. Other sources have reported statistics collected from patients ages 5 through 80, with the peak prevalence around 20 years of age. Estimates of this disease show it occurs in between 2-6.4out of every 100,000 individuals. Aetiology/Causes[edit | edit source] The exact cause of MCTD is unknown, but it has been classified as an autoimmune disorder. Individuals with this disease have high levels of antinuclear antibodies (ANAs) and antibodies to U1 snRNP. A genetic link can be seen in MCTD in that some individuals diagnosed with MCTD report having a family member who also has a connective tissue disease. Also, exposures to certain chemicals or viruses such as silica or polyvinyl chloride have been found as potential causes of MCTD. Characteristics/Clinical Presentation[edit | edit source] The initial sign of MCTD may be shown as a presentation of puffy and swollen hands, Raynaud’s phenomenon, and polyarthritis. Some of the classical conditions or "signs” of MCTD include Raynaud phenomenon Inflammation of muscles and joints Pulmonary hypertension Raynaud phenomenon Swollen fingers, often “sausage-like”, can be a temporary stage of the disease, or may progress into limited movement of the fingers due to thinning of fingers and thickening of the skin The chart below lists some of the symptoms common versus uncommon symptoms in early stages of MCTD. Associated Co-morbidities[edit | edit source] End-stage interstitial lung disease Pulmonary Hypertension Interstitial lung disease (20%) Heart disease Kidney damage (25%) Digestive tract damage , Anemia (75%) ,, Necrosis Hearing loss Sjögren syndrome (25%) Hepatomegaly Neurological involvement Medical Management[edit | edit source] Medical management is undertaken by a range of specialist working together. Since there is no cure yet for the disease, management is focussed on control of symptoms and minimising systemic involvement. A variety of medications will be used to manage the various symptoms associated with the disease (s) Corticosteroids- may be used to manage synovitis ( active, or more severe disease) Anti-malarial drugs- may be used it manage synovitis, may help prevent disease flares Nonsteroidal anti-inflammatory drugs - may be used to manage arthritis/ arthralgia- Immunosuppressive drugs- may be used to manage refractory synovitis and pulmonary hypertension in some patients Calcium channel blockers- may be used to treat Raynaud’s ( Vasodilation and possible antiplatelet effects) Phosphodiesterase inhibitors- may be used to treat pulmonary hypertension Endothelial receptor antagonists - may be used to treat pulmonary hypertension Prostaglandins- may be used to treat pulmonary hypertension Proton pump inhibitors- may be used for heartburn or swallowing problems Diagnostic Tests/Lab Tests/Lab Values[edit | edit source] Laboratory Testing Often Includes: Anti-U1-RNP (ribonucleoprotein) antigens Urinalysis Muscle enzymes (myositis involvement) Complete blood count (CBC) Antinuclear antibodies Lipase and amylase (pancreatitis involvement) Routine blood chemistry To check for systemic involvement, the following imaging testing can be performed Chest radiography- assesses for infiltrates, effusions, and cardiomegaly MRI- assess for neuropsychiatric signs or symptoms CT scan/ultrasound- evaluates abdominal pain in cases of suspected serositis, pancreatitis, and/or visceral perforation related to vasculitis Echocardiography- assesses for effusion, chest pain, pulmonary hypertension, or valvular disease Systemic involvement tests may also include cardiopulmonary testing, such as: Electrocardiography Pulmonary function tests Six-minute walk test Currently, there are three different criteria classification systems that are associated with predicting the probability that an individual may have MCTD. These three classification systems are set forth by Modified Sharp et al (1987), Alarcon Segovia et al (1987), and Kauskawa et al (1987). Listed below are the criteria sets that are presently used in the diagnosing MCTD. Malar rash Physical Therapy Management[edit | edit source] Since there has been limited research regarding physical therapy treatment in patients with MCTD, interventions should be tailored to address the impairments of each individual. Although each person presents differently, there are some common areas that need to be addressed in nearly all cases. Individuals with MCTD often present with decreased aerobic capacity and weakness of the proximal musculature. Physical therapists should treat according to the common deficits seen in the disease, as well as personal impairments that arise with each case. Common areas of focus may include: Patient education regarding joint protection Aerobic and endurance training Range of motion exercises to maintain available range Passive stretching, including splinting for joint protection Strengthening total-body exercises (including proximal musculature) Skin education and management Energy conservation techniques Differential Diagnosis[edit | edit source] Systemic Lupus Erythematosus - a chronic inflammatory disease characterized by protean manifestations with a relapsing and remitting course Scleroderma- progressive skin hardening and induration Dermatomyositis- idiopathic inflammatory myopathy with characteristic signs commonly present in the skin, muscles, and joints Polymyositis- idiopathic inflammatory myopathy which results in symmetrical proximal muscle weakness Primary pulmonary hypertension- elevated pulmonary artery pressure with no known cause, if left untreated will lead to right-sided heart failure Raynaud phenomenon- recurrent vasospasms of the fingers or toes which usually occurs as a result of stress or exposure to the cold Bacterial sepsis- the presence of infection along with the systemic inflammatory response syndrome Pleuritis- inflammation in the lining of the lungs Rheumatoid arthritis- chronic systemic inflammatory disease with an unknown cause Case Reports/ Case Studies[edit | edit source] Karmacharya P, Mainali N, Aryal M, Lloyd B. Recurrent case of ibuprofen-induced aseptic meningitis in mixed connective tissue disease. Case Reports [Internet]. 2013 [2016 Apr 8];2013(apr30 1):bcr2013009571-bcr2013009571. Souto Filho J, de Barros P, da Silva A, Barbosa F, Ribas G. Thrombotic Thrombocytopenic Purpura Associated with Mixed Connective Tissue Disease: A Case Report. Case Reports in Medicine [Internet]. 2011 [Cited 2016 Apr 8];2011:1-5. Fantò M, Salemi S, Socciarelli F, Bartolazzi A, Natale G, Casorelli I et al. A Case of Subacute Cutaneous Lupus Erythematosus in a Patient with Mixed Connective Tissue Disease: Successful Treatment with Plasmapheresis and Rituximab. Case Reports in Rheumatology [Internet]. 2013 [Cited 2016 Apr 8];2013:1-4. Resources[edit | edit source] Lupus Foundation of America, Inc. References[edit | edit source] ↑ 1.0 1.1 Fagundes MN, Caleiro MT, Navarro-rodriguez T, et al. Esophageal involvement and interstitial lung disease in mixed connective tissue disease. Respir Med [Internet]. 2009 [cited 2016 Mar 12];103(6):854-60. Available from: ↑ 2.00 2.01 2.02 2.03 2.04 2.05 2.06 2.07 2.08 2.09 2.10 2.11 2.12 2.13 2.14 Mixed Connective-Tissue Disease: Background, Pathophysiology, Etiology [Internet]. Emedicine.medscape.com. 2016 [cited 8 April 2016]. Available from: ↑ 3.0 3.1 3.2 3.3 3.4 Mixed Connective Tissue Disease (MCTD) - NORD (National Organization for Rare Disorders) [Internet]. NORD (National Organization for Rare Disorders). 2016 [cited 8 April 2016]. Available from: ↑ 4.00 4.01 4.02 4.03 4.04 4.05 4.06 4.07 4.08 4.09 4.10 4.11 4.12 4.13 4.14 Mixed connective tissue disease - Mayo Clinic [Internet]. Mayoclinic.org. 2016 [cited 2016 Apr 8]. Available from: ↑ 5.0 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 Mixed Connective Tissue Disease [Internet]. My.clevelandclinic.org. 2016 [cited 2016 Apr 8]. Available from: ↑ 6.0 6.1 6.2 6.3 Mixed Connective Tissue Disease (MCTD) [Internet]. Merck Manuals Consumer Version. 2016 [cited 2016 Apr 8]. Available from: ↑ 7.0 7.1 Tani C, Carli L, Vagnani S, et al. The diagnosis and classification of mixed connective tissue disease. J Autoimmun [Internet]. 2014 [Cited 2016 Mar 12];48-49:46-9. Available from: ↑ 8.0 8.1 8.2 Cappelli S, Bellando randone S, Martinović D, et al. "To be or not to be," ten years after: evidence for mixed connective tissue disease as a distinct entity. Semin Arthritis Rheum [Internet]. 2012 [cited 2016 Mar 12];41(4):589-98. Available from ↑ 9.0 9.1 9.2 Ungprasert P, Wannarong T, Panichsillapakit T, et al. Cardiac involvement in mixed connective tissue disease: a systematic review. Int J Cardiol [Internet]. 2014 [Cited 2016 Mar 12];171(3):326-30. Available from: ↑ 10.0 10.1 Van der net J, Wissink B, Van royen A, Helders PJ, Takken T. Aerobic capacity and muscle strength in juvenile-onset mixed connective tissue disease (MCTD). Scand J Rheumatol [Internet]. 2010 [Cited 2016 Mar 12];39(5):387-92. Available from: ↑ Marigliano B, Soriano A, Margiotta D, Vadacca M, Afeltra A. Lung involvement in connective tissue diseases: a comprehensive review and a focus on rheumatoid arthritis. Autoimmun Rev [Internet]. 2013[Cited 2016 Mar 12];12(11):1076-84. Available from: Retrieved from " Category: Rheumatology Get Top Tips Tuesday and The Latest Physiopedia updates Email Address I give my consent to Physiopedia to be in touch with me via email using the information I have provided in this form for the purpose of news, updates and marketing. HP Yes please It's free, and you can unsubscribe any time. Privacy policy. Our Partners The content on or accessible through Physiopedia is for informational purposes only. Physiopedia is not a substitute for professional advice or expert medical services from a qualified healthcare provider. Read more pPhysiopedia +Physiopedia Plus Physiopedia About News Donations Shop Contact Content Articles Categories Resources Projects Contribute Courses PAI Legal QA Framework Disclaimer Terms Privacy Cookies Report content AI Licensing Physiopedia available in: French German Italian Spanish Ukrainian © Physiopedia 2025 | Physiopedia is a registered charity in the UK, no. 1173185 Back to top]( [Scleroderma Case Study - Physiopedia Search Search Search Toggle navigation pPhysiopedia pPhysiopedia About News Contribute Courses PAI Resources Shop Contact Login pPhysiopedia About News Contribute Courses PAI Resources Shop Contact p + Contents Editors Categories Cite Contents loading... Editors loading... Categories loading... When refering to evidence in academic writing, you should always try to reference the primary (original) source. That is usually the journal article where the information was first stated. In most cases Physiopedia articles are a secondary source and so should not be used as references. Physiopedia articles are best used to find the original sources of information (see the references list at the bottom of the article). If you believe that this Physiopedia article is the primary source for the information you are refering to, you can use the button below to access a related citation statement. Cite article Scleroderma Case Study Jump to:navigation, search Contents 1 Author/s 2 Patient Characteristics 3 Examination 4 Clinical Impression 5 Summarization of Examination Findings 6 Intervention 7 Outcomes 8 Discussion 9 Related Pages 10 References Author/s[edit | edit source] Catelin Infante, Jessica Frederick, Kristin Casagrand, Jarrod Smith from the Bellarmine University Physical Therapy Program's Pathophysiology of Complex Patient Problems Project. Patient Characteristics[edit | edit source] Demographic Information: (occupation/vocation, gender, age, etc.) Native American Female Born and raised in Oklahoma Works in a factory Thirty-five years old Family history of rheumatoid arthritis on mother's side Medical diagnosis if applicable Scleroderma Co-morbidities Hypertension, Type II Diabetes Previous care or treatment Previous physical therapy for low back pain Examination[edit | edit source] Subjective Ms. Johansson comes into the clinic complaining of joint pain and muscle weakness, especially in her hands, hips and shoulders. She explains that she has been having trouble making it through a full day of work because her body feels limited and she just cannot reach as well for boxes at her job because "her arms won't go up that far". She also has noticed that she gets out of breath and sometimes has chest pain when she lifts a lot too quickly. Ms. Johansson indicates that her fingers often feel very swollen and stiff and sometimes feel "stuck" in a bent position. She reports that for several months she has been getting a really cold, numb feeling in her fingers throughout the day, especially when she is sitting in the air conditioning at her office. Overall, she has just not felt well for the past few months either. When she eats, she has trouble fully opening her mouth to put food in. Then, when she lies down at night after dinner, she gets a feeling of heartburn and sometimes feels she has to vomit. However, she says she has been eating poorly lately, so it's probably her fault she hasn't felt well. Additionally, Ms. Johansson reports losing 20 pounds in the past month without any lifestyle changes. Her goals for therapy are to decrease her joint pain and to get her muscles stronger so she doesn't struggle at work any longer. Her focus is on getting her hands and arms "back to normal" because they are what she uses most in her job. Self-Report Outcome Measures Health Assessment Questionnaire Disability Index (HAQ-DI): 1.1 Disabilities of Arm, Shoulder, and Hand Questionnaire (DASH): 48% Verbal Pain Intensity Scale: "Severe pain" Physical Performance Measures Grip strength (R): 20.7 kg Grip strength (L): 19.18 kg Pinch strength (R): 4.27 kg Pinch strength (L): 3.98 kg Oral aperture: 28 mm Measurement of Oral Aperture Objective MMT: WNL except for the following 4/5 finger and thumb abduction, adduction, flexion, extension 4/5 wrist flexion and extension 3+/5 shoulder flexion and abduction 3+/5 hip flexion and abduction ROM: WNL except for the following Decreased flexion and extension in all finger MCP and IP joints Decreased wrist flexion and extension Decreased hip extension and abduction Reflexes: WNL Sensation: Decreased facial sensation ICF Findings Impairments,, Patient has stiffening around fingers and what appears to be finger flexion contractures developing Patient has noticeable hard deposits palpated around UE joints Patient has slurred speech due to difficulty fully opening mouth. The skin around the face is thickened and has begun to cause puckering Pt has dyspnea and chest pain upon exertion Activity Limitations,, Patient is unable to complete activities in supine secondary to reflux symptoms Patient is unable to complete overhead activities due to decreased shoulder ROM Patient is unable to complete more than 30 minutes of repetitive movements due to fatigue, muscle weakness and joint pain Environmental Factors Patient is employed as a packer at a plastics company Patient is a smoker of 20 years Clinical Impression[edit | edit source] After Ms. Smith's assessment, we have determined that she has decreased range of motion in her fingers and wrists, decreased cardiovascular endurance, decreased muscle mobility, as well as increased muscle weakness and fatigue. All of these impairments have impacted her functional ability to do activities at home and in the work place. Physical therapy could benefit Ms. Smith by improving her impairments that were found and by providing patient education on management of her condition. The interventions during PT would include active joint range of motion, strengthening exercises and aerobic exercises to improve endurance, as well as education on a home exercise program and management of the symptoms of her condition. With the help of physical therapy, Ms. Smith would likely be able to reach her goals for improving her strength and endurance and decreasing her joint pain so that she can particpate in more activities at home, work and in the community. Summarization of Examination Findings[edit | edit source] Targeted intervention for Physical therapy for this patient would be improving available range of motion, joint protection and preservation along with patient education. Based on Evaluation and examination information patient should be educated on proper protection of affected joints with education on temperature monitoring to reduce symptoms. Light range of motion should be built in to the intervention in a safe manor to preserve range that will progress to light strengthening of affected joint as long as joint integrity is not in jeopardy. Intervention[edit | edit source] Phases of Interventions Edematous Stage: Patient Education (About the disease; treatment of Raynaud's disease - dress in layers, monitor temperature changes, avoid air conditioning if possible, avoid things that cause vasoconstriction like smoking, cold temperatures and emotional stress), Orthotic and adaptive equipment needs, Light range of motion exercises,, Joint protection strategies (light weight splints) Fatigue management (Four P's of energy conservation - prioritize, plan activities, pace activities and promote correct posture through proper body mechanics) Sclerotic Stage: Paraffin/heat modalities for hands and fingers (vascular system may be affected with this condition, so the patient might have a higher risk for burns),,, Passive range of motion exercises progressing to active-assisted range of motion, then to active range of motion, can add a sustained stretch for five seconds at end range (finger flexion/extension and abduction/adduction, wrist flexion/extension - can use hand clasp with wrist extension stretch),,,, Soft tissue mobilizations, Joint manipulations (Traction),,, Kabat's method to improve mouth opening (activation of orbicularis oris, zygomaticus, levator labii, nasalis, frontalis and corrugator muscles by exaggerated facial movements) - progress from active-assisted ROM to active ROM,, Soft tissue massage, Aquatic therapy (Upper and lower extremity range of motion and strengthening exercises, walking, squatting, marching, etc), Cycle ergometer Resistive training, Dosage and Parameters SMART framework - combines physical therapy sessions with self-management of care through a home exercise program. The PT session focuses on designing an individualized program for each patient in order to treat their specific impairments and goals. The sessions are planned less frequently during the week, but spaced out over a longer period of time. This is to help shift the focus of the care to self-managed that way the patient can manage their condition for the long-term. The HEP uses exercises, self-stretches and paraffin to help the patient manage their condition at home. The patient is given 4-5 exercises that are performed 1-2 times per day, for 5 days per week. Each physical therapy session is specific for each individual because of the varying presentation of scleroderma. The physician should be consulted frequently about the proper treatment for each patient.,, Rationale for Progression As the same for the dosage and parameters, each individual has a different presentation of scleroderma and the associated symptoms and impairments., Each individual should progress through their rehabilitation according to their reaction to the exercises. Each should be monitored closely, especially with the aerobic exercise, and progressed in a safe manner according to their bodies' response to exercise. The physical therapist must also consider the phase of the disease that the patient is in and their progression through the disease phases. Co-interventions if applicable Occupational Therapy Joint preservation principles Devices for independent living Medications Disease modifying drugs (Penicillamine) - treatment of disease processes ACE inhibitors - treatment of acute hypertension/pulmonary hypertension and renal complications Nonsteroidal anti-inflammatory drugs (NSAIDs) - treatment of joint inflammation Antibiotics/Oral Tetracyclines (Minocycline, doxycycline) - treatment for disease by reduction of pain, severity of condition and quality of life Analgesics - treatment of pain Prazosin, Ilioprost and Cisaprost for Raynaud's Level A Methotrexate and Calcium channel blockers - treatment of Raynaud's Level B, Immunosuppressants - treatment during the acute phase of the disease Outcomes[edit | edit source] Miss Smith participated in a successful physical therapy plan of care related to her medical diagnosis of systemic scleroderma. Miss Smith was able to articulate her understanding of the disease progression of scleroderma, including the different stages of the condition. She was also able to explain to her physical therapist what activities should be performed in each stage, which is an excellent way to ensure patient understanding. The completion of a variety of range of motion activities adapted for Miss Smith's current needs encouraged increased ROM of her hands and wrists along with resistance exercises to increase the strength of her hand intrinsics served to decrease Miss Smith's perceived level of difficulty in completing her ADL's. Aquatic therapy was utilized in a variety of methods. Its greatest impact was seen in Miss Smith's aerobic capacity, which increased and allowed her to become more active at home and in the community. Due to the progressive nature of scleroderma, it was very important for Miss Smith to understand how physical therapy will be important to maintain her abilities to perform ADLs and remain active in the community. She verbally acknowledged that she will need to continue to monitor her condition, and to continue her physical therapy exercises and activities after discharge. Miss Smith was educated on how changes in her symptoms (including decreasing range of motion, strength, and increasing levels of fatigue) may indicate the need for a re-evaluation by her physical therapist to continue to promote her current levels of independence. Discussion[edit | edit source] Despite the low levels of prevalence of scleroderma, additional research is needed related to the physical therapy portion of treatment.As scleroderma is a systemic disorder, it is imperitive that these patients receive interdiscipliary care with consistent communication between healthcare providers. Current literature suggests that specific range of motion activities, resistance exercises, and cardiovascular activities may increase a patient's quality of life and decrease functional limitations in their daily life. However, many of these studies consist of a limited pool of subjects, additional research will help to further develop an evidence based model for physical therapy treatment of scleroderma. Related Pages[edit | edit source] References[edit | edit source] References will automatically be added here, see adding references tutorial. ↑ 1.0 1.1 1.2 1.3 1.4 Goodman CC, Snyder TEK. Differential Diagnosis for physical therapists. 5th ed. St. Louis, MO: Elsevier Saunders, 2013. ↑ 2.00 2.01 2.02 2.03 2.04 2.05 2.06 2.07 2.08 2.09 2.10 Lonneman, E. Connective Tissue Disorders Part II. [unpublished lecture notes]. Bellarmine University DPT Program; notes provided at lecture given 2015 February 23. ↑ Johns Hopkins Scleroderma Center. Understanding scleroderma. (accessed 21 March 2015). ↑ Williams SL. Pain perspective in scleroderma. The Rheumatologist 2011. (accessed 21 March 2015). ↑ Varju C, Balint Z, Solyom AI, Farkas H, Karpati E, Berta B, et al. Cross-cultural adaptation of the disabilities of the arm, shoulder, and hand (DASH) questionnaire into Hungarian and investigation of its validity in patients with systemic sclerosis]. 2008; 776-784. (accessed 21 March 2015). ↑ 6.0 6.1 6.2 6.3 Sawy NE, Suliman I, Nouh M, Naguib A. Hand function in systemic sclerosis: a clinical and ultrasonographic study. The Egyptian Rheumatologist 2012; 34(4): 167-178. (accessed 21 March 2015). ↑ Yuen HK, Marlow NM, Reed SG, Summerlin LM, Leite RS, Mahoney S, et al. Effect of orofacial exercises on oral aperture in adults with systemic sclerosis. Disabil Rehabil 2012; 34(1): 84-89. (accessed 21 March 2015). ↑ Pocket Dentistry. Manual functional analysis. (accessed 21 March 2015). ↑ 9.0 9.1 9.2 National Institute of Arthritis and Musculoskeletal and Skin Diseases. Scleroderma. (accessed 21 March 2015). ↑ 10.00 10.01 10.02 10.03 10.04 10.05 10.06 10.07 10.08 10.09 10.10 Goodman CC, Fuller KS. Pathology: implications for the physical therapist. 3rd ed. St. Louis: Saunders Elsevier, 2009. ↑ 11.00 11.01 11.02 11.03 11.04 11.05 11.06 11.07 11.08 11.09 11.10 Poole JL, MacIntyre NJ, deBoer HN. Evidence-based management of hand and mouth disability in a woman living with diffuse systemic sclerosis (scleroderma). Physiotherapy Canada 2013; 65(4): 317-320. ↑ 12.00 12.01 12.02 12.03 12.04 12.05 12.06 12.07 12.08 12.09 12.10 12.11 12.12 12.13 Wolff A, Weinstock-Zlotnick G, Gordon J. SSc management – In person appointments and remote therapy (SMART): A framework for management of chronic hand conditions. Journal of Hand Therapy 2014; 27: 143-151. ↑ 13.00 13.01 13.02 13.03 13.04 13.05 13.06 13.07 13.08 13.09 13.10 13.11 Attia MW. Evidence-based physical therapy for patients with scleroderma [dissertation]. Ann Arbor: UMI Dissertation Publishing. 2014. ↑ 14.00 14.01 14.02 14.03 14.04 14.05 14.06 14.07 14.08 14.09 14.10 14.11 Bongi SM, Rosso AD, Galluccio F, Tai G, Sigismondi F, Passalacqua M, Landi G, Baccini M, Conforti ML, Miniati I, Matucci-Cerinic M. Efficacy of a tailored rehabilitation program for systemic sclerosis. Experimental Rheumatology 2009; 27(Suppl. 54): S44-S50. Retrieved from " Category: Case Studies Get Top Tips Tuesday and The Latest Physiopedia updates Email Address I give my consent to Physiopedia to be in touch with me via email using the information I have provided in this form for the purpose of news, updates and marketing. HP Yes please It's free, and you can unsubscribe any time. Privacy policy. Our Partners The content on or accessible through Physiopedia is for informational purposes only. Physiopedia is not a substitute for professional advice or expert medical services from a qualified healthcare provider. Read more pPhysiopedia +Physiopedia Plus Physiopedia About News Donations Shop Contact Content Articles Categories Resources Projects Contribute Courses PAI Legal QA Framework Disclaimer Terms Privacy Cookies Report content AI Licensing Physiopedia available in: French German Italian Spanish Ukrainian © Physiopedia 2025 | Physiopedia is a registered charity in the UK, no. 1173185 Back to top]( [Systemic Lupus Erythematosus - Physiopedia Search Search Search Toggle navigation pPhysiopedia pPhysiopedia About News Contribute Courses PAI Resources Shop Contact Login pPhysiopedia About News Contribute Courses PAI Resources Shop Contact p + Contents Editors Categories Cite Contents loading... Editors loading... Categories loading... When refering to evidence in academic writing, you should always try to reference the primary (original) source. That is usually the journal article where the information was first stated. In most cases Physiopedia articles are a secondary source and so should not be used as references. Physiopedia articles are best used to find the original sources of information (see the references list at the bottom of the article). If you believe that this Physiopedia article is the primary source for the information you are refering to, you can use the button below to access a related citation statement. Cite article Systemic Lupus Erythematosus Jump to:navigation, search Original Editors - Alli Christian from Bellarmine University's Pathophysiology of Complex Patient Problems project. Top Contributors - Alli Christian , Elaine Lonnemann , Admin , Kim Jackson , Lucinda hampton , Rachael Lowe , Lotte Vonck , Khloud Shreif , 127.0.0.1 , Laura Ritchie , Wendy Walker and WikiSysop - Elaine Lonnemann Contents 1 Definition/Description 2 Epidemiology 3 Etiology 4 Pathological Process 5 Clinical Presentation 5.1 Systemic Involvement 5.1.1 Musculoskeletal System 5.1.2 Cardiopulmonary/Cardiovascular System 5.1.3 Central Nervous System 5.1.4 Renal System 5.1.5 Cutaneous System 5.1.6 Blood Disorders 5.1.7 Gastrointestinal System 5.2 Associated Co-morbidities 6 Diagnosis 7 Medical Management 8 Physical Therapy Management 9 Occupational Therapy Management 10 Prognosis 11 Resources 12 References Definition/Description[edit | edit source] Systemic lupus erythematosus (SLE) is a chronic autoimmune disease characterized by an inflammation of connective tissue disease with variable manifestations. SLE may affect many organ systems with immune complexes and a large array of autoantibodies, particularly antinuclear antibodies (ANAs). Although abnormalities in almost every aspect of the immune system have been found, the key defect is thought to result from a loss of self-tolerance to autoantigens. It is a disease characterized by relapses, flares, and remissions. Common manifestations, in addition to the malar rash, include cutaneous photosensitivity, nephropathy, serositis, and polyarthritis. The overall outcome of the disease is highly variable with extremes ranging from permanent remission to death. Epidemiology[edit | edit source] There is a strong female predilection in adults, with women affected 9-13 times more than males. In children, this ratio is reversed, and males are affected two to three times more often. Can affect any age group - the peak age at onset is around the 2nd to 4th decades, with 65% of patients presenting between the ages of 16 and 65 years (i.e. during childbearing years). Disease is more common in childbearing age in women however it has been well reported in the pediatric and elderly population. SLE is more severe in children while in the elderly, it tends to be more insidious onset and has more pulmonary involvement and serositis and less Raynaud's, malar rash, nephritis, and neuropsychiatric complications Studies have indicated that although rare, lupus in men tends to be more severe. Prevalence varies according to ethnicity with ratios as high as 1:500 to 1:1000 in Afro-Caribbeans and indigenous Australians, down to 1:2000 in Caucasians. Etiology[edit | edit source] The cause of lupus erythematosus is not known. A familial association has been noted that suggests a genetic predisposition, but a genetic link has not been identified. Approximately 8% of patients with SLE have at least one first-degree family member (parent, sibling, child) with the disease. Environmental factors - Ultraviolet light (increased keratinocyte apoptosis), infection (via molecular mimicry and bacterial CpG motifs), smoking (odds ratio (OR): 1.56 in current smokers, 1.23 in ex-smokers), environmental pollutants (silica) and intestinal dysbiosis (ie digestive disturbances, frequent gas or bloating, feel bloated on most days of the week, abdominal cramping, diarrhea, and constipation) are all known risk factors for SLE. Hormonal abnormality and ultraviolet radiation are considered possible risk factors for the development of SLE. Some drugs have been implicated as initiating the onset of lupus-like symptoms and aggravating existing disease; they include hydralazine hydrochloride, procainamide hydrochloride, penicillin, isonicotinic acid hydrazide, chlorpromazine, phenytoin, and quinidine. Possible childhood risk factors include low birth weight, preterm birth, and exposure to farming pesticides. Pathological Process[edit | edit source] SLE can affect multiple components of the immune system, including the Complement system (i.e. a part of the immune system that enhances the ability of antibodies and phagocytic cells to clear microbes and damaged cells from an organism, promote inflammation, and attack the pathogen's cell membrane), T-suppressor cells: in SLE T cells are key players in causing inflammation and autoimmunity. They release pro-inflammatory cytokines, stimulate B cells to produce harmful autoantibodies and maintain the disease through autoreactive memory T cells. However, SLE patients exhibit abnormal T-cell ratios and functions. T follicular helper (Tfh) cells, essential for immune responses, expand excessively in SLE due to interactions with antigen-presenting cells and TLR7 activation. This leads to heightened antibody production and immune tolerance breakdown. Conversely, regulatory T (Treg) cells, responsible for immune control, are impaired in SLE, partly due to reduced IL-2 levels caused by low activator protein 1 (AP-1) expression. IL-2 also helps restrain the pro-inflammatory cytokine IL-17, elevated in SLE, contributing to tissue damage. Overall, T cell dysregulation is a central feature in SLE pathogenesis. Cytokine production. Emerging evidence has demonstrated a key player in the generation of autoantigens in SLE is the increase in generation (i.e. increased apoptosis) and/or decrease in clearance of apoptotic cell materials (i.e., decreased phagocytosis). Results in the generation of autoantibodies, which may circulate for many years prior to the development of overt clinical SLE. The disease tends to have a relapsing and remitting course. Clinical Presentation[edit | edit source] SLE has a myriad of clinical features: CNS manifestations of systemic lupus erythematosus (CNS lupus): neuropsychiatric events can occur in ~45% (range 14-75%) of cases Gastrointestinal manifestations of systemic lupus erythematosus: there may be GI involvement in ~20% of cases (Ascites, peritonitis, oral ulcers, esophageal dysmotility, and protein-losing enteropathy) Musculoskeletal manifestations of systemic lupus erythematosus (Jaccoud's arthropathy, arthralgia, arthritis, synovitis, tenosynovitis, and myositis) Renal manifestations of systemic lupus erythematosus ( Proteinuria, hematuria, and glomerulonephritis). Cardiovascular manifestation (Libman-sacks-endocarditis, pericarditis, myocarditis). Thoracic manifestations of systemic lupus erythematosus (Pleuritis, pulmonary arterial HTN, interstitial lung disease, pleural effusion). SLE can affect many organs of the body, but it rarely affects them all. The following list includes common signs and symptoms of SLE in order of the most to least prevalent. SLE rash All of the below symptoms might not be present at the initial diagnosis of SLE, but as the disease progresses more of a person’s organ systems become involved. Discoid rash ears, and upper back The most common symptoms associated with SLE are: Constitutional symptoms (fever, malaise, fatigue, weight loss): most commonly fatigue and a low-grade fever. Achy joints (arthralgia) Arthritis (inflamed joints) Skin rashes; top facial rash, and bottom discoid rash Pulmonary involvement (symptoms include: chest pain, difficulty breathing, and cough) Anemia Kidney involvement (lupus nephritis) Sensitivity to the sun or light (photosensitivity) Hair loss Raynaud’s phenomenon CNS involvement (seizures, headaches, peripheral neuropathy, cranial neuropathy, cerebrovascular accidents, neurocognitive disorder, psychosis) Mouth, nose, or vaginal ulcers The most common signs and symptoms of SLE in children and adolescents are fever, fatigue, weight loss, arthritis, rash, and renal disease. Systemic Involvement[edit | edit source] There are many visceral systems that can be affected by SLE, but the extent of the body's involvement differs from person to person. Some people diagnosed with SLE have only a few visceral systems involved, while others have numerous systems that have been affected by the disease. Musculoskeletal System[edit | edit source] Arthritis- typically affects hand, wrists, and knees Arthralgia Tenosynovitis Tendon ruptures Swan-neck deformity Ulnar drift Cardiopulmonary/Cardiovascular System[edit | edit source] Pleuritis Pericarditis Dyspnea Hypertension Myocarditis Endocarditis Tachycarditis Pneumonitis Vasculitis Central Nervous System[edit | edit source] Raynaud phenomenon Emotional instability Psychosis Seizures Cerebrovascular accidents Cranial neuropathy Peripheral neuropathy Organic brain syndrome Renal System[edit | edit source] Glomerulonephritis -an inflammatory disease of the kidneys Hematuria Proteinuria Kidney failure Cutaneous System[edit | edit source] Calcinosis Cutaneous vasculitis Hair loss Raynaud's phenomenon Mucosal ulcers Petechiae Blood Disorders[edit | edit source] Anemia Thrombocytopenia Leukopenia Neutropenia Thrombosis Gastrointestinal System[edit | edit source] Ulcers--Throat & Mouth Ulcerative colitis/Crohn's disease Peritonitis Ascites Pancreatitis Peptic ulcers Autoimmune Hepatitis Associated Co-morbidities[edit | edit source] Include: Fibromyalgia. Atherosclerosis Lupus Nephritis- leads to End Stage Renal Disease (ESRD) Anemia Some types of cancers (especially non-Hodgkin's lymphoma and lung cancer) Infections Hypertension Dyslipidemia Diabetes Mellitus Osteoporosis Avascular Necrosis Diagnosis[edit | edit source] The diagnosis of SLE may be made if four of eleven ACR (American College of Rheumatology) criteria are present, either serially or simultaneously 2. These criteria were initially published in 1982 but were revised in 1997. ACR criteria: Criteria Explanation Serositis Pericarditis, pleurisy on electrocardiogram or imaging scan Oral ulcers Sores, usually painless, on the lips and in the mouth Arthritis Tenderness or swelling of two or more peripheral joints Photosensitivity Unusual skin reaction (skin rash) to sun exposure Blood disorder Leukopenia, lymphopenia, thrombocytopenia, hemolytic anemia Renal involvement Proteinuria, cellular casts Antinuclear antibodies Elevated titers Immune phenomenon Presence of antibodies or lupus erythematosus cells Neurological disorder Seizures or psychosis in absence of other causes Malar rash Fixed erythema over cheeks and nose Discoid rash Raised, red lesions with scaling and follicular plugging Medical Management[edit | edit source] The goal of treatment in SLE is to prevent organ damage and achieve remission. The choice of treatment is dictated by the organ system/systems involved and the severity of involvement and ranges from minimal treatment (NSAIDs, antimalarials) to intensive treatment (cytotoxic drugs, corticosteroids). Patient education, physical and lifestyle measures, and emotional support play a central role in the management of SLE. Patients with SLE should be well educated on the disease pathology, potential organ involvement including brochures, and the importance of medication and monitoring compliance. Stress reduction techniques, good sleep hygiene, exercises, and use of emotional support shall be encouraged. Smoking can worsen SLE symptoms - educated about the importance of smoking cessation. Dietary recommendations - avoiding alfalfa sprouts and echinacea and including a diet rich in vitamin-D. Photoprotection is vital, and all patients with SLE shall avoid direct sun exposure by timing their activities appropriately, light-weight loose-fitting dark clothing covering the maximum portion of the body, and using broad-spectrum (UV-A and UV-B) sunscreens with sun protection factor (SPF) of 30 or more (see image R) Physical Therapy Management[edit | edit source] Exercise is beneficial for patients with SLE because it decreases their muscle weakness; while simultaneously; increasing their muscle endurance. Physical therapists can play an important role for patients with SLE during and between exacerbations. The patient's need for physical therapy will vary greatly depending on the systems involved. Education: It is essential for patients with skin lesions to have appropriate education on the best way to care for their skin and to ensure they do not experience additional skin breakdown. Aerobic Exercise: One of the most common impairments that patients with SLE experience is generalized fatigue that can limit their activities throughout the day. Graded aerobic exercise programs are more successful than relaxation techniques in decreasing the fatigue levels of patients with SLE. Aerobic activity causes many with SLE to feel much better. The aerobic exercise program may consist of 30-50 minutes of aerobic activity (walking/swimming/cycling) with a heart rate corresponding to 60% of the patient's peak oxygen consumption. Both aerobic exercise and range of motion/muscle strengthening exercises can increase the energy level, cardiovascular fitness, functional status, and muscle strength in patients with SLE (aerobic exercise for 20-30 minutes at 70-80% of their maximum heart rate,3 times a week for 50 minutes sessions). Energy Conservation: Physical therapists; can educate patients on appropriate; energy conservation techniques and the best ways to protect joints that; are susceptible to damage. Additionally, physical therapists and patients with SLE should be aware of signs and symptoms that suggest a progression of SLE including those associated with avascular necrosis, kidney involvement, and neurological involvement. Occupational Therapy Management[edit | edit source] During evaluation occupational therapists (OTs) use activity analysis to break down these activities into parts to detect where patient's challenges exist, such as issues with mobility, strength, sensation, pain, or endurance. This assessment helps OTs recommend solutions like splints, home modifications, or post-surgery support and introduce effective adaptations and techniques to enhance independence in daily life. Self-management strategies: involves teaching the patient how to identify and avoid potential triggers for flares, in addition, to recognizing early warning signs, and embracing a healthy lifestyle. This includes adhering to routine medical check-ups, prioritizing rest, minimizing stress, sun protection, maintaining a nutritious diet, regular exercise, and consistent medication adherence, all of which collectively contribute to reducing the likelihood of lupus flares and promoting overall well-being. Education: OTs help to safeguard their joints, reduce tiredness, and minimize pain, individuals with lupus can simplify their daily routines, take things slowly when performing tasks, and think ahead to plan their activities which in turn enhances their ability to perform daily tasks like self-care, work, and leisure activities. Adaptive strategies: OTs can teach them self-management skills to handle daily tasks, and adaptive strategies to help them engage in desired activities. Environmental modifications: OTs help to identify the physical barriers that are limiting the client from engaging in their desired daily activities also suggest the use of adaptive equipment and modifications in the environment, such as easy-to-grip handles and raised toilet seats, to conserve energy and protect joints. These combined strategies aim to meet patients' functional needs and support their independence in various activities despite limitations in joint motion, strength, and endurance. Prognosis[edit | edit source] Recent advances in diagnosis and treatment strategies have resulted in a survival improvement from a 5-year survival at 40-50% in the 1950s up to a 10-year survival now estimated at 80-90%. Nevertheless, patients with SLE have a standardised mortality ratio of 3. Early deaths in the disease course are usually due to active disease and immunosuppression while late deaths tend to arise from coronary artery disease and SLE or treatment complications (e.g. end-stage kidney disease from lupus nephritis). Resources[edit | edit source] Medline Plus: Systemic Lupus WebMD: Systemic Lupus Medicine Net Occupational Therapy and Health Promotion with Lupus References[edit | edit source] ↑ 1.0 1.1 1.2 1.3 1.4 1.5 1.6 Ameer MA, Chaudhry H, Mushtaq J, Khan OS, Babar M, Hashim T, Zeb S, Tariq MA, Patlolla SR, Ali J, Hashim SN. An overview of systemic lupus erythematosus (SLE) pathogenesis, classification, and management. Cureus. 2022 Oct 15;14(10). ↑ Tsai HL, Chang JW, Lu JH, Liu CS. Epidemiology and risk factors associated with avascular necrosis in patients with autoimmune diseases: a nationwide study. Korean J Intern Med. 2022 Jul;37(4):864-876. ↑ 3.0 3.1 Nursing central SLE Available from: (last accessed 5.6.2020) ↑ Giang S, La Cava A. Regulatory T cells in SLE: biology and use in treatment. Current rheumatology reports. 2016 Nov;18:1-9. ↑ Clinical relevance of T follicular helper cells in systemic lupus erythematosus. Nakayamada S, Tanaka Y. Expert Rev Clin Immunol. 2021;17:1143–1150 ↑ 6.0 6.1 6.2 6.3 6.4 Goodman CC, Fuller KS. Pathology: Implications for the Physical Therapist. 3rd edition. St. Louis, Missouri: Saunders Elsevier, 2009. ↑ Cojocaru M, Cojocaru IM, Silosi I, Vrabie CD. Manifestations of systemic lupus erythematosus. Maedica. 2011 Oct;6(4):330. ↑ Tucker LB. Making the diagnosis of systemic lupus erythematosus in children and adolescents. Lupus. 2007 Aug;16(8):546-9. ↑ Lupus Foundation of America. How lupus affects the body page. Website updated: 2010. Website accessed: February 17, 2010. ↑ Goodman CC, Snyder TE. Differential diagnosis for physical therapists screening for referral, Saunders Elsevier, St. Louis, MO. 2007:274-364. ↑ Becker-Merok A, Nossent JC. Prevalence, predictors and outcome of vascular damage in systemic lupus erythematosus. Lupus. 2009 May;18(6):508-15. ↑ 12.0 12.1 12.2 Bertsias G, Gordon C, Boumpas DT. Clinical trials in systemic lupus erythematosus (SLE): lessons from the past as we proceed to the future–the EULAR recommendations for the management of SLE and the use of end-points in clinical trials. Lupus. 2008 May;17(5):437-42. ↑ Wingard R. Increased risk of anemia in dialysis patients with comorbid diseases. Nephrology Nursing Journal. 2004 Mar 1;31(2):211. ↑ Medical Foundation for Medical Education and Research. Mayo Clinic: Lupus Page. www.mayoclinic.com. Updated October 20, 2009. Accessed February 17, 2010. ↑ Bernatsky S, Boivin JF, Joseph L, Rajan R, Zoma A, Manzi S, Ginzler E, Urowitz M, Gladman D, Fortin PR, Petri M. An international cohort study of cancer in systemic lupus erythematosus. Arthritis & Rheumatism. 2005 May;52(5):1481-90. ↑ Tench CM, McCarthy J, McCurdie I, White PD, D'Cruz DP. Fatigue in systemic lupus erythematosus: a randomized controlled trial of exercise. Rheumatology. 2003 Sep 1;42(9):1050-4. ↑ Ramsey‐Goldman R, Schilling EM, Dunlop D, Langman C, Greenland P, Thomas RJ, Chang RW. A pilot study on the effects of exercise in patients with systemic lupus erythematosus. Arthritis Care & Research. 2000 Oct;13(5):262-9. ↑ Baker NA, Carandang K, Dodge C, Poole JL. Occupational Therapy Is a Vital Member of the Interprofessional Team-Based Approach for the Management of Rheumatoid Arthritis: Applying the 2022 American College of Rheumatology Guideline for Exercise, Rehabilitation, Diet, and Additional Integrative Interventions for Rheumatoid Arthritis. Arthritis care & research. 2023 May 25. ↑ Lupus LA. The Benefits of Occupational Therapy in Lupus. Available from: [last accessed 28/9/2023] Retrieved from " Categories: Rheumatology Autoimmune Disorders Bellarmine Student Project Conditions Non Communicable Diseases Get Top Tips Tuesday and The Latest Physiopedia updates Email Address I give my consent to Physiopedia to be in touch with me via email using the information I have provided in this form for the purpose of news, updates and marketing. HP Yes please It's free, and you can unsubscribe any time. Privacy policy. Our Partners The content on or accessible through Physiopedia is for informational purposes only. Physiopedia is not a substitute for professional advice or expert medical services from a qualified healthcare provider. 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https://www.wolframalpha.com/input/?i=integral+of+cos+squared+x
integral of cos squared x - Wolfram|Alpha UPGRADE TO PRO APPS TOUR Sign in Use Wolfram|Alpha Pro for reality-checked results Go Pro Now integral of cos squared x Natural Language Math Input Extended KeyboardExamplesUpload Random Indefinite integral Step-by-step solution Plot Series expansion of the integral at x=0 Big‐O notation» Series expansion of the integral at x=∞ Big‐O notation» Download Page POWERED BY THE WOLFRAM LANGUAGE
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https://www.quora.com/For-which-positive-numbers-is-it-true-that-a-x-1-x-for-all-x
For which positive numbers is it true that a^x >= 1+x for all x? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Exponents Question The Real Numbers Algebraic Proofs Mathematical Inequalities Laws of Exponents Integral Exponents Algebra Mathematical Proof 5 For which positive numbers is it true that a^x >= 1+x for all x? All related (34) Sort Recommended Enrico Gregorio Associate professor in Algebra · Author has 18.4K answers and 16M answer views ·5y The expression a x a x is only defined (for arbitrary x x) when a>0 a>0. Also the inequality is certainly satisfied when x≤−1 x≤−1, so we can study it for x>−1 x>−1. The inequality to be satisfied is then x log a≥log(1+x)x log⁡a≥log⁡(1+x) (natural logarithm). Consider the function f(x)=x log a−log(1+x)f(x)=x log⁡a−log⁡(1+x) which should be everywhere nonnegative and note that lim x→∞f(x)={∞a>1−∞01−∞0<a≤1 so we can restrict the search to the case a>1 a>1. Now we have f′(x)=log a−1 1+x f′(x)=log⁡a−1 1+x which vanishes for x=−1+1/log a x=−1+1/log⁡a. This is the point of minimum, because the function is convex Continue Reading The expression a x a x is only defined (for arbitrary x x) when a>0 a>0. Also the inequality is certainly satisfied when x≤−1 x≤−1, so we can study it for x>−1 x>−1. The inequality to be satisfied is then x log a≥log(1+x)x log⁡a≥log⁡(1+x) (natural logarithm). Consider the function f(x)=x log a−log(1+x)f(x)=x log⁡a−log⁡(1+x) which should be everywhere nonnegative and note that lim x→∞f(x)={∞a>1−∞01−∞0<a≤1 so we can restrict the search to the case a>1 a>1. Now we have f′(x)=log a−1 1+x f′(x)=log⁡a−1 1+x which vanishes for x=−1+1/log a x=−1+1/log⁡a. This is the point of minimum, because the function is convex. The absolute minimum is thus m(a)=1−log a+log log a=log e log a a m(a)=1−log⁡a+log⁡log⁡a=log⁡e log⁡a a which is nonnegative for e log a≥a e log⁡a≥a Consider g(a)=e log a−a g(a)=e log⁡a−a. Then its limit at 1 1 is −1−1 and at ∞∞ is −∞−∞. Moreover g′(a)=−1+e/a g′(a)=−1+e/a, so the function has a maximum at e e. Since g(e)=0 g(e)=0 we have reached our conclusion. The inequality a x≥1+x a x≥1+x holds for every x x if and only if a=e a=e. Upvote · 9 2 Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) ·Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. Continue Reading This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Upvote · 999 485 999 103 99 17 Related questions More answers below How do find all positive real number x,x, such that x+√x(x+1)+√x(x+2)+√(x+1)(x+2)=2 x+x(x+1)+x(x+2)+(x+1)(x+2)=2? For what values of x is x^x^x^x > (x-1)^(x+1)^(x-1)^(x+1)? Is it true that (1+x) ^-1 = 1-x? How do I evaluate lim x→∞(1−x)x(1+x)x lim x→∞(1−x)x(1+x)x? Given x 1,...,x n x 1,...,x n are positive numbers and x 1∗...∗x n=1 x 1∗...∗x n=1 how could I prove that x 1+...+x n≥1 x 1+...+x n≥1? Alexandru Carausu Former Assoc. Prof. Dr. (Ret) at Technical University "Gh. Asachi" Iasi (1978–2010) · Author has 3K answers and 875.8K answer views ·11mo This is a little strange question. The answer depends on the “base” a>0 of the exponential function: If a = 1 then a^x = 1 for any real x and 1 = 1 + x ==> x = 0 : this is the only “all x “ . If 0 <a< 1 then a^x continuously decreases from +∞ to 0 , passing thru the point B ( 0 , 1 ) . if a > 1 then a^x continuously increases from 0 to +∞ to 0 , passing thru the point B ( 0 , 1 ) . Final Notes: 1) It is not very clear what meant the sender of this question by positive numbers : he/she had in view the base a of the exponential function or its argument x ∈ R ? The inequality a^x ≥ 1 is (obviously Continue Reading This is a little strange question. The answer depends on the “base” a>0 of the exponential function: If a = 1 then a^x = 1 for any real x and 1 = 1 + x ==> x = 0 : this is the only “all x “ . If 0 <a< 1 then a^x continuously decreases from +∞ to 0 , passing thru the point B ( 0 , 1 ) . if a > 1 then a^x continuously increases from 0 to +∞ to 0 , passing thru the point B ( 0 , 1 ) . Final Notes: 1) It is not very clear what meant the sender of this question by positive numbers : he/she had in view the base a of the exponential function or its argument x ∈ R ? The inequality a^x ≥ 1 is (obviously) satisfied only for f ( x ) = a^x : f ( 0 ) = a^0 = 1 ==>B ( 0 , 1 ) ∈ G _ f , while ( ∀ x ∈ ( 0 , +∞ ) a^x> 1+ x : the graph G _ f lies above the oblique line ( L ) : y = 1 + x . 2) The sender of this question had the possibility to see how looks the graph of the exponential function f ( x ) = a^x in any manual of CALCULUS / MATHEMATICAL ANALYSIS for high schools / colleges. I cannot attach to my answers (edited with the TEXT format only) any mathematical equations, figures, matrices & determinants, etc. The graph of f ( x ) = a^x can be found in the chapter Elementary Functions of any book like those just mentioned. Upvote · Ron Patey Former R&D Engineer at Experiences in Life (1954–2014) · Author has 1.6K answers and 2.2M answer views ·5y a^x>=1+x: true or false: a can be anything, but kept to positive, as is x. 1 and 0 in any ‘to the power of’ are special cases and limits.. just as like .0000001 is approaching zero. as x gets smaller … 9x9 is like 9^2 which is 81. 9x1 is like 9^1 which is 9 and is the pivot point. but 9^.5 is 3 and 9^.03 is 1.063 and 9^.003 is 1.007 so 9^0 must be 1 Thus (a^0 >= 1 +0) is true means 1=1 which is equal Now when x is small, like 9^.003 is 1.007 then 1 + .003 is less. And when x is a bit more, like 9^.03 is 1.063 then 1 + .03 is less. And when x is 2. like 9^2 is 81, then 1 + 2 is less. I would suspect that the Continue Reading a^x>=1+x: true or false: a can be anything, but kept to positive, as is x. 1 and 0 in any ‘to the power of’ are special cases and limits.. just as like .0000001 is approaching zero. as x gets smaller … 9x9 is like 9^2 which is 81. 9x1 is like 9^1 which is 9 and is the pivot point. but 9^.5 is 3 and 9^.03 is 1.063 and 9^.003 is 1.007 so 9^0 must be 1 Thus (a^0 >= 1 +0) is true means 1=1 which is equal Now when x is small, like 9^.003 is 1.007 then 1 + .003 is less. And when x is a bit more, like 9^.03 is 1.063 then 1 + .03 is less. And when x is 2. like 9^2 is 81, then 1 + 2 is less. I would suspect that the statement is true for all positive or zero x. This is calculus… and a foundational logic. It is a valid method and needs no formulae. Upvote · Thomas Dalton MMath in Mathematics, Durham University (Graduated 2009) · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 12.5K answers and 15M answer views ·Updated 11y Related ∞√x=1, x= all positive numbers x=1^∞ hence x=1 so all x=1 but it was stated that x=all positive numbers, how can all number of x be equal to 1? Please solve this paradox. The problem is that you are assuming that . Unfortunately, expressions with infinity in don't always work the same way as expressions with numbers in. When we write expressions with infinity in like that, it is shorthand for a limit. That means, what you are implicitly relying on is: lim y→∞(y√x)y=lim y→∞(lim z→∞z√x)y lim y→∞(x y)y=lim y→∞(lim z→∞x z)y Unfortunately, that just isn't true. While there are plenty of situations where you can do a limit one bit at a time, it doesn't always work. This is one of the occasions where it doesn't work. If you do t Continue Reading The problem is that you are assuming that . Unfortunately, expressions with infinity in don't always work the same way as expressions with numbers in. When we write expressions with infinity in like that, it is shorthand for a limit. That means, what you are implicitly relying on is: lim y→∞(y√x)y=lim y→∞(lim z→∞z√x)y lim y→∞(x y)y=lim y→∞(lim z→∞x z)y Unfortunately, that just isn't true. While there are plenty of situations where you can do a limit one bit at a time, it doesn't always work. This is one of the occasions where it doesn't work. If you do the limit all at once, you'll get x. If you do it in two stages, you'll get 1. That is proof that that method doesn't work. The problem is that you've lost information at the first stage. The same thing would happen if you multiplied by zero, for example. 05=07 but . As long as you don't lose information (ie. the operation is invertible), then you should be fine using that kind of method. Upvote · 99 23 9 3 Sponsored by CDW Corporation Looking for ways to leverage AI to improve customer experiences? CDW’s AI solutions include 24/7 chatbots and workshops to enhance efficiency and customer satisfaction. Learn More 99 23 Related questions More answers below Is there any set of numbers in which x + 1 < x? If x-1/x=5, what is the value of x⁴+1/X⁴? While 1^1 is 1, will 1^x = 1? (where x is anything)? Is there a value for x for which 1^x is not equal to 1? Are 1 and -1 the only numbers where x^2 = x? How can I prove that lim x→∞[x x+1(x+1)x−(x−1)x x x−1]=1 e lim x→∞[x x+1(x+1)x−(x−1)x x x−1]=1 e? Awnon Bhowmik Studied at University of Dhaka · Author has 3.7K answers and 11.2M answer views ·9y Related For which X is this true? x x>x Γ(x+1)x x>x Γ(x+1) x x>x Γ x+1 x x>x Γ x+1 ⟹x ln x>Γ x+1 ln x⟹x ln⁡x>Γ x+1 ln⁡x Since Γ n=(n−1)!Γ n=(n−1)! ⟹Γ x+1=x!⟹Γ x+1=x! Using this in our inequality x>x!x>x! ⟹x>x(x−1)(x−2)…..⟹x>x(x−1)(x−2)….. ⟹x−x(x−1)(x−2)…..>0⟹x−x(x−1)(x−2)…..>0 ⟹x[1−(x−1)(x−2)…..]>0⟹x[1−(x−1)(x−2)…..]>0 ⟹x>0⟹x>0 Problem is that the graph does not agree with my solution. Maybe x>0 x>0 is the redundant part. We need to solve f(x)=1−(x−1)(x−2)….=0 f(x)=1−(x−1)(x−2)….=0 Putting x=1,f(x)=0 x=1,f(x)=0 Putting x=2,f(x)=1 x=2,f(x)=1 Putting x=3,f(x)=1 x=3,f(x)=1 It is safe to say, that according to intermediate value theorem, root lies in the interval [1,2][1,2] We need another plot, this time, of the function f(x)f(x) And now it is clear that f(x)>0 f(x)>0 in the inter Continue Reading x x>x Γ x+1 x x>x Γ x+1 ⟹x ln x>Γ x+1 ln x⟹x ln⁡x>Γ x+1 ln⁡x Since Γ n=(n−1)!Γ n=(n−1)! ⟹Γ x+1=x!⟹Γ x+1=x! Using this in our inequality x>x!x>x! ⟹x>x(x−1)(x−2)…..⟹x>x(x−1)(x−2)….. ⟹x−x(x−1)(x−2)…..>0⟹x−x(x−1)(x−2)…..>0 ⟹x[1−(x−1)(x−2)…..]>0⟹x[1−(x−1)(x−2)…..]>0 ⟹x>0⟹x>0 Problem is that the graph does not agree with my solution. Maybe x>0 x>0 is the redundant part. We need to solve f(x)=1−(x−1)(x−2)….=0 f(x)=1−(x−1)(x−2)….=0 Putting x=1,f(x)=0 x=1,f(x)=0 Putting x=2,f(x)=1 x=2,f(x)=1 Putting x=3,f(x)=1 x=3,f(x)=1 It is safe to say, that according to intermediate value theorem, root lies in the interval [1,2][1,2] We need another plot, this time, of the function f(x)f(x) And now it is clear that f(x)>0 f(x)>0 in the interval 1<x<2 1<x<2, or (1,2)(1,2) Maybe this answer wasn’t the best I could do, but I tried. Thank you. Please let me know of errors, if any. I would like to advance my knowledge of these type of questions further. Upvote · 9 3 9 4 Alberto Cid M.S.E. in Telecommunications Engineering&Data Transmission, Technical University of Madrid (Graduated 2008) · Author has 2K answers and 3.8M answer views ·1y Related Let x, y and z be positive numbers with 1<=xyz. What's the greatest integer a for which a<=(1+x) (1+y) (1+z)? I think the question is a bit confusing or ambiguous. Example: Let x = 1000; y = 0.001; z = 1 xyz = 1 (1+x) (1+y) (1+z) = 1001•1.001•2 > 2002 So, there are values of (x, y, z) for which the “a” is greater than 8 and greater than 2002. In my opinion the question should say something like: Let S be the set of (x, y, z) such that x > 0; y>0; z>0; 1≤xyz Find the greatest integer “a” such that FOR ALL points (or vectors) in S the following inequality remains true: a ≤ (1+x)(1+y)(1+z) While for some points the “a” can br greater than 2002 or greater than 2 million, the inequality has to be true FOR ALL points in Continue Reading I think the question is a bit confusing or ambiguous. Example: Let x = 1000; y = 0.001; z = 1 xyz = 1 (1+x) (1+y) (1+z) = 1001•1.001•2 > 2002 So, there are values of (x, y, z) for which the “a” is greater than 8 and greater than 2002. In my opinion the question should say something like: Let S be the set of (x, y, z) such that x > 0; y>0; z>0; 1≤xyz Find the greatest integer “a” such that FOR ALL points (or vectors) in S the following inequality remains true: a ≤ (1+x)(1+y)(1+z) While for some points the “a” can br greater than 2002 or greater than 2 million, the inequality has to be true FOR ALL points in S… Consider the point (x, y, z) = P(1, 1, 1). This point P belongs to S and (1+1)(1+1)(1+1) = 8 So, to meet the requirement that the inequality is true for all points in S it must be true also for point P… so “a” must be lower or equal to 8. The next step would be proving the multiplication m = (1+x)(1+y)(1+z) is greater than 8 or equal to 8 in the set S. You can use AM-GM inequality as another answer did… Or you can just notice : 1+w ≥ 2√w (1+w)² ≥ 4w 1+2w+w² ≥ 4w 1–2w+w² ≥ 0 (1-w)² ≥ 0 Of course a square can never be negative and it’s only zero when the base is zero that is when 1-w = 0 if and only if w=1 Then m = (1+x)(1+y)(1+z) ≥ 2√x(1+y)(1+z) only equal when x=1 ≥ 2√x•2√y •(1+z) only equal when x=y=1 ≥ 2√x•2√y•2√z only equal when x=y=z =1 Then m ≥ 8√(xyz) Since f(u) = √u is always increasing, that is, for a greater u’ > u, the √(u’) > √u then for xyz ≥ 1 we have √(xyz) ≥ √1 = 1 Then 8√(xyz) ≥ 8 m will only be 8 when x=y=z=1 For (x, y, z) ≠ (1, 1, 1) we have m > 8√(xyz) and even when xyz=1 (it could be higher), we have m>8 Then for all (x, y, z) only when x=y=z=1 we have that the multiplication m = (1+x)(1+y)(1+z) is exactly 8 And for other points (x, y, z) ≠ (1, 1, 1) we have (1+x)(1+y)(1+z) > 8 Then a = 8 is the greatest real number that is lower or equal to that multiplication for every possible point in the set S. Final notes: The part when I said: “let’s consider the point (x, y, z) = P(1, 1, 1)”. This might seem taken “out of nowhere”… like I “magically” knew the minimum in advance. Well, notice the symmetry. If you have any arbitrary point V 0(x 0,y 0,z 0)V 0(x 0,y 0,z 0) in the set S, that is, positive numbers such that xyz ≥ 1 then also any permutation like V 1(z 0,y 0,x 0)V 1(z 0,y 0,x 0) or V 2(y 0,z 0,x 0)V 2(y 0,z 0,x 0) would both be in S and have the same value of the multiplication function m. There are 3! = 3•2•1 = 6 of such permutations. Then in case there would be a minimum for V_0 then you would have 6 points which give the same minimum value… So, symmetry suggests x=y=z for optimum values. If you make x•x•x = 1 then x=y=z = ³√1 = 1 Also, the step when I said : “notice (1+w) ≥ 2√w “ could also be seen as “magically coming out of nowhere”. An alternative path would be: xyz ≥ 1 z ≥ 1/[xy] > 0 Then, since z>0 and (1+z)>1>0 (1+x)(1+y) (1+z) ≥ (1+x)(1+y) (1+1/[xy]) = (1+x+y+xy)(1+1/[xy]) = 1+1/[xy] + x + 1/y + y +1/x + xy + 1 = 2 + (x+1/x) + (y+1/y) + (xy + 1/[xy]) Notice the 3 parts (w+1/w) You can use Differential Calculus to find the minimum of f(w) = w + 1/w f’(w) = 1 -1/w² = 0 f’’(w) = +2/w³ w = +1 is a minimum [f’’(+1) > 0] f(+1) = 2 then f(w) ≥ 2 for all w>0 and only equal when w=1 Then (1+x)(1+y) (1+z) ≥ (1+x)(1+y) (1+1/[xy]) = 2 + (x+1/x) + (y+1/y) + (xy + 1/[xy]) ≥ 2+2+2+2 = 8 Only equal when x=y=1 =xy In case z = 1/[xy] = 1 too, then we would have the multiplication is exactly 8. But if z > 1/[xy] it could never be 8. Notice f(w) = w + 1/w ≥ 2 is also related to AM-GM : f(w) is twice the AM, so 2AM ≥ 2GM = 2 √ (w•1/w) = 2 √1 = 2 A typical standard method for problems like this is that of Lagrange Multipliers. Lagrange multiplier - Wikipedia The basic idea is to convert a constrained problem into a form such that the derivative test of an unconstrained problem can still be applied. The relationship between the gradient of the function and gradients of the constraints rather naturally leads to a reformulation of the original problem, known as the Lagrangian function or Lagrangian. [ 2 ] In the general case, the Lagrangian is defined as L ( x , λ ) ≡ f ( x ) + ⟨ λ , g ( x ) ⟩ {\displaystyle {\mathcal {L}}(x,\lambda )\equiv f(x)+\langle \lambda ,g(x)\rangle } for functions f , g {\displaystyle f,g} ; the notation ⟨ ⋅ , ⋅ ⟩ {\displaystyle \langle \cdot ,\cdot \rangle } denotes an inner product . The value λ {\displaystyle \lambda } is called the Lagrange multiplier . In simple cases, where the inner product is defined as the dot product , the Lagrangian is L ( x , λ ) ≡ f ( x ) + λ ⋅ g ( x ) {\displaystyle {\mathcal {L}}(x,\lambda )\equiv f(x)+\lambda \cdot g(x)} The method can be summarized as follows: in order to find the maximum or minimum of a function f {\displaystyle f} subject to the equality constraint g ( x ) = 0 {\displaystyle g(x)=0} , find the stationary points of L {\displaystyle {\mathcal {L}}} considered as a function of x {\displaystyle x} and the Lagrange multiplier λ {\displaystyle \lambda ~} . This means that all partial derivatives should be zero, including the partial derivative with respect to λ {\displaystyle \lambda ~} . [ 3 ] ∂ L ∂ x = 0 {\displaystyle {\frac {\partial {\mathcal {L}}}{\partial x}}=0} and ∂ L ∂ λ = 0 ; {\displaystyle {\frac {\ \partial {\mathcal {L}}\ }{\partial \lambda }}=0\ ;} or equivalently ∂ f ( x ) ∂ x + λ ⋅ ∂ g ( x ) ∂ x = 0 {\displaystyle {\frac {\partial f(x)}{\partial x}}+\lambda \cdot {\frac {\partial g(x)}{\partial x}}=0} and g ( x ) = 0 . {\displaystyle g(x)=0~.} The solution corresponding to the original constrained optimization is always a saddle point of the Lagrangian function, [ 4 ] [ 5 ] which can be identified among the stationary points from the definiteness of the bordered Hessian matrix . [ 6 ] The great advantage of this method is that it allows the optimization to be solved without explicit parameterization in terms of the constraints. As a result, the method of Lagrange multipliers is widely used to solve challenging constrained optimization problems. Further, the method of Lagrange multipliers is generalized by the Karush–Kuhn–Tucker conditions , which can also take into account inequality constraints of the form h ( x ) ≤ c {\displaystyle h(\mathbf {x} )\leq c} for a given constant c {\displaystyle c} . The following is known as the Lagrange multiplier theorem. [ 7 ] Let f : R n → R {\displaystyle f\colon \mathbb {R} ^{n}\to \mathbb {R} } be the objective function and let g : R n → R c {\displaystyle g\colon \mathbb {R} ^{n}\to \mathbb {R} ^{c}} be the constraints function, both belonging to C 1 {\displaystyle C^{1}} (that is, having continuous first derivatives). Let x ⋆ {\displaystyle x_{\star }} be an optimal solutio Well, we have an objective function m(x, y, z) = (1+x)(1+y)(1+z) which is continuous and having continuous first derivatives, that is, it belongs to C¹. Also, we have a restriction xyz = K (I will generalize for any K) which we can convert to a function which is always zero: g(x, y, z) = xyz - K This function g also belongs to C¹ The Lagrange Multipliers method considers another function of a parameter c. The Lagrangian Function: h(x, y, z, c) = f(x, y, z) + c•g(x, y, z) This function h will give the same value as f in any point of the restricted set where g is zero. We compute the Gradient of h h = (1+x)(1+y)(1+z) + c(xyz-K) ∂h∂x∂h∂x = (1+y)(1+z) + cyz ∂h∂y∂h∂y = (1+x)(1+z) + cxz ∂h∂z∂h∂z = (1+x)(1+y) + cxy ∂h∂c∂h∂c = xyz-K We equate all of those to zero. 1+y+z +(1+c)yz = 0 1+x+z +(1+c)xz = 0 1+x+y +(1+c)xy = 0 xyz = K We then solve the system of equations. 1st: y = -(1+z)/[1+(1+c)z] 2nd : x = -(1+z)/[1+(1+c)z] That, is, x=y, Usually you could take this relation to the 3rd one: 1+2x + (1+c)x² = 0 x = y = -1/(1+c)± √(1-[1+c])/(1+c) = But, since the equations lead to x=y=z we can take it to the last equation x•x•x = K For any value of K the optimum will be when x=y=z = ³√K And (1+³√K)³ > 8 for K>1 So the absolute minimum of all will be in the case K=1, that is when xyz = 1, not greater than 1 and in that case the minimum is for x=y=z = 1 making the minimum equal to: (1+1)(1+1)(1+1) = 8 I think this method gives a “higher” view of the problem: This finds the minimum for any surface xyz = K … the minimum is (1+³√K)³ It also relates to symmetry and the conclusion x=y=z The x=y=z would be valid for any objective function (v+x)(v+y)(v+z) and many other functions as long as they are symmetric and have some “nice” properties. If they aren't symmetric you’ll probably would not reach the condition x=y=z but, anyway, it provides a general method… as long as the functions belong to C¹. Upvote · 9 1 Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Learn More 999 116 Reuven Harmelin mathematics adict · Author has 2.3K answers and 1.9M answer views ·5y Related If x and y are two different real positive numbers, how do you show that 1/x + 1/y <= x/y^2 + y/x^2? This inequality could be written as follows since x,y are 2 positive numbers, x+y>0 a nd therefore you may write the following equivalent inequality that is which is evidently true, and hence the only cases that there is equality are when and only x=y Continue Reading This inequality could be written as follows since x,y are 2 positive numbers, x+y>0 a nd therefore you may write the following equivalent inequality that is which is evidently true, and hence the only cases that there is equality are when and only x=y Upvote · 9 1 Jan van Delden MSc Math and still interested · Author has 4.8K answers and 6.6M answer views ·4y Related Why is x^(1/x) always positive? I think you have already received nice answers. In order to supplement these answers, I have constructed an extended phase portrait of the complex function: f(z)=z 1/z f(z)=z 1/z on [−1,1]×[−1,1]i[−1,1]×[−1,1]i In the center we have the point z=0 z=0, or the point with coordinates (0,0)(0,0). The first coordinate is the real value of z z, the second the imaginary value of z z. The horizontal line in the middle of the phase portrait describes all values z z such that the imaginary part equals 0 0. In other words, that’s where your x x lives. The colours indicate the phase of f(z)f(z). A phase (or argument) equal to 0 0 corresponds with a re Continue Reading I think you have already received nice answers. In order to supplement these answers, I have constructed an extended phase portrait of the complex function: f(z)=z 1/z f(z)=z 1/z on [−1,1]×[−1,1]i[−1,1]×[−1,1]i In the center we have the point z=0 z=0, or the point with coordinates (0,0)(0,0). The first coordinate is the real value of z z, the second the imaginary value of z z. The horizontal line in the middle of the phase portrait describes all values z z such that the imaginary part equals 0 0. In other words, that’s where your x x lives. The colours indicate the phase of f(z)f(z). A phase (or argument) equal to 0 0 corresponds with a real number, these points are coloured red. Thus red means that the image of f(z)f(z) is real. Since different shades of red are used it is not that easy to spot these points exactly, but you may see that bands of reddish colours pop up in the complex plane. The most obvious location is exactly where x>0 x>0, i.e. from the center of the image to the right. If you check the other side of the origin, where we have x<0 x<0 the colours are all over the place and seem to follow a quite difficult pattern. What’s more there seems to be a line across which these colours jump. Such a line is called a branch line. Near the center of the image you see a black ‘smudge’. This is an artifact that is due to computational limitations and is closely related to the discussion how to define: z 1/z z 1/z when z z tends to 0 0. A point around which the bands of colour get smaller and smaller is called an essential singular point. Which we may translate to: Definitely a problem area. But if we study any other area, besides z=0 z=0 and the branch line I discussed, we may see that the colours gradually change and that the two sets of lines I drew meet at 90 degrees. This means that the given function is analytic, which you may translate to: The function behaves really nice. This statement is also true for the region where the colours fluctuate very fast. All it takes is zooming in to show that the colours do gradually change there as well. You can’t use this phase portrait to decide whether or not the function value is positive, but you do get a fair idea of the complexity of its behaviour. Upvote · 9 4 Sponsored by Zoho One Run your entire small business with one software suite - Zoho One Award-winning business software. 50M+ users across the globe trust Zoho. Start your free trial today! Sign Up 99 22 Sohel Zibara Studied at Doctor of Philosophy Degrees (Graduated 2000) · Author has 5.1K answers and 2.6M answer views ·1y Related Let x, y and z be positive numbers with 1<=xyz. What's the greatest integer a for which a<=(1+x) (1+y) (1+z)? ∀x,y,z∈R∗+,we have :(1+x)(1+y)(1+z)≥AM–GM 2√x⋅2√y⋅2√z=8√x y z≥8 with equality iff x=y=z=1.a=8∀x,y,z∈R+∗,we have :(1+x)(1+y)(1+z)≥⏟AM–GM 2 x⋅2 y⋅2 z=8 x y z≥8 with equality iff x=y=z=1.a=8 Upvote · 9 3 Alexandre Ribeiro Miquilino Number Theory Addict · Upvoted by Nathan Hannon , Ph. D. Mathematics, University of California, Davis (2021) and Jose Arnaldo Dris , M.S. Mathematics & Number Theory, De La Salle University (2008) · Author has 729 answers and 3.1M answer views ·6y Related How do you solve x^(x+1) = (x+1) ^x? Oh boy, this is going to be a good one. Yet, if you jumped the gun and went straight to Wolfram Alpha with this problem, it will kindly give you a numeric solution, one around 2.293166...2.293166..., but that's no fun. Of course, if you want you can stop here and be happy knowing that this is a number that answers the question. The rest of you who want to follow me on this daunting task, come along. Our equation is rather simple: x x+1=(x+1)x x x+1=(x+1)x. At first glance, this looks like a x x x x-like type of equation, usually solvable with our trusty Lambert-W function. Thing is, there is a pitfall that arises Continue Reading Oh boy, this is going to be a good one. Yet, if you jumped the gun and went straight to Wolfram Alpha with this problem, it will kindly give you a numeric solution, one around 2.293166...2.293166..., but that's no fun. Of course, if you want you can stop here and be happy knowing that this is a number that answers the question. The rest of you who want to follow me on this daunting task, come along. Our equation is rather simple: x x+1=(x+1)x x x+1=(x+1)x. At first glance, this looks like a x x x x-like type of equation, usually solvable with our trusty Lambert-W function. Thing is, there is a pitfall that arises in some scenarios, like this one. Let's see what it is. THE WRONG APPROACH Take the equation x x+1=(x+1)x x x+1=(x+1)x, and raise both sides to the 1 x(x+1)1 x(x+1) power. We end up with the following: x 1 x=(x+1)1 x+1 x 1 x=(x+1)1 x+1. Take the natural log on both sides: 1 x ln(x)=1 x+1 ln(x+1)1 x ln⁡(x)=1 x+1 ln⁡(x+1). Multiply by −1−1 and use it as the exponent inside the log: 1 x ln(1 x)=1 x+1 ln(1 x+1)1 x ln⁡(1 x)=1 x+1 ln⁡(1 x+1) Now, take the Lambert-W on both sides. We're going to use the following property: given that W(x)=f−1(x)W(x)=f−1(x) when f(x)=x e x f(x)=x e x, and that f−1(f(x))=x f−1(f(x))=x, we have that W(x e x)=x W(x e x)=x. Now, if x=ln(y)x=ln⁡(y), we have: W(ln(y)e ln(y))=W(y ln(y))=ln(y)W(ln⁡(y)e ln⁡(y))=W(y ln⁡(y))=ln⁡(y). Therefore, W(1 x ln(1 x))=W(1 x+1 ln(1 x+1))W(1 x ln⁡(1 x))=W(1 x+1 ln⁡(1 x+1)), and so ln(1 x)=ln(1 x+1)ln⁡(1 x)=ln⁡(1 x+1). Thus, e ln(1 x)=e ln(1 x+1)e ln⁡(1 x)=e ln⁡(1 x+1), or simply 1 x=1 x+1 1 x=1 x+1. This gives x+1=x x+1=x, and then... 1=0 1=0... Wait… damn. We did something wrong. Lambert, have you failed me? Not quite. THE x 1 x x 1 x FUNCTION By now, after we've cried our tears of frustration, we try to look for some answers. Is there even a number that satisfies this property? Well, there has to be, and more specifically one between 2 2 and 3 3. The reason for this is as follows: consider the function f(x)=x x+1−(x+1)x f(x)=x x+1−(x+1)x. We have f(2)=2 3−3 2=8−9=−1 f(2)=2 3−3 2=8−9=−1, while f(3)=3 4−4 3=81−64=17 f(3)=3 4−4 3=81−64=17. Somewhere on this interval there is a magic number for which f(x)=0 f(x)=0. Ok, so our search is not a futile one - there has to be a value. Where did we go wrong, then? Let's return to the expression we've derived eariler, x 1 x=(x+1)1 x+1 x 1 x=(x+1)1 x+1. This suggests that our problem has an important connection to the function x 1 x x 1 x. Let's take a look at its graph, shall we? This is one of my favorite graphs. It almost starts at the origin (x 1 x x 1 x isn’t defined for x=0 x=0, after all, but the right limit is indeed 0 0), hits its maximum value at x=e x=e, and then decreases asymptotically towards the line y=1 y=1. (All of these can be proved using standard calculus techniques, so I won’t bore you with the details.) If you pay close attention to the graph, you’ll see that for every x>1 x>1, there is always a unique pair of numbers (a,b)(a,b) such that a 1 a=b 1 b a 1 a=b 1 b, and a≠b a≠b (except when x=e x=e). Since x x and x+1 x+1 are different numbers, it follows that x>1 x>1. Now, take a look at the graph when x≥e x≥e; the value of x 1 x x 1 x just keeps decreasing, and thus no two numbers who share the property a 1 a=b 1 b a 1 a=b 1 b can be both greater than e at the same time. By the same token, when 1<x≤e 1<x≤e, the value of x 1 x x 1 x just keeps increasing, and the both numbers cannot also belong to the interval (1,e](1,e] at the same time as well. Therefore, 1<x<e 1<x<e, and e<x+1 e<x+1. We have a single value for x x which is on the interval e−1<x<e e−1<x<e. WHAT WE DID WRONG Recall that when we used the property W(y⋅ln(y))=ln(y)W(y⋅ln⁡(y))=ln⁡(y), we applied it to some value of y=1 x y=1 x. By the inequality derived at the end of the previous section, 1 e<y<1 e−1 1 e<y<1 e−1; therefore, 1 e ln(1 e)<y ln(y)<1 e−1 ln(1 e−1)1 e ln⁡(1 e)<y ln⁡(y)<1 e−1 ln⁡(1 e−1), or also −1 e<y ln(y)<1 e−1 ln(1 e−1)−1 e<y ln⁡(y)<1 e−1 ln⁡(1 e−1). This means that y ln(y)y ln⁡(y) is a negative number, and not just that, but a negative number greater than −1 e−1 e, the cutoff point for the W function, and also the region of the real plane where this function is multi-valued. Therefore, when we did W(y ln(y))=ln(y)W(y ln⁡(y))=ln⁡(y), we got the value for W 0 W 0, without checking instead a possible value for W−1 W−1. Let’s see why this is a problem. What is W(−ln(√2))W(−ln⁡(2))? Well, −ln(√2)=−ln(2 1 2)=−1 2 ln(2)=1 2 ln(1 2)−ln⁡(2)=−ln⁡(2 1 2)=−1 2 ln⁡(2)=1 2 ln⁡(1 2), so W(−ln(√2))=W(1 2 ln(1 2))=ln(1 2)=−ln(2)W(−ln⁡(2))=W(1 2 ln⁡(1 2))=ln⁡(1 2)=−ln⁡(2). Seems simple and straightforward enough… until you consider the following: √2=2 1 2=4 1 4 2=2 1 2=4 1 4. Then, we could have written −ln(2 1 2)=−ln(4 1 4)=1 4 ln(1 4)−ln⁡(2 1 2)=−ln⁡(4 1 4)=1 4 ln⁡(1 4), and then W(−ln(√2))=W(1 4 ln(1 4))=ln(1 4)=−ln(4)=−2 ln(2)W(−ln⁡(2))=W(1 4 ln⁡(1 4))=ln⁡(1 4)=−ln⁡(4)=−2 ln⁡(2). There are two different solutions. Which one is wrong, and which one is right? Answer: Both are. Recall that W(x)W(x) is the inverse to x e x x e x. So what happens when we plug these two values we found into x e x x e x? First, ln(1 2):ln(1 2)e ln(1 2)=ln(1 2)1 2=ln(√1 2)ln⁡(1 2):ln⁡(1 2)e ln⁡(1 2)=ln⁡(1 2)1 2=ln⁡(1 2). Next, ln(1 4):ln(1 4)e ln(1 4)=ln(1 4)1 4=ln(√1 2)ln⁡(1 4):ln⁡(1 4)e ln⁡(1 4)=ln⁡(1 4)1 4=ln⁡(1 2). Let’s graph things to understand what is going on: When x x is −ln(2)−ln⁡(2), we are operating in the W 0 W 0 branch. When x x is −ln(4)−ln⁡(4), we are operating in the W−1 W−1 branch. So basically, our substitution W(y ln(y))=ln(y)W(y ln⁡(y))=ln⁡(y) was invalid. SO HOW DOES THIS ALL WORK? In Alexandre Ribeiro Miquilino's answer to Can I put the relation x^y=y^x at the form of y=f(x)?, I explain how we can find values of x and y that are different and which satisfy this equation at the same time. They both follow a pattern, where x=t 1 t−1 x=t 1 t−1, and y=t t t−1 y=t t t−1 (or the other way around, it doesn’t matter). Not only that, but we also have the relation y=t x y=t x. If we know that y=x+1 y=x+1, then t x=x+1 t x=x+1, and thus x=1 t−1 x=1 t−1. Then: 1 t−1=t 1 t−1 1 t−1=t 1 t−1. which is equivalent to stating t=(t−1)−(t−1)t=(t−1)−(t−1) Substituting t−1=z t−1=z gives us the equation z−z=z+1 z−z=z+1, which as far as I know has no straightforward solving methods other than numerical approximations. We can solve for z z, and since z=t−1 z=t−1, this means that x=1 t−1=1 z x=1 t−1=1 z. So, let’s do this. Define h(z)=z−z−z−1 h(z)=z−z−z−1. We can apply the Newton-Raphson’s method to find a numerical solution; we can start with z 0=1 2 z 0=1 2, and the convergence step is given by z i+1=z i−h(z i)h′(z i)z i+1=z i−h(z i)h′(z i); in our case, we have: z i+1=z i−z−z i i−z i−1−z z i i(1+l n(z i))−1 z i+1=z i−z i−z i−z i−1−z i z i(1+l n(z i))−1 I put this on a spreadsheet for us to see how we can get the value of x x: This actually converged faster than I had expected. And it does match up with the value we gave right at the beginning of our answer. This isn’t quite the type of answer I like to give — I wish I could have found some magical expression involving any sort of complicated (but easily explainable) functions. Who knows, maybe there’s a way, and a more suitable mind than mine might find my answer a good starting point to find it. Upvote · 999 605 99 43 99 14 Mark Gritter recreational mathematician · Author has 5.7K answers and 11.8M answer views ·Updated 7y Related What are the minimum and maximum values of (1+1/x) ^x, if x is a positive integer? The first thing to note is that the expression is monotone increasing over integers. If looking at the graph isn't convincing enough, a short proof can be found here: convergence of the sequence (1+1/n)^n That means the first value is also the minimum value, namely 2. However, there is no maximum value. The limit of this sequence is the mathematical constant e≈2.71828 e≈2.71828, but the limit is never reached for any positive integer. Upvote · 9 7 Francesco Iovine Mathaholic · Author has 2K answers and 5.4M answer views ·9y Related For which X is this true? x x>x Γ(x+1)x x>x Γ(x+1) If you want to consider real numbers, you need to use the Gamma function Being Γ(x+1)=x!Γ(x+1)=x! then the inequality becomes: x x>x Γ(x+1)x x>x Γ(x+1) For x >1, we can take the logarithm on both sides of the inequality as the logarithm is positive and it is an increasing function. So the inequality becomes. x ln(x)>Γ(x+1)ln(x)x ln⁡(x)>Γ(x+1)ln⁡(x) that turns into x>Γ(x+1)x>Γ(x+1) This inequality can be solved graphically. I used sage to plot both the gamma function and the y=x y=x straight line and this is what I get ​ So in the real domain, the inequality holds when 1<x<2 1<x<2 I used sage even to verify the inequality. Wh Continue Reading If you want to consider real numbers, you need to use the Gamma function Being Γ(x+1)=x!Γ(x+1)=x! then the inequality becomes: x x>x Γ(x+1)x x>x Γ(x+1) For x >1, we can take the logarithm on both sides of the inequality as the logarithm is positive and it is an increasing function. So the inequality becomes. x ln(x)>Γ(x+1)ln(x)x ln⁡(x)>Γ(x+1)ln⁡(x) that turns into x>Γ(x+1)x>Γ(x+1) This inequality can be solved graphically. I used sage to plot both the gamma function and the y=x y=x straight line and this is what I get ​ So in the real domain, the inequality holds when 1<x<2 1<x<2 I used sage even to verify the inequality. When x=1.5 we have ​ Additional answers: 0 0 0 0 is an indeterminate form so the inequality has no meaning there. If we pass to the limit, anyway, things are a bit more complex as in my model, you cannot eliminate the logarithm as I did as its value is 0. We know that (n+1)!=(n+1)×n!(n+1)!=(n+1)×n! if n=0 n=0 we have 1!=1×0!1!=1×0! therefore 0!=1!/1=1 0!=1!/1=1 Upvote · 9 5 Nancy Mitchell used to be a teacher. · Upvoted by Michael Naylor , Ph.D. Mathematics Education & Mathematics, Florida State University (1999) · Author has 3.4K answers and 8M answer views ·5y Related For what values of x is it true that: x × x × x ≤ x + x + x? For what values of x is it true that x×x×x<=x+x+x? x×x×x≤x+x+x x×x×x≤x+x+x ⟹x 3≤3 x⟹x 3≤3 x ⟹x 3−3 x≤0⟹x 3−3 x≤0 ⟹x(x 2−3)≤0⟹x(x 2−3)≤0 ⟹x(x−√3)(x+√3)≤0.⟹x(x−3)(x+3)≤0. Solve x(x−√3)(x+√3)=0 x(x−3)(x+3)=0 and get x=0,±√3.x=0,±3. So, the intervals on the x x number line to look at for finding solutions to the inequality x 3<3 x x 3<3 x are (−∞,−√3),(−√3,0),(0,√3)(−∞,−3),(−3,0),(0,3) and(√3,∞).and(3,∞). Choose a value in each interval and see if it so Continue Reading For what values of x is it true that x×x×x<=x+x+x? x×x×x≤x+x+x x×x×x≤x+x+x ⟹x 3≤3 x⟹x 3≤3 x ⟹x 3−3 x≤0⟹x 3−3 x≤0 ⟹x(x 2−3)≤0⟹x(x 2−3)≤0 ⟹x(x−√3)(x+√3)≤0.⟹x(x−3)(x+3)≤0. Solve x(x−√3)(x+√3)=0 x(x−3)(x+3)=0 and get x=0,±√3.x=0,±3. So, the intervals on the x x number line to look at for finding solutions to the inequality x 3<3 x x 3<3 x are (−∞,−√3),(−√3,0),(0,√3)(−∞,−3),(−3,0),(0,3) and(√3,∞).and(3,∞). Choose a value in each interval and see if it solves this inequality. If it does, then all values of x x in that interval do, too. This works because the sign does not change within an interval. Note that √3≈1.732.3≈1.732. x=−2∈(−∞,−√3)x=−2∈(−∞,−3) ⟹(−2)3=−8<3(−2)=−6⟹(−2)3=−8<3(−2)=−6 ⟹x∈(−∞,−√3).⟹x∈(−∞,−3). x=−1∈(−√3,0)x=−1∈(−3,0) ⟹(−1)3=−1≮3(−1)=−3⟹(−1)3=−1≮3(−1)=−3 ⟹x∉(0,−√3).⟹x∉(0,−3). x=1∈(0,√3)x=1∈(0,3) ⟹1 3=1<3(1)=3⟹1 3=1<3(1)=3 ⟹x∈(0,√3).⟹x∈(0,3). x=2∈(√3,∞)x=2∈(3,∞) ⟹2 3=8≮3(2)=6⟹2 3=8≮3(2)=6 ⟹x∉(√3,∞).⟹x∉(3,∞). The endpoints of these intervals, x=0,±√3 x=0,±3 satisfy the given inequality. So, the inequality x×x×x≤x+x+x x×x×x≤x+x+x holds in the intervals (−∞,−√3]and[0,√3].(−∞,−3]and[0,3]. ∴−∞<x≤−√3 or 0≤x≤√3.∴−∞<x≤−3 or 0≤x≤3. Upvote · 9 7 Related questions How do find all positive real number x,x, such that x+√x(x+1)+√x(x+2)+√(x+1)(x+2)=2 x+x(x+1)+x(x+2)+(x+1)(x+2)=2? For what values of x is x^x^x^x > (x-1)^(x+1)^(x-1)^(x+1)? Is it true that (1+x) ^-1 = 1-x? How do I evaluate lim x→∞(1−x)x(1+x)x lim x→∞(1−x)x(1+x)x? Given x 1,...,x n x 1,...,x n are positive numbers and x 1∗...∗x n=1 x 1∗...∗x n=1 how could I prove that x 1+...+x n≥1 x 1+...+x n≥1? Is there any set of numbers in which x + 1 < x? If x-1/x=5, what is the value of x⁴+1/X⁴? While 1^1 is 1, will 1^x = 1? (where x is anything)? Is there a value for x for which 1^x is not equal to 1? Are 1 and -1 the only numbers where x^2 = x? How can I prove that lim x→∞[x x+1(x+1)x−(x−1)x x x−1]=1 e lim x→∞[x x+1(x+1)x−(x−1)x x x−1]=1 e? If x^y=y^x does that mean that x and y are the same number? Let x and y be two real positive numbers with x + 2y = 1. What is the smallest possible value of 1 x+1 y 1 x+1 y? If a^x=a^y and x=y, then what type of numbers are a, x, and y? What is 10 to the power of 3? What is the minimum value of (x+y) /(x-y) if x and y are distinct positive real numbers? Related questions How do find all positive real number x,x, such that x+√x(x+1)+√x(x+2)+√(x+1)(x+2)=2 x+x(x+1)+x(x+2)+(x+1)(x+2)=2? For what values of x is x^x^x^x > (x-1)^(x+1)^(x-1)^(x+1)? Is it true that (1+x) ^-1 = 1-x? How do I evaluate lim x→∞(1−x)x(1+x)x lim x→∞(1−x)x(1+x)x? Given x 1,...,x n x 1,...,x n are positive numbers and x 1∗...∗x n=1 x 1∗...∗x n=1 how could I prove that x 1+...+x n≥1 x 1+...+x n≥1? Is there any set of numbers in which x + 1 < x? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
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https://web.ma.utexas.edu/users/m408s/2016/LM11-11-5.html
Applications of Taylor Polynomials When Functions Are Equal to Their Taylor Series So far we have assumed that we could find a power series representation for functions. However, some functions are not equal to their Taylor series, i.e. are not analytic. How can we tell which are and which aren't? We note that f(x)f(x) is equal to it's Taylor series if lim n→∞T n(x)=f(x)lim n→∞T n(x)=f(x), i.e. the series converges to the limit of its partial sums. We define the remainder of the series by R n(x)R n(x), with R n(x)=f(x)−T n(x)R n(x)=f(x)−T n(x). Then f(x)=T n(x)+R n(x)f(x)=T n(x)+R n(x). We can see by this that a function is equal to its Taylor series if its remainder converges to 0; i.e., if a function f f can be differentiated infinitely many times, and lim n→∞R n(x)=0,lim n→∞R n(x)=0, then f f is equal to its Taylor series. We have some theorems to help determine if this remainder converges to zero, by finding a formula and a bound for R n(x)R n(x). (Remainder) Theorem: Let f(x)=T n(x)+R n(x)f(x)=T n(x)+R n(x). If f(n+1)f(n+1) is continuous on an open interval I I that contains a a and x x, then R n(x)=f(n+1)(z)(n+1)!(x−a)n+1 R n(x)=f(n+1)(z)(n+1)!(x−a)n+1 for some z z between a a and x x. Taylor's Inequality: If the (n+1)(n+1)st derivative of f f is bounded by M M on an interval of radius d d around x=a x=a, then |R n(x)|≤M(n+1)!(x−a)n+1.|R n(x)|≤M(n+1)!(x−a)n+1. From this inequality, we can determine that the remainders for e x e x and sin(x)sin⁡(x), for example, go to zero as k→∞k→∞ (see the video below), so these functions are analytic and are equal to their Taylor series. The Remainder Theorem is similar to Rolle's theorem and the Mean Value Theorem, both of which involve a mystery point between a a and b b. The proof of Taylor's theorem involves repeated application of Rolle's theorem, as is explained in this video.
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https://math.stackexchange.com/questions/1137451/how-to-understand-intuitively-the-stolz-cesaro-theorem-for-sequences
real analysis - How to understand intuitively the Stolz-Cesaro Theorem for sequences? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to understand intuitively the Stolz-Cesaro Theorem for sequences? Ask Question Asked 10 years, 7 months ago Modified3 years, 7 months ago Viewed 8k times This question shows research effort; it is useful and clear 25 Save this question. Show activity on this post. I have to give a presentation on the theorem in Real Analysis with a fellow student. While I've looked over the proof and verified that, yes, step B does indeed follow logically from step A, etc. and have internalized the proof to the extent that I can replicate it myself on paper, I still feel that I have made little progress as to why this theorem works the way it does, i.e. what is the intuition behind the theorem, such that it should make sense that it follows the way it does. Therefore I ask: what kind of intuition is there about this theorem? It would also be helpful to understand how the theorem can be used effectively in analysis. I have seen it referred to as a sort of L'Hopital's rule for sequences, which certainly seems to make sense, but I similarly have little intuitive understanding of how that rule works, either. I am asking this question not only for my personal understanding, but also for the sake of being able to present it in an illuminating manner, such that the rest of the class can also come away with the same sort of intuition of how the theorem works and how it is useful. Any help would be greatly appreciated. EDIT: I should point out that the formulation of the theorem we are being tasked to prove is the following: Let {a n}, {b n} be sequences, b n strictly increasing and unbounded. Then if lim for some l \in \mathbb R, then \lim\limits_{n \to \infty} \frac{a_n}{b_n} = l also. real-analysis sequences-and-series intuition Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Feb 4, 2022 at 7:56 hardmath 37.8k 20 20 gold badges 81 81 silver badges 150 150 bronze badges asked Feb 7, 2015 at 8:12 silvascientistsilvascientist 4,305 2 2 gold badges 23 23 silver badges 52 52 bronze badges \endgroup 5 1 \begingroup Haven't seen it yet, but looks like a discrete l'Hospital.\endgroup Dirk –Dirk 2015-02-07 10:11:51 +00:00 Commented Feb 7, 2015 at 10:11 1 \begingroup Very nicely formulated motivation! +1\endgroup Markus Scheuer –Markus Scheuer 2015-02-10 07:54:26 +00:00 Commented Feb 10, 2015 at 7:54 \begingroup Does this answer help?\endgroup Hans Lundmark –Hans Lundmark 2015-02-12 13:54:19 +00:00 Commented Feb 12, 2015 at 13:54 \begingroup This post came to my attention because of the broken link to PlanetMath.org. The PDF file you linked is not available there but is present on the Wayback archive. Instead I suggest the PlanetMath topic page that is essentially identical. See if you judge that it is an agreeable replacement.\endgroup hardmath –hardmath 2022-01-21 22:36:45 +00:00 Commented Jan 21, 2022 at 22:36 \begingroup I went ahead and made the PlanetMath link replacement. Of course you can rollback the edit if you wish.\endgroup hardmath –hardmath 2022-02-04 07:57:31 +00:00 Commented Feb 4, 2022 at 7:57 Add a comment| 3 Answers 3 Sorted by: Reset to default This answer is useful 21 Save this answer. +50 This answer has been awarded bounties worth 50 reputation by silvascientist Show activity on this post. \begingroup As mentioned in the comments, one way to view it is as a discrete version of L'Hopital's rule. In analogy with a derivative of a function defined in the real numbers, \frac{d}{dx} f(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}, a "discrete derivative" of a sequence a_n should be something like \frac{a_{n+h} - a_n}{h} for some small h. Well the smallest h can be in this case is 1, so this discrete derivative should be \Delta a_n \stackrel{\text{def}}{=} \frac{a_{n+1} - a_n}{1} = a_{n+1} - a_n. Now, the usual L'Hopital's rule says that if \lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \ell then \lim_{x \to \infty} \frac{f(x)}{g(x)} = \ell, and, by analogy, the Stolz-Cesaro theorem says that if \lim_{n \to \infty} \frac{\Delta a_n}{\Delta b_n} = \ell then \lim_{n \to \infty} \frac{a_n}{b_n} = \ell. I also like to view the Stolz-Cesaro theorem in terms of summation. Let a_n = \sum_{k=0}^{n} \alpha_k \qquad \text{and} \qquad b_n = \sum_{k=0}^{n} \beta_k, where \beta_k \geq 0 for all k and \sum_{k=0}^{\infty} \beta_k = \infty. Then the theorem says: If \lim_{n \to \infty} \frac{\alpha_{n}}{\beta_{n}} = \ell, \tag{1} then \lim_{n \to \infty} \frac{\sum_{k=0}^{n} \alpha_k}{\sum_{k=0}^{n} \beta_k} = \ell. \tag{2} For illustrative purposes we will assume from here on that \ell > 0. Let's introduce some new notation that should help illuminate the idea of the theorem. We'll write f_n \sim g_n to mean that \lim_{n \to \infty} \frac{f_n}{g_n} = 1. Intuitively, this notation says that f_n and g_n grow (or shrink) at the same rate. So, in terms of this new notation, the hypothesis (1) becomes \alpha_n \sim \ell \beta_n. That is to say, \alpha_n grows or shrinks at the same rate as a constant times \beta_n. The conclusion, (2), becomes \sum_{k=0}^{n} \alpha_k \sim \ell \sum_{k=0}^{n} \beta_k, or, pulling the constant \ell inside the second sum, \sum_{k=0}^{n} \alpha_k \sim \sum_{k=0}^{n} \ell\beta_k. We can just redefine \beta_n' = \ell\beta_n, so what the theorem is says is really: Let \alpha_n and \beta_n be sequences with \beta_n \geq 0 and \sum_{k=0}^{\infty} \beta_k = \infty. If \alpha_n \sim \beta_n, then \sum_{k=0}^{n} \alpha_k \sim \sum_{k=0}^{n} \beta_k. Intuitively, if the summands \alpha_k and \beta_k grow at the same rate, then the sums \sum_{k=0}^{n} \alpha_k and \sum_{k=0}^{n} \beta_k also grow at the same rate. In other words, the sum we get when we replace \alpha_k in the summand by some other equivalent sequence \beta_k will behave approximately the same. We know in fact that \sum_{k=0}^{\infty} \beta_k diverges, so we can also interpret this as saying that the partial sums diverge at the same rate. Appendix: If \ell = 0 then \alpha_n/\beta_n \to \ell = 0 can be interpreted to mean that \alpha_n is "smaller" than \beta_n in the limit, and the conclusion that \sum_{k=0}^{n}\alpha_k \left/ \sum_{k=0}^{n}\beta_k \right. \to \ell = 0 can be interpreted to mean that \sum_{k=0}^{n}\alpha_k is therefore "smaller" than \sum_{k=0}^{n}\beta_k in the limit. Perhaps this (contrived) example will give some idea of the usefulness of this theorem. Suppose we wish to calculate \lim_{n \to \infty} \log(n)^{-1} \sum_{k=1}^{n} \sin(1/k). We could begin by noticing that \sin(1/k) \sim 1/k as k \to \infty, so by Stolz-Cesaro we know that \sum_{k=1}^{n} \sin(1/k) \sim \sum_{k=1}^{n} \frac{1}{k}. This second sum is much easier to estimate. From this we know that \log(n+1) = \int_1^{n+1} \frac{dx}{x} \leq \sum_{k=1}^{n} \frac{1}{k} \leq 1 + \int_1^n \frac{dx}{x} = 1 + \log(n), so \sum_{k=1}^{n} \frac{1}{k} \sim \log(n). Consequently, \sum_{k=1}^{n} \sin(1/k) \sim \log(n). That is, \lim_{n \to \infty} \log(n)^{-1} \sum_{k=1}^{n} \sin(1/k) = 1. Because of Stolz-Cesaro, we know that instead of having to deal with the troublesome summand \sin(1/k) we could replace it with the much simpler summand 1/k and not change the limiting behavior of the sum. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Feb 12, 2015 at 6:06 answered Feb 10, 2015 at 1:06 Antonio VargasAntonio Vargas 25.6k 2 2 gold badges 71 71 silver badges 161 161 bronze badges \endgroup 3 \begingroup As I mentioned in the question, I can see the connection with L'Hopital, and can follow the logic of the proofs to see that, yes, the theorem must be true, but what I'm lacking is some clear visual picture of why it should work the way it does. Or if maybe a simple picture would be too much to ask for, considering it is a rather complex statement, any sort of intuitive justification, diagrammatic or otherwise, would be satisfactory. But the interpretation in terms of summation is interesting, and does help a bit, though I'm not sure how to see that it's equivalent to the original formulation\endgroup silvascientist –silvascientist 2015-02-10 05:49:55 +00:00 Commented Feb 10, 2015 at 5:49 \begingroup And thanks, by the way, for what is really an excellent answer.\endgroup silvascientist –silvascientist 2015-02-10 06:01:45 +00:00 Commented Feb 10, 2015 at 6:01 \begingroup Sure, glad to help. To see that it's equivalent to the original formulation just plug in the definitions I gave for a_n and b_n into the statement of the theorem you gave; so a_{n+1} - a_n = \sum_{k=0}^{n+1} \alpha_k - \sum_{k=0}^{n} \alpha_k = \alpha_{n+1}, and so on. I've also added an example to illustrate that part of the answer.\endgroup Antonio Vargas –Antonio Vargas 2015-02-12 06:09:18 +00:00 Commented Feb 12, 2015 at 6:09 Add a comment| This answer is useful 17 Save this answer. Show activity on this post. \begingroup Here are some examples which could help to better evaluate the usefulness of the Stolz-Cesàro theorem. But first let's state the theorem and an aspect of its proof. Stolz-Cesàro Theorem: Let \lbrace a_n \rbrace, \lbrace b_n \rbrace be sequences, {b_n} strictly positive, increasing and unbounded. Then if \lim\limits_{n \to \infty} \frac {a_{n+1} - a_n}{b_{n+1} - b_n} = l for some l \in \mathbb R, the limit \lim\limits_{n \to \infty} \frac{a_n}{b_n} = l exists and is equal to l. Hint regarding the proof: If you analyse the proof, you see that differences are summed up so that telescoping leaves only the first and last member of the sum. a_{n+1}-a_n\qquad\rightarrow\qquad\sum_{i=N(\epsilon)}^k\left(a_{i+1}-a_i\right) \qquad\rightarrow\qquad a_k-a_{N(\epsilon)} When considering the quotients \frac{a_{n+1}}{b_{k+1}}-\frac{a_{N(\epsilon)}}{b_{k+1}} we observe that the last member vanishes, since {b_k} is increasing and unbounded. So, telescoping is an essence of the proof and also the vanishing of the last member is a hint for us which problems could be successfully attacked by this theorem. Example 1: Let p be a real number, p\ne 1. Compute \lim_{n\rightarrow\infty}\frac{1^p+2^p+\ldots+n^p}{n^{p+1}} Proof: Let a_n=1^p+2^p+\ldots+n^p and b_n=n^{p+1}. We observe \begin{align} \lim_{n\rightarrow\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}&= \lim_{n\rightarrow\infty}\frac{(n+1)^p}{(n+1)^{p+1}-n^{p+1}} \end{align} We invert the fraction and compute instead \begin{align} \lim_{n\rightarrow\infty}\frac{(n+1)^{p+1}-n^{p+1}}{(n+1)^p} \end{align} Dividing both the numerator and denominator by (n+1)^{p+1}, we obtain \begin{align}\lim_{n\rightarrow\infty}\frac{1-\left(1-\frac{1}{n+1}\right)^{p+1}}{\frac{1}{n+1}}\tag{1} \end{align} Now introducing h=\frac{1}{n+1} and f(x)=(1-x)^{p+1} and (1) becomes -\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h}=-f^{\prime}(0)=p+1 We conclude that the required limit is \begin{align} \frac{1}{p+1}&\ &\qquad\qquad\qquad\qquad\qquad\Box \end{align} Observe, that using the difference a_{n+1}-a_{n} eliminates the sum, leaving a simpler expression (n+1)^{p+1} in the numerator. Example 2: If (u_n)n is a sequence of positive real numbers and if \lim{n\rightarrow\infty}\frac{u_{n+1}}{u_n}=u>0, then \lim_{n\rightarrow\infty}\sqrt[n]{u_n}=u. Proof: This is a also direct application of the Stolz-Cesàro Theorem. Indeed, if we let a_n=\ln u_n and b_n=n we get \begin{align} \ln\frac{u_{n+1}}{u_n}=\ln u_{n+1} - \ln u_n&=\frac{a_{n+1}-a_n}{b_{n+1}-b_n}&\ &&\ \ln\sqrt[n]{u_n}=\frac{1}{n}\ln u_n&=\frac{a_n}{b_n}&\ &&\quad\Box\ \end{align} Example 3: Related with the first example is a nice application, namely Sums of integer powers via the Stolz-Cesàro Theorem by S.H. Kung He starts by mentioning the well known fact, that S_n(k)=1^k+2^k+\ldots+n^k is a polynomial in n of degree k+1. Then he states the known formula 1+2+\ldots+n=\frac{1}{2}n^2+\frac{1}{2}n\qquad\qquad n\geq 1 followed by the general approach for (k \geq 1)S_n(k)=1^k+2^k+\ldots+n^k=c_{k+1}n^{k+1}+c_kn^k\ldots c_1 n+ c_0 where he calculates the c_j, 1\leq j \leq k+1 using the Stolz-Cesàro Theorem. He concludes with c_{k+1}=\frac{1}{k+1} and for 1\leq j \leq k\begin{align} c_j=\frac{1}{j}\left((-1)^{k-j}c_{k+1}\binom{k+1}{j-1}+(-1)^{k-j-1}c_k\binom{k}{j-1}+\ldots +c_{j+1}\binom{j+1}{j-1}\right) \end{align} so that c_{k+1},c_k,\ldots,c_0 can be computed recursively to find \begin{array}{cccccc} k\quad&c_{k+1}&c_k&c_{k-1}&\qquad\cdots\qquad &S_k(n)\ 0\quad&1&&&&n\ 1\quad&\frac{1}{2}&\frac{1}{2}&&&\frac{1}{2}n^2+\frac{1}{2}n\ 2\quad&\frac{1}{3}&\frac{1}{2}&\frac{1}{6}&&\frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n\ &&&&&\qquad\qquad\qquad\qquad\qquad\qquad\Box \end{array} Finally some more examples which can be solved using the Stolz-Cesàro Theorem Example 4: Let 0<x_0<1 and x_{n+1}=x_n-x_n^2 for n \geq 0. Compute \lim_{n\rightarrow\infty}nx_n. Example 5: Let x_0\in [-1,1] and x_{n+1}=x_n-\arcsin(\sin^2 x_n) for n\geq 0. Compute \lim_{n\rightarrow\infty}\sqrt{n}x_n Example 6: For an arbitrary number x_0\in(0,\pi) define recursively the sequence (x_n)n by x{n+1}=\sin x_n,n\geq 0. Compute \lim_{n\rightarrow\infty}\sqrt{n}x_n All the examples above, besides the one by H.S. Kung and some more addressing this theorem can be found as examples 340 - 347 in PUTNAM and BEYOND by Razvan Gelca and Titu Andreescu. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Feb 14, 2015 at 15:49 answered Feb 12, 2015 at 20:31 Markus ScheuerMarkus Scheuer 113k 7 7 gold badges 106 106 silver badges 251 251 bronze badges \endgroup 3 \begingroup Very awesome! Detailed, rigorous, examples, etc. Thank you!\endgroup Procore –Procore 2015-11-22 22:18:38 +00:00 Commented Nov 22, 2015 at 22:18 \begingroup@Procore: You're welcome! Many thanks for your nice comment!\endgroup Markus Scheuer –Markus Scheuer 2015-11-22 22:28:20 +00:00 Commented Nov 22, 2015 at 22:28 1 \begingroup No problem - I'm coincidentally working example 1 now, and your answer is very clear and precise (as well as your entire post in my opinion); moreover, relevant solutions/proofs I found online don't go into the details like you did (or at least what I could find online), and I'm the "stickler-type" when writing my mathematics where I always try to include as much detail as I possibly can.\endgroup Procore –Procore 2015-11-22 22:44:23 +00:00 Commented Nov 22, 2015 at 22:44 Add a comment| This answer is useful -1 Save this answer. Show activity on this post. \begingroup The main (or base) application of Stolz-Cesàro is with sequences that suffer from some oscillation. "Intuitively", any oscillating component that prevents the original sequence from converging is averaged away. Moreover, for a sequence that already converges, the avearaging is not harmful: For large n, the averageis taken over only a few initial sequence terms and an enormous numbre of "late" sequence terms. Since late terms are near the limit and the initial terms are only so few, the average tends to be near the limit as well. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Feb 7, 2015 at 8:21 Hagen von EitzenHagen von Eitzen 384k 33 33 gold badges 379 379 silver badges 686 686 bronze badges \endgroup 1 2 \begingroup It seems like you are explaining why Cesaro summation works. The Stolz-Cesaro theorem makes a slightly different claim.\endgroup PhoemueX –PhoemueX 2015-02-07 08:27:19 +00:00 Commented Feb 7, 2015 at 8:27 Add a comment| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 0Why is Stolz–Cesàro is considered to be the l'Hôpital's rule for sequences? 3Understanding when \lim_\limits{n \to \infty} \frac{a_n}{b_n} = \lim_\limits{n \to \infty} \frac{a_0 + a_1+ \cdots + a_n}{b_0+b_1+ \cdots + b_n}. 2Intuition behind stolz caesaro theorem 16Stolz-Cesàro Theorem 3Examine the convergence of the series a_{n+1}=a_n-\arcsin(\sin^2a_n), where a_o\in[-1,1], and find \lim_{n\to\infty}\sqrt{n}a_n. 2Why viewing sequences as functions allows to find the equivalent? 0Inverting a fraction in a limit Related 16Stolz-Cesàro Theorem 20How to prove the Squeeze Theorem for sequences 0Proof that for two converging sequences and two corresponding unbound sets - the limit is equal. 4Calculting Limit using Stolz-Cesaro theorem 3Summary of my understanding of sequences and series' convergence and divergence? 6Is the product of two Cesaro convergent series Cesaro convergent? 10Can Stolz-Cesaro theorem be applied to this problem? 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10442
http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.1113913.html
SOLUTION: what is the difference between nCk and nPk | | | | | | | | | | | | | | | --- --- --- --- --- --- --- | | | | | | | | | | | | --- --- --- --- | | | | | | | | | | --- --- --- | | | | | | | | --- --- --- | | Algebra: Polynomials, rational expressions and equationsSection | SolversSolvers | LessonsLessons | Answers archiveAnswers | | | | | | | | | --- | | --- | | Click here to see ALL problems on Polynomials-and-rational-expressions --- Question 1113913: what is the difference between nCk and nPk Found 2 solutions by ikleyn, Theo: --- Answer by ikleyn(52786) About Me (Show Source): You can put this solution on YOUR website! . ``` nCk is the number of Combinations of n items taken k at a time. C%5Bn%5D%5Ek = n%21%2F%28k%21%2A%28n-k%29%21%29. nPk is the number of Permutations of k items taken from the set of n elements nPk = n(n-1)(n-2) . . . (n-k+1). We consider combinations when we select the groups of k items from the set of n items without looking on their order (when the order does not matter). We consider permutations when we select the groups of k items from the set of n items and consider these groups as ordered sets (when the order does matter). Again, or one more time: nCk is the number of all subsets of k elements of a given set of n elements. nPk is the number of all ordered subsets of k elements of a given set of n elements. ``` ---------------- On Combinations and Permutations see the lessons - Introduction to Permutations - PROOF of the formula on the number of Permutations - Problems on Permutations - Introduction to Combinations - PROOF of the formula on the number of Combinations - Problems on Combinations - OVERVIEW of lessons on Permutations and Combinations in this site. Also, you have this free of charge online textbook in ALGEBRA-II in this site - ALGEBRA-II - YOUR ONLINE TEXTBOOK. The referred lessons are the part of this online textbook under the topic "Combinatorics: Combinations and permutations". Save the link to this textbook together with its description Free of charge online textbook in ALGEBRA-II into your archive and use when it is needed. Answer by Theo(13342) About Me (Show Source): You can put this solution on YOUR website! nCk is the combination formula. nPk is the permutation formula. combination formula does not take order into consideration. permutation formula does. combination formula is nCk = n! / (k! (n-k)!) permutation formula is nPk - n! / (n-k)!. that k! in the denominator of the nCk formula is what removes consideration of order within each set. here's a simple example: consider the set {abcd} n = 4 k = 2 this means you want to get all possible sets of 2 elements each from a set of 4 elements. 4C2 = 4! / (2! 2!) = 6 4P2 = 4! / 2! = 12 4C2 sets would be: ab ac ad bc bd cd 4P2 sets would be: ab ac ad bc bd cd ba ca da cb db dc the 4C2 results consider ab and ba as members of the same set, since order doesn't matter within each set. the 4P2 results consider ab as one set and ba as another separate set, since order within the set does matter. | | | | | | | | | | | | | | --- --- --- --- | | | | | | | | | | --- --- --- | | | | | | | | --- --- --- | | Algebra: Polynomials, rational expressions and equationsSection | SolversSolvers | LessonsLessons | Answers archiveAnswers | | | | | | | | | | | | | --- --- --- | | | | | | | | --- --- --- | | Algebra: Polynomials, rational expressions and equationsSection | SolversSolvers | LessonsLessons | Answers archiveAnswers | | | | | | | | | | | --- --- --- | | Algebra: Polynomials, rational expressions and equationsSection | SolversSolvers | LessonsLessons | Answers archiveAnswers | | | | | | | --- | | --- | | Click here to see ALL problems on Polynomials-and-rational-expressions --- Question 1113913: what is the difference between nCk and nPk Found 2 solutions by ikleyn, Theo: --- Answer by ikleyn(52786) About Me (Show Source): You can put this solution on YOUR website! . ``` nCk is the number of Combinations of n items taken k at a time. C%5Bn%5D%5Ek = n%21%2F%28k%21%2A%28n-k%29%21%29. nPk is the number of Permutations of k items taken from the set of n elements nPk = n(n-1)(n-2) . . . (n-k+1). We consider combinations when we select the groups of k items from the set of n items without looking on their order (when the order does not matter). We consider permutations when we select the groups of k items from the set of n items and consider these groups as ordered sets (when the order does matter). Again, or one more time: nCk is the number of all subsets of k elements of a given set of n elements. nPk is the number of all ordered subsets of k elements of a given set of n elements. ``` ---------------- On Combinations and Permutations see the lessons - Introduction to Permutations - PROOF of the formula on the number of Permutations - Problems on Permutations - Introduction to Combinations - PROOF of the formula on the number of Combinations - Problems on Combinations - OVERVIEW of lessons on Permutations and Combinations in this site. Also, you have this free of charge online textbook in ALGEBRA-II in this site - ALGEBRA-II - YOUR ONLINE TEXTBOOK. The referred lessons are the part of this online textbook under the topic "Combinatorics: Combinations and permutations". Save the link to this textbook together with its description Free of charge online textbook in ALGEBRA-II into your archive and use when it is needed. Answer by Theo(13342) About Me (Show Source): You can put this solution on YOUR website! nCk is the combination formula. nPk is the permutation formula. combination formula does not take order into consideration. permutation formula does. combination formula is nCk = n! / (k! (n-k)!) permutation formula is nPk - n! / (n-k)!. that k! in the denominator of the nCk formula is what removes consideration of order within each set. here's a simple example: consider the set {abcd} n = 4 k = 2 this means you want to get all possible sets of 2 elements each from a set of 4 elements. 4C2 = 4! / (2! 2!) = 6 4P2 = 4! / 2! = 12 4C2 sets would be: ab ac ad bc bd cd 4P2 sets would be: ab ac ad bc bd cd ba ca da cb db dc the 4C2 results consider ab and ba as members of the same set, since order doesn't matter within each set. the 4P2 results consider ab as one set and ba as another separate set, since order within the set does matter. | | | | | | --- | | Click here to see ALL problems on Polynomials-and-rational-expressions --- Question 1113913: what is the difference between nCk and nPk Found 2 solutions by ikleyn, Theo: --- Answer by ikleyn(52786) About Me (Show Source): You can put this solution on YOUR website! . ``` nCk is the number of Combinations of n items taken k at a time. C%5Bn%5D%5Ek = n%21%2F%28k%21%2A%28n-k%29%21%29. nPk is the number of Permutations of k items taken from the set of n elements nPk = n(n-1)(n-2) . . . (n-k+1). We consider combinations when we select the groups of k items from the set of n items without looking on their order (when the order does not matter). We consider permutations when we select the groups of k items from the set of n items and consider these groups as ordered sets (when the order does matter). Again, or one more time: nCk is the number of all subsets of k elements of a given set of n elements. nPk is the number of all ordered subsets of k elements of a given set of n elements. ``` ---------------- On Combinations and Permutations see the lessons - Introduction to Permutations - PROOF of the formula on the number of Permutations - Problems on Permutations - Introduction to Combinations - PROOF of the formula on the number of Combinations - Problems on Combinations - OVERVIEW of lessons on Permutations and Combinations in this site. Also, you have this free of charge online textbook in ALGEBRA-II in this site - ALGEBRA-II - YOUR ONLINE TEXTBOOK. The referred lessons are the part of this online textbook under the topic "Combinatorics: Combinations and permutations". Save the link to this textbook together with its description Free of charge online textbook in ALGEBRA-II into your archive and use when it is needed. Answer by Theo(13342) About Me (Show Source): You can put this solution on YOUR website! nCk is the combination formula. nPk is the permutation formula. combination formula does not take order into consideration. permutation formula does. combination formula is nCk = n! / (k! (n-k)!) permutation formula is nPk - n! / (n-k)!. that k! in the denominator of the nCk formula is what removes consideration of order within each set. here's a simple example: consider the set {abcd} n = 4 k = 2 this means you want to get all possible sets of 2 elements each from a set of 4 elements. 4C2 = 4! / (2! 2!) = 6 4P2 = 4! / 2! = 12 4C2 sets would be: ab ac ad bc bd cd 4P2 sets would be: ab ac ad bc bd cd ba ca da cb db dc the 4C2 results consider ab and ba as members of the same set, since order doesn't matter within each set. the 4P2 results consider ab as one set and ba as another separate set, since order within the set does matter. | |
10443
https://www.quora.com/How-do-you-show-that-the-GCD-of-a-b-and-a-b-is-either-1-or-2-if-GCD-a-b-1
Something went wrong. Wait a moment and try again. GCD & LCM Greatest Common Divisor Proofs (mathematics) Basic Algebra Theory of Numbers Divisors (math) Mathematical Sciences 5 How do you show that the GCD of a+b and a-b is either 1 or 2, if GCD(a,b) =1? Deb P. Choudhury Former Professor at University of Allahabad · Author has 10K answers and 8M answer views · 3y If m = (a+b, a-b), then either m=1 or it has a prime divisor p. Then p divides (a+b)+(a-b) = 2a as well as p divides (a+b)-(a-b) = 2b. Hence either p divides 2, or else it has to divide both a and b, which is ruled out as (a, b) = 1. This is because p|2a ==> p|2 or p|a etc, as p is a prime. Therefore either m=1 or every prime divisor of m is 2, i.e. m is a power of 2. If m=2^n with n > 1, then as before, 4 is a common divisor of 2a and 2b ==> 2 is a common divisor of a and b #, as (a, b) =1. Hence m = 1 or 2. Related questions If the GCD of (a, b) =1, how do you prove that (a+b, a^2 +b^2) =1 or 2? Gcd (a, b) =gcd (a, b-a) how do I prove it? How do I prove if gcd (a, b) = 1 then gcd (a+b, b) = 1? What is the proof that gcd(a,b) =gcd (a+b, b)? How can I show that gcd(a,b) =gcd (a,a+b)? gcd = greatest common divisor. David Townsend Staff Software Engineer (1982–present) · Author has 29.5K answers and 11M answer views · 3y Originally Answered: How can you show that if a and b are integers with (a, b) = 1, then (a+ b, a - b) =1 or 2? · You can show this by using the properties of the gcd: (a,b) = (a,a+b) and (a,b) = (a,a-b) , that is the gcd is not affected by adding or subtracting a multiple of a to b. and the second property (c.a,b) = (c.b).(a,b) when n and a are relatively prime In our problem we have:- (a+b, a-b) = (2a, a-b) = (2,a)(a,a-b) = (2,a)(a,-b) when a and 2 are relatively prime (2,a) = 1 ( a is odd). If 2 and a are not relatively prime, a is even, and (2,a) = 2. Hence (a+b, a-b) = (2.a)(a,b) = 1(a,b) or 2(a,b) Bernard Leak Firmware Developer (2008–present) · Author has 5.8K answers and 5M answer views · 6y Originally Answered: How do we show that if (a,b) = 1 then (a+b, a-b)=1 or 2? · I'm not sure what I'm allowed to assume (i.e., where I may start from). What factors can a+b, a−b have in common? Plainly all such factors must divide the sum 2a and the difference 2b. The highest common factor either has a factor of 2 or it doesn't. If it doesn't, then we don't need and can't use the factor of 2, and the h.c.f. must also be the h.c.f. of a and b, which we already know is 1. Similarly, if the h.c.f. has a factor 2, then a and b are both even or both odd, and we can simply divide by 2 to get (a+b2,a−b2)=(a,b)=1, and then (a+b,a−b)=2. Enrico Gregorio Associate professor in Algebra · Author has 18.4K answers and 16M answer views · 6y Originally Answered: How do we show that if (a,b) = 1 then (a+b, a-b)=1 or 2? · Suppose d is the greatest common divisor of a+b and a−b. Then d is also a common divisor of (a+b)+(a−b)=2a and of (a+b)−(a−b)=2b. If d=2e is even, then e is a common divisor of a and b, hence e=1, so d=2. If d is odd, then it must be a common divisor of a and b, hence d=1. The case d=2 can actually happen, precisely when a and b are both odd, for instance when a=5 and b=3. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. 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If gcd(a,m) =gcd(b,m)=1, then how do you prove that GCD (ab, m) = 1? How do I prove that gcd (a, b, c) = gcd (a, gcd (b, c))? Note that if you want to use any facts about gcd (a, b, c) beyond the definition, you will need to prove them. Assistant Bot · 1y To show that the GCD of a+b and a−b is either 1 or 2 given that gcd(a,b)=1, we can use properties of the GCD and some algebraic manipulations. Let d=gcd(a+b,a−b). By the properties of GCD, d must divide any linear combination of a+b and a−b. Step 1: Calculate d using linear combinations. We can express a+b and a−b as follows: d∣(a+b)+(a−b)=2a d∣(a+b)−(a−b)=2b Thus, d divides both 2a and 2b. Step 2: Analyze the implications of d dividing 2a and 2b. Since d divides 2a and 2b, it follows that: d∣2⋅gcd(a,b) Given that gcd(a,b)=1, we have: d∣2 This means th To show that the GCD of a+b and a−b is either 1 or 2 given that gcd(a,b)=1, we can use properties of the GCD and some algebraic manipulations. Let d=gcd(a+b,a−b). By the properties of GCD, d must divide any linear combination of a+b and a−b. Step 1: Calculate d using linear combinations. We can express a+b and a−b as follows: d∣(a+b)+(a−b)=2a d∣(a+b)−(a−b)=2b Thus, d divides both 2a and 2b. Step 2: Analyze the implications of d dividing 2a and 2b. Since d divides 2a and 2b, it follows that: d∣2⋅gcd(a,b) Given that gcd(a,b)=1, we have: d∣2 This means that d can only be 1 or 2, as those are the only positive divisors of 2. Step 3: Check if d=2 is possible. To determine if d can actually be 2, we need to check if both a+b and a−b are even. If both a and b are odd, then: a+b is even, a−b is even. If one of a or b is even and the other is odd, then: a+b is odd, a−b is odd. Since gcd(a,b)=1, having both a and b even would contradict this condition (as they would share at least 2 as a common divisor). Therefore, a and b must be of different parity (one is odd, the other is even). Thus, the only possibility is that both a+b and a−b are odd, which leads us to conclude that d must be: d=1 Conclusion: We have shown that d=gcd(a+b,a−b) can only be 1 or 2. However, since a and b cannot both be even, d must be 1. Thus, we conclude: gcd(a+b,a−b) is either 1 or 2. Geo Student at Kendriya Vidyalaya (2019–present) · Author has 109 answers and 60.9K answer views · 5y Let a+b be divisible by n Then a+b/n = s That is a+b= ns So a= number Substituting this in a-b a-b= ns-b-b a-b=ns -2b Dividing by n on both side a-b/n=ns/2- b This can be an integer only if n is equal to one or two That is the GCD is 1 or 2 Krunoslav Popović BS in Mathematics, University of Zagreb, Faculty of Science (Graduated 1990) · Author has 900 answers and 399.4K answer views · 3y Originally Answered: How can you show that if a and b are integers with (a, b) = 1, then (a+ b, a - b) =1 or 2? · Let x,y€Z, then: x|a,x|b→x|(a+b)+(a−b),x|2a,x|2b () Now we have the cases: A)x=2y→y|a,y|b→|y|=1→|x|=2 B)x≠2y→x|a,x|b→|x|=1 GCD is always positive so in total x= 1 or 2. Sponsored by CDW Corporation How can AI help your teams make faster decisions? CDW’s AI solutions offer retrieval-augmented generation (RAG) to expedite info with stronger insights. Enrico Gregorio Associate professor in Algebra · Author has 18.4K answers and 16M answer views · 4y Related How can you show that if m and n are integers such that (m, n) = 1, then (m + n, m − n) =1 or 2? The notation (m,n) for the greatest common divisor is (unfortunately) common. It would be better to use the clearer gcd(m,n). Anyway, suppose d>0 divides both m+n and m−n. Then d also divides (m+n)+(m−n)=2m and (m+n)−(m−n)=2n. If d is odd, then it divides both m and n, so, by assumption, d=1. If m and n are of different parity, then both m+n and m−n are odd, implying that d is odd, hence d=1. If m and n have the same parity, then m+n and m−n are both even. Suppose d=2e is even. Then e divides both m and n, so, by assumption, e=1 and therefore d=2. Thus the only possible cases are d=1 or d=2. The l The notation (m,n) for the greatest common divisor is (unfortunately) common. It would be better to use the clearer gcd(m,n). Anyway, suppose d>0 divides both m+n and m−n. Then d also divides (m+n)+(m−n)=2m and (m+n)−(m−n)=2n. If d is odd, then it divides both m and n, so, by assumption, d=1. If m and n are of different parity, then both m+n and m−n are odd, implying that d is odd, hence d=1. If m and n have the same parity, then m+n and m−n are both even. Suppose d=2e is even. Then e divides both m and n, so, by assumption, e=1 and therefore d=2. Thus the only possible cases are d=1 or d=2. The latter case happens if and only if m and n have the same parity. Thus gcd(m+n,m−n)={1if m and n have different parity2if m and n have the same parity Generalize to gcd(m+n,m−n)={gcd(m,n)if m and n have different parity2gcd(m,n)if m and n have the same parity Brian Sittinger PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Author has 8.5K answers and 21M answer views · 3y Related How can I show that if g c d ( a , b ) = 1 then g c d ( a + b , a 2 − a b + b 2 ) = 1 or 3 ? Let d=gcd(a+b,a2−ab+b2).Then, we see that d∣[(a+b)2−(a2−ab+b2)]=3ab. By using this, we next see that d∣[3b(a+b)−3ab]=3b2 and similarly d∣3a2. Therefore d∣gcd(3a2,3b2). However since gcd(3a2,3b2)=3gcd(a,b) and gcd(a,b)=1, it follows (by prime factorizations or otherwise) that gcd(a2,b2)=1. Hence d∣3, and we conclude that d=gcd(a+b,a2−ab+b2)=1 or 3. Promoted by Spokeo Spokeo - People Search | Dating Safety Tool Dating Safety and Cheater Buster Tool · Apr 16 Is there a way to check if someone has a dating profile? Originally Answered: Is there a way to check if someone has a dating profile? Please be reliable and detailed. · Yes, there is a way. If you're wondering whether someone has a dating profile, it's actually pretty easy to find out. Just type in their name and click here 👉 UNCOVER DATING PROFILE. This tool checks a bunch of dating apps and websites to see if that person has a profile—either now or in the past. You don’t need to be tech-savvy or know anything complicated. It works with just a name, and you can also try using their email or phone number if you have it. It’s private, fast, and really helpful if you’re trying to get some peace of mind or just want to know what’s out there. 🔍 HERE IS HOW IT WORK Yes, there is a way. If you're wondering whether someone has a dating profile, it's actually pretty easy to find out. Just type in their name and click here 👉 UNCOVER DATING PROFILE. This tool checks a bunch of dating apps and websites to see if that person has a profile—either now or in the past. You don’t need to be tech-savvy or know anything complicated. It works with just a name, and you can also try using their email or phone number if you have it. It’s private, fast, and really helpful if you’re trying to get some peace of mind or just want to know what’s out there. 🔍 HERE IS HOW IT WORKS: Start by going to this link 👉 UNCOVER DATING PROFILE Enter the person’s name, email address, or phone number. Name and phone number searches usually give the best and most accurate results The site scans billions of public records in just a few seconds. It also scans over 120 dating and social media websites to see if the person has a profile It will ask you a few quick questions to narrow down the results (like location) Once the search is done, you’ll see blurred preview with: Their full name Dating profiles & social media accounts All known phone numbers Current and past addresses A list of family members Any available court or criminal records And more useful background info ⚠️ KEY CALL OUTS ⚠️ Its not free. You will need to pay to see everything, but its pretty cheap. If nothing shows up, it doesn’t always mean they’re in the clear — some people use fake names or burner emails. So it’s worth digging a little deeper just to be sure. If you’re in a situation where you need to know whether someone is still acting single online, this is one of the most effective and low-stress ways to find out. 👉 Check it out here if you’re ready to start your search. ALSO HERE ARE OTHER HELPFUL TOOLS: Dating Research Tool – Search a large database to learn more about who you’re dating. Who’s Texting Your Partner – Discover who your partner is texting or calling, including their name, age, location, and social profiles. Verify People Tool – Confirm if someone is really who they say they are. Find Social Profiles – Locate someone's social media and dating profiles. People Search Directory – Look up someone's phone number and contact details. Dating Safety Check – Review your date’s background to help keep you safe. Frank Levinson Veterinarian (1985–present) · 5y Let c = GCD((a+b), (a-b)). By definition there are integers d and e such that cd = a+b and ce = a-b. Adding these equations: cd + ce = (a+b) + (a - b). Factor out the c and combine terms: c(d+e) = 2a. Now subtracting the equations instead of adding: cd - ce = (a + b) - (a - b). Simplify these: c(d-e) = 2b. So c is a factor of 2a and 2b. Since a and b are given as having no common factors, this c (the sought GCD) can only be 1 or 2 or else it would divide both 2a and 2b and hence would be a common divisor of a and b. Eli Savransky Lived in Buenos Aires, Argentina (1973–1994) · 4y Related How do you solve gcd (a,b) =1 prove gcd ((a+b), (a-b)) = 1 or 2 (number theory, divisibility, gcd and LCM, math)? By contradiction let’s assume there is a prime number k>2 such that k|a+b and k|a-b. It means k|(a+b)+(a-b), so k|2a. Since k>2, then k|a Also, k|(a+b)-(a-b) and hence k|2b. Since k>2, then k|b But then the gcd of a and b should be a multiple of k. Contradiction…. Nikola Lakić Studied at Mathematical Grammar School · 4y Related How do you let m and n be integers such that gcd (m, n) = 1, then show that gcd (m + n, m - n) = 1 or 2 (discrete mathematics, math)? Let us write the gcd(m+n,m-n)=d which implies d|m+n and d|m-n, now by adding and subtracting the last two facts we get d|2m and d|2n and since gcd(m,n)=1 it must be that d|2 but 2 is a prime so d=1 V d=2. Hope this helps. Brad Ballinger Math teacher · Author has 344 answers and 350K answer views · 3y Related How do I prove GCD(a,b) = 1, ab|c iff a|c and b|c? First, I recommend taking the time to write out the claim properly. The way you've got it written now might be taken to mean that GCD(a,b) = 1 and ab|c iff a|c and b|c. That's incorrect, as shown by the example a=b=2, c=4. How would I write it? Like this: Suppose GCD(a,b) = 1. How do I prove ab|c iff a|c and b|c? The point here is to clearly get the GCD assumption out of the iff. Now we continue. It's a little difficult to say how you can prove it without knowing what facts you have at your disposal, but one useful fact here is that ab=GCD(a,b)LCM(a,b). Since we are given that GCD(a,b)=1, this im First, I recommend taking the time to write out the claim properly. The way you've got it written now might be taken to mean that GCD(a,b) = 1 and ab|c iff a|c and b|c. That's incorrect, as shown by the example a=b=2, c=4. How would I write it? Like this: Suppose GCD(a,b) = 1. How do I prove ab|c iff a|c and b|c? The point here is to clearly get the GCD assumption out of the iff. Now we continue. It's a little difficult to say how you can prove it without knowing what facts you have at your disposal, but one useful fact here is that ab=GCD(a,b)LCM(a,b). Since we are given that GCD(a,b)=1, this implies ab=LCM(a,b). This can help with both directions of your iff, particularly if you know that every common multiple of a and b is also a multiple of their LCM. Related questions If the GCD of (a, b) =1, how do you prove that (a+b, a^2 +b^2) =1 or 2? Gcd (a, b) =gcd (a, b-a) how do I prove it? How do I prove if gcd (a, b) = 1 then gcd (a+b, b) = 1? What is the proof that gcd(a,b) =gcd (a+b, b)? How can I show that gcd(a,b) =gcd (a,a+b)? gcd = greatest common divisor. If gcd(a,b) =1, how do you prove that gcd (b-a,b) =1? How do you solve gcd (a,b) =1 prove gcd ((a+b), (a-b)) = 1 or 2 (number theory, divisibility, gcd and LCM, math)? How can I prove that gcd (a, b) = gcd (a − b, b) (when a ≥ b)? If gcd(a,m) =gcd(b,m)=1, then how do you prove that GCD (ab, m) = 1? How do I prove that gcd (a, b, c) = gcd (a, gcd (b, c))? Note that if you want to use any facts about gcd (a, b, c) beyond the definition, you will need to prove them. How do I prove that gcd (a, b) = gcd (ab, LCM (a, b))? How do you show that GCD (a, b) = 1 is reflexive, symmetric, and transitive? When can we use GCD (b,a) equal to GCD (a, a-b)? How do I prove that if gcd (a, c) = 1 and b | c, then gcd (a, b) = 1? How do you prove that if Gcd (a,b, c) = 1 then Gcd (ab,c) = Gcd (AC,b)? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
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https://courses.lumenlearning.com/suny-beginalgebra/chapter/7-1-1-roots/
Identify and Simplify Roots Learning Objectives Square Roots Use square root notation to write principal square roots Simplify principal square roots using factorization Cube Roots Use cube root notation to write cube roots Simplify cube roots using factorization Simplify Square Roots Simplify square roots with variables Determine when a simplified root needs an absolute value Rational Exponents Convert between radical and exponent notation Use the laws of exponents to simplify expressions with rational exponents Use rational exponents to simplify radical expressions We know how to square a number: 52=25 and (−5)2=25 Taking a square root is the opposite of squaring so we can make these statements: 5 is the nonngeative square root of 25 -5 is the negative square root of 25 Find the square roots of the following numbers: 36 81 -49 0 We want to find a number whose square is 36. 62=36 therefore, the nonnegative square root of 36 is 6 and the negative square root of 36 is -6 We want to find a number whose square is 81. 92=81 therefore, the nonnegative square root of 81 is 9 and the negative square root of 81 is -9 We want to find a number whose square is -49. When you square a real number, the result is always positive. Stop and think about that for a second. A negative number times itself is positive, and a positive number times itself is positive. Therefore, -49 does not have square roots, there are no real number solutions to this question. We want to find a number whose square is 0. 02=0 therefore, the nonnegative square root of 0 is 0. We do not assign 0 a sign, so it has only one square root, and that is 0. The notation that we use to express a square root for any real number, a, is as follows: Writing a Square Root The symbol for the square root is called a radical symbol. For a real number, a the square root of a is written as √a The number that is written under the radical symbol is called the radicand. By definition, the square root symbol, √5 always means to find the nonnegative root, called the principal root. √−a is not defined, therefore √a is defined for a>0 Let’s do an example similar to the example from above, this time using square root notation. Note that using the square root notation means that you are only finding the principal root – the nonnegative root. Example Simplify the following square roots: √16 √9 √−9 √52 Show Solution √16. We are looking for a number whose square is 16, so √16=4. We only write the nonnegative root because that is how the root symbol is defined. √9. We are looking for a number whose square is 9, so √9=3. We only write the nonnegative root because that is how the root symbol is defined. √−9. We are looking for a number whose square is -9. There are no real numbers whose square is -9, so this radical is not a real number. √52. We are looking for a number whose square is 52. We already have the number whose square is 52, it’s 5! The last problem in the previous example shows us an important relationship between squares and square roots, and we can summarize it as follows: The square root of a square For a nonnegative real number, a, √a2=a In the video that follows, we simplify more square roots using the fact that √a2=a means finding the principal square root. What if you are working with a number whose square you do not know right away? We can use factoring and the product rule for square roots to find square roots such as √144, or √225. The Product Rule for Square Roots Given that a and b are nonnegative real numbers, √a⋅b=√a⋅√b In the examples that follow we will bring together these ideas to simplify square roots of numbers that are not obvious at first glance: square root of a square, the product rule for square roots factoring Example Simplify √144 Show Solution Determine the prime factors of 144. √144√2⋅72√2⋅2⋅36√2⋅2⋅2⋅18√2⋅2⋅2⋅2⋅9√2⋅2⋅2⋅2⋅3⋅3 Because we are finding a square root, we regroup these factors into squares. √22⋅22⋅32 Now we can use the product rule for square roots and the square root of a square idea to finish finding the square root. √22⋅22⋅32=√22⋅√22⋅√32=2⋅3⋅2=12 Answer √144=12 Example Simplify √225 Show Solution First, factor 225: √225=√5⋅45=√5⋅5⋅9=√5⋅5⋅3⋅3 Because we are finding a square root, we regroup these factors into squares. Finish simplifying with the product rule for roots, and the square of a square idea. √52⋅32=√52⋅√32=5⋅3=15 Answer √225=15 Caution! The square root of a product rule applies when you have multiplication ONLY under the square root. You cannot apply the rule to sums: √a+b≠√a+√b Prove this to yourself with some real numbers: let a = 64 and b = 36, then use the order of operations to simplify each expression. √64+36=√100=10√64+√36=8+6=1410≠14 So far, you have seen examples that are perfect squares. That is, each is a number whose square root is an integer. But many radical expressions are not perfect squares. Some of these radicals can still be simplified by finding perfect square factors. The example below illustrates how to factor the radicand, looking for pairs of factors that can be expressed as a square. Example Simplify. √63 Show Solution Factor 63 √7⋅3⋅3 Regroup factors into squares √7⋅32 Finish simplifying with the product rule for roots, and the square of a square idea. √7⋅32=√7⋅√32=√7⋅3 Since 7 is prime and we can’t write it as a square, it will have to stay under the radical sign. As a matter of convention, we write the constant, 3, in front of the radical. This helps the reader know that the 3 is not under the radical anymore. 3⋅√7 Answer √63=3√7 The final answer 3√7 may look a bit odd, but it is in simplified form. You can read this as “three radical seven” or “three times the square root of seven.” Shortcut This Way In the next example, we take a bit of a shortcut by making use of the common squares we know, instead of using prime factors. It helps to have the squares of the numbers between 0 and 10 fresh in your mind to make simplifying radicals faster. 02=0 22=4 32=9 42=16 52=25 62=36 72=49 82=64 92=81 102=100 Example Simplify. √2,000 Show Solution Factor 2,000 to find perfect squares. √100⋅20=√100⋅4⋅5 100=102,4=22 \begin{array}\sqrt{100\cdot 4\cdot 5}\\= \sqrt{10^2\cdot 4^2\cdot 5}\\=\sqrt{10^2}\cdot\sqrt{4^2}\cdot\sqrt{5}\\=10\cdot4\cdot\sqrt{5} Multiply. 20⋅√5 Answer √2,000=20√5 In this last video, we show examples of simplifying radicals that are not perfect squares. Cube Roots Rubik’s Cune While square roots are probably the most common radical, you can also find the third root, the fifth root, the 10th root, or really any other nth root of a number. Just as the square root is a number that, when squared, gives the radicand, the cube root is a number that, when cubed, gives the radicand. Find the cube roots of the following numbers: 27 8 -8 0 We want to find a number whose cube is 27. 3⋅9=27 and 9=32, so 3/cdot3/cdot3=33=27 We want to find a number whose cube is 8. 2⋅2⋅2=8 the cube root of 8 is 2. We want to find a number whose cube is -8. We know 2 is the cube root of 8, so maybe we can try -2. −2⋅−2⋅−2=−8, so the cube root of -8 is -2. This is different from square roots because multiplying three negative numbers together results in a negative number. We want to find a number whose cube is 0. 0⋅0⋅0, no matter how many times you multiply 0 by itself, you will always get 0. The cube root of a number is written with a small number 3, called the index, just outside and above the radical symbol. It looks like 3√. This little 3 distinguishes cube roots from square roots which are written without a small number outside and above the radical symbol. Caution! Be careful to distinguish between 3√x, the cube root of x, and 3√x, three times the square root of x. They may look similar at first, but they lead you to much different expressions! We can also use factoring to simplify cube roots such as 3√125. You can read this as “the third root of 125” or “the cube root of 125.” To simplify this expression, look for a number that, when multiplied by itself two times (for a total of three identical factors), equals 125. Let’s factor 125 and find that number. Example Simplify. 3√125 Show Solution 125 ends in 5, so you know that 5 is a factor. Expand 125 into 5⋅25. 3√5⋅25 Factor 25 into 5 and 5. 3√5⋅5⋅5 The factors are 5⋅5⋅5, or 53. 3√53 Answer 3√125=5 The prime factors of 125 are 5⋅5⋅5, which can be rewritten as 53. The cube root of a cubed number is the number itself, so 3√53=5. You have found the cube root, the three identical factors that when multiplied together give 125. 125 is known as a perfect cube because its cube root is an integer. Here’s an example of how to simplify a radical that is not a perfect cube. Example Simplify. 3√32m5 Show Solution Factor 32 into prime factors. 3√2⋅2⋅2⋅2⋅2⋅m5 Since you are looking for the cube root, you need to find factors that appear 3 times under the radical. Rewrite 2⋅2⋅2 as 23. 3√23⋅2⋅2⋅m5 Rewrite m5 as m3⋅m2. 3√23⋅2⋅2⋅m3⋅m2 Rewrite the expression as a product of multiple radicals. 3√23⋅3√2⋅2⋅3√m3⋅3√m2 Simplify and multiply. 2⋅3√4⋅m⋅3√m2 Answer 3√32m5=2m3√4m2 In the example below, we use the following idea: 3√(−1)3=−1 to simplify the radical. You do not have to do this, but it may help you recognize cubes more easily when they are nonnegative. Example Simplify. 3√−27x4y3 Show Solution Factor the expression into cubes. Separate the cubed factors into individual radicals. 3√−1⋅27⋅x4⋅y33√(−1)3⋅(3)3⋅x3⋅x⋅y33√(−1)3⋅3√(3)3⋅3√x3⋅3√x⋅3√y3 Simplify the cube roots. −1⋅3⋅x⋅y⋅3√x Answer 3√−27x4y3=−3xy3√x In the video that follows, we show more examples if simplifying cube roots. You could check your answer by performing the inverse operation. If you are right, when you cube −3xy3√x you should get −27x4y3. (−3xy3√x)(−3xy3√x)(−3xy3√x)−3⋅−3⋅−3⋅x⋅x⋅x⋅y⋅y⋅y⋅3√x⋅3√x⋅3√x−27⋅x3⋅y3⋅3√x3−27x3y3⋅x−27x4y3 You can find the odd root of a negative number, but you cannot find the even root of a negative number. This means you can simplify the radicals 3√−81, 5√−64, and 7√−2187, but you cannot simplify the radicals √−100, 4√−16, or 6√−2,500. Let’s look at another example. Example Simplify. 3√−24a5 Show Solution Factor −24 to find perfect cubes. Here, −1 and 8 are the perfect cubes. 3√−1⋅8⋅3⋅a5 Factor variables. You are looking for cube exponents, so you factor a5 into a3 and a2. 3√(−1)3⋅23⋅3⋅a3⋅a2 Separate the factors into individual radicals. 3√(−1)3⋅3√23⋅3√a3⋅3√3⋅a2 Simplify, using the property 3√x3=x. −1⋅2⋅a⋅3√3⋅a2 This is the simplest form of this expression; all cubes have been pulled out of the radical expression. −2a3√3a2 Answer 3√−24a5=−2a3√3a2 The steps to consider when simplifying a radical are outlined below. Simplifying a radical When working with exponents and radicals: If n is odd, n√xn=x. If n is even, n√xn=|x|. (The absolute value accounts for the fact that if x is negative and raised to an even power, that number will be positive, as will the nth principal root of that number.) Example Simplify. √100x2y4 Show Solution Separate factors; look for squared numbers and variables. Factor 100 into 10⋅10. √10⋅10⋅x2⋅y4 Factor y4 into (y2)2. √10⋅10⋅x2⋅(y2)2 Separate the squared factors into individual radicals. √102⋅√x2⋅√(y2)2 Take the square root of each radical . Since you do not know whether x is positive or negative, use |x| to account for both possibilities, thereby guaranteeing that your answer will be positive. 10⋅|x|⋅y2 Simplify and multiply. 10|x|y2 Answer √100x2y4=10|x|y2 You can check your answer by squaring it to be sure it equals 100x2y4. In the last video, we share examples of finding cube roots with negative radicands. Simplify Square Roots with Variables Radical expressions are expressions that contain radicals. Radical expressions come in many forms, from simple and familiar, such as√16, to quite complicated, as in 3√250x4y. Using factoring, you can simplify these radical expressions, too. Radical Simplifying Square Roots Radical expressions will sometimes include variables as well as numbers. Consider the expression √9x6. Simplifying a radical expression with variables is not as straightforward as the examples we have already shown with integers. Consider the expression √x2. This looks like it should be equal to x, right? Let’s test some values for x and see what happens. In the chart below, look along each row and determine whether the value of x is the same as the value of √x2. Where are they equal? Where are they not equal? After doing that for each row, look again and determine whether the value of √x2 is the same as the value of |x|. | x | x2 | √x2 | |x| | --- --- | | −5 | 25 | 5 | 5 | | −2 | 4 | 2 | 2 | | 0 | 0 | 0 | 0 | | 6 | 36 | 6 | 6 | | 10 | 100 | 10 | 10 | Notice—in cases where x is a negative number, √x2≠x! (This happens because the process of squaring the number loses the negative sign, since a negative times a negative is a positive.) However, in all cases √x2=|x|. You need to consider this fact when simplifying radicals that contain variables, because by definition √x2 is always nonnegative. Taking the Square Root of a Radical Expression When finding the square root of an expression that contains variables raised to a power, consider that √x2=|x|. Examples: √9x2=3|x|, and √16x2y2=4|xy| Let’s try it. The goal is to find factors under the radical that are perfect squares so that you can take their square root. Example Simplify. √9x6 Show Solution Factor to find identical pairs. √3⋅3⋅x3⋅x3 Rewrite the pairs as perfect squares, note how we use the power rule for exponents to simplify x6 into a square: x32 √32⋅(x3)2 Separate into individual radicals. √32⋅√(x3)2 Simplify, using the rule that √x2=|x|. 3∣∣x3∣∣ Answer √9x6=3∣∣x3∣∣ Variable factors with even exponents can be written as squares. In the example above, x6=x3⋅x3=∣∣x3∣∣2 and y4=y2⋅y2=(|y2∣∣)2. Let’s try to simplify another radical expression. Example Simplify. √49x10y8 Show Solution Look for squared numbers and variables. Factor 49 into 7⋅7, x10 into x5⋅x5, and y8 into y4⋅y4. √7⋅7⋅x5⋅x5⋅y4⋅y4 Rewrite the pairs as squares. √72⋅(x5)2⋅(y4)2 Separate the squared factors into individual radicals. √72⋅√(x5)2⋅√(y4)2 Take the square root of each radical using the rule that √x2=|x|. 7⋅∣∣x5∣∣⋅y4 Multiply. 7∣∣x5∣∣y4 Answer √49x10y8=7∣∣x5∣∣y4 You find that the square root of 49x10y8 is 7∣∣x5∣∣y4. In order to check this calculation, you could square 7∣∣x5∣∣y4, hoping to arrive at 49x10y8. And, in fact, you would get this expression if you evaluated (7∣∣x5∣∣y4)2. In the video that follows we show several examples of simplifying radicals with variables. Example Simplify. √a3b5c2 Show Solution Factor to find variables with even exponents. √a2⋅a⋅b4⋅b⋅c2 Rewrite b4 as (b2)2. √a2⋅a⋅(b2)2⋅b⋅c2 Separate the squared factors into individual radicals. √a2⋅√(b2)2⋅√c2⋅√a⋅b Take the square root of each radical. Remember that √a2=|a|. |a|⋅b2⋅|c|⋅√a⋅b Simplify and multiply. The entire quantity ab2c can be enclosed in the absolute value sign because b2 will be positive anyway. ∣∣ab2c∣∣√ab Answer √a3b5c2=∣∣ab2c∣∣√ab In the next section, we will explore cube roots, and use the methods we have shown here to simplify them. Cube roots are unique from square roots in that it is possible to have a negative number under the root, such as 3√−125. Rational Exponents Roots can also be expressed as fractional exponents. The square root of a number can be written with a radical symbol or by raising the number to the 12 power. This is illustrated in the table below. | Exponent Form | Root Form | Root of a Square | Simplified | --- --- | | 2512 | √25 | √52 | 5 | | 1612 | √16 | √42 | 4 | | 10012 | √100 | √102 | 10 | Use the example below to familiarize yourself with the different ways to write square roots. Example Fill in the missing cells in the table. | Exponent Form | Root Form | Root of a Square | Simplified | --- --- | | 3612 | | | | | | √81 | | | | | | √122 | | Show Solution | Exponent Form | Root Form | Root of a Square | Simplified | --- --- | | 3612 | √36 | √62 | 6 | | 8112 | √81 | √92 | 9 | | 14412 | √144 | √122 | 12 | In the following video, we show another example of filling in a table to connect the different notation used for roots. We can extend the concept of writing √x=x12 to cube roots. Remember, cubing a number raises it to the power of three. Notice that in these examples, the denominator of the rational exponent is the number 3. | Radical Form | Exponent Form | Integer | --- | 3√8 | 813 | 2 | | 3√8 | 12513 | 5 | | 3√1000 | 100013 | 10 | These examples help us model a relationship between radicals and rational exponents: namely, that the nth root of a number can be written as either n√x or x1n. | Radical Form | Exponent Form | --- | | √x | x12 | | 3√x | x13 | | 4√x | x14 | | … | … | | n√x | x1n | Convert Between Radical and Exponent Notation When faced with an expression containing a rational exponent, you can rewrite it using a radical. In the table above, notice how the denominator of the rational exponent determines the index of the root. So, an exponent of 12 translates to the square root, an exponent of 15 translates to the fifth root or 5√5, and 18 translates to the eighth root or 8√5 . Example Write 3√81 as an expression with a rational exponent. Show Solution The radical form 4√ can be rewritten as the exponent 14. Remove the radical and place the exponent next to the base. 8113 Answer 3√81=8113 Example Express (2x)13 in radical form. Show Solution Rewrite the expression with the fractional exponent as a radical. The denominator of the fraction determines the root, in this case the cube root. 3√2x The parentheses in (2x)13 indicate that the exponent refers to everything within the parentheses. Answer (2x)13=3√2x Remember that exponents only refer to the quantity immediately to their left unless a grouping symbol is used. The example below looks very similar to the previous example with one important difference—there are no parentheses! Look what happens. Example Express 2x12 in radical form. Show Solution Rewrite the expression with the fractional exponent as a radical. The denominator of the fraction determines the root, in this case the cube root. 2√x The exponent refers only to the part of the expression immediately to the left of the exponent, in this case x, but not the 2. Answer 2x12=2√x The next example is intended to help you practice placing a rational exponent on the appropriate terms in an expression that is written in radical form Example Express 43√xy with rational exponents. Show Solution Rewrite the radical using a rational exponent. The root determines the fraction. In this case, the index of the radical is 3, so the rational exponent will be 13. 4(xy)13 Since 4 is outside the radical, it is not included in the grouping symbol and the exponent does not refer to it. Answer 43√xy=4(xy)13 In the next video, we show examples of converting between radical and exponent form. When converting from radical to rational exponent notation, the degree of the root becomes the denominator of the exponent. If you start with a square root, you will have an exponent of 12 on the expression in the radical (the radicand). On the other hand, if you start with an exponent of 13 you will use a cube root. The following statement summarizes this idea. Writing Fractional Exponents Any radical in the form n√a can be written using a fractional exponent in the form a1n. Simplifying Radical Expressions Using Rational Exponents and the Laws of Exponents Let’s explore some radical expressions now and see how to simplify them. Let’s start by simplifying this expression, 3√a6. One method of simplifying this expression is to factor and pull out groups of a3, as shown below in this example. Example Simplify. 3√a6 Show Solution Rewrite by factoring out cubes. 3√a3⋅a3 Write each factor under its own radical and simplify. 3√a3⋅3√a3a⋅a Answer 3√a6=a2 You can also simplify this expression by thinking about the radical as an expression with a rational exponent, and using the principle that any radical in the form n√ax can be written using a fractional exponent in the form axn. Example Simplify. 3√a6 Show Solution Rewrite the radical using a rational exponent. a63 Simplify the exponent. a2 Answer 3√a6=a2 Note that rational exponents are subject to all of the same rules as other exponents when they appear in algebraic expressions. Both simplification methods gave the same result, a2. Depending on the context of the problem, it may be easier to use one method or the other, but for now, you’ll note that you were able to simplify this expression more quickly using rational exponents than when using the “pull-out” method. Let’s try another example. Example Simplify. 4√81x8y3 Show Solution Rewrite the radical using rational exponents. (81x8y3)14 Use the rules of exponents to simplify the expression. 8114⋅x84⋅y34(3⋅3⋅3⋅3)14x2y34(34)14x2y343x2y34 Change the expression with the rational exponent back to radical form. 3x24√y3 Answer 4√81x8y3=3x24√y3 Again, the alternative method is to work on simplifying under the radical by using factoring. For the example you just solved, it looks like this. Example Simplify. 4√81x8y3 Show Solution Rewrite the expression. 4√81⋅4√x8⋅4√y3 Factor each radicand. 4√3⋅3⋅3⋅3⋅4√x2⋅x2⋅x2⋅x2⋅4√y3 4√34⋅4√(x2)4⋅4√y33⋅x2⋅4√y3 Answer 4√81x8y3=3x24√y3 The following video shows more examples of how to simplify a radical expression using rational exponents. Summary The square root of a number is the number which, when multiplied by itself, gives the original number. Principal square roots are always positive and the square root of 0 is 0. You can only take the square root of values that are nonnegative. The square root of a perfect square will be an integer. Other square roots can be simplified by identifying factors that are perfect squares and taking their square root. A radical expression is a mathematical way of representing the nth root of a number. Square roots and cube roots are the most common radicals, but a root can be any number. To simplify radical expressions, look for exponential factors within the radical, and then use the property n√xn=x if n is odd, and n√xn=|x| if n is even to pull out quantities. All rules of integer operations and exponents apply when simplifying radical expressions. A radical can be expressed as an expression with a fractional exponent by following the convention n√am=amn. Rewriting radicals using fractional exponents can be useful in simplifying some radical expressions. When working with fractional exponents, remember that fractional exponents are subject to all of the same rules as other exponents when they appear in algebraic expressions. Candela Citations CC licensed content, Original Image: Shortcut this way.. Provided by: Lumen Learning. License: CC BY: Attribution Simplify Square Roots (Perfect Square Radicands). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Simplify Square Roots (Not Perfect Square Radicands). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Screenshot: Rubik's Cube. Provided by: Lumen Learning. License: CC BY: Attribution Screenshot: Caution. Provided by: Lumen Learning. License: CC BY: Attribution Simplify Cube Roots (Perfect Cube Radicands). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Simplify Cube Roots (Not Perfect Cube Radicands). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Screenshot: radical. Provided by: Lumen Learning. License: CC BY: Attribution Simplify Square Roots with Variables. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Write Basic Expression in Radical Form and Using Rational Exponents. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Write Expressions Using Radicals and Rational Exponents. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Simplify Radicals Using Rational Exponents. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution CC licensed content, Shared previously Unit 16: Radical Expressions and Quadratic Equations, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. Located at: License: CC BY: Attribution Licenses and Attributions CC licensed content, Original Image: Shortcut this way.. Provided by: Lumen Learning. License: CC BY: Attribution Simplify Square Roots (Perfect Square Radicands). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Simplify Square Roots (Not Perfect Square Radicands). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Screenshot: Rubik's Cube. Provided by: Lumen Learning. License: CC BY: Attribution Screenshot: Caution. Provided by: Lumen Learning. License: CC BY: Attribution Simplify Cube Roots (Perfect Cube Radicands). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Simplify Cube Roots (Not Perfect Cube Radicands). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Screenshot: radical. Provided by: Lumen Learning. License: CC BY: Attribution Simplify Square Roots with Variables. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Write Basic Expression in Radical Form and Using Rational Exponents. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Write Expressions Using Radicals and Rational Exponents. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Simplify Radicals Using Rational Exponents. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution CC licensed content, Shared previously Unit 16: Radical Expressions and Quadratic Equations, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. Located at: License: CC BY: Attribution
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https://pmc.ncbi.nlm.nih.gov/articles/PMC2835894/
Epidermal surface lipids - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Learn more: PMC Disclaimer | PMC Copyright Notice Dermatoendocrinol . 2009 Mar-Apr;1(2):72–76. doi: 10.4161/derm.1.2.7811 Search in PMC Search in PubMed View in NLM Catalog Add to search Epidermal surface lipids Apostolos Pappas Apostolos Pappas 1 The Johnson & Johnson Skin Research Center; CPPW, a Division of Johnson & Johnson Consumer Companies, Inc.; Skillman, New Jersey USA Find articles by Apostolos Pappas 1,✉ Author information Article notes Copyright and License information 1 The Johnson & Johnson Skin Research Center; CPPW, a Division of Johnson & Johnson Consumer Companies, Inc.; Skillman, New Jersey USA ✉ Correspondence to: Apostolos Pappas; The Johnson & Johnson Skin Research Center; CPPW, a Division of Johnson & Johnson Consumer Companies, Inc.; Skillman, New Jersey 08558 USA; Email: APappas@its.jnj.com ✉ Corresponding author. Received 2009 Jan 9; Accepted 2009 Jan 12. © 2009 Landes Bioscience PMC Copyright notice PMCID: PMC2835894 PMID: 20224687 Abstract A layer of lipids, which are of both sebaceous and keratinocyte origin, covers the surface of the skin. The apparent composition of surface lipids varies depending on the selected method of sampling. Lipids produced by the epidermal cells are an insignificant fraction of the total extractable surface lipid on areas rich in sebaceous glands. Due to the holocrine activity of the sebaceous gland, its product of secretion (sebum) is eventually released to the surface of the skin and coats the fur as well. Lipids of epidermal origin fill the spaces between the cells, like mortar or cement. The sebaceous lipids are primarily non polar lipids as triglycerides, wax esters and squalene, while epidermal lipids are a mixture of ceramides, free fatty acids and cholesterol. The composition of the sebaceous lipids is unique and intriguing and elevated sebum excretion is a major factor involved in the pathophysiology of acne. Recent studies have elucidated the roles that epidermal surface lipids have on normal skin functions and acne. Key words: lipid, sebaceous, skin, fatty acid, desaturase, wax, squalene, ceramide Introduction The sebaceous gland is now considered to be an important endocrine organ. The holocrine eruption of the sebaceous cells results in the secretion and release of sebum, which eventually coats the surface of the skin and the fur. The majority of the epidermal surface lipids are in fact of sebaceous origin while the lipids produced by the epidermis are an insignificant fraction of the total extractable surface lipid.1 That is more apparent on areas rich in sebaceous glands, where the epidermal origin lipids average between 5 to 10 µg per sq cm, compared with average recoveries of 150 to 300 µg of sebum per sq cm from the forehead. Since this chapter is part of a sebaceous forum, the focus will include both classes of lipids on the surface of the skin. In addition, areas rich in sebaceous glands are the areas that acne lesions are manifested. Human sebum is a mixture of non-polar lipids, mainly triglycerides, wax esters, squalene, fatty acids and smaller amounts of cholesterol, cholesterol esters and diglycerides.2–5 On the other hand, lipids produced by keratinocytes are a mixture of almost equal proportions of free fatty acids, cholesterol and ceramides.6Figure 1 shows the representative structures of the various lipid classes of epidermal surface lipids. Figure 1. Open in a new tab Representative structures of epidermal surface lipids. Sebaceous Lipids The sebaceous lipids are unique and intriguing. According to Nicolaides:7 “two key words characterize the uniqueness of skin lipids: complexity and perversity”. The relative composition of sebum depends on the sampling method used. In particular, if the major components of sebum, triglycerides, are sampled before or after their modification by bacteria, which hydrolyze them to free fatty acids and glycerol.7–11 The mean weight % that is often cited in the literature is given in Table 1. Table 1. Relative sebum composition Lipid class Range weight, %Mean weight, % Triglycerides 20–60 45 Wax esters 23–29 25 Squalene 10–14 12 Free Fatty Acids 5–40 10 Cholesterol and Sterol esters 1–5 4 Diglycerides 1–2 2 Open in a new tab From references 4, 5, 7, 8 and 11 Interestingly, human sebaceous lipids are significantly different in quantity and quality from sebaceous lipids of other species.12–14 The reason for such a unique sebum composition is not understood; however, one can also consider that human skin has a unique texture. In addition, acne is also unique to humans. These seem to be pieces of the same puzzle, which suggest that the unique sebaceous lipids are associated to this odd and human specific disease. Elevated sebum excretion is clearly a major factor involved in the pathophysiology of acne.15–17 Although the majority of lipids produced by all other organs of the human body are alike, the sebaceous gland produces some unique species that cannot be found in any other organ of the body. The synthesis of sapienic acid or wax esters, the accumulation of squalene and the presence of very long chain branched or hydroxylated fatty acids are uncommon in other organs and unique manifestations in sebum.7,12 In other mammals and rodents even higher levels of unique fatty acids exist and with either odd numbers of carbon atoms or branched chains. It is also possible that some of these molecules are in reality products of the resident skin micro flora, since they are more common to bacterial metabolism.18 However, another possibility is that they could be synthesized from branched precursors, products of essential branched amino acid catabolism.4 Sapienic acid. The predominant fatty acid of sebum is the sapienic acid (16:1, Δ6), which has its single double bond at the sixth position from the carboxyl end.7,19 In nature, long chain fatty acids with similar chain length are abundant but there is a predominant preference for the first double bond to be inserted in the 9 th position from the carboxyl end. The 16-carbon isomer with one double bond at the ninth position is the palmitoleic acid, which is naturally found in many tissues and organisms. Sapienic acid is truly unique to sebum and is not found anywhere else in the human body. In addition, humans do not obtain it from the diet since very few plant species have been reported to manufacture this unusual fatty acid.7,18 The elongation of sapienic acid by two carbons and then an additional insertion of another double bond between the fifth and sixth carbon yields sebaleic acid (18:2Δ5, 8), a reaction and metabolite that occurs only in human sebaceous cells. The levels of sapienic acid are multiple folds higher than any of its derivatives, isomers or other monounsaturated fatty acids found in sebum. However, the potential role of sapienic acid in the etiology of acne is still controversial. It has been argued that its presence in sebum correlates with elevated sebum levels,20 while others report that it can be potent against bacteria commonly associated with acne.21-23 Wax esters. Wax esters are also unique to sebaceous cells and are not produced by any other cell in the body. They account for about 25% of the sebaceous gland lipids and their production correlates with sebaceous gland differentiation.5,19 Animal models demonstrated a strong correlation between atrophic sebaceous gland and impaired wax ester synthesis.24,25 Wax ester synthases26,27 have recently been discovered, however additional recent reports28,29 provided evidence that another family of enzymes can also synthesize waxes. Therefore there is not a unique wax synthase and it is apparent that wax ester biosynthesis is still unexplored in humans. Although active wax synthesis correlates with the differentiation of sebaceous cells, it is still unclear if they are the cause or the effect of the differentiation process. In vitro, this is the pathway that its’ expression is usually downregulated no matter if explanted sebaceous glands, tissues, cell preps or transformed cell lines are used. Age and sex related differences have been reported in wax ester synthesis, which also correlates with total sebum output and activity.30–32 In nature, waxes act as protective layers for leaves and fruits of plants, or skin, feathers and fur of animals. Additionally waxes are also found to coat bacteria, algae and fungi.33 Waxes are more resistant to oxidation, hydrolysis and heat than triglycerides or phospholipids. Besides protection they also serve as lubrication aim. Additionally, they are sealing in the internal moisture of tissues while they are preventing their excessive hydration.33 In certain instances the packing and physicochemical properties of the wax crystals demonstrate unusual surface self cleaning properties that repel not only moisture, but together with water any kind of physical or biological invader. This phenomenon has been termed as the “lotus effect”.34 Squalene. There is nothing unique about the synthesis of squalene, which is a precursor of cholesterol. Most mammalian cells synthesize cholesterol, which is an essential molecule for membrane fluidity and structure. Squalene is a long unsaturated hydrocarbon which other tissues quickly convert to lanosterol and finally to cholesterol.6,7 The uniqueness in human sebum is that this cholesterol precursor accumulates in unusually high levels (∼12%) compared to the levels of any other tissue or organ. On the other hand cholesterol accounts for less than 2% of the total sebaceous gland lipids and about 30–35% of the total epidermal lipids. Squalene synthase is the enzyme responsible for the production of squalene and Squalene epoxidase or monooxygenase for its further metabolism. It is possible that in sebaceous cells the activity of these two enzymes is responsible for the accumulation of squalene. Squalene, as a long and highly unsaturated hydrocarbon, is a natural lubricant and has high penetration efficiency; therefore its role could be more than just a precursor of cholesterol. Past reports demonstrated possible roles of squalene oxidation products on UV protection35 but also irritation.36 These products, together with unsaturated free fatty acids, have been reported to be comedogenic. 37,38 Perhaps that is why human sebum transports lipophilic antioxidants as vitamin E39 or humectants as glycerol,40 which play important roles in protecting skin from lipid oxidation and proper barrier function, respectively. Epidermal (Stratum Corneum) Lipids The epidermal lipids of keratinocyte origin play an essential role in the skin’s barrier function. These lipids provide a barrier against the movement of water and electrolytes as well as a barrier against microorganism invasion.41 Especially the permeability barrier, which limits water and minerals, is localized to the outer layers of the epidermis, the stratum corneum (SC). The SC consists of the upper layers of corneocytes, which are terminally differentiated keratinocytes. These cells are imbedded in a lipophilic extra cellular medium composed of equal proportions of ceramides, cholesterol and free fatty acids, providing fundamental limitations to water and electrolyte movement. The above lipid mixture originates from unique structures found in epidermis, the lamellar bodies. The epidermis has a very active synthesis of cholesterol, fatty acids and ceramides. Disruption of the skin’s barrier function results in a rapid and marked increase in epidermal cholesterol and fatty acid synthesis; furthermore inhibitors of these pathways delay the recovery of the barrier function. The increase of sphingolipid synthesis, which precedes ceramide synthesis, is more delayed than cholesterol and fatty acids, but is equally important for the restoration of skin’s barrier function.6,41 The epidermal surface lipids also include unique ceramide species that cannot be found at any other cell type of the human body. For instance the fatty acid esterified to the amide of the (phyto)sphingosine head group can be either an α-hydroxy or an unusually long chain fatty acid. In some instances the fatty acid chain length could be as long as 34 carbon atoms and in others a substantial portion of the epidermal ceramides contain ω-hydroxy long chain fatty acids, which can be in the form of linoleic acid acyl esters. These molecules are believed to have the best barrier properties.42,43 Importance of Epidermal Surface Lipids and Animal Models Genetic knock out (KO) animal models of lipid synthesis have clearly demonstrated the importance of surface lipids in skin physiology and pathology. In these studies, skin and fur abnormalities became the common denominator, once a certain surface lipid pathway is disturbed. The melanocortin-5 receptor (MC5-R) KO resulted in severe defects in water repulsion and thermoregulation due to decreased production of sebaceous lipids.44 The effect of the MC5-R on sebaceous lipid metabolism shed more light to a different path, besides the anticipated role that melanocortins have on pigmentation, obesity or body weight regulation. Two years later Zheng at al.45 demonstrated by positional cloning that the dramatic alopecia manifested in the asebia mouse is due to the lack of a functional Stearoyl-CoA desaturase (Scd1) enzyme activity. The absence of mature sebaceous glands demonstrated the apparent importance of the SCD1 gene and its products (monounsaturated fatty acids) to normal sebaceous gland function in addition to their role in hair development. The same findings were further confirmed in 2001 by the reverse experiment where the SCD1 KO mice were constructed and bared a similar phenotype.25 The revelation that both the asebia’s and the SCD1’s non-functional SCD1 is solely responsible for scant to absent hair and hypoplastic to absent sebaceous glands58 was further supported by the fact that sebaceous glands are scant in certain forms of alopecias.47 The skin of the SCD1 KO mice has also lower than normal levels of triglycerides, wax esters besides the expected lower than normal levels in monounsaturated fatty acids. A similar phenotype was demonstrated in the Acyl CoA:diacylgylcerol acyltransferase 1(DGAT1) KO mouse, where sebaceous gland atrophy and hair loss were also apparent.24 DGAT is the primary triglyceride synthase and exists in two forms, DGAT1 and DGAT2. The two isoforms differ in sequence and localization. 48 DGAT1 is also involved in the synthesis of wax esters, unlike DGAT2,28 and is expressed in most tissues, including the sebaceous gland.24,48 The apparent involvement of DGAT1 in wax ester synthesis is consistent with the observation that there are no or little wax esters in the fur lipids of the DGAT1 KO mouse. The DGAT2 KO animals,49 similarly to SCD2 KO mice,50 do not survive due to an impaired skin barrier function. The animals deficient in SCD-2 demonstrated abnormal lamellar bodies and epidermal maturation proving that the presence of monounsaturated fatty acids is vital also for skin’s barrier component besides the formation of the sebaceous glands. Another animal model that demonstrated the importance of sebaceous and epidermal lipids to skin function is the ELOVL3 KO mouse.51 The Elovl3 gene product is involved in the formation of very long chain fatty acids (VLCFA) and has a distinct expression in the skin that is restricted to sebaceous glands and epithelial cells of hair follicles. Disruption of that gene impaired the formation of neutral lipids that are necessary for skin functions but also resulted in disturbed water barrier and increased trans epidermal water loss. This was caused partly from a disruption in normal lamellar body formation that the deficiency of Elovl3 has caused. The Elovl3-ablated mice had also sparse hair coats and hyperplastic sebaceous glands with unusual lipid content in monounsaturated fatty acid with 20 carbons. Furthermore the elongation of fatty acids was recently further confirmed in another similar animal model. Mice deficient in elongation of very long chain fatty acid-like 4 (ELOVL4) displayed a scaly and wrinkled skin.43 In addition they demonstrated a severely compromised epidermal permeability barrier function, which results in death within a few hours after birth. Skin histology showed an abnormally compacted outer epidermis (SC) and electron microscopy revealed deficient epidermal lamellar body contents. The KO mice had decreased levels in VLCFA (>C28) in ceramide, glucosylceramide and the free fatty-acid fractions, demonstrating the necessity of VLCFA for the synthesis of skin ceramides. Omega-O-acylceramides, that are key hydrophobic components of the extracellular lamellar membranes in mammalian SC, were also fully depleted in the KO model. That reinforces the notion that skin ceramides ought to have unique structures with either VLCFA or omega-hydroxylation that can be additionally esterified to linoleic acid. Essential Fatty Acids Linoleic (18:2, Δ9,12) and α-linolenic acid (18:2, Δ9,12,15) are well known essential fatty acids that come strictly form the diet. As vitamins and minerals do, they survive the digestive tract and are delivered via the systemic circulation to various organs. Sebum analysis demonstrates that the essential fatty acids that come strictly from the diet and their derivatives constitute small amounts in surface lipid samplings.7 However, two intriguing studies, previously unnoticed by the skin research community,52,53 revealed a tight association of essential fatty acids and skin. When guinea pigs were dosed with radioactively labeled linoleic or linolenic acids, skin and fur were the most heavily labeled tissue, suggesting a necessary role for essential fatty acids in sebaceous gland biology. There is very little known about how these essential nutrients are utilized in human sebaceous cells. However, one study has claimed that many acne patients had a linoleic acid deficiency.54 On the other hand, there is substantial evidence that linoleic acid is an essential structural component of skin ceramides. A recent study55 revealed that linoleic acid undergoes a rapid oxidation and degradation in sebaceous cells. Consequently, this activity will allow palmitic acid to be available as the sole substrate to the delta 6 desaturase of sebaceous cells. This is the predominant desaturase of the human sebaceous cells56 that otherwise would have catalyzed the synthesis of more omega-6 derivatives from linoleic acid, since it is the enzyme’s preferred substrate. In fact, linoleic acid was selectively subjected to β-oxidation and this activity correlated with the ability of sebaceous cells to synthesize wax esters, which as noted above is a differentiation marker for the sebaceous cells. Thus, oxidation of linoleic acid is specific to sebaceous cells and correlates with their differentiation and production of sapienate. Epilogue. Epidermal surface lipids contribute to normal skin functions as the barrier function and the maintenance of healthy skin and fur. Consequently, they contribute to aging and to the conditioning and defense of this organ. The idea that the unusual lipids found on skin’s surface make the skin unfriendly to fungi and bacteria has gained more attention. Even if the major component of sebum, the triglycerides, are hydrolyzed by bacteria to fatty acids; these are unusual enough to orchestrate together with other perverse lipids a unique mechanism that will select, which organism is an enemy and which is desirable on our skin. Additional studies are required before we have a complete understanding of the roles of the epidermal surface lipids on acne. Better analytical techniques would help to increase our understanding on their role and to clarify their complexity. Few years ago the classes of ceramides were expanded from six to nine and with new and modern analytical techniques there is sound evidence that there are more classes of ceramides than previously believed.57,58 The field of epidermal surface lipids is open for many new discoveries and is constantly enhanced by advances in analytical techniques. It is not by coincidence that in an era that genomics is the past, proteomics is the present and metabolomics is the near future the term lipidomics is still a challenging term to most biologists. Undoubtedly future scientists will incorporate all the acquired learning from the various “-omics” fields to advance into the new era of lipidomics. This will eventually shed light on intriguing dermatological diseases such as acne, atopic dermatitis, ichthyosis and many others that scientists do not currently associate with the epidermal surface lipids. Abbreviations SC stratum corneum KO knock out VLCFA very long chain fatty acids Footnotes Previously published online as a Dermato-Endocrinology E-publication: References 1.Greene RS, Downing DT, Pochi PE, Strauss JS. Anatomical variation in the amount and composition of human skin surface lipid. J Invest Dermatol. 1970;54:240–247. doi: 10.1111/1523-1747.ep12280318. [DOI] [PubMed] [Google Scholar] 2.Downing DT, Stewart ME, Strauss JS. Changes in sebum secretion and the sebaceous gland. Clin Geriatr Med. 1989;5:109–114. [PubMed] [Google Scholar] 3.Smith KR, Thiboutot DM. Sebaceous gland lipids: Friend or foe? J Lipid Res. 2008;49:271–281. doi: 10.1194/jlr.R700015-JLR200. [DOI] [PubMed] [Google Scholar] 4.Stewart ME. Sebaceous glands lipids. Seminars in Dermatology. 1992;11:100–105. 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[DOI] [PubMed] [Google Scholar] Articles from Dermato-endocrinology are provided here courtesy of Taylor & Francis ACTIONS View on publisher site PDF (1.1 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction Sebaceous Lipids Epidermal (Stratum Corneum) Lipids Importance of Epidermal Surface Lipids and Animal Models Essential Fatty Acids Abbreviations Footnotes References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
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http://www.econ.ucla.edu/sboard/teaching/econ11_09/econ11_09_handout8.pdf
1 1 PROFIT MAXIMIZATION [See Chap 11] 2 Profit Maximization • A profit-maximizing firm chooses both its inputs and its outputs with the goal of achieving maximum economic profits 3 Model • Firm has inputs (z1,z2). Prices (r1,r2). – Price taker on input market. • Firm has output q=f(z1,z2). Price p. – Price taker in output market. • Firm’s problem: – Choose output q and inputs (z1,z2) to maximise profits. Where: π = pq - r1z1 – r2z2 2 4 One-Step Solution • Choose (z1,z2) to maximise π = pf(z1,z2) - r1z1 – r2z2 • This is unconstrained maximization problem. • FOCs are • Together these yield optimal inputs zi(p,r1,r2). • Output is q(p,r1,r2) = f(z1, z2). This is usually called the supply function. • Profit is π(p,r1,r2) = pq - r1z1 - r2z2 2 2 2 1 1 1 2 1 ) , ( and ) , ( r z z z f p r z z z f p = ∂ = ∂ 5 Example: f(z1,z2)=z1 1/3z2 1/3 • Profit is π = pz1 1/3z2 1/3 - r1z1 – r2z2 • FOCs are • Solving these two eqns, optimal inputs are • Optimal output • Profits 2 3 / 2 2 3 / 1 1 1 3 / 1 2 3 / 2 1 3 1 and 3 1 r z pz r z pz = = − − 2 2 1 3 2 1 2 2 2 1 3 2 1 1 27 1 ) , , ( and 27 1 ) , , ( r r p r r p z r r p r r p z = = 2 1 2 3 / 1 2 3 / 1 1 2 1 9 1 ) ( ) ( ) , , ( r r p z z r r p q = = 2 1 3 2 2 1 1 2 1 27 1 ) , , ( r r p z r z r pq r r p = − − = π 6 Two-Step Solution Step 1: Find cheapest way to obtain output q. c(r1,r2,q) = minz1,z2 r1z1+r2z2 s.t f(z1,z2) ≥ q Step 2: Find profit maximizing output. π(p,r1,r2) = maxq pq - c(r1,r2,q) This is unconstrained maximization problem. • Solving yields optimal output q(r1,r2,p). • Profit is π(p,r1,r2) = pq - c(r1,r2,q). 3 7 Step 2: Output Choice • We wish to maximize pq - c(r1,r2,q) • The FOC is p = dc(r1,r2,q)/dq • That is, p = MC(q) • Intuition: produce more if revenue from unit exceeds the cost from the unit. • SOC: MC’(q)≥0, so MC curve must be upward sloping at optimum. 8 Example: f(z1,z2)=z1 1/3z2 1/3 • From cost slides (p18), c(r1,r2,q) = 2(r1r2)1/2 q3/2 • We wish to maximize π = pq - 2(r1r2)1/2 q3/2 • FOC is p = 3(r1r2)1/2 q1/2 • Rearranging, optimal output is • Profits are 2 1 2 2 1 9 1 ) , , ( r r p r r p q = 2 1 3 2 1 2 1 27 1 ) , , ( ) , , ( r r p q r r c pq r r p = − = π 9 Profit Function • Profits are given by π = pq - c(q) • We can write this as π = pq - AC(q)q = [p-AC(q)]q • We can also write this as where F is fixed cost ∫ ∫ − − = − − = q q F dx x MC p F dx x MC pq 0 0 )] ( [ ) ( π 4 10 Profit Function Left: Profit is distance between two lines. Right: Max profit equals A+B+C. If no fixed cost, this equals A+B+D+E. Supply Functions 11 12 Supply with Fixed Cost output price MC AC p q Maximum profit occurs where p = MC 5 13 Supply with Fixed Cost output price MC AC p q Since p > AC, we have π > 0. 14 Supply with Fixed Cost output price MC AC p q If the price rises to p, the firm will produce q and π > 0 q p 15 Supply with Fixed Cost output price MC AC p = MR q If the price falls to p, we might think the firm chooses q. q p But π < 0 so firm prefers q=0. 6 16 Supply Curve • We can use the marginal cost curve to show how much the firm will produce at every possible market price. • The firm can always choose q=0 – Firm only operates if revenue covers costs. – Firm chooses q=0 if pq<c or p < AC. 17 Supply Function #1 • Increasing marginal costs. • No fixed cost 18 Supply Function #2 • Increasing marginal costs. • Fixed cost 7 19 Supply Function #3 • U-shaped marginal costs. • No fixed cost. 20 Supply Function #4 • Wiggly marginal costs. • No fixed cost 21 Sunk Costs • In the short run, there may be sunk costs (i.e. unavoidable costs). • Let AVC = average variable cost (excluding sunk cost). • Then supply function is given by MC curve if p>AVC. • Firm shuts down if p<AVC. 8 22 Short-Run Supply by a Price-Taking Firm output price SMC SAC SAVC The firm’s short-run supply curve is the SMC curve that is above SAVC 23 Cost Function: Properties 1. π(p,r1,r2) is homogenous of degree 1 in (p,r1,r2) – If prices double, profit equation scales up so optimal choices unaffected and profit doubles. 2. π(p,r1,r2) increases in p and decreases in (r1,r2) 3. Hotelling’s Lemma: – If p rises by ∆p, then π(.) rises by ∆p×q(.) – Optimal output also changes, but effect second order. 4. π(p,r1,r2) is convex in p. ) , , ( ) , , ( 2 1 2 1 r r p q r r p p = ∂ ∂π 24 Convexity and Hotelling’s Lemma • This shows the pseudo-profit functions, when output is fixed, and real profit function. 9 25 Convexity and Hotelling’s Lemma • Hotelling Lemma: B+C ≈ B when ∆p small. • Convexity: C means profit increases more than linearly. When p’ → p’’, profit rises by B+C. 26 Supply Functions: Properties 1. q(p,r1,r2) is homogenous of degree 0 in (p,r1,r2) – If prices double profit equation scales up, so optimal output unaffected. 2. Law of supply – Uses Hotelling’s Lemma and convexity of π(.) – Hence supply curve is upward sloping! 0 ≤       ∂ ∂ ∂ ∂ = ∂ ∂ π p p q p
10447
https://wumbo.net/formulas/angle-between-two-vectors-arc-cosine/
Angle Between Two Vectors (Unsigned) Formula Angle Between Two Vectors (Unsigned) Formula Formula θ=arccos(∥v∥∥w∥v⋅w​) Summary The unsigned angle between two vectors is given by the arc cosine of the result from dividing their dot product by the product of their magnitudes. | Expression | Description | --- | | θ | The angle between the two vectors. | | arccos | The arc cosine function. | | v | The first vector. | | w | The second vector. | How to use This formula calculates the unsigned angle between two vectors using the arc cosine function. For example, the angle between the vectors [4 3​] and [5 0​] is calculated as: arccos(∥v∥∥w∥v⋅w​)=arccos(5 4​)=0.64350… We can visualize the angle between these two vectors like this. The angle returned from this formula is always positive. We can visualize the angles calculated by keeping v in place and visualizing different positions for w like this. The angle returned is unsigned, because it doesn’t describe how to get from the first vector to the second vector. In fact, the order of the vectors doesn’t matter. If you swap v with w, the formula produces the same positive angle. To calculate the signed angle between vectors in two dimensions, see this formula. The advantage of the formula on this page versus the signed version is that it works for higher dimensions. Related Formulas Angle Between Two Vectors (2D) Formula To calculate the signed angle between two vectors you can use the extended arc tangent function. This formula calculates angles between negative 180 degrees and positive 180 degrees. Dot Product Formula The dot product of two vectors is calculated by summing together the product of corresponding elements. Magnitude of Vector Formula The magnitude of a vector is given by the square root of the sum of its components squared. Was this helpful? © 2025 Kurt Bruns
10448
https://pubs.acs.org/doi/10.1021/bi200002a
Arsenate Replacing Phosphate: Alternative Life Chemistries and Ion Promiscuity | Biochemistry Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Manage Preferences Recently Viewedclose modal ACS ACS Publications C&EN CAS Access through institution Log In Arsenate Replacing Phosphate: Alternative Life Chemistries and Ion Promiscuity Cite Citation Citation and abstract Citation and references More citation options Share Share on Facebook X Wechat LinkedIn Reddit Email Bluesky Jump to Abstract Cited By Expand Collapse Back to top Close quick search form clear search Biochemistry All Publications/Website Read Journals Peer-reviewed chemistry research Subjects Specific research topics from across the ACS portfolio Books & References eBooks, monographs, and references C&EN News and analysis in chemistry and chemical engineering Publish Submit a Manuscript Prepare, submit, and track manuscripts Author Resources Author guidelines, templates, publishing policies, and more Purchase Author Services Expert editing, formatting, and graphics support Explore Open Access Open access tools, resources, and news Subscribe Librarians & Account Managers Open Science for Institutions Inquire About Access Help Support FAQ Quick answers to common questions Live Chat with Agent Get real-time help from a support representative Open quick search form Open menu Close menu Read Journals Subjects Books & References C&EN Publish Submit a Manuscript Author Resources Purchase Author Services Explore Open Access Subscribe Librarians & Account Managers Open Science for Institutions Inquire About Access Help Support FAQ Live Chat with Agent Tutorials & Demos Account My Profile Log in Log out Close Publications CONTENT TYPES All Types SUBJECTS Publications: All Types Skip to articleSkip to sidebar Close Next Previous Download Hi-Res ImageDownload to MS-PowerPointCite This:Biochemistry 2011, 50, 7, 1128-1134 ADVERTISEMENT Info Metrics Biochemistry Vol 50/Issue 7 Article Get e-Alerts Cite Citation Citation and abstract Citation and references More citation options Share Share on Facebook X WeChat LinkedIn Reddit Email Bluesky Jump to Abstract Cited By Expand Collapse Current Topic January 7, 2011 Arsenate Replacing Phosphate: Alternative Life Chemistries and Ion Promiscuity Click to copy article link Article link copied! Dan S. Tawfik‡ Ronald E. Viola§ View Author Information View Author Information ‡Department of Biological Chemistry, Weizmann Institute of Science, Rhovoit 76100, Israel §Department of Chemistry, University of Toledo, Toledo, Ohio 43606, United States To whom correspondence should be addressed. E-mail: tawfik@weizmann.ac.il Access Through Access is not provided via Institution Name Loading Institutional Login Options... Access Through Your Institution Add or Change Institution Explore subscriptions for institutions Other Access Options Biochemistry Cite this: Biochemistry 2011, 50, 7, 1128–1134 Click to copy citation Citation copied! Published January 7, 2011 Publication History Received 2 January 2011 Published online 31 January 2011 Published in issue 22 February 2011 review-article Copyright © 2011 American Chemical Society Request reuse permissions Article Views 5564 Altmetric 12 Citations 151 Learn about these metrics close Article Views are the COUNTER-compliant sum of full text article downloads since November 2008 (both PDF and HTML) across all institutions and individuals. These metrics are regularly updated to reflect usage leading up to the last few days. Citations are the number of other articles citing this article, calculated by Crossref and updated daily.Find more information about Crossref citation counts. The Altmetric Attention Score is a quantitative measure of the attention that a research article has received online. Clicking on the donut icon will load a page at altmetric.com with additional details about the score and the social media presence for the given article. Find more information onthe Altmetric Attention Score and how the score is calculated. Abstract Click to copy section link Section link copied! A newly identified bacterial strain that can grow in the presence of arsenate and possibly in the absence of phosphate, has raised much interest, but also fueled an active debate. Can arsenate substitute for phosphate in some or possibly in most of the absolutely essential phosphate-based biomolecules, including DNA? If so, then the possibility of alternative, arsenic-based life forms must be considered. The physicochemical similarity of these two oxyanions speaks in favor of this idea. However, arsenate-esters and arsenate-diesters in particular are extremely unstable in aqueous media. Here, we explore the potential of arsenate to be used as substrate by phosphate-utilizing enzymes. We review the existing literature on arsenate enzymology, that intriguingly, dates back to the 1930s. We address the issue of how and to what degree proteins can distinguish between arsenate and phosphate and what is known in general about oxyanion specificity. We also discuss how phosphate−arsenate promiscuity may affect evolutionary transitions between phosphate- and arsenate-based biochemistry. Finally, we highlight potential applications of arsenate as a structural and mechanistic probe of enzymes whose catalyzed reactions involve the making or breaking of phosphoester bonds. ACS Publications Copyright © 2011 American Chemical Society Subjects what are subjects Article subjects are automatically applied from the ACS Subject Taxonomy and describe the scientific concepts and themes of the article. Anions Genetics Ions Peptides and proteins Phosphates Read this article To access this article, please review the available access options below. Recommended Access through Your Institution You may have access to this article through your institution. 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The number of the statements may be higher than the number of citations provided by ACS Publications if one paper cites another multiple times or lower if scite has not yet processed some of the citing articles. Supporting Supporting 2 Mentioning Mentioning 95 Contrasting Contrasting 0 Explore this article's citation statements onscite.ai powered by This article is cited by 151 publications. Zhan-Biao Ge, Ming-Ming Chen, Wan-Ying Xie, Ke Huang, Fang-Jie Zhao, Peng Wang. Natural Microbial Reactor-Based Sensing Platform for Highly Sensitive Detection of Inorganic Arsenic in Rice Grains. Analytical Chemistry2023, 95 (30) , 11467-11474. Faisal Hossain, Nicholas Balasuriya, M. Mosharraf Hossain, Michael J. Serpe. Orthophosphate Quantification in Water Utilizing an Enzymatic Reaction and a Commercial Glucometer Test Strip. Analytical Chemistry2022, 94 (4) , 2056-2062. J. Michael Sutton, Noha M. El Zahar, Michael G. Bartlett. Oligonucleotide Anion Adduct Formation Using Negative Ion Electrospray Ion-Mobility Mass Spectrometry. Journal of the American Society for Mass Spectrometry2021, 32 (2) , 497-508. Ali Abbaspour Tamijani, Jennifer L. Bjorklund, Logan J. Augustine, Jeffrey G. Catalano, Sara E. Mason. Density Functional Theory and Thermodynamics Modeling of Inner-Sphere Oxyanion Adsorption on the Hydroxylated α-Al2O3(001) Surface. Langmuir2020, 36 (44) , 13166-13180. Samuel I. Etkind, Douglas A. Vander Griend, Timothy M. Swager. Electroactive Anion Receptor with High Affinity for Arsenate. The Journal of Organic Chemistry2020, 85 (15) , 10050-10061. Asma O. Eskhan, Nehal I. Abu-Lail. Force-Averaging DLVO Model Predictions of the Adhesion Strengths Quantified for Pathogenic Listeria monocytogenes EGDe Grown under Variable pH Stresses. Langmuir2020, 36 (30) , 8947-8964. Olushola M. Awoyemi, Seenivasan Subbiah, Kelsey N. Thompson, Anahi Velazquez, Angela Peace, Gregory D. Mayer. Trophic-Level Interactive Effects of Phosphorus Availability on the Toxicities of Cadmium, Arsenic, and Their Binary Mixture in Media-Exposed Scenedesmus acutus and Media and Dietary-Exposed Daphnia pulex. Environmental Science & Technology2020, 54 (9) , 5651-5666. Chenghua Zong, Zijie Zhang, Biwu Liu, Juewen Liu. Adsorption of Arsenite on Gold Nanoparticles Studied with DNA Oligonucleotide Probes. Langmuir2019, 35 (22) , 7304-7311. Rajat K. Doshi, Rajib Mukherjee, and Urmila M. Diwekar . Application of Adsorbate Solid Solution Theory To Design Novel Adsorbents for Arsenic Removal Using CAMD. ACS Sustainable Chemistry & Engineering2018, 6 (2) , 2603-2611. Joseph W. Bennett, Jennifer L. Bjorklund, Tori Z. Forbes, and Sara E. Mason . Systematic Study of Aluminum Nanoclusters and Anion Adsorbates. Inorganic Chemistry2017, 56 (21) , 13014-13028. A. K. Jissy and Ayan Datta . Can Arsenates Replace Phosphates in Natural Biochemical Processes? A Computational Study. The Journal of Physical Chemistry B2013, 117 (28) , 8340-8346. Debasish Jana, Ganesan Mani, and Carola Schulzke . Synthesis of Novel Polyazacryptands for Recognition of Tetrahedral Oxoanions and Their X-ray Structures. Inorganic Chemistry2013, 52 (11) , 6427-6439. Wenhui Zhang and Anthony S. Serianni . Phosphate-Catalyzed Degradation of d-Glucosone in Aqueous Solution Is Accompanied by C1–C2 Transposition. Journal of the American Chemical Society2012, 134 (28) , 11511-11524. Yu Xu, Buyong Ma, and Ruth Nussinov . Structural and Functional Consequences of Phosphate–Arsenate Substitutions in Selected Nucleotides: DNA, RNA, and ATP. The Journal of Physical Chemistry B2012, 116 (16) , 4801-4811. Jitesh Soares (Managing Editor) . We Are Still Alone!. ACS Chemical Biology2011, 6 (2) , 117-117. Nahui Olin Medina-Chávez, Ulises E. Rodriguez-Cruz, Valeria Souza, Susana De la Torre-Zavala, Michael Travisano. 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Heterogeneous Overexpression of Two Oryza sativa Arsenic-Induced RING E3 Ligase4 (OsAIR4.1 and 4.2) Transcripts Enhances Plant Tolerance to Arsenic Stress. Journal of Plant Biology2022, 65 (2) , 121-132. Colin Jackson, Agnes Toth-Petroczy, Rachel Kolodny, Florian Hollfelder, Monika Fuxreiter, Shina Caroline Lynn Kamerlin, Nobuhiko Tokuriki. Adventures on the Routes of Protein Evolution—In Memoriam Dan Salah Tawfik (1955–2021). Journal of Molecular Biology2022, 434 (7) , 167462. Rahul Deogam, Nikhil Kumar Pipil, Navjyoti Chakraborty, Sayan Chatterjee, Ram Singh Purty. In Silico Characterization and Structural Modeling of Proteins Involved in Arsenic Tolerance of Hyper Accumulating Fern Pteris Vittata. 2022, 415-427. Amandeep Cheema, Neera Garg. Differential effectiveness of Arbuscular Mycorrhizae in improving Rhizobial symbiosis by modulating Sucrose metabolism and Antioxidant defense in Chickpea under As stress. Symbiosis2022, 86 (1) , 49-69. Monika Yadav, Nar Singh Chauhan. Spatial distribution of arsenic species in soil ecosystem and their effect on plant physiology. 2022, 197-215. Fang-Jie Zhao, Zhong Tang, Jia-Jun Song, Xin-Yuan Huang, Peng Wang. Toxic metals and metalloids: Uptake, transport, detoxification, phytoremediation, and crop improvement for safer food. Molecular Plant2022, 15 (1) , 27-44. Savan K. Raj, Vartika Sharma, Anshul Yadav, Vaibhav Kulshrestha. Nano-Alumina Wrapped Carbon Microspheres for Ultrahigh Elimination of Pentavalent Arsenic and Fluoride from Water: Experimental and Dft Study. SSRN Electronic Journal2022, 17 . Remediation of heavy metals with nanomaterials. 2022, 97-138. Xianhong Li, Xiaoxia Liu, Neng Cao, Songjun Fang, Caihong Yu. Adaptation mechanisms of arsenic metabolism genes and their host microorganisms in soils with different arsenic contamination levels around abandoned gold tailings. Environmental Pollution2021, 291, 117994. Mohammad Golam Mostofa, Chien Van Ha, Md. 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Environmental toxicants in the brain: A review of astrocytic metabolic dysfunction. Environmental Toxicology and Pharmacology2021, 84, 103608. Srimoyee Koner, Ranjana Pal, Siddhartha Dutta. Arsenic Pollution: An Insight into its Effect in Plant Productivity and Human Health. American Journal of Applied Bio-Technology Research2021, 2 (2) , 22-46. Kanwal Irshad, Kanwal Rehman, Fareeha Fiayyaz, Hina Sharif, Ghulam Murtaza, Shagufta Kamal, Muhammad Sajid Hamid Akash. Role of Heavy Metals in Metabolic Disorders. 2021, 203-219. João Paulo B. Ximenez, Ariane Zamarioli, Melissa A. Kacena, Rommel Melgaço Barbosa, Fernando Barbosa Jr. Association of Urinary and Blood Concentrations of Heavy Metals with Measures of Bone Mineral Density Loss: a Data Mining Approach with the Results from the National Health and Nutrition Examination Survey. Biological Trace Element Research2021, 199 (1) , 92-101. Marcelo Pedrosa Gomes, Raizza Zorman Marques, Clésia Cristina Nascentes, Maria Rita Scotti. Synergistic effects between arbuscular mycorrhizal fungi and rhizobium isolated from As-contaminated soils on the As-phytoremediation capacity of the tropical woody legume Anadenanthera peregrina. International Journal of Phytoremediation2020, 22 (13) , 1362-1371. Manuel Nieto‐Domínguez, Pablo I. Nikel. Intersecting Xenobiology and Neometabolism To Bring Novel Chemistries to Life. ChemBioChem2020, 21 (18) , 2551-2571. Mohamad Baydoun, Anne Da Silva, Raphaël Decou, Maha Krayem, Philippe Hourdin, Karine Cleries, Patrice Fondanèche, Thierry Hak, Véronique Deluchat. Assessment of metal/metalloid occurrence in rivers with their accumulation in macrophyte case study with Myriophyllum alterniflorum. Environmental Monitoring and Assessment2020, 192 (6) Nil Demircan, Gokhan Cucun, Baris Uzilday. Mitochondrial alternative oxidase (AOX1a) is required for the mitigation of arsenic-induced oxidative stress in Arabidopsis thaliana. Plant Biotechnology Reports2020, 14 (2) , 235-245. Oluwatosin Oginni, Gunes A. Yakaboylu, Kaushlendra Singh, Edward M. Sabolsky, Gulcin Unal-Tosun, Deb Jaisi, Samir Khanal, Ajay Shah. Phosphorus adsorption behaviors of MgO modified biochars derived from waste woody biomass resources. Journal of Environmental Chemical Engineering2020, 8 (2) , 103723. Weijia Li, Dengpan Qin. The in situ technique of aqueous binding concentration and diffusion for measurements of arsenate concentrations in lake waters. Analytica Chimica Acta2020, 1106, 139-147. Lakmini P. Egodawatta, Aleicia Holland, Darren Koppel, Dianne F. Jolley. Influence of Soil Phosphate on the Accumulation and Toxicity of Arsenic and Antimony in Choy Sum Cultivated in Individually and Co-contaminated Soils. Environmental Toxicology and Chemistry2020, 39 (6) , 1233-1243. Luis Alberto Saona Acuña, Mariana N. Soria, Patricio G. Villafañe, Tatiana Stepanenko, María Eugenia Farías. Arsenic and Its Biological Role: From Early Earth to Current Andean Microbial Ecosystems. 2020, 275-284. Pinki Sharma, Monika, Khushboo Goyal, Tarun Kumar, Nar Singh Chauhan. Inimical Effects of Arsenic on the Plant Physiology and Possible Biotechnological Solutions to Mitigate Arsenic-Induced Toxicity. 2020, 399-422. Luis A. Saona, Mariana Soria, Patricio G. Villafañe, Agustina I. Lencina, Tatiana Stepanenko, María E. Farías. Andean Microbial Ecosystems: Traces in Hypersaline Lakes About Life Origin. 2020, 167-181. Joseph T. Snow, Philip Holdship, Rosalind E.M. Rickaby. Antagonistic co-limitation through ion promiscuity – On the metal sensitivity of Thalassiosira oceanica under phosphorus stress. Science of The Total Environment2020, 699, 134080. Xiao-Li Zhong, Shao-Hua Wen, Yi Wang, Yu-Xi Luo, Zhi-Mei Li, Ru-Ping Liang, Li Zhang, Jian-Ding Qiu. Colorimetric and electrochemical arsenate assays by exploiting the peroxidase-like activity of FeOOH nanorods. Microchimica Acta2019, 186 (11) Paula A. Lamela, Julio A. Navoni, Roberto D. Pérez, Carlos A. Pérez, Cristian L. 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Are Myriophyllum alterniflorum biomarker responses to arsenic stress differentially affected by hydrodynamic conditions?. Chemosphere2019, 225, 497-506. Wenhong Fan, Dingyuan Liang, Xiangrui Wang, JinQian Ren, Shengtao Xiao, Tingting Zhou. Two-generational effects and recovery of arsenic and arsenate on Daphnia magna in the presence of nano-TiO2. Ecotoxicology and Environmental Safety2019, 172, 136-143. Janusz J. Petkowski, William Bains, Sara Seager. An Apparent Binary Choice in Biochemistry: Mutual Reactivity Implies Life Chooses Thiols or Nitrogen-Sulfur Bonds, but Not Both. Astrobiology2019, 19 (4) , 579-613. Louie L. Wurch, Harriet Alexander, Kyle R. Frischkorn, Sheean T. Haley, Christopher J. Gobler, Sonya T. Dyhrman. Transcriptional Shifts Highlight the Role of Nutrients in Harmful Brown Tide Dynamics. Frontiers in Microbiology2019, 10 Prashanth Suresh Kumar, Leon Korving, Karel J. Keesman, Mark C.M. van Loosdrecht, Geert-Jan Witkamp. Effect of pore size distribution and particle size of porous metal oxides on phosphate adsorption capacity and kinetics. Chemical Engineering Journal2019, 358, 160-169. Shao-Hua Wen, Ru-Ping Liang, Hui-Hui Zeng, Li Zhang, Jian-Ding Qiu. CdSe/ZnS quantum dots coated with carboxy-PEG and modified with the terbium(III) complex of guanosine 5′-monophosphate as a fluorescent nanoprobe for ratiometric determination of arsenate via its inhibition of acid phosphatase activity. Microchimica Acta2019, 186 (1) Rajesh Prajapati, Shivam Yadav, Neelam Atri. Nickel and arsenite‐induced differential oxidative stress and antioxidant responses in two Anabaena species. Journal of Basic Microbiology2018, 58 (12) , 1061-1070. Prashanth Suresh Kumar, Wondesen Workneh Ejerssa, Carita Clarissa Wegener, Leon Korving, Achim Iulian Dugulan, Hardy Temmink, Mark C.M. van Loosdrecht, Geert-Jan Witkamp. Understanding and improving the reusability of phosphate adsorbents for wastewater effluent polishing. Water Research2018, 145, 365-374. Ernest Chi Fru, Nolwenn Callac, Nicole R. Posth, Ariadne Argyraki, Yu-Chen Ling, Magnus Ivarsson, Curt Broman, Stephanos P. Kilias. Arsenic and high affinity phosphate uptake gene distribution in shallow submarine hydrothermal sediments. Biogeochemistry2018, 141 (1) , 41-62. Load all citations Get e-Alerts Get e-Alerts Biochemistry Cite this: Biochemistry 2011, 50, 7, 1128–1134 Click to copy citation Citation copied! Published January 7, 2011 Publication History Received 2 January 2011 Published online 31 January 2011 Published in issue 22 February 2011 Copyright © 2011 American Chemical Society Request reuse permissions Article Views 5564 Altmetric 12 Citations 151 Learn about these metrics close Article Views are the COUNTER-compliant sum of full text article downloads since November 2008 (both PDF and HTML) across all institutions and individuals. These metrics are regularly updated to reflect usage leading up to the last few days. 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Chapter 5 INTRAVENOUS INFUSION: INTRODUCTION Drugs may be administered to patients by one of several routes, including oral, topical, or parenteral routes of administration. Examples of parenteral routes of administration include intravenous, subcutaneous, and intramuscular. Intravenous (IV) drug solutions may be given either as a bolus dose (injected all at once) or infused slowly through a vein into the plasma at a constant or zero-order rate. The main advantage for giving a drug by IV infusion is that IV infusion allows precise control of plasma drug concentrations to fit the individual needs of the patient. For drugs with a narrow therapeutic window (eg, heparin), IV infusion maintains an effective constant plasma drug concentration by eliminating wide fluctuations between the peak (maximum) and trough (minimum) plasma drug concentration. Moreover, the IV infusion of drugs, such as antibiotics, may be given with IV fluids that include electrolytes and nutrients. Furthermore, the duration of drug therapy may be maintained or terminated as needed using IV infusion. Because no drug was present in the body at zero time, drug level rises from zero drug concentration and gradually becomes constant when a plateau or steady-state drug concentration is reached. At steady state, the rate of drug leaving the body is equal to the rate of drug (infusion rate) entering the body. Therefore, at steady state, the rate of change in the plasma drug concentration, dC p/dt = 0, and Based on this simple mass balance relationship, a pharmacokinetic equation for infusion may be derived depending on whether the drug follows one- or two-compartment kinetics. ONE-COMPARTMENT MODEL DRUGS The pharmacokinetics of a drug given by constant IV infusion follows a zero-order input process in which the drug is infused directly into the systemic blood circulation. Equation 5.2, below, gives the plasma drug concentration at any time during the IV infusion, where t is the time for infusion. For most drugs, elimination of drug from the plasma is a first-order process. Therefore, in this one-compartment model, the infused drug follows zero-order input and first-order output. The change in the amount of drug in the body at any time (dD B/dt) during the infusion is the rate of input minus the rate of output. where DB is the amount of drug in the body, R is the infusion rate (zero order), and k is the elimination rate constant (first order). Integration of Equation 5.1 and substitution of DB = Cp VD gives As the drug is infused, the value for time (t) increases in Equation 5.2. At infinite time, t = ∞ , e-kt approaches zero, and Equation 5.2 reduces to Equation 5.4. Steady-State Drug Concentration (C SS) and Time Needed to Reach C SS As stated earlier, the rate of drug leaving the body is equal to the rate of drug entering the body (infusion rate) at steady state. In other words, there is no net change in the amount of drug in the body, DB, as a function of time during steady state. Drug elimination occurs according to first-order elimination rate. Whenever the infusion stops either at steady state or before steady state is reached, the log drug concentration declines according to first-order kinetics with the slope of the elimination curve equal to - k/2.3. If the infusion is stopped before steady state is reached, the slope of the elimination curve remains the same. Mathematically, the time to reach true steady-state drug concentration, C SS, would take an infinite time. The time required to reach the steady state drug concentration in the plasma is dependent on the elimination rate constant of the drug for a constant volume of distribution, as shown in Equation 5.4. Because drug elimination is exponential (first order), the plasma drug concentration becomes asymptotic to the theoretical steady-state plasma drug concentration. For a zero-order elimination process, if the rate of input is greater than the rate of elimination, plasma drug concentration will keep increasing and no steady state will be reached. This is a potentially dangerous situation that will occur when saturation of metabolic process occurs. During the IV infusion, the drug concentration increases in the plasma and the rate of drug elimination increases because rate of elimination is concentration dependent (ie, rate of drug elimination = k Cp). Cp keeps increasing until steady state is reached, at which time the rate of drug input (IV infusion rate) equals the rate of drug output (elimination rate). The resulting plasma drug concentration at steady state (C SS) is related to the rate of infusion and inversely related to the body clearance of the drug, as shown in Equation 5.5. In clinical practice, the activity of the drug will be observed when the drug concentration is close to the desired plasma drug concentration, which is usually the target or desired steady-state drug concentration. The time to reach 90%, 95%, and 99% of the steady-state drug concentration, C SS, may be calculated. For therapeutic purposes, the time for the plasma drug concentration to reach more than 95% of the steady state drug concentration in the plasma is often estimated. As detailed in , after IV infusion of the drug for 5 half-lives, the plasma drug concentration will be between 95% (4.32t 1/2) and 99% (6.65t 1/2) of the steadystate drug concentration. Thus, the time for a drug whose t1/2 is 6 hours to reach at least 95% of the steady state plasma drug concentration will be 5 t1/2, or 5 x 6 hours = 30 hours. An increase in the infusion rate will not shorten the time to reach the steady-state drug concentration. If the drug is given at a more rapid infusion rate, a higher steady-state drug level will be obtained, but the time to reach steady state is the same. This equation may also be obtained with the following approach. At steady state, the rate of infusion equals the rate of elimination. Therefore, the rate of change in the plasma drug concentration is equal to zero. Plasma level-time curve for IV infusions given at rates of R and 2R, respectively. Equation 5.6 shows that the steady-state concentration (C SS) is dependent on the volume of distribution, the elimination rate constant, and the infusion rate. Altering any one of these factors can affect steady-state concentration Notice that in the equation directly above, the time needed to reach steady state is not dependent on the rate of infusion, but only on the elimination half-life. Using similar calculations, the time needed to reach any percentage of the steady-state drug concentration may be obtained. Intravenous infusion may be used to determine total body clearance if the infusion rate and steady-state level are known, as with Equation 5.6 repeated here: INFUSION METHOD FOR CALCULATING PATIENT ELIMINATION HALFLIFE The Cp-versus-time relationship that occurs during an IV infusion (Eq. 5.2) may be used to calculate k, or indirectly the elimination half-life of the drug in a patient. Some information about the elimination half-life of the drug in the population must be known, and one or two plasma samples must be taken at a known time after infusion. Knowing the half-life in the general population helps to determine if the sample is taken at steady state in the patient. To simplify calculation, Equation 5.2 is arranged to solve for k: LOADING DOSE PLUS IV INFUSION: ONE-COMPARTMENT MODEL The loading dose, DL, or initial bolus dose of a drug, is used to obtain desired concentrations as rapidly as possible. The concentration of drug in the body for a one-compartment model after an IV bolus dose is described by Assume that an IV bolus dose DL of the drug is given and that an IV infusion is started at the same time. The total concentration Cp at t hours after the start of infusion is C 1 + C 2, due to the sum contributions of bolus and infusion, or Therefore, if an IV loading dose of R/k is given, followed by an IV infusion, steady-state plasma drug concentrations are obtained immediately and maintained. In this situation, steady state is also achieved in a one-compartment model, since rate in = rate out (R = dD B/dt). IV Infusion with loading dose DL. The loading dose is given by IV bolus injection at the start of the infusion. Plasma drug concentrations decline exponentially after D L whereas they increase exponentially during the infusion. The resulting plasma drug concentration-versus-time curve is a straight line due to the summation of the two curves. In order to maintain instant steady-state level [(dC p/dt) = 0], the loading dose should be equal to R/k. For a one-compartment drug, if the DL and infusion rate are calculated such that C0 and C SS are the same and both DL and infusion are started concurrently, then steady state and C SS will be achieved immediately after the loading dose is administered . Similarly, curve b shows the blood level after a single loading dose of R/k plus infusion from which the concentration desired at steady state is obtained. If the D L is not equal to R/k, then steady state will not occur immediately. If the loading dose given is larger than R/k, the plasma drug concentration takes longer to decline to the concentration desired at steady state (curve a). If the loading dose is lower than R/k, the plasma drug concentrations will increase slowly to desired drug levels (curve c), but more quickly than without any loading dose. Intravenous infusion with loading doses a, b, and c. Curve d represents an IV infusion without a loading dose. Another method for the calculation of loading dose D L is based on knowledge of the desired steady-state drug concentration C SS and the apparent volume of distribution VD for the drug, as shown in Equation 5.18. For many drugs, the desired C SS is reported in the literature as the effective therapeutic drug concentration. The VD and the elimination half-life are also available for these drugs. ESTIMATION OF DRUG CLEARANCE AND V D FROM INFUSION DATA The plasma concentration of a drug during constant infusion was described in terms of volume of distribution and elimination constant k in Equation 5.2. Alternatively, the equation may be described in terms of clearance by substituting for k into Equation 5.2 with k = Cl/V D: The drug concentration in this physiologic model is described in terms of volume of distribution of VD and total body clearance (Cl). The independent parameters are clearance and volume of distribution; k is viewed as a dependent variable that depends on Cl and VD. In this model, the time to reach steady state and the resulting steady-state concentration will be dependent on both clearance and volume of distribution. When a constant volume of distribution is evident, the time to reach steady state is then inversely related to clearance. Thus, drugs with small clearance will take a long time to reach steady state. Although this newer approach is preferred by some clinical pharmacists, the alternative approach to parameter estimation was known for some time in classical pharmacokinetics. Equation 5.21 has been applied in population pharmacokinetics to estimate both Cl and VD in individual patients with one or more data points. However, clearance in patients may differ greatly from subjects in the population, especially subjects with different renal functions. Unfortunately, the plasma samples taken at time equivalent to less than one half-life after infusion was started may not be very discriminating, due to the small change in the drug concentration. Blood samples taken at 3-4 half-lives later are much more reflective of the difference in clearance. LOADING DOSE PLUS IV INFUSION: TWO-COMPARTMENT MODEL Drugs with long half-lives require a loading dose to more rapidly attain steady-state plasma drug levels. It is clinically desirable to achieve rapid therapeutic drug levels by using a loading dose. However, for drugs that follow the two-compartment pharmacokinetic model, the drug distributes slowly into extravascular tissues (compartment 2). Thus, drug equilibrium is not immediate. If a loading dose is given too rapidly, the drug may initially give excessively high concentrations in the plasma (central compartment), which then decreases as drug equilibrium is reached. It is not possible to maintain an instantaneous, stable steady-state blood level for a two compartment model drug with a zero-order rate of infusion. Therefore, a loading dose produces an initial blood level either slightly higher or lower than the steady-state blood level. To overcome this problem, several IV bolus injections given as short intermittent IV infusions may be used as a method for administering a loading dose to the patient. Chapter 8 MULTIPLE-DOSAGE REGIMENS Introduction In earlier chapters of this book we have discussed single-dose drug administration. After single-dose drug administration, the plasma drug level rises above and then falls below the minimum effective concentration (MEC), resulting in a decline in therapeutic effect. To maintain prolonged therapeutic activity, many drugs are given in a multiple-dosage regimen. The plasma levels of drugs given in multiple doses must be maintained within the narrow limits of the therapeutic window (eg, plasma drug concentrations above the MEC but below the minimum toxic concentration or MTC) to achieve optimal clinical effectiveness. Among these drugs are antibacterials, cardiotonics, anticonvulsants, and hormones. Ideally, a dosage regimen is established for each drug to provide the correct plasma level without excessive fluctuation and drug accumulation outside the therapeutic window. For certain drugs, such as antibiotics, a desirable MEC can be determined. Some drugs that have a narrow therapeutic range (eg, digoxin and phenytoin) require definition of the therapeutic minimum and maximum nontoxic plasma concentrations (MEC and MTC, respectively). In calculating a multiple-dose regimen, the desired or target plasma drug concentration must be related to a therapeutic response, and the multiple-dose regimen must be designed to produce plasma concentrations within the therapeutic window. There are two main parameters that can be adjusted in developing a dosage regimen: (1) the size of the drug dose and (2) , the frequency of drug administration (ie, the time interval between doses). DRUG ACCUMULATION To calculate a multiple-dose regimen for a patient or patients, pharmacokinetic parameters are first obtained from the plasma level-time curve generated by single-dose drug studies. With these pharmacokinetic parameters and knowledge of the size of the dose and dosage interval , the complete plasma level-time curve or the plasma level may be predicted at any time after the beginning of the dosage regimen. To calculate multiple-dose regimens, it is necessary to decide whether successive doses of drug will have any effect on the previous dose. The principle of superposition assumes that early doses of drug do not affect the pharmacokinetics of subsequent doses. Therefore, the blood levels after the second, third, or nth dose will overlay or superimpose the blood level attained after the (n -1)th dose. In addition, the AUC (∫∞ 0Cp dt) following the administration of a single dose equals the AUC(∫t2 t1 Cpdt) during a dosing interval at steady state . Simulated data showing blood levels after administration of multiple doses and accumulation of blood levels when equal doses are given at equal time intervals. The principle of superposition allows one to project the plasma drug concentration-time curve of a drug after multiple consecutive doses based on the plasma drug concentration-time curve obtained after a single dose. The basic assumptions are that the drug is eliminated by first-order kinetics and that the pharmacokinetics of the drug after a single dose (first dose) are not altered after taking multiple doses. The plasma drug concentrations after multiple doses may be predicted from the plasma drug concentrations obtained after a single dose. The plasma drug concentrations from 0 to 24 hours are measured after a single dose. A constant dose of drug is given every 4 hours and plasma drug concentrations after each dose are generated using the data after the first dose. Thus, the predicted plasma drug concentration in the patient is the total drug concentration obtained by adding the residual drug concentration obtained after each previous dose. As shown in table below. The superposition principle may be used to predict drug concentrations after multiple doses of many drugs. Because the superposition principle is an overlay method, it may be used to predict drug concentrations after multiple doses given at either equal or unequal dosage intervals. For example, the plasma drug concentrations may be predicted after a drug dose is given every 8 hours, or 3 times a day before meals at 8 AM, 12 noon, and 6 PM. A single oral dose of 350 mg was given and the plasma drug concentrations were measured for 0-24 hr. The same plasma drug concentrations are assumed to occur after doses 2-6. The total plasma drug concentration is the sum of the plasma drug concentrations due to each dose. For this example, V D = 10 L, t 1/2 = 4hr, and k a = 1.5 hr-1. The drug is 100% bioavailable and follows the pharmacokinetics of a one-compartment open model. There are situations, however, in which the superposition principle does not apply. In these cases, the pharmacokinetics of the drug change after multiple dosing due to various factors, including changing pathophysiology in the patient, saturation of a drug carrier system, enzyme induction, and enzyme inhibition. Drugs that follow nonlinear pharmacokinetics, generally do not have predictable plasma drug concentrations after multiple doses using the superposition principle. If the drug is administered at a fixed dose and a fixed dosage interval, as is the case with multiple-dose regimens, the amount of drug in the body will increase and then plateau to a mean plasma level higher than the peak Cp obtained from the initial dose . When the second dose is given after a time interval shorter than the time required to "completely" eliminate the previous dose, drug accumulation will occur in the body. In other words, the plasma concentrations following the second dose will be higher than corresponding plasma concentrations immediately following the first dose. However, if the second dose is given after a time interval longer than the time required to eliminate the previous dose, drug will not accumulate. As repetitive equal doses are given at a constant frequency, the plasma level-time curve plateaus and a steady state is obtained. At steady state, the plasma drug levels fluctuate between C∞ max and C∞ min. Once steady state is obtained, C∞ max and C∞ min are constant and remain unchanged from dose to dose. 1-The C∞ max is important in determining drug safety. The C∞ max should always remain below the minimum toxic concentration. 2-The C∞ max is also a good indication of drug accumulation. If a drug produces the same C∞ max at steady state, compared with the (C n = 1)max after the first dose, then there is no drug accumulation. If C∞ max is much larger than (C n = 1)max, then there is significant accumulation during the multiple-dose regimen. Accumulation is affected by the 1-elimination half-life of the drug and 2-the dosing interval. The index for measuring drug accumulation R is Equation 8.2 shows that drug accumulation measured with the R index depends on the elimination constant and the dosing interval and is independent of the dose. For a drug given in repetitive oral doses, the time required to reach steady state is dependent on the elimination half-life of the drug and is independent of the size of the dose, the length of the dosing interval, and the number of doses. For example, if the 1- dose or 2-dosage interval of the drug is altered as shown below , the time required for the drug to reach steady state is the same, but the final steady- state plasma level changes proportionately. Furthermore, if the drug is given at the same dosing rate but as an infusion (eg, 25 mg/hr), the average plasma drug concentrations (C∞ av) will be the same but the fluctuations between C∞ max and C∞ min will vary . An average steady-state plasma drug concentration is obtained by dividing the area under the curve (AUC) for a dosing period (∫t2 t1Cp dt) by the dosing interval , at steady state. Simulated plasma drug concentration time curves after IV infusion and oral multiple doses for a drug with an elimination half-life of 4 hours and apparent V D of 10 L. IV infusion given at a rate of 25 mg/hr, oral multiple doses are 200 mg every 8 hours, 300 mg every 12 hours, and 600 mg every 24 hours. An equation for the estimation of the time to reach one-half of the steady-state plasma levels or the accumulation half-life has been described by . For IV administration, ka is very rapid (approaches ∞ ); k is very small in comparison to ka and can be omitted in the denominator of Equation 8.3. Thus, Equation 8.3 reduces to Because ka/ka = 1 and log 1 = 0, the accumulation t 1/2 of a drug administered intravenously is the elimination t1/2 of the drug. From this relationship, the time to reach 50% steady-state drug concentrations is dependent on the elimination t 1/2 and not on the dose or dosage interval. As shown in Equation 8.4, the accumulation t 1/2 is directly proportional to the elimination t 1/2. gives the accumulation t1/2 of drugs with various elimination half-lives given by multiple oral doses. From a clinical viewpoint, the time needed to reach 90% of the steady-state plasma concentration is 3.3 times the elimination half-life, whereas the time required to reach 99% of the steady-state plasma concentration is 6.6 times the elimination half-life. It should be noted from that at a constant dose size, the shorter the dosage interval, the larger the dosing rate (mg/hr), and the higher the steady-state drug level. REPETITIVE INTRAVENOUS INJECTIONS The maximum amount of drug in the body following a single rapid IV injection is equal to the dose of the drug. For a one-compartment open model, the drug will be eliminated according to first-order kinetics. If Ƭ is equal to the dosage interval (ie, the time between the first dose and the next dose), then the amount of drug remaining in the body after several hours can be determined with The fraction (f) of the dose remaining in the body is related to the elimination constant (k) and the dosage interval as follows: With any given dose, f depends on k and Ƭ . If Ƭ is large, f will be smaller because DB (the amount of drug remaining in the body) is smaller. tistical_analysis_of_a_new_drug/figures?lo=1 D∞ max can also be calculated directly by the relationship The average amount of drug in the body at steady state, D∞ av , can be found by Equation 8.10 or Equation 8.11. F is the fraction of dose absorbed. For an IV injection, F is equal to 1.0. Equations 8.10 and 8.11 can be used for repetitive dosing at constant time intervals and for any route of administration as long as elimination occurs from the central compartment. where n is the number of doses given and t is the time after the nth dose. At steady state, e -nkƬ approaches zero and Equation 8.20 reduces to MULTIPLE-ORAL-DOSE REGIMEN and present typical cumulation curves for the concentration of drug in the body after multiple oral doses given at a constant dosage interval. The plasma concentration at any time during an oral or extravascular multipledose regimen, assuming a one-compartment model and constant doses and dose interval, can be determined as follows: where n = number of doses, = dosage interval, F = fraction of dose absorbed, and t = time after administration of n doses. LOADING DOSE Since extravascular doses require time for absorption into the plasma to occur, therapeutic effects are delayed until sufficient plasma concentrations are achieved. To reduce the onset time of the drug-that is, the time it takes to achieve the minimum effective concentration (assumed to be equivalent to the C∞ av-a loading (priming) or initial dose of drug is given. The main objective of the loading dose is to achieve desired plasma concentrations, C∞ av as quickly as possible. If the drug follows one-compartment pharmacokinetics, then, steady state is also achieved immediately following the loading dose. Thereafter, a maintenance dose is given to maintain C∞ av and steady state so that the therapeutic effect is also maintained. In practice, a loading dose may be given as a bolus dose or a short-term loading IV infusion. As discussed earlier, the time required for the drug to accumulate to a steady-state plasma level is dependent mainly on its elimination half-life. The time needed to reach 90% of C∞ av is approximately 3.3 half-lives, and the time required to reach 99% of C∞ av is equal to approximately 6.6 half-lives. For a drug with a halflife of 4 hours, it will take approximately 13 and 26 hours to reach 90% and 99% of C∞ av, respectively. For drugs absorbed rapidly in relation to elimination (k a >> k) and are distributed rapidly, the loading dose DL can be calculated as follows: For extremely rapid absorption, as when the product of kaƬ is large or in the case of IV infusion, e -kaƬ becomes approximately zero and Equation 8.42 reduces to The loading dose should approximate the amount of drug contained in the body at steady state. The dose ratio is equal to the loading dose divided by the maintenance dose. As a general rule, the dose ratio should be equal to 2.0 if the selected dosage interval is equal to the elimination half-life.
10450
https://www.unitconverters.net/area/square-millimeter-to-square-meter.htm
Home / Area Conversion / Convert Square Millimeter to Square Meter Convert Square Millimeter to Square Meter Please provide values below to convert square millimeter [mm^2] to square meter [m^2], or vice versa. Square Millimeter to Square Meter Conversion Table | Square Millimeter [mm^2] | Square Meter [m^2] | --- | | 0.01 mm^2 | 1.0E-8 m^2 | | 0.1 mm^2 | 1.0E-7 m^2 | | 1 mm^2 | 1.0E-6 m^2 | | 2 mm^2 | 2.0E-6 m^2 | | 3 mm^2 | 3.0E-6 m^2 | | 5 mm^2 | 5.0E-6 m^2 | | 10 mm^2 | 1.0E-5 m^2 | | 20 mm^2 | 2.0E-5 m^2 | | 50 mm^2 | 5.0E-5 m^2 | | 100 mm^2 | 0.0001 m^2 | | 1000 mm^2 | 0.001 m^2 | How to Convert Square Millimeter to Square Meter 1 mm^2 = 1.0E-6 m^2 1 m^2 = 1000000 mm^2 Example: convert 15 mm^2 to m^2: 15 mm^2 = 15 × 1.0E-6 m^2 = 1.5E-5 m^2 Popular Area Unit Conversions acres to square feet square feet to acres hectare to acres acres to hectare square feet to square meter square meter to square feet acres to square miles square miles to acres square feet to square yards square yards to square feet Convert Square Millimeter to Other Area Units Square Millimeter to Square Kilometer Square Millimeter to Square Centimeter Square Millimeter to Square Micrometer Square Millimeter to Hectare Square Millimeter to Acre Square Millimeter to Square Mile Square Millimeter to Square Yard Square Millimeter to Square Foot Square Millimeter to Square Inch Square Millimeter to Break Square Millimeter to Square Hectometer Square Millimeter to Square Dekameter Square Millimeter to Square Decimeter Square Millimeter to Square Nanometer Square Millimeter to Are Square Millimeter to Barn Square Millimeter to Square Mile (US Survey) Square Millimeter to Square Foot (US Survey) Square Millimeter to Circular Inch Square Millimeter to Township Square Millimeter to Section Square Millimeter to Acre (US Survey) Square Millimeter to Rood Square Millimeter to Square Chain Square Millimeter to Square Rod Square Millimeter to Square Rod (US Survey) Square Millimeter to Square Perch Square Millimeter to Square Pole Square Millimeter to Square Mil Square Millimeter to Circular Mil Square Millimeter to Homestead Square Millimeter to Sabin Square Millimeter to Arpent Square Millimeter to Cuerda Square Millimeter to Plaza Square Millimeter to Varas Castellanas Cuad Square Millimeter to Varas Conuqueras Cuad Square Millimeter to Electron Cross Section
10451
https://www.kodeclik.com/seconds-to-years/
AI Academy Gift Card Learn More Fall 2025 Kodeclik Blog How to convert seconds to years Converting seconds into years involves a straightforward calculation that proves useful in various scientific contexts and everyday scenarios alike. Whether determining the age of celestial bodies or simply gauging the passage of time, mastering the conversion from seconds to years is a fundamental skill. Here’s a comprehensive guide to help you seamlessly make this conversion. Understanding the conversion To convert seconds to years, it's crucial to grasp the relationship between these units of time: Seconds (s): The fundamental SI base unit for measuring time intervals. Years (yr): A unit of time corresponding to the duration of one complete orbit of Earth around the Sun, accounting for leap years and the 365.25 days it encompasses. Conversion Formula This formula breaks down as follows: 60 seconds in a minute 60 minutes in an hour 24 hours in a day 365.25 days in a leap year (accounting for the additional day every four years) By dividing the number of seconds by the total number of seconds in a year, accounting for leap years, you accurately determine the equivalent time in years. Using Conversion Tools: Online calculators and conversion tools simplify this process, ensuring accuracy and efficiency in converting seconds to years without manual calculations. Real-world applications Mastering seconds to years conversion is invaluable in scientific fields such as astronomy and geology. It aids in determining the age of celestial objects, estimating the lifespan of stars based on their radiation periods and transitions between hyperfine levels, and even understanding geological time scales. Use the live form below to explore conversions between seconds and years. Convert Seconds to Years (Type in either textbox to convert) How do we convert seconds to years? Let us think through this in a series of steps. First, you convert seconds to minutes, then minutes to hours, hours to days, and finally days to years. Here's a breakdown of the conversion process. First convert seconds to minutes. There are 60 seconds in a minute. Then, convert minutes to hours. There are 60 minutes in an hour. Next, convert hours to days. There are 24 hours in a day. Finally, convert days to years. In a regular year, there are 365 days. Thus we need to perform a series of divisions, first by 60, then by 60 again, then by 24, and finally by 365. In total we are dividing by 60 times 60 times 24 times 365, or 31,536,000. This relationship can be expressed mathematically as y = s/31,536,000, where s is the number of seconds and y is the number of years. Converting seconds to years in Javascript Let us write a simple Javascript program and embed it in a form so you can use it to do your own calculations! This program is a vanilla one that has the same functionality as the form above but without the styling, error catching, etc. This code creates a simple HTML form with a label, an input textbox, and a button. When the button is clicked, the convert() function is called. This function retrieves the value of the input textbox, converts it to years, and displays the result in a paragraph element with the id "result". As mentioned earlier, the conversion is done by a series of divisions, viz. the number of days in a year (365), the number of hours in a day (24), the number of minutes in an hour (60), and the number of seconds in a minute (60). The form and results will look like: Note that this code assumes that the input is a valid number. It does not perform any input validation or error handling. Don’t try silly things like negative numbers - it will work but the results are not particularly meaningful. Similarly if you give a non-number as input it will result in NaN results (Not a Number). Converting seconds to years in Python Here's a Python function that takes the number of seconds as input and returns the number of years: In this Python function, we divide the input seconds by 60 to get the number of minutes, then divide by 60 again to get the number of hours, then divide by 24 to get the number of days, and finally divide by 365 to get the number of years. Here are some examples of using this Python function: There are other ways to convert seconds to years in Python, such as using the time module or the timedelta class from the "datetime" module. However, the seconds_to_years function we have written is a simple and straightforward way to perform this conversion. Conclusion Converting seconds to years offers insight into time scales ranging from micro to macro. It deepens your understanding of temporal concepts and their applications in scientific research, historical chronology, and everyday life. Whether you're exploring cosmic phenomena or reflecting on personal milestones, knowing how to convert seconds to years enriches your appreciation of time's profound impact on our world. If you liked this blogpost, learn how to convert days to seconds! Want to learn Javascript or Python with us? Sign up for 1:1 or small group classes. Join our mailing list About Kodeclik is an online coding academy for kids and teens to learn real world programming. Kids are introduced to coding in a fun and exciting way and are challeged to higher levels with engaging, high quality content. 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10452
https://www.thesaurus.com/browse/agileness
Advertisement Skip to noun (1) Advertisement agileness noun as in agility Weak matches Advertisement Related Words Words related to agileness are not direct synonyms, but are associated with the word agileness. Browse related words to learn more about word associations. nounas in dexterity From Roget's 21st Century Thesaurus, Third Edition Copyright © 2013 by the Philip Lief Group. Advertisement Advertisement Advertisement Browse Follow us Get the Word of the Day every day! By clicking "Sign Up", you are accepting Dictionary.com Terms & Conditions and Privacy Policies.
10453
https://www.youtube.com/watch?v=OQheVGsD7EY
Coordinate Geometry Triangle Question (2 of 5: Understanding the area) Eddie Woo 1930000 subscribers 69 likes Description 4280 views Posted: 30 Jun 2025 More resources available at www.misterwootube.com 6 comments Transcript: Let's move to part two. Now, this is going to feel really similar, right? Because we are finding the location of another point. We're going to use the equation of a line. In this case, it's the pink line, the normal line. And then we're going to substitute in xals 0 because n just like t is on the y-axis. Okay, so this is hopefully going to feel a bit familiar. Um, but there are some minor tweaks, right? So, remember, it's the equation of the normal that gives us n. By the way, it's N for normal in case you hadn't noticed that. And so to find out the equation of the normal, I'm going to kind of review what I did before. You can see I I use the equation of uh a line using point and then gradient. Now the point is the same. The x1 and the y1 are going to match. The difference is this gradient, right? Remember how we were saying before the normal is perpendicular to the tangent line, right? So therefore, I can say the equation of I'm going to use the same thing I said before. I could just say the equation of the normal, but because I've got names for the points on this line, I'm going to say line a n. When I go and do point gradient form, all the things that I said before about the point will remain the same. So I'm just going to take advantage of the fact that I can write the y - 2 a^ 2. I've got the x - a over here. And the only thing that's different is the gradient of this line, the normal. Right? Now, previously I used 4 A. That was the gradient of the tangent. You may recall going back to coordinate geometry, this is like a couple of years ago. Now, if you're doing a question like this, when you've got two lines that are perpendicular, you can think of it either as you multiply their uh gradients and you get an answer, a product of negative one. Alternatively, I mean, this is a different way of looking at the same thing. The new gradient, the perpendicular gradient is the negative reciprocal. Right? Because if I say m1 m2 =1, their product is indeed -1. Then the gradient of one of them will be the negative reciprocal of the other one. So in this case, I can do that pretty directly. Uh I'll stay with green. I'm going to do this here. I'm going to say it's -1 over 4 a. Uh that's the negative reciprocal. So I'm good to go. And now I just need to expand out and see what happens. Right? Now remember in just like uh we saw before what I'm looking for is to let x equal zero. So in fact I notice at this point I I might not even expand just yet because I'm going to expand use this uh this x term here. But I don't need to worry about that x term. It's going to be zero in a minute. Right? So, I'm just going to go back and I'm going to say, well, even though this is super messy, it's it's in an unsimplified like I haven't manipulated at all from original point gradient form. I can say uh to find n let x= 0 and it will actually do some of the work for me, right? I can go y - 2 a^ 2 = -1 / 4 a. This becomes 0 take away a. So there's one thing one less thing to worry about, right? Uh remembering that I do want the ycoordinate, I'm just going to go ahead and start to uh isolate that and make it the subject. So I'm going to get uh -1 over 4 a a + 2 a 2. You can see me adding that to both sides. And then when I look at this, you can see the negatives are going to cancel and the a's are going to cancel. I've got an a on the numerator and an a on the denominator. So I just end up with 1 over 4 plus 2 a squared. Now pause for a minute. I I would normally instinctively say that looks good. Don't have to do any more simplifying to that. However, I then look back at the question and you can see they've phrased the y-coordinate slightly differently. They've just put it all on one common denominator all over four. So therefore, I'm going to do the same thing. I'm going to say that's 1 over 4 + 8 a^ 2 over 4. You can see I multiplied the 2 a 2 by 4 over 4. So I'm pretty pretty much done at this point, right? So I can say therefore n is at 0 comma 1 + 8 a 2 over 4 full stop. Okay. So we're moving on to part three and this is where it started to introduce this triangle business. Right? So let's look back at the question. It says, show that the area of triangle T A N is equal to and then just like we've seen in parts one and two, they provide us with a solution uh so that we know we don't need to go hunting for what that answer is. Okay, so what's the best way to work out the area triangle T? Well, I'm going to just go ahead and grab uh let's make another copy of this. Move it down a little bit. I can fit it here. All right, let's have a look at triangle TN together. might help if I just make it a little more obvious. Like so. Will that snap? Hooray. There we go. All right. So, this triangle here, T A N. We can work out its area in lots of different ways. By this point in the course, you've learned trigonometry. Um, you've learned all of your formulas related to area um in a triangle, and you can think about all of the stuff you learned even in year seven and 8. There's lots of different, pardon the pun, angles you can take on solving this problem. For me, when I look at this question, and by the way, if you don't want me to spoil it, you should pause and have a think for a minute because once I tell you the way I'm going to do it, uh it's really hard when you see someone or read someone else's solution, it becomes really hard to see your own solution through that and unsee the solution that you looked at that someone else showed you. Right? So I'm I'm kind of delaying here because uh this is actually a point worth highlighting for if you're revising for exams and things like that. You have a go at a past paper. That's a great way to prepare by the way and do it under exam conditions. Absolutely. But if you encounter a question, you're like, "Oh, I don't know how to do this question." A very instinctive thing is, well, let's look up the solutions. This was a past paper after all. The solutions are probably available. What I would say is use the solutions, but use them at your own risk. And definitely don't look at the solutions too quick before you've given it a real go yourself because that can't unsee thing is is a real danger, right? It really eliminates your ability to think carefully through the the problem and uh you you miss uh the value of the problem which is kind of sweating and thinking hard and kind of beating your head against the wall sometimes to try and see the solution. Even though that work is it doesn't feel nice. It sort of there's the word the phrase for it uh is it's cognitively dissonant. Uh it doesn't feel good when you can't work out what the answer is. But that that work that cognitive thinking work is actually what helps you learn. Right? So I think I've stol enough now. I'm going to show you the way that I think is most obvious for this. If you look at this shape and you look at it sideways. Um, I wonder if I don't think Oh, can I? No, I don't think I can rotate this easily. I can flip it. That's not that useful. That's okay. What I'm going to do is um you you might need to turn your head a little bit, but I'm going to think about this side here, T. I'm going to think about that as the base of the triangle. Right? Triangles can be spun around any different direction, right? The base has to be one of the sides, but it really can be any side. uh we're used to thinking of the base as the the bottom side, but of course all you have to do is just look at it from a different direction and you see that any of them can be the base. The reason why this triangle uh rather this side is the best choice I think for the base is because the other piece of information you need to quickly work out the area of the triangle is its height. Now we say height, it's actually shorthand for perpendicular height. Uh we just don't like saying perpendicular because it's such a long word. And if you're looking at this diagram clearly, you can see that the perpendicular height for NT is really easy to work out. It's already been given to us. It's at right angles. This length in here, right? Now, if you can't quite see it, right, to make it really, really obvious, let me just highlight the fact that this length here is equal to the xcoordinate A that we have for capital A. Right? So, this perpendicular height is also little A. All that um is left is to work out what this height is here. Now, uh this is where if if it's not obvious to you how to work out this height, this is where I'm going to pull out a trick, which I often do in questions like this where there's lots of algebra flying around and it's the algebra that makes it hard to to sort of see the logic, right? Let's suppose I wasn't thinking about this algebraically. So, let's suppose I had some values, right? Let's imagine that three is uh sorry n rather is at the um y-coordinate 3 and t is at the y-coordinate 1 or rather negative 1 because it's below the axis. Okay. Now if I'm going from three down to -1 I hope that you can kind of see for yourself that the bottom is just going to be from 1 to 0 that's a unit that's one unit length and then from zero the origin up to three that's three units in length. So it's 3 + 1 which is four. Okay. So what I'm really doing is I'm working out what this length is. Then I'm adding it to this length. Okay. Now the top length is easy to do because all I need is well I need to do the three, right? But for this bottom length, this one down here, I actually need to take the absolute value of -1 or if you like, I'm subtracting the negative which gives me a positive. Okay, all of this is very important because I'm going to use the y-coordinate of n, which I worked out up here. That is equivalent to the the three in this situation. And then I'm going to use the ycoordinate of t. Here it is right here. That's going to be equivalent to the -1 because it's down here, right? So all I have to do is to say length nt. That's what I'm actually working out here. That is the um that's the base of the triangle even though it's oriented vertically. Right? It's going to be equal to what's the top part? It's 1 + 8 a^ 2 on 4 and then I'm going to subtract the y-coordinate of t which we saw before. It's -2 a 2. So I'm going to subtract -2 a 2. Right? What's this going to give us? Well, 1 + 8 a^ 2 over 4 + this is going to be 2 a 2, but I can I can multiply that by 4 over 4. That gives me 1 + 8 a 2 + another 8 a 2 all over 4. So that's giving me 1 + 16 a 2 all over 4. That's the length of nt. And I'm going to use that to work out the area of triangle t a n. It's going to be half times the base, which we just worked out. It's this uh 1 + 16 a 2 all over 4. That's the the base of the triangle. And then I need the height, which we already saw before, is just going to be little a right there. Okay. So, I'm going to multiply that all by a. So on the numerator you can see I'm going to get these terms and on the denominator I'm going to get these terms. So that leaves me with a + 16 a cubed all over 8. That's it. I'm pausing there and going back to the question to see is this kind of what I expected. Now, weirdly a little bit, even though in part two they wanted it all as one fraction, in part three they want these things separate. So fine, it's actually, you know, all good because you can see I've got exactly the answer I was hoping for. There's a over 8 in the first part and then 16 / 8 gives me that 2 a cubed as required. Okay.
10454
https://www.graintab.com/agricultural-economic-outlook-september-2025/
Agricultural Economic Outlook - September 2025 - Graintab 01223 608910 01223 608910 info@crm-group.co.uk info@crm-group.co.uk Subscribe Log in Home ANALYSIS All Analysis & Opinions Growers Strategies Premium Academy Case Study Insights Contact Careers Home ANALYSIS All Analysis & Opinions Growers Strategies Premium Academy Case Study Insights Contact Careers Agricultural Economic Outlook - September 2025 Graintab>Analysis>Agricultural Economic Outlook – September 2025 15th Sep CRM Agri Long term supply and demand trends & funds in focus Long term supply and demand trends –Consumption forecast to outpace production for the remainder of the decade. Fund positions– Funds buy back corn, supporting prices, whilst remaining bearish on wheat. Access all of CRM AgriCommodities’ independent analysis and insights with a subscription. Get research, data and opinions from our team of analysts and advisors to help you make better informed decisions with a unique perspective on markets. If you are already a member and require login details, please get in touch. Log In Subscribe Tags: Energy, Freight, funds, Headline Article, Macroeconomy Categories: Analysis, Grain, Highlight, Opinions & Views ⠀ Graintab is a trading name of CRM Commodities Ltd (CRM Agri) which delivers independent agricultural commodity market research and advisory services. [www.crmagri.co.uk] Read More→ Title Recent post Weekly Oilseed Outlook Weekly Oilseeds Outlook Rapeseed Soybeans Oils & Meals Weekly Grain OutlookWeekly Grain Report Highlights • Wheat - Funds, backed by Australia hopes, maintain short bets • Corn - Shrinking US crop ideas shield market from harvest pressure • Crop condition & weather - EU’s mixed sowings conditions Tags Tags AnalysisBarleyCornGlobal GrainsGlobal OilseedsMacroeconomyMealsOSRpalmRapeseedSoybeanssunflowersUSDAWeekly OpinionsWheat Contact Info Contact Info Phone:01223 608910 Email:info@crm-group.co.uk Daily Grain Markets Weekly highlights - Grain markets again show their knack for surprisesSeptember 26, 2025 Graintab is a trading name of CRM Commodities Ltd (CRM Agri) which delivers independent agricultural commodity market research and advisory services. [www.crmagri.co.uk] Read More→ Recent post Weekly Oilseed Outlook Weekly Oilseeds Outlook Rapeseed Soybeans Oils & Meals Weekly Grain OutlookWeekly Grain Report Highlights • Wheat - Funds, backed by Australia hopes, maintain short bets • Corn - Shrinking US crop ideas shield market from harvest pressure • Crop condition & weather - EU’s mixed sowings conditions Tags AnalysisBarleyCornGlobal GrainsGlobal OilseedsMacroeconomyMealsOSRpalmRapeseedSoybeanssunflowersUSDAWeekly OpinionsWheat Contact Info Phone:01223 608910 Email:info@crm-group.co.uk Daily Grain Markets Weekly highlights - Grain markets again show their knack for surprisesSeptember 26, 2025 © CRM Agri 2024. All rights reserved. Terms and Conditions This website uses cookies to improve your experience. We'll assume you're ok with this, but you can opt-out if you wish. Cookie settingsACCEPT Privacy & Cookies Policy Close Privacy Overview This website uses cookies to improve your experience while you navigate through the website. Out of these cookies, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities... Necessary [x] Necessary Always Enabled Necessary cookies are absolutely essential for the website to function properly. This category only includes cookies that ensures basic functionalities and security features of the website. These cookies do not store any personal information. Non-necessary [x] Non-necessary Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. It is mandatory to procure user consent prior to running these cookies on your website. SAVE & ACCEPT We're offline Leave a message
10455
https://courses.lumenlearning.com/calculus3/chapter/projectile-motion/
Projectile Motion | Calculus III Skip to main content Calculus III Module 3: Vector-Valued Functions Search for: Projectile Motion Learning Outcomes State Kepler’s laws of planetary motion. Now let’s look at an application of vector functions. In particular, let’s consider the effect of gravity on the motion of an object as it travels through the air, and how it determines the resulting trajectory of that object. In the following, we ignore the effect of air resistance. This situation, with an object moving with an initial velocity but with no forces acting on it other than gravity, is known as projectile motion. It describes the motion of objects from golf balls to baseballs, and from arrows to cannonballs. First we need to choose a coordinate system. If we are standing at the origin of this coordinate system, then we choose the positive y-axis to be up, the negative _y-_axis to be down, and the positive _x-_axis to be forward (i.e., away from the thrower of the object). The effect of gravity is in a downward direction, so Newton’s second law tells us that the force on the object resulting from gravity is equal to the mass of the object times the acceleration resulting from to gravity, or Fg\=mgFg\=mg, where FgFg represents the force from gravity and gg represents the acceleration resulting from gravity at Earth’s surface. The value of gg in the English system of measurement is approximately 32 ft/sec232 ft/sec2 and it is approximately 9.8 m/sec29.8 m/sec2 in the metric system. This is the only force acting on the object. Since gravity acts in a downward direction, we can write the force resulting from gravity in the form Fg\=−mgjFg\=−mgj, as shown in the following figure. Figure 1. An object is falling under the influence of gravity. Interactive Visit this website for a video showing projectile motion. Newton’s second law also tells us that F\=maF\=ma, where aa represents the acceleration vector of the object. This force must be equal to the force of gravity at all times, so we therefore know that F\=Fgma\=−mgja\=−gjF\=Fgma\=−mgja\=−gj Now we use the fact that the acceleration vector is the first derivative of the velocity vector. Therefore, we can rewrite the last equation in the form v′(t)\=−gj.v′(t)\=−gj. By taking the antiderivative of each side of this equation we obtain v(t)\=∫ −gjdt\=−gtj+C1v(t)\=∫ −gjdt\=−gtj+C1 for some constant vector C1C1. To determine the value of this vector, we can use the velocity of the object at a fixed time, say at time t\=0t\=0. We call this velocity the initial velocity: v(0)\=v0v(0)\=v0. Therefore, v(0)\=−g(0)j+C1\=v0v(0)\=−g(0)j+C1\=v0 and C1\=v0C1\=v0. This gives the velocity vector as v(t)\=−gtj+v0v(t)\=−gtj+v0. Next we use the fact that velocity v(t)v(t) is the derivative of position s(t)s(t). This gives the equation s′(t)\=−gtj+v0s′(t)\=−gtj+v0. Taking the antiderivative of both sides of this equation leads to s(t)\=∫ −gtj+v0dt\=−12gt2j+v0t+C2,s(t)\=∫ −gtj+v0dt\=−12gt2j+v0t+C2, with another unknown constant vector C2C2. To determine the value of C2C2, we can use the position of the object at a given time, say at time t\=0t\=0. We call this position the initial position: s(0)\=s0s(0)\=s0. Therefore, s(0)\=−(1/2)g(0)2j+v0(0)+C2\=s0s(0)\=−(1/2)g(0)2j+v0(0)+C2\=s0 and C2\=s0C2\=s0. This gives the position of the object at any time as s(t)\=−12gt2j+v0t+s0.s(t)\=−12gt2j+v0t+s0. Let’s take a closer look at the initial velocity and initial position. In particular, suppose the object is thrown upward from the origin at an angle θθ to the horizontal, with initial speed v0v0. How can we modify the previous result to reflect this scenario? First, we can assume it is thrown from the origin. If not, then we can move the origin to the point from where it is thrown. Therefore, s0\=0s0\=0, as shown in the following figure. Figure 2. Projectile motion when the object is thrown upward at an angle θθ. The horizontal motion is at constant velocity and the vertical motion is at constant acceleration. We can rewrite the initial velocity vector in the form v0\=v0cosθi+v0sinθjv0\=v0cos⁡θi+v0sin⁡θj. Then the equation for the position functions s(t)s(t) becomes s(t)\=−12gt2j+v0tcosθi+v0tsinθj\=v0tcosθi+v0tsinθj−12gt2j\=v0tcosθi+(v0tsinθ−12gt2)j.s(t)\=−12gt2j+v0tcos⁡θi+v0tsin⁡θj\=v0tcos⁡θi+v0tsin⁡θj−12gt2j\=v0tcos⁡θi+(v0tsin⁡θ−12gt2)j. The coefficient of ii represents the horizontal component of s(t)s(t) and is the horizontal distance of the object from the origin at time tt. The maximum value of the horizontal distance (measured at the same initial and final attitude) is called the range RR. The coefficient of jj represents the vertical component of s(t)s(t) and is the altitude of the object at time tt. The maximum value of the vertical distance is the height HH. Example: Motion of a Cannonball During an Independence Day celebration, a cannonball is fired from a cannon on a cliff toward the water. The cannon is aimed at an angle of 30∘30∘ above horizontal and the initial speed of the cannonball is 600 ft/sec600 ft/sec. The cliff is 100 ft100 ft above the water (Figure 7). Find the maximum height of the cannonball. How long will it take for the cannonball to splash into the sea? How far out to sea will the cannonball hit the water? Figure 3. The flight of a cannonball (ignoring air resistance) is projectile motion. Show Solution We use the equation s(t)\=v0tcosθi+(v0tsinθ−12gt2)js(t)\=v0tcos⁡θi+(v0tsin⁡θ−12gt2)j with θ\=30∘θ\=30∘, g\=32 ft/sec2g\=32 ft/sec2, and v0\=600 ft/secv0\=600 ft/sec. Then the position equation becomes s(t)\=600t(cos30)i+(600tsin30−12(32)t2)j\=300t√3i+(300t−16t2)j.s(t)\=600t(cos⁡30)i+(600tsin⁡30−12(32)t2)j\=300t3i+(300t−16t2)j. The cannonball reaches its maximum height when the vertical component of its velocity is zero, because the cannonball is neither rising nor falling at that point. The velocity vector is v(t)\=s′(t)\=300√3i+(300−32t)j.v(t)\=s′(t)\=3003i+(300−32t)j. Therefore, the vertical component of velocity is given by the expression 300−32t300−32t. Setting this expression equal to zero and solving tt gives t\=9.375 sect\=9.375 sec. The height of the cannonball at this time is given by the vertical component of the position vector, evaluated at t\=9.375t\=9.375. s(9.375)\=300(9.375)√3i+(300(9.375)−16(9.375)2)j\=4871.39i+1406.25js(9.375)\=300(9.375)3i+(300(9.375)−16(9.375)2)j\=4871.39i+1406.25j Therefore, the maximum height of the cannonball is 1406.39 ft1406.39 ft above the cannon, or 1506.39 ft1506.39 ft above sea level. When the cannonball lands in the water, is is 100 ft100 ft below the cannon. Therefore, the vertical component of the position vector is equal to −100−100. Setting the vertical component of s(t)s(t) equal to −100−100 and solving, we obtain 300t−16t2\=−10016t2−300t−100\=04t2−75t−25\=0t\=75±√(−75)2−4(4)(−25)2(4)\=75±√60258\=75±5√2418.300t−16t2\=−10016t2−300t−100\=04t2−75t−25\=0t\=75±(−75)2−4(4)(−25)2(4)\=75±60258\=75±52418. The positive value of tt that solves this equation is approximately 19.0819.08. Therefore, the cannonball hits the water after approximately 19.08 sec19.08 sec. To find the distance out to sea, we simply substitute the answer from part (b) into s(t)s(t): s(19.08)\=300(19.08)√3i+(300(19.08)−16(19.08)2)j\=9914.26i−100.7424j.s(19.08)\=300(19.08)3i+(300(19.08)−16(19.08)2)j\=9914.26i−100.7424j. Therefore, the ball hits the water about 9914.26 ft9914.26 ft away from the base of the cliff. Notice that the vertical component of the position vector is very close to −100−100, which tells us that the ball just hit the water. Note that 9914.269914.26 feet is not the true range of the cannon since the cannonball lands in the ocean at a location below the cannon. The range of the cannon would be determined by finding how far out the cannonball is when its height is 100 ft100 ft above the water (the same as the altitude of the cannon). try it An archer fires an arrow at an angle of 40∘40∘ above the horizontal with an initial speed of 98 m/sec98 m/sec. The height of the archer is 171.5 cm171.5 cm. Find the horizontal distance the arrow travels before it hits the ground. Show Solution 967.15 m967.15 m Watch the following video to see the worked solution to the above Try It You can view the transcript for “CP 3.16” here (opens in new window). One final question remains: In general, what is the maximum distance a projectile can travel, given its initial speed? To determine this distance, we assume the projectile is fired from ground level and we wish it to return to ground level. In other words, we want to determine an equation for the range. In this case, the equation of projectile motion is s(t)\=v0tcosθi+(v0tsinθ−12gt2)j.s(t)\=v0tcos⁡θi+(v0tsin⁡θ−12gt2)j. Setting the second component equal to zero and solving for tt yields v0tsinθ−12gt2\=0t(v0sinθ−12gt)\=0v0tsin⁡θ−12gt2\=0t(v0sin⁡θ−12gt)\=0 Therefore, either t\=0t\=0 or t\=2v0sinθgt\=2v0sin⁡θg. We are interested in the second value of tt, so we substitute this into s(t)s(t), which gives s(2v0sinθg)\=v0(2v0sinθg)cosθi+(v0(2v0sinθg)sinθ−12g(2v0sinθg)2)j\=(2v20sinθcosθg)i\=v20sin2θgi.s(2v0sin⁡θg)\=v0(2v0sin⁡θg)cos⁡θi+(v0(2v0sin⁡θg)sin⁡θ−12g(2v0sin⁡θg)2)j\=(2v02sin⁡θcos⁡θg)i\=v02sin⁡2θgi. Thus, the expression for the range of a projectile fired at an angle θθ is R\=v20sin2θgi.R\=v02sin⁡2θgi. The only variable in this expression is θθ. To maximize the distance traveled, take the derivative of the coefficient of ii with respect to θθ and set it equal to zero: ddθ(v20tsin2θg)\=02v0cos2θg\=0θ\=45∘.ddθ(v02tsin⁡2θg)\=02v0cos⁡2θg\=0θ\=45∘. This value of θθ is the smallest positive value that makes the derivative equal to zero. Therefore, in the absence of air resistance, the best angle to fire a projectile (to maximize the range) is at a 45∘45∘ angle. The distance it travels is given by s(2v0sin45g)\=v20sin90gi\=v20gj.s(2v0sin⁡45g)\=v02sin⁡90gi\=v02gj. Therefore, the range for an angle of 45∘45∘ is v20/gv02/g. Kepler’s Laws During the early 1600s, Johannes Kepler was able to use the amazingly accurate data from his mentor Tycho Braheto formulate his three laws of planetary motion, now known as Kepler’s laws of planetary motion. These laws also apply to other objects in the solar system in orbit around the Sun, such as comets (e.g., Halley’s comet) and asteroids. Variations of these laws apply to satellites in orbit around Earth. Kepler’s Laws of Planetary Motion Theorem The path of any planet about the Sun is elliptical in shape, with the center of the Sun located at one focus of the ellipse (the law of ellipses). A line drawn from the center of the Sun to the center of a planet sweeps out equal areas in equal time intervals (the law of equal areas) (Figure 8). The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of the lengths of their semi-major orbital axes (the law of harmonies). Figure 4. Kepler’s first and second laws are pictured here. The Sun is located at a focus of the elliptical orbit of any planet. Furthermore, the shaded areas are all equal, assuming that the amount of time measured as the planet moves is the same for each region. Kepler’s third law is especially useful when using appropriate units. In particular, 1 astronomical unit1 astronomical unit is defined to be the average distance from Earth to the Sun, and is now recognized to be 149,597,870,700 m149,597,870,700 m or, approximately 93,000,00093,000,000mi. We therefore write 1 A.U.\=93,000,0001 A.U.\=93,000,000mi. Since the time it takes for Earth to orbit the Sun is 11 year, we use Earth years for units of time. Then, substituting 11 year for the period of Earth and 11A.U. for the average distance to the Sun, Kepler’s third law can be written as T2p\=D3pTp2\=Dp3 for any planet in the solar system, where TpTp is the period of that planet measured in Earth years and DpDp is the average distance from that planet to the Sun measured in astronomical units. Therefore, if we know the average distance from a planet to the Sun (in astronomical units), we can then calculate the length of its year (in Earth years), and vice versa. Kepler’s laws were formulated based on observations from Brahe; however, they were not proved formally until Sir Isaac Newton was able to apply calculus. Furthermore, Newton was able to generalize Kepler’s third law to other orbital systems, such as a moon orbiting around a planet. Kepler’s original third law only applies to objects orbiting the Sun. Proof Let’s now prove Kepler’s first law using the calculus of vector-valued functions. First we need a coordinate system. Let’s place the Sun at the origin of the coordinate system and let the vector-valued function r(t)r(t) represent the location of a planet as a function of time. Newton proved Kepler’s law using his second law of motion and his law of universal gravitation. Newton’s second law of motion can be written as F\=maF\=ma, where FF represents the net force acting on the planet. His law of universal gravitation can be written in the form F\=−GmM∥r∥2⋅r∥r∥F\=−GmM‖r‖2⋅r‖r‖, which indicates that the force resulting from the gravitational attraction of the Sun points back toward the Sun, and has magnitude GmM∥r∥2GmM‖r‖2 (Figure 9). Figure 5. The gravitational force between Earth and the Sun is equal to the mass of the earth times its acceleration. Setting these two forces equal to each other, and using the fact that a(t)\=v′(t)a(t)\=v′(t), we obtain mv′(t)\=−GmM∥r∥2⋅r∥r∥mv′(t)\=−GmM‖r‖2⋅r‖r‖, which can be rewritten as dvdt\=−GM∥r∥3r.dvdt\=−GM‖r‖3r. This equation shows that the vectors dv/dtdv/dt and rr are parallel to each other, so dv/dt×r\=0dv/dt×r\=0. Next, let’s differentiate r×vr×v with respect to time: ddt(r×v)\=drdt×v+r×dvdt\=v×v+0\=0.ddt(r×v)\=drdt×v+r×dvdt\=v×v+0\=0. This proves that r×vr×v is a constant vector, which we call CC. Since rr and vv are both perpendicular to CC for all values of tt, they must lie in a plane perpendicular to CC. Therefore, the motion of the planet lies in a plane. Next we calculate the expression dv/dt×Cdv/dt×C: dvdt×C\=−GM∥r∥3r×(r×v)\=−GM∥r∥3[(r⋅v)r−(r⋅r)v].dvdt×C\=−GM‖r‖3r×(r×v)\=−GM‖r‖3[(r⋅v)r−(r⋅r)v]. The last equality in the above equation is from the triple cross product formula (Introduction to Vectors in Space). We need an expression for r⋅vr⋅v. To calculate this, we differentiate r⋅rr⋅r with respect to time: ddt(r⋅r)\=drdt⋅r+r⋅drdt\=2r⋅drdt\=2r⋅v.ddt(r⋅r)\=drdt⋅r+r⋅drdt\=2r⋅drdt\=2r⋅v. Since r⋅r\=∥r∥2r⋅r\=‖r‖2, we also have ddt(r⋅r)\=ddt∥r∥2\=2∥r∥ddt∥r∥ddt(r⋅r)\=ddt‖r‖2\=2‖r‖ddt‖r‖. Combining the last two equations, we get 2r⋅v\=2∥r∥ddt∥r∥r⋅v\=∥r∥ddt∥r∥.2r⋅v\=2‖r‖ddt‖r‖r⋅v\=‖r‖ddt‖r‖. Substituting into our expression for dvdt×Cdvdt×C gives us dvdt×C\=−GM∥r∥3[(r⋅r)r−(r⋅r)v]\=−GM∥r∥3[∥r∥(ddt∥r∥)r−∥r∥2v]\=−GM[1∥r∥2(ddt∥r∥)r−1∥r∥v]\=GM[v∥r∥−r∥r∥2(ddt∥r∥)].dvdt×C\=−GM‖r‖3[(r⋅r)r−(r⋅r)v]\=−GM‖r‖3[‖r‖(ddt‖r‖)r−‖r‖2v]\=−GM[1‖r‖2(ddt‖r‖)r−1‖r‖v]\=GM[v‖r‖−r‖r‖2(ddt‖r‖)]. However, ddtr∥r∥\=ddt(r)∥r∥−rddt∥r∥∥r∥\=drdt∥r∥−r∥r∥2ddt∥r∥\=v∥r∥−r∥v∥2ddt∥r∥.ddtr‖r‖\=ddt(r)‖r‖−rddt‖r‖‖r‖\=drdt‖r‖−r‖r‖2ddt‖r‖\=v‖r‖−r‖v‖2ddt‖r‖. Therefore, we get dvdt×C\=GM(ddtr∥r∥).dvdt×C\=GM(ddtr‖r‖). Since CC is a constant vector, we can integrate both sides and obtain v×C\=GMr∥r∥+D,v×C\=GMr‖r‖+D, where DD is a constant vector. Our goal is to solve for ∥r∥‖r‖. Let’s start by calculating r⋅(v×C)r⋅(v×C): r⋅(v×C)\=r⋅(GMr∥r∥+D)\=GM∥r∥2∥r∥+r⋅D\=GM∥r∥+r⋅D.r⋅(v×C)\=r⋅(GMr‖r‖+D)\=GM‖r‖2‖r‖+r⋅D\=GM‖r‖+r⋅D. However, r⋅(v×C)\=(r×v)×Cr⋅(v×C)\=(r×v)×C, so (r×v)×C\=GM∥r∥+r⋅D(r×v)×C\=GM‖r‖+r⋅D. Since r×v\=Cr×v\=C, we have ∥C∥2\=GM∥r∥+r⋅D‖C‖2\=GM‖r‖+r⋅D. Note that r⋅D\=∥r∥∥D∥cosθr⋅D\=‖r‖‖D‖cos⁡θ, where θθ is the angle between rr and DD. Therefore, ∥C∥2\=GM∥r∥+∥r∥∥D∥cosθ‖C‖2\=GM‖r‖+‖r‖‖D‖cos⁡θ. Solving for ∥r∥‖r‖, ∥r∥\=∥C∥2GM+∥D∥cosθ\=∥C∥2GM(11+ecosθ)‖r‖\=‖C‖2GM+‖D‖cos⁡θ\=‖C‖2GM(11+ecos⁡θ), where e\=∥D∥/GMe\=‖D‖/GM. This is the polar equation of a conic with a focus at the origin, which we set up to be the Sun. It is a hyperbola if e>1e>1, a parabola if e\=1e\=1, or an ellipse if e<1e<1. Since planets have closed orbits, the only possibility is an ellipse. However, at this point it should be mentioned that hyperbolic comets do exist. These are objects that are merely passing through the solar system at speeds too great to be trapped into orbit around the Sun. As they pass close enough to the Sun, the gravitational field of the Sun deflects the trajectory enough so the path becomes hyperbolic. ■◼ Example: Using Kepler’s Third Law for Nonheliocentric Orbits Kepler’s third law of planetary motion can be modified to the case of one object in orbit around an object other than the Sun, such as the Moon around the Earth. In this case, Kepler’s third law becomes P2\=4π2a3G(m+M),P2\=4π2a3G(m+M), where mm is the mass of the Moon and MM is the mass of Earth, aa represents the length of the major axis of the elliptical orbit, and PP represents the period. Given that the mass of the Moon is 7.35 × 1022 kg7.35 × 1022 kg, the mass of the Earth is 5.97 × 1024 kg5.97 × 1024 kg, G\=6.67 × 10−11m3/kg⋅sec2G\=6.67 × 10−11m3/kg⋅sec2, and the period of the moon is 27.327.3 days, let’s find the length of the major axis of the orbit of the Moon around Earth. Show Solution It is important to be consistent with units. Since the universal gravitational constant contains seconds in the units, we need to use seconds for the period of the Moon as well: 27.3 days×24 hr1 day×3600 sec1 hour\=2,358,720 sec27.3 days×24 hr1 day×3600 sec1 hour\=2,358,720 sec. Substitute all the data into the equation given in the problem and solve for aa: (2,358,720 sec)2\=4π2a3(6.67×10−11m3kg⋅sec2)(7.35×1022 kg+5.97×1024 kg)5.563×1012\=4π2a3(6.67×10−11m3)(6.04×1024)(5.563×1012)(6.67×10−11 m3)(6.04×1024)\=4π2a3a3\=2.241×10274π2m3a\=3.84×108 m≈384,000 km.(2,358,720 sec)2\=4π2a3(6.67×10−11m3kg⋅sec2)(7.35×1022 kg+5.97×1024 kg)5.563×1012\=4π2a3(6.67×10−11m3)(6.04×1024)(5.563×1012)(6.67×10−11 m3)(6.04×1024)\=4π2a3a3\=2.241×10274π2m3a\=3.84×108 m≈384,000 km. Analysis According to solarsystem.nasa.gov, the actual average distance from the Moon to Earth is 384,400 km384,400 km. This is calculated using reflectors left on the Moon by Apollo astronauts back in the 1960s. try it Titan is the largest moon of Saturn. The mass of Titan is approximately 1.35×1023 kg1.35×1023 kg. The mass of Saturn is approximately 5.68×1026 kg5.68×1026 kg. Titan takes approximately 1616 days to orbit Saturn. Use this information, along with the universal gravitation constant G\=6.67×10−11 m3/kg⋅sec2G\=6.67×10−11 m3/kg⋅sec2 to estimate the distance from Titan to Saturn. Show Solution a\=1.224×109≈1,224,000 kma\=1.224×109≈1,224,000 km Watch the following video to see the worked solution to the above Try It You can view the transcript for “CP 3.17” here (opens in new window). Activity: Navigating a Banked Turn How fast can a racecar travel through a circular turn without skidding and hitting the wall? The answer could depend on several factors: The weight of the car; The friction between the tires and the road; The radius of the circle; The “steepness” of the turn; In this project we investigate this question for NASCAR racecars at the Bristol Motor Speedway in Tennessee. Before considering this track in particular, we use vector functions to develop the mathematics and physics necessary for answering questions such as this. A car of mass mm moves with constant angular speed ωω around a circular curve of radius RR (Figure 10). The curve is banked at an angle θθ. If the height of the car off the ground is hh, then the position of the car at time tt is given by the function r(t)\=⟨Rcos(ωt), Rsin(ωt), h⟩r(t)\=⟨Rcos⁡(ωt), Rsin⁡(ωt), h⟩. Figure 6. Views of a race car moving around a track. Find the velocity function v(t)v(t) of the car. Show that vv is tangent to the circular curve. This means that, without a force to keep the car on the curve, the car will shoot off of it. Show that the speed of the car is ωRωR. Use this to show that (2πr)/v\=(2π)/ω(2πr)/v\=(2π)/ω. Find the acceleration aa. Show that this vector points toward the center of the circle and that |a|\=Rω2|a|\=Rω2. The force required to produce this circular motion is called the centripetal force, and it is denoted FcentFcent. This force points toward the center of the circle (not toward the ground). Show that |Fcent|\=(m|v|2)/R|Fcent|\=(m|v|2)/R. As the car moves around the curve, three forces act on it: gravity, the force exerted by the road (this force is perpendicular to the ground), and the friction force (Figure 11). Because describing the frictional force generated by the tires and the road is complex, we use a standard approximation for the frictional force. Assume that |f|\=μ|N||f|\=μ|N|for some positive constant μμ. The constant μμ is called the coefficient of friction. Figure 7. The car has three forces acting on it: gravity (denoted by mgmg), the friction force ff, and the force exerted by the road NN. Let vmaxvmax denote the maximum speed the car can attain through the curve without skidding. In other words, vmaxvmax is the fastest speed at which the car can navigate the turn. When the car is traveling at this speed, the magnitude of the centripetal force is |Fcent|\=mv2maxR|Fcent|\=mvmax2R. The next three questions deal with developing a formula that relates the speed vmaxvmax to the banking angle θθ. Show that |N|cosθ\=mg+|f|sinθ|N|cos⁡θ\=mg+|f|sin⁡θ. Conclude that |bfN|\=(mg)/(cosθ−μsinθ)|bfN|\=(mg)/(cos⁡θ−μsin⁡θ). The centripetal force is the sum of the forces in the horizontal direction, since the centripetal force points toward the center of the circular curve. Show that |Fcent|\=|N|sinθ+|f|cosθ|Fcent|\=|N|sin⁡θ+|f|cos⁡θ. Conclude that |Fcent|\=sinθ+μcosθcosθ−μsinθmg|Fcent|\=sin⁡θ+μcos⁡θcos⁡θ−μsin⁡θmg. Show that v2max\=((sinθ+μcosθ) / (cosθ−μsinθ))gRvmax2\=((sin⁡θ+μcos⁡θ) / (cos⁡θ−μsin⁡θ))gR. Conclude that the maximum speed does not actually depend on the mass of the car. Now that we have a formula relating the maximum speed of the car and the banking angle, we are in a position to answer the questions like the one posed at the beginning of the project. The Bristol Motor Speedway is a NASCAR short track in Bristol, Tennessee. The track has the approximate shape shown in Figure 12. Each end of the track is approximately semicircular, so when cars make turns they are traveling along an approximately circular curve. If a car takes the inside track and speeds along the bottom of turn 1, the car travels along a semicircle of radius approximately 211 ft211 ft with a banking angle of 24∘24∘. If the car decides to take the outside track and speeds along the top of turn 1, then the car travels along a semicircle with a banking angle of 28∘28∘. (The track has variable angle banking.) Figure 8. At the Bristol Motor Speedway, Bristol, Tennessee (a), the turns have an inner radius of about 211 ft and a width of 40 ft (b). (credit: part (a) photo by Raniel Diaz, Flickr) The coefficient of friction for a normal tire in dry conditions is approximately 0.70.7. Therefore, we assume the coefficient for a NASCAR tire in dry conditions is approximately 0.980.98. Before answering the following questions, note that it is easier to do computations in terms of feet and seconds, and then convert the answers to miles per hour as a final step. In dry conditions, how fast can the car travel through the bottom of the turn without skidding? In dry conditions, how fast can the car travel through the top of the turn without skidding? In wet conditions, the coefficient of friction can become as low as 0.10.1. If this is the case, how fast can the car travel through the bottom of the turn without skidding? Suppose the measured speed of a car going along the outside edge of the turn is 105 mph105 mph. Estimate the coefficient of friction for the car’s tires. Candela Citations CC licensed content, Original CP 3.16. Authored by: Ryan Melton. License: CC BY: Attribution CP 3.17. Authored by: Ryan Melton. License: CC BY: Attribution CC licensed content, Shared previously Calculus Volume 3. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at Licenses and Attributions CC licensed content, Original CP 3.16. Authored by: Ryan Melton. License: CC BY: Attribution CP 3.17. Authored by: Ryan Melton. License: CC BY: Attribution CC licensed content, Shared previously Calculus Volume 3. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at Previous Next Privacy Policy
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https://www.youtube.com/watch?v=LpdsraJAt4s
'+C' The Constant of Integration... WHY? #integration Uttkarsh Kohli - Math Channel 482 subscribers 96 likes Description 5909 views Posted: 15 Jul 2020 The constant of integration, conventionally called '+C' is added to maintain numerous solutions to a differential equation. While evaluating an indefinite integral, there is a family of solutions(anti-derivatives) and +C accounts as a general solution to all anti-derivatives. Watch what 0^0 means and how to derive its value: Learn more about Improper Integrals: Watch the proof for Euler's formula and natural log of negative numbers here: My website: Uttkarshkohli.wixsite.com/maths. 0:00 Intro 0:20 Example explaining need for +C 3:30 Explanation through Graphs 4:25 No need for +C in a Definite Integral 6:22 BYE BYE! tags, constant of integration, integration constant, +c, indefinite integration, indefinite integrals, +c constant, calculus, integration by substitution, U-sub, solve integrals with me, math. 13 comments Transcript: Intro hey guys i'm back with another video and in this video we're going to be talking about the integration constant plus c why do we have to put a plus c and what is the use of it Example explaining need for +C okay so now i'm going to show you an example where if we don't put the integration constant our integral answers would differ so let's take the function f of x is equal to tan square x so now let's differentiate this function we will get f prime of x is equal to 2 times tan x times x square x so now if we integrate this we should get back to this answer right and that only works if we put the integration constant so now let's look why that's the reason so let's say we have this function and we don't know this we have to find the integral of f prime of x so that gives us 2 times integral of tan x times sec square x dx so now there's two ways of solving this integral let's look at both the ways so in the first method we can substitute u as tan x and in the second method we will have u is equal to sec square x let's differentiate this we get d u is equal to sec square x dx and this integral just becomes two times udu now let's look at this one we have d u is equal to two times sec x times sec x tan x dx now this is exactly our integral so this just becomes the integral of d u so now if we solve this we get 2 times u square upon 2 plus c and this gives us just u which is sec square x and um this is tan square x plus c so now if we didn't put this c you can see that tan square x and sec square x are not the same function so we will have different answers to the same integral so now how does the constant of integration help so we have two answers one is tan square x plus c and the second one is sec square x plus c so now if we so now let's add plus one and minus one here we get tan square x plus one plus c minus one and c minus 1 will just become another constant and this is the same as x square x so we get sec square x plus c 2 let's say this is the second constant and here we have sec square x plus c or we could do the same and get sec square x plus 1 plus c minus 1 here this time we take 6 square x minus 1 and it becomes tan square x plus c plus 1 and this becomes tan square x plus c2 so now this is why we need the constant of integration so that we don't have different answers to the same integral so now let me explain this using graphs Explanation through Graphs let's look at this graph we have y is equal to x plus five so now the slope of this line is one so we will have d y by d x is equal to one so now from here we get d y is equal to d x so now if we integrate both sides we get y is equal to x but this is without putting the integration constant and we know that this is not our original line in fact if we put the plus c this gives us a family of lines with the same slope but different y-intercept and that is why we need plus c because if we just integrated the derivative we would just get the slope of the line and not its y-intercept No need for +C in a Definite Integral so now let's look at the function f of x is equal to three x square plus six x plus ten so now if we take the derivative f prime of x is equal to six x plus six so now what if we integrate this function but this time we have a boundary let's say from 0 to 10 we integrate 6x plus 6 dx so now this is basically the area under the graph of 6x plus 6. from 0 to 10 this is not a perfect drawing it's supposed to be a straight line but we're basically finding this area so now this won't be a problem even if we have a different constant here let's say we have the constant 3x square plus 6x plus 9 let's say or let's say 100 that wouldn't be a problem because we're still going to be finding this area and it won't change which is because of the constant because when we find the derivative it'll get removed so for all family of functions 3x square plus 6x plus c the definite integral from any boundary would be the same we can also look at it like this let's say we find the indefinite integral of six x plus six dx that will give us six x square upon two plus six x plus c and now we put the boundaries from 0 to 10 so that gives us 3 n squared plus 6 times 10 plus c minus 3 times 0 square my minus three times zero minus c so this just becomes zero and the minus c gets cut out so we don't have to worry about the constant when we're taking a definite integral because when we put the boundaries it will get cut out BYE BYE! hey guys if you've made it till here you're a legend please click here and subscribe for more content and keep watching
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https://www.youtube.com/watch?v=3Yq9r_NIVTM
Dimensional Analysis: Distance from Speed and Time Alan Earhart 2320 subscribers 27 likes Description 4269 views Posted: 21 Mar 2017 This dimensional analysis problem works a distance, rate, and time problem with some units changes. (Chem 1090 DimAnal 2a) 9 comments Transcript: Let's do a more challenging dimensional analysis problem see we're traveling a certain speed we do it in a certain amount of time and we want to know how far we travel maybe you recognize this maybe you can pull a formula out of this okay meters how far that's distance the speed is the rate and the 37 minutes well that's the time distance equals rate times time and you solve for distance piece of cake right and there are some problems look at the units the problem was written in miles per hour but you want meters yeah but also the speed is miles per hour and you're given time in minutes where you could do some conversions ahead of time and then you could solve for distance but I'm not going to do it that way I'm going to show you how to do this problem using the dimensional analysis skills we've been working on now I'll admit when I wrote this problem I deliberately mess with it I mean why ask for meters if it's miles per hour because I want to if you can solve this problem that means you have a very strong grasp of how to do the dimensional analysis because this goes back to the earlier problems we did the relatively simple one-step problems that's what this is this is a series of single step problems but can you put it together if you can put this one together properly and understand the pieces you're ready for more complicated calculations in chemistry well we've got some information here let's start pulling it out see if we can piece it together all right speed well that's what that means miles / hours mean means if I travel for one hour you expect to go 75.0 miles when you work problems with combination units like this do not do the work like this in my experience students don't work the problems properly when they do it like that it's too easy to make mistakes it's too easy to get confused between your numerator and denominator so split it apart the way I have 75.0 miles equals 1 hour let me see minutes and hours okay we need that relationship and we need something to go from meters to miles I can't think of one right off the top of my head if you look at the conversion sheet that I provide students you have one on there that will go directly from miles to meters oh you don't on there do you what do you have what do you have one that bridges between the imperial system miles to the metric system and I put this one on there one mile equals one point six oh nine kilometers and let me see we still have a one more problem we're asked for the different distance in meters but we don't have any meters up there anywhere we have kilometers so we need the relationship between kilometers and meters kilo equals a thousand well I I think we're ready to put this together but boy that's a lot of information what do we do with it what did the question they ask for it asks for a distance now there are a few ways to start this problem and when we're done with this problem I'll at the very end I'll show you another way to assemble it but since I want a distance a common problem solving technique is to take a piece of information that you're given and start with that and we were given the speed the speed is 75.0 miles per one hour equal to one hour okay well we can start conversions one of two ways we can write it this way or we can take the reciprocal depending upon the information that's given and let me remind you of what it means that the piece I just put down there that means they were equivalent to each other so obviously miles and hours aren't equal to each other in that sense you have a distance in you have a time so it's an equivalency and it's because of our speed that makes them equivalent to each other if you drive for one hour you expect to go 75.0 miles we can also take the reciprocal of conversions just what are the units tell us which one this one or the previous one how do you want to start ask yourself where do you want distance units when you're done in the numerator or the denominator if somebody asks you how tall you are you're not going to tell them you're so many 1 over feet or reciprocal feet you're going to say you're so many feet or so many meet or something like that therefore we need our distance unit in the numerator okay i don't like mild I don't like ours I'm going to start working on the miles first you can start working on the hours first but I choose to start working on the miles first I look at the conversions that I wrote down and I see I have one that relates miles and kilometers to each other so when I put that conversion down am I going to go miles over kilometers or kilometers over miles kilometers over miles because allows my miles to cancel so now i have km/h little want kilometers I want meters so I see another conversion that relates kilometers and meters to each other so am I going to go kilometers over meters or meters over kilometers in my conversion meters over kilometers for the same reason as before this allows the kilometers to cancel so not only have units of meters per hour I like the meters hours not so much so let's work on the hours now am I going to go hours over minutes or minutes over hours hours over minutes allows my hours to cancel now I have unit wise let's see I have meters per minute that we're done because we have meters no we want meters not meters per minute what do I have left I didn't pull something I didn't pull something out did I i still have 37 minutes so am i going to multiply or divide by the 37 minutes you're going to multiply so your minutes units will cancel and now i have meters in the numerator nothing else left that's what i want do my multiplication my division and the calculator spits this out now we have to write the answer according to the proper number of significant figures so let's go back and look at all the pieces started on the far left with my speed how many significant figures are in that ignore the one whenever you have one equal something else you ignore the one all the significant figures have been sucked into the other piece three significant figures how about the 1.6 09 does the conversion on the sheet say that its exact it does not does it therefore this has four significant figures now be careful on your conversion choice since I provide a conversion sheet to my students you would not use 1.6 kilometers of the equivalent to one mile because if you use that one that one only has two significant figures the one I provide has four next piece that was in meters and a kilometer well a thousand has one significant figure in it therefore that relationship is its exact it's a metric relationship kilo is a thousand exactly how a significant figures here well that's another exact relationship let's see you count minutes right I count the minutes until I see you again so that's exact no it's not if two significant figures you don't count minutes that way I know people say well I'm counting the minutes you're not really you're measuring the minutes you're using something that measures minutes you can't count minutes like their pencils so with our answer we only get three for exact exact to two significant figures in our answer so we could write it this way no decimal place remember you count left to right we could put it in scientific notation either those bottom two would be correct so there are multiple ways to work this problem and i don't mean whether or not use parentheses what i mean is the order that you place them this is a little crowded i realize that i'm sorry about that but that set up on the bottom is also legitimate in this case which makes the most sense to you heck you could have even started with time remember your dimensional analysis will help you if you don't end up with units of meters you know you did not set it up properly use that to help you
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https://www.jacksonsd.org/cms/lib/NJ01912744/Centricity/Domain/504/BI%206-4.pdf
Section 6.4 Exponential Growth and Decay 313 6.4 Essential Question Essential Question What are some of the characteristics of exponential growth and exponential decay functions? Predicting a Future Event Work with a partner. It is estimated, that in 1782, there were about 100,000 nesting pairs of bald eagles in the United States. By the 1960s, this number had dropped to about 500 nesting pairs. In 1967, the bald eagle was declared an endangered species in the United States. With protection, the nesting pair population began to increase. Finally, in 2007, the bald eagle was removed from the list of endangered and threatened species. Describe the pattern shown in the graph. Is it exponential growth? Assume the pattern continues. When will the population return to that of the late 1700s? Explain your reasoning. 0 2000 4000 6000 8000 10,000 Number of nesting pairs Year 1978 1982 1986 1990 1994 1998 2002 2006 x y Bald Eagle Nesting Pairs in Lower 48 States 9789 6846 1875 1188 3399 5094 Describing a Decay Pattern Work with a partner. A forensic pathologist was called to estimate the time of death of a person. At midnight, the body temperature was 80.5°F and the room temperature was a constant 60°F. One hour later, the body temperature was 78.5°F. a. By what percent did the difference between the body temperature and the room temperature drop during the hour? b. Assume that the original body temperature was 98.6°F. Use the percent decrease found in part (a) to make a table showing the decreases in body temperature. Use the table to estimate the time of death. Communicate Your Answer Communicate Your Answer 3. What are some of the characteristics of exponential growth and exponential decay functions? 4. Use the Internet or some other reference to fi nd an example of each type of function. Your examples should be different than those given in Explorations 1 and 2. a. exponential growth b. exponential decay MODELING WITH MATHEMATICS To be profi cient in math, you need to apply the mathematics you know to solve problems arising in everyday life. Exponential Growth and Decay 314 Chapter 6 Exponential Functions and Sequences 6.4 Lesson Exponential Growth Functions A function of the form y = a(1 + r)t, where a > 0 and r > 0, is an exponential growth function. initial amount time growth factor rate of growth (in decimal form) final amount y = a(1 + r)t What You Will Learn What You Will Learn Use and identify exponential growth and decay functions. Interpret and rewrite exponential growth and decay functions. Solve real-life problems involving exponential growth and decay. Exponential Growth and Decay Functions Exponential growth occurs when a quantity increases by the same factor over equal intervals of time. exponential growth, p. 314 exponential growth function, p. 314 exponential decay, p. 315 exponential decay function, p. 315 compound interest, p. 317 Core Vocabulary Core Vocabulary Using an Exponential Growth Function The inaugural attendance of an annual music festival is 150,000. The attendance y increases by 8% each year. a. Write an exponential growth function that represents the attendance after t years. b. How many people will attend the festival in the fi fth year? Round your answer to the nearest thousand. SOLUTION a. The initial amount is 150,000, and the rate of growth is 8%, or 0.08. y = a(1 + r)t Write the exponential growth function. = 150,000(1 + 0.08)t Substitute 150,000 for a and 0.08 for r. = 150,000(1.08)t Add. The festival attendance can be represented by y = 150,000(1.08)t. b. The value t = 4 represents the fi fth year because t = 0 represents the fi rst year. y = 150,000(1.08)t Write the exponential growth function. = 150,000(1.08)4 Substitute 4 for t. ≈ 204,073 Use a calculator. About 204,000 people will attend the festival in the fi fth year. Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 1. A website has 500,000 members in 2010. The number y of members increases by 15% each year. (a) Write an exponential growth function that represents the website membership t years after 2010. (b) How many members will there be in 2016? Round your answer to the nearest ten thousand. Core Core Concept Concept STUDY TIP Notice that an exponential growth function is of the form y = abx, where b is replaced by 1 + r and x is replaced by t. Section 6.4 Exponential Growth and Decay 315 For exponential growth, the value inside the parentheses is greater than 1 because r is added to 1. For exponential decay, the value inside the parentheses is less than 1 because r is subtracted from 1. Identifying Exponential Growth and Decay Determine whether each table represents an exponential growth function, an exponential decay function, or neither. a. x y 0 270 1 90 2 30 3 10 b. x 0 1 2 3 y 5 10 20 40 SOLUTION a. x y 0 270 1 90 2 30 3 10 + 1 × 1 — 3 × 1 — 3 × 1 — 3 + 1 + 1 b. x 0 1 2 3 y 5 10 20 40 + 1 × 2 + 1 × 2 + 1 × 2 As x increases by 1, y is multiplied by 1 — 3 . So, the table represents an exponential decay function. As x increases by 1, y is multiplied by 2. So, the table represents an exponential growth function. Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Determine whether the table represents an exponential growth function, an exponential decay function, or neither. Explain. 2. x 0 1 2 3 y 64 16 4 1 3. x 1 3 5 7 y 4 11 18 25 Exponential decay occurs when a quantity decreases by the same factor over equal intervals of time. Core Core Concept Concept Exponential Decay Functions A function of the form y = a(1 − r)t, where a > 0 and 0 < r < 1, is an exponential decay function. initial amount time decay factor rate of decay (in decimal form) final amount y = a(1 − r)t STUDY TIP Notice that an exponential decay function is of the form y = abx, where b is replaced by 1 − r and x is replaced by t. 316 Chapter 6 Exponential Functions and Sequences Interpreting and Rewriting Exponential Functions Interpreting Exponential Functions Determine whether each function represents exponential growth or exponential decay. Identify the percent rate of change. a. y = 5(1.07)t b. f (t) = 0.2(0.98)t SOLUTION a. The function is of the form y = a(1 + r)t, where 1 + r > 1, so it represents exponential growth. Use the growth factor 1 + r to fi nd the rate of growth. 1 + r = 1.07 Write an equation. r = 0.07 Solve for r. So, the function represents exponential growth and the rate of growth is 7%. b. The function is of the form y = a(1 − r)t, where 1 − r < 1, so it represents exponential decay. Use the decay factor 1 − r to fi nd the rate of decay. 1 − r = 0.98 Write an equation. r = 0.02 Solve for r. So, the function represents exponential decay and the rate of decay is 2%. Rewriting Exponential Functions Rewrite each function to determine whether it represents exponential growth or exponential decay. a. y = 100(0.96)t/4 b. f (t) = (1.1)t − 3 SOLUTION a. y = 100(0.96)t/4 Write the function. = 100(0.961/4)t Power of a Power Property ≈ 100(0.99)t Evaluate the power. So, the function represents exponential decay. b. f (t) = (1.1)t − 3 Write the function. = (1.1)t — (1.1)3 Quotient of Powers Property ≈ 0.75(1.1)t Evaluate the power and simplify. So, the function represents exponential growth. Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Determine whether the function represents exponential growth or exponential decay. Identify the percent rate of change. 4. y = 2(0.92)t 5. f (t) = (1.2)t Rewrite the function to determine whether it represents exponential growth or exponential decay. 6. f (t) = 3(1.02)10t 7. y = (0.95)t + 2 STUDY TIP You can rewrite exponential expressions and functions using the properties of exponents. Changing the form of an exponential function can reveal important attributes of the function. Section 6.4 Exponential Growth and Decay 317 Writing a Function You deposit $100 in a savings account that earns 6% annual interest compounded monthly. Write a function that represents the balance after t years. SOLUTION y = P ( 1 + r — n ) nt Write the compound interest formula. = 100 ( 1 + 0.06 — 12 ) 12t Substitute 100 for P, 0.06 for r, and 12 for n. = 100(1.005)12t Simplify. Solving a Real-Life Problem The table shows the balance of a money market account over time. a. Write a function that represents the balance after t years. b. Graph the functions from part (a) and from Example 5 in the same coordinate plane. Compare the account balances. SOLUTION a. From the table, you know the initial balance is $100, and it increases 10% each year. So, P = 100 and r = 0.1. y = P(1 + r)t Write the compound interest formula when n = 1. = 100(1 + 0.1)t Substitute 100 for P and 0.1 for r. = 100(1.1)t Add. b. The money market account earns 10% interest each year, and the savings account earns 6% interest each year. So, the balance of the money market account increases faster. Monitoring Progress Help in English and Spanish at BigIdeasMath.com 8. You deposit $500 in a savings account that earns 9% annual interest compounded monthly. Write and graph a function that represents the balance y (in dollars) after t years. Year, t Balance 0 1 2 3 4 5 $100 $110 $121 $133.10 $146.41 $161.05 Solving Real-Life Problems Exponential growth functions are used in real-life situations involving compound interest. Although interest earned is expressed as an annual rate, the interest is usually compounded more frequently than once per year. So, the formula y = a(1 + r)t must be modifi ed for compound interest problems. Core Core Concept Concept Compound Interest Compound interest is the interest earned on the principal and on previously earned interest. The balance y of an account earning compound interest is P = principal (initial amount) r = annual interest rate (in decimal form) t = time (in years) n = number of times interest is compounded per year y = P ( 1 + r — n ) nt . STUDY TIP For interest compounded yearly, you can substitute 1 for n in the formula to get y = P(1 + r)t. 0 25 50 75 100 125 150 175 200 Balance (dollars) Year 0 1 2 3 4 5 6 7 Saving Money t y y = 100(1.1)t y = 100(1.005)12t 318 Chapter 6 Exponential Functions and Sequences Solving a Real-Life Problem The value of a car is $21,500. It loses 12% of its value every year. (a) Write a function that represents the value y (in dollars) of the car after t years. (b) Find the approximate monthly percent decrease in value. (c) Graph the function from part (a). Use the graph to estimate the value of the car after 6 years. SOLUTION 1. Understand the Problem You know the value of the car and its annual percent decrease in value. You are asked to write a function that represents the value of the car over time and approximate the monthly percent decrease in value. Then graph the function and use the graph to estimate the value of the car in the future. 2. Make a Plan Use the initial amount and the annual percent decrease in value to write an exponential decay function. Note that the annual percent decrease represents the rate of decay. Rewrite the function using the properties of exponents to approximate the monthly percent decrease (rate of decay). Then graph the original function and use the graph to estimate the y-value when the t-value is 6. 3. Solve the Problem a. The initial value is $21,500, and the rate of decay is 12%, or 0.12. y = a(1 − r)t Write the exponential decay function. = 21,500(1 − 0.12)t Substitute 21,500 for a and 0.12 for r. = 21,500(0.88)t Subtract. The value of the car can be represented by y = 21,500(0.88)t. b. Use the fact that t = 1 — 12 (12t) and the properties of exponents to rewrite the function in a form that reveals the monthly rate of decay. y = 21,500(0.88)t Write the original function. = 21,500(0.88)(1/12)(12t) Rewrite the exponent. = 21,500(0.881/12)12t Power of a Power Property ≈ 21,500(0.989)12t Evaluate the power. Use the decay factor 1 − r ≈ 0.989 to fi nd the rate of decay r ≈ 0.011. So, the monthly percent decrease is about 1.1%. c. From the graph, you can see that the y-value is about 10,000 when t = 6. So, the value of the car is about $10,000 after 6 years. 4. Look Back To check that the monthly percent decrease is reasonable, multiply it by 12 to see if it is close in value to the annual percent decrease of 12%. 1.1% × 12 = 13.2% 13.2% is close to 12%, so 1.1% is reasonable. When you evaluate y = 21,500(0.88)t for t = 6, you get about $9985. So, $10,000 is a reasonable estimation. Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com 9. WHAT IF? The car loses 9% of its value every year. (a) Write a function that represents the value y (in dollars) of the car after t years. (b) Find the approximate monthly percent decrease in value. (c) Graph the function from part (a). Use the graph to estimate the value of the car after 12 years. Round your answer to the nearest thousand. STUDY TIP In real life, the percent decrease in value of an asset is called the depreciation rate. T th m to S 1 Value of a Car 00 2 4 6 8 5000 10,000 15,000 20,000 Value (dollars) Year t y y = 21,500(0.88)t Section 6.4 Exponential Growth and Decay 319 Dynamic Solutions available at BigIdeasMath.com Exercises 6.4 In Exercises 5–12, identify the initial amount a and the rate of growth r (as a percent) of the exponential function. Evaluate the function when t = 5. Round your answer to the nearest tenth. 5. y = 350(1 + 0.75)t 6. y = 10(1 + 0.4)t 7. y = 25(1.2)t 8. y = 12(1.05)t 9. f (t) = 1500(1.074)t 10. h(t) = 175(1.028)t 11. g(t) = 6.72(2)t 12. p(t) = 1.8t In Exercises 13–16, write a function that represents the situation. 13. Sales of $10,000 increase by 65% each year. 14. Your starting annual salary of $35,000 increases by 4% each year. 15. A population of 210,000 increases by 12.5% each year. 16. An item costs $4.50, and its price increases by 3.5% each year. 17. MODELING WITH MATHEMATICS The population of a city has been increasing by 2% annually. The sign shown is from the year 2000. (See Example 1.) a. Write an exponential growth function that represents the population t years after 2000. b. What will the population be in 2020? Round your answer to the nearest thousand. 18. MODELING WITH MATHEMATICS A young channel catfi sh weighs about 0.1 pound. During the next 8 weeks, its weight increases by about 23% each week. a. Write an exponential growth function that represents the weight of the catfi sh after t weeks during the 8-week period. b. About how much will the catfi sh weigh after 4 weeks? Round your answer to the nearest thousandth. In Exercises 19–26, identify the initial amount a and the rate of decay r (as a percent) of the exponential function. Evaluate the function when t = 3. Round your answer to the nearest tenth. 19. y = 575(1 − 0.6)t 20. y = 8(1 − 0.15)t 21. g(t) = 240(0.75)t 22. f (t) = 475(0.5)t 23. w(t) = 700(0.995)t 24. h(t) = 1250(0.865)t 25. y = ( 7 — 8 ) t 26. y = 0.5 ( 3 — 4 ) t In Exercises 27–30, write a function that represents the situation. 27. A population of 100,000 decreases by 2% each year. 28. A $900 sound system decreases in value by 9% each year. 29. A stock valued at $100 decreases in value by 9.5% each year. Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics 1. COMPLETE THE SENTENCE In the exponential growth function y = a(1 + r)t, the quantity r is called the __. 2. VOCABULARY What is the decay factor in the exponential decay function y = a(1 − r)t? 3. VOCABULARY Compare exponential growth and exponential decay. 4. WRITING When does the function y = abx represent exponential growth? exponential decay? Vocabulary and Core Concept Check Vocabulary and Core Concept Check CITY LIMIT POP . 315,000 BROOKFIELD 320 Chapter 6 Exponential Functions and Sequences 30. A company profi t of $20,000 decreases by 13.4% each year. 31. ERROR ANALYSIS The growth rate of a bacterial culture is 150% each hour. Initially, there are 10 bacteria. Describe and correct the error in fi nding the number of bacteria in the culture after 8 hours. b(t) = 10(1.5)t b(8) = 10(1.5)8 ≈ 256.3 After 8 hours, there are about 256 bacteria in the culture. ✗ 32. ERROR ANALYSIS You purchase a car in 2010 for $25,000. The value of the car decreases by 14% annually. Describe and correct the error in fi nding the value of the car in 2015. v(t) = 25,000(1.14)t v(5) = 25,000(1.14)5 ≈ 48,135 The value of the car in 2015 is about $48,000. ✗ In Exercises 33–38, determine whether the table represents an exponential growth function, an exponential decay function, or neither. Explain. (See Example 2.) 33. x y −1 50 0 10 1 2 2 0.4 34. x y 0 32 1 28 2 24 3 20 35. x y 0 35 1 29 2 23 3 17 36. x y 1 17 2 51 3 153 4 459 37. x y 5 2 10 8 15 32 20 128 38. x y 3 432 5 72 7 12 9 2 39. ANALYZING RELATIONSHIPS The table shows the value of a camper t years after it is purchased. a. Determine whether the table represents an exponential growth function, an exponential decay function, or neither. b. What is the value of the camper after 5 years? 40. ANALYZING RELATIONSHIPS The table shows the total numbers of visitors to a website t days after it is online. t 42 43 44 45 Visitors 11,000 12,100 13,310 14,641 a. Determine whether the table represents an exponential growth function, an exponential decay function, or neither. b. How many people will have visited the website after it is online 47 days? In Exercises 41–48, determine whether each function represents exponential growth or exponential decay. Identify the percent rate of change. (See Example 3.) 41. y = 4(0.8)t 42. y = 15(1.1)t 43. y = 30(0.95)t 44. y = 5(1.08)t 45. r(t) = 0.4(1.06)t 46. s(t) = 0.65(0.48)t 47. g(t) = 2 ( 5 — 4 ) t 48. m(t) = ( 4 — 5 ) t In Exercises 49–56, rewrite the function to determine whether it represents exponential growth or exponential decay. (See Example 4.) 49. y = (0.9)t − 4 50. y = (1.4)t + 8 51. y = 2(1.06)9t 52. y = 5(0.82)t/5 53. x(t) = (1.45)t/2 54. f (t) = 0.4(1.16)t − 1 55. b(t) = 4(0.55)t + 3 56. r(t) = (0.88)4t t Value 1 $37,000 2 $29,600 3 $23,680 4 $18,944 Section 6.4 Exponential Growth and Decay 321 In Exercises 57– 60, write a function that represents the balance after t years. (See Example 5.) 57. $2000 deposit that earns 5% annual interest compounded quarterly 58. $1400 deposit that earns 10% annual interest compounded semiannually 59. $6200 deposit that earns 8.4% annual interest compounded monthly 60. $3500 deposit that earns 9.2% annual interest compounded quarterly 61. PROBLEM SOLVING The cross-sectional area of a tree 4.5 feet from the ground is called its basal area. The table shows the basal areas (in square inches) of Tree A over time. (See Example 6.) Year, t 0 1 2 3 4 Basal area, A 120 132 145.2 159.7 175.7 Tree B Growth rate: 6% Initial basal area: 154 in.2 a. Write functions that represent the basal areas of the trees after t years. b. Graph the functions from part (a) in the same coordinate plane. Compare the basal areas. 62. PROBLEM SOLVING You deposit $300 into an investment account that earns 12% annual interest compounded quarterly. The graph shows the balance of a savings account over time. a. Write functions that represent the balances of the accounts after t years. b. Graph the functions from part (a) in the same coordinate plane. Compare the account balances. 63. PROBLEM SOLVING A city has a population of 25,000. The population is expected to increase by 5.5% annually for the next decade. (See Example 7.) a. Write a function that represents the population y after t years. b. Find the approximate monthly percent increase in population. c. Graph the function from part (a). Use the graph to estimate the population after 4 years. 64. PROBLEM SOLVING Plutonium-238 is a material that generates steady heat due to decay and is used in power systems for some spacecraft. The function y = a(0.5)t/x represents the amount y of a substance remaining after t years, where a is the initial amount and x is the length of the half-life (in years). Plutonium-238 Half-life ≈ 88 years a. A scientist is studying a 3-gram sample. Write a function that represents the amount y of plutonium-238 after t years. b. What is the yearly percent decrease of plutonium-238? c. Graph the function from part (a). Use the graph to estimate the amount remaining after 12 years. 65. COMPARING FUNCTIONS The three given functions describe the amount y of ibuprofen (in milligrams) in a person’s bloodstream t hours after taking the dosage. y ≈ 800(0.71)t y ≈ 800(0.9943)60t y ≈ 800(0.843)2t a. Show that these expressions are approximately equivalent. b. Describe the information given by each of the functions. Savings Account 00 2 4 6 200 400 600 Balance (dollars) Year t y 322 Chapter 6 Exponential Functions and Sequences 66. COMBINING FUNCTIONS You deposit $9000 in a savings account that earns 3.6% annual interest compounded monthly. You also save $40 per month in a safe at home. Write a function C(t) = b(t) + h(t), where b(t) represents the balance of your savings account and h(t) represents the amount in your safe after t years. What does C(t) represent? 67. NUMBER SENSE During a fl u epidemic, the number of sick people triples every week. What is the growth rate as a percent? Explain your reasoning. 68. HOW DO YOU SEE IT? Match each situation with its graph. Explain your reasoning. a. A bacterial population doubles each hour. b. The value of a computer decreases by 18% each year. c. A deposit earns 11% annual interest compounded yearly. d. A radioactive element decays 5.5% each year. A. 0 0 200 400 8 16 t y B. 0 0 200 400 8 16 t y C. 0 0 200 400 8 16 t y D. 0 0 200 400 8 16 t y 69. WRITING Give an example of an equation in the form y = abx that does not represent an exponential growth function or an exponential decay function. Explain your reasoning. 70. THOUGHT PROVOKING Describe two account options into which you can deposit $1000 and earn compound interest. Write a function that represents the balance of each account after t years. Which account would you rather use? Explain your reasoning. 71. MAKING AN ARGUMENT A store is having a sale on sweaters. On the fi rst day, the prices of the sweaters are reduced by 20%. The prices will be reduced another 20% each day until the sweaters are sold. Your friend says the sweaters will be free on the fi fth day. Is your friend correct? Explain. 72. COMPARING FUNCTIONS The graphs of f and g are shown. y g(t) = kf(t) f(t) = 2t t 2 −2 −4 4 2 4 6 8 a. Explain why f is an exponential growth function. Identify the rate of growth. b. Describe the transformation from the graph of f to the graph of g. Determine the value of k. c. The graph of g is the same as the graph of h(t) = f (t + r). Use properties of exponents to fi nd the value of r. Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Solve the equation. Check your solution. (Section 1.3) 73. 8x + 12 = 4x 74. 5 − t = 7t + 21 75. 6(r − 2) = 2r + 8 Find the slope and the y-intercept of the graph of the linear equation. (Section 3.5) 76. y = −6x + 7 77. y = 1 — 4 x + 7 78. 3y = 6x − 12 79. 2y + x = 8 Reviewing what you learned in previous grades and lessons
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Conversion ft^2 to m^2, ft2 to m2, sq ft to sq m. Square feet to square meters. 1 square foot is 0.0929 square meters. Conversion square feet to square meters, ft 2 to m 2. The conversion factor is 0.09290304; therefore, 1 square foot = 0.09290304 square meters. In other words, to convert a value in ft 2 to m 2, divide by the value by 10.76391041671. The calculator answers the questions such as: How many m 2 is 90 ft 2? or How to convert ft 2 to m 2. This is a area unit conversion from ft 2 to m 2. Conversion result: 1 ft 2 = 0.0929 m 2 1 square foot is 0.0929 square meters. Enter area ft 2 Choose other area units Conversion of an area unit in word math problems and questions A pressure How much compressive force does an aquarium with a bottom area of 0.24 m² exert on the table when a pressure of 3 kPa is created? Pascal's law Please calculate according to Pascal's law. Krupp's machines were known for their large size. In 1861, the inventor put a blacksmith's steam hydraulic press into operation in Essen. What was the cross-sectional area of the larger piston if a compressive f Apartment area Anton wanted to measure the area of the apartment. But he bought a meter only 1.5m long. Later, he recalled that he had two meters long at 4.5m and 18m at home. How many times are home gauges longer than the meter he bought? Area Calculate: x = 8 m² + 8 dm² + 8 cm² Compressive force The excavator's area of belts is 5 m 2, and it creates a pressure of 40 kPa. How much compressive force does the excavator exert on the road? more math problems » All maths calculators Math Practice Problems 18530 Worksheets © 2025 HackMath.net | contact | en | cz | sk
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Pythagorean Triples with a Fixed Difference between a Leg and the Hypotenuse Anna Little Advisor: Michael Spivey April 22, 2005 Abstract This project examines sets of Pythagorean triples with a fixed difference k between one of the legs and the hypotenuse. A Pythagorean triple is an ordered triple of positive integers ( x, y, z ) such that x2 + y2 = z2. This paper proposes and proves a formula that will generate every triple of a difference k, that is, every triple where either z − x = k or z − y = k. The proof of this result utilizes the general formula for all primitive Pythagorean triples as well as the prime factorization of the integer k. 1 Introduction Pythagorean triples are a very ancient area in mathematics and throughout the centuries numerous mathe-maticians have deepened our understanding of their characteristics. Sierpinski provides an excellent summary on some of their most well known properties in . By far the most significant discovery in this area was made by Euclid (c. 350 B.C.), who found a formula that characterized every primitive Pythagorean triple . One of the first known references to Pythagorean triples of a fixed difference is found in the writings of Plato. Plato notes that (4 n, 4n2 − 1, 4n2 + 1) yields an infinite set of Pythagorean triples of difference 2 as n ranges over the positive integers . Building on these foundations, this project sought to find a formula that would yield every Pythagorean triple of an arbitrary positive difference k. Numerous formulas for specific k values were found and analyzed until a correct formula was obtained. As noted, the proof of the formula utilizes Euclid’s classification of primitive Pythagorean triples and well-known number theoretic properties. Although similar results have been published , the formula and its proof were arrived at independently. 2 Formula for Pythagorean Triples of a Difference k We have the following theorem as our main result: Theorem 1. Define f (k) = 2 b n02 cpb n12 c 1 pb n22 c 2 ...p b nr 2c r , (2.1) where 2n0 pn1 1 pn2 2 ...p nr r is the prime factorization of 2k. Then every triple of difference k is given by: ( kf (k) (2 m + k mod 2) , 2kf (k)2 (m − b f (k)2 c)( m + d f (k)2 e), 2kf (k)2 (m − b f (k)2 c)( m + d f (k)2 e) + k ) as m runs through all positive integers greater than b f (k)2 c. Now by Euclid’s general formula for all primitive Pythagorean triples , any triple can be written as (l(2 mn ), l (m2 − n2), l (m2 + n2)), where l is some positive integer and m > n > 0 are two relatively prime positive integers of opposite parity. Throughout this paper x will be used to designate the leg that is a multiple of 2 mn , y will be used to designate the leg that is a multiple of m2 − n2, and z will designate the hypotenuse. Thus ( x, y, z ) and ( y, x, z ) are simply two different notations for the same triple and will be used interchangeably. 1We will discuss briefly the outline of the proof of this theorem, and then we will proceed to the actual proof. Let Sk be the set of all triples of difference k, that is, the set of every triple where the hypotenuse is separated by k from a leg. Let Gk be the set of all triples generated by the above formula as m runs over all the positive integers greater than b f (k)2 c. We will show that these sets are equal, that is, that Sk ⊆ Gk and that Gk ⊆ Sk. To show that Sk ⊆ Gk, we will choose an arbitrary triple ( x, y, z ) of difference k and show that it is infact generated by the formula given in Theorem 1. In doing this we will examine two cases: k is even and k is odd. When k is even, we will have two subcases: z − x = k and z − y = k. When k is odd, however, we will have only z − x = k. After showing that in every case the triple ( x, y, z ) is generated by the formula given in Theorem 1, we conclude that ( x, y, z ) ∈ Gk and thus that Sk ⊆ Gk. To show that Gk ⊆ Sk, we pick an arbitrary element ( q, r, s ) that is generated by the formula given in Theorem 1, and we show that it is infact a Pythagorean triple of difference k. To do this, we have to verify that q, r, and s are all positive integers, that q2 + r2 = s2, and that either s − q = k or s − r = k. After showing that ( q, r, s ) is indeed a Pythagorean triple of difference k, we conlcude that ( q, r, s ) ∈ Sk and therefore that Gk ⊆ Sk. We will then have shown that the sets are equal and thus that the formula given in Theorem 1 produces every triple of differnce k and nothing else. We will now begin the actual proof. First of all, we show that Sk ⊆ Gk. Let k be an arbitrary positive integer and let ( x, y, z ) ∈ Sk. We will look at the cases where k is even and where k is odd, and we will show that in both cases ( x, y, z ) ∈ Gk. Case 1. Let k be an even integer. By the general formula for primitive Pythagorean triples, ( x, y, z ) is of the form: x = l(2 mn ) y = l(m2 − n2) z = l(m2 + n2), (2.2) where l is some positive integer and m > n > 0 are two relatively prime positive integers of opposite parity. Now since this triple has a difference k, either z − x = k or z − y = k. We will examine both of these two subcases. Subcase 1: Suppose that z − y = k. Then l(m2 + n2) − l(m2 − n2) = k, which yields k = 2 ln 2. Thus n = √ k 2l , and substituting this expression for n into equation 2.2, we get: x = m√2lk y = lm 2 − k 2 z = lm 2 + k 2 . (2.3) We will use the following lemma to show ( x, y, z ) ∈ Gk: Lemma 1. If (x, y, z ) is a triple where z − y = k, and thus has the form given above, it can be written in the form x = m1 √2l1ky = l1m21 − k 2 z = l1m21 + k 2 , (2.4) for some integers l1 and m1, where l1 is the smallest positive integer such that 2l1k is a perfect square and m1 = md , where d = √2lk √2l1k .Proof. Let q = √2l1k, where l1 is the smallest positive integer such that 2 l1k is a perfect square. Let p = √2lk , which must be an integer since x is an integer. We will show that q divides p. Now q = √2l1k =√2n0 pn1 1 ...p nr r l1, where 2 n0 pn1 1 ...p nr r is the prime factorization of 2 k. Now let po1, p o2, ..., p oj be the primes 2with odd exponents and let pe1, p e2, ..., p es be the primes with even exponents. Since l1 is the smallest integer such that q ∈ Z, l1 = po1po2...p oj . Now let no1, n o2, ..., n oj be the odd exponents and let ne1, n e2, ..., n es be the even exponents. Then q = √ pe1ne1 pe2ne2 ...p es nes po1no1+1 po2no2+1 ...p oj noj +1 . (2.5) Now consider p = √2lk . We have: p = √ pe1ne1 pe2ne2 ...p es nes po1no1 po2no2 ...p oj noj l. (2.6) Thus in order for p to be an integer, l must contain at least one copy of each of the primes with odd exponents. Thus po1po2...p oj |l. Let r = lpo1po2...p oj . Thus: p = √ pe1ne1 pe2ne2 ...p es nes po1no1+1 po2no2+1 ...p oj noj +1 r. (2.7) Now since every prime exponent is even, every exponent in the prime factorization of r must also be even, or p won’t be an integer. Thus r must be a perfect square, and √r = d, for some positive integer d. Thus: pq = √pe1ne 1 pe2ne 2 ...p es nes po1no 1+1 po2no 2+1 ...p oj noj +1 r √pe1ne 1 pe2ne 2 ...p es nes po1no 1+1 po2no 2+1 ...p oj noj +1 = √r = d. Thus p = qd , and q|p. Now let m1 = md . Then: m1 √2l1k = md √2l1k = m√2l1kr = m√2lk = x (since l = po1po2...p oj r = l1r) l1m21 − k 2 = l1(md )2 − k 2 = l1m2r − k 2 = lm 2 − k 2 = yl1m21 + k 2 = l1(md )2 + k 2 = l1m2r + k 2 = lm 2 + k 2 = z. (2.8) Thus we have shown that the triple ( x, y, z ) we selected, where z − y = k, can be written in the form: x = m1 √2l1ky = l1m21 − k 2 z = l1m21 + k 2 , (2.9) for some integers l1 and m1, where l1 is the smallest positive integer such that 2 l1k is a perfect square. Now we need to show that this triple is in the set Gk. We will do this by showing that there exists some integer m2 > b f (k)2 c such that: x = kf (k) (2 m2 + k mod 2) y = 2kf (k)2 (m2 − b f (k)2 c)( m2 + d f (k)2 e) z = 2kf (k)2 (m2 − b f (k)2 c)( m2 + d f (k)2 e) + k. (2.10) Before we can show this, however, we must prove an additional lemma, which holds regardless of the parity of k.3Lemma 2. If l1 is the smallest positive integer such that 2l1k is a perfect square and f (k) is as defined in Theorem 1, then l1 = 2kf (k)2 .Proof. Let 2 n0 pn1 1 pn2 2 ...p nr r be the prime factorization of 2 k. Now let po1, p o2, ..., p oj be the primes with odd exponents and let pe1, p e2, ..., p es be the primes with even exponents; also, let ne1, n e2, ..., n es be the even exponents and let no1, n o2, ..., n oj be the odd exponents. Then: 2kf (k)2 = 2n0 pn11 pn22 ...p nr r (2 b n02 cpb n12 c 1 pb n22 c 2 ...p b nr 2c r )2 = pe1ne 1 pe2ne 2 ...p es nes po1no 1 po2no 2 ...p oj noj (pe bne 12c 1 pe bne 22c 2 ...p e bnes 2c s )2(po bno 12c 1 po bno 22c 2 ...p o bnoj 2c j )2 = pe1ne 1 pe2ne 2 ...p es nes po1no 1 po2no 2 ...p oj noj pene 11 pene 22 ...p enes s pono 1−11 pono 2−12 ...p onoj −1 j = po1po2...p oj = l1.(Since l1 is the smallest integer such that 2 l1k is a perfect square, it contains exactly one copy of each prime that has an odd exponent in the prime factorization of 2 k.) Now we will proceed to show that the integer m1, as described in Lemma 1, is the integer that will generate the triple ( x, y, z ) when inserted into our formula for Gk, as given in Theorem 1. Since y > 0 and y = l1m21 − k 2 , we have: l1m21 − k 2 0 l1m21 > k 22kf(k)2 m21 > k 2 (using the substitution from Lemma 2) 4m21 > f (k)2 2m1 > f (k) m1 > f (k)2 m1 > b f (k)2 c. (2.11) Thus since m1 is an integer and m1 > b f (k)2 c, m1 is in the domain of the formula given in Theorem 1. Now 4since k is even, we have: x = m1 √2l1k = m1 √ 2k2kf (k)2 = 2 m1 kf (k) = kf (k) (2 m1 + k mod 2) y = l1m21 − k 2 = 2kf (k)2 m21 − 2k 4 = 2kf (k)2 (m21 − f (k)2 4 )= 2kf (k)2 (m1 − f (k)2 )( m1 + f (k)2 )= 2kf (k)2 (m1 − b f (k)2 c)( m1 + d f (k)2 e) z = l1m21 + k 2 = l1m21 − k 2 k = 2kf (k)2 (m1 − b f (k)2 c)( m1 + d f (k)2 e) + k. (2.12) Note that since k is even, f (k) is also clearly even, since 2 k has at least two factors of 2 in it. Thus f(k)2 = b f (k)2 c = d f (k)2 e, which was used in the above derivation. Thus we have shown that when k is even and when z − y = k, ( x, y, z ) ∈ Gk because the integer m1 produces it. Subcase 2: Now suppose that z − x = k (k still even). We will make use of the following lemma to show that ( x, y, z ) ∈ Gk. Note, however, that this lemma does not require k to be even. Lemma 3. If (x, y, z ) is a triple such that z − x = k, then there exist integers m2 and l2 such that x = 2 m2(m2l2 − √kl 2) y = 2 m2 √kl 2 − kz = 2 m2(m2l2 − √kl 2) + k, (2.13) where l2 is the smallest positive integer such that kl 2 is a perfect square. Proof. Let ( x, y, z ) be a triple where z − x = k. By the general formula for primitive Pythagorean triples, x = l(2 mn ) y = l(m2 − n2) z = l(m2 + n2), (2.14) where l is some positive integer and m > n > 0 are two relatively prime positive integers of opposite parity. Since z − x = k, we have l(m2 + n2) − l(2 mn ) = k, which yields n = m − √ kl . Substituting this expression 5for n into equation 2.14, we get: x = 2 m(ml − √kl ) y = 2 m√kl − kz = 2 m(ml − √kl ) + k. (2.15) Now we need to show that there exists some integer m2 such that: x = 2 m2(m2l2 − √kl 2) y = 2 m2 √kl 2 − kz = 2 m2(m2l2 − √kl 2) + k, (2.16) where l2 is the smallest positive integer such that kl 2 is a perfect square. Let q = √kl 2 and let p = √kl (note that p must be an integer for x, y, and z to be integers). Let 2 a0 pa1 1 pa2 2 ...p ar r be the prime factorization of k.Also, let po1, p o2, ..., p oj be the primes with odd exponents and let pe1, p e2, ..., p es be the primes with even exponents. Now let ao1, a o2, ..., a oj be the odd exponents and let ae1, a e2, ..., a es be the even exponents. We will show that q divides p. Now, q = √kl 2 = √ pa0 0 pa1 1 pa2 2 ...p ar r l2 = √ pe1ae1 pe2ae2 ...p es aes po1ao1 po2ao2 ...p oj aoj l2. (2.17) Since l2 is the smallest integer that will make every exponent under the radical even, l2 = po1po2...p oj , and √kl 2 = √ pe1ae1 pe2ae2 ...p es aes po1ao1+1 po2ao2+1 ...p oj aoj +1 . (2.18) Now, p = √kl = √ pa0 0 pa1 1 pa2 2 ...p ar r l = √ pe1ae1 pe2ae2 ...p es aes po1ao1 po2ao2 ...p oj aoj l. (2.19) Now in order for p to be an integer, l must contain as factors at least one copy of every prime with an odd exponent in the prime factorization of k; thus l2|l, so l = l2r, for some integer r. The integer r must also contain only even exponents in its prime factorization, or p will again not be an integer. Thus √r = d, for some integer d. So we have: √kl = √ pe1ae1 pe2ae2 ...p es aes po1ao1+1 po2ao2+1 ...p oj aoj +1 r, (2.20) and therefore pq = √kl √kl 2 = √ pe1ae1 pe2ae2 ...p es aes po1ao1+1 po2ao2+1 ...p oj aoj +1 r √ pe1ae1 pe2ae2 ...p es aes po1ao1+1 po2ao2+1 ...p oj aoj +1 = √r = d. (2.21) Thus q|p. Now let m2 = md . Then: 2m2(m2l2 − √kl 2) = 2 md (mdl 2 − √kl 2)= 2 m2rl 2 − 2md √kl 2 = 2 m2l − 2m√kl 2r = 2 m2l − 2m√kl = 2 m(ml − √kl )= x (2.22) 62m2 √kl 2 − k = 2 md √kl 2 − k = 2 m√kl 2r − k = 2 m√kl − k = y 2m2(m2l2 − √kl 2) + k = 2 md (mdl 2 − √kl 2) + k = 2 m(ml − √kl ) + k = z. (2.23) Thus we have shown that for any triple ( x, y, z ) where z − x = k, there exist integers m2 and l2 such that x = 2 m2(m2l2 − √kl 2) y = 2 m2 √kl 2 − kz = 2 m2(m2l2 − √kl 2) + k, (2.24) where l2 is the smallest positive integer such that kl 2 is a perfect square. Now we are trying to show that when k is even and z − x = k, our triple ( x, y, z ) ∈ Gk. In Lemma 3, we established the general form that any triple ( x, y, z ) where z − x = k must have. We now state an additional lemma that provides a simplification of this form when k is required to be even. Lemma 4. Suppose (x, y, z ) is a triple such that z − x = k for some even k. Then there exists some integer m1 such that x = l1m21 − k 2 y = m1 √2l1kz = l1m21 + k 2 , (2.25) where l1 is the smallest integer such that √2kl 1 is a perfect square. Proof. Let ( x, y, z ) be a triple such that z − x = k for some even k. Now by Lemma 3, ( x, y, z ) has the following form: x = 2 m2(m2l2 − √kl 2) y = 2 m2 √kl 2 − kz = 2 m2(m2l2 − √kl 2) + k, (2.26) for some integers m2 and l2, where l2 is the smallest positive integer such that kl 2 is a perfect square. Now k is even, and it has either an even or an odd number of 2’s in its prime factorization. We will show that in either case, an integer m1 can be found that will allow us to write the triple in the form given in Lemma 4. Case 1: Suppose k has an odd number of 2’s in its prime factorization. Then k = pe1ne1 pe2ne2 ...p es nes 2no1 po2no2 ...p oj noj , where 2 , p o2, ..., p oj are the prime factors with odd exponents and pe1, p e2, ..., p es are the prime factors with even exponents. Recall that since l1 is the smallest integer such that 2 kl 1 is a perfect square, l1 = po2...p oj (since k contains an odd number of 2’s, there is no need for an additional factor of 2 to make 2 kl 1 a perfect square; thus 2 is not a factor of l1). However, l2 must have a factor of 2 in order for kl 2 to be a perfect square; namely, l2 = 2 po2...p oj . Therefore, l1 = l2 2 , that is l2 = 2 l1. Thus: 2kl 1 = pe1ne1 pe2ne2 ...p es nes 2no1+1 po2no2+1 ...p oj noj +1 , (2.27) 7which implies that k√2kl 1 = pe1ne 1 pe2ne 2 ...p es nes 2no 1 po2no 2 ...p oj noj pe1 ne 12pe2 ne 22...p es nes 22no 1+1 2po2 no 2+1 2...p oj noj +1 2 = pe1 ne 12 pe2 ne 22 ...p es nes 2 2 no 1−12 po2 no 2−12 ...p oj noj −12 . (2.28) Now since ne1, n e2, ..., n es and no1 − 1, n o2 − 1, ..., n oj − 1 are all even, nonnegative integers, ne1 2 , ne2 2 ..., nes 2 and no1−12 , no2−12 , ..., noj −12 are all nonnegative integers, and thus the above product is an integer. Namely, k√2kl 1 is an integer. Now let m1 = 2 m2 − k√2kl 1 . We will show that this is the integer we need to write the triple ( x, y, z ) in the form given in Lemma 4. We have: x = 2 m2(m2l2 − √kl 2)= 2( m1 2 k 2√2kl 1 )( 2l1m1 2 2l1k 2√2kl 1 − √2kl 1)= ( m1 + k√2kl 1 )( l1m1 + l1k√2kl 1 − √2kl 1)= ( m1 + √k√2l1 )( l1m1 + √l1k√2 − √2kl 1)= l1m12 + m1 √l1k√2 − m1 √2kl 1 + m1 √l1k√2 k√l1 2√l1 − k = l1m12 + m1( √l1k√2 − √2√2kl 1√2 √l1k√2 ) − k 2 = l1m12 + m1( 2√l1k−2√l1k√2 ) − k 2 = l1m12 − k 2 y = 2 m2 √kl 2 − k = 2( m1 2 k 2√2kl 1 )( √2kl 1) − k = m1 √2kl 1 z = 2 m2(m2l2 − √kl 2) + k = l1m12 − k 2 k = l1m12 + k 2 , (2.29) since we just showed that x = 2 m2(m2l2 − √kl 2) = l1m12 − k 2 . Thus when there are an odd number of 2’s in the prime factorization of k, a triple ( x, y, z ) where z − x = k can be written in the form given in Lemma 4. Now we proceed to the second case in the proof of Lemma 4. Case 2. Suppose k has an even number of 2’s in its prime factorization. We will show that m1 = m2− k√2kl 1 is the integer that will allow us to write ( x, y, z ) in the desired form. First of all, however, we must show that k√2kl 1 is indeed an integer. Since k has an even number of 2’s in its prime factorization, k = 2 ne1 pe2ne2 ...p es nes po1no1 po2no2 ...p oj noj ,where po1, p o2, ..., p oj are the prime factors with odd exponents and 2 , p e2, ..., p es are the prime factors with 8even exponents. Now since k contains an even number of 2’s, l1 must contain a factor of 2 in order for 2 kl 1 to be a perfect square. However, l2 does not need a factor of 2 in order for l2k to be a perfect square, and thus l1 = 2 l2. So we get l1 = 2 po1po2...p oj and 2 kl 1 = 2 ne1+2 pe2ne2 ...p es nes po1no1+1 po2no2+1 ...p oj noj +1 . Thus: k√2kl 1 = 2ne 1 pe2ne 2 ...p es nes po1no 1 po2no 2 ...p oj noj √2ne 1+2 pe2ne 2 ...p es nes po1no 1+1 po2no 2+1 ...p oj noj +1 = 2ne 1 pe2ne 2 ...p es nes po1no 1 po2no 2 ...p oj noj 2 ne 1+2 2 pe2 ne 22 ...p es nes 2 po1 no 1+1 2 po2 no 2+1 2 ...p oj noj +1 2 = 2 ne 1−22 pe2 ne 22 ...p es nes 2 po1 no 1−12 po2 no 2−12 ...p oj noj −12 . Now ne1 − 2, n e2, ..., n es and no1 − 1, n o2 − 1, ..., n oj − 1 are all even; thus all the exponents are integers. Note that since k is even, ne1 ≥ 2, and thus ne1 − 2 ≥ 0. Thus we know that all of the exponents are in fact nonnegative integers, and thus the above product is an integer; namely, k√2kl 1 is an integer. Now let m1 = m2 − k√2kl 1 . Then x = 2 m2(m2l2 − √kl 2)= 2( m1 + k√2kl 1 )( l1m1 2 l1k 2√2kl 1 − √ kl 1 2 )= (2 m1 + √2k√l1 )( l1m1 2 √l1k 2√2 − √l1k√2 )= l1m21 + m1 √l1k√2 − m1 √2kl 1 + m1 √2kl 1 2 k 2 − k = l1m21 + m1( √2l1k 2 − 2√2kl 1 2 √2kl 1 2 ) − k 2 = l1m21 − k 2 y = 2 m2 √kl 2 − k = 2( m1 + k√2kl 1 )( √ kl 1 2 ) − k = (2 m1 + 2k√2kl 1 )( √ kl 1 2 ) − k = m1 √2kl 1 + k − k = m1 √2kl 1 z = 2 m2(m2l2 − √kl 2) + k = l1m21 − k 2 k = l1m21 + k 2 , (2.30) since we already showed in our analysis of x that 2 m2(m2l2 − √kl 2) = l1m21 − k 2 .Thus we have shown that when ( x, y, z ) is a triple where z − x = k for some even k, there exists some 9integer m1 such that the triple can be written as: x = l1m21 − k 2 y = m1 √2l1kz = l1m21 + k 2 , (2.31) where l1 is the smallest integer such that 2 kl 1 is a perfect square. We are trying to show that when ( x, y, z )is a triple where z − x = k for some even k, ( x, y, z ) ∈ Gk. Having shown in Lemma 4 that ( x, y, z ) has the above form, we can now use a derivation almost identical to the one in 2.12 to show that ( x, y, z ) ∈ Gk;namely, we will show that when the integer m1 is inserted into the formula given in Theorem 1, the triple (x, y, z ) is produced. Note that since x > 0 and x = l1m21 − k 2 , we have l1m21 − k 2 0, and an identical argument to the one given in 2.11 shows that m1 > b f (k)2 c. Thus m1 is in the domain of the formula given in Theorem 1, and we have: x = l1m21 − k 2 = 2kf (k)2 m21 − 2k 4 = 2kf (k)2 (m21 − f (k)2 4 )= 2kf (k)2 (m1 − f (k)2 )( m1 + f (k)2 )= 2kf (k)2 (m1 − b f (k)2 c)( m1 + d f (k)2 e) y = m1 √2l1k = m1 √ 2k2kf (k)2 = m1 2kf (k) = kf (k) 2m1 = kf (k) (2 m1 + k mod 2) z = l1m21 + k 2 = l1m21 − k 2 k = 2kf (k)2 (m1 − b f (k)2 c)( m1 + d f (k)2 e) + k. (2.32) Thus we see that when the integer m1 is inserted into the formula given in Theorem 1, the triple ( y, x, z ) is produced. Since we are considering the triple ( y, x, z ) to be the same triple as ( x, y, z ), ( x, y, z ) ∈ Gk. In the above derivation we once again made use of the fact that l1 = 2kf (k)2 , which we proved in Lemma 2. Thus we have shown that when k is even, ( x, y, z ) ∈ Gk for the subcases z − y = k and z − x = k. Since these are the only possible subcases, we have shown that when k is even, ( x, y, z ) ∈ Gk. We now proceed to Case 2, where k is odd. Case 2: Suppose that k is an odd integer. Now by the general formula for primitive Pythagorean triples, 10 we know that ( x, y, z ) has the following form: x = l(2 mn ) y = l(m2 − n2) z = l(m2 + n2), (2.33) where l is some positive integer and m > n > 0 are two relatively prime positive integers of opposite parity. Now since we are supposing this triple has difference k, either z − x = k or z − y = k. Observe that z − y = l(m2 + n2) − l(m2 − n2) = 2 ln 2; thus z − y is even. Since k is odd, z − y 6 = k, and we must have z − x = k. Thus for k odd, we do not have two subcases like we did for k even. Now by Lemma 3 (since Lemma 3 holds for both even and odd k), the triple ( x, y, z ) can be written in the following form for some integer m2: x = 2 m2(m2l2 − √kl 2) y = 2 m2 √kl 2 − kz = 2 m2(m2l2 − √kl 2) + k, (2.34) where l2 is the smallest positive integer such that kl 2 is a perfect square. We need to show that this triple is in the set Gk. We will proceed by showing that the triple ( y, x, z ) ∈ Gk, which clearly implies that (x, y, z ) ∈ Gk. To show ( y, x, z ) ∈ Gk, we must show that there exists some integer m1 greater than b f (k)2 c such that: y = kf (k) (2 m1 + k mod 2) x = 2kf (k)2 (m1 − b f (k)2 c)( m1 + d f (k)2 e) z = 2kf (k)2 (m1 − b f (k)2 c)( m1 + d f (k)2 e) + k. (2.35) Recall that l1 is the smallest positive integer such that 2 kl 1 is a perfect square and that l2 is the smallest integer such that kl 2 is a perfect square. Thus l2 contains one copy of every prime factor of k that has an odd exponent, while l1 contains one copy of every prime factor of 2 k that has an odd exponent. Since 2 is not a factor of k, l2 does not contain a factor of 2, but since 2 is a factor of 2 k and has an odd exponent in the prime factorization of 2 k, l1 does have a factor of 2. However, all the other factors of l1 and l2 are clearly the same. Thus we have l2 = l1 2 . Recall that l1 = 2kf (k)2 , as shown in Lemma 2, which holds regardless of the parity of k. So: l2 = l1 2 = 2k 2f (k)2 = kf (k)2 . (2.36) Now let m1 = m2 − f (k)+1 2 . We will show that m1 is the integer that will produce the triple ( y, x, z ) when inserted into the formula given in Theorem 1. First of all, however, we must show that m1 is in the domain of the formula given in Theorem 1; namely, we must verify that m1 is an integer and that m1 > b f (k)2 c.Now since k is odd, there is only one 2 in the prime fctorization of 2 k, that is 2 k = 2 1pn1 1 pn2 2 ...p nr r , where 2, p 1, p 2, ..., p r are the distinct prime factors of 2 k. Thus f (k) = 2 b 12 cpb n12 c 1 pb n22 c 2 ...p b nr 2c r , which shows that f (k) has no factors of 2. Thus f (k) is odd, making f (k) + 1 even and f (k)+1 2 an integer. Therefore, m1 = m2 − f (k)+1 2 is also an integer. 11 Note also that since x > 0 and x = 2 m2(m2l2 − √kl 2), we must have m2l2 − √kl 2 > 0. Thus: (m1 + f (k)+1 2 )l2 > √kl 2 (m1 + f (k)+1 2 ) kf (k)2 > √ k2 f (k)2 m1 + f (k)+1 2 > f (k) m1 > 2f (k)2 − f (k)+1 2 m1 > f (k)−12 m1 > b f (k)2 c (since k is odd, f (k) is odd, and thus f (k)−12 = b f (k)2 c). (2.37) Thus m1 is an integer greater than b f (k)2 c and is thus in the domain of the formula given in Theorem 1. Since m2 = m1 + f (k)+1 2 , we get: y = 2 m2 √kl 2 − k = 2 m2 √ k2 f (k)2 − k = 2 m2 kf (k) − k = 2( m1 + f (k)+1 2 )( kf (k) ) − k = (2 m1 + f (k) + 1)( kf (k) ) − k = 2kf (k) m1 + k + kf (k) − k = kf (k) (2 m1 + 1) = kf (k) (2 m1 + k mod 2) x = 2 m2(m2l2 − √kl 2)= 2( m1 + f (k)+1 2 )[( m1 + f (k)+1 2 ) kf (k)2 − √ k2 f (k)2 ]= (2 m1 + f (k) + 1)( m1kf (k)2 + k 2f (k) + k 2f (k)2 − kf (k) )= (2 m1 + f (k) + 1)( m1kf (k)2 + k 2f (k)2 − k 2f (k) )= m21 2kf (k)2 + m1 kf (k)2 − m1 kf (k) + m1 kf (k) + k 2f (k) − k 2 + m1 kf (k)2 + k 2f (k)2 − k 2f (k) = m21 2kf (k)2 + m1 2kf (k)2 − k 2 + k 2f (k)2 = 2kf (k)2 (m21 + m1 − f (k)2 4 + 14 )= 2kf (k)2 (m21 + m1( f (k)+1 2 − f (k)−12 ) − ( f (k)2−14 )) = 2kf (k)2 (m21 − m1( f (k)−12 ) + m1( f (k)+1 2 ) − ( f (k)−12 )( f (k)+1 2 )) (2.38) 12 Now since k is odd, f (k) is odd, and thus we have that b f (k)2 c = f (k)−12 and d f (k)2 e = f (k)+1 2 . Substituting this we get: x = 2kf (k)2 (m21 − m1b f (k)2 c + m1d f (k)2 e − d f (k)2 eb f (k)2 c)= 2kf (k)2 (m1 − b f (k)2 c)( m1 + d f (k)2 e). (2.39) Likewise, we see that: z = x + k = 2kf (k)2 (m1 − b f (k)2 c)( m1 + d f (k)2 e) + k. (2.40) Thus when ( x, y, z ) is a triple where z − x = k for some odd integer k, ( y, x, z ) is clearly an element of Gk because it is the triple that is generated when the integer m1 is inserted into the formula given in Theorem 1. Since, again, we are considering ( y, x, z ) to be the same triple as ( x, y, z ), we have shown that ( x, y, z ) ∈ Gk.Thus we have shown that for an arbitrary triple ( x, y, z ) of difference k, that is for an arbitrary element of Sk, ( x, y, z ) ∈ Gk both when k is even and when k is odd. Therefore we have shown that Sk ⊆ Gk. For k even, we examined two subcases: z − y = k and z − x = k, while for k odd we had only z − x = k.To conclude the proof of our main theorem, we must now show that Gk ⊆ Sk. Let ( q, r, s ) be an arbitrary element of Gk and let m and k be the integers required by Theorem 1 to produce ( q, r, s ). In order for (q, r, s ) ∈ Sk, we need to verify that ( q, r, s ) is a Pythagorean triple of difference k. Note that if ( q, r, s ) is a Pythagorean triple, then it is clearly a Pythagorean triple with a difference of k. For since ( q, r, s ) has the following form: q = kf (k) (2 m + k mod 2) r = 2kf (k)2 (m − b f (k)2 c)( m + d f (k)2 e) s = 2kf (k)2 (m − b f (k)2 c)( m + d f (k)2 e) + k, (2.41) clearly s − r = k. Therefore, all that needs to be shown is that ( q, r, s ) is a Pythagorean triple, namely that q, r, and s are positive integers and that q2 + r2 = s2.Now since m > b f (k)2 c, clearly q, r, and s must all be positive. Thus if kf (k) and 2kf (k)2 are integers, then q, r, and s must in fact be positive integers. It is clear from Lemma 2 that 2kf (k)2 is an integer; thus we will simply show that kf (k) is an integer. We will look at the k even and k odd cases. Case 1: Suppose k is even and let 2 n0 pn1 1 ...p nr r be the prime factorization of k. Then: kf(k) = 2n0 pn11 ...p nr r 2bn0+1 2cpbn12c 1pbn22c 2...p bnr 2c r = 2 n0−b n0+1 2 cpn1−b n12 c 1 pn2−b n22 c 2 ...p nr −b nr 2c r . (2.42) Now since every exponent in the above expression is clearly nonnegative, kf (k) is an integer. Case 2: Suppose k is odd and that k = pn1 1 pn2 2 ...p nr r is the prime factorization of k. Then: kf(k) = pn11 pn22 ...p nr r 2b12cpbn12c 1pbn22c 2...p bnr 2c r = pn11 pn22 ...p nr r 20pbn12c 1pbn22c 2...p bnr 2c r = pn1−b n12 c 1 pn2−b n22 c 2 ...p nr −b nr 2c r (2.43) Once again, since every exponent is clearly a nonnegative integer, kf (k) is also an integer. Thus we have verified that q, r, and s are indeed positive integers. Now we will show that q2 + r2 = s2.Once again, we will look at two cases. 13 Case 1: Suppose that k is even. Then since k is even, f (k) is also even because it clearly contains a factor of 2. Thus b f (k)2 c = d f (k)2 e = f (k)2 . Also, k mod 2 = 0. Thus our formulas for q, r, and s can be simplified as follows: q = kf (k) (2 m) r = 2kf (k)2 (m − f (k)2 )( m + f (k)2 ) s = 2kf (k)2 (m − f (k)2 )( m + f (k)2 ) + k. (2.44) Thus: q2 + r2 = (2 m kf (k) )2 + [ 2kf (k)2 (m − f (k)2 )( m + f (k)2 )] 2 = (2 m kf (k) )2 + [ 2kf (k)2 (m2 − f (k)2 4 )] 2 = (2 m kf (k) )2 + ( 2kf (k)2 m2 − k 2 )2 = 4 m2 k2 f(k)2 4k2 f(k)4 m4 − 2k2 f(k)2 m2 + k2 4 = 4k2 f(k)4 m4 + 2k2 f(k)2 m2 + k2 4 = ( 2kf (k)2 m2 + k 2 )2 = ( 2kf (k)2 m2 − k 2 k)2 = [ 2kf (k)2 (m2 − f (k)2 4 ) + k]2 = [ 2kf (k)2 (m − f (k)2 )( m + f (k)2 ) + k]2 = s2 (2.45) Case 2: Suppose that k is odd. Then f (k) is also odd since it does not contain a factor of 2. Thus b f (k)2 c = f (k)−12 and d f (k)2 e = f (k)+1 2 . Also, k mod 2 = 1. Thus our formulas for q, r, and s simplify as follows: q = kf (k) (2 m + 1) r = 2kf (k)2 (m − f (k)−12 )( m + f (k)+1 2 ) s = 2kf (k)2 (m − f (k)−12 )( m + f (k)+1 2 ) + k. (2.46) Thus: q2 + r2 = [ kf (k) (2 m + 1)] 2 + [ 2kf (k)2 (m − f (k)−12 )( m + f (k)+1 2 )] 2 = k2 f(k)2 (4 m2 + 4 m + 1) + 4k2 f(k)4 [( m − f (k)−12 )( m + f (k)+1 2 )] 2 = 4k2 f(k)2 m2 + 4k2 f(k)2 m + k2 f(k)2 4k2 f(k)4 [( m − f (k)−12 )( m + f (k)+1 2 )] 2 = 4k2 f(k)2 m2 + 4k2 f(k)2 m − k2 + k2 f(k)2 k2 + 4k2 f(k)4 [( m − f (k)−12 )( m + f (k)+1 2 )] 2 (2.47) 14 = 4k2 f(k)2 (m2 + m − f (k)2 4 14 ) + k2 + 4k2 f(k)4 [( m − f (k)−12 )( m + f (k)+1 2 )] 2 = 4k2 f(k)2 (m − f (k)−12 )( m + f (k)+1 2 ) + k2 + 4k2 f(k)4 [( m − f (k)−12 )( m + f (k)+1 2 )] 2 = 4k2 f(k)4 [( m − f (k)−12 )( m + f (k)+1 2 )] 2 + 4k2 f(k)2 (m − f (k)−12 )( m + f (k)+1 2 ) + k2 = [ 2kf (k)2 (m − f (k)−12 )( m + f (k)+1 2 ) + k]2 = s2 (2.48) So we see that ( q, r, s ), both for k even and k odd, is indeed a Pythagorean triple with a difference of k,and so ( q, r, s ) ∈ Sk. Thus we have shown that Gk ⊆ Sk. Finally, since Sk ⊆ Gk and Gk ⊆ Sk, Gk = Sk,which concludes the proof of our main theorem. Thus the formula given in Theorem 1 produces every triple of difference k for any positive integer k. 3 Alternate Formula for Pythagorean Triples of Difference k Having proved that Gk = Sk, we can now prove some additional properties about Sk using this equivalence. In particular, we have the following result: Theorem 2. The set of all triples of difference k can be generated by the union of f (k) disjoint, infinite subsets, where the following property holds for each of these subsets: ddn Leg 2( n) = 2( Leg 1( n)) , (3.1) where (Leg 1( n), Leg 2( n), Leg 2( n) + k) denote the generating formulas of the subset. In particular, for k even: Sk = ⋃f (k)−1 i=0 {(2 kn + 2ki f (k) , 2kn 2 + 4ki f (k) n + 2ki 2 f(k)2 − k 2 , 2kn 2 + 4ki f (k) n + 2ki 2 f(k)2 k 2 ) : n ∈ Z, n > 12 − if (k) }. (3.2) Similarly, for k odd: Sk = ⋃f (k)−1 i=0 {(2 kn + 2ik f (k) + kf (k) , 2kn 2 + 4ki f (k) n + 2kf (k) n + 2ki 2 f(k)2 2ki f (k)2 − k 2 14 , 2kn 2 + 4ki f (k) n + 2kf (k) n + 2ki 2 f(k)2 2ki f (k)2 + k 2 14 ) : n ∈ Z, n > 12 − 12f (k) − if (k) }. (3.3) The proof of this theorem is a clear application of Theorem 1. Once again, we will examine both the k even and k odd cases. Proof. Let k be some positive integer. Now by Theorem 1, every triple of difference k is generated by the following formula as m ranges over the positive integers greater than b f (k)2 c:( kf (k) (2 m + k mod 2) , 2kf (k)2 (m − b f (k)2 c)( m + d f (k)2 e), 2kf (k)2 (m − b f (k)2 c)( m + d f (k)2 e) + k). (3.4) Case 1: Suppose k is even. Then our generating formula simplifies to: ( kf (k) (2 m), 2kf (k)2 (m − f (k)2 )( m + f (k)2 ), 2kf (k)2 (m − f (k)2 )( m + f (k)2 ) + k). (3.5) Thus as m ranges over the positive integers greater than b f (k)2 c every triple of difference k is produced. It is easy to see that every distinct positive integer produces a unique triple when inserted into the formula. Now 15 by the division algorithm , we can partition all the positive integers greater than b f (k)2 c into their residue classes mod f (k). If we were partitioning all of the positive integers into their residue classes mod f (k), the ith residue class would be given by f (k)n + i, where n is any nonnegative integer. However, to eliminate producing integers in the ith residue class that are smaller than b f (k)2 c, we must have f (k)n + i > b f (k)2 c,which happens exactly when n > 12 − if (k) . Thus the set {f (k)n + i : n ∈ Z, n > 12 − if (k) } is exactly the ith residue class mod f (k) of the positive integers greater than b f (k)2 c. The restriction on n causes n to start at 0 or 1 as necessary, so that an integer smaller than b f (k)2 c is never produced. Now running m over this particular residue class will produce a subset of Sk. In order to only produce the triples generated by the ith residue class, we must substitute f (k)n + i for m in the formula given in Theorem 1 and allow n to range over the integers greater than 12 − if (k) . After making this substitution into our formula, we find that (2 kn + 2ki f (k) , 2kn 2 + 4ki f (k) n + 2ki 2 f (k)2 − k 2 , 2kn 2 + 4ki f (k) n + 2ki 2 f (k)2 + k 2 ) (3.6) gives every triple generated by the ith residue class of the domain of Theorem 1 as n ranges over the integers greater than 12 − if (k) . Now since every triple produced by our original formula is unique and the residue classes are disjoint, the subsets produced by these residue classes will clearly be disjoint. Also, if we take the union of the triples generated by every residue class we will have the triples generated by every positive integer greater than b f (k)2 c, that is, we will have the set of all triples of difference k. Thus we find that, for k even: Sk = ⋃f (k)−1 i=0 {(2 kn + 2ki f (k) , 2kn 2 + 4ki f (k) n + 2ki 2 f(k)2 − k 2 , 2kn 2 + 4ki f (k) n + 2ki 2 f(k)2 k 2 ) : n ∈ Z, n > 12 − if (k) }, (3.7) as stated in Theorem 2. Thus we have rewritten Sk as the union of f (k) disjoint, infinite subsets. Notice also that, allowing Leg 1( n) = 2 kn + 2ki f (k) and Leg 2( n) = 2 kn 2 + 4ki f (k) n + 2ki 2 f(k)2 − k 2 to denote the generating formulas for the two legs in the ith residue class, the following property holds for each of the subsets in the above union: ddn Leg 2( n) = ddn (2 kn 2 + 4ki f (k) n + 2ki 2 f(k)2 − k 2 )= 4 kn + 4ki f (k) = 2(2 kn + 2ki f (k) )= 2( Leg 1( n)) . (3.8) Thus we have seen that Theorem 2 holds for even k.Case 2: Let k be an odd positive integer. Then our generating formula for Sk simplifies as follows: ( kf (k) (2 m + 1) , 2kf (k)2 (m − f (k) − 12 )( m + f (k) + 1 2 ), 2kf (k)2 (m − f (k) − 12 )( m + f (k) + 1 2 ) + k). Thus as m ranges over the positive integers greater than b f (k)2 c, every triple of difference k is produced. Once again, we partition the positive integers greater than b f (k)2 c into their residue classes mod f (k). We 16 must ensure that only those integers greater than b f (k)2 c are produced, namely, we must have: f (k)n + i > b f (k)2 c f (k)n + i > f (k)−12 2f (k)n + 2 i > f (k) − 12f (k)n > f (k) − 1 − 2in > 12 − 12f (k) − if (k) . (3.9) Thus f (k)n+i, as n ranges over the integers greater than 12 − 12f (k) − if (k) , produces the ith residue class mod f (k) of the integers greater than b f (k)2 c. Therefore, substituting f (k)n + i for m in the Theorem 1 formula and allowing n to range over the integers greater than 12 − 12f (k) − if (k) produces the triples generated by the ith residue class. Making this substitution and simplifying, we get: (2 kn + 2ik f (k) + kf (k) , 2kn 2+ 4ki f (k) n+ 2kf (k) n+ 2ki 2 f (k)2 + 2ki f (k)2 − k 2 + 14 , 2kn 2+ 4ki f (k) n+ 2kf (k) n+ 2ki 2 f (k)2 + 2ki f (k)2 + k 2 + 14 )(3.10) as the generating fomula for the triples produced by the ith residue class. Thus, once again, since every triple produced by the Theorem 1 formula is unique and the residue classes are clearly disjoint, the subsets produced by the residue classes are disjoint. If we take the union of these subsets, we will get every triple generated by the Theorem 1 formula, that is, every triple of a difference k. Thus we can rewrite the set of all triples of difference k as: Sk = ⋃f (k)−1 i=0 {(2 kn + 2ik f (k) + kf (k) , 2kn 2 + 4ki f (k) n + 2kf (k) n + 2ki 2 f(k)2 2ki f (k)2 − k 2 14 , 2kn 2 + 4ki f (k) n + 2kf (k) n + 2ki 2 f(k)2 2ki f (k)2 − k 2 14 + k) : n ∈ Z, n > 12 − 12f (k) − if (k) }. (3.11) Once again allowing Leg 1( n) = 2 kn + 2ik f (k) + kf (k) and Leg 2( n) = 2 kn 2 + 4ki f (k) n+ 2kf (k) n+ 2ki 2 f(k)2 2ki f (k)2 − k 2 14 to be the generating formulas for the legs of the triples produced by the ith residue class, we find that: ddn Leg 2( n) = ddn (2 kn 2 + 4ki f (k) n + 2kf (k) n + 2ki 2 f(k)2 2ki f (k)2 − k 2 14 )= (4 kn + 4ki f (k) + 2kf (k) )= 2(2 kn + 2ki f (k) + kf (k) )= 2( Leg 1( n)) . (3.12) Thus for k odd we can very similarly write Sk as the union of f (k) disjoint, infinite subsets where the property ddn Leg 2( n) = 2( Leg 1( n)) holds for the generating formulas of the legs of each subset. It is interesting to note that this derivative property in general does not hold for the generating formulas of the legs of the larger set Sk. 4 Conclusion In summary, we have found and proved two separate but connected formulas that will generate every Pythagorean triple of a fixed difference k. First of all, we found that when when we define f (k) = 2 b n02 cpb n12 c 1 pb n22 c 2 ...p b nr 2c r , (4.1) 17 where 2 n0 pn1 1 pn2 2 ...p nr r is the prime factorization of 2 k, then every triple of a difference k is given by: ( kf (k) (2 m + k mod 2) , 2kf (k)2 (m − b f (k)2 c)( m + d f (k)2 e), 2kf (k)2 (m − b f (k)2 c)( m + d f (k)2 e) + k)as m runs through the positive integers greater than b f (k)2 c. To prove this result, we let ( x, y, z ) be an arbitrary Pythagorean triple with a difference of k and examined both the k even and k odd cases. For k even, we had two subcases: z − x = k and z − y = k. For k odd, we had only z − x = k. We showed that in every case the triple would be generated by our formula. We also showed that everything generated by the above formula is indeed a Pythagorean triple with difference k. Having completed the proof of our main theorem, we then used the theorem to find an alternate formula that also generates every triple of difference k. Namely, we found that, for even k, every triple of a difference k is given by the following: f(k)−1 ⋃ i=0 {(2 kn + 2ki f (k) , 2kn 2 + 4ki f (k) n + 2ki 2 f (k)2 − k 2 , 2kn 2 + 4ki f (k) n + 2ki 2 f (k)2 + k 2 ) : n ∈ Z, n > 12 − if (k) }, (4.2) and that for odd k, every triple of difference k is given by: ⋃f (k)−1 i=0 {(2 kn + 2ik f (k) + kf (k) , 2kn 2 + 4ki f (k) n + 2kf (k) n + 2ki 2 f(k)2 2ki f (k)2 − k 2 14 , 2kn 2 + 4ki f (k) n + 2kf (k) n + 2ki 2 f(k)2 2ki f (k)2 + k 2 14 ) : n ∈ Z, n > 12 − 12f (k) − if (k) }. (4.3) For each subset in the above unions we noted that the following property holds for the generating formulas of the legs of the triples: ddn Leg 2( n) = 2( Leg 1( n)) . (4.4) A better understanding of why the above property holds for the given subsets but not for the entire set of Pythagorean triples of difference k, as well as an investigation into other unions that yield the entire set, are areas of future research. Also, the formula in Theorem 1 appears to produce the negative Pythagorean triples. The proof of the theorem, however, made use of Euclid’s classification, which only holds for positive triples. Thus another area of future research is investigating whether the formula given in Theorem 1 and its proof can be extended to include all negative Pythagorean triples. 18 References Apostol, Tom M. Introduction to Analytic Number Theory. Springer-Verlag, Inc., New York, 1976. Burton, David M. Elementary Number Theory. McGraw-Hill, New York, 2002. Konvalina, John. A characterization of the Pythagorean triples. A collection of manuscripts related to the Fibonacci sequence, pp. 160-162, Fibonacci Assoc., Santa Clara, CA, 1980. McCullough, Darryl. Height and Excess of Pythagorean Triples. Mathematics Magazine. Vol. 78(2005), no. 1, pp. 26-44. Wade, Peter W.; Wade, William R. Recursions that produce Pythagorean triples. College Math Journal. Vol. 31(2000), no.2, pp. 98-101. Sierpinski, Waclaw. Pythagorean Triples . Graduate School of Science, Yeshiva University. New York, 1962. Turnbull, Herbert Westren. The Great Mathematicians. Barnes and Noble Books, New York, 1993. 19
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https://cs.brown.edu/courses/cs100/lectures/lecture15a.pdf
Plan for the week ●M: Classical Statistics ○ Confidence Intervals ●W: Classical Statistics (cont’d) ○ Hypothesis Testing ●F: Introduction to Machine Learning ○ Supervised Learning (Regression + Classification) ○ Socially Responsible Computing Confidence Intervals Point Estimate Given a statistical model of a population, a point estimate is a single value used to estimate a model parameter. Examples: ● A sample mean is often used to estimate the mean (i.e., the “true” mean) of a normal distribution. ● Likewise, a sample proportion is often used to estimate the probability of success of a binomial distribution. But even the very best, data-driven estimate is usually wrong! ● Goal is to find not just a single point estimate, but an interval estimate ● An interval estimate is delimited by an upper and lower bound. As such, it specifies an interval where a parameter of interest might fall ● Further, it quantifies the uncertainty in the estimate, via a confidence level, α ○ α is usually small ○ α = 0.05 implies a 95% confidence interval ○ α = 0.10 implies a 90% confidence interval Interval Estimate Image Source Central Limit Theorem By the CLT, a sampling distribution approximates a normal with mean 𝝁 and sd σ. So it gives us an interval around an estimate: Pr[𝝁 - zloσ ≤ estimate ≤ 𝝁 + zhiσ] = 1 - α. Likewise, an interval around the mean: Pr[estimate - zloσ ≤ 𝝁 ≤ estimate + zhiσ] = 1 - α A confidence interval is a pair θlo ≤ θhi such that Pr[θlo ≤ θ ≤ θhi] ≥ 1 - α, where the randomness stems from the sampling process and impacts θlo and θhi A confidence interval is a NOT pair θlo ≤ θhi such that Prθ[θlo ≤ θ ≤ θhi] ≥ 1 - α, because θ is NOT a random quantity (in classical statistics, anyway!) Potential Pitfall Confidence Intervals ● Our goal is to build a confidence interval around the true mean 𝝁: Pr[estimate - zloσ ≤ 𝝁 ≤ estimate + zhiσ] = 1 - α ● The CLT gives us an interval around our estimate: Pr[𝝁 - zloσ ≤ estimate ≤ 𝝁 + zhiσ] = 1 - α ● Rearranging yields Pr[estimate - zloσ ≤ 𝝁 ≤ estimate + zhiσ] = 1 - α ● The choice of α dictates values for zlo and zhi ○ E.g., α = 0.5 implies zlo = 1.96 and zhi = -1.96 Standard Normal Table Image Source In the olden days (back when I was a student), we used the standard normal table to answer queries. Find zlo and zhi s.t. P[zlo ≤ Z ≤ zhi] = 1 - α ● If 1 - α = 90%, then |zα/2| = z1-α/2 = 1.645 ● If 1 - α = 95%, then |zα/2| = z1-α/2 = 1.96 ● If 1 - α = 98%, then |zα/2| = z1-α/2 = 2.33 ● If 1 - α = 99%, then |zα/2| = z1-α/2 = 2.58 zlo = -zα/2 & zhi = z1-α/2 are called critical values E.g., P[-1.96 ≤ Z ≤ 1.96] = .95 Standard Normal Table Image Source These days, we use R: ● qnorm(0.975)= 1.959964 Find zlo and zhi s.t. P[zlo ≤ Z ≤ zhi] = 1 - α ● If 1 - α = 90%, then |zα/2| = z1-α/2 = 1.645 ● If 1 - α = 95%, then |zα/2| = z1-α/2 = 1.96 ● If 1 - α = 98%, then |zα/2| = z1-α/2 = 2.33 ● If 1 - α = 99%, then |zα/2| = z1-α/2 = 2.58 zlo = -zα/2 & zhi = z1-α/2 are called critical values E.g., P[-1.96 ≤ Z ≤ 1.96] = .95 Let’s say you are running for mayor! ● You hire a polling agency to determine if you are likely to win or not. ● After sampling 100 likely voters, the agency reports that 55 of those 100 support you. ● But there are more than 100 likely voters! ● Is .55, the sample mean, also the population mean? ○ The polling agency also reports a 95% confidence interval around the number 55: e.g., (45, 65). ■ Polling agencies tend to report margins of error (MoE = 10). ■ MoE and confidence intervals are related. ○ But wait…how did the polling agency come up with this confidence interval/MoE? Two Brown alumni were recently elected mayor in Greece, in Athens and in Thessaloniki. Let’s say you are running for mayor! ● You hire a polling agency to determine if you are likely to win or not. ● After sampling 100 likely voters, the agency reports that 55 of those 100 support you. ● But there are more than 100 likely voters! ● Is .55, the sample mean, also the population mean? ○ The polling agency also reports a 95% confidence interval around the number 55: e.g., (45, 65). ■ Polling agencies tend to report margins of error (MoE = 10). ■ MoE and confidence intervals are related. ○ Here’s what this doesn’t quite mean: ■ There is a 95% chance the proportion of the population who support you lies in this interval. ○ Here’s what this does mean: ■ If the polling agency were to take this poll for you repeatedly, then the true proportion of voters who support you would fall within this interval 95% of the time. ○ But wait…how did the polling agency come up with this confidence interval/MoE? Two Brown alumni were recently elected mayor in Greece, in Athens and in Thessaloniki. ● Let X be the number of people who support you in a poll, and let phat = X/n. ● By the central limit theorem (YAY!), for large n, phat is approximately normal, with mean 𝝁 and standard deviation (a.k.a. standard error) σ. ● Pr[𝝁 - zloσ ≤ phat ≤ 𝝁 + zhiσ] = 1 - α ● Choose α = 0.05, and rearrange: ● Pr[phat - 1.96σ ≤ 𝝁 ≤ phat + 1.96σ] = 0.95 ● Sounds good, but what is the standard error of phat? Building a Confidence Interval ● Sounds good, but what is the standard error of phat? ○ 𝝁 = E[phat] = E[X/n] = 1/n E[X] ○ Var[phat] = Var[X/n] = 1/n2 Var[X] ● This begs the question: what is the variance of X? ● Well, how is it distributed? Easy: X is binomially distributed. ○ Mean = np ○ Variance = np(1-p) ● The mean of phat is p. ○ 𝝁 = E[phat] = E[X/n] = np/n = p ● The variance of phat is p(1 - p)/n: ○ Var[phat] = Var[X/n] = 1/n2 Var[X] = 1/n2 (np)(1 - p) = p(1 - p)/n ○ SE = √p(1 - p)/n The mean and variance of X ● Let X be the number of people who support you in a poll, and let phat = X/n. ● By the central limit theorem (YAY!), for large n, phat is approximately normal, with mean 𝝁 and standard deviation (a.k.a. standard error) σ. ● Pr[𝝁 - zloσ ≤ phat ≤ 𝝁 + zhiσ] = 1 - α ● Choose α = 0.05, and rearrange: ● Pr[phat - 1.96σ ≤ 𝝁 ≤ phat + 1.96σ] = 0.95 ● Sounds good, but what is the standard error of phat? ● Now we know σ: SE = √p(1 - p)/n ● Or do we? Building a Confidence Interval (cont’d) New problem: We don’t know p! ● We estimate p by phat. ○ N.B. This is cheating, but only negligibly so. Hence, we build our 95% confidence interval as follows: ● Let σhat = √phat(1 - phat)/n = sqrt{(.55)(.45)/100} = 0.05 ● Pr[phat - zloσhat ≤ 𝝁 ≤ phat + zhiσhat] = .95 ○ Lower Bound: 0.55 + (-1.96)(0.05) = .45 ○ Upper Bound: 0.55 + (1.96)(0.05) = .65 ● The value (1.96)(0.05) ≈ 0.1 is called the margin of error. Building a Confidence Interval (cont’d) Deriving Standard Error Standard Error ● The sample mean is the average of all the sample values. ● The sample variance is the average of the squared deviations from the mean. ● The sample standard deviation is the square root of the sample variance. ● The standard error is the standard deviation of the sampling distribution. ● In this class, you will learn to build confidence intervals for sample means and sample proportions. ● To do so requires the standard error of sample means and sample proportions. ● Let XM represent the sample mean: ○ XM = (X1 + … + XN)/n ● First, we need the variance of the sample mean: ○ Xi are normally distributed with variance σ2 ○ Var[XM] = Var[(X1 + … + XN)/n] = (1/n2) (n) Var[X1] = σ2/n ● We now have a formula for the standard error of the sample mean: ○ SE[XM] = σ/sqrt{n} The Standard Error of the Sample Mean ● Let Phat represent the sample proportion: ○ Phat = X/n, where X is a binomial random variable distributed according to (n, p) ● First, we need the variance of the sample proportion: ○ X ~ B(n, p) ○ Var[Phat] = Var[X/n] = (1/n2)(np)(1 - p) = p(1 - p)/n ● We now have a formula for the standard error of the sample proportion: ○ SE[Phat] = sqrt{p(1 - p)/n} The Standard Error of the Sample Proportion ● Let’s find the standard error of the difference between two sample means. ○ Let XM - YM represent the difference between two sample means. ● XM - YM = (X1 + … + XN)/n - (Y1 + … + YM)/m ○ Xi and Yi are normally distributed with variance σX and σY ○ Var[XM - YM] = Var[XM] + Var[YM] = Var[(X1 + … + XN)/n] + Var[(Y1 + … + YM)/m] = (1/n2)n Var[X1] + (1/m2)m Var[Y1] = σX 2/n + σY 2/m ● We now have a formula for the standard error of the difference between two sample means: ○ SE[XM] = sqrt{σX 2/n + σY 2/m} Difference of Two Sample Means ● Let’s find the standard error of the difference between two sample proportions. ○ Let P1 represent one proportion, and P2, a second proportion. ● P1 = X/n and P2 = Y/m ○ X ~ B(n, p1) and Y ~ B(m, p2) ○ Var[P1- P2] = Var[X/n] + Var[Y/m] = (1/n2)(np1)(1 - p1) + (1/m2)(mp2)(1 - p2) = p1(1 - p1)/n + p2(1 - p2)/m ● We now have a formula for the standard error of the difference between two sample proportions: ○ SE[P1 - P2] = sqrt{p1(1 - p1)/n + p2(1 - p2)/m} Difference of Two Sample Proportions Student t-Distribution When does the CLT kick in? ● The distribution looks roughly normal with a sample size of 30. ● So, as a rule of thumb, people often say that 30 is a large enough sample size for the central limit theorem to apply. ● However, the histogram becomes more and more bell-shaped as the sample size increases. ● So all things being equal, larger sample sizes are always better than smaller ones! What is a large enough sample size? What if the sample size is not large enough? ● We can extend our methodologies by introducing a new distribution, called the (Student) t-distribution, which approximates the normal. ○ The student in question was one William S. Gosset. ○ He discovered this (family of) distribution(s) in 1908, while employed as a statistician by the Guinness brewing company, who forbid him from publishing under his own name. ○ He wrote under the pen name “Student” instead. ● The t-distribution allows us to perform statistical inference even when the sample size is not large enough to apply the central limit theorem. The t-Distribution ● The t-distribution has 1 parameter: the degrees of freedom. ● As n goes to infinity, the t-distribution converges to the normal. ● Thus, using the t-distribution when the sample size is small is consistent with using the normal distribution when the sample size is large. Image source Building Confidence Intervals ● The only difference between building confidence intervals using the t-distribution and building them using the normal is: the critical values for the normal distribution are familiar numbers to most statisticians. Recall: ○ Find zlo and zhi s.t. P[zlo ≤ Z ≤ zhi] = 1 - α% ■ If 1 - α = 90%, then |zα/2| = z1-α/2 = 1.645 ■ If 1 - α = 95%, then |zα/2| = z1-α/2 = 1.96 ■ If 1 - α = 98%, then |zα/2| = z1-α/2 = 2.33 ■ If 1 - α = 99%, then |zα/2| = z1-α/2 = 2.58 ○ zlo = zα/2 & zhi = z1-α/2 are called critical values. ● For the t-distribution, we have to look up the critical values. ○ There is more uncertainty because their are fewer samples. ○ So fixing α, the magnitude of these values is higher. An Example ● Let’s say we polled 30 people to see who they thought would win the 2016 presidency, and 60% of them said Clinton. ● We can build a 95% confidence interval as follows: ○ The estimate is 0.6. ○ The SE = sqrt{(.6)(.4)/30} = 0.09. ○ We look up the critical t-values: ■ qt(0.025, 29) = -2.05 and qt(0.975, 29) = 2.05 ■ The number 29 is the degrees of freedom, which is the number of values that are free to vary ● The 95% CI is (0.6 - (2.05)(0.09), 0.6 + (2.05)(0.09)) = (0.42, 0.78) An Example ● Let’s say we polled 20 people to see who they thought would win the 2016 presidency, and 60% of them said Clinton. ● We can build a 95% confidence interval as follows: ○ The estimate is 0.6. ○ The SE = sqrt{(.6)(.4)/20} = 0.11. ○ We look up the critical t-values: ■ qt(0.025, 19) = -2.09 and qt(0.975, 19) = 2.09 ■ The number 19 is the degrees of freedom, which is the number of values that are free to vary ● The 95% CI is (0.6 - (2.09)(0.11), 0.6 + (2.09)(0.11)) = (0.37, 0.83) An Example ● If we poll 30 people, the 95% CI is (0.6 - (2.05)(0.09), 0.6 + (2.05)(0.09)) = (0.42, 0.78) ● If we poll 20 people, the 95% CI is (0.6 - (2.09)(0.11), 0.6 + (2.09)(0.11)) = (0.37, 0.83) ● If we poll 10 people, the 95% CI is (0.6 - (2.26)(0.155), 0.6 + (2.26)(0.155)) = (0.25, 0.95) ● Notice how the width of the confidence interval increases as the sample size decreases, holding the significance level α constant. Back to your run for mayor, using Student t ● The polling agency polled 100 people to see if they support you for mayor, and 55 people said they did. ● We can build a 95% confidence interval as follows: ○ The estimate is 0.55. ○ The SE = sqrt{(.55)(.45)/100} = 0.05. ○ We look up the critical t-values: ■ qt(0.025, 99) = -1.98 and qt(0.975, 99) = 1.98 ■ The number 99 is the degrees of freedom, which is the number of values that are free to vary ● The 95% CI is (0.55 - (1.98)(0.05), 0.55 + (1.98)(0.05)) = (0.451, 0.649) ● The 95% CI was (0.55 - (1.96)(0.05), 0.55 + (1.96)(0.05)) = (0.45, 0.65) Inference with Confidence Intervals ● A Harris Poll surveyed 2242 adults in the United States about their favorite flavor of ice cream. ○ 27% of people said chocolate was their favorite flavor. ○ 23% of people said vanilla was their favorite flavor. ● Is the preference for chocolate over vanilla significant? That is, do people really prefer chocolate? ● Plan of attack: ○ Compare the confidence intervals for chocolate and vanilla. ○ If the confidence intervals overlap, we cannot conclude that there is a difference between the proportion of people who prefer chocolate and the proportion who prefer vanilla. Favorite Ice Cream Flavors ● We need to calculate the standard error to find a confidence interval: ○ (.27)(2242) = 605 people preferred chocolate ○ (.23)(2242) = 516 people preferred vanilla ○ Var[pc-hat] = (0.27)(1 - 0.27) / 605 = 0.000326 ○ Var[pv-hat] = (0.23)(1 - 0.23) / 516 = 0.000343 ○ The standard error is the square root of the variance: ■ SE[pc-hat] = sqrt(0.000326) = 0.018 / 2242 ■ SE[pv-hat] = sqrt(0.000343) = 0.019 / 2242 ● So, the confidence intervals, at the 95% level, are: ○ For chocolate: [0.27 - (1.96)(0.018), 0.27 + (1.96)(0.018)] = [0.234, 0.305] ○ For vanilla: [0.23 - (1.96)(0.019), 0.23 + (1.96)(0.019)] = [0.193, 0.267] ● The confidence intervals overlap! ○ This means we cannot conclude, at the 95% confidence level, that people prefer chocolate to vanilla. ○ But maybe people prefer chocolate to vanilla at the 90% confidence level. (Do they? Find out for yourself.) Favorite Ice Cream Flavors ● A Harris Poll surveyed 2242 adults in the United States about their favorite flavor of ice cream. ○ 27% of people said chocolate was their favorite flavor. ○ 23% of people said vanilla was their favorite flavor. ● Is the preference for chocolate over vanilla significant? That is, do people really prefer chocolate? ● Plan of attack: ○ Look at the difference in the proportions. It is 4%. ○ Calculate the confidence interval around this difference. ○ If it contains 0, then we cannot conclude that there is a difference. Favorite Ice Cream Flavors, Revisited ● We need to calculate the standard error to find a confidence interval: ○ (.27)(2242) = 605 people preferred chocolate ○ (.23)(2242) = 516 people preferred vanilla ○ Var[P1 - P2] = (0.27)(1 - 0.27) / 605 + (0.23)(1 - 0.23) / 516 = 0.00067 ○ The standard error is the square root of the variance: sqrt(0.00067) = 0.026 ● So, the confidence interval, at the 95% level, is: ○ [0.04 - (1.96)(0.026), 0.04 + (1.96)(0.026)] ○ [-0.011, 0.09] ○ [-1.1%, 9%] ● 0% is included in this interval! ○ This means we cannot conclude, at the 95% confidence level, that people prefer chocolate to vanilla. ○ But maybe people prefer chocolate to vanilla at the 90% confidence level. (Check this!) Favorite Ice Cream Flavors, Revisited ● YouGov surveyed 1300 people in the United States and asked “Who will you vote for in the election for President in November?” ○ The poll was conducted on October 7th and 8th ○ 44% of people planned to vote for Clinton ○ 38% of people planned to vote for Trump ● Given these results, how confident are you that Clinton will win the election? ● Plan of attack: ○ Compare the confidence intervals for Clinton and Trump. ○ If the confidence intervals overlap, we cannot conclude that there is a difference between the proportion of people who prefer Clinton and the proportion who prefer Trump. The 2016 Presidential Election ● We need to calculate the standard error to find a confidence interval: ○ (.44)(1300) = 572 people preferred Clinton ○ (.38)(1300) = 494 people preferred Trump ○ Var[Pc-hat] = (0.44)(1 - 0.44) / 572 = 0.000431 ○ Var[Pt-hat] = (0.38)(1 - 0.38) / 494 = 0.000477 ○ The standard error is the square root of the variance: ■ SE[Pc-hat] = sqrt(0.00431) / 1300 = 0.021 ■ SE[Pt-hat] = sqrt(0.00477) / 1300 = 0.022 ● So, the confidence intervals, at the 95% level, are: ○ For Clinton: [0.44 - (1.96)(0.021), 0.44 + (1.96)(0.021)] = [0.40, 0.48] ○ For Trump: [0.38 - (1.96)(0.022), 0.38 + (1.96)(0.022)] = [0.34, 0.42] ● The confidence intervals overlap! (Don’t say we didn’t warn you!) ○ Actually, this tells us nothing: if confidence intervals don’t overlap, then we can conclude that the difference is statistically significant; but if they do overlap, then we cannot draw any statistical conclusions. The 2016 Presidential Election ● YouGov surveyed 1300 people in the United States and asked “Who will you vote for in the election for President in November?” ○ The poll was conducted on October 7th and 8th ○ 44% of people planned to vote for Clinton ○ 38% of people planned to vote for Trump ● Given these results, how confident are you that Clinton will win the election? ● Plan of attack: ○ Look at the difference in the proportions. It is 6%. ○ Calculate the confidence interval around this difference. ○ If it contains 0, then we cannot conclude that there is a difference. The 2016 Presidential Election, Revisited ● We need to calculate the standard error to find a confidence interval: ○ (.44)(1300) = 572 people preferred Clinton ○ (.38)(1300) = 494 people preferred Trump ○ Var[P1 - P2] = (0.44)(1 - 0.44)/572 + (0.38)(1 - 0.38)/494 = 0.0009 ○ The standard error is the square root of the variance: sqrt(0.0009) = 0.0301 ● So, the confidence interval, at the 95% level, is: ○ 0.06 - (1.96)(0.0301) to 0.06 + (1.96)(0.0301) ○ 0.001 to 0.119 ○ 0.1% to 11.9% ● 0% is not included in this interval! ○ This means we can conclude, at the 95% confidence level, that Clinton is preferred to Trump. ○ But people probably don’t prefer Clinton to Trump at the 99% confidence level. (Check this!) The 2016 Presidential Election, Revisited ● Our methods make use of the central limit theorem. ○ In the examples, we do not assume anything about how preferences are distributed. ○ However, by the CLT, we know the sampling distribution is approximately normal. ○ The surveys were large enough for the CLT to apply. However, the CLT may not have applied if the sample size were smaller. (Rule of thumb: CLT applies whenever n ≥ 30.) ● Very common mistake: a 95% confidence interval between 1 and 2 is often interpreted as a 95% chance the parameter lies between 1 and 2. ○ This is a misconception! ○ A 95% interval means that if we repeated the experiment 100 times, the parameter would be contained in 95 of the resulting 100 intervals. ○ These two interpretations are not the same. Be careful to avoid this potential pitfall! Some words of warning! Extras ● Physics: number of directions in which independent motion can occur ○ Elbows have one degree of freedom ○ Shoulders and wrists have three ● Chemistry: number of independent factors required to describe equilibrium ● Statistics: number of values in a calculation that are free to vary ○ Often, one less than the number of observations Degrees of Freedom
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https://courses.lumenlearning.com/wmopen-collegealgebra/chapter/introduction-probability/
Probability | College Algebra Skip to main content College Algebra Module 17: Probability and Counting Principles Search for: Probability Learning Outcomes Construct probability models. Compute probabilities of equally likely outcomes. Compute probabilities of the union of two events. Use the complement rule to find probabilities. Compute probability using counting theory. Residents of the Southeastern United States are all too familiar with charts, known as spaghetti models, such as the one above. They combine a collection of weather data to predict the most likely path of a hurricane. Each colored line represents one possible path. The group of squiggly lines can begin to resemble strands of spaghetti, hence the name. In this section, we will investigate methods for making these types of predictions. An example of a “spaghetti model,” which can be used to predict possible paths of a tropical storm. Construct Probability Models Suppose we roll a six-sided number cube. Rolling a number cube is an example of an experiment, or an activity with an observable result. The numbers on the cube are possible results, or outcomes, of this experiment. The set of all possible outcomes of an experiment is called the sample space of the experiment. The sample space for this experiment is {1,2,3,4,5,6}{1,2,3,4,5,6}. An event is any subset of a sample space. The likelihood of an event is known as probability. The probability of an event p p is a number that always satisfies 0≤p≤1 0≤p≤1, where 0 indicates an impossible event and 1 indicates a certain event. A probability model is a mathematical description of an experiment listing all possible outcomes and their associated probabilities. For instance, if there is a 1% chance of winning a raffle and a 99% chance of losing the raffle, a probability model would look much like the table below. | Outcome | Probability | --- | | Winning the raffle | 1% | | Losing the raffle | 99% | The sum of the probabilities listed in a probability model must equal 1, or 100%. How To: Given a probability event where each event is equally likely, construct a probability model. Identify every outcome. Determine the total number of possible outcomes. Compare each outcome to the total number of possible outcomes. Example: Constructing a Probability Model Construct a probability model for rolling a single, fair die, with the event being the number shown on the die. Show Solution Begin by making a list of all possible outcomes for the experiment. The possible outcomes are the numbers that can be rolled: 1, 2, 3, 4, 5, and 6. There are six possible outcomes that make up the sample space. Assign probabilities to each outcome in the sample space by determining a ratio of the outcome to the number of possible outcomes. There is one of each of the six numbers on the cube, and there is no reason to think that any particular face is more likely to show up than any other one, so the probability of rolling any number is 1 6 1 6. OutcomeRoll of 1 Roll of 2 Roll of 3 Roll of 4 Roll of 5 Roll of 6 Probability1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 Q & A Do probabilities always have to be expressed as fractions? No. Probabilities can be expressed as fractions, decimals, or percents. Probability must always be a number between 0 and 1, inclusive of 0 and 1. Try It Construct a probability model for tossing a fair coin. Show Solution | Outcome | Probability | --- | | Heads | 1 2 1 2 | | Tails | 1 2 1 2 | Computing Probabilities of Equally Likely Outcomes Let S S be a sample space for an experiment. When investigating probability, an event is any subset of S S. When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the event by the total number of outcomes in S S. Suppose a number cube is rolled, and we are interested in finding the probability of the event “rolling a number less than or equal to 4.” There are 4 possible outcomes in the event and 6 possible outcomes in S S, so the probability of the event is 4 6=2 3 4 6=2 3. A General Note: Computing the Probability of an Event with Equally Likely Outcomes The probability of an event E E in an experiment with sample space S S with equally likely outcomes is given by P(E)=number of elements in E number of elements in S=n(E)n(S)P(E)=number of elements in E number of elements in S=n(E)n(S) E E is a subset of S S, so it is always true that 0≤P(E)≤1 0≤P(E)≤1. Example: Computing the Probability of an Event with Equally Likely Outcomes A number cube is rolled. Find the probability of rolling an odd number. Show Solution The event “rolling an odd number” contains three outcomes. There are 6 equally likely outcomes in the sample space. Divide to find the probability of the event. P(E)=3 6=1 2 P(E)=3 6=1 2 Try It A number cube is rolled. Find the probability of rolling a number greater than 2. Show Solution 2 3 2 3 Probability for Multiple Events We are often interested in finding the probability that one of multiple events occurs. Suppose we are playing a card game, and we will win if the next card drawn is either a heart or a king. We would be interested in finding the probability of the next card being a heart or a king. The union of two eventsE and F,written E∪F E and F,written E∪F, is the event that occurs if either or both events occur. P(E∪F)=P(E)+P(F)−P(E∩F)P(E∪F)=P(E)+P(F)−P(E∩F) Suppose the spinner below is spun. We want to find the probability of spinning orange or spinning a b b. There are a total of 6 sections, and 3 of them are orange. So the probability of spinning orange is 3 6=1 2 3 6=1 2. There are a total of 6 sections, and 2 of them have a b b. So the probability of spinning a b b is 2 6=1 3 2 6=1 3. If we added these two probabilities, we would be counting the sector that is both orange and a b b twice. To find the probability of spinning an orange or a b b, we need to subtract the probability that the sector is both orange and has a b b. 1 2+1 3−1 6=2 3 1 2+1 3−1 6=2 3 The probability of spinning orange or a b b is 2 3 2 3. A General Note: Probability of the Union of Two Events The probability of the union of two events E E and F F (written E∪F E∪F ) equals the sum of the probability of E E and the probability of F F minus the probability of E E and F F occurring together (( which is called the intersection of E E and F F and is written as E∩F E∩F ). P(E∪F)=P(E)+P(F)−P(E∩F)P(E∪F)=P(E)+P(F)−P(E∩F) Example: Computing the Probability of the Union of Two Events A card is drawn from a standard deck. Find the probability of drawing a heart or a 7. Show Solution A standard deck contains an equal number of hearts, diamonds, clubs, and spades. So the probability of drawing a heart is 1 4 1 4. There are four 7s in a standard deck, and there are a total of 52 cards. So the probability of drawing a 7 is 1 13 1 13. The only card in the deck that is both a heart and a 7 is the 7 of hearts, so the probability of drawing both a heart and a 7 is 1 52 1 52. Substitute P(H)=1 4,P(7)=1 13,and P(H∩7)=1 52 P(H)=1 4,P(7)=1 13,and P(H∩7)=1 52 into the formula. P(E∪F)=P(E)+P(F)−P(E∩F)=1 4+1 13−1 52=4 13 P(E∪F)=P(E)+P(F)−P(E∩F)=1 4+1 13−1 52=4 13 The probability of drawing a heart or a 7 is 4 13 4 13. Try It A card is drawn from a standard deck. Find the probability of drawing a red card or an ace. Show Solution 7 13 7 13 Computing the Probability of Mutually Exclusive Events Suppose the spinner from earlier is spun again, but this time we are interested in the probability of spinning an orange or a d d. There are no sectors that are both orange and contain a d d, so these two events have no outcomes in common. Events are said to be mutually exclusive events when they have no outcomes in common. Because there is no overlap, there is nothing to subtract, so the general formula is P(E∪F)=P(E)+P(F)P(E∪F)=P(E)+P(F) Notice that with mutually exclusive events, the intersection of E E and F F is the empty set. The probability of spinning an orange is 3 6=1 2 3 6=1 2 and the probability of spinning a d d is 1 6 1 6. We can find the probability of spinning an orange or a d d simply by adding the two probabilities. P(E∪F)=P(E)+P(F)=1 2+1 6=2 3 P(E∪F)=P(E)+P(F)=1 2+1 6=2 3 The probability of spinning an orange or a d d is 2 3 2 3. A General Note: Probability of the Union of Mutually Exclusive Events The probability of the union of two mutually exclusive events E E and F F is given by P(E∪F)=P(E)+P(F)P(E∪F)=P(E)+P(F) How To: Given a set of events, compute the probability of the union of mutually exclusive events. Determine the total number of outcomes for the first event. Find the probability of the first event. Determine the total number of outcomes for the second event. Find the probability of the second event. Add the probabilities. Example: Computing the Probability of the Union of Mutually Exclusive Events A card is drawn from a standard deck. Find the probability of drawing a heart or a spade. Show Solution The events “drawing a heart” and “drawing a spade” are mutually exclusive because they cannot occur at the same time. The probability of drawing a heart is 1 4 1 4, and the probability of drawing a spade is also 1 4 1 4, so the probability of drawing a heart or a spade is 1 4+1 4=1 2 1 4+1 4=1 2 Try It A card is drawn from a standard deck. Find the probability of drawing an ace or a king. Show Solution 2 13 2 13 Find the Probability That an Even Will Not Happen We have discussed how to calculate the probability that an event will happen. Sometimes, we are interested in finding the probability that an event will not happen. The complement of an eventE E, denoted E′E′, is the set of outcomes in the sample space that are not in E E. For example, suppose we are interested in the probability that a horse will lose a race. If event W W is the horse winning the race, then the complement of event W W is the horse losing the race. To find the probability that the horse loses the race, we need to use the fact that the sum of all probabilities in a probability model must be 1. P(E′)=1−P(E)P(E′)=1−P(E) The probability of the horse winning added to the probability of the horse losing must be equal to 1. Therefore, if the probability of the horse winning the race is 1 9 1 9, the probability of the horse losing the race is simply 1−1 9=8 9 1−1 9=8 9 A General Note: The Complement Rule The probability that the complement of an event will occur is given by P(E′)=1−P(E)P(E′)=1−P(E) Example: Using the Complement Rule to Calculate Probabilities Two six-sided number cubes are rolled. Find the probability that the sum of the numbers rolled is less than or equal to 3. Find the probability that the sum of the numbers rolled is greater than 3. Show Solution The first step is to identify the sample space, which consists of all the possible outcomes. There are two number cubes, and each number cube has six possible outcomes. Using the Multiplication Principle, we find that there are 6×6 6×6, or 36 36 total possible outcomes. So, for example, 1-1 represents a 1 rolled on each number cube. 1 - 1 1 - 1 1 - 2 1 - 2 1 - 3 1 - 3 1 - 4 1 - 4 1 - 5 1 - 5 1 - 6 1 - 6 2 - 1 2 - 1 2 - 2 2 - 2 2 - 3 2 - 3 2 - 4 2 - 4 2 - 5 2 - 5 2 - 6 2 - 6 3 - 1 3 - 1 3 - 2 3 - 2 3 - 3 3 - 3 3 - 4 3 - 4 3 - 5 3 - 5 3 - 6 3 - 6 4 - 1 4 - 1 4 - 2 4 - 2 4 - 3 4 - 3 4 - 4 4 - 4 4 - 5 4 - 5 4 - 6 4 - 6 5 - 1 5 - 1 5 - 2 5 - 2 5 - 3 5 - 3 5 - 4 5 - 4 5 - 5 5 - 5 5 - 6 5 - 6 6 - 1 6 - 1 6 - 2 6 - 2 6 - 3 6 - 3 6 - 4 6 - 4 6 - 5 6 - 5 6 - 6 6 - 6 We need to count the number of ways to roll a sum of 3 or less. These would include the following outcomes: 1-1, 1-2, and 2-1. So there are only three ways to roll a sum of 3 or less. The probability is 3 36=1 12 3 36=1 12 Rather than listing all the possibilities, we can use the Complement Rule. Because we have already found the probability of the complement of this event, we can simply subtract that probability from 1 to find the probability that the sum of the numbers rolled is greater than 3. P(E′)=1−P(E)=1−1 12=11 12 P(E′)=1−P(E)=1−1 12=11 12 Try It Two number cubes are rolled. Use the Complement Rule to find the probability that the sum is less than 10. Show Solution 5 6 5 6 Computing Probability Using Counting Theory Many interesting probability problems involve counting principles, permutations, and combinations. In these problems, we will use permutations and combinations to find the number of elements in events and sample spaces. These problems can be complicated, but they can be made easier by breaking them down into smaller counting problems. Assume, for example, that a store has 8 cellular phones and that 3 of those are defective. We might want to find the probability that a couple purchasing 2 phones receives 2 phones that are not defective. To solve this problem, we need to calculate all of the ways to select 2 phones that are not defective as well as all of the ways to select 2 phones. There are 5 phones that are not defective, so there are C(5,2)C(5,2) ways to select 2 phones that are not defective. There are 8 phones, so there are C(8,2)C(8,2) ways to select 2 phones. The probability of selecting 2 phones that are not defective is: ways to select 2 phones that are not defective ways to select 2 phones=C(5,2)C(8,2)=10 28=5 14 ways to select 2 phones that are not defective ways to select 2 phones=C(5,2)C(8,2)=10 28=5 14 Example: Computing Probability Using Counting Theory A child randomly selects 5 toys from a bin containing 3 bunnies, 5 dogs, and 6 bears. Find the probability that only bears are chosen. Find the probability that 2 bears and 3 dogs are chosen. Find the probability that at least 2 dogs are chosen. Show Solution We need to count the number of ways to choose only bears and the total number of possible ways to select 5 toys. There are 6 bears, so there are C(6,5)C(6,5) ways to choose 5 bears. There are 14 toys, so there are C(14,5)C(14,5) ways to choose any 5 toys. C(6,5)C(14,5)=6 2,002=3 1,001 C(6,5)C(14,5)=6 2,002=3 1,001 We need to count the number of ways to choose 2 bears and 3 dogs and the total number of possible ways to select 5 toys. There are 6 bears, so there are C(6,2)C(6,2) ways to choose 2 bears. There are 5 dogs, so there are C(5,3)C(5,3) ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are C(6,2)⋅C(5,3)C(6,2)⋅C(5,3) ways to choose 2 bears and 3 dogs. We can use this result to find the probability. C(6,2)C(5,3)C(14,5)=15⋅10 2,002=75 1,001 C(6,2)C(5,3)C(14,5)=15⋅10 2,002=75 1,001 It is often easiest to solve “at least” problems using the Complement Rule. We will begin by finding the probability that fewer than 2 dogs are chosen. If less than 2 dogs are chosen, then either no dogs could be chosen, or 1 dog could be chosen.When no dogs are chosen, all 5 toys come from the 9 toys that are not dogs. There are C(9,5)C(9,5) ways to choose toys from the 9 toys that are not dogs. Since there are 14 toys, there are C(14,5)C(14,5) ways to choose the 5 toys from all of the toys. C(9,5)C(14,5)=63 1,001 C(9,5)C(14,5)=63 1,001 If there is 1 dog chosen, then 4 toys must come from the 9 toys that are not dogs, and 1 must come from the 5 dogs. Since we are choosing both dogs and other toys at the same time, we will use the Multiplication Principle. There are C(5,1)⋅C(9,4)C(5,1)⋅C(9,4) ways to choose 1 dog and 1 other toy. C(5,1)C(9,4)C(14,5)=5⋅126 2,002=315 1,001 C(5,1)C(9,4)C(14,5)=5⋅126 2,002=315 1,001 Because these events would not occur together and are therefore mutually exclusive, we add the probabilities to find the probability that fewer than 2 dogs are chosen. 63 1,001+315 1,001=378 1,001 63 1,001+315 1,001=378 1,001 We then subtract that probability from 1 to find the probability that at least 2 dogs are chosen. 1−378 1,001=623 1,001 1−378 1,001=623 1,001 Try It A child randomly selects 3 gumballs from a container holding 4 purple gumballs, 8 yellow gumballs, and 2 green gumballs. Find the probability that all 3 gumballs selected are purple. Find the probability that no yellow gumballs are selected. Find the probability that at least 1 yellow gumball is selected. Show Solution a.1 91 b.5 91 c.86 91 a.1 91 b.5 91 c.86 91 Key Equations probability of an event with equally likely outcomes P(E)=n(E)n(S)P(E)=n(E)n(S) probability of the union of two events P(E∪F)=P(E)+P(F)−P(E∩F)P(E∪F)=P(E)+P(F)−P(E∩F) probability of the union of mutually exclusive events P(E∪F)=P(E)+P(F)P(E∪F)=P(E)+P(F) probability of the complement of an event P(E')=1−P(E)P(E')=1−P(E) Key Concepts Probability is always a number between 0 and 1, where 0 means an event is impossible and 1 means an event is certain. The probabilities in a probability model must sum to 1. When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the event by the total number of outcomes in the sample space for the experiment. To find the probability of the union of two events, we add the probabilities of the two events and subtract the probability that both events occur simultaneously. To find the probability of the union of two mutually exclusive events, we add the probabilities of each of the events. The probability of the complement of an event is the difference between 1 and the probability that the event occurs. In some probability problems, we need to use permutations and combinations to find the number of elements in events and sample spaces. Glossary complement of an event the set of outcomes in the sample space that are not in the event E E event any subset of a sample space experiment an activity with an observable result mutually exclusive events events that have no outcomes in common outcomes the possible results of an experiment probability a number from 0 to 1 indicating the likelihood of an event probability model a mathematical description of an experiment listing all possible outcomes and their associated probabilities sample space the set of all possible outcomes of an experiment union of two events the event that occurs if either or both events occur Candela Citations CC licensed content, Original Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution CC licensed content, Shared previously College Algebra. Authored by: Abramson, Jay et al.. Provided by: OpenStax. Located at: License: CC BY: Attribution. License Terms: Download for free at Question ID 838. Authored by: Eldridge, Jeff. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL Question ID 7089, 34682. Authored by: Wallace, Tyler. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL CC licensed content, Specific attribution Precalculus. Authored by: OpenStax College. Provided by: OpenStax. Located at: License: CC BY: Attribution The figure is for illustrative purposes only and does not model any particular storm. ↵ Licenses and Attributions CC licensed content, Original Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution CC licensed content, Shared previously College Algebra. Authored by: Abramson, Jay et al.. Provided by: OpenStax. Located at: License: CC BY: Attribution. License Terms: Download for free at Question ID 838. Authored by: Eldridge, Jeff. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL Question ID 7089, 34682. Authored by: Wallace, Tyler. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL CC licensed content, Specific attribution Precalculus. Authored by: OpenStax College. Provided by: OpenStax. Located at: License: CC BY: Attribution PreviousNext
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https://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1240&context=nasapub
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10464
https://phys.libretexts.org/Bookshelves/University_Physics/Physics_(Boundless)/18%3A_Electric_Potential_and_Electric_Field/18.3%3A_Point_Charge
18.3: Point Charge - Physics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 18: Electric Potential and Electric Field Physics (Boundless) { } { "18.1:_Overview" : "property get Map 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18.3: Point Charge 14552 14552 admin { } Anonymous Anonymous 2 false false [ "article:topic", "conductor", "Static Equilibrium", "equipotential", "authorname:boundless", "neuron", "axon", "cell membrane", "flux density", "showtoc:no" ] [ "article:topic", "conductor", "Static Equilibrium", "equipotential", "authorname:boundless", "neuron", "axon", "cell membrane", "flux density", "showtoc:no" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Bookshelves 3. University Physics 4. Physics (Boundless) 5. 18: Electric Potential and Electric Field 6. 18.3: Point Charge Expand/collapse global location Physics (Boundless) Front Matter 1: The Basics of Physics 2: Kinematics 3: Two-Dimensional Kinematics 4: The Laws of Motion 5: Uniform Circular Motion and Gravitation 6: Work and Energy 7: Linear Momentum and Collisions 8: Static Equilibrium, Elasticity, and Torque 9: Rotational Kinematics, Angular Momentum, and Energy 10: Fluids 11: Fluid Dynamics and Its Applications 12: Temperature and Kinetic Theory 13: Heat and Heat Transfer 14: Thermodynamics 15: Waves and Vibrations 16: Sound 17: Electric Charge and Field 18: Electric Potential and Electric Field 19: Electric Current and Resistance 20: Circuits and Direct Currents 21: Magnetism 22: Induction, AC Circuits, and Electrical Technologies 23: Electromagnetic Waves 24: Geometric Optics 25: Vision and Optical Instruments 26: Wave Optics 27: Special Relativity 28: Introduction to Quantum Physics 29: Atomic Physics 30: Nuclear Physics and Radioactivity Back Matter 18.3: Point Charge Last updated Nov 5, 2020 Save as PDF 18.2: Equipotential Surfaces and Lines 18.4: Capacitors and Dielectrics Page ID 14552 Boundless Boundless ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Electric Potential Due to a Point Charge 2. Overview 3. Point Charges 4. Superposition of Electric Potential 1. Superposition of Electric Potential Key Points Key Terms learning objectives Express the electric potential generated by a single point charge in a form of equation Electric Potential Due to a Point Charge Overview Recall that the electric potential is defined as the electric potential energy per unit charge (18.3.1)V=PE q The electric potential tells you how much potential energy a single point charge at a given location will have. The electric potential at a point is equal to the electric potential energy (measured in joules) of any charged particle at that location divided by the charge (measured in coulombs) of the particle. Since the charge of the test particle has been divided out, the electric potential is a “property” related only to the electric field itself and not the test particle. Another way of saying this is that because PE is dependent on q, the q in the above equation will cancel out, so V is not dependent on q. The potential difference between two points ΔV is often called the voltage and is given by (18.3.2)Δ⁢V=V B−V A=Δ⁢PE q Point Charges Point charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (like on a metal sphere, see figure below) create external electric fields exactly like a point charge. The electric potential due to a point charge is, thus, a case we need to consider. Using calculus to find the work needed to move a test charge q from a large distance away to a distance of r from a point charge Q, and noting the connection between work and potential (W=–qΔV), it can be shown that the electric potential V of a point charge is V=k⁢Q r(point charge) where k is a constant equal to 9.0×10 9 N⋅m 2/C 2 . Van de Graaff Generator: The voltage of this demonstration Van de Graaff generator is measured between the charged sphere and ground. Earth’s potential is taken to be zero as a reference. The potential of the charged conducting sphere is the same as that of an equal point charge at its center. The potential at infinity is chosen to be zero. Thus V for a point charge decreases with distance, whereas E for a point charge decreases with distance squared: (18.3.3)E=F q=k⁢Q r 2 The electric potential is a scalar while the electric field is a vector. Note the symmetry between electric potential and gravitational potential – both drop off as a function of distance to the first power, while both the electric and gravitational fields drop off as a function of distance to the second power. Superposition of Electric Potential To find the total electric potential due to a system of point charges, one adds the individual voltages as numbers. learning objectives Explain how the total electric potential due to a system of point charges is found Superposition of Electric Potential We’ve seen that the electric potential is defined as the amount of potential energy per unit charge a test particle has at a given location in an electric field, i.e. (18.3.4)V=PE q We’ve also seen that the electric potential due to a point charge is where k is a constant equal to 9.0×10 9 N⋅m 2/C 2. The equation for the electric potential of a point charge looks similar to the equation for the electric field generated for a point particle (18.3.5)E=F q=kQ r 2 with the difference that the electric field drops off with the square of the distance while the potential drops off linearly with distance. This is analogous to the relationship between the gravitational field and the gravitational potential. Superposition of Electric Potential: The electric potential at point L is the sum of voltages from each point charge (scalars). Recall that the electric potential V is a scalar and has no direction, whereas the electric field E is a vector. To find the voltage due to a combination of point charges, you add the individual voltages as numbers. So for example, in the figure above the electric potential at point L is the sum of the potential contributions from charges Q 1, Q 2, Q 3, Q 4, and Q 5 so that (18.3.6)V L=k⁢[Q 1 d 1+Q 2 d 2+Q 3 d 3+Q 4 d 4+Q 5 d 5] To find the total electric field, you must add the individual fields as vectors, taking magnitude and direction into account. This is consistent with the fact that V is closely associated with energy, a scalar, whereas E is closely associated with force, a vector. The summing of all voltage contributions to find the total potential field is called the superposition of electric potential. Summing voltages rather than summing the electric simplifies calculations significantly, since addition of potential scalar fields is much easier than addition of the electric vector fields. Note that there are cases where you might need to sum potential contributions from sources other than point charges; however, that is beyond the scope of this section. Key Points Recall that the electric potential is defined as the potential energy per unit charge, i.e. V=PE q. The potential difference between two points ΔV is often called the voltage and is given b Δ⁢V=V B−V A=Δ⁢PE q. The potential at an infinite distance is often taken to be zero. The case of the electric potential generated by a point charge is important because it is a case that is often encountered. A spherical sphere of charge creates an external field just like a point charge, for example. The equation for the electric potential due to a point charge is V=kQ r, where k is a constant equal to 9.0×10 9 N⋅m 2/C 2. The electric potential V is a scalar and has no direction, whereas the electric field E is a vector. To find the voltage due to a combination of point charges, you add the individual voltages as numbers. So for example, in the electric potential at point L is the sum of the potential contributions from charges Q 1, Q 2, Q 3, Q 4, and Q 5 so that V L=k⁢[Q 1 d 1+Q 2 d 2+Q 3 d 3+Q 4 d 4+Q 5 d 5]. To find the total electric field, you must add the individual fields as vectors, taking magnitude and direction into account. This is consistent with the fact that V is closely associated with energy, a scalar, whereas E is closely associated with force, a vector. The summing of all voltage contributions to find the total potential field is called the superposition of electric potential. It is much easier to sum scalars than vectors, so often the preferred method for solving problems with electric fields involves the summing of voltages. Key Terms electric potential: The potential energy per unit charge at a point in a static electric field; voltage. voltage: The amount of electrostatic potential between two points in space. vector: A directed quantity, one with both magnitude and direction; the between two points. scalar: A quantity that has magnitude but not direction; compare vector. superposition: The summing of two or more field contributions occupying the same space. LICENSES AND ATTRIBUTIONS CC LICENSED CONTENT, SHARED PREVIOUSLY Curation and Revision. Provided by: Boundless.com. License: CC BY-SA: Attribution-ShareAlike CC LICENSED CONTENT, SPECIFIC ATTRIBUTION electric potential. Provided by: Wiktionary. Located at: en.wiktionary.org/wiki/electric_potential. License: CC BY-SA: Attribution-ShareAlike OpenStax College, College Physics. September 17, 2013. Provided by: OpenStax CNX. Located at: License: CC BY: Attribution Electric potential. Provided by: Wikipedia. Located at: en.Wikipedia.org/wiki/Electric_potential. License: CC BY-SA: Attribution-ShareAlike OpenStax College, College Physics. September 17, 2013. Provided by: OpenStax CNX. Located at: License: CC BY: Attribution voltage. Provided by: Wiktionary. Located at: en.wiktionary.org/wiki/voltage. License: CC BY-SA: Attribution-ShareAlike OpenStax College, College Physics. December 13, 2012. Provided by: OpenStax CNX. Located at: License: CC BY: Attribution OpenStax College, Electric Potential in a Uniform Electric Field. September 18, 2013. Provided by: OpenStax CNX. Located at: License: CC BY: Attribution Electric potential. Provided by: Wikipedia. Located at: en.Wikipedia.org/wiki/Electric_potential. License: CC BY-SA: Attribution-ShareAlike OpenStax College, College Physics. September 18, 2013. Provided by: OpenStax CNX. Located at: License: CC BY: Attribution scalar. Provided by: Wiktionary. Located at: en.wiktionary.org/wiki/scalar. License: CC BY-SA: Attribution-ShareAlike Boundless. Provided by: Boundless Learning. Located at: www.boundless.com//physics/definition/superposition. License: CC BY-SA: Attribution-ShareAlike vector. Provided by: Wiktionary. Located at: en.wiktionary.org/wiki/vector. License: CC BY-SA: Attribution-ShareAlike OpenStax College, College Physics. December 13, 2012. Provided by: OpenStax CNX. Located at: License: CC BY: Attribution Potencial eletrico resultante. Provided by: Wikimedia. Located at: commons.wikimedia.org/wiki/File:Potencial_eletrico_resultante.PNG. License: Public Domain: No Known Copyright This page titled 18.3: Point Charge is shared under a not declared license and was authored, remixed, and/or curated by Boundless. Back to top 18.2: Equipotential Surfaces and Lines 18.4: Capacitors and Dielectrics Was this article helpful? Yes No Recommended articles 18.2: Equipotential Surfaces and LinesAn ideal conductor exists only in the world of theory; it has “ideal” properties that make calculations easy to perform. 17.4: The Electric Field RevisitedA point charge creates an electric field that can be calculated using Coulomb’s law. 6.5: Permanent Magnet DevicesThis page explores the properties of permanent magnets, focusing on residual flux density and magnetic energy density. It describes how external field... 18.1: OverviewElectric potential and field are related in that potential is a property of the field that describes the field’s action. 19.4: Equipotential LinesAn equipotential line is a line along which the electric potential is constant.An equipotential surface is a three-dimensional version of equipotenti... Article typeSection or PageAuthorBoundlessShow TOCno Tags axon cell membrane conductor equipotential flux density neuron Static Equilibrium © Copyright 2025 Physics LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status× contents readability resources tools ☰ 18.2: Equipotential Surfaces and Lines 18.4: Capacitors and Dielectrics
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https://mathoverflow.net/questions/51054/what-does-log-convexity-mean
ca.classical analysis and odes - What does log convexity mean? - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community MathOverflow helpchat MathOverflow Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more What does log convexity mean? Ask Question Asked 14 years, 9 months ago Modified14 years, 3 months ago Viewed 9k times This question shows research effort; it is useful and clear 24 Save this question. Show activity on this post. The Bohr–Mollerup theorem characterizes the Gamma function Γ(x) as the unique function f(x) on the positive reals such that f(1)=1, f(x+1)=x f(x), and f is logarithmically convex, i.e. log(f(x)) is a convex function. What meaning or insight do we draw from log convexity? There's two obvious but less than helpful answers. One is that log convexity means exactly what the definition says, no more and no less. The other is the more or less circular one that since the Gamma function is so important, any property that characterizes it is also significant. The wikipedia article point out that "a logarithmically convex function is a convex function, but the converse is not always true" with the counterexample of f(x)=x 2. The only logarithmically convex examples in the article come trivially from exponentiating convex functions, and the example Γ(x). Let me say in advance that I'm less interested in the Gamma function than I am in the notion of log convexity, so this question is not a duplicate of Importance of Log Convexity of the Gamma Function A thoughtful answer by Andrey Rekalo to that question, is that functions which can be realized as finite of moments of Borel measures are log convex functions. But I'm more interested in things that are implied by (v. imply) log convexity. My real motivation is the fact that the Riemann Hypothesis implies that the Hardy function is log convex for sufficiently large t. (The Hardy function Z(t) is just ζ(1/2+i t) with the phase taken out, so Z(t) is real valued and |Z(t)|=|ζ(1/2+i t)|.) This is in Edward's book 'Riemann's Zeta Function' Section 8.3, in the language that RH ⇒Z′/Z is monotonic. This says that between consecutive real zeros, −log|Z(t)| is convex. Any insight would be welcome. ca.classical-analysis-and-odes analytic-number-theory soft-question Share Share a link to this question Copy linkCC BY-SA 2.5 Cite Improve this question Follow Follow this question to receive notifications edited Apr 13, 2017 at 12:58 CommunityBot 1 2 2 silver badges 3 3 bronze badges asked Jan 3, 2011 at 20:04 StoppleStopple 11.7k 3 3 gold badges 45 45 silver badges 72 72 bronze badges 4 2 Maybe this paper is relevant: inf.ucv.ro/~niculescu/articles/2000/…weakstar –weakstar 2011-01-03 20:28:19 +00:00 Commented Jan 3, 2011 at 20:28 2 log convexity leads one to conclude that the Riemann Hypothesis implies the Lindelof hypothesis. Are you perhaps thinking about a converse to Hardy's Three Circle Theorem?Alex R. –Alex R. 2011-01-03 20:34:28 +00:00 Commented Jan 3, 2011 at 20:34 @Alex - Thanks! This is probably something I should have already known, but it's the kind of answer I was looking for.Stopple –Stopple 2011-01-03 20:35:53 +00:00 Commented Jan 3, 2011 at 20:35 Whoops not Hardy, Hadamard three circle...Alex R. –Alex R. 2011-01-03 20:46:20 +00:00 Commented Jan 3, 2011 at 20:46 Add a comment| 4 Answers 4 Sorted by: Reset to default This answer is useful 5 Save this answer. Show activity on this post. From the comments, log convexity leads one to conclude that Riemann Hypothesis implies Lindelof hypothesis. The implication of log convexity comes from Hadamard Three Circle Theorem Share Share a link to this answer Copy linkCC BY-SA 2.5 Cite Improve this answer Follow Follow this answer to receive notifications answered Jan 3, 2011 at 20:49 Alex R.Alex R. 5,012 2 2 gold badges 42 42 silver badges 67 67 bronze badges Add a comment| This answer is useful 12 Save this answer. Show activity on this post. Log convexity pops up in lots of different places (eg, the determinant of positive definite matrices is log-concave, the perron-frobenius eigenvalue of positive matrices is log convex, volumes of convex bodies are log concave, etc, etc). For a discussion having absolutely nothing to do with the gamma function, see S. Boyd and L. Vanderberghe's book "Convex optimization" (page 105). PDF available for free on Boyd's web site (google "Stephen Boyd Stanford") Share Share a link to this answer Copy linkCC BY-SA 2.5 Cite Improve this answer Follow Follow this answer to receive notifications answered Jan 3, 2011 at 20:42 Igor RivinIgor Rivin 1 2 7 Igor, you are not responsible for that of course, but these log-convexity statements do not refer to the same definition of LC. In general, a log-convex function is a function x↦f(x)>0 such that x↦log f(x) is convex (as stated in the question). Instead, the log-convexity of the Perron-Frobenius eigenvalue λ p f is the convexity of log λ p f(A) in terms of the variables log a i j. Something significantly different!Denis Serre –Denis Serre 2011-01-04 06:58:45 +00:00 Commented Jan 4, 2011 at 6:58 I guess Denis then the lc of the Perron-Frobenius eigenvalue should be actually called "multiplicative convexity".Suvrit –Suvrit 2011-01-04 09:25:57 +00:00 Commented Jan 4, 2011 at 9:25 Add a comment| This answer is useful 9 Save this answer. Show activity on this post. In my opinion, some of the reasons that make log-convexity (l.c.) nice / useful are: If f and g are l.c., then f+g and f g are both log-convex (note log-concavity on the other hand is not closed under addition, though as Allen commented above, it is closed under convolution (Prekopá's theorem)) L.c. functions are related to completely monotonic functions (i.e., functions for which (−1)n f(n)(x)≥0 for all x>0, and n∈N), in that any completely monotonic function is also l.c.; If I recall correctly, one place completely monotonic functions come up naturally is when considering Laplace transforms. Another example: the Laplace transform of any non-negative function is l.c. L.c. is closely related to multiplicatively convex functions (i.e., f(√x y)≤√f(x)f(y)), and it is known that the logarithmic integral is multiplicatively convex. Many more nice properties may be found in the book: Convex functions and their applications: A contemporary approach by C. P. Niculescu and L.-E. Persson. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jun 15, 2020 at 7:27 CommunityBot 1 2 2 silver badges 3 3 bronze badges answered Jan 4, 2011 at 9:16 SuvritSuvrit 28.7k 7 7 gold badges 83 83 silver badges 152 152 bronze badges Add a comment| This answer is useful 8 Save this answer. Show activity on this post. Many common probability densities are log-concave, which turns out to be a very useful property. For one thing, it's a way to quantify that a density is something like a normal density, perhaps sufficiently like a normal for a theorem to generalize. Also, og-concave densities satisfy nice theorems. For example, if the PDF of a random variable is log-concave, the CDF is as well. Also, the convolution of two log-concave densities is also log-concave. Share Share a link to this answer Copy linkCC BY-SA 2.5 Cite Improve this answer Follow Follow this answer to receive notifications answered Jan 3, 2011 at 23:03 John D. CookJohn D. Cook 5,307 1 1 gold badge 48 48 silver badges 72 72 bronze badges 2 3 Also, if you take a log-concave function of many variables and integrate some out, the result is again log-concave. This gives the convolution statement above and also Igor's one about convex bodies. Which prompts the question: is there a natural log-concave function of n variables, integrable in the first n−1, whose integral is Γ?Allen Knutson –Allen Knutson 2011-01-04 05:31:15 +00:00 Commented Jan 4, 2011 at 5:31 1 @Allen. I had this question in mind. Could be a valuable MO question.Denis Serre –Denis Serre 2011-01-04 06:27:23 +00:00 Commented Jan 4, 2011 at 6:27 Add a comment| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions ca.classical-analysis-and-odes analytic-number-theory soft-question See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Linked 24Importance of Log Convexity of the Gamma Function Related 42Is Li(x) the best possible approximation to the prime-counting function? 49The Hardy Z-function and failure of the Riemann hypothesis 5How naturally can functions defined by parametric integrals be interpolated from N to R+? 12Fourier transform of the critical line of zeta? 3An integral identity evaluating the gamma function 10Statement of the pair correlation conjecture 5Uniformity in Wirsing's Mean Value Theorems 2Gamma function and the somewhat extended version of Bohr-Mollerup theorem Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? 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https://www.youtube.com/watch?v=OBD_EliDAZk
Homogeneous first order ordinary differential equations Professor M does Science 28700 subscribers 64 likes Description 1077 views Posted: 28 Feb 2024 📚 Homogeneous first order ordinary differential equations consist of a derivative equal to a homogeneous function of degree zero. In this video, we explain the general strategy to solve homogeneous differential equations, which involves a simple substitution that turns them into separable differential equations. We also solve, step by step, an example of a homogeneous equation, and we then plot the resulting solution. 0:00 Introduction 0:24 Homogeneous first order differential equations 2:43 Solution strategy 6:45 Alternative forms for homogeneous equations 9:24 Example 17:15 Wrap-up ⏭️ WHAT NEXT? Differential equation for tangent circles: [COMING SOON] Exact differential equations: [COMING SOON] Inexact differential equations: [COMING SOON] Linear differential equations: [COMING SOON] ~ Director and writer: BM Producer and designer: MC Thumbnail image credits: Vecteezy.com 18 comments Transcript: Introduction hi everyone this is professor and Science and this video is part of our series on Maths for Science and Engineering this video will cover homogeneous differential equations spoiler to solve them there is a very ingenious trick that allows us to transform them into separable differential equations which we know how to solve let's go in this video we Homogeneous first order differential equations consider homogeneous first order ordinary differential equation we consider a differential equation of this form and we call such an equation homogeneous if the function f is a homogeneous function of degree zero homogeneous functions are a topic in themselves but for the purposes of this video all we need to know is the definition of a homogeneous function of degree n we say that a function of two variables f of x y is homogeneous of degree n if when we multiply every argument of the function by any scalar Lambda the result is equal to Lambda to the^ n multiplying the original function going back to our differential equation at the top we say that this equation is homogeneous if the function f here is a homogeneous function of degree Z big warning here pay close attention this is where the definition gets a little bit convoluted a homogeneous function can in general be of any degree say n here however the definition of a homogeneous first order differential equation states that this function f of XY has to be of degree zero if F of XY was a general homogeneous function of a degree other than zero say degree 2 then this would not be a homogeneous first order differential equation I know it may initially appear confusing and it kind of is but it's the just the definition to make things explicit the function f must be such that F of Lambda X Lambda Y is equal to Lambda to the power of 0 time F of XY as Lambda to the power of 0 is simply 1 we can rewrite this as simply F of XY in the rest of this video we will explore how to solve differential equations of this form let's write down a general Solution strategy homogeneous equation again and let's also write down the general condition that F must be a homogeneous function of degree zero we're going to try to solve this General OD by making some clever substitutions first let's choose Lambda to be 1 /x this means that we can write F of lamb X Lambda y as F of 1 /x x 1 /x y we can further simplify this expression as F of 1 y /x as f is a homogeneous function of degree zero then we also have that F of Lambda X Lambda y must be equal to F of XY and so bringing these results together implies that F of XY is equal to F of 1 Y /x in turn this implies that we can rewrite our original differential equation as dydx equal to f of1 y /x let's write down the latest result for the next step we need to realize that the right hand side no longer depends on X and Y separately but only depends on their ratio this motivates us to introduce a new variable v as equal to y/x and this new variable V allows us to write the function f in terms of V only now this looks promising but we also need to deal with the left hand side of the equation which involves the derivative of y with respect to X using this definition we can rewrite Y in terms of V and X and our derivative becomes this we can now use the chain rule to write it as the first term times the derivative of the second term plus the second term times the derivative of the first term this derivative is equal to one so we end up with v + x dvdx inserting the top expression into the right hand side and inserting the bottom expression into the left hand side we can rewrite our original differential equation like this this is still the same equation but now written in terms of X and V rather than the original X and Y so what have we accomplished with this new form for our equation well this equation is now separable to see it more clearly let's copy it down here we can move this V term to the other side and we obtain this and it is now clear that we can move all V dependent terms to one side and all X dependent terms to the other side and this is now clearly a separable equation and we know how to solve separable equations all we need to do is to integrate both sides like this if you do need a refresher about seable equations you can go back to the corresponding video we've linked it in the description once we've evaluated these integrals we will have the solution to our differential equation in terms of V and X X we can then take a final step and use the fact that V is equal to y/x to get our solution in terms of y and x and we will be done we will look at an example of Alternative forms for homogeneous equations solving a homogeneous equation in a moment but before we do that let me briefly discuss notation as you'll often encounter homogeneous equations discussed in different forms confusingly this is going to involve homogeneous functions of different orders but at the end of the day all the definitions are mathematically equivalent let's start with the definition we've used so far of a homogeneous equation where f is a homogeneous function of degree zero a very common alternative is to replace a function f on the right hand side by a ratio of two functions A and B like this in this this case the equation is homogeneous if a is a homogeneous function of some degree n and B is also a homogeneous function of the same degree n let me emphasize again that A and B must be homogeneous functions of the same degree n here and here and you should convince yourself that this alternative form is entirely equivalent to the form that we've been using up here another very common way to write down homogeneous equations is to start with this second form and then rewrite it as B of XY Dy equal to a of XY DX as another alternative this last form is often reared ranged so that everything is moved to one side which is then set to zero yet another alternative form this minus sign here is often removed by defining new functions m equal to a and N = minus B with these we can rewrite our differential equation as M of XY DX plus n of XY y d y equal to Zer all these forms for a homogeneous equation are equivalent and all can be solved in the same way you will most certainly encounter homogeneous equations written in any of these forms in your work so you should be comfortable with all of them to finish let's look at an example Example dydx = to y - x /x first we need to check that this is indeed a homogeneous equation to do so we identify the function f of XY as equal to y - x /x next we need to evaluate F of Lambda X Lambda y we get Lambda Yus Lambda X all over Lambda X and we can then cancel the Lambda in the numerator here and here with the Lambda in the denominator here we end up with y - x /x and this is our original function f this means f is a homogeneous function of degrees zero so we are indeed dealing with a homogeneous equation let's make some room now that we know we have a homogeneous equation the key step is to introduce the new function v as equal to y/x to rewrite our equation in terms of this new function V let's first consider the derivative on the left hand side we've already done this before but very quickly we can rewrite it as the derivative of VX and then using the chain rule we end up with v+ X dvdx next let's consider the right hand side we can replace y by V X to get this expression and we can now cancel the X in the numerator here and here with the X in the denominator here and we end up with v minus1 bringing everything together we can use this expression for the left hand side and we get V + x dvdx and this expression for the right hand side and we get V minus 1 the V cancels here and here so we end up with X DVD x equal to -1 let's make Su room again and let's copy the latest differential equation we've got this is now trivially separable so we can move all the dependent terms to one side and all X dependent terms to the other side I'm leaving some spaces so that in the next step we can integrate the left hand side with respect to V and the right hand side with respect to X the integral with respect to V is a standard one that gives V for the right hand side we have the integral of 1 /x and this is again a standard integral which you will remember is equal to the natural logarithm of X so overall we can rewrite this as V = minus logi of X and we add an integration constant C so this is our solution in terms of V and X but to really be able to finish we want to go back to Y considering the definition of v up here we can rewrite our solution as y/x equal to this expression and as a final step we can write y of x as equal to x C minus the logarithm of X and this is our solution for the homogeneous differential equation continuing with our example we have now solved this homogeneous differential equation and found that the solution takes this form whenever we find a solution to a differential equation we should always always always check that we have the correct Solution by inserting it back into the equation and confirming that the equation is satisfied so let's do it our proposed solution is y = x multiplying C minus the logarithm of X and to simplify things we can first multiply through with X to end up with this new expression now to confirm that this is a valid solution we can first evaluate the left hand side of the equation which is the derivative of y with respect to X inserting this solution into the derivative we get the derivative with respect to X of this expression the derivative of the first term is simply C and for the second term we need to use the chain rule we get the derivative of the first term times the second term plus the first term times the derivative of the second term and this is trivially equal to 1 and the derivative of the logarithm is 1 /x putting everything together we end up with C minus logarithm of x- one in parallel to this we can evaluate the right hand side of the equation which we first copy down substituting our proposed solution into this expression we get this very long new expression the factors of X cancel here here and here in the numerator with the X here in the denominator so that we end up with cus logi of x - 1 and comparing this expression for the left hand side with this expression for the right hand side we confirm that our solution is indeed correct hooray as a very final step let's draw the solution we will draw the function on this graph and we're focusing on the positive X AIS because the logarithm function up here is undefined for negative values of X the solution of a differential equation involves the constant C so we really have a family of solutions to make things a bit simpler with a drawing let's first consider the value = 0 in this case our solution becomes y x = - x the logarithm of X and this blue curve represents that solution it starts at the origin here then reaches a maximum value at this point whose value is the inverse of the number e and then crosses the x axis again at 1 after which it becomes negative now for positive C values we get this family of green curves they have a similar shape to the blue curve but the maximum of the function and its crossing point with the horizontal axis increasingly move towards larger values of X for negative C values we get this family of red orange curves and again we have a similar shape to the others but the maximum and crossing point of the function go in the opposite direction with decreasing value for C overall for any value of C we get a curve of a similar shape and each one of these curves is a solution to our differential Wrap-up equation you can check out additional examples of homogeneous differential equations Linked In the description I hope that you liked the video and please subscribe
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https://www.vocabineer.com/different-types-of-balls/
Different Types of Balls Used in Sports with Pictures Sports around the world use many kinds of balls, each designed for the way the game is played. Some balls are round for kicking, others are oval for carrying, and some are small and light for hitting with bats or rackets. Knowing the different types of sports balls with their names and pictures makes it easier to talk about games, learn new rules, and even shop for equipment. By learning the names of balls in sports, students, fans, and players can connect better with sports in school, professional matches, or casual play. In This Page Common Balls Used in Sports Worldwide Balls are at the center of many sports, and some types are played in almost every country. These balls are designed to match the rules and style of each game, making them easy to recognize around the world. Below is a list of the most common balls used in sports worldwide: Soccer Ball Basketball Tennis Ball Cricket Ball Baseball Volleyball Golf Ball Rugby Ball Table Tennis Ball Softball American Football Handball Netball Squash Ball Billiard Ball Bowling Ball Water Polo Ball Lacrosse Ball Dodgeball Exercise Ball Yoga Ball Medicine Ball Polo Ball Bocce Ball Rounders Ball Racquetball Snooker Ball Beach Ball Futsal Ball Kickball List of Different Types of Balls in Sports with Names Sports use many kinds of balls, and each game has its own design and purpose. From basketball to cricket, every ball is made to fit the way the sport is played. Below are the different types of balls in sports with details of each. Ball Games Played with the Foot (Football Sports Balls) Sports played with the foot use round or oval balls for kicking, passing, or running. The size, bounce, and material depend on the sport. Football Ball – Oval, leather or synthetic, used in American football for throwing and kicking. Soccer Ball – Round, stitched in panels, made for passing, dribbling, and goal scoring. Rugby Ball – Larger than a football, oval with rounded ends, designed for running and ground passes. Australian Rules Football Ball – Oval, rounder ends, created for long-distance kicking. Futsal Ball – Smaller than a soccer ball, low bounce, made for indoor court control. Ball Games Played with the Hand (Hand Sports Balls) Hand-played sports need balls that are light enough to throw but durable for fast movements. Basketball – Large, textured surface for dribbling, passing, and shooting into hoops. Volleyball – Lightweight, soft panels, made for spiking and passing over nets. Handball – Smaller ball, grippy surface, built for quick passes and fast goals. Netball – Similar to basketball but lighter, used in passing and shooting without a backboard. Bat-and-Ball Sports (Striking Balls) These sports need hard, dense balls that can handle fast hits and field play. Cricket Ball – Hard leather, stitched seam, made for bowling and batting. Baseball – White with red stitching, used for pitching, batting, and fielding. Softball – Larger and softer than baseball, pitched underhand. Rounders Ball – Small, lightweight, played with wooden bats in schools or parks. Polo Ball – Hard, smooth, designed for striking with mallets while riding horses. Racket and Paddle Sports Balls These games use light balls or ball-like objects for striking with rackets or paddles. Tennis Ball – Pressurized, covered in felt, bouncy for outdoor and indoor courts. Table Tennis Ball (Ping Pong Ball) – Very small, hollow, fast-moving on tables. Squash Ball – Rubber, low bounce, made for indoor wall rallies. Racquetball – Bigger and bouncier than squash balls, used indoors. Badminton Shuttlecock – Feathered or plastic, not a ball but plays the same role. Precision and Target Sports Balls These games need small, heavy, or uniquely designed balls for accuracy and control. Golf Ball – Small, dimpled surface, built for long-distance shots. Billiard Ball / Pool Ball – Hard, shiny, used with cues on green tables. Snooker Ball – Similar to billiards but slightly smaller and used in professional snooker. Bowling Ball – Very heavy, with finger holes, rolled to knock down pins. Bocce Ball – Solid, smooth, thrown to get close to a target ball. Water and Beach Sports Balls These balls float, resist water, and are easy to grip when wet. Water Polo Ball – Textured for grip, made to float in pools. Beach Ball – Inflatable, colorful, used for fun on sand and beaches. Lacrosse Ball – Small, solid rubber ball used with sticks to pass and score. Winter and Ice Sports Balls Cold-weather sports use balls or ball-like objects for sliding on ice. Ice Hockey Puck – Flat rubber disk, used as a ball-equivalent on ice. Broomball Ball – Round, rubbery, made for playing on ice with sticks. Recreational and Training Balls These balls are used for casual games, fitness, or practice. Dodgeball – Soft and safe, used for throwing games indoors or outdoors. Kickball – Larger than soccer balls, played in playgrounds with bases. Exercise Ball – Large, inflatable, used for workouts, stretching, and balance. Yoga Ball – Similar to exercise ball, helps with posture and stability training. Medicine Ball – Heavy, solid, made for strength workouts and athletic drills. FAQs Different Types of Balls Used in Sports How many different types of sports balls are there? There are dozens of sports balls, including round, oval, and specialized ones. Common examples include soccer balls, basketballs, cricket balls, tennis balls, golf balls, and many more designed for specific games. What is the difference between a soccer ball and a futsal ball? A soccer ball is larger, lighter, and designed for outdoor play, while a futsal ball is smaller, heavier, and has less bounce, making it better for indoor courts and close control. Why do different sports need different types of balls? Each sport has unique rules and playing conditions, so balls are designed for the right weight, bounce, size, and material to suit the game. For example, tennis balls bounce, cricket balls are hard, and volleyballs are lightweight. What is the heaviest sports ball? The bowling ball is one of the heaviest, weighing between 6–16 pounds (2.7–7.3 kg), depending on the player’s choice. Medicine balls used in training can be even heavier. Is a hockey puck considered a ball? In ice hockey, the puck is not a ball but is often grouped as a “ball-equivalent” because it serves the same function—being struck, passed, and scored with during the game. Read More 3 Letter Words Vehicles Name in English Animals Name in English FacebookXPinterestWhatsApp About the author Muhammad Qasim Muhammad Qasim is an English language educator and ESL content creator with a degree from the University of Agriculture Faisalabad and TEFL certification. He has over 5 years of experience teaching grammar, vocabulary, and spoken English. Muhammad manages several educational blogs designed to support ESL learners with practical lessons, visual resources, and topic-based content. He blends his teaching experience with digital tools to make learning accessible to a global audience. He’s also active on YouTube (1.6M Subscribers), Facebook (1.8M Followers), Instagram (100k Followers) and Pinterest( (170k Followers), where he shares bite-sized English tips to help learners improve step by step. View all posts You may also like Picture Vocabulary Tropical Fruits Names in English with Pictures Picture Vocabulary Flying Animals Names in English with Pictures Picture Vocabulary 21 Animals Names from Q Letter
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https://www.scribd.com/presentation/589224938/Divisibility-Test-Modular-Design-Presentation
Divisibility Test Modular Design Presentation | PDF | Division (Mathematics) | Abstract Algebra Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 359 views 38 pages Divisibility Test Modular Design Presentation The document discusses modular arithmetic and divisibility rules. It explains that divisibility rules are shortcuts to determine if a number is divisible by another number by examining its d… Full description Uploaded by Glenn Michael Pascua AI-enhanced description Go to previous items Go to next items Download Save Save Divisibility Test Modular Design Presentation For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Divisibility Test Modular Design Presentation For Later You are on page 1/ 38 Search Fullscreen Application of Congruence Divisibility test Modular Design Prepared by:JOHN RAYMOND D. PERGIS MaEd-Mathematics adDownload to read ad-free Divisibility Rules adDownload to read ad-free Divisibility Rules In this mini-lesson, we will explore about divisibility rules by learning how to apply divisibility rules with examples, and the divisibility rules of specific numbers while discovering the interesting facts around them.In a 1962 Scientific American article, the popular science writer, Martin Gardner, discusses divisibility rules for 2–12, where he explains that the rules were widely known during the renaissance and used to reduce fractions with large numbers down to the lowest terms. adDownload to read ad-free Since every number is not completely divisible by every other number, they may leave a remainder other than zero. There are certain rules which help us determine the actual divisor of a number just by considering the digits of that number. These are called divisibility rules.Do you think, 375 is divisible 6? adDownload to read ad-free What Ar e Divisibility Rules? A d ivisibility rule is a kind of shortcut to figure out if a given integer is divisible by a divisor, without performing the whole division process but by examining its digits.Multiple divisibility rules can be applied to the same number which can quickly determine its prime factorization. Divisibility rules are a set of general rules or heuristics that are used to figure out if a number is wholly divisible by another number. adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Class 3 Imo Papers No ratings yet Class 3 Imo Papers 35 pages BMA Class 6 CH 1 Knowing Our Numbers 100% (1) BMA Class 6 CH 1 Knowing Our Numbers 8 pages Nlmo - 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AP Language & Composition Tone Words Positive lighthearted confident amused complimentary amiable relaxed soothing jubilant encouraging reverent hopeful cheery elated passionate whimsical romantic calm enthusiastic elevated exuberant optimistic sympathetic proud fanciful appreciative consoling ecstatic jovial loving compassionate friendly pleasant brave joyful energetic Negative angry wrathful threatening agitated obnoxious insulting choleric disgusted bitter accusing arrogant quarrelsome surly outraged irritated condemnatory belligerent disgruntled furious indignant inflammatory aggravated brash testy Humor/Irony/Sarcasm scornful bantering disdainful irreverent condescending pompous mocking ridiculing wry sarcastic taunting cynical insolent patronizing whimsical malicious droll critical ironic facetious flippant mock-heroic teasing quizzical comical satiric amused sardonic contemptuous caustic ribald irreverent Sorrow/Fear/Worry somber mournful concerned morose hopeless remorseful poignant melancholy solemn fearful pessimistic grave staid ominous sad serious despairing sober solemn resigned horror disturbed apprehensive gloomy foreboding mournful regretful Neutral formal objective questioning learned authoritative disbelieving sentimental pretentious apathetic conventional judgmental reflective ceremonial candid instructive factual incredulous urgent fervent histrionic callous forthright lyrical sincere restrained clinical matter-of-fact didactic shocked nostalgic earnest resigned contemplative haughty objective detached admonitory informative baffled reminiscent patriotic meditative intimate obsequious
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https://bmcanesthesiol.biomedcentral.com/articles/10.1186/s12871-023-02233-7
Advertisement Optimal dose of neostigmine antagonizing cisatracurium-induced shallow neuromuscular block in elderly patients: a randomized control study BMC Anesthesiology volume 23, Article number: 269 (2023) Cite this article 2015 Accesses 4 Citations Metrics details Abstract Background Residual neuromuscular block after using neuromuscular blocking agents is a common and potentially harmful complication of general anesthesia. Neostigmine is a widely used antagonist, but its optimal dose for elderly patients is unclear. Objectives To compare the optimal dosage and safety of neostigmine for reversing shallow residual block in elderly patients after cisatracurium-induced neuromuscular block. Methods A randomized controlled trial was conducted in 196 elderly patients undergoing non-cardiac surgery under general anesthesia with cisatracurium. Patients were assigned to receive either no neostigmine (control group) or neostigmine at 20 µg/kg, 40 µg/kg or 50 µg/kg when train-of-four (TOF) ratio reached 0.2 at the end of surgery. The primary outcome was the time to reach TOF ratio of 0.9 after administration. Secondary outcomes included TOF ratio at 10 min after administration, postoperative nausea and vomiting, postoperative cognitive impairment and post-anesthesia care unit (PACU) stay time. Results The time to reach TOF ratio of 0.9 in the 20 µg/kg, 40 µg/kg and 50 µg/kg groups was significantly shorter than the control group (H = 104.257, P < 0.01), and the time of 40 µg/kg group and 50 µg/kg group was significantly shorter than the 20 µg/kg group (P < 0.001). There was no significant difference between 40 µg/kg and 50 µg/kg groups (P = 0.249). The TOF ratio at 10 min after administration showed similar results. There were no significant differences among groups in postoperative nausea and vomiting, postoperative cognitive impairment or post-operation hospital stay. Conclusions Timely use of neostigmine after general anesthesia in elderly patients can significantly shorten time of TOF value reaching 0.9, among which 40 µg/kg dosage may be a more optimized choice. Trial registration this study was registered on chictr.org.cn (ChiCTR2100054685, 24/12/2021). Peer Review reports Background Neuromuscular blocking agents (NMBAs) are widely used to provide muscle relaxation for endotracheal intubation, certain modes of mechanical ventilation and surgical procedures. However, with the widespread application of neuromuscular blocking agents in general anesthesia, the residual effects have become one of the important factors of postoperative complications [1,2,3]. At present, the most widely used muscle relaxation antagonists include acetylcholinesterase inhibitors and sugammadex. These intermediate-acting NMBAs, such as cisatracurium, can be metabolized quickly and consequently reduce the incidence of postoperative residual neuromuscular block . However, the pooled rate of residual blockade, defined as a TOF ratio less than 0.90, was 41% with Confidence Interval (25-58%) when studies using intermediate-acting NMBDs were analyzed , leading to significant respiratory events (e.g. severe hypoxemia, airway obstruction), pharyngeal function impairment, and even increased mortality [4, 6,7,8,9]. Previous studies have shown that the optimal dose of neostigmine for reversal of minimal NMB (TOFr = 0.5) in adults is 40 µg/kg . Nevertheless, compared to the adult, the elderly changes physiologically, including a reduced glomerular filtration rate, increased body fat, decrease in lean muscle mass, and decrease in total body water [11, 12], which may alter the pharmacokinetics and metabolism of drugs. A larger dose may be required to antagonize the residual block of muscle relaxants in the elderly . However, a larger dose of neostigmine may increase the occurrence of side effects such as bradycardia and increased secretions. Thus, it is urgent to recommend the optimal dose of neostigmine for antagonizing the neuromuscular blocking effect of cisatracurium in the elderly with general anesthesia. Accordingly, we aimed to explore the optimal dosage and safety of neostigmine to reverse shallow residual block from a TOF ratio of 0.2 in elderly patients. Methods Study design and patient selection This study was approved by the Ethics Committee of the Third Xiangya Hospital. Written informed consent was obtained from all patients. Inclusion criteria were: aged 60 to 85 years, American Society of Anesthesiology (ASA) physical status 1 to 3, and scheduled for elective surgery under general anesthesia with cisatracurium for tracheal intubation. Exclusion criteria were: BMI < 18.5 kg/m2 or BMI ≥ 28 kg/m2, significant hepatic or renal dysfunction (glutamic-pyruvic transaminase/glutamic oxaloacetic transaminase > 80 U/L, creatinine > 104 umol/L), family history of malignant hyperthermia, known allergy to one of the drugs used in this protocol, or recent use of sedatives, anti-depressants. Patients were randomly assigned into four groups according to the dose of neostigmine used, using a computer generated list of random numbers. Neuromuscular block was induced with cisatracurium and neostigmine 0 µg/kg (Normal Saline, NS), 20 µg/kg, 40 µg/kg or 50 µg/kg was administered at a TOF ratio of 0.2 to reverse the block. Anaesthesia nurses who were not involved in the care of the patients helped prepare the study drug according to randomisation. Procedure On arrival at the operating room, an intravenous cannula was inserted in the forearm vein of the patient, and standard anesthesia monitoring (noninvasive blood pressure, electrocardiogram, and oxygen saturation) were established and anesthesia depth was monitored using Bispectral Index (BIS, Medtronic, Minneapolis, MN, USA). Anesthesia was induced with sufentanil (0.5 µg/kg) and etomidate (0.2–0.3 mg/kg), cisatracurium (0.15–0.2 mg/kg). Anaesthesia was maintained by a continuous inhalation of 1% sevoflurane and infusion of propofol (3–4 mg/kg/h) and remifentanil (0.5-1.0 µg/kg/min); the infusion speed was adjusted to maintain the BIS between 40 and 60. Cisatracurium was added as required to facilitate the completion of the surgery. After tracheal intubation, ventilation was controlled to maintain arterial oxygen saturation at 96% or higher and normocapnia. Body temperature was maintained at 36.0 °C or higher. Sevoflurane was stopped about 1 h before the end of the surgery and the flow of fresh gas was increased to ensure no residual of inhalational anesthetics at the end of the surgery. Ondansetron 4 mg was administered intravenously 0.5 h before the end of the surgery to prevent nausea and vomiting. An additional 10–15 µg of sufentanil was administered at the end of surgery for postoperative pain management. Neuromuscular function was evaluated by TOF-Watch SX acceleration muscle relaxation monitor (Organon, Dublin, Ireland). After the skin had been cleaned carefully, two surface skin electrodes were attached over the ulnar nerve proximal to the wrist at a distance of 3–6 cm. After immobilising the forearm, the acceleration transducer was fixed firmly to the volar side of the distal phalanx of the thumb on a small elastic hand adapter (TOF-Watch Handadapter; Schering-Plough, Swords, Ireland). The monitoring was initiated after induction, but before cisatracurium administration. The device was calibrated automatically after a 50-Hz tetanic stimulation for 5 s and a stable baseline (< 5% change in TOF ratio) was recorded. Normalized TOF values were calculated and recorded according to baseline TOF values. TOF stimulus was applied every 60 s until a TOF ratio greater than or equal to 0.9. When a TOF ratio of 0.2 was reached two times consecutively, a pre-determined dose of neostigmine provided by Sine Jinzhu Pharmaceutical company (Shanghai, China) was administered according to the patient group (NS, 20 µg/kg, 40 µg/kg or 50 µg/kg), accompanied by glycopyrrolate in an 1:5 ratio. The time from the administration of neostigmine to a TOF ratio of 0.9 was recorded. The proportion of patients with TOF value ≥ 0.9 10 min after neostigmine administration; the incidence of nausea and vomiting in patients on postoperative day 1-day 3; the proportion of patients with Mini-Mental State Examination (MMSE) score decrease ≥ 3 points on postoperative day 3; and PACU stay time after neostigmine administration were followed up. Statistical analysis A sample size was calculated based on the primary outcome variable (the time from the administration of neostigmine to a TOF ratio of 0.9). To achieve a power of 0.8 with an alpha level of 0.05 and an effect size of f = 0.25, and considering a 10% drop out rate, a total sample size of N = 200 was needed. SPSS 22.0 software was used for statistical analysis. Count data were expressed as composition ratio; normally distributed measures were expressed as mean ± standard deviation, and t-test was used for comparison between two groups and one-way ANOVA for comparison between multiple groups; non-normally distributed measures were expressed as median (interquartile range), and Kruskal-Wallis H rank sum test was used for comparison between multiple groups; while categorical data were expressed as numbers with percentages and were compared using the Chi-square test. P < 0.05 was considered to indicate statistical difference. Results From January 2022 to November 2022, 240 ASA I-III patients aged from 60 to 85 who were scheduled for elective non-cardiac surgery under general anesthesia were enrolled, of which 40 patients were excluded due to contraindications (n = 18), allergies (n = 12) or declined to participate (n = 10). 200 patients completed randomization and participated in the study. 4 patients were excluded due to protocol violation (3 in NS group and 1 in 20 µg/kg group B). The number of patients included in the final statistical analysis was 196 (Fig. 1). (adapted from CONSORT 2010 flow diagram. Comparison of general data and perioperative data between the groups Table 1 summarized baseline characteristics and perioperative data of patients. There were no significant differences in the age, gender, BMI between four groups (P > 0.05, Table 1), which means baseline characteristics were generally consistent between groups. As for the perioperative data, there was no significant difference in the amount of muscle relaxant cisatracurium, surgery duration, anesthesia duration among the four groups (P > 0.05, Table 1). Intergroup comparison suggests that patients from 40 µg/kg group had more blood loss than those from control group (P < 0.05). The type of surgery is comparable among the groups. Comparison of results of postoperative muscle relaxation monitoring As listed in Table 2, the time to reach TOF value of 0.9 after administration of neostigmine, TOF value at 10 min after administration of neostigmine and PACU stay time were significantly different among the four groups of patients (H = 104.257, P < 0.01). In detail, intergroup comparison showed that the time to reach TOF value of 0.9 in 20 µg, 40 and 50 µg groups was significantly shorter than that in NS group after administration of neostigmine (all P < 0.01); meanwhile, the time to reach TOF value of 0.9 in 40 and 50 µg groups were further significantly reduced compared with that in 20 µg group (P = 0.032 and P < 0.001, respectively), but there was no significant difference between 40 and 50 µg groups (P = 0.249) (Fig. 2). Similar to the results of time to reach TOF value of 0.9, TOF value at 10 min after administration in 20 µg, 40 and 50 µg groups was significantly increased compared with that in control group (H = 93.351, P < 0.01); meanwhile, TOF value at 10 min after administration in 40 and 50 µg groups was further significantly increased compared with that in 20 µg group (P = 0.049 and P < 0.001,respectively), but there was no significant difference between 40 and 50 µg groups (P = 0.345) (Fig. 3). In addition, we evaluated the postoperative outcomes of each group. As shown in the Table 2, PACU stay time after neostigmine was different among the four groups (H = 10.671, P < 0.01). Intergroup analysis suggests that 50 µg/kg may shortern PACU stay time compared with normal saline (P = 0.024). We found no difference among the groups in the incidence of cognitive decline, postoperative nausea and vomiting; the length of postoperative hospital stay showed no difference either. Analysis of time of TOF ≥ 90% after neostigmine (min). Time of 20 µg, 40 and 50 µg groups was significantly shorter than that in NS group after administration of neostigmine (all P < 0.001); Time of 40 and 50 µg groups was further significantly reduced compared with that in 20 µg (P = 0.032 and P < 0.001,respectively);yet the time between 40 and 50 µg was comparable (P = 0.249); P < 0.05 Analysis of TOF value 10 min after neostigmine (%). TOF value of 20 µg, 40 and 50 µg groups was significantly higher than that in NS group after administration of neostigmine (P < 0.001); value of 40 and 50 µg groups was further significantly increased compared with that in 20 µg (P = 0.049 and P < 0.001,respectively);yet the difference between 40 and 50 µg was not significant (P = 0.345);P < 0.05 Discussion This study explored the optimal dose of neostigmine in elderly patients to reverse the TOF value to 0.9. The results indicated that the dose of 40 µg/kg may be an optimized choice and did not increase the incidence of postoperative cognitive impairment or PONV. Incomplete neuromuscular recovery after general anesthesia is a common complication in patients after general anesthesia, with an incidence ranging from 31 to 64% [5, 14,15,16]. Neostigmine, as an acetylcholinesterase inhibitor, can bind to acetylcholinesterase like acetylcholine, but the binding is firm and hydrolysis is slow, which makes acetylcholinesterase lose its activity. The released neurotransmitter acetylcholine accumulates in the synaptic gap, thereby exciting skeletal muscle and achieving the effect of antagonizing non-depolarizing muscle relaxants. Elderly patients are prone to residual muscle relaxation due to organ function degeneration and physiological changes that lead to slow metabolism of neuromuscular blockers . Related guidelines and consensus also recommend intravenous administration of neostigmine for antagonism under the premise of no contraindications; however, the recommended dose is no different from that of normal adults, while large doses may lead to an increase in adverse effects of neostigmine (e.g. cause nausea, vomiting, convulsions, coma, slurred speech, anxiety, bradycardia and other symptoms). Therefore, clinical anesthesiologists maybe reduce or even avoid using neostigmine from a safety reason. The results of this study show that neostigmine was used for antagonism at the end of surgery, and even a small dose (20 µg/kg) can significantly accelerate muscle function recovery; but 40 µg/kg can achieve the purpose of rapid reversal of muscle blockade effect, further increasing the dose (50 µg/kg) cannot significantly shorten the time for muscle recovery. It suggested that a dose of 40 µg/kg of neostigmine may be an optimized choice to reverse the effects of muscle relaxants quickly, without increasing postoperative complications. Previous studies have explored the dose of neostigmine required for reversal of neuromuscular blockade. Preault A et al. concluded that the neostigmine 20 µg/kg was sufficient to successfully reverse the TOF ratio from 0.4 to 0.9 within 10 min . Similar result was also observed in another study conducted by Fuchs-Buder T . However, these studies did not evaluate the postoperative outcome after the use of neostigmine, the conclusions still need to be further verified. Furthermore, consistent with our study, the results of E. S. Choi et al. also showed that 40 µg/kg dosage of neostigmine may be a better choice. Nevertheless, the population of their study were adults aged 20–70 years (while our study was 60–85 years), the results may not be applicable to older patients. Meanwhile, our study also shows that in balanced anesthesia, 40 µg/kg of neostigmine is still the appropriate dosage to reverse neuromuscular blockade after adequate sevoflurane clearance. Additionally, our study found that the use of neostigmine to antagonize muscle relaxation at the end of surgery did not significantly shorten the postoperative PACU stay time of patients, which may be affected by confounding factors such as surgical type, analgesic sedative drugs and wound drainage at the end of surgery. Subsequent further trials should limit and standardize trial conditions, reduce related confounding factors and increase sample size to clarify the role of neostigmine in shortening PACU stay. Conclusion Timely use of neostigmine after general anesthesia in elderly patients can significantly shorten time of TOF value reaching 0.9, among which 40 µg/kg dosage may be a more optimized choice. Availability of data and materials The datasets used and/or analyzed during the current study are available from the corresponding author on reasonable request. References Kopman AF, Yee PS, Neuman GG. Relationship of the train-of-four fade ratio to clinical signs and symptoms of residual paralysis in awake volunteers. Anesthesiology. 1997;86(4):765–71. Article CAS PubMed Google Scholar Sundman E, Witt H, Olsson R, Ekberg O, Kuylenstierna R, Eriksson LI. 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Article CAS PubMed Google Scholar Raval AD, Uyei J, Karabis A, Bash LD, Brull SJ. Incidence of residual neuromuscular blockade and use of neuromuscular blocking agents with or without antagonists: a systematic review and meta-analysis of randomized controlled trials. J Clin Anesth. 2020;64:109818. Article CAS PubMed Google Scholar Xará D, Santos A, Abelha F. Adverse respiratory events in a post-anesthesia care unit. Arch Bronconeumol. 2015;51(2):69–75. English, Spanish. Article PubMed Google Scholar Murphy GS, Szokol JW, Avram MJ, Greenberg SB, Shear TD, Vender JS, Parikh KN, Patel SS, Patel A. Residual neuromuscular block in the Elderly: incidence and clinical implications. Anesthesiology. 2015;123(6):1322–36. Article PubMed Google Scholar Broens SJL, Boon M, Martini CH, Niesters M, van Velzen M, Aarts LPHJ, Dahan A. Reversal of partial neuromuscular block and the ventilatory response to Hypoxia: a randomized controlled trial in healthy volunteers. Anesthesiology. 2019;131(3):467–76. Article PubMed Google Scholar Choi ES, Oh AY, Seo KS, Hwang JW, Ryu JH, Koo BW, Kim BG. Optimum dose of neostigmine to reverse shallow neuromuscular blockade with rocuronium and cisatracurium. Anaesthesia. 2016;71(4):443–9. Article CAS PubMed Google Scholar McLean AJ, Le Couteur DG. Aging biology and geriatric clinical pharmacology. Pharmacol Rev. 2004;56(2):163–84. Article CAS PubMed Google Scholar Thürmann PA. Pharmacodynamics and pharmacokinetics in older adults. Curr Opin Anaesthesiol. 2020;33(1):109–13. Article PubMed Google Scholar McCarthy GJ, Cooper R, Stanley JC, Mirakhur RK. Dose-response relationships for neostigmine antagonism of vecuronium-induced neuromuscular block in adults and the elderly. Br J Anaesth. 1992;69(3):281–3. Article CAS PubMed Google Scholar Stewart PA, Liang SS, Li QS, Huang ML, Bilgin AB, Kim D, Phillips S. The impact of residual neuromuscular blockade, oversedation, and hypothermia on adverse respiratory events in a Postanesthetic Care Unit: a prospective study of prevalence, predictors, and outcomes. Anesth Analg. 2016;123(4):859–68. Article PubMed Google Scholar Yu B, Ouyang B, Ge S, Luo Y, Li J, Ni D, Hu S, Xu H, Liu J, Min S, Li L, Ma Z, Xie K, Miao C, Wu X. RECITE–China investigators. Incidence of postoperative residual neuromuscular blockade after general anesthesia: a prospective, multicenter, anesthetist-blind, observational study. Curr Med Res Opin. 2016;32(1):1–9. Article PubMed Google Scholar Thilen SR, Weigel WA, Todd MM, et al. 2023 American Society of Anesthesiologists practice guidelines for monitoring and antagonism of neuromuscular blockade: a report by the American Society of Anesthesiologists Task Force on Neuromuscular Blockade. Anesthesiology. 2023;138(1):13–41. Article PubMed Google Scholar Fuchs-Buder T, Meistelman C, Alla F, Grandjean A, Wuthrich Y, Donati F. Antagonism of low degrees of atracurium-induced neuromuscular blockade: dose-effect relationship for neostigmine. Anesthesiology. 2010;112(1):34–40. Article PubMed Google Scholar Preault A, Capron F, Chantereau C, Donati F, Dimet J. Under sevoflurane anaesthesia, a reduced dose of neostigmine can antagonize a shallow neuromuscular block: a double-blind, randomised study. Anaesth Crit Care Pain Med. 2016;35(4):269–73. Article PubMed Google Scholar Download references Funding Not applicable. Author information Mengya Cao and Huifan Huang are the first authors Authors and Affiliations Department of Anesthesiology, Third Xiangya Hospital, Central South University, Changsha, Hunan, 410013, P.R. China Mengya Cao, Jianbin Tong, Yangwen Ou & Yan Liao Department of Anesthesiology, The First Affiliated Hospital of Xiamen University, Xiamen University, Xiamen, P.R. China Huifan Huang Hunan Province Key Laboratory of Brain Homeostasis, Third Xiangya Hospital, Central South University, Changsha, Hunan, 410013, P.R. China Jianbin Tong Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Contributions Yan Liao and Jianbin Tong designed the study and collected the data. Mengya Cao, Yangwen Ou analyzed the data and Huifan Huang, Yan Liao drafted the manuscript. Mengya Cao and Jianbin Tong provided critical feedback on the study design, data analysis, and manuscript revisions.All authors have read and approved the final manuscript. Corresponding author Correspondence to Yan Liao. Ethics declarations Ethics approval and consent to participate The study was approved by the ethics committee of the Ethics Committee of the Third Xiangya Hospital (reference number [R22008]) and written informed consent was obtained from all participants. The authors declare that all experiments were performed in accordance with relevant guidelines and regulations (such as the Declaration of Helsinki). Consent for publication Not applicable. Competing interests The authors declare that they have no competing interests. Additional information Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Rights and permissions Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit The Creative Commons Public Domain Dedication waiver ( applies to the data made available in this article, unless otherwise stated in a credit line to the data. Reprints and permissions About this article Cite this article Cao, M., Huang, H., Tong, J. et al. Optimal dose of neostigmine antagonizing cisatracurium-induced shallow neuromuscular block in elderly patients: a randomized control study. BMC Anesthesiol 23, 269 (2023). Download citation Received: 19 March 2023 Accepted: 04 August 2023 Published: 10 August 2023 DOI: Share this article Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. 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https://andthentheresphysics.wordpress.com/2018/05/29/initial-value-problem-vs-boundary-value-problem/
Initial value problem vs boundary value problem I haven’t actually looked at Judith Curry’s blog for a while, but popped across there and noticed a guest post about energy budgets, climate system domains, and internal variability. One reason why we think that we can actually do long-term climate modelling is that the evolution of our climate depends mostly on the boundary conditions, rather than the initial conditions. The basic suggestion in the guest post on Judith Curry’s blog is that this is wrong. It’s essentially a convoluted but chaos argument. I probably shouldn’t bother rebutting this, but I’m waiting for someone to come and service my boiler, and I’ve been thinking about this a little, so thought I would write a quick post. Essentially, if we want to make predictions about the weather a few days in advance, then the initial conditions are important. These are things like temperatures, pressures, winds, clouds, etc. You put these initial conditions, which you get from actual measurements, into the simulation and run it forward in time. You might also perturb these slightly to see how this influences the output, but you keep it close to the known initial conditions. Climate modelling, on the other hand, is not trying to make predictions about the weather, but is trying to understand what is typical. We would generally regard the climate as being an average of some property (temperature, for example) over a suitably large region and a suitably large time interval. It turns out that this depends less on the initial values of the system, than on the boundary values. The boundary values are the conditions that constrain the climate over the long-term and are things like how much energy we get from the Sun, how much is reflected back into space, how much energy is radiated from the surface, and how much of this escapes into space. The latter depends on the composition of the atmosphere, and so this is often more associated with a boundary value, rather than being regarded as an initial value. A key point is that the system will always tend towards a state in which the amount of energy coming in, matches the amount going out into space, and that this state depends mostly on the boundary conditions. This quasi-equilibrium state will then set things like the surface temperatures, latitudonal temperature gradients, large-scale circulation patterns, and how much energy is in the system. Hence, it will determine the typical properties of the climate. The counter-argument is that the system is inherently chaotic and, therefore, we cannot make long-term predictions. This is true for weather predictions, but not for climate modelling. Even if we could get very accurate initial conditions, there would still be a limit to how far in advance we could predict the weather. The climate, however, doesn’t depend very strongly on the initial conditions, and so this property doesn’t impact climate modelling in the same way as it does weather modelling. A few additional comments. Even though climate modelling is more a boundary value problem, than an initial value problem, doesn’t mean that the initial conditions don’t matter. The exact path that we follow will depend on the initial conditions. However, if we were to consider numerous simulations with different initial condition, but the same boundary conditions, then we would expect the typical climate to be similar. This also doesn’t mean that the non-linear, chaotic nature of the system can’t have an impact on climate. It is possible that the non-linear dynamics could lead to some big change in some circulation pattern that could substantially influence the climate. Dansgaard Oeschger events may be an example of exactly this (although these are still ultimately associated with a change in one of the boundary conditions). It’s just that this is probably unlikely; we don’t often see shifts in climate that we can’t associate with a change in one of the boundary conditions. I’ve written this quite quickly, so may not have explained it as well as I could have. Probably also worth reading the posts I link to below. I will add that part of the problem may be that when communicating publicly it’s often necessary to provide reasonably simple explanations. You can’t possibly provide all the details and complexities when trying to explain something like this to a non-expert audience. A consequence of this, though, is that people can then pick holes in the explanation, if they’re not willing to accept that it’s been intentionally simplified. Links: Initial value vs boundary value problems (Steve Easterbrook). Chaos and climate (James Annan and William Connolley). Share this: Click to share on X (Opens in new window) X Click to share on Facebook (Opens in new window) Facebook Click to share on LinkedIn (Opens in new window) LinkedIn Click to share on Pinterest (Opens in new window) Pinterest Click to share on Tumblr (Opens in new window) Tumblr Click to share on Reddit (Opens in new window) Reddit Related There is no tribe!In "Climate change" Judith Curry says it’s okayIn "Climate change" David Rose on Judith Curry’s stadium waveIn "Climate change" This entry was posted in Climate change, ClimateBall, Science and tagged Chaos, Climate etc., Dan Hughes, James Annan, Judith Curry, Non-linear dynamics, Steve Easterbrook, William Connolley. Bookmark the permalink. ← The adults in the room Something about Tropical Cyclones → Ken Rice Astronomy and some other things ...and Then There's Physics likhipa inhlanzi emanzini Neverending Audit Due Diligence James' Empty Blog likhipa inhlanzi emanzini Isaac Held's Blog Isaac Held's Blog EssaysConcerning Essays, musings & missives - concerning science, art & society ClimateBall The Only Losing Move is Not to Play Vitamin CCS Environmental research snippets from Scotland rachel.blog Eats plants Lawrence Torcello Associate Professor of Philosophy, Rochester Institute of Technology Small Epiphanies A BLOG BY G.P. WAYNE Real Skeptic Ventures into skepticism Critical Angle Reflections on the refractory problems of climate and energy Gra Machree Heather Brach & John Samuel Open Parachute Simple Climate Straightforwardly explaining climate change, so you can read, react and then get on with your life. 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A response to Climate Change disinformation at wattsupwiththat.com Open Mind KIDS' LIVES MATTER so let's stop climate change Comment Reblog Subscribe Subscribed ...and Then There's Physics Already have a WordPress.com account? Log in now. ...and Then There's Physics Subscribe Subscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar Design a site like this with WordPress.com Get started
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https://en.wikipedia.org/wiki/Dithionite
Dithionite - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Production and reactions 2 Use and occurrenceToggle Use and occurrence subsection 2.1 Chemical analyses 2.2 Harmful properties 3 References 4 Further reading Dithionite [x] 7 languages Français Italiano Português Suomi தமிழ் Tiếng Việt 中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item From Wikipedia, the free encyclopedia Oxyanion of sulfur at the oxidation number +3 Not to be confused with dithionate. The unusual structure of the dithionite anion. It has a remarkably long sulfur-sulfur bond. A ball-and-stick model of the dithionite ion. The dithionite is the oxyanion with the formula [S 2 O 4]2−. It is commonly encountered as the salt sodium dithionite. For historical reasons, it is sometimes called hydrosulfite, but it contains no hydrogen and is not a sulfite. The dianion has a steric number of 4 and trigonal pyramidal geometry. Production and reactions [edit] In its main applications, dithionite is generally prepared in situ by reduction of sulfur dioxide by sodium borohydride, described by the following idealized equation: NaBH 4 + 8 SO 2 + 8 NaOH → 4 Na 2 S 2 O 4 + NaBO 2 + 6 H 2 O Dithionite is a reducing agent. At pH7, its reduction potential is −0.66 V vs SHE. Its oxidation occurs with formation of sulfite: S 2 O 2− 4 + 2 H 2 O → 2 HSO− 3 + 2 e− + 2 H+ Dithionite undergoes acid hydrolyticdisproportionation to thiosulfate and bisulfite: 2 S 2 O 2− 4 + H 2 O → S 2 O 2− 3 + 2 HSO− 3 It also undergoes alkaline hydrolytic disproportionation to sulfite and sulfide: 3 Na 2 S 2 O 4 + 6 NaOH → 5 Na 2 SO 3 + Na 2 S + 3 H 2 O It is formally derived from dithionous acid (H 2 S 2 O 4), but this acid does not exist in any practical sense. Use and occurrence [edit] Sodium dithionite finds widespread use in industry as a reducing agent. It is for example used in bleaching of wood pulp and some dyes. Chemical analyses [edit] Dithionite is used in conjunction with complexing agents (for example, citric acid) to reduce iron(III) oxy-hydroxide into soluble iron(II) compounds and to remove amorphous iron(III)-bearing mineral phases in soil analyses (selective extraction). Harmful properties [edit] The decomposition of dithionite produces reduced species of sulfur that can be very aggressive for the corrosion of steel and stainless steel. Thiosulfate (S 2 O 2− 3) is known to induce pitting corrosion, whereas sulfide (S 2−, HS−) is responsible for stress corrosion cracking (SCC). References [edit] ^International Union of Pure and Applied Chemistry (2005). Nomenclature of Inorganic Chemistry (IUPAC Recommendations 2005). Cambridge (UK): RSC–IUPAC. ISBN0-85404-438-8. p.130. Electronic version. ^ Jump up to: abcJosé Jiménez Barberá; Adolf Metzger; Manfred Wolf (2000). "Sulfites, Thiosulfates, and Dithionites". Ullmann's Encyclopedia of Industrial Chemistry. Weinheim: Wiley-VCH. doi:10.1002/14356007.a25_477. ISBN978-3527306732. ^ Jump up to: abWietelmann, Ulrich; Felderhoff, Michael; Rittmeyer, Peter (2016-09-29) . "Hydrides". Ullmann's Encyclopedia of Industrial Chemistry. Weinheim, Germany: Wiley-VCH Verlag GmbH & Co. KGaA. doi:10.1002/14356007.a13_199.pub2. ISBN978-3-527-30673-2. OCLC751968805. ^Mayhew, S. G. (2008). "The Redox Potential of Dithionite and SO 2− from Equilibrium Reactions with Flavodoxins, Methyl Viologen and Hydrogen plus Hydrogenase". European Journal of Biochemistry. 85 (2): 535–547. doi:10.1111/j.1432-1033.1978.tb12269.x. PMID648533. Further reading [edit] Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.). Butterworth-Heinemann. doi:10.1016/C2009-0-30414-6. ISBN978-0-08-037941-8. Wikimedia Commons has media related to Dithionite ion. This corrosion-related article is a stub. You can help Wikipedia by expanding it. v t e Retrieved from " Categories: Reducing agents Corrosion Dithionites Sulfur oxyanions Electrochemistry stubs Chemical process stubs Metal stubs Hidden categories: Articles with short description Short description matches Wikidata Commons category link is on Wikidata All stub articles This page was last edited on 22 November 2024, at 08:34(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Dithionite 7 languagesAdd topic
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https://www.youtube.com/watch?v=O5_18wjClo4
Applied Calculus - Critical Points of Multivariable Functions Nathaniel Johnston 6890 subscribers 22 likes Description 1365 views Posted: 23 Sep 2020 We learn about critical points of differentiable functions 2 or more variables. We see that critical points are points where the graph of the function is flat, and that they come in 3 types: maximums, minimums, and saddle points. Previous video: Next video: IMPORTANT NOTE: This video only considers differentiable functions. If you consider more general functions, then points where f'(x) does not exist (single variable case), or the gradient of f does not exist (multivariable case), are also critical points. Timestamps: 00:00 - Introduction 02:23 - First example 03:43 - Types of critical points 05:36 - Second example 2 comments Transcript: Introduction hey folks my name is nathan johnston and today we're going to learn about critical points for multi-variable functions we're going to focus particularly on functions of two variables but basically everything that we say will extend naturally to functions of three or more variables as well all right so think back for now back to functions of just one variable though okay we already learned about critical points for one variable functions and all the critical point was back in that setting was it was a point where if you plugged it into the derivative of the function you got zero okay so it was a value of x for which f prime of x equals zero okay and what this meant geometrically was it was a flat spot on the graph of the original function right it was a spot where it had a horizontal tangent line okay well a critical point for a multi-variable function it's going to be basically the same thing it's a spot where on the graph of that function it's flat it's horizontal okay so how do we sort of formalize this how do we make this sort of proper and and talk about it mathematically well for a function of two variables a critical point it's a pair x y right it's a set of inputs for which the gradient equals zero okay so this time you don't just need the derivative because well there's a whole bunch of derivatives now right you have partial derivatives in the different directions this time we need the entire gradient to equal zero in other words we need the x partial derivative to equal zero and the y partial derivative t equals zero okay and sort of geometrically what this means is it means that the the graph of the function it's flat it's horizontal in the x direction and it's also flat or horizontal in the y direction okay but also very importantly if you think about directional derivatives remember directional derivatives the way we computed them was in terms of the gradient okay directional derivatives every directional derivative is just the gradient dotted with the direction vector okay well if the gradient equals zero then this dot product always equals zero so in particular if your x partial is zero and your y partial is zero therefore your gradient is zero therefore every directional derivative equals zero so it's not just flat in the x direction in the y direction it's actually flat in every direction if it's flat in those two directions okay but for the purposes of computation it suffices just to check the x and y directions okay you only have to check that the x partial is zero and the y partial is zero that gets you all of the other directions for free all right so let's go through a quick First example example here to get a feeling for critical points of multivariable functions okay so let's start off just with this simple function f of x y equals x squared plus y squared let's find all of its critical points okay and the way that you do this is you just take the partial derivatives and you set them equal to zero so the x partial derivative derivative of x squared is 2x derivative of y squared is just zero because with respect to x y is a constant okay and similarly similarly the y partial derivative is just 2y because x squared's a constant with respect to y and derivative of y squared is 2y with respect to y okay and what we need is we need both of these partial derivatives to be zero okay to get a critical point of a two variable function you need both of the partial derivatives to be zero not just one or the other okay so we need two x equals zero and two y equals zero in other words we need both x and y to be zero okay so what this tells us is that zero zero is the only critical point okay that's the only pair of x y values that make both of these equal to zero okay and geometrically what's going on here is that means on the graph of this function there's only one spot where the graph of that function's flat or horizontal and yeah that spot it's right down here if you plot this function it looks like a bowl okay it's called a paraboloid but it just looks like a bowl and then down at the very tippy bottom of the bowl that's where the critical point is okay Types of critical points so this raises a very natural question what can critical points of two variable functions look like okay we just saw an example f of x equals x squared plus y squared where the critical point it was the bottom of a bowl it was a minimum okay so that's the first possibility critical points they can be minimums just like this you know paraboloid bowl example okay but there are other possibilities as well and the the first other one that we're going to talk about is not too surprising it could also be a maximum critical points can be maximums so for example if we just stick a minus sign in front of the function that we were just working with and instead consider minus x squared minus y squared well all that is is that's the upside down paraboloid now now it's opening to the bottom rather than opening up okay so this function again is gonna have a unique critical point at zero zero but this time that critical point is up at the top of the hill okay so the critical point is a maximum in this case back in the one variable setting those are basically the only possibilities critical points were maxes or mins but in the multi-variable setting there's another possibility that results from the fact that there are lots of different directions there are also what are called saddle points okay and these are points where the function it increases in some directions and decreases in others okay so for example the function f of x equals x times y if you plot that well it's got a unique critical point at zero zero so again at the origin but if you sort of trace along in this direction you're gonna see oh yeah coming away from zero zero it goes up like a parabola but if you go in the perpendicular direction you find that it goes sort of down like an upside down parabola okay so it's not a max or a minute depends on which direction you walk in it's sort of a max in one direction but a min in another direction okay so those are called saddle points basically because the graph it looks kind of like a saddle Second example okay so let's go through another example now to sort of highlight these different types of critical points and in particular we're going to see that this function has a whole bunch of different critical points okay the previous example we went through there was just one critical point but now we're going to have a whole bunch and they're going to be of different types okay so the function that we're going to go through now is 3xy over e to the x squared plus y squared and so the way that you find the critical points is the same as what we did before you take the partial derivatives and you set them both equal to zero okay so take its x partial derivative and you get this junk here and if you take the y partial derivative you get this junk here via similar calculation but both of those can be found via the quotient rule or via the product rule combined with the chain rule okay so find those partial derivatives however you like and i've done something here i factored both of these partial derivatives as much as i can right i factored that numerator as much as i can and i simplified things down you're gonna want to do that before this next step you want to factor things as much as possible because now you're gonna set these both equal to zero okay and setting something equal to zero you wanna have it factored before you try to do that okay so i need to set both of those equal to zero let's do it i'm gonna focus on one of them at a time okay and then i'll combine my answers at the end okay so let's start off with this partial derivative with respect to x let's set that equal to zero okay well the way that you set a fraction equal to zero is you just set the numerator equal to zero don't worry about the e to the x squared plus y squared in the denominator that has no effect on whether or not the whole expression equals zero so just forget about it okay so that partial derivative equals zero if and only if the numerator equals zero and when does that numerator equal to zero well because it's factored we know it equals zero if and only if one of the factors equals zero so i know either three y equals zero or one minus two x squared equals zero okay one of the factors has to equal zero okay now i'm just going to simplify the second part a little bit one minus two x squared equals zero well when does that equal zero i'll just rearrange and solve for x x is plus or minus one over root two okay so those are the two possibilities those are the ways that the x partial derivative can equal zero either y equals zero or x is plus or minus 1 over root 2. okay and now we're just going to do the same thing with the other partial derivative do the same thing with the y partial set the numerator equal to zero okay so 3x times 1 minus 2y squared equals 0 that means that hey at least one of those factors equals 0. so either x equals 0 or 1 minus 2y squared equals 0. and again just solve for y in that second expression and you're going to get y equals plus or minus 1 over root 2 again okay it's very similar to the expression on the left just the roles of x and y i've swapped okay now be careful here with your and and or quantifiers okay so we need this to hold and this to hold okay so it's a weird combination of ands and ors be a little bit careful so what i'm gonna do is i'm gonna go through and ask okay well what happens if y equals zero and then i need one of these to hold and i'm gonna go through separately well what if what happens if x equals plus or minus one over root two and then i also need one of these to hold okay so one way for both of these expressions to hold is if y equals zero and now i jump over here and well x could equal zero okay so i get zero zero as one of my critical points another possibility i could think okay well what if y equals zero and y equals plus or minus one over root two oh that can't happen i can't have y equal to two different things okay so that's sort of splitting up of things doesn't get me a valid critical point all right so the next case i'm going to consider is what if x equals plus one over root two then x could equal zero oh no no it couldn't it can't equal two different things okay so x equals plus one over root two and y equals plus 1 over root 2 that's the thing that could happen okay so that's another critical point or i could also have x equals plus 1 over root 2 and y equals minus 1 over root 2. that's another critical point or i could do a similar thing with the negative x branch over here i could have x is minus 1 over root 2 and y is plus or i could have them both be negative okay and then i'm out of cases there's no other ways for this expression to be true and this expression to be true there are only those five critical points okay our follow-up question once we've found the critical points is usually well what types of critical points are they are they maxes are they mins or are they saddle points okay and we don't have the tools to answer that problem yet one thing that we could do is we could graph this function and just plot where the critical points are and here they are okay so 0 0 is this one in the middle and then 1 over root 2 1 over root 2 is over there and then minus minus is over there and then the ones with the mixed terms are there and there okay so what that's telling us what it looks like from the graph is we've got two maxes here and here and it also looks like we've got two mins here and here right those guys are down to bottoms of valleys and it also looks like we've got a saddle point in the middle it looks like zero zero is a saddle point because if you sort of go in this direction you're going to increase away from zero zero if you go in that direction you're going to decrease away from zero zero okay so it looks like we've got a saddle two maxes and two mins but how could we prove that how could we really sort sort of show that mathematically well that's what we're gonna do next lecture so i will see you then
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https://www.quora.com/How-many-four-digit-numbers-without-repetition-can-be-formed-using-the-digits-1-2-3-4-and-5
How many four digit numbers without repetition can be formed using the digits 1, 2, 3, 4, and 5? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Combinations Ma Four-digit Numbers Permutations Number Theory Number Combinations Permutations Combination Permutations and Com Permutations and Combinat... 5 How many four digit numbers without repetition can be formed using the digits 1, 2, 3, 4, and 5? All related (52) Sort Recommended Marianne Bauer Author has 456 answers and 326K answer views ·2y 5 options for 1st position, then one less option for each of the following places: 2nd, 3rd & 4th, thus 5432=120 four-digit numbers can be formed. These are: 1234 1235 1243 1245 1253 1254 1324 1325 1342 1345 1352 1354 1423 1425 1432 1435 1452 1453 1523 1524 1532 1534 1542 1543 2134 2135 2143 2145 2153 2154 2314 2315 2341 2345 2351 2354 2413 2415 2431 2435 2451 2453 2513 2514 2531 2534 2541 2543 3124 3125 3142 3145 3152 3154 3214 3215 3241 3245 3251 3254 3412 3415 3421 3425 3451 3452 3512 3514 3521 3524 3541 3542 4123 4125 4132 4135 4152 4153 4213 4215 4231 4235 4251 4253 4312 4315 4321 4325 Continue Reading 5 options for 1st position, then one less option for each of the following places: 2nd, 3rd & 4th, thus 5432=120 four-digit numbers can be formed. These are: 1234 1235 1243 1245 1253 1254 1324 1325 1342 1345 1352 1354 1423 1425 1432 1435 1452 1453 1523 1524 1532 1534 1542 1543 2134 2135 2143 2145 2153 2154 2314 2315 2341 2345 2351 2354 2413 2415 2431 2435 2451 2453 2513 2514 2531 2534 2541 2543 3124 3125 3142 3145 3152 3154 3214 3215 3241 3245 3251 3254 3412 3415 3421 3425 3451 3452 3512 3514 3521 3524 3541 3542 4123 4125 4132 4135 4152 4153 4213 4215 4231 4235 4251 4253 4312 4315 4321 4325 4351 4352 4512 4513 4521 4523 4531 4532 5123 5124 5132 5134 5142 5143 5213 5214 5231 5234 5241 5243 5312 5314 5321 5324 5341 5342 5412 5413 5421 5423 5431 5432 Upvote · Related questions More answers below How many numbers greater than 150 can be formed using the digits 1,2,3,4,5,6 without repetition? How many 4 digit numbers can be formed from the digits 1, 2, 3, 4, 5, and 6 without repetition of digits? How many 3-digit numbers can be formed using only 1, 2, 3,4? How many 3-digit numbers can be formed with the digits 1, 2, 3, 4, and 5 without repetition allowed? How many 5 digit numbers from 1, 2, 3, 4? Eric Jacobsen 6y Originally Answered: How many four-digit numbers can be made using the digits 1, 2, 3, 4 and 5 without repetition? · There are 3 answers depending on what “without repetition” covers. “without repetition” means only repeating of digits with each four-digit number. Then the solution set contains { 1234, 1234, 1234, … }, and the answer is infinity. “without repetition” means only repeating of the four-digit numbers. This gives us the set { 1111, 1112, 1113, 1114, 1115, 1121, 1122, … }. Each of the four digits can have 5 possible values. 5^4 = 625. “without repetition” means both of the above cases. Then the others who answered 120 are correct. The first digit can be one of the five { 1, 2, 3, 4, 5 }, the Continue Reading There are 3 answers depending on what “without repetition” covers. “without repetition” means only repeating of digits with each four-digit number. Then the solution set contains { 1234, 1234, 1234, … }, and the answer is infinity. “without repetition” means only repeating of the four-digit numbers. This gives us the set { 1111, 1112, 1113, 1114, 1115, 1121, 1122, … }. Each of the four digits can have 5 possible values. 5^4 = 625. “without repetition” means both of the above cases. Then the others who answered 120 are correct. The first digit can be one of the five { 1, 2, 3, 4, 5 }, then the next digit can only have four possible values, then three, etc. 5432 = 5! = 120 Upvote · 99 17 9 1 Tej Kaur Former Retired Assistant Director at Department of Higher Education (2004–2007) · Author has 6.9K answers and 1.4M answer views ·2y 1234,1235, , 1243,1245, 1253, 1254 1324, 1325, 1342, 1345, 1352, 1354 1423, 1425, 1432, 1435, 1452, 1453, 1523, 1524, 1532, 1534, 1542, 1543, 2134, 2143, 2135, 2145, 2153, 2154 2314, 2315, 2341, 2345, 2351, 2354, 2413, 2415, 2431, 2435 , 2451,2453, 2513, 2514, 2534, 2541, 2543, 2545, 3124, 3125, ,3142, 3145, 3152, 3154 3214, 3215, 3241, 3251, 3245, 32,54 3412, 3415, 3421, 3425, 3451,3452 3512, 3514, 3521, 3524, 3541, 3542, 4123,4125, 4132, 4135,4152, 4153 4213, 4215, 4231, 4235, 4215, 4251, 4312, 4315, 4321,4325, 43 51, 4352, 4512, 4513, 4521, 4523, 4531,4532, 5123, 5124, 5132, ,5134, 5142, 5143 5213, 5234, 52 Continue Reading 1234,1235, , 1243,1245, 1253, 1254 1324, 1325, 1342, 1345, 1352, 1354 1423, 1425, 1432, 1435, 1452, 1453, 1523, 1524, 1532, 1534, 1542, 1543, 2134, 2143, 2135, 2145, 2153, 2154 2314, 2315, 2341, 2345, 2351, 2354, 2413, 2415, 2431, 2435 , 2451,2453, 2513, 2514, 2534, 2541, 2543, 2545, 3124, 3125, ,3142, 3145, 3152, 3154 3214, 3215, 3241, 3251, 3245, 32,54 3412, 3415, 3421, 3425, 3451,3452 3512, 3514, 3521, 3524, 3541, 3542, 4123,4125, 4132, 4135,4152, 4153 4213, 4215, 4231, 4235, 4215, 4251, 4312, 4315, 4321,4325, 43 51, 4352, 4512, 4513, 4521, 4523, 4531,4532, 5123, 5124, 5132, ,5134, 5142, 5143 5213, 5234, 5235, 5241, 5234, 5243, 5312,. 5314, 5321,,5324, 5341,5342 5412, 5413, 5421, 5423, 5431, 5432, ( 120 ) Upvote · 9 2 Sanjiv Soni Former Pvt Tutor · Author has 9.7K answers and 1.9M answer views ·5y Originally Answered: How many 4-digit numbers can be formed from 1,2,3,4, and 5 if no repetition is allowed, and if repetition is allowed? · A four digit number has digitsvin the thousands,hundreds tens and unit places. Digitsgiven are 1,3,3,4and 5 Case 1:when repetition is not allowedDigit in thousands place can be chosen from the given five digits in 5 ways. As repetition is not allowed digits in hundreds,tens and unit place can be chosen in 4,3and 2 ways respectively. 4-digit number formed without repetition=5432=120ways Answer. 120 n Continue Reading A four digit number has digitsvin the thousands,hundreds tens and unit places. Digitsgiven are 1,3,3,4and 5 Case 1:when repetition is not allowedDigit in thousands place can be chosen from the given five digits in 5 ways. As repetition is not allowed digits in hundreds,tens and unit place can be chosen in 4,3and 2 ways respectively. 4-digit number formed without repetition=5432=120ways Answer. 120 numbers. Case:2 when re... Upvote · 9 3 9 1 Related questions More answers below How many 4-digit even numbers can be formed from digits {1,2,3,4,5,6} without repetition? How many 2 digit numbers can you make using 5, 1, 3, and 2 without repetition? Using 1,2,3,4,5,6 how many 4 digit combinations are there? How many numbers of 5 digits can be formed using the digits 1, 2, 3, 4, 5, and 6? How many four digits number can be formed using the digits 0,1,2,3,4,5,6? Ayush Priya Trust me.. I know this stuff.. ;) ·9y Originally Answered: How many 4 digit numbers can be formed using the numbers 1,2,3,4,5 with digits not repeated? · This question can be answered using two methods. Lets start with the most simplest one. Method 1: The number is four digits, so for them lets take four blanks _ _ _ _ The first blank can be filled using any of the five digits given hence we have 5 ways to fill the first blank. The second blank can now be filled by four remaining digits because repetition is not allowed and hence the digit selected for the first blank cannot be selected. So four ways. Similarly 3 for the third blank and two for the last blank. So total number of combinations become 5x4x3x2=120 Hence the answer is 120 such number can b Continue Reading This question can be answered using two methods. Lets start with the most simplest one. Method 1: The number is four digits, so for them lets take four blanks _ _ _ _ The first blank can be filled using any of the five digits given hence we have 5 ways to fill the first blank. The second blank can now be filled by four remaining digits because repetition is not allowed and hence the digit selected for the first blank cannot be selected. So four ways. Similarly 3 for the third blank and two for the last blank. So total number of combinations become 5x4x3x2=120 Hence the answer is 120 such number can be formed. Method 2: We have a set of five digits which are all distinct. Now, we have to select 4 digits from the given set of 5 digits. This can be done in 5 C 4 5 C 4 ways. Hence 5 ways. These 4 digits selected can be arranged in 4! ways. Hence total number of combinations become 5x4!=120 ways Hence the answer is 120 such numbers can be formed. Hope it helped you! :) Upvote · 99 10 9 2 Ayush Singh Patel Ph.D in Civil Engineering&Environmental Engineering, Integral Universty Lucknow (Expected 2027) · Author has 395 answers and 1M answer views ·Jun 18 Originally Answered: How many four-digit numbers can be formed using the digits 1, 2, 3, 4, and 5 without repetition, such that the number is even? · For the number to be even the last digit should be 2 or 4 Rest we have 3 more places so the number is _ _ _ (2 or 4). For the first place we have 4 options, For the second place we have 3 options, For the third place we have 2 options, so 4322 options in total = 48 numbers in total that are even “Hope this helps “ Upvote · Andrew Droffner Studied Mathematics at Rutgers University (Graduated 1995) · Author has 8.8K answers and 5.7M answer views ·5y Originally Answered: How many 4-digit numbers can be formed from the digits 1, 2, 3, 4, and 5 if no digits are repeated? · This is a basic combinatorics problem to select N distinct objects from a set S without replacement. In this case, the set S has more elements, 5, than need to be selected. N=4,S={1,2,3,4,5}N=4,S={1,2,3,4,5} Select a number from the set S S. There are c a r d(S)=5 c a r d(S)=5 choices the first time. Select again from the remaining set S 4 S 4 that has c a r d(S 4)=4 c a r d(S 4)=4 choices left. Repeat this a total of four times to get the combinations C C. C=5×4×3×2=120 C=5×4×3×2=120 ways to do it Example Selection Here is an example combination to generate the number 3514 3514. N=4,S={1,2,3,4,5}N=4,S={1,2,3,4,5} is the initial state Continue Reading This is a basic combinatorics problem to select N distinct objects from a set S without replacement. In this case, the set S has more elements, 5, than need to be selected. N=4,S={1,2,3,4,5}N=4,S={1,2,3,4,5} Select a number from the set S S. There are c a r d(S)=5 c a r d(S)=5 choices the first time. Select again from the remaining set S 4 S 4 that has c a r d(S 4)=4 c a r d(S 4)=4 choices left. Repeat this a total of four times to get the combinations C C. C=5×4×3×2=120 C=5×4×3×2=120 ways to do it Example Selection Here is an example combination to generate the number 3514 3514. N=4,S={1,2,3,4,5}N=4,S={1,2,3,4,5} is the initial state. c a r d(S 4)=4,S 4={1,2,4,5}c a r d(S 4)=4,S 4={1,2,4,5} c a r d(S 3)=3,S 3={1,2,4}c a r d(S 3)=3,S 3={1,2,4} c a r d(S 2)=2,S 2={2,4}c a r d(S 2)=2,S 2={2,4} c a r d(S 1)=1,S 1={2}c a r d(S 1)=1,S 1={2} Upvote · 9 2 9 1 Ricky Author has 417 answers and 1.4M answer views ·8y Originally Answered: How many 4 digit numbers can be formed using 1, 2, 3 and 4 without repeating a digit? · The answer is the number of ways to permute (1,2,3,4).(1,2,3,4). This is given by the size of the permutation group S 4 S 4 which is equal to 4!=4⋅3⋅2⋅1=24.4!=4⋅3⋅2⋅1=24. Upvote · 99 19 Chad Hanna B.Sc(Eng) in Electrical and Electronics Engineering, Imperial College London (Graduated 1974) · Author has 2.8K answers and 1.3M answer views ·5y Originally Answered: How many 4 digit numbers can be named using the digits 2,3,4, and 5 without repetition? · Ho hum, 4! (4 factorial), that is 4 x 3 x 2 x 1 or 24 Why? There are four ways of choosing the first digit, then three ways of choosing the second digit, two ways of choosing the third digit, and only one option for the last digit. Upvote · 9 1 John Chadwick Former Head of Electrical, Mechanical & MV Engineering at South Cheshire College (1972–2007) · Author has 1K answers and 609.8K answer views ·Updated 2y Originally Answered: How many four digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6, and 7 if repetition is not allowed? · FOUR digit integers without repetition: Digit 1 ==> 7 choices. Digit 2 ==> 6 choices. Digit 3 ==> 5 choices. Digit 4 ==> 4 choices. Number of integers with NO repeated digits = 7 6 5 4 = 840 OR using nPr to find 4 permutations possible from 7 different items = 7P4 = 7!/(7 - 4)! = 7!/3! = 840 Upvote · Rajendra Kumar Phophalia BE from Birla Institute of Technology and Science, Pilani · Author has 5.9K answers and 2.6M answer views ·Jun 15 Originally Answered: How many four-digit numbers can be formed using the digits 1, 2, 3, 4, and 5 without repetition, such that the number is even? · Last (ones units)digit should be 2 or4 to be even numbers,tens digits can be remaing 4 digits,100 digit will be remaing 3 digit sand 1000s digit can be remaing 2 digits ,si total 4 digit can be formed will be 2×3×4×2=48 Upvote · Khaliquzzaman Khan Taught Secondary School level classes · Author has 1.2K answers and 1.1M answer views ·3y Originally Answered: How many 4 digit numbers can be formed from the digits 1, 2, 3, 4, 5, and 6 without repetition of digits? · Digits = 1,2,3,4,5,6 Number of digits in number formed =4 No repetition of digits is allowed. First place may be filled in 6 ways second place may be filled in 5 ways third place may be filled in 4 ways fourth place may be filled in 3 ways Total number formed = 6543= 3012= 360 Upvote · Related questions How many numbers greater than 150 can be formed using the digits 1,2,3,4,5,6 without repetition? How many 4 digit numbers can be formed from the digits 1, 2, 3, 4, 5, and 6 without repetition of digits? How many 3-digit numbers can be formed using only 1, 2, 3,4? How many 3-digit numbers can be formed with the digits 1, 2, 3, 4, and 5 without repetition allowed? How many 5 digit numbers from 1, 2, 3, 4? How many 4-digit even numbers can be formed from digits {1,2,3,4,5,6} without repetition? How many 2 digit numbers can you make using 5, 1, 3, and 2 without repetition? Using 1,2,3,4,5,6 how many 4 digit combinations are there? How many numbers of 5 digits can be formed using the digits 1, 2, 3, 4, 5, and 6? How many four digits number can be formed using the digits 0,1,2,3,4,5,6? How many 5-digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if repetitions of digits are allowed? How many four digit numbers are formed with digits 1, 2, 3, 4, and 5, if the repetition of digits is allowed? How many 5-digit numbers can be formed with no digit repetition using digits 1, 2, 3, 4, and 5 using the listing method in illustrating permutation? How many odd three digit numbers can be formed using the numbers 0,1,2,3,4,5,6? How many two digit numbers can be formed using 1 2 3 4 5 6? Related questions How many numbers greater than 150 can be formed using the digits 1,2,3,4,5,6 without repetition? How many 4 digit numbers can be formed from the digits 1, 2, 3, 4, 5, and 6 without repetition of digits? How many 3-digit numbers can be formed using only 1, 2, 3,4? How many 3-digit numbers can be formed with the digits 1, 2, 3, 4, and 5 without repetition allowed? How many 5 digit numbers from 1, 2, 3, 4? How many 4-digit even numbers can be formed from digits {1,2,3,4,5,6} without repetition? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
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https://www.biologyonline.com/dictionary/hnrna
Heterogeneous nuclear Ribonucleic Acid (HnRNA) - Definition and Examples Skip to content Main Navigation Search Dictionary Articles Tutorials Dictionary> HnRNA HnRNA Definition noun, plural: hnRNAs The abbreviated form for heterogeneous nuclear ribonucleic acid: an extrachromosomal RNA molecule in the nucleus that serve as primary transcript from DNA Supplement The hnRNA is the collective term for the unprocessed mRNA (pre-mRNA) molecules in the nucleus. It is largely comprised of the pre-mRNA molecules that require extensive processing to become mature mRNA molecules. The hnRNA that is associated with proteins form the heterogenous nuclear ribonucleoprotein (hnRNP). The initial step of synthesizing mRNA is transcription. During transcription, the DNA sequence is read by RNA polymerase in order for the latter to create a complementary, antiparallel RNA strand (called primary transcript). The primary transcript that later becomes a messenger RNA (mRNA) is referred to as a pre-mRNA. The pre-mRNA, in turn, needs to undergo further processing to become a functional mature mRNA. A pre-mRNA molecule becomes a mature mRNA molecule after having undergone the following processes: 5′ cap addition – the addition of a 5’cap to the front or 5′ end of the pre-mRNA molecule at the start of transcription splicing – removal of any introns (non-coding sequences) and the splicing together of exons (protein-coding sequences) editing – in some instances, the nucleotide composition is altered polyadenylation – a poly a tail is added to the free 3′ end at the cleavage site Once these processes are completed the mature mRNA can then be transported from the nucleus into the cytoplasm for protein translation. Although the hnRNA serve as source of mRNA, not all hnRNA become cytoplasmic mRNA. See also: pre-mRNA Last updated on October 12th, 2020 You will also like... Animal Water Regulation Animals adapt to their environment in aspects of anatomy, physiology, and behavior. This tutorial will help you understa.. Plant Water Regulation Plants need to regulate water in order to stay upright and structurally stable. Find out the different evolutionary adap.. The Water Cycle The water cycle (also referred to as the hydrological cycle) is a system of continuous transfer of water from the air, s.. Adaptation Tutorial Adaptation, in biology and ecology, refers to the process or trait through which organisms or the populations in a habit.. Plant Biology Plantlife can be studied at a variety of levels, from the molecular, genetic and biochemical level through organelles, c.. The Origins of Life This tutorial digs into the past to investigate the origins of life. The section is split into geological periods in the.. Related Articles... No related articles found See all Related Topics Home Dictionary Articles Tutorials About Us Contact Us Editorial Guidelines The content on this website is for information only. It is not intended to provide medical, legal, or any other professional advice. Any information here should not be considered absolutely correct, complete, and up-to-date. Views expressed here do not necessarily reflect those of Biology Online, its staff, or its partners. Before using our website, please read our Privacy Policy. © 2001-2025 BiologyOnline. All Rights Reserved
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https://www.britannica.com/video/Just-the-facts-Apollo-11-moon-landing/-246490
Apollo 11's incredible journey to the Moon Transcript
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https://www.sciencedirect.com/topics/engineering/dimensionless-number
Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Review article On dimensionless numbers The dimensionless numbers are useful for several reasons. They reduce the number of variables needed for description of the problem. They can thus be used for reducing the amount of experimental data and at making correlations. They simplify the governing equations, both by making them dimensionless and by neglecting ‘small’ terms with respect ‘large’ terms. They produce valuable scale estimates, whence order-of-magnitude estimates, of important physical quantities. When properly formed, they have clear physical interpretation and thus contribute to physical understanding of the phenomenon under study. Also, choosing the relevant scales, they indicate the dominant processes. There are two main sources of DN: dimensional analysis and scaling of governing equations. View article Read full article URL: Self-Assembly of Nano- and Micro-structured Materials using Colloidal Engineering 6.3.2.5 Dimensionless Number Summary As we can see from Section 6.3.2, it is possible to describe the fluid flow conditions in a microfluidic system using four main dimensionless numbers. These are briefly summarized in Table 1. Table 1. Table of relevant dimensionless numbers for microfluidic flows. | | | | | | --- --- | Re | Reynolds | ρvL/μ | Inertial/viscous | Eq. (3) | | Pe | Péclet | | Convection/diffusion | Eq. (4) | | Ca | Capillary | | Viscous/interfacial | Eq. (5) | | We | Weber | | Inertial/interfacial | Eq. (6) | View chapterExplore book Read full chapter URL: Book series2019, Frontiers of NanoscienceMax Meissner, ... C. Patrick Royall Chapter Fluid Mechanics and Biofluids Principles 2016, Biofluid MechanicsAli Ostadfar PhD 1.2.4 Dimensionless Numbers of Biofluid Mechanics Dimensionless (pure) numbers represent a property of a physical phenomenon, without any physical units such as distance, time or mass. These numbers are a useful tool to characterize the mechanical behavior of fluids in dynamic conditions. Dimensionless numbers help us to understand correlation between some parameters or forces in the phenomenon and also rescaling of a fluidic experience. Mechanical engineering has more than 40 dimensionless numbers. Some of these numbers are very essential in biofluid mechanics, such as: ▪ : Reynolds number (Re) ▪ : Womersley number (α) ▪ : Strouhal number (St) ▪ : Dean number (De) ▪ : Stokes number (Stk) Reynolds number (Re), is defined as the ratio of inertial forces to viscose forces when a fluid flows in a channel. Reynolds number is also an assessment number to evaluate flow regimes within a similar fluid. The Reynolds number for a channel (or pipe) is defined as: (1.17) where is hydraulic diameter (m), is density (kg/m3), is fluid velocity (m/s) and is dynamic viscosity (Pa·s). Womersley number (α), is defined as the ratio of oscillatory inertia to viscose effects for a fluid in pulsatile flow. The Womersley number in the aorta and arteries is large and in arterioles it becomes very small (less than 1). The Womersley number for a channel (or pipe) is defined as: (1.18) where is hydraulic diameter (m), is density (kg/m3), is fluid frequency (Hz) and is dynamic viscosity (Pa·s). In the human body, heart beat rate can be considered as frequency. Strouhal number (St), is a dimensionless number to explain oscillatory flow mechanisms. This number is the ratio of inertial forces as a result of oscillatory flow conditions to inertial forces because of velocity changing between two points of a flow field. The Strouhal number for a channel (or pipe) is defined as: (1.19) where is characteristic length (for instance hydraulic diameter ) (m), is fluid frequency (Hz) and is velocity of the fluid. There is a relation among Womersley, Reynolds and Strouhal numbers : (1.20) Dean number (De), is the ratio of viscous force on a fluid flowing in a curved pipe to the centrifugal force. The Dean number for a channel (or pipe) is defined as: (1.21) where is Reynolds number, is hydraulic diameter (m) and R is the radius of curvature of the channel (m). A schematic of a curved pipe (vessel) is shown in Fig. 1.14. Stokes number (Stk), is defined as the ratio of the characteristic reaction time of a suspended particle in fluid flow to a characteristic time of the flow. The reaction time is the time that the particle takes to respond to velocity changes in flow. The Stokes number is defined as: (1.22) where is particle density, is particle diameter, is fluid velocity, is dynamic viscosity of fluid and D is diameter of the obstacle (typically for channels is inner diameter). View chapterExplore book Read full chapter URL: Book2016, Biofluid MechanicsAli Ostadfar PhD Chapter Dimensionless Characterization Number for the Heat Transfer 2016, Heat Exchanger Design GuideM. Nitsche, R.O. Gbadamosi Abstract The following dimensionless numbers are calculated: Reynolds number, Prandtl number, Nusselt number, Stanton number, Colburn Factor, Kern Factor, and Graßhof number View chapterExplore book Read full chapter URL: Book2016, Heat Exchanger Design GuideM. Nitsche, R.O. Gbadamosi Chapter Introduction 2015, Biofluid Mechanics (Second Edition)David A. Rubenstein, ... Mary D. Frame 1.6 Salient Biofluid Mechanics Dimensionless Numbers Dimensionless numbers tend to be very useful in characterizing many types of engineering systems. You may be familiar with this type of analysis from different classes. For instance, you may have encountered the engineering parameter of strain, which is a dimensionless number that relates the percent of stretch that a material experiences to the resting length of that same material (note that some prefer to report strain in dimensions of length/length, e.g., mm/mm, but these dimensions do cancel out and thus you would remain with a dimensionless quantity). This type of dimensionless number helps us to scale a parameter across multiple types of scenarios that engineers may come across. In fluid mechanics, you may also encounter this type of dimensionless number to simplify the analysis. Some fluid mechanics engineers will report variables divided by some characteristics or constant value. For instance, some will divide the velocity (which can be a variable of space and time) by the inflow velocity (if it is uniform) or the centerline velocity at a given point of interest. Therefore, all of the velocity values become related to this value and the velocity profile will be scaled by this constant value. Fundamentally, this does not change the fluid properties or analysis of the problem, however, it may be easier to report data in this manner or analyze the problem under these conditions. There is a second type of dimensionless number that exists in engineering fields. This second type of dimensionless number provides a measure of the importance of phenomenon that plays a role in dictating how an event occurs. This second type of dimensionless number also helps us to rescale problems as needed. Since, this type of dimensionless number provides important information about the flow conditions, many different dimensionless quantities have been developed. We will discuss many of these dimensionless numbers and one method to derive these dimensionless numbers in Section 14.3. However, due to the importance of two dimensionless numbers in biofluids mechanics phenomena, we will briefly discuss them here, and leave the more thorough discussion for Section 14.3. The Reynolds number (Re) is the first dimensionless number that is important to nearly all biofluid mechanics flows. As stated above, dimensionless numbers relate two (or more) important phenomenon that play a role in the flow that is being analyzed. The Reynolds number relates the overall inertial forces that govern the flow to the viscous forces that will impede the flow. This is important because for any flow to occur, enough force must be present to overcome the fluids resistance to flow. This does not mean that the Reynolds number must always be greater than one, because the Reynolds number does not relate the driving forces to the resistance to flow; it only relates some of the inertial forces to some of the viscous forces (e.g., adhesion of the fluid to do the bounding surface is not in this quantification). Recall that inertia is defined as the objects resistance to change its velocity. Objects with a very high inertia resist changes to velocity very strongly, whereas objects with a low inertia do not resist changes to velocity very strongly. A simple experiment to observe inertia is to take a pen and hold it at its center between your forefinger and thumb. Now move your fingers to cause the ends of the pen to wiggle up and down. With a typical pen, these constant changes in motion are relatively easy to accomplish. Now conduct the same experiment, with the same pen, however, hold the pen at one of the ends. The pen resists the changes in motion much greater under the second conditions. The moment of inertia of the pen in the second scenario is much larger than the first scenario. Fluid viscosity is simplistically defined as the internal resistance of a fluid to deform under shear loading conditions (we will discuss the more stringent definition of viscosity in later sections). A fluid with a large viscosity, requires a large shearing force to deform the sample (or a small force applied for a very lengthy time), whereas a fluid with a lower viscosity, requires a smaller shearing force to deform the sample. Another simple experiment can illustrate viscosity. Hold a glass of water and tip it at an angle (make sure not to spill the water!). Did the water deform? It should have and the force was very low on the water. Now conduct the same experiment with syrup, toothpaste or honey. Did the second fluid deform as easily; probably not. In this case, water has a lower viscosity as compared with the second fluid. A fluid that will experience a change in its velocity will have to balance both the inertial forces and the viscous forces, which resist changes to velocity or deformation, respectively, since when a fluid proceeds to a new velocity, the inertia will play a role in resisting this change and the viscous interactions will play a role in the deformation changes. Thus, the Reynolds number provides a measure of which forces dominate changes to a fluids velocity. To relate these important parameters, the Reynolds number is defined as (1.1) where ρ is the density of the fluid, v is some characteristic velocity (e.g., centerline velocity, max velocity, or other), d is a characteristic length of the flow (e.g., channel length, radius, diameter, or other), and μ is the dynamic viscosity of the fluid. Even though there are choices inherent within the Reynolds number formulation, there are some conventional choices for the characteristic velocity and length. If your flow can be approximated to be passing through a perfect tube with a constant cross-section, v is conventionally chosen as the spatial mean flow velocity over the circular cross-section and d is conventionally chosen as the diameter of the tube. The Reynolds number also provides a measure of flow characteristics. For instance, low Reynolds number flows tend to be laminar whereas high Reynolds number flows tend to be turbulent. The transition between laminar and turbulent flows tends to be hard to strictly define, since there are many properties that affect the overall laminar versus turbulent flow properties. However, for perfect tubes, the flow begins to transit to turbulence when the flow exceeds a Reynolds number of approximately 2,300. However, true turbulence (as defined in Section 2.12), will only be found in flows with a Reynolds number of approximately 10,000. Flows in-between these two values are said to be transitioning and exhibit both laminar and turbulent flow properties. If the geometry of the flow changes, these transitioning values will also change; typical Reynolds number values, to describe laminar or turbulent flows can be found in fluid mechanics textbooks. A second salient dimensionless number for biofluid mechanics flows is the Womersley number, which is related to the pulsatility of the flow. Flows that have regular oscillating time-dependent components (e.g., are not steady), are said to have pulsatility. The Womersley number is a ratio of the fluids oscillatory inertia to the viscous momentum. The oscillatory inertia is a measure of the forces that are governing the pulsatile flow, whereas the viscous forces are again a measure of the fluids’ overall resistance to changes in its velocity. The Womersley parameter can be defined as (1.2) where ρ is the fluid density, v is a characteristic velocity, ω is the angular frequency associated with the oscillation, μ is the dynamic viscosity, d is a characteristic length, and ν is the kinematic viscosity. A fluid with a low Womersley number is characterized by having essentially no phase difference between the pulsatile pressure waveform that is driving the flow and the pulsatile velocity waveform associated with the flow. With increasing Womersley numbers, a phase difference between the pressure and velocity waveform can be observed; which is due to the inertia of the fluid resisting changes governed by the pressure waveform and that the pulse frequency is relatively high. Other important dimensionless numbers and how dimensionless numbers can be used in various biofluid mechanics applications will be discussed in Section 14.3. End of Chapter Summary 1.1 : This textbook will discuss basic fluid mechanics principles, flows within the macrocirculation, flows within the microcirculation, specialty circulations, and experimental techniques. 1.2 : The National Institutes of Health working definition of biomedical engineering is “Biomedical engineering integrates physical, chemical, mathematical, and computational sciences and engineering principles to study biology, medicine, behavior, and health. It advances fundamental concepts; creates knowledge from the molecular to the organ systems level; and develops innovative biologics, materials, processes, implants, devices and informatics approaches for the prevention, diagnosis, and treatment of disease, for patient rehabilitation, and for improving health.” 1.3 : Fluid mechanics is useful for the analysis of anything that includes an interaction with a liquid or gas. This includes traditional engineering applications, as well as many biological applications. 1.4 : Biofluid mechanics is focused on how biological systems interact with and/or use liquids/gases. For humans, this includes obtaining and transporting oxygen, maintaining body temperature, and regulating homeostasis. Cardiovascular diseases account for nearly one of three deaths in the United States; the highest prevalence regions of cardiovascular diseases are along the lower Mississippi River valley. 1.5 : Dimensions are physical properties, whereas units are arbitrary names that correlate to a measurement. There are seven base dimensions, including time, length, mass, temperature, electric current, amount of substance, and luminous intensity. All other physical parameters can be related to these base units. Depending on which system of units are chosen, the base units can change, but again all dimensions/units can be defined from one another. 1.6 : Dimensionless numbers are important for either scaling fluid properties, relating important parameters that govern fluid flows or both. Two of the most widely used biofluid mechanics dimensionless numbers are the Reynolds number and the Womersley number which relate the inertial forces to the viscous forces and the pulsatility to the viscous forces, respectively. View chapterExplore book Read full chapter URL: Book2015, Biofluid Mechanics (Second Edition)David A. Rubenstein, ... Mary D. Frame Chapter Microfluidic devices for cell manipulation 2013, Microfluidic Devices for Biomedical ApplicationsH.O. Fatoyinbo 8.3.1 Dimensionless numbers Dimensionless numbers reduce the number of variables that describe a system, thereby reducing the amount of experimental data required to make correlations of physical phenomena to scalable systems. The most common dimensionless group in fluid dynamics is the Reynolds number (Re), named after Osborne Reynolds who published a series of papers describing flow in pipes (Reynolds, 1883). It represents the ratio of inertial forces to viscous forces (Equation [8.1]), where ρ is the fluid density, u is average fluid velocity, Dh is cross-sectional length of the system, and μ is the dynamic fluid viscosity. Gravesen et al. (1993) indicated that flow regimes within 32 different microfluidic devices analysed were not simply laminar or turbulent but had a transitional Reynolds number, Ret = 30 × (L/Dh), accounting for flow development, varying as a function of entrance length (L) and the hydraulic diameter (Dh). The three regimes described were based on differences in pressure drops due to inertial forces and viscous forces. Large length to hydraulic diameter ratios, greater than 70, give Ret > 2300, although since none of the microsystems operated in fully developed turbulent regimes (Re ≈ 4000) the value had little significance. Typically, as a system is scaled down the influences of inertial forces decreases, while viscous forces become more dominant. Thus, microflows are generally characterised as laminar, with transitions to turbulent flow (Re ≥ 2300) rarely developing (Brody et al., 1996; Schulte et al., 2002). Table 8.1 defines some other common dimensionless numbers used in describing Newtonian and non-Newtonian fluid characteristics in microfluidic systems (Ruzicka, 2008). View chapterExplore book Read full chapter URL: Book2013, Microfluidic Devices for Biomedical ApplicationsH.O. Fatoyinbo Review article On dimensionless numbers 2008, Chemical Engineering Research and DesignM.C. Ruzicka The dimensionless equations have certain advantages. They are independent of the system of units. The dimensionless numbers are relevant for the problem. The proportion between individual terms can be seen. These equations apply to all physically similar systems, so they are useful for scale-up/down. View article Read full article URL: Journal2008, Chemical Engineering Research and DesignM.C. Ruzicka Review article Mathematical characterization of scenarios of fluid flow and solute transport in porous media by discriminated nondimensionalization 2012, International Journal of Engineering ScienceI. Alhama MantecaJr., ... F. Alhama Highlights ► Flow and solute transport scenarios are characterized by dimensionless numbers. ► Discriminated nondimensionalization applies to obtain the dimensionless numbers. ► The less number of dimensionless groups are deduced. ► Resulting dimensionless groups are different from those of classical studies. ► Dimensionless numbers are interpreted in terms of balances in the domain. View article Read full article URL: Journal2012, International Journal of Engineering ScienceI. Alhama MantecaJr., ... F. Alhama Chapter Introduction 2022, Biofluid Mechanics (Third Edition)David A. Rubenstein, ... Mary D. Frame 1.6 Salient biofluid mechanics dimensionless numbers Dimensionless numbers tend to be very useful in characterizing many types of engineering systems. You may be familiar with this type of analysis from previous engineering courses. We will briefly introduce dimensionless numbers at this point, but more fully complete the analysis at a later time. For instance, you may have encountered the engineering parameter of strain, which is a dimensionless number that relates the percent of stretch that a material experiences to the resting length of that same material (note that some prefer to report strain in dimensions of length/length, e.g., mm/mm, but these dimensions do cancel out and thus you would remain with a dimensionless quantity). This type of dimensionless number helps us scale a parameter across multiple types of scenarios that engineers may come across, and typically these relationships become independent of other parameters (e.g., strain is largely independent of cross-sectional area geometry). In fluid mechanics, you may also encounter this type of dimensionless number to simplify the analysis. Some fluid mechanics engineers will report variables divided by some characteristics parameter or constant value to help relate the scale or parameters across various dimensions. For instance, some will divide the velocity (which can be a variable of space and time) by the inflow velocity (if it is uniform) or perhaps the centerline velocity at a given point of interest. Therefore, all of the velocity values become relative to this value and the velocity profile will be scaled by this constant value. Fundamentally, this does not change the fluid properties or analysis of the problem; however, it may be easier to report, analyze, or interpret the data in this manner. There is a second type of dimensionless number that exists in engineering fields. This second type of dimensionless number provides a measure of the importance of the phenomenon that plays a role in dictating how an event occurs. This type of dimensionless number also helps us to rescale problems as needed. Because this type of dimensionless number provides important information about the flow conditions, many different dimensionless quantities of importance to fluid mechanics and biofluid mechanics have been developed. We will discuss many of these dimensionless numbers and one method to derive these dimensionless numbers in Section 15.3. However, because of the importance of two dimensionless numbers in biofluids mechanics phenomena, we will briefly discuss them here, while leaving a more thorough discussion for Section 15.3. The Reynolds number (Re) is the first dimensionless number that is important to nearly all biofluid mechanics flows. As stated earlier, dimensionless numbers relate two (or more) important phenomenon that play a role in the flow being analyzed. The Reynolds number relates the overall inertial forces that govern the flow to the viscous forces that may impede the flow. This is important because for any flow to occur, enough force must be present to overcome the fluid’s resistance to flow. This does not mean that the Reynolds number must always be greater than 1, because the Reynolds number does not relate the driving forces to the resistance to flow; it only relates some of the inertial forces to some of the viscous forces (e.g., adhesion of the fluid to the bounding surface is not in this quantification). Recall that inertia is defined as the object’s resistance to change its velocity. Objects with a very large inertia resist changes to velocity very easily, whereas objects with a low inertia do not resist changes to velocity as strongly as high-inertia materials do. A simple experiment to observe inertia is to take a pen and hold it at its center between your forefinger and thumb. Now move your fingers to cause the ends of the pen to wiggle up and down. With a typical pen, these constant changes in motion are relatively easy to accomplish. Now conduct the same experiment, with the same pen, but hold the pen at one of the ends. The pen resists the changes in motion much greater under the second conditions. The moment of inertia of the pen in the second scenario is much larger than the first scenario. Fluid viscosity is simplistically defined as the internal resistance of a fluid to deform under shear loading conditions (we will discuss a more stringent definition of viscosity in later sections). A fluid with a large viscosity requires a large shearing force to deform the sample (or a small force applied for a very lengthy time), whereas a fluid with a lower viscosity requires a smaller shearing force to deform the sample. Another simple experiment can illustrate viscosity. Hold a glass of water and tip it at an angle (make sure not to spill the water!). Did the water deform? It should have, and the force was very low on the water. Now conduct the same experiment with syrup, toothpaste, or honey. Did the second fluid deform as easily; probably not. In this case water has a lower viscosity compared with the second fluid. A fluid that will experience a change in its velocity will have to balance changes to both the inertial forces and the viscous forces, which resist changes to velocity or deformation, respectively. This is true because when a fluid proceeds to a new velocity, the inertia will play a role in resisting this change and the viscous interactions will play a role in the deformation changes. Thus, the Reynolds number provides a measure of which forces dominate changes to a fluids velocity. To relate these important parameters, the Reynolds number is defined as (1.6) where ρ is the density of the fluid, ν is some characteristic velocity (e.g., centerline velocity, max velocity, or other), d is a characteristic length of the flow (e.g., channel length, radius, diameter, or other), and μ is the dynamic viscosity of the fluid. Even though there are choices inherent within the Reynolds number formulation, there are some conventional choices for the characteristic velocity and length. If your flow can be approximated to be passing through a perfect tube with a constant cross section, v is conventionally chosen as the spatial mean flow velocity over the circular cross section and d is conventionally chosen as the diameter of the tube. The Reynolds number also provides a measure of flow characteristics. For instance, low Reynolds number flows tend to be laminar whereas high Reynolds number flows tend to be turbulent. The transition between laminar and turbulent flows tends to be hard to strictly define, since there are many properties that affect the overall laminar versus turbulent flow properties. However, for perfect tubes with a constant cross-sectional area, the flow begins to transition to turbulence when the flow exceeds a Reynolds number of approximately 2300. However, true turbulence (as defined in Section 2.12) will only be found in flows with a Reynolds number of approximately 10,000. Flows in-between these two values are said to be transitioning and exhibit both laminar and turbulent flow properties. If the geometry of the flow changes, these transitioning values will also change; typical Reynolds number values, to describe laminar or turbulent flows can be found in fluid mechanics textbooks. A second salient dimensionless number for biofluid mechanics flows is the Womersley number, which is related to the pulsatility of the flow. Flows that have regular oscillating time-dependent components (e.g., are not steady flows), are said to have pulsatility. The Womersley number is a ratio of the fluids oscillatory inertia to the viscous momentum. The oscillatory inertia is a measure of the forces that govern the pulsatile flow, whereas the viscous forces are again a measure of the fluids’ overall resistance to changes in its velocity. The Womersley parameter can be defined as (1.7) where ρ is the fluid density, v is a characteristic velocity, ω is the angular frequency associated with the oscillation, μ is the dynamic viscosity, d is a characteristic length, and ν is the kinematic viscosity. A fluid with a low Womersley number is characterized by having essentially no phase difference between the pulsatile pressure waveform that is driving the flow and the pulsatile velocity waveform associated with the flow. With increasing Womersley numbers, a phase difference between the pressure and velocity waveform can be observed, which is due to the inertia of the fluid resisting changes governed by the pressure waveform and that the pulse frequency is relatively high. Other important dimensionless numbers and how dimensionless numbers can be used in various biofluid mechanics applications will be discussed in Section 15.3. View chapterExplore book Read full chapter URL: Book2022, Biofluid Mechanics (Third Edition)David A. Rubenstein, ... Mary D. Frame Chapter Hydraulic and thermal analysis 2019, Handbook of Multiphase Flow AssuranceTaras Y. Makogon Dimensionless numbers The dimensionless numbers more commonly used in flow assurance as in multiphase flow, fluid interfaces or solids deposition modeling include: The use of dimensionless numbers may be useful in understanding flow characteristics. For example Beggs and Brill used Froude number in coordinates to present a flow regime map. Someday multiphase software tools will be able to plot each of these dimensionless values, which may lead to new understanding of multiphase flow phenomena and improved machine learning pattern recognition. View chapterExplore book Read full chapter URL: Book2019, Handbook of Multiphase Flow AssuranceTaras Y. Makogon Related terms: Energy Engineering Heat Transfer Coefficient Nusselt Number Dimensional Analysis Nanofluid Phase Change Material Reynolds' Number Viscous Force Tension Surface Fluid Viscosity View all Topics
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https://www.johndcook.com/blog/2021/06/21/reversed-cauchy-schwarz-inequality/
Reversed Cauchy-Schwarz inequality Posted on by John This post will state a couple forms of the Cauchy-Schwarz inequality and then present the lesser-known reverse of the Cauchy-Schwarz inequality due to Pólya and Szegö. Cauchy-Schwarz inequality The summation form of the Cauchy-Schwarz inequality says that for sequences of real numbers xn and yn. The integral form of the Cauchy-Schwarz inequality says that for any two real-valued functions f and g over a measure space (E, μ) provided the integrals above are defined. You can derive the sum form from the integral form by letting your measure space be the integers with counting measure. You can derive the integral form by applying the sum form to the integrals of simple functions and taking limits. Flipping Cauchy-Schwarz The Cauchy-Schwarz inequality is well known . There are reversed versions of the Cauchy-Schwarz inequality that not as well known. The most basic such reversed inequality was proved by Pólya and Szegö in 1925 and many variations on the theme have been proved ever sense. Pólya and Szegö’s inequality says for some constant C provided f and g are bounded above and below. The constant C does not depend on the functions per se but on their upper and lower bounds. Specifically, assume Then where Sometimes you’ll see C written in the equivalent form This way of writing C makes it clear that the constant only depends on m and M via their ratio. Note that if f and g are constant, then the inequality is exact. So the constant C is best possible without further assumptions. The corresponding sum form follows immediately by using counting measure on the integers. Or in more elementary terms, by integrating step functions that have width 1. Sum example Let x = (2, 3, 5) and y = (9, 8, 7). The sum of the squares in x is 38 and the sum of the squares in y is 194. The inner product of x and y is 18+24+35 = 77. The product of the lower bounds on x and y is m = 14. The product of the upper bounds is M = 45. The constant C = 59²/(4×14×45) = 1.38. The left side of the Pólya and Szegö inequality is 38×194 = 7372. The right side is 1.38×77²= 8182.02, and so the inequality holds. Integral example Let f(x) = 3 + cos(x) and let g(x) = 2 + sin(x). Let E be the interval [0, 2π]. The following Mathematica code shows that the left side of the Pólya and Szegö inequality is 171π² and the right side is 294 π². The function f is bound below by 2 and above by 4. The function g is bound below by 1 and above by 3. So m = 2 and M = 12. ``` In:= f[x_] := 3 + Cos[x] In:= g[x_] := 2 + Sin[x] In:= Integrate[f[x]^2, {x, 0, 2 Pi}] Integrate[g[x]^2, {x, 0, 2 Pi}] Out= 171 π² In:= {m, M} = {2, 12}; In:= c = (m + M)^2/(4 m M); In:= c Integrate[f[x] g[x], {x, 0, 2 Pi}]^2 Out= 294 π² ``` Related posts Three books on inequalities Sum and mean inequalities move in opposite directions The classic book on inequalities by Hardy, Littlewood, and Pólya mentions the Pólya-Szegö inequality on page 62, under “Miscellaneous theorems and examples.” Maybe Pólya was being inappropriately humble, but it’s odd that his inequality isn’t more prominent in his book.
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https://www.thorlabs.com/navigation.cfm?guide_id=8
Thank you. × Products Home / Polarization Optics Polarization Optics Thorlabs offers polarization optics and polarization instruments. Polarization optics change the state of polarization of incident radiation, whereas polarization instruments control and measure the polarization state. Thorlabs' polarization optics operate over the UV, visible, or IR spectral ranges and include polarizers, wave plates / retarders, quartz-wedge depolarizers, liquid crystal polymer depolarizers, and vortex retarders. Our polarization instruments include fiber polarization management, polarimeters, and liquid crystal devices. Solutions are available for both free space and fiber optic polarization applications. Polarizers Wave Plates & Retarders Vortex Retarders Quartz-Wedge Depolarizers Liquid Crystal Depolarizers Faraday Rotators Liquid Crystal Devices Fiber Polarization Management Polarization Instrumentation
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https://jmedicalcasereports.biomedcentral.com/articles/10.1186/1752-1947-4-269
Advertisement Agranulocytosis and hepatic toxicity with ticlopidine therapy: a case report Journal of Medical Case Reports volume 4, Article number: 269 (2010) Cite this article 4482 Accesses 4 Citations Metrics details Abstract Introduction Ticlopidine is a platelet inhibitor used to prevent thrombosis in patients with cerebrovascular or coronary artery disease. The most common side effects are mild and transitory: diarrhea, dyspepsia, nausea and rashes. More serious, but less frequent, adverse effects are hematological dyscrasia and cholestatic hepatitis. We report a rare case of agranulocytosis associated with hepatic toxicity, probably related to the use of ticlopidine. Case presentation A 70-year-old Caucasian woman, with no previous history of hematological or liver diseases, was treated with ticlopidine 250 mg twice daily immediately after a vertebrobasilar stroke. Upon admission, her blood tests were normal. About four weeks later she developed agranulocytosis and hepatic toxicity. Ticlopidine was discontinued immediately, and aspirin 25 mg and dipyridamole 200 mg were given twice daily. She was treated with hematopoietic growth factors (granulocyte colony stimulating factor), with a rapidly increased white blood count and progressive normalization of liver tests as a result. Conclusion In the first three months following initiation of ticlopidine therapy, regular monitoring of complete blood cell count and of liver function tests is essential for the early detection of serious and unpredictable side effects. Peer Review reports Introduction Ticlopidine is a thienopyridine derivative with platelet inhibitor capability. It acts by inhibiting the platelet aggregation induced by adenosine diphosphate and by blocking the membrane receptors of fibrinogen. It is used to prevent thrombosis in patients with cerebrovascular or coronary artery disease. Two randomized clinical studies [1, 2] proved the drug's efficacy versus placebo and aspirin in reducing the risk of transient ischemic attack and stroke in patients with a history of cerebrovascular events. Because of its adverse effects, the use of this drug is reserved for patients in whom aspirin is contraindicated, not tolerated, or when treatment with aspirin fails. The most common side effects are mild and transitory: diarrhea, dyspepsia, nausea and rashes. More serious, but less frequent, adverse effects are hematological dyscrasia (particularly agranulocytosis, aplastic anemia, neutropenia, pancytopenia, thrombocytopenia and thrombotic thrombocytopenia purpura) and cholestatic hepatitis. However, to our knowledge, there are only a few published reports of the simultaneous occurrence of hematological and hepatic toxicity induced by ticlopidine. We report a case of agranulocytosis associated with cholestatic hepatitis related to the use of ticlopidine. Case presentation A 70-year-old Caucasian woman was admitted to our Rehabilitation Ward (San Paolo Hospital, Milan) because of gait ataxia after right bulbar stroke, which occurred 10 days previously. Her medical history pointed out hypertension and hypercholesterolemia. She had no prior history with regard to hematological or liver diseases, alcohol abuse or blood transfusion. Her habitual medications were aspirin 100 mg/day, atorvastatin 20 mg/day and amlodipine 5 mg/day. Immediately after her stroke, she discontinued aspirin and started therapy with ticlopidine 250 mg twice daily. Upon admission, her blood tests were normal. About four weeks later, she developed agranulocytosis. Her white blood count was 2600 cells/μL (reference range: 4000 to 10,000 cells/μL), neutrophil count was 100 cells/μL (reference range: 2000 to 7000 cells/μL), and liver function tests revealed a mixed cholestasis and hepatocellular injury. She had no fever and she was asymptomatic. She had elevated levels of alanine aminotransferase (560 U/L, reference range 5 to 41 U/L), of aspartate aminotransferase (551 U/L, reference range 5 to 41 U/L), of γ-glutamyl transpeptidase (449 U/L, reference range 11 to 50 U/L), of alkaline phosphatase (821 U/L, reference range 98 to 279 U/L). Total and direct bilirubin and the coagulation tests were normal. Serology tests for hepatitis A, B and C, and for Epstein-Barr virus (EBV) and cytomegalovirus (CMV) were negative. The anti-nuclear antibodies (ANA), the anti-mitochondrial antibodies (AMA) and anti-smooth muscle antibodies (LKM) were all negative. Cobalamin and folate dosages were normal. An abdominal ultrasound scan showed liver steatosis but did not highlight any alterations in the intra-hepatic and extra-hepatic biliary pathways and, in particular, no sign of dilatation emerged. A bone marrow aspirate showed myeloid maturation arrest, with decreased myeloid precursors and immature forms, like by iatrogenic attack. A cytogenetic analysis on bone marrow blood was 46, XX. Ticlopidine was immediately discontinued. Aspirin 25 mg and dipyridamole 200 mg twice daily were started . She was treated with granulocyte colony stimulating factor (Filgastrim, 0.3 mg/day) with an excellent evolution. On the second day, her white blood count was normal (white blood cells: 5300 cells/μL; neutrophil count 1500 cells/μL); her liver function tests progressively got better with normalization after four weeks. A liver biopsy was not performed because of her serious hemathological dyscrasia and the self-limiting nature of liver disorder. The pathogenesis of the various types of toxic effects associated with ticlopidine therapy is unclear. There is no test available that can confirm the diagnostic hypothesis of the drug toxicity apart from the exclusion of other possible causes and the normalization of the blood tests after the drug discontinuation. In our patient, ticlopidine may have been responsible of concomitant hematological and hepatic toxicity. In fact, other diagnostic hypothesises were excluded and when the drug was discontinued, the blood cell count and the liver function tests rapidly normalized. The onset of hematological dyscrasia is temporally related to the initiation of ticlopidine therapy, generally occurring within the first three months, and the dyscrasia resolves within three weeks after discontinuation of therapy . The latent period between the introduction of ticlopidine and the appearance of hepatotoxicity is variable, ranging from one week to six months, but it is in the range of two to 12 weeks in most patients . Hepatic toxicity is not dose dependent and is not related to the treatment duration . When ticlopidine is discontinued, symptoms and liver abnormalities usually resolve within one to three months. In the cases of drug-induced hepatotoxicity, the liver biopsy can suggest but not establish the diagnosis, and is mainly directed to exclude other diagnosis. While severe neutropenia is a life-threatening adverse effect due to the occurrence of fatal infections, there are no fatal cases and no irreversible hepatic damages. The only reported fatal case was due to the co-occurrence of neutropenia, which led to septic shock . We emphasize that, in the first three months following initiation of ticlopidine therapy, besides a complete blood cell count, periodic checks of liver function are recommended. Hepatic toxicity induced by ticlopidine is underestimated. Regular monitoring of complete blood cell count and of liver function tests is important for prompt detection and treatment of adverse reactions but is unlikely to prevent their occurrence altogether. Conclusions In the first three months after starting ticlopidine therapy, regular monitoring of complete blood cell counts and of liver function tests should be recommended for the early detection of serious side effects, even if infrequent. Consent Written informed consent was obtained from the patient for publication of this case report and any accompanying images. A copy of the written consent is available for review by the Editor-in-Chief of this journal. References Gent M, Blakely JA, Easton JD, Ellis DJ, Hachinski VC, Harbison JW, Panak E, Roberts RS, Sicurella J, Turpie AG: The Canadian American Ticlopidine Study (CATS) in the thromboembolic stroke. Lancet. 1989, 1: 1215-1220. 10.1016/S0140-6736(89)92327-1. Article CAS PubMed Google Scholar Hass WK, Easton JD, Adams HP, Pryse-Phillips W, Molony BA, Anderson S, Kamm B: A randomized trial comparing ticlopidine hydrochloride with aspirin for the prevention of stroke in high-risk patients. Ticlopidine Aspirin Stroke Study Group. N Engl J Med. 1989, 321: 501-507. 10.1056/NEJM198908243210804. Article CAS PubMed Google Scholar SPREAD Stroke Prevention And Educational Awerness Diffusion - V Edizione 2007 - Ictus cerebrale: linee guida italiane di prevenzione e trattamento, 319. Paradiso-Hardy F, Angelo MC, Lanctot KL, Cohen EA: Hematologic dyscrasia associated with ticlopidine therapy: evidence for casuality. CMAJ. 2000, 163: 1441-1448. CAS PubMed PubMed Central Google Scholar Grieco A, Vecchio FM, Greco AV, Gasbarrini G: Cholestatic hepatitis due to ticlopidine: clinical and histological recovery after drug withdrawal. Case report and review of the literature. Eur J Gastroenterol Hepatol. 1998, 10: 713-715. CAS PubMed Google Scholar Alberti L, Alberti-Flor JJ: Ticlopidine-induced cholestatic hepatitis successfully treated with corticosteroids (letter). Am J Gastroenterol. 2002, 97: 107-10.1111/j.1572-0241.2002.05642.x. Article Google Scholar Kubin Cj, Shermann O, Hussain KB, Feinman L: Delayed onset ticlopidine induced cholestatic jaundice. Pharmacotherapy. 1999, 18: 1006-1010. 10.1592/phco.19.11.1006.31567. Article Google Scholar Celyan C, Kirimli O, Akarsu M, Under B, Guneri S: Early ticlopidine-induced hepatic dysfunction, dermatitis and irreversible aplastic anemia after coronary artery stenting. Am J Hematol. 1998, 59: 260-264. Google Scholar Download references Acknowledgements We wish to thank Dr Giuseppina Frigo who helped with data collection. Author information Authors and Affiliations University of Milan, Department of Medicine, Surgery and Dentistry, Rehabilitation Unit, San Paolo Hospital, Milan, Italy Antonino M Previtera Rehabilitation Unit, San Paolo Hospital, Milan, Italy Rossella Pagani Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Corresponding author Correspondence to Antonino M Previtera. Additional information Competing interests The authors declare that they have no competing interests. Authors' contributions AMP designed the study, treated our patient, drafted the manuscript, and contributed to the data collection. RP helped to design the study, treated our patient, contributed to manuscript drafts, and contributed to the data collection. All authors have read and approved the final version of the manuscript. Rights and permissions This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Reprints and permissions About this article Cite this article Previtera, A.M., Pagani, R. Agranulocytosis and hepatic toxicity with ticlopidine therapy: a case report. J Med Case Reports 4, 269 (2010). Download citation Received: 26 January 2010 Accepted: 12 August 2010 Published: 12 August 2010 DOI: Share this article Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. Provided by the Springer Nature SharedIt content-sharing initiative Keywords Advertisement Journal of Medical Case Reports Journal of Medical Case Reports ISSN: 1752-1947 Contact us Read more on our blogs Receive BMC newsletters Manage article alerts Language editing for authors Scientific editing for authors Policies Accessibility Press center Support and Contact Leave feedback Careers Follow BMC BMC Twitter page BMC Facebook page BMC Weibo page By using this website, you agree to our Terms and Conditions, Your US state privacy rights, Privacy statement and Cookies policy. Your privacy choices/Manage cookies we use in the preference centre. 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https://study.com/skill/learn/how-to-graph-an-exponential-function-and-its-asymptote-in-the-form-f-x-b-x-or-f-x-bax-explanation.html
How to Graph an Exponential Function and its Asymptote in the Form F(x)=(B)^-X or F(x)=-A(B)^X | Precalculus | Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up How to Graph an Exponential Function and its Asymptote in the Form F(x)=(B)^-X or F(x)=-A(B)^X Precalculus Skills Practice Click for sound 5:29 You must c C reate an account to continue watching Register to access this and thousands of other videos Are you a student or a teacher? I am a student I am a teacher Try Study.com, risk-free As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it risk-free It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time. Already registered? Log in here for access Back What teachers are saying about Study.com Try it risk-free for 30 days Already registered? Log in here for access 1:04 Graph y=-3(1/2)^x - Ex1 3:10 Graph y=(4)^-x - Ex2 Jump to a specific example Speed Normal 0.5x Normal 1.25x 1.5x 1.75x 2x Speed Kayla Zeliff, Lynn Ellis Instructors Kayla Zeliff Kayla has a Bachelor’s in Mathematics and a Master’s in Mechanical Engineering. As a mathematician, her work has focused on mission assurance and machine learning View bio Lynn Ellis Lynn Ellis has taught mathematics to high school and community college students for over 13 years. She has a Bachelor's degree in Mathematics from Middlebury College and a Master's Degree in Education from the University of Phoenix. View bio Example SolutionsPractice Questions Steps for How to Graph an Exponential Function and its Asymptote in the Form f(x)=(b)−x or f(x)=−a(b)x Step 1: Determine the y-intercept and horizontal asymptote of the function. If the function is of the form f(x)=(b)−x, then the y-intercept is (0,1) as this function can also be written as f(x)=1(b)−x. If the function is of the form f(x)=−a(b)x, then the y-intercept is −a. The horizontal asymptote for both forms is y=0. Step 2: Determine whether the given equation is reflected over the x- or y-axis from the parent function y=(b)x. If the function is of the form y=(b)−x, then the graph is reflected across the y-axis from the parent function y=a(b)x. If the function is of the form y=−a(b)x, the then graph is reflected across the x-axis from the parent function y=a(b)x. Step 3: Using the location of the horizontal asymptote and the identified reflection, select the graph that most accurately depicts the given function. Equations and Definitions for How to Graph an Exponential Function and its Asymptote in the Form f(x)=(b)−x or f(x)=−a(b)x Horizontal Asymptote: A horizontal asymptote is a horizontal line, y=c, in which the graph of a function approaches, but never touches. Exponential Function: An exponential function is a function in which the variable x is the exponent. The parent function is expressed as f(x)=a(b)x, where a is the y-intercept and b describes the rate of change of the function. Reflection: A reflection is a mirroring of a graph. Most commonly, functions are reflected over the x- or y-axis. Let's practice identifying the graph of exponential functions by working through two examples. Example Problem 1 - How to Graph an Exponential Function and its Asymptote in the Form f(x)=(b)−x Select the graph of the exponential function f(x)=3−x. A. B. C. D. Step 1: We must first identify the locations of the y-intercept and horizontal asymptote of the function. The function can be written as f(x)=1(3)−x so the y-intercept will occur at (0,1). The graph of the function y=b x has a horizontal asymptote at y=0. Since our function is of the form y=b−x, the horizontal asymptote is still the line y=0. Step 2: Next, we determine whether the graph is reflected over the x-axis or y-axis. Since the negative is in the exponent, we know that the reflection is over the y-axis. This means that when when x<0 the graph will fall very steeply toward x=0 and begin to level out as the value of x increases. Step 3: In Step 1 we determined the location of the y-intercept and the equation of the horizontal asymptote. Since all of the graphs have a horizontal asymptote of y=0, this criteria does not help to determine the correct graph. However, we noted in Step 2 that the graph will fall steeply from left to right and begin to approach the horizontal asymptote as the values of x increase. From the possible options, only C has the correct behavior. Therefore, the correct graph of the function is C. Example Problem 2 - How to Graph an Exponential Function and its Asymptote in the Form f(x)=−a(b)x Select the graph of the exponential function f(x)=−2(2)x. A. B. C. D. Step 1: The y-intercept of the function is (0,−2). The horizontal asymptote of the function is y=0. Step 2: Since the function y=2 x is multiplied by the coefficient −2, we know that the graph is reflected over the x-axis. This means that as the value of x increases, the value of y decreases. Step 3: Of the given options, the only graph that is reflected over the x-axis and has a horizontal asymptote of y=0 is A. Therefore, the correct graph is A. Get access to thousands of practice questions and explanations! Create an account Table of Contents Steps for How to Graph an Exponential Function and its Asymptote in the Form f(x)=(b)−x or f(x)=−a(b)x Equations and Definitions for How to Graph an Exponential Function and its Asymptote in the Form f(x)=(b)−x or f(x)=−a(b)x Example Problem 1 - How to Graph an Exponential Function and its Asymptote in the Form f(x)=(b)−x Example Problem 2 - How to Graph an Exponential Function and its Asymptote in the Form f(x)=−a(b)x Test your current knowledge Practice Graphing an Exponential Function and its Asymptote in the Form F(x)=B-X or F(x)=-BAX Recently updated on Study.com Videos Courses Lessons Articles Quizzes Concepts Teacher Resources The First & Second Balkan Wars | Background & Consequences Ethnic Groups in Indonesia | Demographics & People Gods of the Winter Solstice Holes by Louis Sachar | Themes, Quotes & Analysis Aurangzeb | Empire, Achievements & Failures Libya Ethnic Groups | Demographics, Population & Cultures Cold War Lesson for Kids: Facts & Timeline The Chronicles of Narnia Series by C.S. 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Continue Can't find your university or library? See options to log in as a guest Log In Options Page Sequence /518 View Page Format Plain Text Image View Mode Scroll Flip Thumbnails Grinstead and Snell's Introduction to Probability About This Item Grinstead and Snell's Introduction to Probability 518 page scans Catalog Record Text-Only View Rights Creative Commons Attribution-NoDerivatives. Download Format Ebook (PDF) Ebook (EPUB) Text (.txt) Text (.zip) Image (JPEG) Image (TIFF) Range Current page scan (#1) Whole item Selected page scans Download Download Help Building your PDF Close modal Close modal Please wait while we build your PDF. What affects the download speed? CancelDownload Search in This Text Search in this text [x] Open results in a new tab Jump to Section Jump to page... Title Page(p. iii,scan #1) Table of Contents(p. v,scan #3) Preface(p. vii,scan #5) Section 1(p. 1,scan #9) Section 2(p. 41,scan #49) Section 3(p. 75,scan #83) Section 4(p. 133,scan #141) Section 5(p. 183,scan #191) Section 6(p. 225,scan #233) Section 7(p. 285,scan #293) Section 8(p. 305,scan #313) Section 9(p. 325,scan #333) Section 10(p. 365,scan #373) Section 11(p. 405,scan #413) Section 12(p. 471,scan #479) Index(p. 503,scan #511) Jump to page scan Close modal Close modal Jump to a page scan by page number orpage scan sequence. Page number or sequence Hints follow. Hints Page numbers are entered as number, e.g. 10 Page scan sequences are entered as #number, e.g. #10 Use a page scan sequence between #1-#518 Use a page number between iii-510 Use + to jump ahead by a number of pages, e.g. +10 Use - to jump back by a number of pages, e.g. -10 Cancel Jump Collections Log in to make your personal collections permanent. In these featured collections: University Press of Florida Add this item to a collection: Add Shared Collections New Collection Close modal Close modal Collection Name (required) Collection names can be 100 characters long (100 characters remaining). Description Descriptions can be 255 characters long (255 characters remaining). Contributor Name Optional. Set a public contributor name (255 characters remaining). Is this collection visible to others? Private This collection will be temporary. Log in to create permanent and public collections. Close Save Changes Share Permanent link to this item Link to this page scan Embed this item Embed this item Close modal Close modal Copy the code below and paste it into the HTML of any website or blog. Code Block Scroll View Flip View More information Version 2013-11-06 14:13 UTC 1 (p.iii) Grinstead and Snell's Introduction to Probability The CHANCE Project1 Version dated 4 July 2006 'Copyright (C) 2006 Peter G. Doyle. This work is a version of Grinstead and Snell's 'Introduction to Probability, 2nd edition', published by the American Mathematical So- ciety, Copyright (C) 2003 Charles M. Grinstead and J. Laurie Snell. This work is freely redistributable under the terms of the GNU Free Documentation License. 2 (p.iv) To our wives and in memory of Reese T. Prosser 3 (p.v) Contents Preface 1 Discrete Probability Distributions 1.1 Simulation of Discrete Probabilities . . . . . . . . . . . . . . . . . . . 1.2 Discrete Probability Distributions . . . . . . . . . . . . . . . . . . . . 2 Continuous Probability Densities 2.1 Simulation of Continuous Probabilities . . . . . . . . . . . . . . . . . 2.2 Continuous Density Functions . . . . . . . . . . . . . . . . . . . . . . vii 1 1 18 41 41 55 75 75 92 120 3 Combinatorics 3.1 Permutations . . . . . . . . . . . . . . . . . . . . 3.2 Combinations . . . . . . . . . . . . . . . . . . . . 3.3 Card Shuffling . . . . . . . . . . . . . . . . . . . . 4 Conditional Probability 4.1 Discrete Conditional Probability . . . . . . . . . 4.2 Continuous Conditional Probability . . . . . . . . 4.3 Paradoxes . . . . . . . . . . . . . . . . . . . . . . 5 Distributions and Densities 5.1 Important Distributions . . . . . . . . . . . . . . 5.2 Important Densities . . . . . . . . . . . . . . . . 6 Expected Value and Variance 6.1 Expected Value . . . . . . . . . . . . . . . . . . . 6.2 Variance of Discrete Random Variables . . . . . . 6.3 Continuous Random Variables . . . . . . . . . . . 7 Sums of Random Variables 7.1 Sums of Discrete Random Variables . . . . . . . 7.2 Sums of Continuous Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 . . . . . . . . . . . 133 . . . . . . . . . . . 162 . . . . . . . . . . . 175 183 . . . . . . . . . . . 183 . . . . . . . . . . . 205 225 . . . . . . . . . . . 225 . . . . . . . . . . . 257 . . . . . . . . . . . 268 285 285 291 8 Law of Large Numbers 8.1 Discrete Random Variables . . . . . . . . . . . 8.2 Continuous Random Variables . . . . . . . . . . 305 ................ 305 ................ 316 V 4 (p.vi) vi CONTENTS 9 Central Limit Theorem 325 9.1 Bernoulli Trials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 9.2 Discrete Independent Trials . . . . . . . . . . . . . . . . . . . . . . . 340 9.3 Continuous Independent Trials . . . . . . . . . . . . . . . . . . . . . 356 10 Generating Functions 365 10.1 Discrete Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . 365 10.2 Branching Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . 376 10.3 Continuous Densities . . . . . . . . . . . . . . . . . . . . . . . . . . . 393 11 Markov Chains 405 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405 11.2 Absorbing Markov Chains . . . . . . . . . . . . . . . . . . . . . . . . 416 11.3 Ergodic Markov Chains . . . . . . . . . . . . . . . . . . . . . . . . . 433 11.4 Fundamental Limit Theorem . . . . . . . . . . . . . . . . . . . . . . 447 11.5 Mean First Passage Time . . . . . . . . . . . . . . . . . . . . . . . . 452 12 Random Walks 471 12.1 Random Walks in Euclidean Space . . . . . . . . . . . . . . . . . . . 471 12.2 Gambler's Ruin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486 12.3 Arc Sine Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493 Appendices 499 Index 503 5 (p.vii) 6 (p.viii) 7 (p.ix) 8 (p.x) 9 (p.1) 10 (p.2) 11 (p.3) 12 (p.4) 13 (p.5) 14 (p.6) 15 (p.7) 16 (p.8) 17 (p.9) 18 (p.10) 19 (p.11) 20 (p.12) 21 (p.13) 22 (p.14) 23 (p.15) 24 (p.16) 25 (p.17) 26 (p.18) 27 (p.19) 28 (p.20) 29 (p.21) 30 (p.22) 31 (p.23) 32 (p.24) 33 (p.25) 34 (p.26) 35 (p.27) 36 (p.28) 37 (p.29) 38 (p.30) 39 (p.31) 40 (p.32) 41 (p.33) 42 (p.34) 43 (p.35) 44 (p.36) 45 (p.37) 46 (p.38) 47 (p.39) 48 (p.40) 49 (p.41) 50 (p.42) 51 (p.43) 52 (p.44) 53 (p.45) 54 (p.46) 55 (p.47) 56 (p.48) 57 (p.49) 58 (p.50) 59 (p.51) 60 (p.52) 61 (p.53) 62 (p.54) 63 (p.55) 64 (p.56) 65 (p.57) 66 (p.58) 67 (p.59) 68 (p.60) 69 (p.61) 70 (p.62) 71 (p.63) 72 (p.64) 73 (p.65) 74 (p.66) 75 (p.67) 76 (p.68) 77 (p.69) 78 (p.70) 79 (p.71) 80 (p.72) 81 (p.73) 82 (p.74) 83 (p.75) 84 (p.76) 85 (p.77) 86 (p.78) 87 (p.79) 88 (p.80) 89 (p.81) 90 (p.82) 91 (p.83) 92 (p.84) 93 (p.85) 94 (p.86) 95 (p.87) 96 (p.88) 97 (p.89) 98 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DISEÑO DE EJES DE TRANSMISIÓN DE POTENCIA | DISEÑO MECÁNICO | MOTT | EJERCICIO 12.5 dcahue-ingeniería 24600 subscribers 148 likes Description 11223 views Posted: 5 Dec 2021 📕📗 Tema: DISEÑO DE EJES DE TRANSMISIÓN DE POTENCIA. 🔴 EJERCICIO 12.5. PROBLEMA Vea la figura P12-5. El eje gira a 240 rpm y sostiene un engrane recto D con 48 dientes y paso diametral 6. Los dientes tienen perfil de involuta de 20°, a profundidad completa. El engrane recibe 15 HP del piñón Q, cuya ubicación se indica. Calcule el par torsional entregado al eje y las fuerzas tangencial y radial que ejerce el engrane sobre el eje. Descomponga esas fuerzas en sus componentes horizontal y vertical, y determine las fuerzas netas que actúan sobre el eje en D, en direcciones horizontal y vertical. SOLUCIÓN ⏱️📍 00:00 INTRODUCCIÓN. ⏱️📍 05:00 CÁLCULO DEL TORQUE DE ENTRADA. ⏱️📍 07:00 FUERZAS EN EL ENGRANE D. ⏱️📍 11:30 FUERZAS EN LA POLEA E. ✅ SUSCRÍBETE 💬 BLOG: 🔴 YOUTUBE: 👥 Facebook: 📸 Instagram: 📕📗 📘 REFERENCIAS:📕📗 📘 -Norton, R. (2009), -Diseño de Maquinas, -Editorial Pearson. México,. -4th ed. Ingeniería mecánica, mecatrónica. DiseñoMecanico #MecanicaDeMateriales Transcript: en el problema que se presenta a continuación nos están solicitando determinar las fuerzas que están actuando en el eje así como los torques en este caso tenemos que determinar las fuerzas verticales y las fuerzas horizontales que están actuando sobre este nos están indicando la velocidad angular del eje y el ángulo de presión del mismo además también nos están indicando la potencia que está saliendo a través de dicho eje podemos apreciar que la potencia está entrando a partir del engrane de y que nosotros vamos a tener algunas otras poleas y catarinas entonces lo que vamos a tener es lo siguiente para comenzar vamos a determinar el torque que se está llevando a cabo en el engrane de de esta manera pues nosotros vamos a poder determinar las fuerzas que en este estarían actuando como quedaría dicho torque muy bien primero es muy importante determinar o más bien analizar la posición que tiene de la línea de centros de los engranes ya que hay que recordar que vamos a tener derivado de este contacto una fuerza normal y una fuerza tangencial entonces como quedarían estas fuerzas como pueden apreciar a nosotros ya nos están indicando el sentido de giro que va a tener el engrane entonces nosotros lo que vamos a hacer es poner el sentido de giro del otro en grande que va a venir en esta dirección ahora como ya sabemos en qué dirección van hay que prestar mucha atención a lo siguiente ingés resulta de que nosotros hasta ahorita sabemos que el engrane que está impulsando al engrane de eso quiere decir que la fuerza que va a entrar al en grande tiene que venir en sentido del engrane q qué quiere decir esto que para que este engrane d se pueda mover en sentido horario necesitamos aplicarle una fuerza que venga más o menos así miren esta fuerza va a venir así y le vamos a llamar fuerza tangencial observen la fuerza tangencial viene de esta manera ingenieros imagínense que aquí nosotros tenemos un diente a la hora de que nosotros aplicamos esta fuerza tangencial al diente pues vamos a hacer girar al engrane de y esto se debe atención a que el engrane que está impulsando al engrane de y que otra cosa sabemos que como q impulsa a d pues va a haber una fuerza que se va a dirigir hacia el engrane de porque lo va a empujar así es aquí lo están impulsando pero va a haber una fuerza que va a ir hacia el centro o va a venir mejor dicho hacia el engrane de que es el que está empujando y esta fuerza que nosotros tenemos que empujar en grande le vamos a llamar nosotros fuerza radial listo hasta que ustedes pueden apreciar como ya están nuestras fuerzas pero para que quede más claro los voy a pasar a un diagrama en esta parte aquí nosotros vamos a poner al engrane impulsor que es el engrane de y vamos a dibujar en esta otra parte al engrane impulsado que sería nuestro engrane de y este es el engrane que vamos a ponerle sus nombres aquí este es el engrane de y este va a ser el engrane q entonces cómo les había mencionado pues los giros como vienen este va a girar más o menos en este sentido así como se aprecia y éste como girar y ah pues éste va a girar ingenieros en el sentido así está en sentido antihorario entonces las fuerzas que nosotros representábamos en la parte de abajo son así aquí nosotros vamos a tener la fuerza tangencial vamos a llamarle ft y vamos a tener una fuerza en la dirección del en grande porque está empujando al en grande y esta fuerza es lo que nosotros vamos a llamar la fuerza radial ahora sí recuerden que el la línea de centros de estos 12 engranes está a 30 grados con respecto a la horizontal así que bien ahorita solamente nos vamos a concentrar en los engranes posteriormente estaremos sacando las cargas para las bandas o en este caso para la banda y para la catarina así que vamos a encontrar estas fuerzas y qué es lo que tenemos pues bien como ustedes pueden apreciar nosotros tenemos la velocidad angular y la potencia que se está suministrando a estos engranes oa este en grande eso quiere decir que nosotros a través de la ecuación de potencia vamos a decir que la potencia es igual al torque por la velocidad angular pero en este caso nosotros lo que necesitamos es el torque porque ya tenemos la potencia que es de 20 15 hp y tenemos la velocidad angular entonces nosotros vamos a tener que el torque es igual a la potencia entre la velocidad angular pero presten atención a lo siguiente porque como saben la potencia está en hp las revoluciones o más bien la velocidad está en revoluciones por minuto pues vamos a aplicar la siguiente ecuación ya de una vez con el factor de conversión para obtener como resultado libra por pulgada aplicando este factor de conversión vamos a tener entonces 63 mil que va a multiplicar a la potencia en hp y vamos a tener que son 15 hp eso y esto nosotros lo vamos a dividir entre la velocidad angular y esta velocidad angular ingenieros pues nosotros la vamos a poner como omega y tiene un valor de 240 rpm y el torque que nosotros vamos a tener en ese engrane que le vamos a llamar torrente es de 3000 900 37 punto 5 y las unidades van a ser libras por pulgada y listo hasta aquí yo ya tengo el primer torque que es el torque que se está aplicando en d una vez que nosotros ya tenemos el torque podemos proceder a calcular las fuerzas y vamos a comenzar con la fuerza tangencial esa fuerza tangencial hay que recordar que la podríamos obtener a través de la siguiente ecuación nosotros sabemos que el torque sobre todo si estamos trabajando con magnitudes es igual a la fuerza por una distancia perpendicular a esa fuerza sin embargo observen como mi fuerza viene ft viene direccionada más o menos en esta dirección y qué pasa que nosotros vamos a tener una distancia perpendicular a esa fuerza esa distancia es la que viene desde el centro hasta el punto de aplicación de la fuerza y vemos que sería el radio de el engrane de y este radio tiene un valor de 8 o más bien el diámetro de este engrane de acuerdo a lo que nos solicita el problema tiene un valor de 8 pulgadas eso quiere decir que el radio que nosotros vamos a tener es de 4 pulgadas de este modo es que nosotros vamos a encontrar la fuerza y sería la fuerza tangencial entonces esta fuerza tangencial a que sería igual la vamos a despejar de esta ecuación y vamos a tener que es igual al torque el torque como ya lo dijimos es de 3000 930 y 7.5 libras por pulgada y esto lo vamos a dividir nosotros entre el radio y el radio que damos que era de 4 pulgadas vamos a obtener una fuerza tangencial de 984 punto 375 y las unidades son en libras esta es la fuerza tangencial pero mucha atención porque esta fuerza tangencial viene en una dirección más o menos así y qué ángulo lleva nosotros podemos apreciar que está este ángulo de cuánto sería sería de 60 grados y por qué decimos que es de 60 grados si ustedes observan la fuerza tangencial es perpendicular a la línea de centros entonces eso quiere decir que vamos a tener 90 grados entre la línea de centros y la fuerza si yo prolongó la dirección de la fuerza nos vamos a dar cuenta de que hay 90 grados entre la línea de centros y esta línea y como hay 90 grados pues el ángulo complementario sería este de 60 grados y listo por eso es que decimos que la fuerza tangencial va a tener ese ángulo de 60 grados ahora bien vamos a calcular la fuerza radial vamos a ver qué pasa como ustedes recordarán nosotros tenemos un ángulo de contacto de 20 grados eso quiere decir que para calcular la fuerza radial va a ser necesario la fuerza tangencial porque la vamos a multiplicar por la tangente del ángulo de presión en este caso pues vamos a tener que es d 980 y 4.375 libras y esto nosotros lo vamos a multiplicar por la tangente de 20 grados y el valor de esta fuerza radial va a ser de 358 punto 283 y qué dirección llevaría bueno pues va a llevar una dirección desde luego más o menos así y sabemos que el ángulo sería de 30 grados porque esta fuerza radial recuerden que va está empujando al en grande pero actúa en la dirección de la línea de centros entonces sabemos que va a 30 grados hasta aquí nosotros ya contamos con las fuerzas que están actuando en el engrane al final es que las vamos a descomponer mejor vamos a centrarnos en obtener las otras fuerzas que serían la fuerza de la pole en b y sería la fuerza de la catarina entonces lo que vamos a hacer a continuación es descomponer estas fuerzas en la componente x y en la componente i y como nos quedaría vamos a comenzar escribiendo para x y para la componente en x que tendríamos vamos a decir que las fuerzas en este sentido van a ser positivas que tendríamos nosotros bueno le podríamos poner para ahí vamos a tener que es d menos novecientos ochenta y 4.375 vean cómo va esta carga hacia la izquierda y yo dije que hacia la derecha y van a ser positivos por eso es que estoy poniendo el signo negativo y como yo deseo la componente en x de esta fuerza pues la voy a multiplicar por el coseno de 60 grados después voy a tener que es más y vamos a tener que es más la fuerza de 350 y 8.283 y digo que es positivo porque vean como esta fuerza en la componente en x va hacia la derecha y esto lo vamos a multiplicar por el coseno de 30 grados observen como esto aquí nosotros le vamos a llamar fuerzas en x y el valor de esta fuerza en x tiene un es de 100 menos 180 y 1.90 y esto me va a quedar en libras presten atención que este signo negativo lo único que me está indicando es la dirección quiere decir que las fuerzas netas que están actuando en x van hacia la izquierda ahora vamos a hacer algo similar pero para calcular las fuerzas en que íbamos a tener que las fuerzas en que quedan de la siguiente forma vamos a decir además que las cargas hacia arriba son positivas y que tenemos pues vamos a tener novecientos ochenta y 4.375 y podemos apreciar que ambos ambas cargas van hacia arriba vamos a multiplicar entonces por el seno de 60 grados y vamos a tener que es más la otra carga que es de 350 y 8.283 vamos a multiplicar esto por el seno de 30 grados y el valor que vamos a obtener para esta fuerza es de 1000 31.63 libras como podemos apreciar bueno pues nos acaba de dar un valor positivo la fuerza en ye de acuerdo a nuestra convención sabemos que la fuerza en lleva irá hacia arriba la fuerza resultante y la fuerza en x va a venir hacia la izquierda con esto ingenieros damos por terminado todo lo que el problema nos había estado solicitando por lo tanto te invito a suscribirte a este canal y ver cada una de las listas de reproducción que aparecen en el seguramente van a ser de gran utilidad en tu estancia como estudiante de ingeniería así que regalarnos un like comparte los vídeos y te deseamos mucho éxito que la fuerza te acompañe
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https://www.geeksforgeeks.org/engineering-mathematics/discrete-mathematics-applications-of-propositional-logic/
Discrete Mathematics - Applications of Propositional Logic Last Updated : 23 Jul, 2025 Suggest changes 20 Likes A proposition is an assertion, statement, or declarative sentence that can either be true or false but not both. For example, the sentence "Ram went to school." can either be true or false, but the case of both happening is not possible. So we can say, the sentence "Ram went to school." is a proposition. But, the sentence "N is greater than 100" is not a proposition as we cannot state whether it is true or false unless the value of N is given. A few more examples of propositions are "12 - 10 = 3", "The library is open.", etc. The area of mathematical logic that deals with propositions is called Propositional Logic or Propositional Calculus. It is also known as sentential logic or sentential calculus. It studies the logical values of propositions when taken as a whole and logical relationships when connected using logical connectives(such as logical AND, logical OR, etc.). Table of Content About Propositional Logic in Discrete Mathematics Importance of Propositional Logic Applications of Propositional Logic 1) Translating English Sentences into logical statements 2) System Specifications 3) Logical Puzzles 4) Boolean Searches 5) Logic/Computer Circuits 6) Inference and Decision Making 7) Artificial Intelligence - Fuzzy Logic Solved Examples on Propositional Logic in Discrete Mathematics About Propositional Logic in Discrete Mathematics Propositional Logic is a fundamental area of discrete mathematics that deals with propositions, which are declarative statements that can either be true or false but not both. It forms the basis for various areas of logic and reasoning in mathematics, computer science, and related fields. Importance of Propositional Logic Propositional Logic plays an important role in computer science as well as in a person's daily life. The main benefits of studying and using propositional logic are that it prevents us from making inconsistent inferences and incautious decisions. It incorporates reasoning and thinking abilities in one's daily life. Applications of Propositional Logic In the computer science field, propositional logic has a wide variety of applications and hence is very important. It is used in system specifications, circuit designing, logical puzzles, etc. Apart from this, it can also be used in translating English sentences to mathematical statements and vice-versa. Let us look at this vast variety of applications in detail. 1) Translating English Sentences into logical statements Like any other human language, English sentences can be ambiguous. This ambiguity might lead to uninformed decision-making and other fatal errors. To remove this ambiguity, we can translate these English sentences into logical expressions with the help of Propositional Logic. Note that sometimes this may include making a few assumptions based on the sentence's intended meaning. Example: Given a sentence "You can purchase this book if you have $20 or $10 and a discount coupon." Now, this is a bit complex to be understood at once. So we translate this into a logical expression that will make it simple to understand. Let a, b, c, and d representthe sentences "You can purchase this book.", "You have $20.", "You have $10.", and "You have a discount coupon." respectively. Then the given sentence can be translated to (b ∨ (c ∧ d) -> a, which simply means that "if you either have $20 or $10, along with a discount coupon, then you can purchase the book." 2) System Specifications When developing/manufacturing a system(Software or Hardware), the developers/manufacturers have to meet certain needs and specifications, which are usually stated in English. But as English sentences can be ambiguous, developers/engineers translate these system specifications into logical expressions to state specifications rigorously and unambiguously. Example: Let a, b, c, and d represent the sentences "The computer is within the local network.", "The computer has a valid login id.", "The computer is under the use of administrator.", and "Internet is accessible to the computer." So the complex sentence, "If the computer is within the local network or it is not within the local network but has a valid login id or it is under the use of administrator, then the Internet is accessible to the computer." can be expressed as (a ∨ (¬a ∧ b) ∨ c) -> d. 3) Logical Puzzles Puzzles that are solved using reasoning and logic are called logical puzzles. They can be used for brain exercises, recreational purposes, and for testing a person's reasoning capabilities. Solving such puzzles is generally tricky, but it can be done easily using propositional logic. Some of the famous logic puzzles are the muddy children puzzle, Smullyan's puzzles about knights and knaves, etc. Example: Problem Statement: There is an island that has two kinds of inhabitants, knights, who always tell the truth, and their opposites, knaves, who always lie. You encounter two people A and B. Determine what are A and B if A says "B is a knave" and B says "Both of us are of same types"? Solution: Let p and q be the statements that A is a knight and B is a knight, respectively, so ¬p and ¬q are the statements that A is a knave and B is a knave, respectively. Let's assume A is a knight, i.e., p is true. So A is telling the truth, which means ¬q is true. Now as B is a knave whatever it is saying is a lie, i.e., (p ∧ q) ∨ (¬p ∧ ¬q) is false, which simply means if either of them is a knave then the other one is a knight or vice-versa. Now, as per the assumption, this statement is true. So we can conclude that A is a knight and B is a knave. 4) Boolean Searches Another important application of propositional logic is Boolean searches. These searches use techniques from propositional logic. Logical connectives are used extensively in searches of large collections of information, such as indexes of Web Pages. In Boolean searches, the connective AND is used to find records that contain both of the two terms, the connective OR is used to find records that have either one or both of the terms, and the connective NOT, also written as AND NOT, is used when we need to exclude a particular search term. Example: Some web search engines support web page searching that uses Boolean techniques. For instance, if we want to look for web pages about hiking in West Alps, then we can look for pages matching WEST AND ALPS AND HIKING. And if we want web pages about hiking in the Alps, but not in West Alps, then we can search for pages matching the record HIKING AND ALPS AND NOT(WEST AND ALPS). 5) Logic/Computer Circuits Logic gates or circuits are electronic devices that implement Boolean functions, i.e. it does a logic operation on one or more bits of input and gives a bit as an output. They are the basic building blocks of any digital system. The relationship between the input and output is based on a certain propositional logic. Example: Logic gates are named as AND gate, OR gate, NOT gate, etc. based on the names of logical connectives AND, OR, NOT, etc. The output truth values for the given input bits for these gates are the same as that those returned by the logical connectives. 6) Inference and Decision Making Propositional Logic is widely used in the making rules of inference and decision making. These rules of inferences can then be used to build arguments. When several premises are given, it is hard to tell if a given argument is valid. Thus, we use these rules of inference to validate an argument and make a decision. Example: We can prove that the following premises build up a valid argument using the rule of inference. "If today is Tuesday, I have a test in English or Science. If my English Professor is absent, then I will not have a test in English. Today is Tuesday and my English Professor is absent. Therefore I have a test in Science." T: Today is Tuesday E: I have a test in English S: I have a test in Science A: My English Professor is absent Hence, the argument stated is valid. 7) Artificial Intelligence - Fuzzy Logic AI algorithms make use of fuzzy logic. In fuzzy logic, there is no logic for the absolute truth and absolute false value. But in fuzzy logic, there is an intermediate value too present which is partially true and partially false. For Example: The truth value 0.6 can be assigned to the statement "Fred is happy", because Fred is happy slightly more than most of the time, and the truth value 0.5 can be assigned to the statement "Percy is happy" because Percy is happy half of the time. Solved Examples Here are a few solved examples on propositional logic: Q1: Consider these statements, P: It will rain today. Q: I shall go to the party. Solve these propositions with reference to the above statements: (i) (~P V~Q) (ii) (~P ^ Q) (iii) (P V Q) Solution: (i) (~P V~Q) : It will not rain today or I shall not go to the party. (ii) (~P ^ Q) : It will not rain today and I shall go to the party. (iii) (P V Q) : It will rain today or I shall go to the party. Q2: Write the following in symbolic form "If either Jennie eats pie or Minnie eats cake then Sherry will eat pudding" Solution: Let, p: Jennie eats pie q: Minnie eats cake r: Sherry will eat pudding The symbolic form of the above compound statement is (p v q) -> r. Q3: Construct a truth table for (P ->Q) V (~Q ->P) Solution: The output is all true. This is also known as a Tautology. Also Check: Problems on Tautology Q4: Show that statements P <-> Q and (P^Q)V(~P^~Q) are equivalent. We're going to prove this statement with the help of a truth table: Practice Problems on Discrete Mathematics - Applications of Propositional Logic Here are some practice problems for you solve: 1. The meaning of p -> q is "q is necessary condition for p" or "If p then q" or "p only if q". Write the following statement for given values of p and q. i.) p: x3 = 8 q: x = 2 ii.) p: She is a good student. q: She studies hard. 2. Let, p, q, r are propositions : p: Sia cooks well q: Sia wins the cooking competition r: Sia becomes a chef Express the following propositions as English sentences: i) (p -> q) ^ r ii) (~p -> -r) iii) (~p -> ~q) V (q -> r) 3. Show that: i. P -> Q ≣ ~P V Q ii. P <-> Q ≣ (P -> Q) ∧ (Q -> P) iii. P ⊕ Q ≣ (P ∧ ~Q) V (~P ∧ Q) 4. Write down the dual of this expression: ~p ->(q -> r) ≣ q -> (p V r) 5. Construct the truth table of: i. (p -> ~q) V (q -> ~r) ii. (~p <-> q) -> (p->q) 6. Using propositional algebra, solve the following questions: i. (p V q) ∧ p ii. (p ∧ q) v (p ∧ ~q) iii. ~(p v (~p ∧ q)) Next Article Matrices A aijazhera2762 Improve Article Tags : Computer Subject Engineering Mathematics Discrete Mathematics Similar Reads Engineering Mathematics Tutorials Engineering mathematics is a vital component of the engineering discipline, offering the analytical tools and techniques necessary for solving complex problems across various fields. Whether you're designing a bridge, optimizing a manufacturing process, or developing algorithms for computer systems, 3 min read Linear Algebra Matrices Matrices are key concepts in mathematics, widely used in solving equations and problems in fields like physics and computer science. A matrix is simply a grid of numbers, and a determinant is a value calculated from a square matrix.Example: \begin{bmatrix} 6 & 9 \ 5 & -4 \ \end{bmatrix}_{2 3 min readRow Echelon Form Row Echelon Form (REF) of a matrix simplifies solving systems of linear equations, understanding linear transformations, and working with matrix equations. A matrix is in Row Echelon form if it has the following properties:Zero Rows at the Bottom: If there are any rows that are completely filled wit 4 min readEigenvalues and Eigenvectors Eigenvalues and eigenvectors are fundamental concepts in linear algebra, used in various applications such as matrix diagonalization, stability analysis and data analysis (e.g., PCA). They are associated with a square matrix and provide insights into its properties.Eigen value and Eigen vectorTable 10 min readSystem of Linear Equations A system of linear equations is a set of two or more linear equations involving the same variables. Each equation represents a straight line or a plane and the solution to the system is the set of values for the variables that satisfy all equations simultaneously.Here is simple example of system of 5 min readMatrix Diagonalization Matrix diagonalization is the process of reducing a square matrix into its diagonal form using a similarity transformation. This process is useful because diagonal matrices are easier to work with, especially when raising them to integer powers.Not all matrices are diagonalizable. A matrix is diagon 8 min readLU Decomposition LU decomposition or factorization of a matrix is the factorization of a given square matrix into two triangular matrices, one upper triangular matrix and one lower triangular matrix, such that the product of these two matrices gives the original matrix. 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Cauchy's Mean Value theorem is also called the Extended Mean Value Theorem or the Second Mean Value Theorem.According 7 min readTaylor Series A Taylor series represents a function as an infinite sum of terms, calculated from the values of its derivatives at a single point.Taylor series is a powerful mathematical tool used to approximate complex functions with an infinite sum of terms derived from the function's derivatives at a single poi 8 min readInverse functions and composition of functions Inverse Functions - In mathematics a function, a, is said to be an inverse of another, b, if given the output of b a returns the input value given to b. Additionally, this must hold true for every element in the domain co-domain(range) of b. In other words, assuming x and y are constants, if b(x) = 3 min readDefinite Integral | Definition, Formula & How to Calculate A definite integral is an integral that calculates a fixed value for the area under a curve between two specified limits. The resulting value represents the sum of all infinitesimal quantities within these boundaries. i.e. if we integrate any function within a fixed interval it is called a Definite 8 min readApplication of Derivative - Maxima and Minima Derivatives have many applications, like finding rate of change, approximation, maxima/minima and tangent. In this section, we focus on their use in finding maxima and minima.Note: If f(x) is a continuous function, then for every continuous function on a closed interval has a maximum and a minimum v 6 min read Probability & Statistics Mean, Variance and Standard Deviation Mean, Variance and Standard Deviation are fundamental concepts in statistics and engineering mathematics, essential for analyzing and interpreting data. These measures provide insights into data's central tendency, dispersion, and spread, which are crucial for making informed decisions in various en 10 min readConditional Probability Conditional probability defines the probability of an event occurring based on a given condition or prior knowledge of another event. It is the likelihood of an event occurring, given that another event has already occurred. In probability, this is denoted as A given B, expressed as P(A | B), indica 12 min readBayes' Theorem Bayes' Theorem is a mathematical formula used to determine the conditional probability of an event based on prior knowledge and new evidence. It adjusts probabilities when new information comes in and helps make better decisions in uncertain situations.Bayes' Theorem helps us update probabilities ba 13 min readProbability Distribution - Function, Formula, Table A probability distribution is a mathematical function or rule that describes how the probabilities of different outcomes are assigned to the possible values of a random variable. It provides a way of modeling the likelihood of each outcome in a random experiment.While a Frequency Distribution shows 13 min readCovariance and Correlation Covariance and correlation are the two key concepts in Statistics that help us analyze the relationship between two variables. Covariance measures how two variables change together, indicating whether they move in the same or opposite directions. Relationship between Independent and dependent variab 6 min read Practice Questions Last Minute Notes - Engineering Mathematics GATE CSE is a national-level engineering entrance exam in India specifically for Computer Science and Engineering. It's conducted by top Indian institutions like IISc Bangalore and various IITs. In GATE CSE, engineering mathematics is a significant portion of the exam, typically constituting 15% of 15+ min readEngineering Mathematics - GATE CSE Previous Year Questions Solving GATE Previous Year's Questions (PYQs) not only clears the concepts but also helps to gain flexibility, speed, accuracy, and understanding of the level of questions generally asked in the GATE exam, and that eventually helps you to gain good marks in the examination. Previous Year Questions h 4 min read We use cookies to ensure you have the best browsing experience on our website. 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https://www.khanacademy.org/math/differential-calculus/dc-diff-intro/dc-secant-lines/v/simplified-secant-line-arbitrary-difference
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https://fdotwww.blob.core.windows.net/sitefinity/docs/default-source/roadway/drainage/files/drainagedesignguide/chapter3.pdf?sfvrsn=4ea60bff_4
January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-i CHAPTER 3: OPEN CHANNEL OPEN CHANNEL .....................................................................................................3-1 3.1 Open Channel Flow Theory ............................................................................... 3-1 3.1.1 Mass, Energy, and Momentum ..................................................................... 3-1 3.1.1.1 Mass ...........................................................................................................3-1 3.1.1.2 Energy ........................................................................................................3-1 3.1.1.3 Momentum..................................................................................................3-4 3.1.2 Uniform Flow ..................................................................................................3-5 3.1.2.1 Manning’s Equation.................................................................................... 3-5 3.1.3 Critical Flow ..................................................................................................3-10 3.1.3.1 Specific Energy and Critical Depth ..........................................................3-10 3.1.3.2 Critical Velocity .........................................................................................3-14 3.1.3.3 Super-Critical Flow ...................................................................................3-14 3.1.3.4 Sub-Critical Flow ......................................................................................3-14 3.1.3.5 Theoretical Considerations ......................................................................3-15 3.1.4 Non-Uniform Flow ........................................................................................3-15 3.1.4.1 Gradually Varied Flow ..............................................................................3-17 3.1.4.2 Gradually Varied Flow Profile Computation ............................................3-17 3.1.4.3 Rapidly Varied Flow .................................................................................3-24 3.1.5 Channel Bends ............................................................................................3-27 3.2 Open Channel Design .......................................................................................3-29 3.2.1 Types of Open Channels for Highways ......................................................3-29 3.2.2 Roadside Ditches .........................................................................................3-31 3.2.3 Median Ditches ............................................................................................3-35 3.2.4 Interceptor Ditches .......................................................................................3-36 3.2.5 Outfall Ditches ..............................................................................................3-37 3.2.6 Hydrology .....................................................................................................3-37 3.2.6.1 Frequency .................................................................................................3-38 3.2.6.2 Time of Concentration ..............................................................................3-38 3.2.7 Tailwater and Backwater .............................................................................3-38 3.2.8 Side Drains ...................................................................................................3-47 3.2.8.1 Design Analysis Requirements for Side Drains ......................................3-47 3.2.8.2 Material Requirements .............................................................................3-48 3.2.8.3 End Treatment ..........................................................................................3-48 3.3 Channel Linings ................................................................................................3-51 3.3.1 Flexible Linings ............................................................................................3-51 3.3.1.1 Vegetation ................................................................................................3-51 3.3.1.2 Other Flexible Linings ..............................................................................3-52 3.3.2 Rigid Linings.................................................................................................3-54 3.3.2.1 Cast-in-Place Concrete ............................................................................3-55 3.3.2.2 Fabric Formed Revetment .......................................................................3-55 January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-ii 3.3.3 Velocity and Shear Stress Limitations ........................................................3-55 3.3.4 Application Guidance for Some Common Channel Linings .......................3-58 3.3.4.1 Rubble Riprap ..........................................................................................3-58 3.3.4.2 Fabric Formed Revetments .....................................................................3-59 3.3.4.3 Gabions ....................................................................................................3-61 3.3.4.4 Soil Stabilizers ..........................................................................................3-65 3.4 Drainage Connection Permitting and Maintenance Concerns ...................3-67 3.4.1 Drainage Connection Permitting .................................................................3-67 3.4.1.1 Roadside Ditch Impacts ...........................................................................3-67 3.4.1.2 Median Ditch Impacts...............................................................................3-69 3.4.1.3 Outfall Ditch Impacts ................................................................................3-71 3.4.2 Maintenance Concerns ................................................................................3-71 3.4.2.1 Ditch Closures ..........................................................................................3-71 3.4.2.2 Acquisition of Ditches from Local Ownership ..........................................3-77 3.4.2.3 Addition of Sidewalks to Roadway Projects ............................................3-77 January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-1 3. OPEN CHANNEL 3.1 OPEN CHANNEL FLOW THEORY 3.1.1 Mass, Energy, and Momentum The three basic principles that generally apply to flow analysis, including open channel flow evaluations, are: • Conservation of mass • Conservation of energy • Conservation of linear momentum 3.1.1.1 Mass You can mathematically express the conservation of mass for continuous steady flow in the Continuity Equation as: AvQ ×= (3.1-1) where: Q = Discharge, in cubic feet per second A = Cross-sectional area, in square feet v = Average channel velocity, in feet per second For continuous unsteady flow, the Continuity Equation must include time as a variable. For additional information on unsteady flow, see Chow (1959) or Henderson (1966). 3.1.1.2 Energy The total energy head at a point in an open channel is the sum of the potential and kinetic energy of the flowing water. The potential energy is represented by the elevation of the water surface. The water surface elevation is the depth of flow, d, defined in Section 1.4, added to the elevation of the channel bottom, z. The water surface elevation is a measure of the potential work that the flow can do as it transitions to a lower elevation. The kinetic energy is the energy of motion as measured by the velocity, v. If a straight tube is inserted down into the flow, the water level in the tube will rise to the water surface elevation in the channel. If a tube with a 90-degree elbow is inserted into the flow with the open end pointing into the flow, then the water level will rise to a level higher than the water surface elevation in the channel—this distance is a measure of the ability of the water velocity to do work. Using Newton’s Laws of Motion, this distance is January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-2 v 2/2g, where g is the acceleration due to gravity. Therefore, the total energy head at a point in an open channel is: d + z + v 2/2g . As water flows down a channel, the flow loses energy because of friction and turbulence. The total energy head between two points in a channel reach may be set equal to one another if the losses between the sections are added to the downstream total energy head. This equality is commonly known as the Energy Equation, which is expressed as: loss hzgvdzgvd +++=++ 22221211 22 (3.1-2) where: d1, d2 = Depth of open channel flow at channel sections 1 and 2, respectively, in feet v 1, v 2 = Average channel velocities at channel sections 1 and 2, respectively, in feet per second z 1, z 2 = Channel elevations above an arbitrary datum at channel sections 1 and 2, respectively, in feet hloss = Head or energy loss between channel sections 1 and 2, in feet g = Acceleration due to gravity, 32.174 ft/sec 2 A longitudinal profile of total energy head elevations is called the energy grade line (gradient). The longitudinal profile of water surface elevations is called the hydraulic grade line (gradient). The energy and hydraulic grade lines for uniform open channel flow are illustrated in Figure 3.1-1. For flow to occur in an open channel, the energy grade line must have a negative slope in the direction of flow. A gradual decrease in the energy grade line for a given length of channel represents the loss of energy caused by friction. When considered together, the hydraulic and energy grade lines reflect not only the loss of energy by friction, but also the conversion between potential and kinetic forms of energy. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-3 Figure 3.1-1: Characteristics of Uniform Open Channel Flow Figure 3.1-2: Definition Sketch for Specific Head and Sub-Critical and Super-Critical Flow January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-4 For uniform flow conditions, the energy grade line is parallel to the hydraulic grade line, which is parallel to the channel bottom (see Figure 3.1-1). Thus, for uniform flow, the slope of the channel bottom becomes an adequate basis for the determination of friction losses. During uniform flow, no conversions occur between kinetic and potential forms of energy. If the flow is accelerating, the hydraulic grade line would be steeper than the energy grade line, while decelerating flow would produce an energy grade line steeper than the hydraulic grade line. The Energy Equation presented in Equation 3.1-2 ignores the effect of a non-uniform velocity distribution on the computed velocity head. The actual distribution of velocities over a channel section are non-uniform (i.e., slow along the bottom and faster in the middle). The velocity head for actual flow conditions generally is greater than the value computed using the average channel velocity. Find guidance on kinetic energy coefficients that account for non-uniform velocity conditions in Chapter 5 (Bridge Hydraulics). For typical prismatic channels with a fairly straight alignment, the effect of disregarding the existence of a non-uniform velocity distribution is negligible, especially when compared to other uncertainties involved in such calculations. Therefore, Equation 3.1-2 is appropriate for most open channel problems. However, if velocity distributions are non-typical, obtain additional information related to velocity coefficients, as presented by Chow (1959) or Henderson (1966). Equation 3.1-2 also assumes that the hydrostatic law of pressure distribution is applicable. This law states that the distribution of pressure over the channel cross section is the same as the distribution of hydrostatic pressure; that is, that the distribution is linear with depth. The assumption of a hydrostatic pressure distribution for flowing water is valid only if the flow is not accelerating or decelerating in the plane of the cross section. Thus, restrict the use of Equation 3.1-2 to conditions of uniform or gradually varied non-uniform flow. If the flow will be varying rapidly, obtain additional information, as presented by Chow (1959) or Henderson (1966). 3.1.1.3 Momentum According to Newton's Second Law of Motion, the change of momentum per unit of time is equal to all the resultant external forces applied to the moving body. Applying this principle to open channel flow produces a relationship that is virtually the same as the Energy Equation expressed in Equation 3.1-2. Theoretically, these principles of energy and momentum are unique, primarily because energy is a scalar quantity (magnitude only), while momentum is a vector quantity (magnitude and direction). In addition, the head loss determined by the Energy Equation measures the internal energy dissipated in a particular channel reach, while the Momentum Equation measures the losses due to external forces exerted on the water by the walls of the channel. However, for uniform flow, since the losses due to external forces and internal energy dissipation are equal, the Momentum and Energy Equations give the same results. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-5 Applying the momentum principle has certain advantages for problems involving substantial changes of internal energy, such as a hydraulic jump. Thus, the momentum principle for evaluating rapidly varied non-uniform flow conditions should be used. Theoretical details of the momentum principle applied to open channel flow are presented by Chow (1959) and Henderson (1966). Section 3.1.4.3 provides a brief presentation of hydraulic jump fundamentals. 3.1.2 Uniform Flow Although steady uniform flow is rare in drainage facilities, it is practical in many cases to assume that steady uniform flow occurs in appropriate segments of an open channel system. The results obtained from calculations based on this assumption will be approximate and general, but still can provide satisfactory solutions for many practical problems. 3.1.2.1 Manning’s Equation Determine the hydraulic capacity of an open channel by applying Manning's Equation, which determines the average velocity when given the depth of flow in a uniform channel cross section. Given the velocity, calculate the capacity (Q) as the product of velocity and cross-sectional area (see Equation 3.1-1). Manning's Equation is an empirical equation with values of constants and exponents derived from experimental data of turbulent flow conditions. According to Manning's Equation, the mean velocity of flow is a function of the channel roughness, the hydraulic radius, and the slope of the energy gradient. As noted previously, for uniform flow, assume that the slope of the energy gradient is equal to the channel bottom slope. Manning's Equation is expressed mathematically as follows: 2132 486 .1 SRnv = (3.1-3) or 2132 486 .1 SAR nQ = (3.1-4) where: v = Average channel velocity, in feet per second Q = Discharge, in cubic feet per second n = Manning's roughness coefficient R = Hydraulic radius of the channel, in feet, calculated: 𝑅𝑅 = 𝐴𝐴 𝑃𝑃 January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-6 P = Wetted perimeter of channel, in feet S = Slope of the energy gradient, in feet per feet A = Cross-sectional area of the open channel, in square feet Values for Manning’s roughness coefficient for artificial channels (i.e., roadside, median, interceptor, and outfall ditches) are listed in Chapter 2 (Section 2.7) of the Drainage Manual . Guidance on methods for estimate Manning’s roughness coefficient for natural channels is found in Chapter 5 (Bridge Hydraulics). Example 3.1-1—Discharge given Normal Depth Given: Depth = 0.6 ft Longitudinal Slope = 0.005 ft/ft Trapezoidal Cross Section shown below Manning’s Roughness = 0.06 Calculate: Discharge, assuming normal depth Not to scale Note: To make things easier, try breaking the drawing into three parts: two triangles and a rectangle. Step 1: Calculate Wetted Perimeter and Cross-Sectional Area Wetted Perimeter (P): Solve for the left triangle’s hypotenuse 22 )6.04(6.0 ×+=x x = 2.474 ft Solve for the right triangle’s hypotenuse 22 )6.06(6.0 ×+=x x = 3.650 ft Wetted Perimeter (P) = 2.474 + 3.650 + 5 = 11.124 ft Cross-Sectional Area (A): Solve for the left triangle’s area )6.0)( 6.04(211 ×=A 5 ft 4 1 6 10.6 ft January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-7 72 .01 =A ft 2 Solve for the right triangle’s area )6.0)( 6.06(212 ×=A 08 .12 =A ft 2 Solve for the rectangle’s area 6.053 ×=A 33 =A ft 2 Cross-Sectional Area (A) = 8.4308 .172 .0 =++ ft 2 Step 2: Calculate Hydraulic Radius Hydraulic Radius (R) = PA Hydraulic Radius (R) = 4315 .0124 .11 8.4 = ft Step 3: Calculate Average Velocity Average Velocity (v) = 2132)()(486 .1 SRn Average Velocity (v) = 00 .1)005 .0()4315 .0(06 .0486 .1 2132 = ft/sec Step 4: Calculate the Discharge Discharge (Q) = Av × Discharge (Q) = 00 .1 ft/sec × 8.4 ft 80 .42 = ft 3 /sec As an alternative approach, Example C.1 of Appendix C solves this example problem using equations from Figure C-4. Example 3.1-1 has a direct solution because the depth is known. The next problem will be more difficult to solve because the discharge will be given and the normal depth must be calculated. The equations cannot be solved directly for depth, so an iterative process is used to solve for normal depth. You also can solve Example 3.1-1 using the charts in Appendix C. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-8 Example 3.1-2—Normal Depth given Discharge Given: Discharge = 9 ft 3 /sec Use the channel cross section shape, slope, and Manning’s roughness coefficient given in Example 3.1-1 Calculate: Normal Depth Note: The solution must use trial and error since you cannot solve the equations implicitly for depth. Perform the first trial in the steps below and the remaining trials will be shown in a table. The initial trial depth (i.e., the first guess) should be greater than the depth given previously in Example 3.1-1 because the discharge is greater. So we will perform our trial with an estimated depth of flow of 0.8 ft. Step 1: Calculate Wetted Perimeter and Cross-Sectional Area Wetted Perimeter (P): Solve for the left triangle’s hypotenuse 22 )8.04(8.0 ×+=x 298 .3=x ft Solve for the right triangle’s hypotenuse 22 )8.06(8.0 ×+=x 866 .4=x ft Wetted Perimeter (P) = 164 .13 5866 .4298 .3 =++ ft Cross-sectional Area (A): Solve for the left triangle’s area )8.0)( 8.04(211 ×=A 28 .11 =A ft 2 Solve for the right triangle’s area )8.0)( 8.06(212 ×=A 92 .12 =A ft 2 5 ft 4 1 6 1January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-9 Solve for the rectangle’s area 8.053 ×=A 43 =A ft 2 Cross-Sectional Area (A) = 2.7492 .128 .1 =++ ft 2 Step 2: Calculate Hydraulic Radius Hydraulic Radius (R) = PA Hydraulic Radius (R) = 547 .164 .13 2.7 = ft Step 3: Calculate Average Velocity Average Velocity (v) = 2132)()(486 .1 SRn Average Velocity (v) = 171 .1)005 .0()547 .0(06 .0486 .1 2132 = ft/sec Step 4: Calculate the Discharge Discharge (Q) = Av × Discharge (Q) = 171 .1 ft/sec × 20 .7 ft 43 .82 = ft 3 /sec The discharge calculated in Step 4 is still less than 9 ft 3/sec, so normal depth is greater than 0.8 feet. Use a slightly higher depth of flow for the next guess. The following table summarizes subsequent trials. The trial-and-error process continues until you achieve the ideal level of accuracy. Depth (ft) Area Perimeter Radius Velocity Discharge 0.8 7.2 13.16469 0.546917 1.171 8.433 0.85 7.8625 13.67499 0.574955 1.211 9.521 0.82 7.462 13.36881 0.558165 1.187 8.859 0.826 7.54138 13.43005 0.56153 1.192 8.989 The normal depth for the given channel and flow rate is 0.83 feet. You should perform intermediate calculations using more significant digits than needed, and then round in the last step to avoid rounding errors. The Drainage Manual recommends that, where the flow depth is greater than 0.7 feet, reduce the roughness value to 0.042. However, the normal depth using n = 0.042 is 0.69 feet. The recommended roughness for flow depths less than 0.7 feet is 0.06. The abrupt change in the recommended roughness values causes this anomaly. If the flow depth is January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-10 the primary concern, then using n = 0.06 will give a conservative answer. However, if the velocity is the primary concern, then using n = 0.042 is conservative. 3.1.3 Critical Flow The energy content of flowing water with respect to the channel bottom often is referred to as the specific energy head, which is expressed by the equation: gvdE 22 += (3.1-5) where: E = Specific energy head, in feet d = Depth of open channel flow, in feet v = Average channel velocity, in feet per second g = Acceleration due to gravity, 32.174 ft/sec 2 Considering the relative values of potential energy (depth) and kinetic energy (velocity head) in an open channel can help you with the hydraulic analysis of open channel flow problems. Usually, you will perform these analyses using a curve that shows the relationship between the specific energy head and the depth of flow for a given discharge in a given channel that you can place on various slopes. Generally, you will use the curve representing specific energy head for an open channel to identify regions of super-critical and sub-critical flow conditions. This information usually is necessary to properly perform hydraulic capacity calculations and evaluate the suitability of channel linings and flow transition sections. 3.1.3.1 Specific Energy and Critical Depth Figure 3.1-2 (Part B) illustrates a typical curve representing the specific energy head of an open channel. The straight diagonal line on this figure represents points where the depth of flow and specific energy head are equal. At these points, the kinetic energy is zero; therefore, this diagonal line is a plot of the potential energy, or energy due to depth. The ordinate interval between the diagonal line of potential energy and the specific energy curve for the ideal discharge is the velocity head, or kinetic energy, for the depth in question. The lowest point on the specific energy curve represents flow with the minimum content of energy. The depth of flow at this point is known as the critical depth. Express the general equation for determining the critical depth as: TAgQ 32 = (3.1-6) where: January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-11 Q = Discharge, in cubic feet per second g = Acceleration due to gravity, 32.174 ft/sec 2 T = Top width of water surface, in feet A = Cross-sectional area, in square feet You can calculate critical depth for a given channel through trial and error by using Equation 3.1-6. Chow (1959) presents a procedure for the analysis of critical flow that uses the Critical Flow Section Factor (Z), defined as the ratio of the cross-sectional area and the square root of the hydraulic depth, expressed mathematically as: TAADAZ == (3.1-7) where: Z = Critical flow section factor A = Cross-sectional area of the flow perpendicular to the direction of flow, in square feet D = Hydraulic depth, in feet T = Top width of the channel, in feet Using the definition of the critical section factor and a velocity distribution coefficient of one, the equation for critical flow conditions is: gQZ = (3.1-8) where: Z = Critical flow section factor Q = Discharge, in cubic feet per second g = Acceleration due to gravity, 32.174 ft/sec 2 When you know the discharge, Equation 3.1-8 gives the critical section factor and, thus, by substitution into Equation 3.1-6, the critical depth. Conversely, when you know the critical section factor, you can calculate the discharge with Equation 3.1-8. It is important to note that the determination of critical depth is independent of the channel slope and roughness, since critical depth simply represents a depth for which the specific energy head is at a minimum. According to Equation 3.1-6, the magnitude of critical depth depends only on the discharge and the shape of the channel. Thus, for any given size and shape of channel, there is only one critical depth for the given discharge, which is independent of the channel slope or roughness. However, if Z is not a single-valued January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-12 function of depth, it is possible to have more than one critical depth. For a given value of specific energy, the critical depth results in the greatest discharge, or conversely, for a given discharge, the specific energy is a minimum for the critical depth. Example 3.1-3—Critical Depth given Discharge Given: Discharge = 9 ft 3 /sec Cross Section and Roughness from Example 3.1-1 Calculate: Critical Depth Note: The solution must use trial and error since you cannot implicitly solve the equations for depth. You can perform the first trial as shown in the steps below, with the remaining trials shown in a table. Typically, the slope of a roadside ditch channel must exceed 2 percent to have a normal depth that is super-critical. Since the slope in Example 3.1-1 and Example 3.1-2 is 0.5 percent, the critical depth is probably much less than the normal depth of 0.83 feet calculated in Example 3.1-2 for 9 cfs. So, we will perform our trial with an estimated depth of flow of 0.4 ft. Step 1: Calculate Cross-Sectional Area Cross-Sectional Area (A): Solve for the left triangle’s area )4.0)( 4.04(211 ×=A 32 .01 =A ft 2 Solve for the right triangle’s area )4.0)( 4.06(212 ×=A 48 .02 =A ft 2 Solve for the rectangle’s area 4.053 ×=A 23 =A ft 2 Cross-Sectional Area (A) = 8.2248 .032 .0 =++ ft 2 Step 2: Calculate Top Width Top Width (T): 5 ft 4 1 6 1January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-13 Base Length of Left Triangle + Bottom Width + Base Length of Right Triangle 9)4.06(5)4.04( =×++× ft Step 3: Rearrange Equation 3.1-6 to Solve for Discharge TAgQ 32 = gTAQ ×= 32 gTAQ ×= 3 86 .8174 .32 98.2 3 =×=Q ft 3 /sec The discharge calculated in Step 3 is less than 9 ft 3/sec, so critical depth is greater than 0.4 feet. Use a slightly higher depth of flow for the next guess. The following table summarizes subsequent trials. The trial-and-error process continues until you achieve the ideal level of accuracy. Depth (ft) Area (sq. ft.) Top Width Discharge (cfs) 0.4 2.8 9 8.858665864 0.45 3.2625 9.5 10.84467413 0.41 2.8905 9.1 9.2404111 0.404 2.83608 9.04 9.010440628 You also can solve this problem by determining the minimum specific energy, as discussed in the previous section. The following table solves Equation 3.1-5 for depths bracketing the critical depth determined above and shows that the critical depth has the minimum specific energy. Depth (ft) Area (sq. ft.) Perimeter (ft) Velocity (ft/sec) V 2 /2g Specific Energy 0.403 2.827045 9.112965 3.18354 0.1575 0.560501438 0.404 2.83608 9.123171 3.17339 0.1565 0.560499521 0.405 2.845125 9.133377 3.16331 0.15551 0.56050604 Most computer programs that solve water surface profiles for natural channels use the minimum specific energy approach. For more information, refer to Chapter 5 (Bridge Hydraulics). January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-14 3.1.3.2 Critical Velocity The velocity at critical depth is called the critical velocity. An equation for determining the critical velocity in an open channel of any cross section is: mC gd v = (3.1-9) where: v c = Critical velocity, in feet per second g = Acceleration due to gravity, 32.174 ft/sec 2 dm = Mean depth of flow, in feet, calculated from: TAd m = (3.1-10) where: A = Cross-sectional area, in square feet T = Top width of water surface, in feet 3.1.3.3 Super-Critical Flow For conditions of uniform flow, the critical depth, or point of minimum specific energy, occurs when the channel slope equals the critical slope (i.e., the normal depth of flow in the channel is critical depth). When channel slopes are steeper than the critical slope and uniform flow exists, the specific energy head is higher than the critical value due to higher values of the velocity head (kinetic energy). The specific head curve segment to the left of critical depth in Figure 3.1-2 (Part B) illustrates this characteristic of open channel flow, which is known as super-critical flow. Super-critical flow is characterized by relatively shallow depths and high velocities, as shown in Figure 3.1-2 (Part A). If the natural depth of flow in an open channel is super-critical, you can influence the depth of flow at any point in the channel by an upstream control section. The relationship of super-critical flow to the specific energy curve is shown in Figure 3.1-2 (Parts A and B). 3.1.3.4 Sub-Critical Flow When channel slopes are flatter than the critical slope and uniform flow exists, the specific energy head is higher than the critical value due to higher values of the normal depth of flow (potential energy). The specific head curve segment to the right of critical depth in Figure 3.1-2 (Part B) illustrates this characteristic of open channel flow, which is known as sub-critical flow. Sub-critical flow is characterized by relatively large depths with low velocities, as shown in Figure 3.1-2 (Part C). If the natural depth of flow in an open channel is sub-critical, a downstream control section can influence the depth of flow at any point in the channel. The relationship of sub-critical flow to the specific energy curve is shown in Figure 3.1-2 (Parts B and C). January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-15 3.1.3.5 Theoretical Considerations There are several noteworthy points about Figure 3.1-2. First, at depths of flow near the critical depth for any discharge, a minor change in specific energy will cause a much greater change in depth. Second, the velocity head for any discharge in the sub-critical portion of the specific energy curve in Figure 3.1-2 (Parts B and C) is relatively small when compared to specific energy. For this sub-critical portion of the specific energy curve, changes in depth of flow are approximately equal to changes in specific energy. Finally, the velocity head for any discharge in the super-critical portion of the specific energy curve increases rapidly as depth decreases. For this super-critical portion of the specific energy curve, changes in depth are associated with much greater changes in specific energy. 3.1.4 Non-Uniform Flow In locations where changes in the channel section or slope will cause non-uniform flow profiles, you cannot directly solve Manning's Equation since the energy gradient for this situation does not equal the channel slope. Three typical examples of non-uniform flow are illustrated in Figures 3.1-3 through 3.1-5, below. The following sections describe these non-uniform flow profiles and briefly explain how to use the total head line for approximating these water surface profiles in a qualitative manner. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-16 Figure 3.1-3: Non-Uniform Water Surface Profile for Downstream Control Caused by a Flow Restriction Figure 3.1-4: Non-Uniform Water Surface Profile Caused by a Change in Slope Conditions Figure 3.1-5: Non-Uniform Water Surface Profile Caused by a Hydraulic Jump January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-17 3.1.4.1 Gradually Varied Flow Figure 3.1-3 illustrates a channel on a mild slope (sub-critical) discharging into a reservoir or pool. The figure exaggerates the vertical scale for clearer illustration. Cross Section 1 is upstream of the pool, where uniform flow occurs in the channel. Cross Section 2 is at the beginning of a level pool. The depth of flow between Sections 1 and 2 is changing, and the flow is non-uniform. The water surface profile between the sections is known as a backwater curve and is characteristically very long. Figure 3.1-4 illustrates a channel in which the slope changes from sub-critical (mild) to super-critical (steep). The flow profile passes through critical depth near the break in slope (Section 1). This is true whether the upstream slope is mild, as in the sketch, or the water above Section 1 is ponded, as would be the case if Section 1 were the crest of a dam spillway. If, at Section 2, you were to compute the total head, assuming normal depth on the steep slope, it would plot above the elevation of total head at Section 1 (Point “a” in Figure 3.1-4). This is physically impossible, because the total head line must slope downward in the direction of flow. The actual total head line will take the position shown and have a slope approximately equal to S o, the slope of the channel bottom, at Section 1 and approaching S o farther downstream. The drop in the total head line (h loss ) between Sections 1 and 2 represents the loss in energy due to friction. At Section 2, the actual depth (d 2) is greater than normal depth (d n) because sufficient acceleration has not occurred, and the assumption of normal depth at this point would clearly be in error. As you move Section 2 downstream, so that the total head for normal depth drops below the pool elevation above Section 1, the actual depth quickly approaches the normal depth for the steep channel. This type of water surface curve (Section 1 to Section 2) is characteristically much shorter than the backwater curve discussed previously. Another common type of non-uniform flow is the drawdown curve to critical depth that occurs upstream from Section 1 (Figure 3.1-4) where the water surface passes through critical depth. The depth gradually increases upstream from critical depth to normal depth, provided that the channel remains uniform over a sufficient distance. The length of the drawdown curve is much longer than the curve from critical depth to normal depth in the steep channel. 3.1.4.2 Gradually Varied Flow Profile Computation Typically, you can compute water surface profiles using the Energy Equation (Equation 3.1-2). Given the channel geometry, flow, and the depth at one of the cross sections, compute the depth at the other cross section. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-18 The losses between cross sections include friction, expansion, contraction, bend, and other form losses. Expansion, contraction, bend, and other form losses will be neglected in the computations presented in this design guide. Refer to Chapter 5 (Bridge Hydraulics) for more information. Determine the remaining loss—the friction loss—which is express as: LSh ff = (3.1-11) where: hf = Friction head loss, in feet Sf = Slope of the energy grade line, in feet per feet L = Flow length between cross sections, in feet Calculate the slope of the energy grade line at each cross section by rearranging Manning’s Equation (Equation 3.1-4) into the following expression: 232 49 .1 = AR Qn S (3.1-12) For uniform flow, the slope of the channel bed, the slope of the water surface (hydraulic grade line), and the slope of the energy grade line are all equal. For non-uniform flow, including gradually varied flow, each slope is different. Use the slope determined at each cross section to estimate the average slope for the entire flow length between the cross sections. You can use several different averaging schemes to estimate the average slope, and these techniques are discussed in more detail in the Chapter 5 (Bridge Hydraulics). The simplest estimate of slope of the energy gradient between two sections is: 221 SSS f += (3.1-13) where: S 1, S 2 = Slope of the energy gradient at Sections 1 and 2, in feet per feet Computing backwater curves in a quantitative manner can be quite complex. If you require a detailed analysis of backwater curves, consider using computer software for this purpose. Typical computer programs used for water surface profile computations include HEC-RAS by the U.S. Army Corps of Engineers, HEC-2 by the U.S. Army Corps of Engineers (1991), E431 by the USGS (1984), and WSPRO by the USGS (1986). In January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-19 addition, textbooks by Chow (1959), Henderson (1966), or Streeter (1971), and publications by the USGS (1976b), Brater and King (1976), or the USDA, SCS (NEH-5, 2008) may be useful. Example 3.1.4—Gradually Varied Flow Example Upon consultation, the District Drainage Engineer approved an exception to the minimum ditch bottom width (5.0 ft.) due to a right-of-way constraint. The ditch cross section previously used must be reduced to a 3.5-foot bottom width and a 1:3 back slope for a distance of 100 feet. The transition length between the two ditch shapes is 15 feet. Given: Discharge = 25 ft 3 /sec Roughness = 0.04 Cross Section from Example 3.1-1 Slope = 0.005 ft/ft Calculate: Depth of flow in narrower cross section Figure 3.1-6: Plan View You can estimate the flow depths in the two cross sections using the slope conveyance method, which solves Manning’s Equation and assumes that the ditch is flowing at normal depth. Example C.2 (Appendix C) shows the computation of the normal depths for the 100’ 15’ 15’ R/W January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-20 ditch in this problem using the nomographs in Appendix C. The normal depth in the standard ditch is 1.12 feet, and the normal depth in the narrowed ditch is 1.25 feet. Although it is not standard practice to perform a standard step backwater analysis in a roadside ditch, solving this example will illustrate how a gradually varied profile can be computed using Equations 3.1-2 and 3.1-10 through 3.1-12. The Froude Number (Fr) for normal depth flow at the first section is: ..87 .11 )12 .1)( 12 .14(21)12 .1)( 12 .16(21)512 .1( ft sq Area =×+×+×= .2.16 12 .1)46(5 ft T =++= 733 .02.16 87 .11 === TAD fps AQv 11 .287 .11 25 === 43 .0)733 .0174 .32 (11 .2)( 2121 =×== gD vFr Because Fr is less than one, the flow in the channel will be sub-critical. Therefore, you will start the analysis at the downstream cross section and proceed upstream. Assume normal depth in the standard ditch at a point just downstream of the downstream transition (Section 1 in the figure above). This assumes that the ditch downstream is uniform for a sufficient distance to establish normal depth at Section 1. The water depth at Section 1 is 1.12 feet, as determined in Example C.2 (Appendix C). The first row of the table on the next page shows this depth, along with other geometric and hydraulic values needed for the computations. The elevation, z, is arbitrarily taken as zero. Next, you will determine the depth at Section 2 from a trial-and-error procedure. The first trial depth will be the normal depth at Section 2, which is 1.25 feet. Use Equations 3.1-10, 3.1-11, 3.1-12, and 3.1-2 to back calculate the depth at Section 2. The back-calculated depth of 1.11 feet is shown in the last column. You can assume additional trial depths until the trial and the back-calculated depths agree at the chosen level of accuracy. After you have calculated the depth at Section 2, then calculate the depth at Section 3 using the same trial-and-error process. Repeat the same process to solve for the depth at Section 4. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-21 1 2 3 4 5 6 7 8 9 10 11 XS Depth Guess (ft) Area (ft 2 ) Perimeter (ft) Radius (ft) Velocity (ft /s) V 2 / 2g (ft) Z (ft) EGL (ft) Slope Loss (ft) Depth (ft) 1 1.12 11.872 16.43057 0.72256 2.105795 0.068912 0 1.188912 0.005 1.25 11.40625 15.0563 0.75757 2.191781 0.074655 0.075 1.399655 0.00504 0.075301 1.114558 1.1 9.295 13.66954 0.67998 2.689618 0.112421 0.075 1.287421 0.008766 0.103245 1.104737 1.104 9.348672 13.70652 0.68206 2.674177 0.111134 0.075 1.290134 0.00863 0.102228 1.105007 2 1.105 9.362113 13.71577 0.68258 2.670337 0.110815 0.075 1.290815 0.008597 0.101977 1.105075 1.25 11.40625 15.0563 0.75757 2.191781 0.074655 0.575 1.899655 0.00504 0.681854 1.323014 1.29 12.00345 15.4261 0.77812 2.082735 0.067411 0.575 1.932411 0.004392 0.649423 1.297827 1.296 12.09427 15.48157 0.78120 2.067094 0.066403 0.575 1.937403 0.004303 0.645002 1.294414 3 1.295 12.07911 15.47233 0.78069 2.069688 0.066569 0.575 1.936569 0.004318 0.645732 1.294977 1.12 11.872 16.43057 0.72256 2.105795 0.068912 0.65 1.838912 0.004955 0.069549 1.287206 1.28 14.592 18.06351 0.80781 1.713268 0.045616 0.65 1.975616 0.002827 0.053585 1.294538 1.294 14.84218 18.20639 0.81522 1.684389 0.044091 0.65 1.988091 0.002699 0.052628 1.295107 4 1.295 14.86013 18.2166 0.81575 1.682355 0.043985 0.65 1.988985 0.002691 0.052562 1.295147 Column 2. Use Area formula for trapezoid with the depth guessed in Column 1 Column 3. Use Wetted Perimeter formula for trapezoid with depth guessed in Column 1 Column 4. Column 2 ÷ Column 3 Column 5. Q ÷ Column 2 Column 8. Column 1 + Column 6 + Column 7 Column 9. Solve Equation 3.1-12 using Column 2 and Column 4 values Column 10. Calculate S fwith Equation 3.1-13 using Column 9 from this row and last row of previous section. Calculate the loss with Equation 3.1-11 by multiplying S fby the distance to the previous cross section. Column 11. Back calculate Depth by calculating the Total Energy (Col. 8 of previous cross section + Col. 10) and subtracting the Datum and the Velocity Head (Col. 7 + Col. 6). January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-22 Looking at the results of the profile analysis on the previous page, there are several things you might not expect. First, the flow depth at Section 2 (1.105 feet) is less than the flow depth at Section 1 (1.12 feet), which might be unexpected because the normal depth of Section 2 is greater than Section 1. However, this is not an unusual occurrence in contracted sections. The reason that the flow depth decreases is because the velocity, and, therefore, the velocity head, increases. The increase in the velocity head is greater than the losses between the sections; therefore, the depth must decrease to balance the energy equation. The opposite can occur in an expanding reach, resulting in an unexpected rise in the flow depth even though the normal depth decreases. The next unusual result is that the flow depth at Section 3 is greater than the normal depth in the narrow section. Since the flow depth is less than normal depth at Section 2, the water surface profile should approach normal depth from below as the calculations proceed upstream. Therefore, the flow depth at Section 3 should be less than the normal depth. The reason that the profile jumps over the normal depth line is because of numerical errors introduced by Equation 3.1-13. When the change in the energy gradient between two cross sections is too large, Equation 3.1-13 does not accurately estimate the average energy gradient between the sections. Cross sections must be added between these cross sections to reduce the numerical errors to an acceptable amount. This example was solved using HEC-RAS with the extra cross sections added. The details are described below, but the results indicate that the flow depth essentially converges to normal depth within the 100-foot distance between Sections 2 and 3. The normal depth is 1.25 feet compared to the 1.24 feet computed by HEC-RAS at Section 3. This Illustrates one of the primary reasons that water surface profiles are not necessary in the typical roadside ditch design. The water depth does not significantly vary from normal depth at any location. So, assuming that the design includes some freeboard, the ditch will operate adequately when designed by assuming normal depth. HEC-RAS Solution: Four cross sections with the trapezoidal ditch shapes and slope were input into the program. The expansion and contraction coefficients were changed to zero so that the only the friction loss will be calculated. The friction loss method also was changed to the Average Friction Loss to match Equation 3.1-13. The results of the analysis are shown below. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-23 To compare the results with the spreadsheet solution, the depth of flow must be calculated from the water surface elevation. Section River Station Water Surface Z Flow Depth (Ft.) 1 0 1.12 0 1.12 2 15 1.18 0.075 1.11 3 115 1.87 0.575 1.30 4 130 1.94 0.65 1.29 The flow depths match the solution in Section 2. However, a conveyance ratio warning at Section 3 indicates a possible error at that location. To improve the analysis, extra cross sections were inserted between Section 2 and 3. Four cross sections are added by interpolation and the profile is recomputed. The results are shown below: January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-24 The new flow depth at Section 3 is 1.81 – 0.575 = 1.24 feet. The profile in the narrow section has essentially converged to normal depth (1.25 feet). The depth of the complete profile is shown below: Section River Station Water Surface Z Flow Depth (Ft.) 1 0 1.12 0 1.12 2 15 1.18 0.075 1.11 35 1.35 0.175 1.18 55 1.48 0.275 1.21 75 1.60 0.375 1.23 95 1.71 0.475 1.24 3 115 1.81 0.575 1.24 4 130 1.90 0.65 1.25 3.1.4.3 Rapidly Varied Flow A hydraulic jump occurs as an abrupt transition from super-critical to sub-critical flow. You should consider the potential for a hydraulic jump in all cases where the Froude Number is close to 1.0 and/or where the slope of the channel bottom changes abruptly from steep to mild. For grass-lined channels, unless the erosive forces of the hydraulic jump are controlled, serious damage may result. It is important to know where a hydraulic jump will form, since the turbulent energy released in a jump can cause extensive scour in an unlined channel. For simplicity, you can assume that the flow in the channel is uniform except in the reach between the jump and the break in the channel slope. The jump may occur in either the steep channel or the mild channel, depending on whether the downstream depth is greater or less than the depth sequent to the upstream depth. Using the equation below, you can calculate the sequent depth: gdvddd 112112242 ++−= (3.1-14) where: d2 = Depth below jump, in feet d1 = Depth above jump, in feet v 1 = Velocity above jump, in feet per second g = Acceleration due to gravity, 32.174 ft/sec 2January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-25 If the downstream depth is greater than the sequent depth, the jump will occur in the steep region. If the downstream depth is lower than the sequent depth, the jump will move into the mild channel (Chow). For more discussion on the location of hydraulic jumps, refer to Open-Channel Hydraulics, by V.T. Chow, PhD. When you have determined the location of the jump, you can determine the length using Figure 3.1-7. This figure plots the Froude Number of the upstream flow against the dimensionless ratio of jump length to downstream depth. The curve was prepared by V.T. Chow from data gathered by the Bureau of Reclamation for jumps in rectangular channels. You also can use the curve for approximate results for jumps formed in trapezoidal channels. Figure 3.1-7: Lengths of Hydraulic Jumps When you have determined the location and the length of the hydraulic jump, you can determine the need for alternative channel lining, as well as the limits the alternative lining will need to be applied. Detailed information on the quantitative evaluation of hydraulic jump conditions in open channels is available in publications by Chow (1959), Henderson (1966), and Streeter (1971), and in HEC-14 from USDOT, FHWA (1983). In addition, handbooks by Brater and King (1976) and the USDA, SCS (NEH-5, 2008) may be useful. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-26 Example 3.1-5—Hydraulic Jump Example Given: Q = 60.23 cfs V 1 = 13.81 fps g = 32.2 ft/s 2 d1 = 0.33 ft d2 = 6.74 ft You calculated the depths above using Manning’s Equation. The ditch has a 12.5-foot bottom width with 1:2 side slopes. The longitudinal slopes are 10 percent and 0.001 percent, respectively. The roughness value for the proposed rubble riprap is 0.035. Calculate: Hydraulic Jump and the extent of rubble needed. Step 1: Calculate Froude Number and the Length of the Hydraulic Jump Froude Number, F 1: 111 gd VF = )33 .0)( 2.32 (81 .13 1 =F 24 .41 =F Length of the Hydraulic Jump, L: From Figure 3.1-7, 85 .52 = dL Therefore, )74 .6)( 85 .5(85 .5 2 == dL ft ft L 40 4.39 ≈= Step 2: Calculate the Upstream Sequent Depth Upstream Sequent Depth, d 1 ’: 422211211'1 dgdVdd ++−= 4)33 .0(2.32 )33 .0()81 .13 (2233 .0 22'1 ++−=d ft d 81 .1'1 =January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-27 Since the downstream depth d 2 (6.74 ft) is greater than the upstream sequent depth d 1’(1.81 ft), the hydraulic jump occurs in the steep region. Assuming a more conservative approach, you can split the length of the hydraulic jump between the two regions and provide rubble riprap ditch protection for 20 feet downstream. 3.1.5 Channel Bends At channel bends, the water surface elevation increases at the outside of the bend because of the super-elevation of the water surface. Additional freeboard is necessary in bends, and you can calculate it using the following equation: C gR TVd 2 =∆ (3.1-15) where: ∆d = Additional freeboard required because of super-elevation, in feet V = Average channel velocity, in feet per second T = Water surface top width, in feet g = Acceleration due to gravity, in feet per second squared R C = Radius of curvature of the bend to the channel centerline, in feet d2 d1’ d1 LJanuary 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-28 Example 3.1-6—Channel Bend Example The channel of Example 3.1-2 takes a 45-degree bend with a radius of 30 feet. What is the increased depth on the outside of the channel at the bend? From Example 3.1-2, V = 1.192 ft/sec Calculate Top Width .26 .13 )64(826 .05 ft T =++= ft gR TVdC 02 .0)30 (174 .32 )26 .13 (192 .1 22 ===∆ The depth of flow on the outside edge of the ditch is 0.86 + 0.02 = 0.88 ft. The super-elevation is insignificant for this example problem, as it is for many ditches in Florida. The variable that affects water surface super-elevation the most is the velocity because it is squared in Equation 3.1-15. Ditches with a high velocity at a bend with a small radius will have greater super-elevations. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-29 3.2 OPEN CHANNEL DESIGN Channel shape, slope, and roughness were given in the previous example problems. From these example problems, the flow depths and velocities were determined using the analysis methods described in this chapter. If a project incorporates existing channels, then apply the analysis methods to those channels similar to the example problems. However, many projects will require designing new channels. This section discusses how to select the channel geometry and channel linings for FDOT projects. 3.2.1 Types of Open Channels for Highways You can classify open channels generally as those that occur naturally and those that are manmade, including improved natural channels. The latter, called artificial channels, are used on most roadway projects. The types of channels commonly used on FDOT projects are listed in Chapter 2 of the Drainage Manual : • Roadside Ditch • Median Ditch • Interceptor Ditch • Outfall Ditch • Canals Section 2.2 of the Drainage Manual recommends design frequencies for each of these channel types. The roadside ditch receives runoff from the roadway pavement and shoulders as directed by the cross slope and shoulder slopes. The roadside ditch also may receive flow from offsite drainage areas on adjacent properties. The roadside ditch also may intercept ground water to protect the base of the roadway. The roadside ditch conveys the flow to an outfall point, although the ditch may flow into other ditches or components of the stormwater management system before reaching the ultimate outfall point from FDOT right of way. Depressed medians will collect runoff and a median ditch will be needed to convey runoff to an outfall point. In general, roadside and median ditches are relatively shallow trapezoidal channels, while swales are shallow, triangular, zero-bottom-width channels. Interceptor ditches have various purposes. They provide a method for intercepting offsite flow above cut slopes, thereby controlling slope erosion. They can also collect offsite flow and keep it separate from the project stormwater. This flow can bypass the stormwater treatment facilities, reducing their size and cost. Design outfall ditches, in most cases, to receive runoff from numerous secondary drainage facilities, such as roadside ditches or storm drains. The delineation between a roadside ditch and an outfall ditch can become blurred. If the discharge from a stormwater January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-30 management facility is brought back to the roadside ditch to convey the flow to another point on the project for ultimate discharge, then consider the roadside ditch to be an outfall ditch for the purpose of selecting the design frequency. If you combine considerable flows from offsite areas and onsite project flows together in the roadside ditch to become a significant discharge, then consider the roadside ditch to be an outfall ditch for the purpose of selecting the design frequency. It is unwise to use a roadside ditch as an outfall ditch, since its probable depth and size could create a potential hazard. Canals, like outfalls, also are large artificial channels that accept flows from other drainage components. The added connotation of a canal is that there is always water in the channel, unlike many outfalls that only flow immediately after a rainfall event. If the canal, which always has water, is close to the road, then it can be a potential hazard. For the purpose of identifying a hazard, the FDM defines a canal as an open ditch parallel to the roadway for a minimum distance of 1,000 feet, and with a seasonal water depth in excess of three feet for extended periods of time (24 hours or more). Water Management Districts and local agencies may have a different definition for canals when determining regulatory jurisdiction. Other FDOT publications mention other types of ditches. Right-of-way ditches are mentioned in the Standard Specifications and a detail is given on Standard Plans, Index 524-001 . The right-of-way ditch often functions as a type of relief ditch, handling drainage needs other than those for the roadway and thus freeing roadside ditches from carrying anything except roadway runoff. You usually can consider right-of-way ditches as interceptor ditches when selecting the design frequency. The term “lateral ditch” is used in the FDM and the Standard Specifications . The term is used to determine: • How the ditch excavation will be paid for • How the ditch is shown in the plans A lateral ditch generally is perpendicular to the roadway and can flow either toward or away from the road. However, a lateral ditch also can run parallel to the road right of way if the ditch or channel is separate from the roadway template. Refer to the FDM for guidance on selecting the excavation pay item. Consider the purpose of the lateral ditch and associate it with one of the ditch types listed above to select the design frequency. Several FDOT publications use the term roadway ditch rather than roadside ditch. These two terms are interchangeable. Other FDOT publications or engineers performing work for the Department also may use many other terms to refer to open channels. The definitions of most of these terms are self-explanatory because of their descriptive names. Some examples are: January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-31 • Drainage ditch • Stormwater ditch • Bypass ditch • Diversion ditch • Conveyance channel • Agricultural ditch A swale is a special kind of artificial ditch that has become important in Florida. The following legal definition of a swale as it relates to the regulation and treatment of stormwater discharge is from section 403.803(11), Florida Statutes: "Swale" means a manmade trench which: a) has a top width-to-depth ratio of the cross section equal to or greater than 6:1, or side slopes equal to or greater than 3 feet horizontal to one-foot vertical; and b) contains contiguous areas of standing or flowing water only following a rainfall event; and c) is planted with or has stabilized vegetation suitable for soil stabilization, stormwater treatment, and nutrient uptake; and d) is designed to take into account the soil erodibility, soil percolation, slope, slope length, and drainage area so as to prevent erosion and reduce pollutant concentration of any discharge. 3.2.2 Roadside Ditches You can design roadside ditches using the following steps: Step 1—Establish a Preliminary Drainage Plan. Roadside ditches will be components of an overall drainage system. Since the roadside ditch generally will follow the grade of the road, the high points in the roadway grade will be initial drainage boundaries. However, you can adjust these boundaries by using special ditch grades so that the ditch flows in a different direction than the roadway grade. You also can adjust the boundaries significantly for projects in flat terrain. It is, however, best to keep existing drainage patterns if possible. You also can adjust low points with special ditch grades if the ideal discharge point is not at the low point of the roadway grade. Most projects will have stormwater management facilities, so the roadside ditches will connect with the conveyance components to the various facilities. Not all portions of the roadside ditch can physically be directed to a stormwater management facility, so short segments may need to discharge to other points, such as streams or ditches near cross drains and bridges, or other points along the roadway. When determining initial ditch grades, provide a ditch slope with sufficient grade to minimize ponding and sediment accumulation. The Drainage Manual requires a minimum physical slope of 0.0005 feet/feet for ditches where positive flow is required. These flat January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-32 slopes are difficult to grade during construction and clumps of grass left behind by mowers easily impede the flow. Existing utilities also may control the grade of the ditch to maintain minimum cover over the utility. Step 2—Select Standard Ditch Components. The standard roadside ditch will be shown in the plans on the typical section. You can find standard ditch sections in the FDM for several roadway types, and in Figures 3.2-1 and 3.2-2, below. You may need to adjust the standard ditch due to peculiarities that are consistent throughout the project. An example might be a narrow border width and limited right of way. The typical ditch shown in Figure 3.2-1 for two-lane roads is narrower than most mitered end sections. In some situations, you can use a wider typical ditch section. If the wider ditch is not used, then check the right of way at each mitered end to be sure the right of way will be adequate to accommodate a wider ditch at the mitered end section. Figure 3.2-1: Typical Ditch for Two-Lane Rural Roadway January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-33 Figure 3.2-2: Typical Roadside and Median Ditches If the ditch size needs to be reduced due to right-of-way limitations, you can consider the following options: • Vary the front slope as noted in the FDM Section 215.2.7.1. • You can narrow the bottom width. Five feet is an ideal minimum, but Maintenance and Construction may have equipment to build and maintain a two-foot bottom width. Avoid V-bottomed ditches with steep side slopes. Refer to Chapter 2 of the Drainage Manual for criteria regarding V-bottomed ditches. Avoid using a bottom width narrower than the side drain endwalls. • You can steepen the back slope if the following is considered: o Steeper slopes are harder to maintain, especially 1:3 and steeper o Check the soils for stability o Significant offsite drainage down a steep back slope will cause erosion on the slope • You can reduce the depth to the shoulder point if the following is considered: o Check the ditch capacity o Consider the type of facility and base clearance needs • You also can enclose the ditch with a pipe system, although a ditch or swale usually still is needed to collect the roadway runoff into inlets. Enclosing the system will increase construction costs, but may be less expensive than obtaining more right of way. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-34 Step 3—Check for locations where the standard ditch will not work. A good way to check is to plot the standard ditch on the cross sections. Look for places where the ditch extends beyond the right of way or conflicts with utilities and other obstructions. Also look in the Plan View to check for obstructions between the cross sections. You can adjust the size of the ditch while also considering the same issues identified in the previous step. If the grade of the ditch must be adjusted, then you must develop a special ditch profile and plot it in the plans. Some locations where the ditch grade may need to be adjusted include: • Outfall locations—The grade of the standard ditch will follow the grade of the road. If the outfall location is not at the lowest point in the roadway profile, then you need to develop a special ditch profile. • Locations of high water table—These areas may require feedback to the roadway designer to raise the roadway grade. • Cross drains, median drains, and side drains—These structures may need to be at a lower elevation than the standard ditch elevation. If the entrance end of the culvert is depressed below the stream bed, more head is exerted on the inlet for the same headwater elevation. Usually, the sump is paved, but for small depressions, an unpaved excavation may be adequate. • Locations where the top of the back slope creates a ditch that is too shallow— Sometimes, you can use a berm to contain the ditch instead of changing the grade. Be careful that offsite drainage is not blocked. If you use a berm, provide an adequate top width and side slopes for ease of maintenance. A suggested minimum top width is three feet, but five feet is ideal. You will need to develop special ditch profiles if the profile grade is less than the minimum ditch slope. Refer to the Drainage Manual for minimum ditch slope criteria. At vertical curve crests, the ditch grade will be less than the minimum ditch grade criteria given in the Drainage Manual . (In fact, the ditch grade will go to zero at the high point.) A special ditch grade is not necessary at a vertical curve crest. Step 4—Compute the Flow Depths and Velocities. Although some designers check the ditch at regular intervals, it is not necessary. Checking at critical locations is adequate. Check the ditch at the outfall point. The discharge will be greatest at this location, so it may represent the worst-case conditions for the entire ditch. Other critical locations to check are: • Changes in slope, specifically steeper slopes • Changes in shape, specifically narrower sections • Shallowest ditch depths • Changes in lining (roughness) • Changes in flow January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-35 Determine the maximum allowable depth of the ditch at these sections, including freeboard. Section 2.4.5 of the Drainage Manual provides freeboard requirements. If the actual depth exceeds the maximum allowable depth in the ditch, then the ditch does not have enough capacity. Possible ways to increase the ditch capacity include: • Increase bottom width • Make ditch side slopes flatter • Make longitudinal ditch slope steeper • Provide a smoother ditch lining • Install drop inlets and a storm drain pipe beneath the ditch • Berm up the back slope of the ditch Step 5—Check Lining Requirements. When the ditch geometry components are set and the depth of flow is determined to be adequate, then the ditch needs to be checked to determine if you need a ditch lining. Check the maximum velocity in the ditch against the allowable velocities for bare earth shown in Table 2.4 of the Drainage Manual . If these velocities are met, then you can use the standard treatment of grassing and mulching. If the maximum ditch velocity exceeds the allowable velocity for bare earth, then you should provide sodding, ditch paving, or other forms of ditch lining. See Section 3.3 for more discussion of ditch linings. 3.2.3 Median Ditches The design steps for median ditches are similar to those for roadside ditches. Step 1—Establish a Preliminary Drainage Plan. As with roadside ditches, median ditches also will be components of an overall drainage system. The grade of the median ditch generally will follow the grade of the road. Generally, curbs are not provided on the edge of the pavement and the median ditch drains part or all of the shoulder area in addition to the median itself. Even where curbs are provided, it is preferable to slope medians wider than 15 feet to a ditch. This keeps water in the median off the pavement. Medians less than 15 feet wide generally are crowned for drainage, and, if they are less than six feet in width, they usually are paved. Permitting agencies may request that the median ditch be depressed. When the width of the median ditch is established, locate outfall points from the median. If the travel lanes slope to the outside and the median is impervious, then the median runoff may not need to be conveyed to a stormwater treatment facility. The median may be able to discharge directly into cross drains via inlets. Median cross overs, bridge piers, or other structures often interrupt continuous flow in medians. Decide whether to convey around the obstruction or to one side of the roadway. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-36 Consider the flow depth in the median, feasible means to convey the flow around the obstruction, the size of pipe to convey the flow to the outside, the cover available, and the elevation of the roadside ditch to which the flow will be conveyed. Also consider the actual low point of the median ditch, which is usually at the low point of the roadway grade. This may be affected by guardrail, turn lanes, etc. Turn lanes and other non-typical roadway configurations also may create a depressed gore area. You will need to analyze these areas with methods similar to those used for roadside ditches. Considerations to determine which side of the roadside to discharge to include: • Maintenance of traffic phasing and construction sequencing • Which side the outfall or stormwater facility is located on • Commingling with offsite runoff Step 2—Select Standard Ditch Components. The standard median ditch will be shown in the Plans on the Typical Section. Standard ditch sections are given in the FDM for several roadway types, and one is shown in Figure 3.2-2. Step 3—Compute the Flow Depths and Velocities. Determine critical locations to check depth of flow and velocities, as outlined above. In addition to the critical areas for the roadside ditch, you also should evaluate the median ditch in gore areas caused by turn lanes or additional pavement. If the actual depth exceeds the maximum allowable depth, then you will need to increase the capacity of the ditch. Use methods similar to those for increasing the capacity of a roadside ditch. Be mindful of the additional clear zone requirements for median ditches. Step 4—Check Lining Requirements. After you establish the section of the ditch, check the maximum velocities against the allowable velocities for bare soil. If those velocities are exceeded, then you need to research further to determine the appropriate lining for the ditch. See Section 3.3 of this design guide for further discussion. 3.2.4 Interceptor Ditches Interceptor ditches run along the natural ground near the top edge of a cut slope or along the edge of the right of way to intercept the runoff before it reaches the roadway. Interceptor ditches along the edge of the right of way are commonly referred to as right-of-way ditches. The interceptor ditch generally will follow the grade of the natural ground adjacent to the project, not the profile grade of the road. If possible, locate the high points in an interceptor ditch at the drainage divides of the adjacent property to maintain existing drainage patterns. Low points also typically follow the adjacent terrain, allowing the interceptor ditch to discharge to points such as streams near cross drains and bridges. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-37 Most projects will have stormwater management facilities. These facilities often are set off from the project area, so it is important to consider conflicts that may arise where the outfall ditch intersects the interceptor ditch. The design steps for interceptor ditches are the same as those for the roadside ditch. See Section 3.2.2 for the design procedure. 3.2.5 Outfall Ditches Since outfall ditches receive runoff from numerous secondary drainage facilities, including stormwater management facilities, design the standard ditch section for a larger capacity. You should evaluate the standard ditch section against the clear zone criteria for the project. Even though outfall ditches have a larger design event and carry larger flows, the design steps are the same as those for the roadside ditch. See Section 3.2.2 for the design procedure. The design also should include consideration of the following: • The drainage area that flows into the outfall ditch by overland flow. Designers often forget to include this area in the total drainage area when determining the design flow rates for the outfall ditch. Another concern is erosion down the side slope from the sheet flow from these areas. You can use spoil from the ditch construction to create berms to block and collect the flow in inlets to prevent this erosion. • Check for existing outfall easements. Some easements may require a specific type of conveyance, such as a ditch or a pipe system. 3.2.6 Hydrology As stated in Section 2.3 of the Drainage Manual , hydrologic data used for the design of open channels will be based on one of the following methods, as appropriate for the particular site: • Use a frequency analysis of observed (gage) data when available • Use the regional or local regression equation developed by the USGS • Use the Rational Equation for drainage areas up to 600 acres • Use the method applied for the design of the stormwater management facility in the design of the outfall from this facility • Request hydrologic data from the controlling entity for regulated or controlled canals For a more detailed discussion on procedure selection and method for calculating runoff rates, refer to Chapter 2 (Hydrology). January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-38 3.2.6.1 Frequency Roadside or median ditches or swales, including bypass and interceptor ditches, usually are designed to convey a 10-year frequency storm without damage; outfall ditches or canals should convey a 25-year frequency storm without damage. However, because the risks and drainage requirements for each project are unique, site-specific factors may warrant the use of an atypical design frequency. Regardless of the frequency selected, you should always consider the potential for flooding that exceeds standard criteria. Pre-development stages for all frequencies up to and including the 100-year event must not be exceeded unless flood rights are obtained or the flow is contained within the ditch. It also is important to consider sediment transport requirements for conditions of flow below the design frequency. A low flow channel component within a larger channel can reduce the maintenance effort by improving sediment transport in the channel. Design temporary open channel facilities for use during construction to handle flood flows commensurate with risks. The recommended minimum frequency for temporary facilities and the temporary lining of permanent facilities is 20 percent of the standard frequency for permanent facilities, which extrapolates as a two-year frequency for roadside ditches and a five-year frequency for outfall ditches. 3.2.6.2 Time of Concentration The time of concentration is defined as the time it takes runoff to travel from the most remote point in the watershed to the point of interest. When using the Velocity Method, calculate the time of travel for main channel flow using the velocity in the section and the channel length. Segments used to determine the velocity should have uniform characteristics. Use a new segment each time there is a change in the channel geometry, such as cross section or channel slope. Calculate the time for each segment and then add them together to determine the total time of concentration for the channel. See Chapter 2 (Hydrology) for a discussion of methods and procedures to determine the time of concentration. 3.2.7 Tailwater and Backwater The water depth at the downstream end of the ditch will affect the flow depth and velocities in the ditch for some distance upstream. The downstream water depth, or tailwater, may cause a backwater condition with a gradually varied water surface profile. In roadside ditches, you can approximate the water surface profile as a flat water surface at the tailwater (T w ) elevation that intercepts the normal depth (d n) of flow in the ditch, as shown in Figure 3.2-3. If the tailwater depth is less than the normal depth in the ditch, then you can approximate the water surface profile in the ditch as the normal depth in the ditch, as shown in Figure 3.2-4. For the low tailwater condition, perform the velocity check for lining requirements using the velocity for the tailwater depth, not the normal depth. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-39 Figure 3.2-3: Assumed Water Surface for Tw > dn Figure 3.2-4: Assumed Water Surface for T w < d n To summarize the water surface approximation, the water surface elevation at any point in the ditch is the higher of the normal depth elevation or the tailwater elevation. You can determine the frequency of the design tailwater elevation using the same recommendations for storm drains in Section 3.4 of the Drainage Manual . The same water surface profile assumptions illustrated above also apply to other backwater conditions in the ditch. Side drains are an example. The water surface elevation in the ditch at any point upstream of a side drain should be the greater of the normal depth elevation or the headwater elevation of the culvert. The normal depth in the ditch changes if the ditch slope, cross section, or roughness changes. If the downstream normal depth is greater, then the assumed water surface is shown in Figure 3.2-5. dn Actual water surface fil Assumed water surface Tw dn Actual water surface fil Assumed water surface TwJanuary 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-40 Figure 3.2-5: Assumed Water Surface for change in d n dn Actual water surface fil Assumed water f dnJanuary 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-41 Example 3.2-1—Roadside Ditch Design Example The figures below show the plan and profile views of a proposed four-lane roadway. Complete the design of the left roadside ditch. Step 1—Drainage Plan. On the left side of the roadway near Station 3125+00A, there is a stormwater pond to treat and attenuate the roadway runoff. Roadside ditches will collect the runoff from the roadway and convey it to the cross drain, which empties into the pond. The offsite drainage area is small; therefore, dual ditches are not needed to reduce the size of the pond. 3125 3130 3135 3140 3145 Pond Mitered End Section 3125 3130 3135 3140 3145 400’ V.C. 1,500’ V.C. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-42 The left roadside ditch will discharge into a mitered end section at Station 3126+50. The design frequency for the ditch will be 10 years (refer to the Drainage Manual for the design frequency). The pipe system and the pond may have different design frequencies than the ditch, but you can determine a 10-year elevation in the pond and the 10-year hydraulic grade line for the pipe system at the mitered end section. The hydraulic grade line of the pipe system at this headwall will be the tailwater elevation for the ditch. The design of the overall drainage system may be iterative. The design of one component, such as the pond, can affect the design of other components, such as the left and right roadside ditches, the cross drain, and even the median ditch. To simplify this example, the tailwater elevation for the ditch will be given as 76.52 feet. Step 2—Standard Ditch Components. The standard ditch shown in Figure 3.2-2 will be used. The vertical distance from the profile grade line (PGL) to the ditch bottom elevation of the standard ditch will be: Elevation Difference = (24 ft. x 0.02) + (12 ft. x 0.06) + 3.5 ft. = 4.7 ft. Step 3—Check for locations where the standard ditch will not work. Three reasons why the standard ditch will not work are: • The backslope tie in to natural ground extends beyond the right-of-way line and acquiring additional right of way is not prudent. • The natural ground elevation is lower than the standard ditch bottom elevation, or low enough that the standard ditch is too shallow. • The profile grade is less than the minimum ditch slope. Plotting the standard ditch on the roadway cross sections is a good way to look for locations where the standard ditch will not work. Also, starting at the downstream end of the ditch and working upstream will afford an orderly approach to design the ditch. For this example, the profile grade elevation will be 79.00 and the bottom of the standard ditch will be 74.3 feet at Station 3127+00, as shown in the figure below. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-43 The PGL is flat (0.000 percent) between this cross section and the end section at Station 3126+50. The minimum slope of the ditch is 0.05 percent, and the ideal slope is at least 0.1 percent. Therefore, you will need a special ditch grade between these stations. If the flowline at the headwall (Station 3126+50) is set at 74.2 feet, the ditch grade between these stations will be 0.1/50 = 0.002, or 0.2 percent. At this point in the design process, calculate the discharge at the downstream end of the ditch. For this example, the discharge will be given as 12.7 cfs at the end section. Refer to the Chapter 2 (Hydrology) for an explanation of how to calculate the discharge. Solving Manning’s Equation with the standard ditch shape (five-foot bottom width, 1:6 front slope, 1:4 back slope), the slope of 0.2 percent, n = 0.042, and the discharge of 12.7 cfs gives a flow depth in the ditch of 1.03 feet. At the headwall, the normal depth elevation would be 74.2 + 1.03 = 75.23 feet. This elevation is less than the tailwater elevation. Therefore, the flow depth in the ditch is the tailwater elevation of 76.52 feet. The outside edge of the shoulder elevation is lower than the back of the ditch elevation at this location and will, therefore, control the allowable flow depth in the ditch. Since the tailwater elevation is lower than the allowable flow depth, the ditch depth is adequate. Proceed upstream to continue the design. Looking at the cross sections between Stations 3133+00 and 3136+00, the standard ditch bottom elevation will be higher than the natural ground elevation for several hundred feet, as typified by the cross section shown below for Station 3134+00. El. 74.3 El. 79.00 Sta. 3127+00 January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-44 The standard ditch could be used if a berm was constructed. However, there are at least two reasons not to construct the berm. First, some offsite flow to the ditch would be blocked. Second, the cost of constructing the berm is unnecessary since you can use a special ditch profile to lower the ditch into the natural ground. The discharge needs to be determined at this point to continue the design. A conservative assumption would be to use the discharge at the downstream end of the ditch. In this case, the designer judges that the discharge might be significantly different and calculates the discharge at this point. To simplify the example, the discharge at this location is given as 10.2 cfs. Assuming a ditch bottom elevation of about 79.3 ft (2 feet below natural ground), the slope to Station 3127+00 would be (79.3 – 74.3)/700 = 0.007, or 0.07 percent. Selecting the value of 2 feet was based on some preliminary calculations of the flow depth and including some freeboard. Solving Manning’s Equation with the standard ditch shape, the slope of 0.7 percent, n = 0.042, and the discharge of 10.2 cfs gives a flow depth in the ditch of 0.68 feet. This would leave a freeboard of approximately 1.3 feet at this location, which is more than needed. The flow depth of 0.68 feet is close enough to 0.7 feet that using n of 0.042 is reasonable given the amount of freeboard provided. A special ditch grade of 0.07 percent will be used between Stations 3127+00 and 3134+00. The special ditch grade has to tie back into the standard ditch grade someplace further upstream. The standard ditch bottom will return to an adequate depth into natural ground to contain the flow at Station 3137+00. The PGL at Station 3137+00 is 91.17 feet. The ditch bottom elevation for the standard ditch is 86.47 feet. The ditch grade will be (86.47 – 79.3)/300 = 0.0239, or 2.39 percent. Solving Manning’s Equation with the standard ditch shape, the slope of 2.39 percent, n = 0.06, and the discharge of 10.2 cfs gives a flow depth in the ditch of 0.59 feet and a velocity of 2.2 fps. Note that the roughness changes because the flow depth is less than 0.7 feet. The velocity is low enough that ditch lining Sta. 3134+00 El. 87.08 El. 81.3 January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-45 will not be needed. However, sod will be needed, instead of seed and mulch, to establish grass during construction. Checking the cross sections between 3134+00 and 3137+00, the ditch depth is at least 1.5 feet, which will provide acceptable freeboard. To summarize, the special ditch grades will be: • 0.2 percent from Station 3126+50 to 3127+00 • 0.07 percent from Station 3127+00 to 3134+00 • 2.39 percent from Station 3134+00 to 3137+00 The standard ditch will provide an adequate depth from 3137+00 to the top of the hill. Checking the cross section plots shows that the earthwork to construct the standard ditch will not extend beyond the proposed right-of-way line. Step 4—Compute the Flow Depths and Velocities. These values were calculated in the description of the previous step. In most cases, the designer will be iterating through Steps 3 and 4 as the ditch is designed. Figure 3.2-6 shows the ditch checks appropriate for including in the Drainage Documentation to prove the design. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-46 HYDRAULIC WORKSHEET FOR ROADSIDE DITCHES LT 0.20 2.61 0.75 15 6.5 12.7 6 5.0 4 0.042 1.03 1.2 SOD LT 0.70 12.7 6 5.0 4 0.042 0.75 1.9 SOD LT 0.70 1.79 0.75 10 7.6 10.2 6 5.0 4 0.042 0.68 1.87 SDO LT 2.39 10.2 6 5.0 4 0.6 0.59 2.2 SOD Note: F.S. = Front Slope B.W. = Bottom Width B.S. = Back Slope Manning "N" is Transitioning as the depth Approaches 0.7' Project Number: 1234567_ STATION TO STATION SIDE %Slope Drain Area "C" Tc Q(cfs) Ditch Section F.S. B.W. Sheet 1_ of 1_ Prepared by: XXX_ Date: 4/1/09 Checked by: YYY_ Date: 4/1/09 "d allowed " Calculated Freeboard B.S. 3134+00 3134+00 Road: New Road_______ 3126+50 TW El. will control 3127+00 TW El. will control Remarks "n" "d" Vel (fps) Ditch Lining Side Drain Pipe Dia I10 Figure 3.2-6: Roadside Ditch Design Example January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-47 3.2.8 Side Drains Continuous flow in a roadside ditch can be interrupted by side street/road connections and/or driveway connections to the project roadway. Even a limited access roadway, such as an interstate highway, may have an occasional access driveway that will impede roadside ditch flow, especially at or near adjacent stormwater pond locations. You can maintain ditch flow continuity through such obstructions via roadside ditch culverts or side drains. A side drain is a class of culvert pipe that can transport flow through fill placed in a roadside ditch. A side drain is normally aligned parallel or nearly parallel to the project roadway and along the flowline of the ditch. Side drains located under public roads connecting to the project roadway, are identified and hydraulically sized as a cross drains (see Chapter 4, Culverts). Side drains and cross drains are similar in many ways, but there are some differences in design analysis requirements, materials, and end treatment. Cross drains have to meet more rigorous criteria for some parameters. 3.2.8.1 Design Analysis Requirements for Side Drains You size a side drain for the storm frequency required to design the roadside ditch that contains the side drain (usually the 10-year frequency, as mentioned in Section 3.2.6.1). You can determine the side drain design flow by applying the same hydrologic method used to compute the corresponding ditch design flows (usually the Rational Equation, described in Section 2.2.3). Then, you can determine the side drain pipe dimensions via the inlet-control/outlet-control procedure described in Section 4.5. (Note: The FHWA HY-8 computer software is one of several computer programs capable of applying this procedure to the side drain design data.) You will normally develop the design flow for a side drain in the design calculations spreadsheet or worksheet for the roadside ditch that contains the side drain. (Figure 2-1 of the Drainage Manual depicts such a ditch design worksheet.) The design flow and surface water depth for the ditch section at the upstream end of the side drain are determined in the ditch calculations, and this ditch flow is the side drain design inflow as well. This flow typically is also the design flow for the ditch section at the downstream end of the side drain, and must be accounted for in the calculations for the remainder of the downstream ditch length. Of course, if additional flow enters the side drain between its upstream and downstream ends, this additional flow also must be appropriately accounted for in both the side drain hydraulic design and in the downstream ditch design calculations. Determine the tailwater elevation at the culvert outlet. Since the culvert usually is placed through fill in the roadside ditch, the ditch calculations downstream of the culvert are used to determine the tailwater. The culvert tailwater will be the normal depth in the January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-48 downstream ditch unless the tailwater for the ditch controls the water surface elevation at the side drain outlet. Refer to Section 3.2.7 for more discussion on tailwater. Then you can generate the hydraulic calculations for a side drain, using the procedure described above to determine the pipe dimensions needed to safely pass the design flow to the downstream ditch segment. Include these side drain calculations in the Drainage Documentation Report as either a separate section or as part of the Ditch Calculations section. Note that the surface water depth computed for culvert flow at the upstream end of a side drain generally will be larger than the depth computed for ditch flow at that location. If the difference in this flow depth is not significant, evaluate the ditch flow depths upstream from the side drain and adjust (if appropriate) for the “flat pool” that will be established in the ditch by the higher of the two water surface elevations. If the difference in surface water flow depth at the side drain is substantial and the ditch design is sensitive to actual flow depths, a backwater analysis may be needed rather than the “flat pool” approximation in determining the actual flow depth estimates. 3.2.8.2 Material Requirements In general, side drains are not considered to be as critical as cross drains. Therefore, material service life requirements for side drains are less stringent than for cross drains. Consult Chapter 6 of the Drainage Manual , the FDOT Standard Specifications , Chapter 8 (Optional Pipe Materials) in this handbook, and the appropriate District Drainage Engineer for any clarification needed on pipe materials acceptable for use as side drains. Culvert and ditch calculations may show the need for two allowable pipe sizes, depending on the Manning’s roughness coefficients of the optional pipe materials for the side drain. 3.2.8.3 End Treatment The only allowable side drain end treatment is the mitered end section ( Standard Plans, Index 430-022 ). Due to the normal side drain alignment and close proximity to the project roadway (usually within the clear zone), Standard Plans, Index 430-022 specifies that grates be installed for the larger pipe sizes. The grates are intended to provide a measure of safety for errant vehicles that encounter the end treatment. The grates, however, will potentially collect debris and will increase the entrance loss coefficient, Ke, from 0.7 to 1.0 for the mitered end section. When a grate is likely to be used, consider the following items: • Recognize that the specification of a grate could increase the required side drain size (due to the increase in Ke). • In critical hydraulic locations, evaluate the potential debris transport prior to using grates. Vegetated ditch grades in excess of 3 percent, pipe with less than 1.5 feet of cover, or paved ditch grades in excess of 1 percent will require such an evaluation. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-49 • Determine highly corrosive locations and specify in the plans when the grates need to be hot-dipped galvanized after fabrication. Example 3.2-2 – Side Drain Design Problem Statement: A driveway is included in the design of the left roadside ditch for a new two-lane rural roadway segment. Figure 3.2-1 depicts the typical section for the left side of the roadway. The ditch extends and flows from Station 10+00 to Station 45+00, with the centerline of the driveway located at Station 40+00. The width of the proposed driveway base at the ditch flowline is 40 feet, and the ditch section is uniform throughout its length with a 2-foot allowable depth below the left top-of-bank. At its upstream and downstream ends, the ditch flowlines must match elevations of 100.0 feet and 96.0 feet, respectively. The following sketch shows the ditch longitudinal slopes are 0.1 percent from Station 10+00 to Station 35+00, and 0.15 percent from Station 35+00 to Station 45+00. The natural ground slopes away from the left top-of-bank of the ditch section. Determine the required side drain diameter. Design Approach: First, develop the ditch design calculations to determine the side drain design inflow at Station 39+80. These calculations are shown on Figure 3.2-7, and identify a side drain design flow of 4.60 cfs. Next, refer to Section 4.5 for the side drain hydraulic design procedure. Use either the inlet control and outlet control nomographs from FHWA HDS-5, or software such as HY-8, to develop the required side drain size. Culvert 0.1% 0.15% 10+00 35+00 45+00 40+00 40’ El. 100 El. 96 January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-50 HYDRAULIC WORKSHEET FOR ROADSIDE DITCHES LT 0.10 2.75 0.47 60.1 3.24 4.19 6 5.0 4 0.042 0.7 0.7 Seed & Mulch LT 0.15 3.31 0.47 69.5 2.96 4.60 6 5.0 4 0.042 0.67 0.83 Seed & Mulch LT 4.60 18" LT 0.15 3.86 0.47 78.6 2.72 4.93 6 5.0 4 0.042 0.69 0.85 Seed & Mulch Note: F.S. = Front Slope B.W. = Bottom Width B.S. = Back Slope Manning "N" is Transitioning as the depth Approaches 0.7' 39+80 -40+20 See Side Drain Calcs for details 40+20 -45+00 Drain Area includes 1/2 of driveway width Road: New Road__ 10+00 -35+00 35+00 -39+80 Drain Area includes 1/2 of driveway width Remarks "n" "d" Vel (fps) Ditch Lining Side Drain Pipe Dia I10 Q(cfs) Ditch Section F.S. B.W. Sheet 1 of ___1_ Prepared by: _XXX_ Date: 4/1/09 Checked by: YYY_ Date: 4/1/09 "d allowed " Calculated Freeboard B.S. Project Number: 1234567_ STATION TO STATION SIDE %Slope Drain Area "C" Tc Figure 3.2-7: Side Drain Design Example January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-51 3.3 CHANNEL LININGS As stated in Section 2.4.3 of the Drainage Manual , when designing open channels, determine channel lining requirements. Erosion and sloughing cause most maintenance problems in channels. Channel linings often solve these problems. The Standard Plans, and the Standard Specifications identify standard lining types. The two main classifications of open channel linings are flexible and rigid. Flexible linings include vegetative linings such as grass, rubble riprap, and geotextile or interlocking concrete grids. Rigid linings include concrete, asphalt, and soil-cement. From an erosion control standpoint, the primary difference between rigid and flexible channel linings is their response to changes in channel shape (i.e., width, depth, and alignment). For most artificial channels, the ideal lining is natural, emerging vegetation, with grass used to provide initial and long-term erosion resistance. The following are examples of lining materials in each classification. Flexible Linings: a. Grasses or natural vegetation b. Rubble riprap c. Wire-enclosed riprap (gabions) d. Turf reinforcement (non-biodegradable) Rigid Linings: a. Cast-in-place concrete or asphaltic concrete b. Soil cement and roller-compacted concrete c. Fabric formed revetment d. Partially grouted riprap e. Articulated concrete blocks 3.3.1 Flexible Linings Flexible linings have several advantages compared to rigid linings. They generally are less expensive, permit infiltration and exfiltration, and can be vegetated to have a natural appearance. Flow in channels with flexible linings is similar to that found in natural small channels. Natural conditions offer better habitat opportunities for local flora and fauna. In many cases, flexible linings are designed to provide only transitional protection against erosion while vegetation establishes and becomes the permanent lining of the channel; flexible channel linings are best suited to conditions of moderate shear stresses. Channel reaches with accelerating or decelerating flow (expansions, contractions, drops, and backwater) and waves (transitions, flows near critical depth, and shorelines) will require special analysis and may not be suitable for flexible channel linings. 3.3.1.1 Vegetation Vegetative linings consist of seeded or sodded grasses placed in and along the channel, as well as naturally occurring vegetation. Vegetation is one of the most common and most January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-52 ideal channel linings for an artificial channel. It stabilizes the body of the channel, consolidates the soil mass of the bed, checks erosion on the channel surface, and controls the movement of soil particles along the channel bottom. Vegetative channel lining also is recognized as a best management practice for stormwater quality design in highway drainage systems. The slower flow of a vegetated channel helps the uptake of highway runoff contaminants (particularly suspended sediments) before they leave the highway right of way and enter streams. There are conditions for which vegetation may not be acceptable, so you will need to consider other linings. These conditions include, but are not limited to: • Standing or continuous flowing water • Areas which do not receive the regular maintenance necessary to prevent domination by taller vegetation • Lack of nutrients and excessive soil drainage • Areas where sod will be excessively shaded The Department operates on the premise that, with proper seeding and mulching during construction, maintenance of most ditches on normal sections and grades can be handled economically until a growth of grass becomes established. The use of temporary erosion control measures in ditches with low velocities will provide time for grassing and mulching to establish a vegetative ditch. When velocities exceed those for bare soils, seeding and mulching should not be used. Sodding is recommended when the design velocity exceeds the value permitted for the bare base soil conditions but is less than 4 feet per second. Lapped or shingle sod is recommended when the design velocity exceeds that for sod (4 feet per second), and is suitable with velocities up to 5.5 feet per second. 3.3.1.2 Other Flexible Linings Flexible linings usually are less expensive than rigid linings, provide a safer roadside, and have self-healing qualities that reduce maintenance. They also allow the infiltration and exfiltration of water. (A) Rubble Riprap After grass, rubble riprap is the most common type of flexible lining. It presents a rough surface that can dissipate energy and mitigate velocity increases. There are two standard types of rubble riprap. Use ditch lining rubble riprap in standard or typical ditches or channels. It consists of smaller stone sizes, which reduces construction costs over bank and shore rubble. Limit bank and shore rubble riprap to uses such as revetments and linings along stream banks and shorelines where extreme flows or wave action occurs. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-53 Limited right of way and availability of material may restrict the use of this type of flexible lining. Place rubble riprap on a filter blanket and prepared slope to form a well-graded mass with a minimum of voids. Riprap and gabion linings can perform in the initial range of hydraulic conditions where you would use rigid linings. Stones used for riprap and gabion installations preferably have an angular shape that allow them to interlock. These linings usually require a filter material between the stone and the underlying soil to prevent soil washout and migration of fine grained soils. Sometimes you will need a bedding stone layer to protect the filter fabric from larger stone. (B) Gabion Mats Gabions are made of riprap enclosed in a wire container or closed structure that binds units of the riprap lining together. The wire enclosure normally consists of a rectangular container made of steel wire woven in a uniform pattern and reinforced on corners and edges with heavier wire. The containers are filled with stone, connected together, and anchored to the channel side slope. The forms of wire-enclosed riprap vary from thin mattresses to boxlike gabions. Use gabions typically when rubble riprap is either not available or not large enough to be stable. Although flexible, wire mesh restricts gabion movement. The wire mesh must provide an adequate service life. If the wire mesh fails, the individual stones will migrate. (C) Articulating Concrete Block (ACB) Revetment Systems ACB systems consist of a precast block matrix connected together by cables. The articulating properties of the matrix allow the system to accommodate changes in the ground surface that may occur due to settling. The block configuration varies with the manufacturer. The systems typically are manufactured in units of multiple precast blocks that can be lifted easily and placed with construction equipment. HEC-23 and the National Concrete Masonry Association’s Design Manual for Articulating Concrete Block Revetment Systems provide guidance for the design of these systems. (D) Turf Reinforcement Depending on the application, materials, and method of installation, turf reinforcement may serve a transitional or long-term function. The concept of turf reinforcement is to provide a structure to the soil/vegetation matrix that will both assist in the establishment of vegetation and provide support to mature vegetation. Two types of turf reinforcement commonly are available: soil/gravel methods and turf reinforcement mats (TRMs). To create soil/gravel turf reinforcement, you mix gravel mulch into on-site soils and seed the soil-gravel layer. The rock products industry provides a variety of uniformly graded gravels for use as mulch and soil stabilization. A gravel/soil mixture provides a non-degradable lining that is created as part of the soil preparation and is followed by seeding. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-54 A TRM is a non-degradable rolled erosion control product (RECP) composed of UV-stabilized synthetic fibers, filaments, netting, and/or wire mesh processed into a three-dimensional matrix. TRMs provide sufficient thickness, strength, and void space to permit soil filling and establishment of grass roots within the matrix. One limitation to the use of TRMs is in areas where siltation is a problem. When the ditch is cleaned by maintenance, it is likely that the geofabric will be snagged and pulled out by the equipment. 3.3.2 Rigid Linings Rigid linings generally are constructed of concrete, asphalt, or soil-cement pavement whose smoothness offers a higher capacity for a given cross-sectional area. Higher velocities, however, create the potential for scour at channel lining transitions from the rigid lining back to the grass lining. A rigid lining can be destroyed by flow undercutting the lining, channel headcutting, or the buildup of hydrostatic pressure behind the rigid surfaces. When properly designed, rigid linings may be appropriate where the channel width is restricted. Rigid linings are useful in flow zones where high shear stress or rapidly varied or turbulent flow conditions exist, such as at transitions in channel shape or at an energy dissipation structure. Rigid linings are particularly vulnerable to a seasonal rise in the water table that can cause a static uplift pressure on the lining. If you need a rigid lining in such conditions, incorporate a reliable system of under drains and weep holes as a part of the channel design. Evaluate the migration of fine grained soils into filter layers to ensure that the ground water is being discharged without filter clogging or collapse of the underlying soil. A related case is the buildup of soil pore pressure behind the lining when the flow depth in the channel drops quickly. Using watertight joints and backflow preventers on weep holes can help to reduce the buildup of water behind the lining. Section 2.4.3.1.2 of the Drainage Manual requires the design for the potential for buoyancy due to the uplift water pressure when concrete linings are to be used where soils may become saturated. The total upward force is equal to the weight of the water displaced by the channel. The total weight of the lining helps to resist the uplift pressure. When the weight of the lining is less than the uplift pressure, the channel is unstable. Acceptable countermeasures include: • Increasing the thickness of the lining to add additional weight • For sub-critical flow conditions, specifying weep holes at appropriate intervals in the channel bottom to relieve the upward pressure on the channel • For super-critical flow conditions, using sub-drains in lieu of weep holes January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-55 3.3.2.1 Cast-in-Place Concrete Refer to Standard Plans, Index 524-001 for typical ditch pavement details. Asphalt linings have limited use since routine maintenance activities often damage or destroy them. Use filter fabric to prevent soil loss through pavement cracks. Despite the non-erodible nature of concrete linings, they are susceptible to failure from foundation instability. The major cause of failure is undermining that can occur in a number of ways. Inadequate erosion protection at the outfall, at the channel edges, and on bends can initiate undermining by allowing water to carry away the foundation material and leaving the channel to break apart. Concrete linings also may break up and deteriorate due to conditions such as a high water table or swelling soils that exert an uplift pressure on the lining. When a rigid lining breaks and displaces upward, the lining continues to move due to dynamic uplift and drag forces. The broken lining typically forms large, flat slabs that are particularly susceptible to these forces. 3.3.2.2 Fabric Formed Revetment Fabric formed revetments, also known as grout-filled mattresses, are the result of pumping a concrete mix into fabric envelopes or cases. The advantage of using fabric formed revetments is that they reduce construction time by eliminating the need for wooden forms and expensive lifting machines and also allow the concrete to be pumped and cured below the water line. Filter point fabric formed revetments consist of a dual wall fabric that is injected with concrete. This type of fabric formed revetment is characterized by a deeply cobbled surface. The filter points woven into the fabric provide a means for groundwater to escape and to provide release for the hydrostatic pressure. Filter point fabrics provide a higher coefficient of friction to promote energy dissipation. As of June 2020, FDOT has Developmental Specification 531 for fabric formed revetment systems available. 3.3.3 Velocity and Shear Stress Limitations HEC-15 provides a detailed presentation of stable channel design concepts for roadside and median channels. This section provides a brief summary of significant concepts. Stable channel design concepts provide a means of evaluating and defining channel configurations that will perform within acceptable limits of stability. Most highway drainage channels cannot tolerate bank instability and lateral migration. When the material forming the channel boundary effectively resists the erosive forces of the flow, then you have achieved stability. You can apply principles of rigid boundary hydraulics to evaluate this type of system. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-56 Apply both velocity and tractive force methods to help determine channel stability. Permissible velocity procedures are empirical in nature, so they have been used to design numerous channels in Florida and throughout the world. However, tractive force methods consider actual physical processes occurring at the channel boundary and represent a more realistic model of the detachment and erosion processes. The hydrodynamic force that water flowing in a channel creates causes a shear stress on the channel bottom. The bed material, in turn, resists this shear stress by developing a tractive force. Tractive force theory states that the flow-induced shear stress should not produce a force greater than the tractive resisting force of the bed material. This tractive resisting force of the bed material creates the permissible or critical shear stress of the bed material. In a uniform flow, the shear stress is equal to the effective component of the gravitational force acting on the body of water parallel to the channel bottom. The average shear stress is equal to: τ = γ R S (3.3-1) where: τ = Average shear stress, in pounds per square feet γ = Unit weight of water,62.4 lb/ft 3 R = Hydraulic radius, in feet S = Average bed slope or energy slope, in feet per feet The maximum shear stress for a straight channel occurs on the channel bed and is less than or equal to the shear stress at maximum depth. Compute the maximum shear stress as follows: τd =γ d S (3.3-2) where: τd = Maximum shear stress, in pounds per square feet d = Maximum depth of flow, in feet S = Channel bottom slope, in feet per feet Velocity limitations for artificial open channels should be consistent with stability requirements for the selected channel lining. As indicated above, use seed and mulch only when the design velocity does not exceed the allowable velocity for bare soil. Table 2.3 of the Drainage Manual presents maximum shear stress values and allowable velocities for different soils. When design velocities exceed those acceptable for bare soil, sod, or lapped sod, consider flexible or rigid linings. Table 2.4 of the Drainage Manual summarizes maximum velocities for these lining types. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-57 Side Slope Stability The shear stress on the channel sides generally is less than the maximum shear stress calculated on the channel bottom, but you should consider this issue when determining the height of a channel lining along the side slope of the channel. The maximum shear stress on the side of a channel is given by: τs= Κ 1τd (3.3-3) where: τs = Side shear stress on the channel, in pounds per square feet Κ 1 = Ratio of channel side to bottom shear stress τd = Shear stress in channel at maximum depth, in pounds per square feet The value K 1 depends on the size and shape of the channel. For parabolic channels, the shear stress at any point on the side slope is related to the depth at that point and you can calculate it using Equation 3.3-2. For trapezoidal and triangular channels, K 1 is based on the horizontal dimension 1: Z (V: H) of the side slopes. K1 = 0.77 Z ≤ 1.5 K 1 = 0.066Z + 0.67 1.5 < Z < 5 K1 = 1.0 5 ≤ Z Avoid using side slopes steeper than 1:3 for flexible linings other than riprap or gabions because of the potential for erosion at the side slopes. Steep side slopes are allowable within a channel if cohesive soil conditions exist. Maintenance Considerations Also consider maintenance of the channel when choosing a channel lining. The channel will need to be accessible by mowers and trucks. Mowing Side slopes of vegetated channels will need to be traversable for mowing equipment and crews. The maximum traversable slope for this equipment is 1:4. Access Across Channel If there is rubble riprap lining the channel and a vegetated buffer on the backside of the channel along the right of way, the irregularity of the riprap typically prevents access. In this situation, it may become impractical to maintain the vegetation. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-58 3.3.4 Application Guidance for Some Common Channel Linings 3.3.4.1 Rubble Riprap Types • Ditch Lining —Flexible layer or facing of rock placed on a filter blanket and prepared slope used to line a ditch or channel for protection from erosion. • Bank and Shore —Flexible layer or facing of rock placed on a bank or shore to prevent erosion or scour of the embankment or a structure. What is its purpose? Use rubble riprap in channels, along embankments, or around structures that are vulnerable to erosion or scour. Where and how is it commonly used? • Ditch Lining —In this case, use rubble riprap to line ditches and channels to protect slopes from erosion. • Bank and Shore —In this case, use it as a flexible revetment to line banks and shores subject to erosion. When should it be installed? • Ditch Lining —Install rubble riprap in channels with moderate shear stresses. To prevent uplifting forces on the lining, the filter requires adequate permeability. • Bank and Shore —Use rubble riprap to protect banks or shores with flows that generally are greater than 50 ft 3/s or that are subject to wave action. When should it not be installed? • Bank and Shore —Do not install rubble riprap when ditch lining methods are applicable. Advantages and disadvantages ADVANTAGES • Flexible • Not weakened by minor shifting caused by settlement • Easily repaired by additional rock placement • Simple construction method • Recoverable/reusable • Long-term or temporary installations DISADVANTAGES • Hauling and installation costs • Prohibits maintenance equipment from traversing channels • If hand placement is required, then labor is intensive January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-59 • Vegetation growth can hinder inspections North Carolina Erosion and Sediment Planning and Control Manual Figure 3.3-1: Riprap-lined Channel Cross Sections 3.3.4.2 Fabric Formed Revetments Types Fabric formed revetments for concrete with filtering points that provide for the relief of hydrostatic pressures. What is the purpose? Use fabric formed revetments—filter point or articulating—for slopes or areas that are subject to severe to moderate erosion problems. Where and how are they commonly used? Use fabric formed revetments in ditches, channels, canals, streams, rivers, ponds, lakes, reservoirs, marinas, and ports/harbors to reduce the impact of erosion. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-60 When should they be installed? Install fabric formed revetments where there are moderate to severe erosion problems and where the channel is subjected to hydrostatic uplift pressures. Also, install these where there is a need to allow water to permeate into the soil and not remain wet. When should they not be installed? Do not use fabric formed revetments in ditches or channels that are subject to changes in soil conditions such as erosion under the mat or consolidation. Advantages and disadvantages ADVANTAGES • Adapts easily to contours • Easy to install • Permeable • Reduces uplift pressure • Can be installed under the water line DISADVANTAGES • Needs to be installed on a prepared slope • Not aesthetically pleasing • Easily undermined if not toed properly January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-61 (Source: ) Construction Techniques, Inc. Figure 3.3-2: Fabric Formed Revetment with Filter Point Linings 3.3.4.3 Gabions Types • Gabion Mats —Wire mesh mats filled with stones • Gabion Baskets —Wire mesh baskets filled with stones What is the purpose? Rock-filled baskets or mattresses that are used to line large ditches, channels, canals, and coastal shores for stabilization and protection. Where and how are they commonly used? Gabion Mats —Use gabion mats in ditches, channels, canals, streams, rivers, ponds, lakes, reservoirs, marinas, and ports/harbors to reduce the impact of erosion. When should they be installed? Gabion Mats —Install gabion mats in large areas where there are moderate to severe erosion problems due to extreme velocities. Also where there is a need to allow water to permeate into the soil and not remain wet. When should they not be installed? Gabion Baskets —Small areas subject to low velocities and when a temporary situation exists. Advantages and disadvantages ADVANTAGES • Protects seed mix from eroding when used • Permeable • Increases retention of soil moisture • Permits the growth of vegetation • Able to span minor pockets of bank subsidence without failure January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-62 DISADVANTAGES • Cost of installation • Susceptibility of the wire baskets to corrosion and abrasion damage • More difficult and expensive to repair • Less flexible than standard riprap January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-63 (Source: ) Modular Gabion Systems, a division of C.E. Shepherd Company Figure 3.3-3: Gabion Dimensions January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-64 (Source: ) Modular Gabion Systems, a division of C.E. Shepherd Company Figure 3.3-4: Gabion Binding January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-65 3.3.4.4 Soil Stabilizers Types • Turf Reinforcement Mats —A long-term non-degradable mat composed of UV stabilized synthetic fibers, nettings, and/or filaments. • Erosion Control Blankets —A temporary degradable mat composed of processed natural or polymer fibers mechanically, structurally, or chemically bound together to form a continuous matrix. What is the purpose? To protect disturbed slopes and channels from wind and water erosion. The blanket materials are natural materials, such as straw, wood excelsior, coconut, or are geotextile synthetic woven materials, such as polypropylene. Where and how are they commonly used? • Turf Reinforcement Mats —Use them on ditch slopes and fill slopes to reduce the impact of erosion for long periods of construction. • Erosion Control Blankets —Use them on ditch slopes and fill slopes to reduce the impact of erosion during short periods of construction. When should they be installed? • Turf Reinforcement Mats —Where there are low velocities of flow • Erosion Control Blankets —Where there are low velocities of flow and where there are sensitive environmental areas When should they not be installed? • Turf Reinforcement Mats —Do not install for permanent situations and where there are high velocities of flow. • Erosion Control Blankets —Do not install for permanent situations and where there are high velocities of flow. Advantages and disadvantages ADVANTAGES • Adapts easily to contours • Easy to install • Permeable • Reduced uplift pressure DISADVANTAGES • Cost • Maintenance equipment can damage or pull out January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-66 (Source: )(Source: ) Propex Geosynthetics Propex Geosynthetics Figure 3.3-5: Erosion Control Mat in Channel Figure 3.3-6: Initial Anchor (Downstream) (Source: ) Propex Geosynthetics Figure 3.3-7: Longitudinal Anchor Trench Detail (Trapezoidal Channel) January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-67 3.4 DRAINAGE CONNECTION PERMITTING AND MAINTENANCE CONCERNS 3.4.1 Drainage Connection Permitting Adjacent property owners must obtain a Drainage Connection Permit from FDOT according to Section 334.044(15), Florida Statute (F.S.), Chapter 14-86, Florida Administrative Code (F.A.C)., Rules of the Department of Transportation, when developing their property. In general terms, the Drainage Connection Permit ensures that the development will not overload the Department’s stormwater conveyance systems and cause flooding on either the roadway or other downstream properties. For more information on Drainage Connection Permits, refer to the Drainage Connection Permitting Handbook . This section will discuss several aspects of the Department’s ditches that you should consider during the Drainage Connection Permitting process. 3.4.1.1 Roadside Ditch Impacts Discharges to the roadside ditch from the proposed development will be limited by the Permit so that the ditch flow will not be increased. However, the proposed development can physically impact the roadside ditch by placing or widening driveways to the property or by widening the roadway to add turn lanes. If the roadside ditch is a linear treatment pond, then any reduction in the volume of the ditch could violate the conditions of the permit obtained for the facility. The simplest way to resolve this issue is to rework the ditch so that any volume lost as a result of the development is replaced. This may require that the property owner donate some property to the Department to provide an area to rework the ditch. Even if the roadside ditch is not a linear treatment facility, you must maintain the capacity of the ditch. Include a side drain to convey the ditch flow from one side of the turnout to the other, unless the turnout is located at a high point in the ditch and the flow is away from the turnout in both directions. An added turn lane may require that the roadside ditch be relocated. The relocated portion of the ditch should have the same capacity or more than the existing ditch. If the existing right of way is not wide enough to accommodate the relocated ditch, then right of way may need to be donated to FDOT for the ditch. A turnout requiring a side drain and a turn lane requiring donated right of way for the ditch relocation are shown in Figure 3.4-1. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-68 Figure 3.4-1: Effect of Adjacent Development on a Roadside Ditch In some cases, the developer may need to add a left-turn lane. Widening the road to accommodate the left-turn lane also may affect the ditch on the opposite side of the road from the development. Often, the developer will not own the property on both sides of the road. In this case, the roadside ditches and roadway must be redesigned to accommodate the new turn lanes in such a way as to require donated right of way on the new development’s side of the road. The flow lines of the side drain should match the existing ditch. Also ensure that the flow lines of the new side drain are higher than the next side drain downstream and lower than the next side drain upstream to avoid temporary ponding in the ditch. Make sure to size the side drain properly. You can make some judgments about the size of the pipe by looking at the side drains upstream and downstream of the new drive. Analyze the side drain to ensure the new pipe does not cause the water levels to pop out of the ditch. In some cases, you can obtain the design discharge for the ditch from the old plans for the roadway. Or you can calculate the flow by determining the drainage area and performing the proper hydrologic calculations; typically, the Rational Equation. You can find more details on these hydrology calculations in Chapter 2. Calculate the losses through the pipe using methods given in Chapter 4. Additional sizing considerations are discussed in Section 3.2.8. When adding new side drains, another consideration is the proximity of other existing side drains. If side drains are too close to each other, then the hydraulic losses can be too large. The general requirement is that the end sections of two side drains in series should be at least 25 feet apart. If the distance is less than 25 feet, then you should enclose the area and add an inlet to collect the runoff from the area between the driveways. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-69 Evaluate potential erosion at the infall point of the connection, especially for pipe connections. Chapter 4 explains how to calculate the outlet velocity from a pipe. Refer to Section 3.3 for channel linings. You can find outlet erosion protection criteria in the Drainage Manual . 3.4.1.2 Median Ditch Impacts A new development can impact the median ditch if the Department allows a new median opening or left-turn lane. Unless you can place a new median opening at the high point in the median ditch, or close enough to the high point that it is possible to regrade the ditch to flow away from the new median opening in both directions, then the new opening will block the flow in the ditch. Figure 3.4-2 shows a typical situation where there is an existing median opening at the high point in the median ditch and the ditch flows to a median drain, which consists of a ditch bottom inlet, pipe, and endwall. The median drain discharges runoff from the median to keep the median from filling with water and spilling across the roadway. Figure 3.4-2: Existing Median Ditch If you add a new median opening to accommodate an adjacent development, the opening may block the flow in the median ditch. Include a new drainage structure with the opening to discharge the flow from the median. Figure 3.4-3 shows a side drain included to convey the ditch flow from one side of the new median opening to the other. This often will be the most economical method to provide adequate drainage for the median. However, in many cases, the median ditch will be too shallow and the side drain will not have adequate cover over the pipe. Refer to Appendix C of the Drainage Manual for the minimum cover needed over the pipe. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-70 Figure 3.4-3: New Median Opening with Side Drain Figure 3.4-4 includes a new median drain to accommodate the median flow. If you choose to use this option, check the capacity of the roadside ditch with the added discharge from the median. Unless you jack and bore the pipe, the existing pavement would have to be cut and patched to install the pipe. Make sure to consider the cutting and patching operations in maintenance of traffic plans. Figure 3.4-4: New Median Drain Another option that might avoid the expense of jacking and boring or the concerns of cutting and patching the existing roadway is shown in Figure 3.4-5. You could connect the new ditch bottom inlet (DBI) to the existing median drain with a pipe beneath the new median opening. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-71 Figure 3.4-5: New Median Drainage System Adding a turn lane in the median often will reduce the size of the median ditch adjacent to the new turn lane. Check the reduced ditch for capacity, and add extra median drainage structures if needed. Super-elevated roadways that drain to the median can worsen the capacity problems in areas where the ditch has been reduced. 3.4.1.3 Outfall Ditch Impacts Requested connections or crossing may physically impact outfall ditches. Usually, the permitted flow will not be greater than the existing flow rate because of the requirements of the connection permit. However, you need to evaluate losses associated with the physical impacts to ensure there is no compromise to the capacity of the outfall ditch. Overland flow connections can cause bank erosion and sloughing if the flow becomes concentrated. To avoid this problem, use point connections through pipes or ditches. Erosion problems also can occur at the connections to an outfall ditch. Refer to Section 3.4.1.1 for guidance to protect the infall point. 3.4.2 Maintenance Concerns 3.4.2.1 Ditch Closures Residents or other property owners occasionally will request that the roadside ditch in front of their property be filled and replaced with a pipe system. Piping a ditch can increase the energy loss and reduce infiltration. Under storm conditions, open ditching is an efficient method of accommodating a significantly greater quantity of drainage than a pipe. Therefore, any piping or filling of a roadside ditch generally is of no benefit to the Department and may reduce operational and maintenance aspects of the road. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-72 Drainage connection applicants should perform a hydraulic assessment to determine ditch piping or filling impacts on the area drainage system. These impacts should adhere to Rule 14-86 requirements, as consistent with the Drainage Manual . Unless you acquire flood rights, any increase over pre-development stages should not change land use values significantly. Do not consider filling an open ditch if the basis for the modification is for aesthetic purposes, for landscaping, or to benefit the abutting private property owner only. Table 3.4-1 lists criteria and other considerations for converting existing drainage ditches to closed drainage systems. Table 3.4-1: Capacity of Closed System Criteria Comments Design Storms: The more stringent of: • Rule 14-86, F.A.C. Storms: • Original Ditch Design Storms: • Drainage Manual Design Storms: o Evacuation route? o Upstream owner constraints?  Potential for flooding upstream? o Downstream constraints?  Tailwater • Planned work program improvements: Primary considerations: • Minimize adverse impact on Department & other facility users • Maximize capacity of facility • Maximize life of facility o Avoid need to reconstruct for later foreseeable projects • Minimize maintenance cost Pipe Size: The more stringent of: • Rule 14-86, F.A.C. Criteria: • Original Ditch Design Criteria: • Drainage Manual Criteria: • Future Work Program Requirements: Check various scenarios and use the criteria that most satisfies the Department’s interests. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-73 Table 3.4-1 (continued) Capacity of Closed System (continued) Criteria Comments Method: Prove that the headwater elevation for the design storms shall not be increased immediately upstream of the proposed system. Base design on hydrologic conditions in the field, not the size of existing pipe systems. Base design on condition that entire length of the ditch will eventually have a closed system. Do not rely solely on the size of existing upstream systems for designing capacity of ditch systems downstream. While knowledge of upstream systems is useful in many ways, these existing systems: • May be undersized due to: o Design errors o Under estimated watershed area o Subsequent land development activity o Subsequent system changes or diversion • May not reflect current design standards • May not be adequate for current or future needs o Existing flooding conditions o Future road improvements Other considerations: Remember that the Department owns not only the current capacity of its outfall easements, but also the right to use any potential excess capacity available in the outfall. Any proposed piped outfall must be sized for the Design Frequency noted in Chapter 2 of the Drainage Manual . Select solutions that maximize preservation of the Department’s ability to expand its system to the full use of its facility for future needs. Consider the consequences that result when the proposed system fails and make any reasonable adjustments to minimize damage and liability for the Department. The applicant usually hopes to reduce the Department’s easement area by closing the open ditch with pipe or other structures. This usually represents a false economy when one adds the requirements necessary to maintain the closed system at minimum expense. Oftentimes, you can eliminate or greatly reduce major risk of damage due to system failure by careful attention to the failure mode and addition of details to re-route overflows or provide protective measures such as curbs, berms, emergency spillways, etc. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-74 Table 3.4-1 (continued) Work Program Criteria Comments Considerations: • In Work Program: o If already designed & approved – use the design o If not designed – coordinate design for approval by DOT project engineer • Not in Work Program: o Route design submittal for review and approval by District Drainage Engineer among others The possibility exists that the applicant can simply build the outfall already under design by the Department, especially if the applicant cannot wait for the Department’s future construction job to complete the work. Erosion Control Considerations: • Erosion at outlet • Erosion when flows exceed system capacity • Soils • Flow velocity • Slopes May result in failure of the pipe outfall system. Possible turbid discharge downstream. Methods: • Drainage Manual • Erosion and Sediment Control Designer and Reviewer Manual • Protective measures o Structural solutions o Non-structural methods Maintenance Responsibility: • Applicant (local government) responsible o When concession needed from Department in negotiation o When special structures require more maintenance attention or expense • DOT responsible o At DOT discretion Define this carefully in the agreement. January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-75 Table 3.4-1 (continued) Maintenance (continued) Criteria Comments Considerations: • Reasonable & Safe Access o For equipment o For personnel o For operations – spoil, staging, etc. • Other facilities in easement o Above ground - trees, fences, sheds, etc. o Underground - utilities, drainage, etc. • Potential to damage adjacent facilities o Above ground structures, buildings, etc. o Overhanging structures, utilities, etc. • Limitations: o Depth of work - shoring needed? o Groundwater Consider these factors when negotiating the terms of agreement. Remember: If the new facility cannot be reasonably maintained in a safe and cost effective manner, then perhaps the easement should remain an open ditch. Right-of-way Considerations: • Additional right-of-way required: o To maintain access o To enable maintenance o To minimize Cost of Maintenance o To preserve or secure drainage rights • Donation of right-of-way • Reduction of right-of-way: o Only when fully justified o Must meet Drainage Manual requirements for dimension, etc. Consult with right -of -way attorney to determine: • the appropriate style of easement • relation to downstream owners not involved in the transaction o Where to end the easement when drainage exits applicant’s property and falls onto another person’s property? • special terms to add into the easement document January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-76 Table 3.4.1 (continued) Permitting Criteria Comments Document: • Contractual Agreement • Easement Agreement • Easement Donation / Exchange • Drainage Connection Permit A Drainage Connection Permit is not the appropriate form for approval of this category of work, unless the work is performed as part of a larger scope of property improvements that require the permit and there is no need to alter the existing easement in any way. A contractual agreement with appropriate terms and conditions is the preferred method of approval. Process: • If easement relocation or exchange required: o Follow “Property Management Related Reconstruction Process” chart • If no change needed to existing easement : Consult early with Legal Department to determine form of agreement Perform review proposed design to determine any special conditions or terms required in the agreement Legal Department to draft agreement Maintenance to review draft agreement and resolve any issues. Deliver agreement to applicant for signature. Obtain Department signature Administer terms of agreement Some typical contract terms: • Review and approval of plans • Party responsible for maintenance • Failure-to-perform provisions • Responsibility to obtain all required permits • Review of plans • Notice of changes • As-built plans & computations • Final certification by engineer • May waive need for other permits, if practicable • Other conditions as needed Construction Considerations: • Pre-construction meeting • All permits in hand • Erosion control measures in place • Oversight & Inspection January 1, 2024 Drainage Design Guide Chapter 3: Open Channel Chapter 3: Open Channel 3-77 Table 3.4-1 (continued) Construction (continued) Inspection: •Administer contract •Obtain approval from engineer for changes •Erosion control Acceptance: •Follow contract terms for completion of contract •File as-built plans & design computations 3.4.2.2 Acquisition of Ditches from Local Ownership When roadways pass from local ownership to FDOT, it is not unusual for issues to arise. Often, the roadside ditches on these roadways do not meet FDOT standards. They often were designed for a lesser design frequency and do not contain enough capacity. Other ditches have substandard slopes located within the clear zone. When safety concerns force these roadways to be updated, evaluate the existing conditions to bring the ditches up to current standards. In some cases, there may be enough right of way available to reconstruct the ditch to standards. More frequently, though, right of way is not sufficient to provide these upgrades. Then, it may be practical to purchase additional right of way or drainage easements in which to upgrade the current ditch system. If additional right of way proves to be too costly, consider a closed system with a series of inlets and storm drain pipes. The least ideal but often unavoidable option will consist of obtaining exceptions or variances of the current standards for the existing ditch. 3.4.2.3 Addition of Sidewalks to Roadway Projects In an ongoing attempt to connect communities with pedestrian walkways, existing roadways often have sidewalks added. The sidewalks often are located outside of the existing ditch system along the right-of-way line. When designing these sidewalks, ensure that the sidewalk does not impede flow from offsite runoff. Place it so that offsite runoff can sheet flow over the sidewalk into the existing ditch or that the system can collect runoff and pipe it under the sidewalk into the ditch or an existing storm drain system. In many cases, you can construct a simple pedestrian bridge to cross over existing ditches without impacts to the ditch.
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https://engineering.purdue.edu/~wassgren/teaching/ME30800/NotesAndReading/PipeFlows_Losses_Reading.pdf
Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics CHAPTER 11 Pipe Flows 11.1. Entrance Region The flow in the entrance region is complex (Figure 11.1) and will not be investigated here. Experiments have shown that the dimensionless length of the entrance region depends on whether the entering flow is laminar or turbulent, with, laminar flow: L D ⇡0.06ReD, (11.1) turbulent flow: L D ⇡4.4Re1/6 D , (11.2) where L is the length of the entrance region and D is the pipe diameter. For many engineering flows, 1 ⇥104 < ReD < 1 ⇥105 = ) 20 < L/D < 30. The shorter entrance region length for turbulent flows is due to the fact that turbulent mixing rapidly averages the flow speeds across the pipe cross-section. Figure 11.1. The structure of a pipe flow entrance region. 11.2. Fully Developed Laminar Circular Pipe Flow (Poiseuille Flow) The derivation in this section was previously covered in Chapter 8 and is repeated here, in a slightly condensed form, for convenience. Consider the steady flow of an incompressible, constant viscosity, Newtonian fluid within an infinitely long, circular pipe of radius, R (Figure 11.2). Make the following assumptions, (1) The flow is axi-symmetric and there is no “swirl” velocity. = ) @ @✓(. . . ) = 0 and u✓= 0 (2) The flow is at steady state. = ) @ @t (. . . ) = 0 (3) The flow is fully-developed in the z-direction. = ) @ur @z = @uz @z = 0 (4) There are no body forces. = ) fr = f✓= fz = 0 C. Wassgren 1029 2021-12-15 Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics Figure 11.2. A schematic of flow through a circular pipe. Let’s first examine the Continuity Equation, 1 r @(rur) @r + 1 r @u✓ @✓+ @uz @z = 0. (11.3) From assumptions #1 and #3, rur = constant. (11.4) Since there is no flow through the walls, the constant must be zero and, thus, ur = 0 (Call this condition # 5.) (11.5) Now let’s examine the Navier-Stokes equation in the z-direction, ⇢ ✓@uz @t + ur @uz @r + u✓ r @uz @✓+ uz @uz @z ◆ = −@p @z + µ 1 r @ @r ✓ r@uz @r ◆ + 1 r2 @2uz @✓2 + @2uz @z2 $ + ⇢fz. (11.6) We can simplify this equation using our assumptions, ⇢ 0 B B @ @uz @t |{z} =0(#2) + ur |{z} =0(#5) @uz @r + u✓ r @uz @✓ | {z } =0(#1) +uz @uz @z |{z} =0(#3) 1 C C A = −@p @z + µ 2 6 6 4 1 r @ @r ✓ r@uz @r ◆ + 1 r2 @2uz @✓2 | {z } =0(#1) + @2uz @z2 | {z } =0(#3) 3 7 7 5 + ⇢ fz |{z} =0(#4) , (11.7) = ) d dr ✓ rduz dr ◆ = r µ dp dz , (11.8) = ) rduz dr = r2 2µ dp dz + c1, (11.9) = ) uz = r2 4µ dp dz = c1 ln r + c2. (11.10) Note that in the previous derivation the fact that uz is a function only of r has been used to change the partial derivatives to ordinary derivatives. Furthermore, examining the Navier-Stokes equations in the r and ✓directions demonstrates that the pressure, p, is a function only of z and, thus, ordinary derivatives can be used when di↵erentiating the pressure with respect to z. Now let’s apply boundary conditions to determine the unknown constants c1 and c2. First, note that the fluid velocity in a pipe must remain finite as r ! 0 so the constant c1 must be zero (this is a type of kinematic boundary condition). Also, the pipe wall is fixed so we have uz(r = R) = 0 (no-slip condition). After applying boundary conditions we have, uz = R2 4µ ✓ −dp dz ◆ 1 − ⇣r R ⌘2$ . (11.11) This is known as Poiseuille Flow in a Circular Pipe. Notes: C. Wassgren 1030 2021-12-15 Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics (1) The velocity profile is a paraboloid with the maximum velocity occurring along the centerline. The average velocity in the pipe is found from, ¯ u = 1 ⇡R2 ˆ r=R r=0 uz(2⇡rdr) = R2 8µ ✓ −dp dz ◆ = D2 32µ ✓ −dp dz ◆ = 1 2umax, (11.12) where umax is the maximum fluid speed and D is the pipe diameter. (2) The volumetric flow rate through the pipe is, Q = ¯ u ⇣⇡ 4 D2⌘ = ⇡D4 128µ ✓ −dp dz ◆ . (11.13) (3) We can determine stresses using the constitutive relations for a Newtonian fluid. The shear stress that the pipe walls apply to the fluid, ⌧w, is, ⌧w = R 2 ✓dp dz ◆ = −4µ¯ u R , (11.14) where ¯ u is the average speed in the pipe. Note that an alternate method for determining the average wall shear stress, which in this case is equal to the exact wall shear stress, is to balance shear forces and pressure forces on a small slice of the flow as shown in Figure 11.3. Figure 11.3. A free body diagram showing the forces on a thin disk of fluid in the pipe. In engineering applications, it is common to express the average shear stress in terms of a dimen-sionless (Darcy) friction factor, fD, which is defined as, fD := 7 7 7 7 4⌧w 1 2⇢¯ u2 7 7 7 7 = 64 ✓µ ⇢¯ uD ◆ = 64 ReD , (11.15) where D = 2R is the pipe diameter and ReD is the Reynolds number based on the pipe diameter. The Darcy friction factor appears in the Moody chart for incompressible, viscous pipe flow. Note again that this solution is only valid only for a laminar flow. The condition for the flow to remain laminar is found experimentally to be, ReD = ⇢¯ uD µ < 2300. (11.16) (4) Re-write Eq. (11.12), ¯ u = D2 32µ ✓ −dp dz ◆ = ) |¯ u| = D2 32µ ✓∆p L ◆ , (11.17) where, in the fully developed region, the pressure gradient remains constant and we may write dp/dz as ∆p/L where ∆p is the pressure drop over a length L of the pipe (Figure 11.4). Re-arranging Eq. (11.17) and dropping the absolute value symbol for convenience, ∆p = 32µ¯ uL D2 . (11.18) C. Wassgren 1031 2021-12-15 Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics Make the previous equation dimensionless by dividing through by the dynamic pressure based on the average flow speed, ∆p 1 2⇢¯ u2 = 64µ ⇢¯ uD ✓L D ◆ = ✓64 ReD ◆✓L D ◆ . (11.19) The dimensionless pressure drop is also referred to as a loss coefficient, k. Hence, for a laminar flow, the loss coefficient corresponding to the viscous stresses at the pipe walls is, k laminar, wall stresses = ✓64 ReD ◆✓L D ◆ = fD ✓L D ◆ . (11.20) Figure 11.4. A schematic showing the change in pressure over the pipe length. C. Wassgren 1032 2021-12-15 pipe_13 A liquid with a specific gravity of 0.95 flows steadily at an average velocity of 10 m/s through a horizontal, smooth tube of diameter 5 cm. The fluid pressure is measured at 1 m intervals along the pipe as follows: x [m] 0 1 2 3 4 5 6 p [kPa] 304 273 255 240 226 213 200 a. Estimate the average wall shear stress, in Pa, in the fully developed region of the pipe. b. What is the approximate wall shear stress between stations 1 and 2? State any significant assumptions you make. SOLUTION: First determine the fully developed region by examining the pressure gradient in the pipe. The pressure gradient is constant in the fully developed region. x [m] 0 1 2 3 4 5 6 p [kPa] 304 273 255 240 226 213 200 dp/dx [kPa/m] -31 -18 -15 -14 -13 -13 Hence, the fully developed region starts at x = 4 m where the pressure drop remains constant at dp/dx = -13 kPa/m. To determine the average wall shear stress in the pipe, apply the linear momentum equation in the x direction to the control volume shown in the figure below. , (1) where, (steady flow), (2) (no body forces in x-direction), (3) , (4) (5) (since for a fully-developed flow, the inlet and outlet velocity profiles are identical) Substitute and simplify, , (6) . (7) d dt uxρ dV CV ∫ + ux ρurel ⋅dA ( ) CS ∫ = F S,x + F B,x d dt uxρ dV CV ∫ = 0 F B,x = 0 F S,x = pπ R2 − p + dp dx dx ⎛ ⎝ ⎜ ⎞ ⎠ ⎟π R2 −τ w 2π Rdx ( ) = −dp dx dxπ R2 −τ w 2π Rdx ( ) ux ρurel ⋅dA ( ) CS ∫ = 0 0 = −dp dx dxπ R2 −τ w 2π Rdx ( ) τ w = −R 2 dp dx x ppR2 [p + (dp/dx)dx]pR2 tw(2pRdx) dx Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics Video solution: C. Wassgren 1033 2021-12-15 pipe_13 Using the given data, R = (0.05/2) m = 0.025 m dp/dx = -13 kPa/m For part (b), apply the same linear momentum equation, except that between stations 1 and 2, the velocity profile is not fully developed, hence the momentum flux term in the linear momentum equation (Eq. (5)) won’t be zero. However, if the flow is turbulent, as would be expected for such a large velocity and assuming a liquid viscosity similar to that of water, the velocity profile will not change considerably as the flow continues downstream in the entrance region. The reason for this is that a turbulent velocity profile already looks like an average velocity profile due to the radial mixing associated with turbulence. Hence, although the momentum flux term isn’t exactly zero, it is expected to be small in comparison to the pressure gradient term. As a result, even in the entrance region the average wall shear stress may be found using, . (8) Using the given data between stations 1 and 2, R = (0.05/2) m = 0.025 m dp/dx = -18 kPa/m ⇒τ w = 163 Pa τ w ≈−R 2 dp dx ⇒τ w = 225 Pa Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics Video solution: C. Wassgren 1034 2021-12-15 Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics Equation (11.29) is implicit in fD, which means that an iterative approach must be used to solve for fD as a function of ReD. A number of approximations to this relation have been proposed that are easier to solve. For example, Blasius, a student of Prandtl’s, suggested the following approximation, fD ≈0.316 Re1/4 D , (11.30) which is valid for 4000 < ReD < 1 × 105. 11.3.2. Turbulent Flow in a Very Rough Pipe The roughness of the pipe walls can significantly affect the friction factor for turbulent flows (roughness has a negligible effect on the friction factor for laminar flows). Recall from the Law of the Wall that the time averaged velocity in the laminar sub-layer is, ¯ u u∗= yu∗ ν for 0 ≤yu∗ ν ≤5. (11.31) Thus, the thickness of the laminar sub-layer, δLSL, is, δLSLu∗ ν = 5 = ⇒δLSL = 5ν u∗. (11.32) Since, u∗= rτw ρ = ¯ u r fD 8 (refer to Eq. (11.24)), (11.33) we have, δLSL = 5ν ¯ u r 8 fD = ⇒δLSL D = 5ν ¯ uD r 8 fD = 5 ReD r 8 fD , (11.34) ∴δLSL D = 14.1 ReD √fD . (11.35) Thus, if the wall roughness, ϵ, is much smaller than the laminar sub-layer thickness, then we’ll still have a laminar sub-layer and the flow won’t be significantly affected by the wall roughness, i.e., we may treat the wall as being smooth (but still frictional). However, if ϵ ≫δLSL, then the laminar sub-layer will be destroyed and the wall roughness becomes the new length scale for use in the Law of the Wall, i.e., ¯ u u∗= f y ϵ  . (11.36) Following the same analysis as that for turbulent flow in a smooth pipe, but using y/ϵ in place of yu∗/ν, we obtain, r 1 fD ≈−2.0 log10 ϵ/D 3.7  . (11.37) This is the friction factor for turbulent flow in a very rough pipe. The term ϵ/D is known as the relative roughness. Note that this equation is independent of the Reynolds number. 11.3.3. Turbulent Flow in a Rough Pipe For the transitional regime where ϵ/D is between “smooth” and “very rough”, empirical formulas in which the friction factor is a function of both ϵ/D and ReD have been developed, r 1 fD ≈−2.0 log10 ϵ/D 3.7 + 2.51 ReD √fD  ReD > 4000 Colebrook Formula, (11.38) C. Wassgren 1036 2021-12-15 Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics which is implicit in fD, or the explicit empirical formula, r 1 fD ⇡−1.8 log10 " 6.9 ReD + ✓✏/D 3.7 ◆1.11# ReD > 4000 Haaland Formula. (11.39) The Haaland formula isn’t as accurate as the Colebrook formula, but it’s easier to calculate since it’s explicit in fD. To solve the Colebrook formula for fD, an iterative algorithm must be used. An initial first guess for fD using the Haaland formula usually results in convergence in the Colebrook formula within one or two iterations. 11.4. The Moody Plot The previous friction factor relations have been summarized into a single plot known as the Moody Plot, which is shown in Figure 11.6. Figure 11.6. The Moody plot, which plots the (Darcy) friction factor as a function of Reynolds number for di↵erent relative roughnesses. This figure is from Pritchard, P.J. and Mitchell, J.W., Fox and McDonald’s Introduction to Fluid Mechanics, 9th ed., Wiley. Notes: (1) For Reynolds numbers less than 2,300, one may use either the analytical expression for the friction factor, fD = 64 ReD , (11.40) or the Moody plot. C. Wassgren 1037 2021-12-15 Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics (2) Reynolds numbers between approximately 2,300 and 4,000 correspond to the transitional regime between laminar and turbulent flow. The gray region in the Moody plot reflects the fact that the friction factor can vary significantly in this region. At best, bounds can be determined for the friction factor in this region rather than a specific value. (3) The fully rough zone (aka wholly turbulent zone, fully turbulent zone) in the Moody plot is a region where the friction factor is a weak function of the Reynolds number, but a strong function of the relative roughness. If the Reynolds number of a flow is unknown, but is expected to be large, it is often helpful to assume that the flow is in the fully rough zone as an initial first guess. (4) The roughnesses of various types of pipe materials have been compiled into tables such as Table 11.1. Note that “smooth” in the table does not mean frictionless. Table 11.1. A table of pipe material wall roughnesses. Material (new) ✏(ft) ✏(mm) riveted steel 0.003 - 0.03 0.9 - 9.0 concrete 0.001 - 0.01 0.3 - 3.0 wood stave 0.0006 - 0.003 0.18 - 0.9 cast iron 0.0085 0.26 galvanized iron 0.0005 0.15 asphalted cast iron 0.0004 0.12 commercial steel or wrought iron 0.00015 0.046 drawn tubing 0.000005 0.0015 glass smooth smooth C. Wassgren 1038 2021-12-15 pipe_49 Page 1 of 1 1. Using the Moody chart, determine the friction factor for a Reynolds number of 105 and a relative roughness of 0.001. 2. What is the friction factor for a Reynolds number of 1000? 3. What is the friction factor for a Reynolds number of 106 in a smooth pipe? SOLUTION: 1. The friction factor is fD » 0.0225 (Follow the red lines in the following figure.) 2. Since the Reynolds number is less than 2300, we can use the exact laminar flow relation: Þ fD = 0.064. Alternately, we could use the Moody chart by following the blue lines in the following figure. 3. The friction factor is fD » 0.012. (Follow the green lines in the following figure.) 64 Re D D f = Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics C. Wassgren 1039 2021-12-15 pipe_50 Page 1 of 3 Create a computer program that uses the Colebrook formula to calculate a friction factor. The program inputs should be the Reynolds number and the relative roughness. The output should be the friction factor. Use your program to create a copy of the Moody plot. Plot the friction factor for Reynolds numbers between 4000 and 1108 for 10 relative roughness values between 110-6 and 110-2. Use logarithmic horizontal and vertical axes. SOLUTION: The Colebrook formula is, ! ! " ≈−2.0 log!# $ % ⁄ '.) + .+! ,-!.",. (1) This function is implicit in f so an iterative scheme must be used to solve it. There are various algorithms that can be used to solve for f. In this solution, the following algorithm is used: 1. Guess a value for the friction factor f. Use the Haaland formula for this first guess, ! ! " ≈−1.8 log!# / /.0 ,-! + 0 $ % ⁄ '.) 1 !.!! 2. (2) 2. Solve for the friction factor on the left-hand side of the Colebrook formula (Eq. (1)), call this f’, using the guessed value for f on the right-hand side. 3. Is the value of f’ equal to f within an acceptable tolerance (tol), i.e., Is 1""2"1 " < 789? (3) If not, then let f = f’ and repeat step 2. If so, then we now have our value for f. A counter is also included in the program to ensure that we don’t iterate an unacceptably large number of times. Usually this algorithm solve for f within a few iterations, but the counter is a fail-safe measure since the iterative scheme isn’t guaranteed to converge. This algorithm is implemented here using the Python programming language. The plot is shown at the end of this document. # pipe_50.py # Import some helpful Python libraries. import numpy as np # used for numerical routines import matplotlib.pyplot as plt # used for making plots # Create a function for the Haaland formula. def f_Haaland(Re, e_D): return (-1.8np.log10(6.9/Re+(e_D/3.7)1.11))-2 # Create a function for solving the Colebrook formula. def f_Colebrook(Re, e_D): # The first guessed value for the friction factor uses the Haaland formula. fprime = f_Haaland(Re, e_D) # Set the relative difference between f and fprime large initially # to start the loop. freldiff = 1 # Set the acceptable tolerance to be <= 0.1 percent. tol = 0.001 # Only go up to 1000 iterations. max_counter = 1000 # Initialize the counter. counter = 0 # Loop until the relative difference is less than the tolerance or # until the maximum number of iterations is reached. while ((freldiff > tol) and (counter < max_counter)): counter = counter + 1 # Update the counter. Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics C. Wassgren 1040 2021-12-15 pipe_50 Page 2 of 3 f = fprime # Set f equal to fprime for solving the Colebrook formula. # Colebrook formula. fprime = (-2.0np.log10(e_D/3.7 + 2.51/Re/np.sqrt(f)))-2 # Calculate the relative difference freldiff = np.absolute((fprime-f)/f) # The maximum number of iterations was reached. Print a warning # and exit the program. if (counter == max_counter): print("The maximum number of iterations was reached. Did not converge on a value for f.") exit(1) # Return the converged value for the friction factor. return f # Make arrays of Reynolds numbers and relative roughnesses values. Re = np.geomspace(4e3, 1e8, num=100) e_D = np.geomspace(1e-6, 1e-2, num=10) # Create an array of friction factor values the same size as the # Reynolds number array. Initialize the array with zero values. f = np.full_like(Re,0) # Calculate the friction factor for all combinations of the Reynolds # numbers and relative roughnesses. for j in range(len(e_D)): for i in range(len(Re)): f[i] = f_Colebrook(Re[i], e_D[j]) # Create some text for the plot legend. legendtext = '$\epsilon/D = %.3e$' % e_D[j] # Plot these friction factors as a function of Reynolds number for # this particular relative roughness. plt.plot(Re, f, "-", label=legendtext) # Label the plot, change the axes to be logarithmic, and show the legend. plt.xlabel(r"Reynolds number, $Re_D$") plt.ylabel(r"friction factor, $f$") plt.xscale("log") plt.yscale("log") plt.legend() plt.show() Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics C. Wassgren 1041 2021-12-15 pipe_50 Page 3 of 3 Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics C. Wassgren 1042 2021-12-15 Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics 11.5. Other Losses The loss due to the viscous resistance caused by the pipe walls is referred to as a major loss. Pressure losses may occur due to viscous dissipation resulting from fluid interactions with other parts of a pipe system such as valves, bends, contractions/expansions, inlets, and connectors. These losses are known as minor losses. The names can be misleading since it’s not uncommon in pipe systems to have most of the pressure loss resulting from the minor losses, e.g., a pipe system with a large number of bends and valves, but short sections of straight pipe. What causes these minor losses? The pressure loss results primarily from viscous dissipation in regions with large velocity gradients, such as in a recirculation zone as shown in Figure 11.7. Figure 11.7. A sketch showing where viscous losses occur in a sudden pipe expansion. A closely related phenomenon known as the vena contracta acts to e↵ectively reduce the diameter at entrances and bends (Figure 11.8). The recirculation zone also results in a pressure loss. Figure 11.8. A sketch illustrating a vena contracta. Although minor loss coefficients can be determined analytically for certain situations, most frequently the loss coefficient for a particular device is found experimentally. Essentially, one measures the pressure drop across the device, ∆p, and forms the loss coefficient, k, using, k = ∆p 1 2⇢¯ u2 , (11.41) where ⇢is the fluid density and ¯ u is the average speed through the device. Many tables with experimentally determined loss coefficients are available. Notes: (1) When using a loss coefficient, it is important to know what velocity has been used to form the coeffi-cient. For example, the loss coefficient for a contraction is typically based on the speed downstream of the contraction, while the loss coefficient for an expansion is based on the speed upstream of the expansion. (2) Minor losses are sometimes given in terms of equivalent lengths of pipe. An equivalent minor loss of 10 pipe diameters worth of a particular type of pipe means that the major loss caused by a pipe of that type, 10 diameters in length will give the same pressure loss as the minor loss. Thus, a loss coefficient and equivalent pipe length, Le, can be related by, k = fD ✓Le D ◆ . (11.42) C. Wassgren 1043 2021-12-15 Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics (3) For non-circular pipes or pipes that are not completely filled, the same methods of determining the friction factor and loss coefficients are used, except that a hydraulic diameter, Dh, is used in place of the diameter. The hydraulic diameter is defined as, Dh := 4A Pw , (11.43) where A and Pw are the cross-sectional flow area (not necessarily the pipe cross-sectional area) and wetted perimeter of the pipe, i.e., the part of the pipe that is in contact with the fluid. For example, consider a completely filled square pipe of side length L (Figure 11.9). The hydraulic diameter for such a pipe is, Dh = 4(L2) 4L = L. (11.44) Now consider a completely filled annular pipe with outer diameter Do and inner diameter Di Figure 11.9. A square cross-sectioned pipe filled with fluid. (Figure 11.10). The hydraulic diameter for this case is, Dh = 4 & ⇡ 4 D2 o −⇡ 4 D2 i ' ⇡Do −⇡Di = D2 o −D2 i Do + Di = Do −Di. (11.45) Now consider a half-filled circular pipe of diameter D (Figure 11.11). The hydraulic diameter for Figure 11.10. An annular cross-sectioned pipe filled with fluid. this case is, Dh = 4 & 1 2 ⇡ 4 D2' & 1 2⇡D ' = D. (11.46) Notes: Figure 11.11. A circular cross-sectioned pipe half filled with fluid. (a) Often a hydraulic radius, Rh, is used instead of a hydraulic diameter for flows in conduits with a free surface. The hydraulic radius is defined as, Rh := A Pw . (11.47) Using this definition, Dh 6= 2Rh, but is instead, Dh = 4Rh, which can be confusing. The Manning Formula (not covered in these notes) is frequently used in the analysis of free surface conduit flows. C. Wassgren 1044 2021-12-15
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Planning Week: Your 2025 guide to incentive & sales planning. Register now.  Intro text here, With custom blockquotes, I can add a bunch of optional fields. There's the intro text, and all sorts of information about the author. The coolest part is that any element is optional! Author N. Head of Placeholders Massa tincidunt dui ut ornare. Habitasse platea dictumst vestibulum rhoncus est pellentesque elit ullamcorper dignissim  Named a Strong Performer in “The Forrester Wave™: Sales Performance Management Platforms, Q1 2023” Tour the Product Explore our new thought leadership hub for all things incentive compensation management See what's new Sales Commission Calculator (Plus Formulas for Common Structures!) Is the art of sales commission calculations leaving you perplexed? We can help take away some of that frustration and break it down, providing simple or more complex formulas for when your sales team needs to get paid. Book a Meeting Do you dread the end of your sales cycle when it's time to calculate commissions? Calculating sales commissions can be a daunting task, especially if you're trying to figure out complex sales structures and rewards programs. Not only are these figures critical in determining performance incentives, but they also provide valuable insight into sales trends within an organization's portfolio. To make these calculations more bearable, sales teams should consider using automation tools that can quickly and accurately compute sales commission figures. This will help save time, reduce stress levels, and ensure the accuracy of sales information. Don’t have a spreadsheet template or automated system in place to handle sales commission calculations? Fear not. We’ll cover the basics in this guide and provide formulas for common commission structures to help get you started on those manual calculations. Let’s get crunching! Common Sales Commission Structures: An Overview There is no single commission structure that works best for all teams. As a manager, it’s up to you to choose the one that best supports business needs and motivates reps for the foreseeable future. Some of the most common sales commission structures include: Base salary plus commission: Reps receive a base salary independent of quota in addition to commissions for each sale. Commission only: Reps only receive commission payments for the sales they make. Tiered commission: Reps earn a variable commission rate depending on how much they’ve sold, with different sales performance tiers offering different payouts. Draw against commission: Reps receive a set amount of their monthly commission payments upfront that they must repay through future commission payouts. Territory volume commission: Reps earn commissions based on the total sales for their region(s). In shared territories with more than one rep, the commissions may be divided between them. Gross profit commission: Reps earn commissions based on the profit from products and services rather than the total amount sold. Multiplier commission: Reps earn a standard commission based on the percentage of each sale, but this amount is multiplied by factors like exceeding quota or selling special products. As a sales manager, you may use a combination of these commission structures to incentivize different behaviors and outcomes within your team. For example, you might offer a base salary plus commission to provide stability for new hires while using tiered commissions to motivate experienced reps to exceed their quotas. How to Calculate Sales Commissions At a fundamental level, there is the sales commission rate: the percentage of a sales price paid to an individual for their efforts in selling a product or service. This rate is typically based on the amount of revenue or profit generated by the sale, and it can vary depending on the industry, company, and type of sale. Basic Formula (Straight Commission) The basic formula for a percentage-based sales commission: Total Sales ($) x Commission Rate (%) = Total Commission ($) Example: A salesperson receives a 5% commission on all sales they make, so if they sell $10,000 worth of products in a month, they will earn an additional $500 in commission. This commission is in addition to their regular salary or wages, so if they make $2,000 per month, their total monthly earnings would be $2,500. But again, that’s the basic formula. There are more! Before we go into some of the different commission structures, know that all are variable commissions. Oxford Languages defines a variable as “an element, feature, or factor that is liable to vary or change.” Variable commission is a “catch-all” for any type of commission based on the amount of revenue or products/services sold rather than a fixed salary. As this infographic highlights, more than half of reps receive 25 to 49% variable pay. However, they’d prefer that number to be at least a 50/50 split (or even higher variable proportion). The good news? Most managers (69%) agree! So, what are the common types of variable sales commissions? Graduated or tiered commissions: This type of commission structure rewards salespeople for reaching specific sales targets, with higher commissions paid for higher levels of sales. Gross profit (or gross margin) commission: Commission is calculated based on the percentage of the gross profit margin of each sale. Commission bonuses: Some companies offer bonuses or incentives for salespeople who exceed sales targets or milestones. In short: Different commission rate formulas can be used, such as graduated commission rates that pay commissions based on sales tiers, gross profit commission rates that pay commissions depending on the amount of profit a product brings in, and bonus commission rates that are paid when specific sales goals are met. Each commission rate formula has pros and cons, so it’s important to understand how each type works before deciding which rate is best for a business. Some commission structures have a maximum amount of commission that a salesperson can earn, called commission caps. With commission caps, businesses can adjust their commission structures as needed to make sure they stay within budget. For example, a salesperson might have a commission cap of $500 per month, so even if they make more than $10,000 in sales, they would still only earn $500 in commission. Now, let’s look at each of these commission rate examples. Graduated or Tiered Commission In a graduated or tiered (often used interchangeably) commission structure, the percentage of commission you earn increases as you sell more products. Multiple commission levels or "tiers" correspond to different sales volume milestones. Are you looking to motivate your sales team? Hit certain sales milestones? Launch a new product or service? Enter a new market? If so, a graduated commission structure may be a good option. This model is often used in industries where salespeople have the potential to earn significant amounts of money through commissions, such as real estate, automobile, insurance, and financial services. Here’s an example of what this could look like: For sales between $0 and $10,000, the commission rate is 5% For sales between $10,001 and $20,000, the commission rate is 7% For sales over $20,001, the commission rate is 10% As you can see, there’s an incentive to sell more because the commission rate increases at each level. So, if you crack the $10,000 level, you’ll bump to a 7% commission rate. There are often “additional incentives” associated with a tiered commission structure — bumps for selling new products or services, selling add-ons, upgrades, and so on. In layman’s terms, the formula for calculating tiered commission is multiplying the salesperson's sales by the appropriate percentage for each tier. Using the example above, let’s say they made $25,000 in sales. Commission is calculated as follows: For the first $10,000 in sales, the commission is $10,000 x 5% = $500 For the next $10,000 in sales (between $10,000 and $20,000), the commission is $10,000 x 7% = $700 For the remaining $5,000 in sales (over $20,000), the commission is $5,000 x 10% = $500 The total commission is $500 + $700 + $500 = $1,700 To calculate tiered commission in Excel, you can use the following formula: =SUM(IF(A1:A10<=B1,A1:A10C1,0), IF(A1:A10>B1,(A1:A10-B1)D1,0), IF(A1:A10>B2,(A1:A10-B2)E1,0)) In this formula, A1:A10 is the range of cells containing the sales amounts, B1 is the upper limit of the first tier, C1 is the percentage for the first tier, D1 is the percentage for the second tier, B2 is the upper limit of the second tier, and E1 is the percentage for the third tier. Again, using the example above, you could use the following formula: =SUM(IF(A1:A10<=10000,A1:A100.05,0), IF(A1:A10>10000,(A1:A10-10000)0.07,0), IF(A1:A10>20000,(A1:A10-20000)0.1,0)) This formula will calculate the commission for each sale in the range A1:A10, based on the tiered commission structure. The SUM function is then used to add the commissions for all the sales in the range to give the total commission earned. Gross Profit Commission As indicated by this commission structure, gross profit (or gross margin) is tied to the profitability of a specific sale. Not all sales are created equal. Some have higher profit margins compared to others. If you sell more high-profit margin deals, you’ll make more commission. More low-profit margin deals ... less commission. Note: the commission rate is usually the same, but the total commission will vary. A gross profit commission calculator could look like this: Revenue ($) - Costs ($) = Gross Profit Margin ($) x Commission Rate (%) = Total Commission ($) Example: You close a $100,000 deal. However, the cost of that business was $10,000. The gross profit margin is $90,000. If you receive a 10% commission, that would equal $9,000 in commission. Note: If this were a “straight” commission model, you’d earn $10,000 in commission (10% of $100,000). To calculate gross profit commission in a spreadsheet, you can use the following formula: =(SUM(A1:A10)-SUM(B1:B10))C1 In this formula, A1:A10 is the range of cells containing the sales amounts, B1:B10 is the range of cells containing the cost of goods sold, and C1 is the commission percentage. So, if you want to calculate a 10% commission on the gross profit of sales in the range A1:A10, with the cost of goods sold in the range B1:B10, you could use the following formula: =(SUM(A1:A10)-SUM(B1:B10))0.1 This formula will first calculate the gross profit for each sale in the range A1:A10, by subtracting the cost of goods sold from the sales amount. It will then multiply the gross profit by the commission percentage to calculate the commission for each sale. The SUM function is then used to add up the commissions for all the sales in the range to give the total commission earned. Commission Bonuses Commission bonuses are typically based on performance or delivery of expected objectives and results, giving employees a tangible reward for their hard work beyond a salary increase. They are typically leveraged to motivate employees to focus on important areas of the business and drive positive results. For example, a salesperson might earn a $1,000 bonus if they exceed their sales target by a certain amount. To calculate a commission bonus in a spreadsheet, you can use the following formula: =IF(A1>B1,C1,0) In this formula, A1 is the sales amount, B1 is the threshold sales amount at which the bonus is earned, and C1 is the bonus amount. For example, if you want to calculate a $1,000 bonus for sales over $20,000, you could use the following formula: =IF(A1>20000,1000,0) This formula will check whether the sales amount in cell A1 is greater than $20,000. If it is, the formula will return the bonus amount of $1,000. If it is not, the formula will return 0. You can use this formula for each sales amount in a range of cells and then use the SUM function to add up the bonuses for all the sales in the range to give the total bonus earned. For example, if you have sales amounts in the range A1:A10, you could use the following formula to calculate the total bonus earned: =SUM(IF(A1:A10>20000,1000,0)) This formula will apply the IF formula to each sales amount in the range and then use the SUM function to add up the bonuses for all the sales in the range to give the total bonus earned. Revenue-Based Commission A revenue‑based commission is one of the simplest structures—sales reps earn a commission purely as a percentage of the revenue generated. For example, say a sales rep receives a 10% commission on the revenue generated from their sales actions. If they sell products that produce $50,000 in revenue, they will see a $5,000 commission on their next payment because revenue multiplied by the commission rate equals the commission. However, you may want to account for things like revenue from multiple product lines, the possibility of returns or cancellations, or even splitting revenue among several reps. For example, if you’re tracking revenue per transaction or per product category, you might calculate commissions on each revenue line before summing them up. The basic formula is simple: Revenue ($) x Commission Rate (%) = Total Commission ($) If you have revenues from different sources (say, Regions or Product Lines), you could write: (Rev1×Rate1) + (Rev2×Rate2)+…= Total Commission ($) In Excel, if cell A2 contains revenue from Product Line 1 and B2 holds the commission rate (expressed as a decimal), then the commission for that line is: =A2B2 If you have multiple lines (say, revenue in cells A2 through A10 and a common commission rate in cell B2), you might use: =SUM(A2:A10)B2 Or if each product has its own commission rate in B2:B10: =SUMPRODUCT(A2:A10, B2:B10) Residual Commission Not all products or services are one-time sales. Subscriptions or Software-as-a-Service (SaaS) services are billed to customers regularly, typically every month or every year. In a residual commission agreement, sales reps earn a commission each time the customer is billed. This commission formula recognizes future potential, so reps who keep customers on longer earn more than reps who make a sale but can’t keep clients. It’s one of the few commission structures directly tied to customer lifetime value (LTV) and recurring revenue. Let’s take the example of a sales rep who brings on a new SaaS client. The software's monthly subscription cost is $5,000. If the rep’s commission is 5%, they can earn $250 monthly without pitching another lead or closing a new sale. They only need to keep that existing client happy and maintain their agreement to be billed for future months. Here’s the formula: Monthly Revenue ($) x Commission Rate (%) = Total Commission ($) For this formula, the monthly revenue ($5,000) gets multiplied by the commission rate (5%) to get the residual commission earned each month ($250). In a spreadsheet, simply put the following formula into cell C2: =A2B2 In this case, cell A2 represents the monthly revenue amount, and B2 represents the commission percentage. Multiply them to get the commission earned, showing up in cell C2. Accelerated Commission An accelerated commission structure works like a standard, tiered commission but with accelerators that reward higher performance milestones. Rather than simply paying a different flat commission rate for each tier, it may only reward higher performance above certain sales quotas but at a much more aggressive rate. It’s designed to reward highly motivated reps to drive their performance even higher over time. So, if a rep earns a base commission of 10% but has an accelerated rate of 15% on all sales above their $100,000 monthly quota, they could see payouts like this: For the first $100,000 (under quota), they get 10%, or $10,000. But on the next $20,000 (above quota), they earn 15%, or $3,000. The Excel formula would look something like this: =IF(Revenue > Threshold, (Threshold Base_Rate) + ((Revenue - Threshold) Accelerated_Rate), Revenue Base_Rate) For the example we shared above, you could put the following formula into the E2 cell: =IF(A2>B2, (B2C2)+((A2-B2)D2), A2C2) So, a revenue of $120,000 would be entered into the A2 cell, and the threshold of $100,000 would be entered into the C2 cell. Next, add the 15% accelerated rate in cell D2. Your result of the $13,000 commission should show up in cell E2. Territory-Based Commission Territory-based commission, also called a territory volume commission, rewards sales reps based on the total sales volume for a territory, regardless of their role in revenue. This method promotes teamwork among reps, as it pays everyone the same amount for shared territories. It can also be used for territories with just one rep or for reps with more than one territory at a time. Let's say a rep manages Region A, which had $30,000 in sales last month, and Region B, which had $20,000. If the same rep has an 8% commission rate, they will earn $2,400 and $1,600, respectively, for a total of $4,000. (If this territory is shared between two reps, they may both earn that commission amount or divide it, depending on how their sales compensation plan is set up.) Here is an example formula for territory sales in an Excel spreadsheet: =SUM(Sales_Region) Commission_Rate In our example above, the total of all territory earnings (SUM)(Sales_Region) gets multiplied by the commission rate to get our $4,000 earnings. For our Excel spreadsheet, we’ll put the following formula in cell D2: =(A2+B2)C2 In this formula, Region A’s earnings are put into cell A2, Region B’s into B2, and so on to total them together. This total amount is multiplied by the amount in C2, or the commission rate. The total commission appears as the product in D2. Draw Against Commission In a draw against commission, sales reps receive a set amount of their commission before they earn it. If they fail to meet the quota needed to cover what they’ve been paid, the overdrawn amount will come from future commissions. For example, say a rep earned $2,500 in commissions for the month but was already paid $2,000 as a set monthly draw against those commissions. They will receive the remaining $500 in commissions at the next payout. The formula for drawing against commissions looks like this: =MAX(Commission - Draw, 0) For our example, the rep’s $2000 already-paid commissions (or draw) get subtracted from the earned commissions to find the amount left to be paid. In an Excel document, the below formula would be entered into cell C2: =MAX(A2-B2, 0) The commission earned goes into cell A2, with the draw amount in cell B2. The resulting commission left to be paid then appears in cell C2. Also, the converse can happen when a sales rep doesn’t make their quota. For example, a rep may only earn a $2000 commission (A2) but take a $2500 draw (B2). In this case, the draw is more than the commission, resulting in -$500 in the C2 cell. Automate Sales Commission Calculations With CaptivateIQ While most sales commission calculations might not seem too complicated, they quickly add up, as you have to input, monitor, and adjust Excel sheets constantly. As you scale, the complexity grows exponentially with more sales reps, varied commission structures, and multiple product lines to track. CaptivateIQ simplifies the entire process. It integrates with your existing tech stack (CRM, ERP, HRIS) to automatically pull in sales data and calculate commissions in real-time. Using an intuitive visual interface, you can easily design and modify commission plans, from simple percentage-based structures to complex multi-tier systems with accelerators and special incentives. The result is a transparent, error-free commission process that saves hours of admin work while keeping your sales team motivated. Sales reps get instant visibility into their earnings through personalized dashboards, while sales leaders and finance teams get powerful analytics to optimize compensation plans and forecast costs. No more spreadsheet nightmares—just accurate, automated commission management that scales with your business. CaptivateIQ helped us automate portions of our commission processes, freeing up time to focus on streamlining and exploring new ways to manage the parts that couldn't be automated. A process that was once completely manual and took 5-6 hours a month to prepare statements has been trimmed to 2.5 hours each month. — Lynn Bell, Vice President, Revenue Enablement, DataBank IMX Ready to get started? Sign up for a demo today. FAQ What is a sales commission? A sales commission is a variable payment paid to sales reps based on performance. Reps can earn more or less depending on how much they sell. Most sales commissions are paid as a percentage of revenue or profit made from each sale, motivating them to perform better and earn more over time. How do I choose the right commission structure? Picking the right commission structure depends on what you sell, your company’s business model, and your commission budget. Ideally, you want to choose a structure that encourages your sales teams to perform, but you must balance this with your cost to reward them. It should also be considered part of your company's larger sales compensation plan, which can be planned, tested, and optimized easily using an incentive compensation management (ICM) tool like CaptivateIQ. How often should commissions be paid? Commissions should be paid as often as needed to motivate reps and meet your company's revenue goals. Larger companies that take longer to get client payments or who need more liquidity may opt for more time between commission payments (monthly or quarterly). Smaller companies or those with shorter sales cycles may prefer to pay as often as bi-weekly. Reps with lower base salaries or who are paid on a higher, commission-only basis may need to be paid more often than those with a generous base salary and lower commissions, as they rely on those commissions to meet monthly financial obligations. What’s the best way to track sales commissions? Modern sales organizations are moving away from manual spreadsheets to automated ICM systems. The best practice is to implement a dedicated ICM platform that automatically pulls data from your CRM and other systems, calculates commissions in real-time, and provides transparency to all stakeholders. Your ICM should handle the heavy lifting of commission tracking by automating calculations, providing real-time visibility into earnings, validating data accuracy, and generating detailed reports. When properly set up, the system can manage complex commission structures, track multiple quotas and goals, and scale effortlessly as your team grows. How can I improve my sales team’s performance with commissions? Any type of commission can encourage better performance, but some commission structures motivate better than others. Those that work well for top-performing reps include: Tiered commission structures that start with lower commission rates and increase as reps sell more. Multiplier commission structures that increase the rate as reps engage in specific behaviors, like meeting 1.5x the quota or closing higher-valued tickets. One agile platform from planning to payout Talk to our sales performance experts to learn how you can make sales planning and compensation a strategic growth driver.
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https://impactfactor.org/PDF/IJPCR/16/IJPCR,Vol16,Issue5,Article188.pdf
e-ISSN: 0975-1556, p-ISSN:2820-2643 Available online on www.ijpcr.com International Journal of Pharmaceutical and Clinical Research 2024; 16(5); 1116-1123 Arasi et al. International Journal of Pharmaceutical and Clinical Research 1116 Original Research Article Evaluation of ER, PR, P53 and HER2 Neu Biomarkers in Endometrial Cancers Arasi Rajesh1, Dina Mary M2, Maheswari S3 1Professor and Head, Department of Pathology, Government Thoothukudi Medical College, Thoothukudi 2Associate Professor, Department of Pathology, Government Tirunelveli Medical College, Tirunelveli 3Associate Professor, Department of Pathology, Government Thoothukudi Medical College, Thoothukudi Received: 25-02-2024 / Revised: 23-03-2024 / Accepted: 26-04-2024 Corresponding Author: Dr. Maheswari S Conflict of interest: Nil Abstract: Introduction: Endometrial cancer (EC) is the most common gynecological malignancy. Factors reported to be predictive of response to endocrine therapy include low grade, endometrioid histology, and positive estrogen receptor (ER)/progesterone receptor (PR) status. The prognostic value of ER/PR is well established, with higher levels of ER and PR expression associated with longer overall Survival, longer cancer-specific survival, and longer Progression Free Survival. Methods: In our study, a panel of immunohistochemical markers ER, PR, Her-2, and p53 were done in 50 cases of endometrial carcinoma and their relationships with the histopathological and prognostic parameters were analysed. Results: in our study ER, PR expression was noted commonly in Grade I, Grade II endometrioid carcinoma whereas Type II endometrial carcinoma was negative for these markers. p53 mutation and Her2 neu overexpression was found commonly in Type II endometrial carcinoma. Conclusion: The absence of hormone receptors, Her2 Neu overexpression, p53 mutation indicates aggressive tumor and poor prognosis. Keywords: endometrial carcinoma, ER, PR, HER2 NEU, P53. This is an Open Access article that uses a funding model which does not charge readers or their institutions for access and distributed under the terms of the Creative Commons Attribution License ( and the Budapest Open Access Initiative ( which permit unrestricted use, distribution, and reproduction in any medium, provided original work is properly credited. Introduction Endometrial cancer (EC) is the most common gynecological malignancy in developed countries like USA. In India, it ranks third after Carcinoma cervix and Carcinoma ovary. The burden of endometrial cancer is increasing worldwide and hence there is increased need to investigate its causes to improve prevention and for early diagnosis and treatment. The escalation in the number of women entering menopause in addition to risk factors, such as obesity and diabetes, may explain a fraction of the increased incidence of endometrial cancer 3. The median age at diagnosis is 61 years with approximately 85% of the cases being diagnosed after 50 years of age. Consequently, this is generally a disease of postmenopausal women Most cases are diagnosed in early stages owing to the clinical symptoms of postmenopausal bleeding and abnormal discharge. Endometrial carcinomas are divided into two broad histologic types. Type 1 includes Endometrioid and mucinous carcinoma accounting for about 80% of the cases wherein there is unopposed estrogen stimulation and is associated with precursor lesions such as Atypical Endometrial Hyperplasia (AEH)/Endometrial Intraepithelial Neoplasia (EIN). Type I tumor presents with low tumour grade and show distinct genetic abnormalities such as PTEN, PAX2and k-ras mutation. Type 2 includes Serous Carcinoma, Clear Cell Carcinoma, undifferentiated carcinoma and carcinosarcoma accounting for about 10% of the cases, less associated with estrogen stimulation, presenting with higher tumour grade and stage. Serous Carcinoma exhibit early TP53 mutations and serous intraepithelial carcinoma is proposed as its preinvasive precursor The Undifferentiated Endometrioid Carcinoma (UEC) is a solid-pattern tumor without specific morphologic evidence of epithelial differentiation. It has an aggressive growth pattern and tends to be diagnosed at an advanced International Federation of Gynecology and Obstetrics (FIGO) stage and is resistant to conventional chemotherapy. Dedifferentiated Endometrioid Carcinoma (DEC) is characterized by the coexistence of low-grade EC and UEC. DEC has not been widely recognized due to its solid part usually being misdiagnosed as a International Journal of Pharmaceutical and Clinical Research e-ISSN: 0975-1556, p-ISSN: 2820-2643 Arasi et al. International Journal of Pharmaceutical and Clinical Research 1117 grade 3 EC, but has a worse outcome than grade 3 EC. In a recent population-based study using the National Cancer Database of the United States (2004–2013), 1.1% of all ECs met the criteria of UEC, which may reflect an underdiagnosed in earlier years . Loss of PAX 8, E-cadherin ER and PR, focal expression of cytokeratin, and EMA can support a diagnosis of undifferentiated/ dedifferentiated carcinoma over Grade 3 or Grade 2 endometrial carcinoma. The dedifferentiated rhabdoid variant is characterised by the presence of an undifferentiated component which shows rhabdoid cells embedded in myxoid stroma. This variant is often misdiagnosed as a Mixed Mullerian tumour (MMMT). UEC harbors specific genetic features different from endometrioid carcinoma. Generally, endometrioid carcinoma is a hormone-dependent tumor that expresses hormone receptors that may respond to hormone therapy. Unfortunately, UEC seldom has detectable hormone receptors and its tumorigenesis pathway has distinct features, such as microsatellite instability (MSI-H)/mismatch repair (MMR) protein and the genomic inactivation of core components of the SWI/SNF chromatin-remodeling complex . In the regular progression of the menstrual cycle, the lining of the uterus is subject to pair of steroid hormones, estrogen and progesterone, that each exerts an opposing effect on the endometrial glandular epithelium. In particular, estrogen has a mitogenic effect that drives the proliferation of the endometrial epithelium via Estrogen Receptor (ER). Left unopposed, estrogen can lead to the rapid onset of endometrial hyperplasia and consequently, the development of Endometrial Carcinoma. Progesterone, however, acts as an antagonist to estrogen by down regulating ER expression, inhibiting active cell division, and promoting cell differentiation through Progesterone Receptor (PR). As the endometrium expresses both ER and PR, the lining of the uterus is highly sensitive to hormone activity. Therefore, any shift to the endocrine balance in favor of high estrogen level will ultimately stimulate oncogenesis. Such overexposure to estrogen arises in the majority of type I tumors. This is high risk factor among women undergoing estrogen-only hormonal therapy, using tamoxifen as adjunct therapy for breast cancer and in obese women as adipose tissue releases estrone, which is converted into estradiol in the uterus. HER2, a well characterized oncogene in the pathogenesis of breast cancer, has also been implicated as a potential biomarker for type II endometrial tumors. Overexpression of HER2 results in sustained cell proliferation via constitutive activation of the kinase domain in a ligand-independent manner. HER2 expression is mostly associated with a poor prognosis in type II lesions. Recent studies suggest HER2 overexpression is also found in advanced and recurrent type I endometrioid Carcinomas. Hormone receptor status may therefore be a valuable prognostic marker for EC development and progression. The tumor suppressor gene p53 is activated in response to various stress signals in the cell and it acts on several pathways leading to inhibition of growth, cell cycle arrest and apoptosis. In type II endometrial cancers, the most common mutation identified have been in p53 with mutations Iin over 90% of serous carcinomas compared with only 20% of type I cancers. Some studies suggest that loss-of-function mutations in p53 may be an early event in serous carcinogenesis since it is found in approximately 75% of precursor lesions. In addition to the association with type II histology, p53 mutations are also associated with poor clinical outcome3.In a multivariate analysis adjusting for histology grade, FIGO stage and lymph nodes metastasis, there was an 11-fold increased risk of death in patients with p53 mutations compared to those without. Factors reported to be predictive of response to endocrine therapy include low grade, endometrioid histology, and positive estrogen receptor (ER)/progesterone receptor (PR) status The prognostic value of ER/PR is well established, with higher levels of ER and PR expression associated with longer overall Survival, longer cancer-specific survival, and longer Progression Free Survival. In our study, a panel of immunohistochemical markers ER, PR, Her-2, and p53 were done in 50 cases of endometrial carcinoma and their relationships with the histopathological and prognostic parameters were analysed Aim of the study: 1. To evaluate the expression of ER, PR, Her2neu, p53 in endometrial adenocarci-nomas. 2. To analyse ER, PR, Her-2neu & P53 ex-pression with histological Type and grade. Materials and Methods The study includes analysis of 50 cases of endometrial carcinoma received in the Department of Pathology, Government Tirunelveli Medical College, Tirunelveli from 2018- 2020. This study included 50 histopathologically confirmed cases of endometrial carcinoma diagnosed from endometrial biopsy or hysterectomy specimens. The biopsy tissue and hysterectomy specimens were fixed in 10% formalin solution. Sections were stained with routine Haematoxylin and Eosin (H and E) and were examined under the microscope For IHC, International Journal of Pharmaceutical and Clinical Research e-ISSN: 0975-1556, p-ISSN: 2820-2643 Arasi et al. International Journal of Pharmaceutical and Clinical Research 1118 sections of 4 microns thickness were cut, immunohistochemical staining for ER, PR, HER-2/neu and p53 was performed. Her2 score was assessed using guidelines by the American Society of Clinical Oncology and the College of American Pathologists. Scoring of HER2/neu was done as follows: Score 0: no immunostaining/membrane staining in less than 10% of neoplastic cells; Score 1+: weak staining in more than 10% of neoplastic cells in only portions of the membrane; Score 2+: weak/ moderate circumferential membranous staining in >10% of tumor cells; Score 3+: strong complete membranous staining in more than 10% of tumor cells. Meanwhile, scores equal to 2+ and 3+ were considered as HER2/neu positive. HER2 expression was scored on epithelial component in carcinosarcoma case. Figure 1: Photomicrograph showing ER expression (40X) Figure 2: Photomicrograph showing PR expression (40X) Figure 3: Photomicrograph showing Her2 overexpression (40 X) International Journal of Pharmaceutical and Clinical Research e-ISSN: 0975-1556, p-ISSN: 2820-2643 Arasi et al. International Journal of Pharmaceutical and Clinical Research 1119 Figure 4: Photomicrograph showing mutant p53 (40 X) Results: Total of 50 cases were included in our study, of which majority of cases (38%) were in the age group of 51-60 years followed by 61-70 years (28%). Only one patient was diagnosed before 40 years. Mean age at diagnosis is 56.5 years. Table 1: Age Distributions of Patients Age in years No of patients Percentage <40 1 2% 40-50 12 24% 51-60 19 38% 61-70 14 28% >70 4 8% Total 50 100% Among 50 cases studied 42 cases (84%) were endometrioid carcinoma, of which 21 cases were grade I, 18 cases were grade II, 3 cases were grade III. The other 16 % included 8 cases of serous carcinoma, 2 cases of clear cell carcinoma, 1 case of carcinosarcoma and 2 cases of dedifferentiated carcinoma. Table 2: Histological Type and Grade Tumor Type No of cases Percentage Endometrioid carcinoma grade I 21 42% Endometrioid carcinoma grade II 18 36% Endometrioid carcinoma grade III 3 6% Serous carcinoma 3 6% Clear cell carcinoma 2 4% Carcinosarcoma 1 2% Dedifferentiated carcinoma 2 4% Total 50 100% Figure 5: 86% of grade I endometrioid carcinoma, 89% of grade II endometrioid carcinoma and 33% of garde III endometrioid carcinoma showed positivity for ER. All cases of non endometrioid cancers were ER negative. histologic type endometrioid serous clear cell carcinosarcoma UDC/DDC International Journal of Pharmaceutical and Clinical Research e-ISSN: 0975-1556, p-ISSN: 2820-2643 Arasi et al. International Journal of Pharmaceutical and Clinical Research 1120 Table 3: ER Expression in Tumors Tumor ER Positive ER Negative Total Endometrioid carcinoma grade I 18(86%) 3(14%) 21(100%) Endometrioid carcinoma grade II 16(89%) 2(11%) 18(100%) Endometrioid carcinoma grade III 1(33%) 2(67%) 3(100%) Serous carcinoma 0 3(100%) 3(100%) Clear cell carcinoma 0 2(100%) 2(100%) carcinosarcoma 0 1(100%) 1(100%) Undifferentiated/ dedifferentiated carcinoma 0 2(100%) 2(100%) Total 35(70 %) 15(30%) 50(100%) Figure 6: 76% cases of grade I Endometrioid carcinoma, 83% of grade II Endometrioid carcinoma were PR positive. All cases of grade III Endometrioidcarcinoma, serous carcinoma, clear cell carcinoma. Carcinosarcoma, dedifferentiated carcinoma was negative for PR. Table 4: PR Expression in Tumors Tumor PR Positive PR Negative Total Endometrioid carcinoma grade I 16(76%) 5(24%) 21(100%) Endometrioid carcinoma grade II 15(83%) 3(17%) 18(100%) Endometrioid carcinoma grade III 0 3(100%) 3(100%) Serous carcinoma 0 3(100%) 3(100%) Clear cell carcinoma 0 2(100%) 2(100%) carcinosarcoma 0 1(100%) 1(100%) Undifferentiated/ dedifferentiated carcinoma 0 2(100%) 2(100%) Total 31(62%) 19(38%) 50(100%) 0 5 10 15 20 25 Endometrioid GI Endometrioid GII Endometrioid GIII Serous Clear cell carcinosarcoma UDC/DDC er negative er positive International Journal of Pharmaceutical and Clinical Research e-ISSN: 0975-1556, p-ISSN: 2820-2643 Arasi et al. International Journal of Pharmaceutical and Clinical Research 1121 Figure 7: In contrast to ER and PR HER2Neu overexpression was commonly found in serous carcinoma (100%), clear cell carcinoma (50%) and carcinosarcoma (100%). Among endometrioid carcinomas grade III tumors show over expression in 67% cases, grade II tumors show over expression in 11% cases and grade I tumors show over expression in 5% cases. Table 5: Her 2 over Expression In Tumors Tumor Her2 (2+&3+) Her2 Negative (0 &1+) Total Endometrioid carcinoma grade I 1(5%) 20(95 %) 21(100%) Endometrioid carcinoma grade II 2(11%) 16(89%) 18(100%) Endometrioid carcinoma grade III 2(67%) 1(33%) 3(100%) Serous carcinoma 3(100%) 0 3(100%) Clear cell carcinoma 1(50%) 1(50%) 2(100%) carcinosarcoma 1(100%) 0 1(100%) Dedifferentiated carcinoma 0 2(100%) 2(100%) Total 10(20%) 40(80%) 50(100%) Figure 8: All cases of serous carcinoma, clear cell carcinoma, carcinosarcoma showed mutant p 53 expression, whereas all cases of undifferentiated carcinoma, grade I and II endometrioid carcinoma showed wild type P53. 33% of grade III endometrioid carcinoma showed mutant P 53 and 67% showed wild type P 53. 0 5 10 15 20 25 Endometrioid GI Endometrioid GII Endometrioid GIII Serous Clear cell carcinosarcoma UDC/DDC PR NEGATIVE PR POSITIVE 0 5 10 15 20 25 Endometrioid G I Endometrioid G II Endometrioid GIII Serous carcinoma Clear cell carcinoma carcinosarcoma UDC/DDC Her 2 negative Her 2 positive International Journal of Pharmaceutical and Clinical Research e-ISSN: 0975-1556, p-ISSN: 2820-2643 Arasi et al. International Journal of Pharmaceutical and Clinical Research 1122 Table 6: P53 Expression in Tumors Tumor P53 mutant P53wild type Total Endometrioid carcinoma grade I 0 21(100%) 21(100%) Endometrioid carcinoma grade II 0 18(100%) 18(100%) Endometrioid carcinoma grade III 1(33%) 2(67%) 3(100%) Serous carcinoma 3(100%) 0 3(100%) Clear cell carcinoma 2(100%) 0 2(100%) carcinosarcoma 1(100%) 0 1(100%) Undifferentiated/ dedifferentiated carcinoma 0 2(100%) 2(100%) Total 7(14%) 43(86%) 50(100%) Discussion We have included 50 cases of endometrial carcinomas in our study with mean age of 56.5 years. Highest incidence of endometrial carcinoma was found in 51 - 60 years in our study. This correlated with studies by Bhawani Shekar et al who reported a mean age of 58.4 years and Afaf.T.Elnashar et al whose study showed mean age of 59.4 years. AMANY SALAMA et al reported a mean age of 59.8 years, Nayar Musfera Abdul Masjeed et al reported mean age of 58.14 years, Samina Wagar et al reported mean age of 58.3 years which also correlated with our study. In our study 84% of cases were endometrioid carcinoma and 16% were non endometrioidcarcinoma. Our results were compared with other studies. Table 7: Study Endometrioid Carcinoma Nonendometrioid Bhawani Shekar et al 66.67% 33.33% Afaf.T.Elnashar et al 74% 26% Jasmine Kaur et al 84% 16% Sunamchoksrijaipracharoen et al 86.1% 13.9% Our Study 84% 16% In our study all cases of nonendometrioid carcinoma, were negative for ER, whereas 86% of grade I Endometrioid carcinoma, 89% of grade II Endometrioid carcinoma and 33% of grade III Endometrioid carcinoma were positive for ER. Similarly PR also showed negativity in all non endometrioid carcinoma and also in grade III endometrioid carcinoma. 76% of grade I and 83% of grade II showed positivity for PR. Overall 70% cases were positive for ER and 62% cases were positive for PR. Our study was comparable with Caifeng Wang Davis A. Tranz et al whose study showed ER positivity in 59.8% of cases and PR positivity in 75% of cases. Jasmine Kaur et al showed ER positivity in 64% of cases and PR positivity in 60% of cases; Nayar Musfera Abdul Masjeed et al reported ER positivity in 60.7% of cases and PR positivity in 64.28% of cases. They also reported all cases of grade III endometrioid carcinoma and all non endometrioid carcinoma were ER, PR negative. Also most ER, PR expression was seen in Grade II endometrioid carcinoma, which was true in our study too. Samina Waqar et al noted PR positivity in 66.1% of cases which correlated with our study, but ER positivity was seen in 55% of cases which was little low compared to our study. Her2Neu overexpression was noted in 100% of cases of Serous carcinoma, Carcinosarcoma and 50% cases of Clear Cell Carcinoma. Among Endometrioid carcinoma grade III showed 67% cases with overexpression. This is comparable with Bhawani Shekar et al whose study also revealed non endometrioid carcinoma was more common with Her 2 Neuoverexpression. Similarly Samina Waqar et al noted Her2 Neu overexpression in grade III endometrioid carcinoma and Serous carcinoma with no cases of Grade I, Grade II tumor showing overexpression. In our study all cases of serous carcinoma, clear cell carcinoma, carcinosarcoma and 33% of grade III endometrioid carcinoma showed muthant p53. Our results were comparable with RAVI M SWAMI et al who also showed 100% p53 positivity in grade III endometrioid and serous carcinoma. Conclusion ER,PR expression decreased with increase in grade, Her-2neu and P53 expression were present in higher grade tumour. ER, PR, Her-2Neu & P53 status if included in pathology reports, will improve the understanding of behaviour of tumors and aid in management of patients. (The absence of hormone receptors, Her2 Neu overexpression, and p53 mutation indicates aggressive tumor and poor prognosis). International Journal of Pharmaceutical and Clinical Research e-ISSN: 0975-1556, p-ISSN: 2820-2643 Arasi et al. International Journal of Pharmaceutical and Clinical Research 1123 References 1. Guan J, Xie L, Luo X, Yang B, Zhang H, Zhu Q, Chen X. The prognostic significance of es-trogen and progesterone receptors in grade I and II endometrioid endometrial adenocarci-noma: hormone receptors in risk stratification. J GynecolOncol. 2019 Jan; 30(1):e13. 2. Shekhar B, Kumar S, Mathur S, Singhal S, Meena J, Sharma DN, Singh N. Evaluation of prevalence and prognostic significance of hu-man epidermal growth factor receptor 2/neuexpression in carcinoma endometrium. International Journal of Molecular and Immu-no Oncology. 2020 Jan 21; 5(1):8-13. 3. Binder PS, Mutch DG. Update on Prognostic Markers for Endometrial Cancer. Women’s Health. 2014; 10(3):277-288. 4. Masjeed NMA, Khandeparkar SGS, Joshi AR, Kulkarni MM, Pandya N. Immunohistochemi-cal Study of ER, PR, Ki67 and p53 in Endome-trial Hyperplasias and Endometrial Carcino-mas. J ClinDiagn Res. 2017 Aug; 11(8):EC31-EC34. 5. Tung HJ, Wu RC, Lin CY, Lai CH. Rare Sub-type of Endometrial Cancer: Undifferentiat-ed/Dedifferentiated Endometrial Carcinoma, from Genetic Aspects to Clinical Practice. Int J Mol Sci. 2022 Mar 30; 23(7):3794. 6. Giordano G, Ferioli E, Guareschi D, Tafuni A. Dedifferentiated Endometrial Carcinoma: A Rare Aggressive Neoplasm-Clinical, Morpho-logical and Immunohistochemical Features. Cancers. 2023 Oct 26; 15(21):5155. 7. Wang C, Tran DA, Fu MZ, Chen W, Fu SW, Li X. Estrogen receptor, progesterone receptor, and HER2 receptor markers in endometrial cancer. Journal of Cancer. 2020;11(7):1693. 8. MacKay HJ, Freixinos VR, Fleming GF. Ther-apeutic Targets and Opportunities in Endome-trial Cancer: Update on Endocrine Therapy and Nonimmunotherapy Targeted Options. Am Soc Clin OncolEduc Book. 2020 Mar;40:1-11. 9. Noha Ed. Hassab Elnaby, M.D. Atem, Mo-hamed N. Salem, M.D. Aaam. Her-2 Neu Ex-pression in Endometrial Carcinoma. The Med-ical Journal of Cairo University. 2018; 860:69-6. 10. Salama A, Arafa M, ElZahaf E, Shebl AM, Awad AAE, Ashamallah SA, Hemida R, Gamal A, Foda AA, Zalata K, Abdel-Hady EM. Potential Role for a Panel of Immuno-histochemical Markers in the Management of Endometrial Carcinoma. J PatholTransl Med. 2019 May; 53(3):164-172. 11. Waqar S, Khan SA, Sarfraz T, Waqar S. Ex-pression of Estrogen Receptors (ER), Proges-terone Receptors (PR) and HER-2/neu recep-tors in Endometrial Carcinoma and their asso-ciations with histological types, grades and stages of the tumor. Pak J Med Sci. 2018 Mar-Apr; 34(2):266-271. 12. Jasmeen Kaur, Anil Suri, Manjit Kaur. A Clinicopathological Study: Expression of ER, PR and HER/2neu in Endometrial Carcinoma. Obstetrics and Gynecology Research.2022; 5: 41-53. 13. Srijaipracharoen S, Tangjitgamol S, Tanvanich S, Manusirivithaya S, Khunnarong J, Thava-ramara T, Leelahakorn S, Pataradool K. Ex-pression of ER, PR, and Her-2/neu in endome-trial cancer: a clinicopathological study. Asian Pac J Cancer Prev. 2010; 11(1):215-20. 14. Swami R M, Lakhe R, Doshi P, Karandikar M N, Nimbargi R, Immunohistochemical study of ER, PR, p53 and Ki67 expression in patients with endometrial adenocarcinoma and atypical endometrial hyperplasia. IP Arch CytolHisto-pathol Res 2020; 5(4):274-279.
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CE Activity | Respiratory Syncytial Virus Infection in Children | Pharmacist PhysicianPhysician Board ReviewsPhysician Associate Board ReviewsCMELifetime CMEFree CMEMATE and DEA ComplianceCPD StudentUSMLE® Step 1USMLE® Step 2USMLE® Step 3COMLEX-USA® Level 1COMLEX-USA® Level 2COMLEX-USA® Level 396 Medical School ExamsStudent Resource Center NCLEX - RNNCLEX - LPN/LVN/PN24 Nursing Exams Nurse PractitionerAPRN/NP Board ReviewsCNS Certification ReviewsCE - Nurse PractitionerFREE CE NurseRN Certification ReviewsCE - NurseFREE CE PharmacistPharmacy Board Exam PrepCE - Pharmacist AlliedAllied Health Exam PrepDentist ExamsCE - Social WorkerCE - Dentist Point of Care Institutional Sales Student Resources Search Sign-upLogin PhysicianPhysician Board ReviewsPhysician Associate Board ReviewsCMELifetime CMEFree CMEMATE and DEA ComplianceCPD StudentUSMLE® Step 1USMLE® Step 2USMLE® Step 3COMLEX-USA® Level 1COMLEX-USA® Level 2COMLEX-USA® Level 396 Medical School ExamsStudent Resource Center NCLEX - RNNCLEX - LPN/LVN/PN24 Nursing Exams Nurse PractitionerAPRN/NP Board ReviewsCNS Certification ReviewsCE - Nurse PractitionerFREE CE NurseRN Certification ReviewsCE - NurseFREE CE PharmacistPharmacy Board Exam PrepCE - Pharmacist AlliedAllied Health Exam PrepDentist ExamsCE - Social WorkerCE - Dentist Point of Care Free CME/CE Respiratory Syncytial Virus Infection in Children Home Pharmacist CE Respiratory Syncytial Virus Infection in Children Overview 4.7 out of 5 (810 Reviews) Credits 1.50 Post-Assessment Questions 13 Start Date 1 Sep 2023 Last Review Date 1 Sep 2023 Expiration Date 31 Aug 2026 Estimated Time To Finish 90 Minutes Start This Activity Unlimited Pharmacist CE Stay up to date on the latest medical knowledge with 6632 CE activities. In these online self-assessment activities, read our reference articles and test your knowledge with more than 7896.5 hours of CE. Single Activity Take this single activity $43.50 1 activity Buy Now 6 Month Unlimited Pharmacist CE Access to all the Unlimited Pharmacist CE activities in all specialties. $89 per half year per user Buy Now 1 Year Unlimited Pharmacist CE Access to all the Unlimited Pharmacist CE activities in all specialties. $139 per 1 year per user Buy Now Need Help?If you have a system or content concerns, please contactsupport@statpearls.com Activity Description The human respiratory syncytial virus (RSV) is one of the most common viruses to infect children worldwide and increasingly is recognized as an important pathogen in adults, especially the elderly. The most common clinical scenario encountered in RSV infection is an upper respiratory infection, but RSV commonly presents in young children as bronchiolitis, a lower respiratory tract illness with small airway obstruction, and can rarely progress to pneumonia, respiratory failure, apnea, and death. This activity reviews the pathophysiology of respiratory syncytial virus infection and highlights the role of the interprofessional team in its management. UAN: JA4008338-0000-23-3951-H01-P Target Audience This activity has been designed to meet the educational needs of physicians, physician associates, nurses, pharmacists, and nurse practitioners. Learning Objectives At the conclusion of this activity, the learner will be better able to: Describe the pathophysiology of respiratory syncytial virus infection. Review the presentation of respiratory syncytial virus infection. Outline the treatment and management options available for the respiratory syncytial virus. Describe interprofessional team strategies for improving care coordination and communication to advance the treatment of respiratory syncytial virus and improve outcomes. Pharmacy Activity Type: Disease/Drug Therapy Related Disclosures StatPearls, LLC requires everyone who influences the content of an educational activity to disclose relevant financial relationships with ineligible companies that have occurred within the past 24 months. Ineligible companies are organizations whose primary business is producing, marketing, selling, re-selling, or distributing healthcare products used by or on patients. All relevant conflict(s) of interest have been mitigated.Hover over contributor names for financial disclosures. Others involved in planning this educational activity have no relevant financial relationships to disclose. Commercial Support: This activity has received NO commercial support. Authors: Hanish Jain Editors: Nathaniel A. 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Reviews Gabrial L. on 5/9/2021 Laura C. on 9/3/2021 Elyshia L. on 10/8/2021 David M. on 1/7/2022 Christine U. on 9/5/2022 Christine G. on 11/2/2022 Noelle R. on 11/3/2022 Sheila V. on 11/4/2022 Casey B. on 11/4/2022 kathleen b. on 11/5/2022 Christina M. on 11/4/2022 Audi A. on 11/4/2022 Robert S. on 11/4/2022 Jennifer C. on 11/4/2022 Morgan P. on 11/4/2022 Jana D. on 11/5/2022 Beranda T. on 11/6/2022 Yvonne C. on 11/6/2022 Amilda D. on 11/6/2022 Thank you Am\ B. on 11/6/2022 Theophilous B. on 11/6/2022 Sarina D. on 11/6/2022 MARIE T. on 11/6/2022 Linda Y. on 12/23/2022 Meghan H. on 11/6/2022 Alison G. on 11/6/2022 S D. on 11/6/2022 Repetition of the topic presentation was an effective way to improve critical thinking skills. Tracy D. on 11/6/2022 Dianne T. on 11/6/2022 Kristin H. on 11/6/2022 Katherine R. on 11/6/2022 good and useful as usual. larry h. on 11/7/2022 Winifred F. on 11/7/2022 Shannon C. on 11/7/2022 Alexandra P. on 11/7/2022 I appreciate free educational materials. -Nurses salaries are limited :) Mary M. on 11/7/2022 Olga W. on 11/7/2022 Kim M. on 11/7/2022 Lori B. on 11/7/2022 Raafat I. on 11/7/2022 Sonja P. on 11/7/2022 Florencia L. on 11/7/2022 andrea w. on 11/7/2022 Mary Ellen C. on 11/7/2022 Mital P. on 11/7/2022 lori d. on 12/5/2022 Catherine B. on 11/7/2022 Ashley C. on 11/7/2022 Erick Roberto . on 11/7/2022 Ryan K. on 11/8/2022 Katherine M. on 11/8/2022 Thank you. Dr.Aditi S. on 11/9/2022 Daie M. on 11/9/2022 Dotan S. on 11/9/2022 David T. on 11/9/2022 Cortney C. on 11/9/2022 Jordan A. on 11/9/2022 Robie O. on 11/9/2022 Katie O. on 11/9/2022 Nancy C. on 11/9/2022 Shauna P. on 11/10/2022 Brenda A. on 11/10/2022 Sarah J. on 11/10/2022 HEATHER O. on 11/10/2022 Briana L. on 11/10/2022 Overall a very nice and complete review of RSV. Julia M. on 11/11/2022 Margaret C. on 11/11/2022 Jeromy P. on 11/11/2022 simply put together. I liked the clinical example questions! Natalie L. on 11/11/2022 Desiree W. on 11/11/2022 Muhammad J. on 11/12/2022 CYNTHIA S. on 11/12/2022 james s. on 11/12/2022 Bryce E. on 11/17/2022 Christine K. on 11/12/2022 Shah Z. on 11/13/2022 Lisa D. on 11/13/2022 Hassen C. on 11/14/2022 Rebecca F. on 11/14/2022 john n. on 11/14/2022 Mala Dasari N. on 11/14/2022 Joyce S. on 11/14/2022 Elizabeth K. on 11/14/2022 Jason M. on 11/14/2022 NGAN Q. on 11/14/2022 Iris P. on 11/14/2022 Tim H. on 11/15/2022 Jennifer B. on 11/15/2022 Miranda W. on 11/15/2022 Crystal B. on 11/15/2022 Tammy D. on 11/15/2022 Durdana I. on 11/15/2022 Tricia F. on 11/15/2022 Patricia R. on 11/15/2022 Brian G. on 11/16/2022 Eye Cancer Gary D. on 11/16/2022 christina r. on 12/27/2022 Jenna S. on 11/16/2022 Kimberly H. on 11/17/2022 Monica D. on 11/16/2022 Shelby W. on 11/16/2022 Racine J. on 11/16/2022 Rafael C. on 11/17/2022 andrea b. on 11/17/2022 Natasha A. on 11/18/2022 Deanna H. on 11/18/2022 Donna P. on 11/18/2022 Sheree M. on 11/18/2022 susan s. on 11/19/2022 Beatrice M. on 11/19/2022 Kimberly K. on 11/19/2022 Jessica C. on 11/20/2022 Kathleen M. on 11/20/2022 Michele P. on 12/1/2022 Jamie K. on 11/20/2022 Marjorie B. on 11/20/2022 Kaitlin R. on 11/20/2022 Imren O. on 11/20/2022 Albert O. on 11/20/2022 Maureen G. on 11/20/2022 Rhonda E. on 11/20/2022 Angela B. on 11/20/2022 thank you Mayory R. on 12/3/2022 Catherine M. on 11/21/2022 Aliya F. on 11/21/2022 Gary M. on 11/21/2022 Gwendolyn W. on 11/22/2022 Allison S. on 11/21/2022 Angelia M. on 11/21/2022 Thomas Z. on 11/22/2022 Abigail B. on 11/23/2022 Coneca G. on 11/22/2022 Teri T. on 11/22/2022 Virginia N. on 11/22/2022 Laurel S. on 11/22/2022 Bridget I. on 11/27/2022 Marilyn G. on 11/22/2022 Ravi J. on 11/23/2022 Daisy L. on 11/23/2022 Jennifer M. on 11/23/2022 Karletta C. on 11/23/2022 Dr. Juan C. on 11/25/2022 BENEECE D. on 11/25/2022 Michele D. on 11/25/2022 daniel u. on 11/25/2022 Siomara S. on 11/25/2022 Chelynne C. on 11/26/2022 Ines M. on 11/26/2022 Tamara C. on 11/26/2022 Rhonda S. on 11/29/2022 It is good Mohamed Kamal E. on 11/27/2022 Jennifer L. on 11/27/2022 Sharon W. on 11/27/2022 Sara E. on 11/27/2022 Robert K. on 11/27/2022 Jennifer H. on 11/27/2022 Linda A. on 11/27/2022 Candace C. on 11/27/2022 Lauren G. on 11/27/2022 Ciara C. on 11/27/2022 Carol C. on 11/28/2022 erica P. on 11/28/2022 Kelly S. on 11/28/2022 Rachael F. on 11/28/2022 SHALVI S. on 11/29/2022 Sean H. on 11/29/2022 Donna F. on 11/29/2022 Tiwanna L. on 11/29/2022 MARY R. on 11/29/2022 bridget w. on 11/29/2022 Karen P. on 11/30/2022 Nicole R. on 11/30/2022 Michelle R. on 11/30/2022 Needed pediatric cardiology topics Roland C. on 12/1/2022 Douglas M. on 12/1/2022 Patsy B. on 12/1/2022 Cathleen W. on 12/1/2022 Leanna M. on 12/1/2022 Stephen L. on 12/1/2022 susan o. on 12/1/2022 Tifanny D. on 12/1/2022 Belinda L. on 12/1/2022 Melekte A. on 12/2/2022 Marissa H. on 12/2/2022 Kayli C. on 12/2/2022 Andrea C. on 12/2/2022 ERNEST E. on 12/2/2022 Natalie F. on 12/2/2022 Roseline T. on 12/2/2022 Anna F. on 12/3/2022 judy l. on 12/3/2022 Jennifer D. on 12/3/2022 Koco C. on 12/3/2022 ryan g. on 12/3/2022 Della A. on 12/3/2022 Olanrewaju P. on 12/3/2022 Marc H. on 12/3/2022 Lisa W. on 12/3/2022 Marlene O. on 12/3/2022 Linda D. on 12/3/2022 It would look much more professional if you were consistent in your descriptions of the child. For instance, “A young boy was brought in…and she was…” or “A female infant was brought in, treated, and he was discharged.” It’s not a huge deal but it makes it look less professionally when the grammar is poor. Carey N. on 12/3/2022 Nicole Q. on 12/4/2022 Stephanie L. on 12/4/2022 Amy W. on 12/3/2022 Gary P. on 12/3/2022 Andrea N. on 12/3/2022 Konstantinos S. on 12/4/2022 Well done presentation! 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Joanna F. on 12/5/2022 John C. on 12/5/2022 Perri M. on 12/5/2022 None Steve N. on 12/5/2022 Puteri Zaharatulnur Fatehah B. on 12/5/2022 Michele H. on 12/5/2022 Imayarze C. on 12/8/2022 gabriella l. on 12/6/2022 Sarah S. on 12/19/2022 Amy P. on 12/6/2022 clare r. on 12/6/2022 Elaine Y. on 12/6/2022 Heather B. on 12/6/2022 This info provided is very useful Sharron J. on 12/6/2022 Connie P. on 12/7/2022 Marissa V. on 12/7/2022 Kaiah F. on 12/7/2022 Shante H. on 12/7/2022 Victoria L. on 12/7/2022 Natasha S. on 12/7/2022 Gregory L. on 12/7/2022 Ivy B. on 12/8/2022 Concise and informative. 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Sharon L. on 12/19/2022 Richard K. on 12/19/2022 Selena R. on 12/19/2022 Catherine B. on 12/19/2022 Lynne S. on 12/19/2022 Informative and engaging Camile C. on 12/19/2022 Kayla S. on 12/20/2022 Jeanine H. on 12/20/2022 Lin S. on 12/20/2022 Kelli M. on 12/20/2022 Bergen S. on 12/20/2022 John G. on 12/21/2022 Lacey H. on 12/21/2022 Meredith R. on 12/22/2022 neha b. on 12/22/2022 Madeleine A. on 12/22/2022 Congratulations! I has been so informative! 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https://chem.libretexts.org/Courses/Anoka-Ramsey_Community_College/Introduction_to_Chemistry/12%3A_Organic_Chemistry/12.04%3A_Branched_Alkanes
12.4: Branched Alkanes - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 12: Organic Chemistry Introduction to Chemistry { } { "12.01:_Organic_Chemistry" : "property get Map 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"license:ccbync", "source-chem-53995", "program:ck12", "source-chem-53995" ] [ "article:topic", "showtoc:no", "license:ccbync", "source-chem-53995", "program:ck12", "source-chem-53995" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Campus Bookshelves 3. Anoka-Ramsey Community College 4. Introduction to Chemistry 5. 12: Organic Chemistry 6. 12.4: Branched Alkanes Expand/collapse global location Introduction to Chemistry Front Matter 1: What is Chemistry? 2: Measurements 3: Matter and Energy 4: Atoms and Elements 5: Chemical Nomenclature 6: Chemical Composition 7: Chemical Reactions 8: Stoichiometry 9: Gases 10: Electrons in Atoms 11: Chemical Bonding 12: Organic Chemistry 13: States of Matter 14: Solutions 15: Acids and Bases 16: Environmental Chemistry 17: Appendix Back Matter 12.4: Branched Alkanes Last updated Apr 13, 2023 Save as PDF 12.3: Alkanes 12.5: Alkenes and Alkynes picture_as_pdf Full Book Page Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report Submit a Peer Review View on CommonsDonate Page ID 355088 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. IUPAC Nomenclature 2. Branched Alkanes 3. 4. Summary The ball-and-stick models you have seen thus far in this text use a rendering tool for viewing molecules online called JSmol. JSmol was designed to replicate molecular model kits (see Figure 12.4.1)that have long been used in chemistry classes to represent three-dimensional structures of molecules. Figure 12.4.1: A molecular model kit used to build three-dimensional molecular models. [Sonia(left) and Bin im Garten (right), via Wikimedia Commons] A common hands-on activity is to provide students with molecular model kits to build models that have a specified chemical formula and to compare the models. There are millions of examples that may be selected for comparison, but if that formula was C 6 H 14, students would soon notice that there are five different ways to build a molecule of C 6 H 14. Substances that have the same molecular formula, but a different arrangement of atoms are called isomers. The five isomers of C 6 H 14 are shown inFigure 12.4.2: Jmol._Canvas2D (Jmol) "jmolApplet95473"[x] script 1 started zoomLarge = false antialiasDisplay = true callback set for errorCallback myCallback ERROR errorCallback = "myCallback" Jmol JavaScript applet jmolApplet45770__1394187012889534__ initializing Jmol getValue debug null Jmol getValue logLevel null Jmol getValue allowjavascript null AppletRegistry.checkIn(jmolApplet45770__1394187012889534__) vwrOptions: { "name":"jmolApplet45770","applet":true,"documentBase":" object]","syncId":"1394187012889534","bgcolor":"white" } setting document base to " (C) 2015 Jmol Development Jmol Version: 14.29.3 2018-02-03 09:25 java.vendor: Java2Script (HTML5) java.version: 2018-01-28 23:38:52 (JSmol/j2s) os.name: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/140.0.0.0 Safari/537.36 Access: ALL memory: 0.0/0.0 processors available: 1 useCommandThread: false appletId:jmolApplet45770 (signed) Jmol getValue emulate null defaults = "Jmol" Jmol getValue boxbgcolor null Jmol getValue bgcolor white backgroundColor = "white" Jmol getValue ANIMFRAMECallback null Jmol getValue APPLETREADYCallback Jmol._readyCallback APPLETREADYCallback = "Jmol._readyCallback" callback set for APPLETREADYCallback Jmol._readyCallback APPLETREADY Jmol getValue ATOMMOVEDCallback null Jmol getValue AUDIOCallback null Jmol getValue CLICKCallback null Jmol getValue DRAGDROPCallback null Jmol getValue ECHOCallback null Jmol getValue ERRORCallback null Jmol getValue EVALCallback null Jmol getValue HOVERCallback null Jmol getValue IMAGECallback null Jmol getValue LOADSTRUCTCallback null Jmol getValue MEASURECallback null Jmol getValue MESSAGECallback null Jmol getValue MINIMIZATIONCallback null Jmol getValue SERVICECallback null Jmol getValue PICKCallback null Jmol getValue RESIZECallback null Jmol getValue SCRIPTCallback null Jmol getValue SYNCCallback null Jmol getValue STRUCTUREMODIFIEDCallback null Jmol getValue doTranslate null language=en_US Jmol getValue popupMenu null Jmol getValue script null Jmol getValue loadInline null Jmol getValue load null Jmol applet jmolApplet45770__1394187012889534__ ready Jmol._Canvas2D (Jmol) "jmolApplet45770"[x] script 1 started zoomLarge = false antialiasDisplay = true callback set for errorCallback myCallback ERROR errorCallback = "myCallback" Jmol JavaScript applet jmolApplet44980__1394187012889534__ initializing Jmol getValue debug null Jmol getValue logLevel null Jmol getValue allowjavascript null AppletRegistry.checkIn(jmolApplet44980__1394187012889534__) vwrOptions: { "name":"jmolApplet44980","applet":true,"documentBase":" object]","syncId":"1394187012889534","bgcolor":"white" } setting document base to " (C) 2015 Jmol Development Jmol Version: 14.29.3 2018-02-03 09:25 java.vendor: Java2Script (HTML5) java.version: 2018-01-28 23:38:52 (JSmol/j2s) os.name: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/140.0.0.0 Safari/537.36 Access: ALL memory: 0.0/0.0 processors available: 1 useCommandThread: false appletId:jmolApplet44980 (signed) Jmol getValue emulate null defaults = "Jmol" Jmol getValue boxbgcolor null Jmol getValue bgcolor white backgroundColor = "white" Jmol getValue ANIMFRAMECallback null Jmol getValue APPLETREADYCallback Jmol._readyCallback APPLETREADYCallback = "Jmol._readyCallback" callback set for APPLETREADYCallback Jmol._readyCallback APPLETREADY Jmol getValue ATOMMOVEDCallback null Jmol getValue AUDIOCallback null Jmol getValue CLICKCallback null Jmol getValue DRAGDROPCallback null Jmol getValue ECHOCallback null Jmol getValue ERRORCallback null Jmol getValue EVALCallback null Jmol getValue HOVERCallback null Jmol getValue IMAGECallback null Jmol getValue LOADSTRUCTCallback null Jmol getValue MEASURECallback null Jmol getValue MESSAGECallback null Jmol getValue MINIMIZATIONCallback null Jmol getValue SERVICECallback null Jmol getValue PICKCallback null Jmol getValue RESIZECallback null Jmol getValue SCRIPTCallback null Jmol getValue SYNCCallback null Jmol getValue STRUCTUREMODIFIEDCallback null Jmol getValue doTranslate null language=en_US Jmol getValue popupMenu null Jmol getValue script null Jmol getValue loadInline null Jmol getValue load null Jmol applet jmolApplet44980__1394187012889534__ ready Jmol._Canvas2D (Jmol) "jmolApplet44980"[x] script 1 started zoomLarge = false antialiasDisplay = true callback set for errorCallback myCallback ERROR errorCallback = "myCallback" Jmol JavaScript applet jmolApplet12621__1394187012889534__ initializing Jmol getValue debug null Jmol getValue logLevel null Jmol getValue allowjavascript null AppletRegistry.checkIn(jmolApplet12621__1394187012889534__) vwrOptions: { "name":"jmolApplet12621","applet":true,"documentBase":" object]","syncId":"1394187012889534","bgcolor":"white" } setting document base to " (C) 2015 Jmol Development Jmol Version: 14.29.3 2018-02-03 09:25 java.vendor: Java2Script (HTML5) java.version: 2018-01-28 23:38:52 (JSmol/j2s) os.name: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/140.0.0.0 Safari/537.36 Access: ALL memory: 0.0/0.0 processors available: 1 useCommandThread: false appletId:jmolApplet12621 (signed) Jmol getValue emulate null defaults = "Jmol" Jmol getValue boxbgcolor null Jmol getValue bgcolor white backgroundColor = "white" Jmol getValue ANIMFRAMECallback null Jmol getValue APPLETREADYCallback Jmol._readyCallback APPLETREADYCallback = "Jmol._readyCallback" callback set for APPLETREADYCallback Jmol._readyCallback APPLETREADY Jmol getValue ATOMMOVEDCallback null Jmol getValue AUDIOCallback null Jmol getValue CLICKCallback null Jmol getValue DRAGDROPCallback null Jmol getValue ECHOCallback null Jmol getValue ERRORCallback null Jmol getValue EVALCallback null Jmol getValue HOVERCallback null Jmol getValue IMAGECallback null Jmol getValue LOADSTRUCTCallback null Jmol getValue MEASURECallback null Jmol getValue MESSAGECallback null Jmol getValue MINIMIZATIONCallback null Jmol getValue SERVICECallback null Jmol getValue PICKCallback null Jmol getValue RESIZECallback null Jmol getValue SCRIPTCallback null Jmol getValue SYNCCallback null Jmol getValue STRUCTUREMODIFIEDCallback null Jmol getValue doTranslate null language=en_US Jmol getValue popupMenu null Jmol getValue script null Jmol getValue loadInline null Jmol getValue load null Jmol applet jmolApplet12621__1394187012889534__ ready Jmol._Canvas2D (Jmol) "jmolApplet12621"[x] script 1 started zoomLarge = false antialiasDisplay = true callback set for errorCallback myCallback ERROR errorCallback = "myCallback" Jmol JavaScript applet jmolApplet62901__1394187012889534__ initializing Jmol getValue debug null Jmol getValue logLevel null Jmol getValue allowjavascript null AppletRegistry.checkIn(jmolApplet62901__1394187012889534__) vwrOptions: { "name":"jmolApplet62901","applet":true,"documentBase":" object]","syncId":"1394187012889534","bgcolor":"white" } setting document base to " (C) 2015 Jmol Development Jmol Version: 14.29.3 2018-02-03 09:25 java.vendor: Java2Script (HTML5) java.version: 2018-01-28 23:38:52 (JSmol/j2s) os.name: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/140.0.0.0 Safari/537.36 Access: ALL memory: 0.0/0.0 processors available: 1 useCommandThread: false appletId:jmolApplet62901 (signed) Jmol getValue emulate null defaults = "Jmol" Jmol getValue boxbgcolor null Jmol getValue bgcolor white backgroundColor = "white" Jmol getValue ANIMFRAMECallback null Jmol getValue APPLETREADYCallback Jmol._readyCallback APPLETREADYCallback = "Jmol._readyCallback" callback set for APPLETREADYCallback Jmol._readyCallback APPLETREADY Jmol getValue ATOMMOVEDCallback null Jmol getValue AUDIOCallback null Jmol getValue CLICKCallback null Jmol getValue DRAGDROPCallback null Jmol getValue ECHOCallback null Jmol getValue ERRORCallback null Jmol getValue EVALCallback null Jmol getValue HOVERCallback null Jmol getValue IMAGECallback null Jmol getValue LOADSTRUCTCallback null Jmol getValue MEASURECallback null Jmol getValue MESSAGECallback null Jmol getValue MINIMIZATIONCallback null Jmol getValue SERVICECallback null Jmol getValue PICKCallback null Jmol getValue RESIZECallback null Jmol getValue SCRIPTCallback null Jmol getValue SYNCCallback null Jmol getValue STRUCTUREMODIFIEDCallback null Jmol getValue doTranslate null language=en_US Jmol getValue popupMenu null Jmol getValue script null Jmol getValue loadInline null Jmol getValue load null Jmol applet jmolApplet62901__1394187012889534__ ready Jmol._Canvas2D (Jmol) "jmolApplet62901"[x] script 1 started zoomLarge = false antialiasDisplay = true callback set for errorCallback myCallback ERROR errorCallback = "myCallback" Jmol JavaScript applet jmolApplet99424__1394187012889534__ initializing Jmol getValue debug null Jmol getValue logLevel null Jmol getValue allowjavascript null AppletRegistry.checkIn(jmolApplet99424__1394187012889534__) vwrOptions: { "name":"jmolApplet99424","applet":true,"documentBase":" object]","syncId":"1394187012889534","bgcolor":"white" } setting document base to " (C) 2015 Jmol Development Jmol Version: 14.29.3 2018-02-03 09:25 java.vendor: Java2Script (HTML5) java.version: 2018-01-28 23:38:52 (JSmol/j2s) os.name: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/140.0.0.0 Safari/537.36 Access: ALL memory: 0.0/0.0 processors available: 1 useCommandThread: false appletId:jmolApplet99424 (signed) Jmol getValue emulate null defaults = "Jmol" Jmol getValue boxbgcolor null Jmol getValue bgcolor white backgroundColor = "white" Jmol getValue ANIMFRAMECallback null Jmol getValue APPLETREADYCallback Jmol._readyCallback APPLETREADYCallback = "Jmol._readyCallback" callback set for APPLETREADYCallback Jmol._readyCallback APPLETREADY Jmol getValue ATOMMOVEDCallback null Jmol getValue AUDIOCallback null Jmol getValue CLICKCallback null Jmol getValue DRAGDROPCallback null Jmol getValue ECHOCallback null Jmol getValue ERRORCallback null Jmol getValue EVALCallback null Jmol getValue HOVERCallback null Jmol getValue IMAGECallback null Jmol getValue LOADSTRUCTCallback null Jmol getValue MEASURECallback null Jmol getValue MESSAGECallback null Jmol getValue MINIMIZATIONCallback null Jmol getValue SERVICECallback null Jmol getValue PICKCallback null Jmol getValue RESIZECallback null Jmol getValue SCRIPTCallback null Jmol getValue SYNCCallback null Jmol getValue STRUCTUREMODIFIEDCallback null Jmol getValue doTranslate null language=en_US Jmol getValue popupMenu null Jmol getValue script null Jmol getValue loadInline null Jmol getValue load null Jmol applet jmolApplet99424__1394187012889534__ ready Figure 12.4.2: The five isomers of C 6 H 14. Molecules rendered in JSmol may be rotated in all directions to examine molecules in different orientations. Simply "grab" onto the molecule and drag in the desired direction(s) as shown inFigure 12.4.3. Try rotating the molecules in Figure 12.4.2above to make all 14 hydrogen atoms visible in each isomer. Go ahead...try it! Figure 12.4.3: Rotating a hexane molecule rendered with JSmol. IUPAC Nomenclature While molecules are written or drawn in two dimensions, it is important to remember that molecules are three-dimensional. Since isomers of each other are unique, they have different chemical and physical properties.The differences are sometimes small and at other times, quite significant. To distinguish isomers from one another, it is important that they be assigned a unique name. The International Union of Pure and Applied Chemistry (IUPAC) has devised a system of nomenclature that begins with the names of the alkanes and can be adjusted from there to account for more complicated structures. Success with organic nomenclature begins with the ability to count carbon atoms. Similar to howGreek prefixes were used to designate a specific number of atoms in binary molecular compounds, IUPAC nomenclature uses prefixes based on the alkanes to designate a specific number of carbon atoms. Notice that Greek prefixes are used for specifying five or more carbon atoms. Table 12.4.1: IUPAC Prefixes Used to Specify the Number of Carbon Atoms in Organic Compounds| Number of Carbon Atoms | IUPAC Prefix | Number of Carbon Atoms | IUPAC Prefix | --- --- | | 1 | meth- | 6 | hex- | | 2 | eth- | 7 | hept- | | 3 | prop- | 8 | oct- | | 4 | but- | 9 | non- | | 5 | pent- | 10 | dec- | Branched Alkanes The isomers of C 6 H 14 shown inFigure 12.4.2above are shown again inTable 12.4.2below along with their condensed structural formulas, line formulas, and IUPAC names. While we already learned how to name hexane and other straight-chained alkanes in the previous section, the remaining isomers of C 6 H 14 are branched. How are the names for these branched molecules determined? Here are some basic guidelines: ☞ IUPAC Nomenclature of Branched Alkanes Identify and name the parent chain. The parent chain is the longest consecutive chain of carbon atoms in the molecule. Identify and name all branches off the parent chain as alkyl groups. An alkyl group is an alkane that has had one H atom removed. If the H atom is removed from the end carbon of a straight-chain alkane, the alkyl group is named by changing the –ane suffix of the alkane to –yl. The name(s) of the alkyl group(s) precedes the name of the parent chain. Common alkyl groups: methyl CH 3— ethyl CH 3⁢CH 2— propyl CH 3⁢CH 2⁢CH 2— isopropyl CH 3⁢C│⁢H 2⁡CH 3 butyl CH 3⁢CH 2⁢CH 2⁢CH 2— Locate and assign a number to the position of each branch off the parent chain. For a single branch off the parent chain: Count in the direction that gives the lowest number. Place the number in front of the name of the alkyl group, separating the number and name with a hyphen (Example: 3-methyl). For multiple branches off the parent chain: Count in the direction that gives the lowest number for any branch off the parent chain. For branches of the same type: Indicate the number of that type by using the Greek prefixes di–, tri–, tetra–, etc. in front of the name of the alkyl group. Write the number of the position of each branch in ascending order, separating the numbers by commas. Example: 2,3,3-trimethylheptane For branches of different types: The alkyl groups are named in alphabetical order, with a number indicating the position of each alkyl group off the parent chain. Example: 3-ethyl-2-methylhexane Additional Notes: The name of an alkane has no spaces. Numbers are separated by commas. Numbers and letters are separated by hyphens. Whether branched or unbranched, alkanes always have a general formula of C n H 2n +2. Table 12.4.2: The Five Isomers of C 6 H 14.| Three-Dimensional Structure | Condensed Structural Formula | Line Formula | IUPAC Name | --- --- | | Jmol._Canvas2D (Jmol) "jmolApplet99424"[x] script 1 started zoomLarge = false antialiasDisplay = true callback set for errorCallback myCallback ERROR errorCallback = "myCallback" Jmol JavaScript applet jmolApplet2483__1394187012889534__ initializing Jmol getValue debug null Jmol getValue logLevel null Jmol getValue allowjavascript null AppletRegistry.checkIn(jmolApplet2483__1394187012889534__) vwrOptions: { "name":"jmolApplet2483","applet":true,"documentBase":" object]","syncId":"1394187012889534","bgcolor":"white" } setting document base to " (C) 2015 Jmol Development Jmol Version: 14.29.3 2018-02-03 09:25 java.vendor: Java2Script (HTML5) java.version: 2018-01-28 23:38:52 (JSmol/j2s) os.name: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/140.0.0.0 Safari/537.36 Access: ALL memory: 0.0/0.0 processors available: 1 useCommandThread: false appletId:jmolApplet2483 (signed) Jmol getValue emulate null defaults = "Jmol" Jmol getValue boxbgcolor null Jmol getValue bgcolor white backgroundColor = "white" Jmol getValue ANIMFRAMECallback null Jmol getValue APPLETREADYCallback Jmol._readyCallback APPLETREADYCallback = "Jmol._readyCallback" callback set for APPLETREADYCallback Jmol._readyCallback APPLETREADY Jmol getValue ATOMMOVEDCallback null Jmol getValue AUDIOCallback null Jmol getValue CLICKCallback null Jmol getValue DRAGDROPCallback null Jmol getValue ECHOCallback null Jmol getValue ERRORCallback null Jmol getValue EVALCallback null Jmol getValue HOVERCallback null Jmol getValue IMAGECallback null Jmol getValue LOADSTRUCTCallback null Jmol getValue MEASURECallback null Jmol getValue MESSAGECallback null Jmol getValue MINIMIZATIONCallback null Jmol getValue SERVICECallback null Jmol getValue PICKCallback null Jmol getValue RESIZECallback null Jmol getValue SCRIPTCallback null Jmol getValue SYNCCallback null Jmol getValue STRUCTUREMODIFIEDCallback null Jmol getValue doTranslate null language=en_US Jmol getValue popupMenu null Jmol getValue script null Jmol getValue loadInline null Jmol getValue load null Jmol applet jmolApplet2483__1394187012889534__ ready | | | hexane | | Jmol._Canvas2D (Jmol) "jmolApplet2483"[x] script 1 started zoomLarge = false antialiasDisplay = true callback set for errorCallback myCallback ERROR errorCallback = "myCallback" Jmol JavaScript applet jmolApplet37258__1394187012889534__ initializing Jmol getValue debug null Jmol getValue logLevel null Jmol getValue allowjavascript null AppletRegistry.checkIn(jmolApplet37258__1394187012889534__) vwrOptions: { "name":"jmolApplet37258","applet":true,"documentBase":" object]","syncId":"1394187012889534","bgcolor":"white" } setting document base to " (C) 2015 Jmol Development Jmol Version: 14.29.3 2018-02-03 09:25 java.vendor: Java2Script (HTML5) java.version: 2018-01-28 23:38:52 (JSmol/j2s) os.name: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/140.0.0.0 Safari/537.36 Access: ALL memory: 0.0/0.0 processors available: 1 useCommandThread: false appletId:jmolApplet37258 (signed) Jmol getValue emulate null defaults = "Jmol" Jmol getValue boxbgcolor null Jmol getValue bgcolor white backgroundColor = "white" Jmol getValue ANIMFRAMECallback null Jmol getValue APPLETREADYCallback Jmol._readyCallback APPLETREADYCallback = "Jmol._readyCallback" callback set for APPLETREADYCallback Jmol._readyCallback APPLETREADY Jmol getValue ATOMMOVEDCallback null Jmol getValue AUDIOCallback null Jmol getValue CLICKCallback null Jmol getValue DRAGDROPCallback null Jmol getValue ECHOCallback null Jmol getValue ERRORCallback null Jmol getValue EVALCallback null Jmol getValue HOVERCallback null Jmol getValue IMAGECallback null Jmol getValue LOADSTRUCTCallback null Jmol getValue MEASURECallback null Jmol getValue MESSAGECallback null Jmol getValue MINIMIZATIONCallback null Jmol getValue SERVICECallback null Jmol getValue PICKCallback null Jmol getValue RESIZECallback null Jmol getValue SCRIPTCallback null Jmol getValue SYNCCallback null Jmol getValue STRUCTUREMODIFIEDCallback null Jmol getValue doTranslate null language=en_US Jmol getValue popupMenu null Jmol getValue script null Jmol getValue loadInline null Jmol getValue load null Jmol applet jmolApplet37258__1394187012889534__ ready | | | 2-methylpentane | | Jmol._Canvas2D (Jmol) "jmolApplet37258"[x] script 1 started zoomLarge = false antialiasDisplay = true callback set for errorCallback myCallback ERROR errorCallback = "myCallback" Jmol JavaScript applet jmolApplet29447__1394187012889534__ initializing Jmol getValue debug null Jmol getValue logLevel null Jmol getValue allowjavascript null AppletRegistry.checkIn(jmolApplet29447__1394187012889534__) vwrOptions: { "name":"jmolApplet29447","applet":true,"documentBase":" object]","syncId":"1394187012889534","bgcolor":"white" } setting document base to " (C) 2015 Jmol Development Jmol Version: 14.29.3 2018-02-03 09:25 java.vendor: Java2Script (HTML5) java.version: 2018-01-28 23:38:52 (JSmol/j2s) os.name: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/140.0.0.0 Safari/537.36 Access: ALL memory: 0.0/0.0 processors available: 1 useCommandThread: false appletId:jmolApplet29447 (signed) Jmol getValue emulate null defaults = "Jmol" Jmol getValue boxbgcolor null Jmol getValue bgcolor white backgroundColor = "white" Jmol getValue ANIMFRAMECallback null Jmol getValue APPLETREADYCallback Jmol._readyCallback APPLETREADYCallback = "Jmol._readyCallback" callback set for APPLETREADYCallback Jmol._readyCallback APPLETREADY Jmol getValue ATOMMOVEDCallback null Jmol getValue AUDIOCallback null Jmol getValue CLICKCallback null Jmol getValue DRAGDROPCallback null Jmol getValue ECHOCallback null Jmol getValue ERRORCallback null Jmol getValue EVALCallback null Jmol getValue HOVERCallback null Jmol getValue IMAGECallback null Jmol getValue LOADSTRUCTCallback null Jmol getValue MEASURECallback null Jmol getValue MESSAGECallback null Jmol getValue MINIMIZATIONCallback null Jmol getValue SERVICECallback null Jmol getValue PICKCallback null Jmol getValue RESIZECallback null Jmol getValue SCRIPTCallback null Jmol getValue SYNCCallback null Jmol getValue STRUCTUREMODIFIEDCallback null Jmol getValue doTranslate null language=en_US Jmol getValue popupMenu null Jmol getValue script null Jmol getValue loadInline null Jmol getValue load null Jmol applet jmolApplet29447__1394187012889534__ ready | | | 3-methylpentane | | Jmol._Canvas2D (Jmol) "jmolApplet29447"[x] script 1 started zoomLarge = false antialiasDisplay = true callback set for errorCallback myCallback ERROR errorCallback = "myCallback" Jmol JavaScript applet jmolApplet96420__1394187012889534__ initializing Jmol getValue debug null Jmol getValue logLevel null Jmol getValue allowjavascript null AppletRegistry.checkIn(jmolApplet96420__1394187012889534__) vwrOptions: { "name":"jmolApplet96420","applet":true,"documentBase":" object]","syncId":"1394187012889534","bgcolor":"white" } setting document base to " (C) 2015 Jmol Development Jmol Version: 14.29.3 2018-02-03 09:25 java.vendor: Java2Script (HTML5) java.version: 2018-01-28 23:38:52 (JSmol/j2s) os.name: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/140.0.0.0 Safari/537.36 Access: ALL memory: 0.0/0.0 processors available: 1 useCommandThread: false appletId:jmolApplet96420 (signed) Jmol getValue emulate null defaults = "Jmol" Jmol getValue boxbgcolor null Jmol getValue bgcolor white backgroundColor = "white" Jmol getValue ANIMFRAMECallback null Jmol getValue APPLETREADYCallback Jmol._readyCallback APPLETREADYCallback = "Jmol._readyCallback" callback set for APPLETREADYCallback Jmol._readyCallback APPLETREADY Jmol getValue ATOMMOVEDCallback null Jmol getValue AUDIOCallback null Jmol getValue CLICKCallback null Jmol getValue DRAGDROPCallback null Jmol getValue ECHOCallback null Jmol getValue ERRORCallback null Jmol getValue EVALCallback null Jmol getValue HOVERCallback null Jmol getValue IMAGECallback null Jmol getValue LOADSTRUCTCallback null Jmol getValue MEASURECallback null Jmol getValue MESSAGECallback null Jmol getValue MINIMIZATIONCallback null Jmol getValue SERVICECallback null Jmol getValue PICKCallback null Jmol getValue RESIZECallback null Jmol getValue SCRIPTCallback null Jmol getValue SYNCCallback null Jmol getValue STRUCTUREMODIFIEDCallback null Jmol getValue doTranslate null language=en_US Jmol getValue popupMenu null Jmol getValue script null Jmol getValue loadInline null Jmol getValue load null Jmol applet jmolApplet96420__1394187012889534__ ready | | | 2,3-dimethylbutane | | Jmol._Canvas2D (Jmol) "jmolApplet96420"[x] Jmol JavaScript applet jmolApplet95473__1394187012889534__ initializing Jmol getValue debug null Jmol getValue logLevel null Jmol getValue allowjavascript null AppletRegistry.checkIn(jmolApplet95473__1394187012889534__) vwrOptions: { "name":"jmolApplet95473","applet":true,"documentBase":" object]","syncId":"1394187012889534","bgcolor":"white" } setting document base to " (C) 2015 Jmol Development Jmol Version: 14.29.3 2018-02-03 09:25 java.vendor: Java2Script (HTML5) java.version: 2018-01-28 23:38:52 (JSmol/j2s) os.name: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/140.0.0.0 Safari/537.36 Access: ALL memory: 0.0/0.0 processors available: 1 useCommandThread: false appletId:jmolApplet95473 (signed) Jmol getValue emulate null defaults = "Jmol" Jmol getValue boxbgcolor null Jmol getValue bgcolor white backgroundColor = "white" Jmol getValue ANIMFRAMECallback null Jmol getValue APPLETREADYCallback Jmol._readyCallback APPLETREADYCallback = "Jmol._readyCallback" callback set for APPLETREADYCallback Jmol._readyCallback APPLETREADY Jmol getValue ATOMMOVEDCallback null Jmol getValue AUDIOCallback null Jmol getValue CLICKCallback null Jmol getValue DRAGDROPCallback null Jmol getValue ECHOCallback null Jmol getValue ERRORCallback null Jmol getValue EVALCallback null Jmol getValue HOVERCallback null Jmol getValue IMAGECallback null Jmol getValue LOADSTRUCTCallback null Jmol getValue MEASURECallback null Jmol getValue MESSAGECallback null Jmol getValue MINIMIZATIONCallback null Jmol getValue SERVICECallback null Jmol getValue PICKCallback null Jmol getValue RESIZECallback null Jmol getValue SCRIPTCallback null Jmol getValue SYNCCallback null Jmol getValue STRUCTUREMODIFIEDCallback null Jmol getValue doTranslate null language=en_US Jmol getValue popupMenu null Jmol getValue script null Jmol getValue loadInline null Jmol getValue load null Jmol applet jmolApplet95473__1394187012889534__ ready script 1 started zoomLarge = false antialiasDisplay = true callback set for errorCallback myCallback ERROR errorCallback = "myCallback" Viewer cachePut cache://localLOAD_ Viewer cachePut Viewer cachePut cache://localLOAD_ Viewer cachePut Viewer cachePut cache://localLOAD_ Viewer cachePut Viewer cachePut cache://localLOAD_ Viewer cachePut Viewer cachePut cache://localLOAD_ Viewer cachePut Viewer cachePut cache://localLOAD_ Viewer cachePut Viewer cachePut cache://localLOAD_ Viewer cachePut Viewer cachePut cache://localLOAD_ Viewer cachePut Viewer cachePut cache://localLOAD_ Viewer cachePut Viewer cachePut cache://localLOAD_ Viewer cachePut FileManager.getAtomSetCollectionFromFile( The Resolver thinks Mol 6403 -OEChem-09292507413D 6403 pubchem_compound_cid:"6403" pubchem_conformer_rmsd:"0.4" pubchem_conformer_diverseorder:"1" pubchem_mmff94_partial_charges:"0" 0 partial charges read pubchem_effective_rotor_count:"1" pubchem_pharmacophore_features:"2\n1 6 hydrophobe\n4 1 3 4 5 hydrophobe" pubchem_heavy_atom_count:"6" pubchem_atom_def_stereo_count:"0" pubchem_atom_udef_stereo_count:"0" pubchem_bond_def_stereo_count:"0" pubchem_bond_udef_stereo_count:"0" pubchem_isotopic_atom_count:"0" pubchem_component_count:"1" pubchem_cactvs_tauto_count:"1" pubchem_conformer_id:"0000190300000001" pubchem_mmff94_energy:"10.8061" pubchem_feature_selfoverlap:"10.15" pubchem_shape_fingerprint:"137420 1 16743923505276132339\n139733 1 12180113263932609231\n20096714 4 17684931302065791716\n21040471 1 17906449926867356884\n24536 1 17459210520101090254\n29004967 10 18337113478827805443\n5943 1 17698761242850914825" pubchem_shape_multipoles:"123.48\n2.1\n1.15\n1.13\n1.03\n0.45\n0.51\n-0.34\n-0.31\n-0.27\n-0.22\n-0.26\n-0.1\n-0.01" pubchem_shape_selfoverlap:"215.439" pubchem_shape_volume:"85.1" pubchem_coordinate_type:"2\n5\n10" Time for openFile(:6403): 15 ms reading 20 atoms ModelSet: haveSymmetry:false haveUnitcells:false haveFractionalCoord:false 1 model in this collection. Use getProperty "modelInfo" or getProperty "auxiliaryInfo" to inspect them. Default Van der Waals type for model set to Babel 20 atoms created ModelSet: not autobonding; use forceAutobond=true to force automatic bond creation Time for creating model: 8 ms 6403 spinY = 10.0 Script completed Jmol script terminated FileManager.getAtomSetCollectionFromFile( The Resolver thinks Mol 6403 -OEChem-09292507413D 6403 pubchem_compound_cid:"6403" pubchem_conformer_rmsd:"0.4" pubchem_conformer_diverseorder:"1" pubchem_mmff94_partial_charges:"0" 0 partial charges read pubchem_effective_rotor_count:"1" pubchem_pharmacophore_features:"2\n1 6 hydrophobe\n4 1 3 4 5 hydrophobe" pubchem_heavy_atom_count:"6" pubchem_atom_def_stereo_count:"0" pubchem_atom_udef_stereo_count:"0" pubchem_bond_def_stereo_count:"0" pubchem_bond_udef_stereo_count:"0" pubchem_isotopic_atom_count:"0" pubchem_component_count:"1" pubchem_cactvs_tauto_count:"1" pubchem_conformer_id:"0000190300000001" pubchem_mmff94_energy:"10.8061" pubchem_feature_selfoverlap:"10.15" pubchem_shape_fingerprint:"137420 1 16743923505276132339\n139733 1 12180113263932609231\n20096714 4 17684931302065791716\n21040471 1 17906449926867356884\n24536 1 17459210520101090254\n29004967 10 18337113478827805443\n5943 1 17698761242850914825" pubchem_shape_multipoles:"123.48\n2.1\n1.15\n1.13\n1.03\n0.45\n0.51\n-0.34\n-0.31\n-0.27\n-0.22\n-0.26\n-0.1\n-0.01" pubchem_shape_selfoverlap:"215.439" pubchem_shape_volume:"85.1" pubchem_coordinate_type:"2\n5\n10" Time for openFile(:6403): 5 ms reading 20 atoms ModelSet: haveSymmetry:false haveUnitcells:false haveFractionalCoord:false 1 model in this collection. Use getProperty "modelInfo" or getProperty "auxiliaryInfo" to inspect them. Default Van der Waals type for model set to Babel 20 atoms created ModelSet: not autobonding; use forceAutobond=true to force automatic bond creation Time for creating model: 2 ms 6403 spinY = 10.0 Script completed Jmol script terminated FileManager.getAtomSetCollectionFromFile( The Resolver thinks Mol 7892 -OEChem-09292507413D 7892 pubchem_compound_cid:"7892" pubchem_conformer_rmsd:"0.4" pubchem_conformer_diverseorder:"1\n2" pubchem_mmff94_partial_charges:"0" 0 partial charges read pubchem_effective_rotor_count:"2" pubchem_pharmacophore_features:"2\n1 6 hydrophobe\n3 1 4 5 hydrophobe" pubchem_heavy_atom_count:"6" pubchem_atom_def_stereo_count:"0" pubchem_atom_udef_stereo_count:"0" pubchem_bond_def_stereo_count:"0" pubchem_bond_udef_stereo_count:"0" pubchem_isotopic_atom_count:"0" pubchem_component_count:"1" pubchem_cactvs_tauto_count:"1" pubchem_conformer_id:"00001ED400000001" pubchem_mmff94_energy:"1.1461" pubchem_feature_selfoverlap:"10.149" pubchem_shape_fingerprint:"139733 1 9150908169265567743\n14390081 3 18202275875583584392\n16714656 1 18272090530151213903\n20096714 4 18266743471840046620\n21040471 1 18051418370278651853\n23552423 10 18188496774394857150\n23552449 1 18341892974143002409\n24536 1 18339634542978485262\n29004967 10 16200157603219033622\n5460574 1 9871754598137484054" pubchem_shape_multipoles:"123.48\n3.58\n1.16\n0.67\n1.79\n0.29\n-0.01\n-0.56\n0.25\n-0.66\n0.09\n0.01\n0.01\n-0.08" pubchem_shape_selfoverlap:"209.726" pubchem_shape_volume:"85.2" pubchem_coordinate_type:"2\n5\n10" Time for openFile(:7892): 9 ms reading 20 atoms ModelSet: haveSymmetry:false haveUnitcells:false haveFractionalCoord:false 1 model in this collection. Use getProperty "modelInfo" or getProperty "auxiliaryInfo" to inspect them. Default Van der Waals type for model set to Babel 20 atoms created ModelSet: not autobonding; use forceAutobond=true to force automatic bond creation Time for creating model: 2 ms 7892 spinY = 10.0 Script completed Jmol script terminated FileManager.getAtomSetCollectionFromFile( The Resolver thinks Mol 7892 -OEChem-09292507413D 7892 pubchem_compound_cid:"7892" pubchem_conformer_rmsd:"0.4" pubchem_conformer_diverseorder:"1\n2" pubchem_mmff94_partial_charges:"0" 0 partial charges read pubchem_effective_rotor_count:"2" pubchem_pharmacophore_features:"2\n1 6 hydrophobe\n3 1 4 5 hydrophobe" pubchem_heavy_atom_count:"6" pubchem_atom_def_stereo_count:"0" pubchem_atom_udef_stereo_count:"0" pubchem_bond_def_stereo_count:"0" pubchem_bond_udef_stereo_count:"0" pubchem_isotopic_atom_count:"0" pubchem_component_count:"1" pubchem_cactvs_tauto_count:"1" pubchem_conformer_id:"00001ED400000001" pubchem_mmff94_energy:"1.1461" pubchem_feature_selfoverlap:"10.149" pubchem_shape_fingerprint:"139733 1 9150908169265567743\n14390081 3 18202275875583584392\n16714656 1 18272090530151213903\n20096714 4 18266743471840046620\n21040471 1 18051418370278651853\n23552423 10 18188496774394857150\n23552449 1 18341892974143002409\n24536 1 18339634542978485262\n29004967 10 16200157603219033622\n5460574 1 9871754598137484054" pubchem_shape_multipoles:"123.48\n3.58\n1.16\n0.67\n1.79\n0.29\n-0.01\n-0.56\n0.25\n-0.66\n0.09\n0.01\n0.01\n-0.08" pubchem_shape_selfoverlap:"209.726" pubchem_shape_volume:"85.2" pubchem_coordinate_type:"2\n5\n10" Time for openFile(:7892): 5 ms reading 20 atoms ModelSet: haveSymmetry:false haveUnitcells:false haveFractionalCoord:false 1 model in this collection. Use getProperty "modelInfo" or getProperty "auxiliaryInfo" to inspect them. Default Van der Waals type for model set to Babel 20 atoms created ModelSet: not autobonding; use forceAutobond=true to force automatic bond creation Time for creating model: 2 ms 7892 spinY = 10.0 Script completed Jmol script terminated FileManager.getAtomSetCollectionFromFile(cache://localLOAD_ The Resolver thinks Mol C6H14 APtclcactv09282519373D 0 0.00000 0.00000 C6H14 Time for openFile(cache://localLOAD_ 4 ms reading 20 atoms ModelSet: haveSymmetry:false haveUnitcells:false haveFractionalCoord:false 1 model in this collection. Use getProperty "modelInfo" or getProperty "auxiliaryInfo" to inspect them. Default Van der Waals type for model set to Babel 20 atoms created ModelSet: not autobonding; use forceAutobond=true to force automatic bond creation Time for creating model: 2 ms C6H14 spinY = 10.0 Script completed Jmol script terminated FileManager.getAtomSetCollectionFromFile( The Resolver thinks Mol 6589 -OEChem-09292507413D 6589 pubchem_compound_cid:"6589" pubchem_conformer_rmsd:"0.4" pubchem_conformer_diverseorder:"1\n2\n3" pubchem_mmff94_partial_charges:"0" 0 partial charges read pubchem_effective_rotor_count:"1" pubchem_pharmacophore_features:"2\n3 1 3 4 hydrophobe\n3 2 5 6 hydrophobe" pubchem_heavy_atom_count:"6" pubchem_atom_def_stereo_count:"0" pubchem_atom_udef_stereo_count:"0" pubchem_bond_def_stereo_count:"0" pubchem_bond_udef_stereo_count:"0" pubchem_isotopic_atom_count:"0" pubchem_component_count:"1" pubchem_cactvs_tauto_count:"1" pubchem_conformer_id:"000019BD00000001" pubchem_mmff94_energy:"6.4127" pubchem_feature_selfoverlap:"10.182" pubchem_shape_fingerprint:"137420 1 14210207548903878447\n16714656 1 17531538572250223695\n20096714 4 18341602690404418173\n21040471 1 17895747405347166253\n24536 1 18042116725187677885\n29004967 10 18060424577569916894\n5943 1 16099473149988438040" pubchem_shape_multipoles:"123.48\n2.2\n1.33\n0.97\n0\n0\n0.14\n0\n-0.27\n0\n0.28\n0\n0\n0.54" pubchem_shape_selfoverlap:"212.669" pubchem_shape_volume:"85.1" pubchem_coordinate_type:"2\n5\n10" Time for openFile(:6589): 5 ms reading 20 atoms ModelSet: haveSymmetry:false haveUnitcells:false haveFractionalCoord:false 1 model in this collection. Use getProperty "modelInfo" or getProperty "auxiliaryInfo" to inspect them. Default Van der Waals type for model set to Babel 20 atoms created ModelSet: not autobonding; use forceAutobond=true to force automatic bond creation Time for creating model: 2 ms 6589 spinY = 10.0 Script completed Jmol script terminated FileManager.getAtomSetCollectionFromFile(cache://localLOAD_ The Resolver thinks Mol C6H14 APtclcactv09282519373D 0 0.00000 0.00000 C6H14 Time for openFile(cache://localLOAD_ 4 ms reading 20 atoms ModelSet: haveSymmetry:false haveUnitcells:false haveFractionalCoord:false 1 model in this collection. Use getProperty "modelInfo" or getProperty "auxiliaryInfo" to inspect them. Default Van der Waals type for model set to Babel 20 atoms created ModelSet: not autobonding; use forceAutobond=true to force automatic bond creation Time for creating model: 1 ms C6H14 spinY = 10.0 Script completed Jmol script terminated FileManager.getAtomSetCollectionFromFile( The Resolver thinks Mol 6589 -OEChem-09292507413D 6589 pubchem_compound_cid:"6589" pubchem_conformer_rmsd:"0.4" pubchem_conformer_diverseorder:"1\n2\n3" pubchem_mmff94_partial_charges:"0" 0 partial charges read pubchem_effective_rotor_count:"1" pubchem_pharmacophore_features:"2\n3 1 3 4 hydrophobe\n3 2 5 6 hydrophobe" pubchem_heavy_atom_count:"6" pubchem_atom_def_stereo_count:"0" pubchem_atom_udef_stereo_count:"0" pubchem_bond_def_stereo_count:"0" pubchem_bond_udef_stereo_count:"0" pubchem_isotopic_atom_count:"0" pubchem_component_count:"1" pubchem_cactvs_tauto_count:"1" pubchem_conformer_id:"000019BD00000001" pubchem_mmff94_energy:"6.4127" pubchem_feature_selfoverlap:"10.182" pubchem_shape_fingerprint:"137420 1 14210207548903878447\n16714656 1 17531538572250223695\n20096714 4 18341602690404418173\n21040471 1 17895747405347166253\n24536 1 18042116725187677885\n29004967 10 18060424577569916894\n5943 1 16099473149988438040" pubchem_shape_multipoles:"123.48\n2.2\n1.33\n0.97\n0\n0\n0.14\n0\n-0.27\n0\n0.28\n0\n0\n0.54" pubchem_shape_selfoverlap:"212.669" pubchem_shape_volume:"85.1" pubchem_coordinate_type:"2\n5\n10" Time for openFile(:6589): 4 ms reading 20 atoms ModelSet: haveSymmetry:false haveUnitcells:false haveFractionalCoord:false 1 model in this collection. Use getProperty "modelInfo" or getProperty "auxiliaryInfo" to inspect them. Default Van der Waals type for model set to Babel 20 atoms created ModelSet: not autobonding; use forceAutobond=true to force automatic bond creation Time for creating model: 1 ms 6589 spinY = 10.0 Script completed Jmol script terminated FileManager.getAtomSetCollectionFromFile( The Resolver thinks Mol 7282 -OEChem-09292507413D 7282 pubchem_compound_cid:"7282" pubchem_conformer_rmsd:"0.4" pubchem_conformer_diverseorder:"1\n6\n7\n2\n3\n4\n5" pubchem_mmff94_partial_charges:"0" 0 partial charges read pubchem_effective_rotor_count:"2" pubchem_pharmacophore_features:"3\n1 4 hydrophobe\n1 5 hydrophobe\n1 6 hydrophobe" pubchem_heavy_atom_count:"6" pubchem_atom_def_stereo_count:"0" pubchem_atom_udef_stereo_count:"0" pubchem_bond_def_stereo_count:"0" pubchem_bond_udef_stereo_count:"0" pubchem_isotopic_atom_count:"0" pubchem_component_count:"1" pubchem_cactvs_tauto_count:"1" pubchem_conformer_id:"00001C7200000001" pubchem_mmff94_energy:"1.5689" pubchem_feature_selfoverlap:"15.225" pubchem_shape_fingerprint:"139733 1 9439404626941083391\n16714656 1 18412826858707038430\n20096714 4 18195251237553861989\n23552423 10 18043256730047065462\n23552449 1 18264482884871725788\n29004967 10 18334862722569769523\n5460574 1 8358260341012732419\n5943 1 11246165242289545169" pubchem_shape_multipoles:"123.48\n3.34\n1.16\n0.67\n0\n0.46\n0\n-0.55\n-0.37\n0\n0.12\n0\n0\n0" pubchem_shape_selfoverlap:"209.738" pubchem_shape_volume:"85.2" pubchem_coordinate_type:"2\n5\n10" Time for openFile(:7282): 4 ms reading 20 atoms ModelSet: haveSymmetry:false haveUnitcells:false haveFractionalCoord:false 1 model in this collection. Use getProperty "modelInfo" or getProperty "auxiliaryInfo" to inspect them. Default Van der Waals type for model set to Babel 20 atoms created ModelSet: not autobonding; use forceAutobond=true to force automatic bond creation Time for creating model: 2 ms 7282 spinY = 10.0 Script completed Jmol script terminated FileManager.getAtomSetCollectionFromFile( The Resolver thinks Mol 7282 -OEChem-09292507413D 7282 pubchem_compound_cid:"7282" pubchem_conformer_rmsd:"0.4" pubchem_conformer_diverseorder:"1\n6\n7\n2\n3\n4\n5" pubchem_mmff94_partial_charges:"0" 0 partial charges read pubchem_effective_rotor_count:"2" pubchem_pharmacophore_features:"3\n1 4 hydrophobe\n1 5 hydrophobe\n1 6 hydrophobe" pubchem_heavy_atom_count:"6" pubchem_atom_def_stereo_count:"0" pubchem_atom_udef_stereo_count:"0" pubchem_bond_def_stereo_count:"0" pubchem_bond_udef_stereo_count:"0" pubchem_isotopic_atom_count:"0" pubchem_component_count:"1" pubchem_cactvs_tauto_count:"1" pubchem_conformer_id:"00001C7200000001" pubchem_mmff94_energy:"1.5689" pubchem_feature_selfoverlap:"15.225" pubchem_shape_fingerprint:"139733 1 9439404626941083391\n16714656 1 18412826858707038430\n20096714 4 18195251237553861989\n23552423 10 18043256730047065462\n23552449 1 18264482884871725788\n29004967 10 18334862722569769523\n5460574 1 8358260341012732419\n5943 1 11246165242289545169" pubchem_shape_multipoles:"123.48\n3.34\n1.16\n0.67\n0\n0.46\n0\n-0.55\n-0.37\n0\n0.12\n0\n0\n0" pubchem_shape_selfoverlap:"209.738" pubchem_shape_volume:"85.2" pubchem_coordinate_type:"2\n5\n10" Time for openFile(:7282): 5 ms reading 20 atoms ModelSet: haveSymmetry:false haveUnitcells:false haveFractionalCoord:false 1 model in this collection. Use getProperty "modelInfo" or getProperty "auxiliaryInfo" to inspect them. Default Van der Waals type for model set to Babel 20 atoms created ModelSet: not autobonding; use forceAutobond=true to force automatic bond creation Time for creating model: 2 ms 7282 spinY = 10.0 Script completed Jmol script terminated | | | 2,2-dimethylbutane | Let's look at how the name 2-methylpentane is derived. Following the guidelines for the IUPAC Nomenclature of Branched Alkaneslisted above: The parent chain has five carbon atoms, or pentane. There is a one-carbon alkyl group that is branched off of the parent chain. If a one-carbon alkane is called methane, then a one-carbon alkyl group is called a methyl group. There are no spaces in the name, so methyl is written in front of pentane, giving us methylpentane. The methyl group is branched off of the second carbon atom in the parent chain, so 2 is written in front of methylpentane. Letters and numbers are separated with a hyphen, making it 2-methylpentane. Next, lets look at how the name 2,3-dimethylbutane is derived. Following the guidelines for the IUPAC Nomenclature of Branched Alkaneslisted above: The parent chain has four carbon atoms, or butane. There are two methyl groups branched off of the parent chain, which means that dimethyl is written in front of butane, making it dimethylbutane. One methyl group is branched off the second carbon atom of the parent chain and one is branched off the third carbon atom,so 2,3- is written in front of dimethylbutane, making it 2,3-dimethylbutane. ✅Example 12.4.1 Write the IUPAC name for each molecule. Solution The parent chain has seven carbon atoms, yielding a name of heptane. There is one ethyl group and one methyl group branched off of the parent chain. The parent chain is numbered from right to left, since this gives the lowest number for any branch off of the parent chain. The methyl group is branched off the 2nd carbon on the parent chain (making it 2-methyl) and the ethyl group is branched off the 5th carbon atom on the parent chain (making it 5-ethyl). Alphabetically, ethyl comes before methyl, so 5-ethyl comes first in the name, making it 5-ethyl-2-methylheptane. The parent chain has seven carbon atoms, which is longer than the six consecutive carbon atoms obtained by tracing straight across the molecule. Therefore, the parent chain is named heptane. There are three methyl groups branched off of the parent chain, which means that trimethyl is written in front of heptane, yielding trimethylheptane. The parent chain is numbered from right to left, since this gives the lowest number for any branch off of the parent chain. The methyl groups are branched off the second, fourth, and fifth carbon atoms of the parent chain, so 2,4,5- is written in front of trimethylheptane, making it 2,4,5-trimethylheptane. ✏️Exercise 12.4.1 Write the IUPAC name for each molecule. Answer A3,4-dimethyloctaneAnswer B3-ethylhexane ✅Example 12.4.2 Write the condensed structural formula and line formula for each. 3-ethyl-4-propyloctane 3,3-dimethylhexane Solution It's often the easiest to begin by drawing out the skeleton structure. The name octane indicates the parent chain has eight carbon atoms. An ethyl group has two carbon atoms and is branched off the 3rd carbon atom on the parent chain.A propyl group has three carbon atoms and is branched off the 4th carbon atom on the parent chain. From here, fill in the H atoms. Remember that hydrogen can only make one bond, while carbon makes four. If a carbon atom is bonded to one other atom by a single bond, it can hold three hydrogen atoms. If a carbon atom makes three bonds to other atoms, it can only hold one hydrogen atom. Once all of the hydrogen atoms have been added, you can check to see if the structure conforms to the general formula for alkanes, C n H 2n +2. Since there are 13 carbon atoms, there should be 2(13) + 2 = 28 hydrogen atoms. Summing together all of the hydrogen atoms in the condensed structural formula shows the formula is indeed C 13 H 28. The line formula is shown above. The parent chain shows a total of eight ends or bends. There are two carbon atoms branched off the third carbon of the parent chain.There are three carbon atoms branched off the fourth carbon of the parent chain. Begin by drawing the skeleton structure. The name hexane indicates the parent chain has six carbon atoms. A methyl group has one carbon atom. The prefix di– indicates there are two methyl groups.Both methyl groups are branched off the 3rd carbon atom on the parent chain. From here, fill in the H atoms. Each hydrogen makes one bond, while carbon makes four.Once all of the hydrogen atoms have been added, you can check to see if the structure conforms to the general formula for alkanes, C n H 2n +2. Since there are 8 carbon atoms, there should be 2(8) + 2 = 18 hydrogen atoms. Summing together all of the hydrogen atoms in the condensed structural formula shows the formula is indeed C 8 H 18. The line formula is shown above. The parent chain shows a total of six ends or bends. There are two one-carbon atom branches off the third carbon of the parent chain. ✏️Exercise 12.4.2 Draw the line structure of 2,4-dimethylpentane. Write the condensed structural formula of 4-ethyl-2,3-dimethyloctane. Answer AAnswer B Summary A structural isomer is one of multiple molecules that have the same molecular formula, but different structural formulas. Nomenclature rules for branched hydrocarbons are given. This pageis shared under a CK-12 license and was authored, remixed, and/or curated by Lance S. Lund (Anoka-Ramsey Community College).Original source: LICENSED UNDER 12.4: Branched Alkanes is shared under a CC BY-NC license and was authored, remixed, and/or curated by LibreTexts. 25.3: Branched Alkanes by CK-12 Foundation is licensed CK-12. Original source: Toggle block-level attributions Back to top 12.3: Alkanes 12.5: Alkenes and Alkynes Was this article helpful? Yes No Recommended articles 12: Organic ChemistryOrganic chemistry involving the scientific study of the structure, properties, and reactions of organic compounds and organic materials, i.e., matter ... Article typeSection or PageLicenseCC BY-NCOER program or PublisherCK-12Show Page TOCno on page Tags source-chem-53995 source-chem-53995 © Copyright 2025 Chemistry LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status× Application loaded. contents readability resources tools ☰ 12.3: Alkanes 12.5: Alkenes and Alkynes Complete your gift to make an impact
10492
https://en.wikipedia.org/wiki/Pinacoderm
Pinacoderm - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Structure 2 References Pinacoderm [x] 3 languages Deutsch Polski தமிழ் Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia The pinacoderm is the outermost layer of body cells (pinacocytes) of organisms of the phylum Porifera (sponges), equivalent to the epidermis in other animals. Structure [edit] The pinacoderm is composed of pinacocytes, flattened epithelial cells that can expand or contract to slightly alter the size and shape of the sponge. It also contains porocytes, oval-shaped cells extending from the pinacoderm to the choanoderm (the body layer containing choanocytes). References [edit] ^Charles F. Lytle; John R. Meyer (2005). General Zoology Laboratory Guide (Fourteenth ed.). This poriferan- (or sponge-) related article is a stub. You can help Wikipedia by expanding it. v t e This animal anatomy–related article is a stub. You can help Wikipedia by expanding it. v t e Retrieved from " Categories: Sponge anatomy Poriferan stubs Animal anatomy stubs Hidden category: All stub articles This page was last edited on 6 May 2023, at 22:39(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Pinacoderm 3 languagesAdd topic
10493
https://dictionary.cambridge.org/us/dictionary/english/petulant
Cambridge Dictionary +Plus My profile +Plus help Log out {{userName}} Cambridge Dictionary +Plus My profile +Plus help Log out Log in / Sign up English (US) Meaning of petulant in English Add to word list Add to word list easily annoyed and complaining in a rude way like a child Synonyms cranky (BAD-TEMPERED) mainly US informal Thesaurus: synonyms, antonyms, and examples getting angry easily bad-temperedHe's so bad-tempered! He never should have become a teacher. irritableBe careful what you say; he's very irritable today. grumpyI'm just a little tired and grumpy today. crankyUS She's very cranky because she has a toothache. impatientI'm too impatient for fishing. See more results » SMART Vocabulary: related words and phrases Bad-tempered argumentative bad-tempered be hell on wheels idiom be like a bear with a sore head idiom be spoiling for a fight idiom grouchily grouchiness grouchy grumpily grumpiness peevishly peppery pettish pettishly petulance stroppiness stroppy surlily surliness testily See more results » You can also find related words, phrases, and synonyms in the topics: Physically and mentally mature & immature Related words petulance petulantly (Definition of petulant from the Cambridge Advanced Learner's Dictionary & Thesaurus © Cambridge University Press) petulant | Intermediate English petulant adjective us /ˈpetʃ·ə·lənt/ Add to word list Add to word list easily angered or annoyed, esp. in a rude way: He plays the part of a petulant young man in the film. (Definition of petulant from the Cambridge Academic Content Dictionary © Cambridge University Press) Examples of petulant petulant The petulant pitcher then fled the ballpark as a clubhouse staffer called police. From New York Post And yet these very same people can at times be boring, irritating, petulant, selfish, angry, or depressed. From Huffington Post Empathy is perhaps why the grackle feels the need to approach humans with the aggression of a petulant and hungry child. From The Verge The internet, in its petulant and infantile glory, is slowly moving toward that ideal. From TechCrunch But he came across as angry, whiny and petulant in the familiar manner of losers who know they're losing. From New York Post They are entirely consistent with the thin-skinned, petulant character of the president-elect. From Washington Post His tone turns petulant, and he begins to issue orders to follow him. From Heritage.org As a result players often behave like petulant psychopaths, and accommodating them can feel disappointing to the game designer. From TechCrunch Can't argue the merits of an issue--so like petulant children, they start calling names. From Washington Post He has absolutely trounced every argument you have made, and even answered your petulant appeals to authority by showing your misunderstanding of that same authority. From Phys.Org For now the main countervailing force is to put a spotlight on the petulant men behaving this way. From The Atlantic It's like a petulant child agreeing to do the bare minimum to get out of a chore. From Gizmodo It's worth noting that both "get a life" and "get over it" are expressions added to our language by petulant 13-year-olds. From The Atlantic Both sides got increasingly, exponentially petulant and combative. From NPR These examples are from corpora and from sources on the web. Any opinions in the examples do not represent the opinion of the Cambridge Dictionary editors or of Cambridge University Press or its licensors. What is the pronunciation of petulant? Translations of petulant in Chinese (Traditional) 耍孩子脾氣的，任性的, 脾氣暴躁的… See more in Chinese (Simplified) 耍孩子脾气的，任性的, 脾气暴躁的… See more in Spanish de mal genio… See more in Portuguese caprichoso, mal-humorado… See more in more languages in Polish in Turkish in Russian kapryśny… See more huysuz, hırçın, ters… See more капризный, вздорный… See more Need a translator? Get a quick, free translation! Translator tool Browse petty thief petty thievery petulance petulancy petulant petulantly petunia pew pewter Word of the Day take something back to admit that something you said was wrong About this Blog Calm and collected (The language of staying calm in a crisis) Read More New Words vibe coding More new words has been added to list To top Contents EnglishIntermediateExamplesTranslations Cambridge Dictionary +Plus My profile +Plus help Log out English (US) Change English (UK) English (US) Español Português 中文 (简体) 正體中文 (繁體) Dansk Deutsch Français Italiano Nederlands Norsk Polski Русский Türkçe Tiếng Việt Svenska Українська 日本語 한국어 ગુજરાતી தமிழ் తెలుగు বাঙ্গালি मराठी हिंदी Follow us Choose a dictionary Recent and Recommended English Grammar English–Spanish Spanish–English Definitions Clear explanations of natural written and spoken English English Learner’s Dictionary Essential British English Essential American English Grammar and thesaurus Usage explanations of natural written and spoken English Grammar Thesaurus Pronunciation British and American pronunciations with audio English Pronunciation Translation Click on the arrows to change the translation direction. Bilingual Dictionaries English–Chinese (Simplified) Chinese (Simplified)–English English–Chinese (Traditional) Chinese (Traditional)–English English–Dutch Dutch–English English–French French–English English–German German–English English–Indonesian Indonesian–English English–Italian Italian–English English–Japanese Japanese–English English–Norwegian Norwegian–English English–Polish Polish–English English–Portuguese Portuguese–English English–Spanish Spanish–English English–Swedish Swedish–English Semi-bilingual Dictionaries English–Arabic English–Bengali English–Catalan English–Czech English–Danish English–Gujarati English–Hindi English–Korean English–Malay English–Marathi English–Russian English–Tamil English–Telugu English–Thai English–Turkish English–Ukrainian English–Urdu English–Vietnamese Dictionary +Plus Word Lists Contents English Adjective Examples Translations Grammar All translations My word lists To add petulant to a word list please sign up or log in. 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10494
https://www.ck12.org/section/analyzing-rational-functions/
Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up HomeMathematicsAnalyzing Rational Functions Analyzing Rational Functions Difficulty Level: Basic | Created by: CK-12 Last Modified: Dec 29, 2014 Learning Objectives Apply properties of rational functions to find roots and asymptotes Graph rational functions using roots, asymptotes, and behavior of the function around vertical asymptotes Summarize Analysis of Rational Functions Recall that rational functions are defined as where and are polynomials. Rational functions can be slightly more complicated to analyze than polynomials, mostly due to the fact that anything divided by zero is undefined. In this section we review major points about analyzing rational functions, and show how a graphing calculator or other computer tool can be used to analyze rational functions. Finding Vertical Asymptotes and Breaks in the Graph of a Rational Function To find vertical asymptotes and breaks in the domain of a rational function, set the denominator equal to zero and solve for . Given , set and solve for . Note, however, that some rational functions do not have vertical asymptotes and they are well-defined for all real numbers . Other rational functions have a break in the function, but no vertical asymptote. This usually happens when one term in the numerator cancels with one term in the denominator. Below we illustrate examples of each of these cases. Example 1 Consider the following rational function. Find all restrictions on the domain and asymptotes Solution Factoring the numerator Canceling Notice that there is no asymptote in this function, but rather a break in the graph at . Example 2 Find the restrictions on the domain of Solution Setting the denominator equal to 0, Thus, the domain of is the set all real numbers with the restriction . has two vertical asymptotes, one at and one at . Example 3 Find the vertical asymptotes of Solution Setting the denominator equal to zero, There are no real solutions, so there are no vertical asymptotes and no restrictions on the domain of this function. Example 4 What are the vertical asymptotes of the function Solution Setting the denominator equal to zero, Again there are no real solutions. The horizontal asymptote is . End Behavior of Rational Function Previously you have analyzed the end behavior of polynomials. You may recall that, as a general rule, as you “zoom out,” the graph of a polynomial looks like the graph of the power function made of the leading term of the polynomial. For instance, looks like the function if you choose a sufficiently large range for and . There is a similar rule for rational functions. The End Behavior of a rational function can often be identified by the horizontal asymptote. That is, as the values of get very large or very small, the graph of the rational function will approach (but not reach) the horizontal asymptote. In Example 4, we could write as and as . This can be shortened by writing as . Finding Oblique Asymptotes Not all asymptotes of rational functions are vertical or horizontal. If we look at the graph of the rational function in Example 3 from above, , we can see that there is no horizontal asymptote of this function. There is no horizontal asymptote in this function because the degree of the numerator is greater than the degree of the denominator. As a reminder, the following guidelines can help identify the asymptotes of a rational function : If the degree of the denominator is greater than the degree of the numerator, then the line is a horizontal asymptote. If the degree of the numerator and the denominator are equal, then the line is a horizontal asymptote, where is the leading coefficient of , the numerator, and is the leading coefficient of , the denominator. If the degree of the numerator is larger than the degree of the denominator, then the quotient function, , found by dividing the numerator and denominator of the rational function is an oblique asymptote. Recall that for any rational function , you can use polynomial division to re-write that function in the form where is the quotient and is the remainder. To illustrate this last point, look at . By polynomial division we have, So . This tells us that the line is an oblique asymptote of . Notice that the oblique asymptotes of a rational function also describe the end behavior of the function. That is, as you “zoom out” from the graph of a rational function it looks like a line or the function defined by in . Example 5 Find the oblique asymptote of . Sketch a graph of . Solution Using polynomial long division, . Thus, the line is an oblique asymptote of . To sketch the graph we can find the vertical asymptotes by setting the denominator equal to zero, So the two vertical asymptotes are and . is undefined, so there is no intercept. Also, there is no simple way to solve for the roots (setting the numerator equal to zero), but we can see by inspection that . To get an idea of the shape of the graph we will make a table of a few test points. We used a calculator to evaluate decimal values of in . Finally we use all of this information to make a sketch of the graph of : Applications, Technological Tools The last example illustrates the difficulty of graphing any rational function by hand. A graphing calculator can be a valuable tool to help with graphing and analyzing rational functions, provided you know how to use the calculator effectively. Below we illustrate a few of the benefits and potential pitfalls of graphing rational functions using TI-83/84 Calculators. Most of the calculator analysis tools that we have reviewed for quadratic and polynomial functions also work with rational functions. You can use the and GRAPH menus to view the graph of a rational function. Once you have graphed a rational function you can use the CALC menu to find the roots and the min/max values of the function. Potential Pitfalls of Graphing Rational Functions There are two common pitfalls when using a calculator to graph a rational function: When graphing a rational function by entering the function in the screen, remember that you need to use parenthesis to group the numerator and denominator of the rational function. Also, the independent variable of a rational function is always in the TI-83/84. Thus, for example, you would enter the function by typing “” in the slot. Vertical asymptotes are sometimes graphed as vertical lines. Graphs of rational functions can be difficult to interpret if the window settings are not chosen carefully. The following example illustrates the second and third problems. Example 6 Graph on the window . (This means and ) Solution ((insert screen shot of the graph of showing that vertical line at )) is undefined and has a vertical asymptote at , but the way the graphing calculator draws the graph, it shows a vertical line at . One way to “fix” this problem is to press MODE and select the option “Dot” rather than “Connected”. However, dot graphs can be hard to interpret as well. Using the Table to Examine Behavior of a Rational Function Despite the potential pitfalls of using a graphing calculator to graph rational functions, the TABLE function of the graphing calculator is one very effective tool for analyzing the behavior of a rational function near its vertical asymptotes. Let's consider the function from Example 5 above, . The vertical asymptotes of are and . To analyze the behavior of near these asymptotes, first enter the equation of in the menu. Next, press 2ND WINDOW to go to the TABLSET menu. There, change the setting of Independent to ASK. ((Calculator Image: Show TABLE SETUP menu)) Press 2ND GRAPH to enter the TABLE screen. In the table you can enter values in the first column, and the second column will display corresponding values. To analyze the behavior of near , enter several values for getting progressively closer to 0. It looks like as from the left that becomes large. (In calculus you will use more rigorous language to describe this situation). Let's look at values larger than zero, that is look at how behaves as from the right. This shows that as from the right, . Likewise, you can use the table to show that as from the left, , and as from the right, . While using the table to sketch a graph by hand may be redundant (after all, you have a graphing calculator!), this information helps you interpret the output of the graphing tool, which is an important skill to develop as you learn to use a graphing calculator. Exercises For each function in questions 1-5, describe the behavior of the rational function. a) What are its asymptotes? b) What is the end behavior of the function? c) How does the function behave as approaches each vertical asymptote form the left and from the right? d) Sketch a graph of the function an an appropriate domain Notes/Highlights | Color | Highlighted Text | Notes | | --- --- | | | Please Sign In to create your own Highlights / Notes | | | Currently there are no resources to be displayed. Description No description available here... Difficulty Level Basic Tags polynomials, rational functions, polynomial, synthetic division, fundamental theorem of algebra, CK.MAT.ENG.SE.1.Math-Analysis.2, (3 more) Subjects mathematics Grades 11, 12 Standards Correlations - Concept Nodes License CC BY NC Language English | Cover Image | Attributions | --- | | | License: CC BY-NC | Date Created Feb 23, 2012 Last Modified Dec 29, 2014 . | Image | Reference | Attributions | --- | | | License: CC BY-NC | | | | License: CC BY-NC | | | | License: CC BY-NC | Show Attributions Show Details ▼ Related Content Simplifying Rational Expressions: Sliding Solver PLIX + Graphing Rational Functions Video Reviews Was this helpful? Yes No 0% of people thought this content was helpful. 00 Back to the top of the page ↑ Oops, looks like cookies are disabled on your browser. Click on this link to see how to enable them. Student Sign Up Are you a teacher? Having issues? 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10495
https://www.nist.gov/programs-projects/acoustic-thermometry
Acoustic Thermometry | NIST Skip to main content An official website of the United States government Here’s how you know Here’s how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you’ve safely connected to the .gov website. Share sensitive information only on official, secure websites. Search NIST Menu Close Publications What We Do All Topics Advanced communications Artificial intelligence Bioscience Buildings and construction Chemistry Cybersecurity and Privacy Electronics Energy Environment Fire Forensic science Health Information technology Infrastructure Manufacturing Materials Mathematics and statistics Metrology Nanotechnology Neutron research Performance excellence Physics Public safety Quantum information science Resilience Standards Transportation Labs & Major Programs Assoc Director of Laboratory Programs Laboratories Communications Technology Laboratory Engineering Laboratory Information Technology Laboratory Material Measurement Laboratory Physical Measurement Laboratory User Facilities NIST Center for Neutron Research CNST NanoFab Research Test Beds Research Projects Tools & Instruments Major Programs Baldrige Performance Excellence Program CHIPS for America Initiative Manufacturing Extension Partnership (MEP) Office of Advanced Manufacturing Special Programs Office Technology Partnerships Office Services & Resources Measurements and Standards Calibration Services Laboratory Accreditation (NVLAP) Quality System Standard Reference Materials (SRMs) Standard Reference Instruments (SRIs) Standards.gov Time Services Office of Weights and Measures Software Data Chemistry WebBook National Vulnerability Database Physical Reference Data Standard Reference Data (SRD) Storefront License & Patents Computer Security Resource Center (CSRC) NIST Research Library News & Events News Events Blogs Feature Stories Awards Video Gallery Image Gallery Media Contacts About NIST About Us Leadership Organization Structure Budget & Planning Contact Us Visit Careers Student programs Work with NIST History NIST Digital Archives NIST Museum NIST and the Nobel Educational Resources PROJECTS/PROGRAMS Acoustic Thermometry Share Facebook Linkedin X.com Email Summary By measuring the speed of sound in argon gas, we have determined thermodynamic temperature with unprecedented accuracy in the range 0 °C to 552 °C. Our results are used to improve the accuracy of platinum resistance thermometry. Description The basketball-sized resonator for the NIST Acoustic Thermometer, partially enclosed in its pressure vessel and inner oven. Data from the NIST acoustic thermometer will form an improved basis for the equations used with platinum resistance thermometry, which is the primary mechanism to disseminate accurate temperature standards. Objective: To obtain measurements of thermodynamic temperature in the range between 273 K and 700 K with an uncertainty five times smaller than 1990 benchmark values. Technical approach: Temperature defined on the International Temperature Scale of 1990 (ITS-90) approximates the thermodynamic temperature. Unfortunately, in the range 273 K to 730 K, the ITS-90 was constructed by necessity from the average of two sets of constant volume gas thermometry data that are inconsistent, for unknown reasons, by much more than their combined uncertainties. The NIST acoustic thermometer is based on the simple relation between the speed of sound of a monatomic gas and the thermodynamic temperature. By measuring ratios of the acoustic resonance frequencies within a nearly spherical cavity at an unknown temperature and at a temperature close to the triple point of water (defined as 273.16 K), we obtain the ratio of the argon speed of sound at these two temperatures. With simultaneous measurements of microwave resonance frequencies in the same cavity, we can account for the thermal expansion of the stainless-steel shell enclosing the argon. Our thermometer has several novel features construction from inert metals and ceramics only, continuous gas purging of the resonator to minimize gas impurities, a gas-chromatography system to measure the impurities in the gas exiting the resonator, measurements on the ITS-90 with up to five long-stem standard platinum resistance thermometers, simultaneous measurement of microwave and acoustic resonances, and high-temperature acoustic transducers. We have achieved our goal of reducing the uncertainty of thermodynamic temperature by a factor of five in the range 273 K to 552 K. Our data is in excellent agreement with recent radiometric measurements (at 700 K and above), other acoustic measurements (at 390 K and below), and one of the past gas thermometry results. In 2008, we successfully developed techniques to perform precision acoustic measurements with a microphone at room temperature. The technique has pushed the upper limit of our data to 650 K. Major Accomplishments In 2007, published results up to 552 K, resolving previous discrepancy that limited accuracy of the international temperature scale In 2008, completed measurements up to 650 K at unprecedented accuracy. Thermometry metrology and Thermodynamics Organizations NIST Headquarters Laboratory Programs Physical Measurement Laboratory Sensor Science Division Temperature & Humidity Group NIST Staff Weston L. Tew Gregory F. Strouse Contact Weston L. Tew weston.tew@nist.gov (301) 975-4811 Project Status Completed Related Publications Acoustic Thermometry Results From 271 K to 552 K Progress in Primary Acoustic Thermometry at NIST: 273 K to 505 K Techniques for Primary Acoustic Thermometry to 800 K Created August 28, 2015, Updated March 26, 2025 Was this page helpful? HEADQUARTERS 100 Bureau Drive Gaithersburg, MD 20899 301-975-2000 Webmaster | Contact Us | Our Other Offices X.comFacebookLinkedInInstagramYouTubeGiphyRSS FeedMailing List How are we doing? 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10496
https://d9-wret.s3.us-west-2.amazonaws.com/assets/palladium/production/mineral-pubs/sulfur/sulfumyb03.pdf
SULFUR—2003 74.1 SULFUR By Joyce A. Ober Domestic survey data and tables were prepared by Brian W. Jaskula, statistical assistant, and the world production table was prepared by Glenn J. Wallace, international data coordinator. The United States was once again the world’s leading sulfur producer in 2003 with 9.6 million metric tons (Mt). The sulfur was produced as a byproduct of efforts to meet environmental requirements that limit atmospheric emissions of sulfur dioxide. Worldwide, regulations forced increased sulfur recovery for environmental reasons, resulting in a continued decline in the production of native sulfur and pyrites. Production outpaced sulfur demand, which resulted in increased stocks at some operations, especially at a few in remote locations from which it is difficult and costly to ship the product to market. Through its major derivative, sulfuric acid, sulfur ranks as one of the most important elements used as an industrial raw material and is of prime importance to every sector of the world’s fertilizer and manufacturing industries. Sulfuric acid production is the major end use for sulfur, and consumption of sulfuric acid has been regarded as one of the best indices of a nation’s industrial development. More sulfuric acid is produced in the United States every year than any other chemical; 41.0 Mt, which is equivalent to about 13.3 Mt of elemental sulfur, was produced in 2003; this was 4.6% more than that of 2002 (U.S. Census Bureau, 2004). In 2003, domestic production and shipments of sulfur in all forms were 3.5% and 3.7% higher, respectively, than those of 2002. Consumption increased, as did imports and prices (table 1; figures 1-4). The United States maintained its position as the leading world consumer of sulfur and sulfuric acid. The quantity of sulfur recovered domestically during the refining of petroleum continued the upward trend established in 1939, the second year that such production was reported, by increasing by 3.3%. Sulfur recovered from natural gas processing increased by 11.0%. Byproduct sulfuric acid from the Nation’s nonferrous smelters and roasters, produced as a result of laws restricting sulfur dioxide emissions, supplied a significant quantity of sulfuric acid to the domestic merchant (commercial) acid market. Production from this sector decreased by 11.5% as a result of decreased recovery at copper smelters. Three copper smelters, one lead smelter, one molybdenum smelter, and one zinc smelter reported production of byproduct sulfuric acid. Estimated world sulfur production was slightly higher in 2003 than it was in 2002 (table 1). Recovered elemental sulfur is produced primarily during the processing of natural gas and crude petroleum. For the third consecutive year, about 90% of the world’s elemental sulfur production came from recovered sources. Some sources of byproduct sulfur are unspecified, which means that the material could be elemental or byproduct sulfuric acid. The quantity of sulfur produced from recovered sources was dependent on the world demand for fuels, nonferrous metals, and petroleum products, not for sulfur. World sulfur consumption was slightly higher than it was in 2002; about 50% was used in fertilizer production, and the remainder, in myriad other industrial uses. World trade of elemental sulfur increased by 10% from the levels recorded in 2002. Worldwide inventories of elemental sulfur were slightly higher. Legislation and Government Programs Late in 2003, the U.S. Environmental Protection Agency (EPA) issued a report finding that U.S. refiners were on target for meeting the 2006 deadline for low-sulfur diesel fuel. By that time, an estimated 96% of diesel fuel produced in the United States will meet the 15-part-per-million (ppm) standard (Oil & Gas Journal, 2003b). Earlier in the year, the EPA proposed standards intended to reduce pollution from diesel-powered off-road vehicles to match the requirements for on-road vehicles. Included in the rules are changes to the design for new engines for off-road vehicles that would limit emissions, including those of nitrogen oxides, particulates, and sulfur dioxide. In addition to changes in diesel engine design, fuel specifications will also change. The sulfur limit in fuels for off-road vehicles will be reduced in two steps. The sulfur limit will be reduced to 500 ppm from 3,400 ppm by 2007, and further reductions will take it to 15 ppm in 2010 (Fialka, 2003). Production Elemental Sulfur.—U.S. production statistics were collected on a monthly basis and published in the U.S. Geological Survey (USGS) monthly sulfur Mineral Industry Surveys. All of the 108 operations to which survey requests were sent responded; this represented 100% of the total production listed in table 1. In 2003, production and shipments were about 5% higher than those of 2002. The value of shipments was 2.5 times higher than in 2002 owing to a similar increase in the average unit value of elemental sulfur. Trends in sulfur production are shown in figures 1 and 3. Frasch.—Until 2000, native sulfur associated with the caprock of salt domes and in sedimentary deposits in the United States was mined by the Frasch hot-water method in which the native sulfur was melted underground with super-heated water and brought to the surface by compressed air. Freeport-McMoRan Sulphur Inc. (a subsidiary of McMoRan Exploration Co.) closed the last domestic Frasch mine, Main Pass, in 2000 (Fertilizer Markets, 2000). Recovered.—Recovered elemental sulfur, which is a nondiscretionary byproduct from petroleum-refining, natural-gas-processing, and coking plants, was produced primarily to comply with environmental regulations that were applicable directly to emissions from the processing facility or indirectly 74.2 U.S. GEOLOGICAL SURVEY MINERALS YEARBOOK—2003 by restricting the sulfur content of the fuels sold or used by the facility. Recovered sulfur was produced by 38 companies at 108 plants in 26 States and 1 plant in the U.S. Virgin Islands. Most of these plants were small with only 33 reporting production that exceeded 100,000 metric tons per year (t/yr). By source, 78.2% of recovered elemental sulfur production came from petroleum refineries or satellite plants that treated refinery gases and coking plants, and the remainder was produced at natural-gas treatment plants (table 3). The leading producers of recovered sulfur, all with more than 500,000 metric tons (t) of sulfur production, in descending order of production, were Exxon Mobil Corp., BP p.l.c., ChevronTexaco Corp., ConocoPhillips Co., Shell Oil Co. (including its joint-venture and subsidiary operations), and CITGO Petroleum Corp. (including its joint-venture refinery). The 53 plants owned by these companies accounted for 68.4% of recovered sulfur output during the year. Recovered sulfur production by State and district is listed in tables 2 and 3. Four of the world’s 16 largest refineries, each with capacity of at least 400,000 barrels per day (bbl/d), are in the United States. They are, in decreasing order of production, ExxonMobil’s Baytown, TX, refinery; Hovensa LLC’s St. Croix, VI, refinery; ExxonMobil’s Baton Rouge, LA, refinery; and BP’s Texas City, TX, refinery (Nakamura, 2003). Refining capacity does not necessarily mean that these refineries were the leading producers of refinery sulfur. Sulfur production depends on installed sulfur recovery capacity as well as the types of crude oil that are refined at the specific refineries. Major refineries that process low-sulfur crudes may have relatively low sulfur production. Byproduct Sulfuric Acid.—Sulfuric acid production at copper, lead, molybdenum, and zinc roasters and smelters accounted for about 7.1% of the total domestic production of sulfur in all forms; this was a decrease of 11.5% compared with that of 2002 (table 4). Three acid plants operated in conjunction with copper smelters, and three were accessories to lead, molybdenum, and zinc smelting and roasting operations. The three leading sulfuric acid plants were associated with copper mines and accounted for 86.4% of the output. The copper producers—ASARCO Incorporated, Kennecott Utah Copper Corp., and Phelps Dodge Corp.—each operated a sulfuric acid plant at their primary copper smelters. Consumption Apparent domestic consumption of sulfur in all forms was 4.6% higher than that of 2002 (table 5). Of the sulfur consumed, 73.5% was obtained from domestic sources—elemental sulfur (68.4%) and byproduct acid (5.1%)—compared with 74.6% in 2002 and 79.9% in 2001. The remaining 26.5% was supplied by imports of recovered elemental sulfur (24.0%) and sulfuric acid (2.5%). The USGS collected end-use data on sulfur and sulfuric acid according to the standard industrial classification of industrial activities (table 6). Sulfur differs from most other major mineral commodities in that its primary use is as a chemical reagent rather than as a component of a finished product. This use generally requires that it be converted to an intermediate chemical product prior to its initial use by industry. The leading sulfur end use, sulfuric acid, represented 62.7% of reported consumption with an identified end use. Some identified sulfur end uses were tabulated in the “Unidentified” category because these data were proprietary. Data collected from companies that did not identify shipment by end use also were tabulated as “Unidentified.” A significant portion of the sulfur in the “Unidentified” category may have been shipped to sulfuric acid producers or exported, although data to support such an assumption were not available. Because of its desirable properties, sulfuric acid retained its position as the most universally used mineral acid and the most produced and consumed inorganic chemical, by volume. Data based on USGS surveys of sulfur and sulfuric acid producers showed that reported U.S. consumption of sulfur in sulfuric acid (100% basis) increased by 6%. Data from that survey, however, showed total sulfur consumption was more than 20% higher than that of 2002, a figure that is much higher than reasonable expectations would warrant. Reported consumption figures do not correlate with calculated apparent consumption owing to reporting errors and possible double counting in some data categories. Significant increases in industrial end use data in 2003 are a result of more complete reporting from companies than in 2002. These data are considered independently from apparent consumption as an indication of market shares rather than actual consumption totals. Agriculture was the leading sulfur-consuming industry; consumption increased slightly to 8.51 Mt compared with 8.46 Mt in 2002. Reported consumption of sulfur in the production of phosphatic fertilizers was about the same as that of 2002. According to export data from the U.S. Census Bureau (2004), the estimated quantity of sulfur needed to manufacture exported phosphatic fertilizers increased by 9.3% to 5.2 Mt. The second leading end use for sulfur was in petroleum refining and other petroleum and coal products. Producers of sulfur and sulfuric acid reported a 55% increase in the consumption of sulfur in that end use. Changes in the refining industry indicate increases in refinery processes that require sulfur and sulfuric acid, but the dramatic increases are probably also owing to improved survey response in addition to increased consumption. Demand for sulfuric acid in copper ore leaching, which was the third leading end use, decreased by 40% as a result of continued low copper production from leaching operations and limited sulfuric acid availability in the regions of the United States where these operations are located. The U.S. Census Bureau (2004) also reported that 2.85 Mt of sulfuric acid was produced as a result of recycling spent and contaminated acid from petroleum alkylation and other processes. Two types of companies recycle this material—companies that produce acid for consumption in their own operations and also recycle their own spent acid and companies that provide acid regeneration services to sulfuric acid users. The petroleum refining industry was believed to be the leading source and consumer of recycled acid for use in its alkylation process. Stocks Yearend inventories held by recovered elemental sulfur producers decreased to 206,000 t, or about 14% more than that of 2002 (table 1). Based on apparent consumption of all forms SULFUR—2003 74.3 of sulfur, combined yearend stocks amounted to about a 6-day supply in 2003, compared with a 6-day supply in 2002, an 8-day supply in 2001, a 6-day supply in 2000, and a 12-day supply in 1999. Final stocks in 2003 represented 3.6% of the quantity held in inventories at the end of 1976 when sulfur stocks peaked at 5.65 Mt; this was a 7.4-month supply at that time (Shelton, 1978, p. 1296). Although markets were favorable throughout the year, U.S. producers on the Gulf of Mexico were planning for the possibility of excess supplies in the future. Most refineries face difficult choices when sulfur production exceeds demand and could be forced to curtail refining without an outlet for the sulfur produced. For this reason, ExxonMobil, the leading sulfur producer in the United States, was considering the construction of a sulfur-forming plant somewhere on the Gulf Coast, providing the possibility of exporting formed sulfur, if the need should arise. The company was seeking support from other producers and contemplating the best site for the plant (North American Sulphur Review, 2003b). Prices The contract prices for elemental sulfur at terminals in Tampa, FL, which are reported weekly in Green Markets, began the year at $56.50 to $59.50 per metric ton. In April, prices increased to $68.50 to $71.50 per ton and remained there until July when they fell to $64.50 to $67.50 per ton. Contract prices rose in October to $67.50 to $70.50 per ton and remained at that level through the remainder of the year. Based on total shipments and value reported to the USGS, the average value of shipments for all elemental sulfur was estimated to be $28.71 per ton, which was 142% higher than that of 2002. This dramatic increase was a result of increased demand worldwide. Prices continued to vary greatly on a regional basis, which caused the price discrepancies between Green Markets and USGS data. Tampa prices were usually the highest reported because of the large sulfur demand in the central Florida area. At the beginning of 2003, U.S. west coast prices were listed at $0 per ton, although, in reality, west coast producers can often face negative values as a result of costs incurred at forming plants. These costs were necessary to make solid sulfur in acceptable forms, often known as prills, to be shipped overseas. The majority of west coast sulfur was sent to prillers who may have been subsidized by the refineries, and the formed sulfur was shipped overseas. By March, however, increased international demand spurred the increase of west coast prices to between $15 and $20 per ton. Foreign Trade Exports of elemental sulfur from the United States, which included the U.S. Virgin Islands, as listed in table 7, were 8.0% higher in quantity than those of 2002 and 15.1% higher in value because the average unit value of U.S. export material increased to $62.08 per ton. Exports from the west coast were 651,000 t, or 87.7% of total U.S. exports. The United States continued to be a net importer of sulfur. Imports of elemental sulfur exceeded exports by more than 2 Mt. Recovered elemental sulfur from Canada and Mexico delivered to U.S. terminals and consumers in the liquid phase furnished about 91.2% of all U.S. sulfur import requirements. Total elemental sulfur imports increased by about 12.0% in quantity and higher prices resulted in the value being more than 2.5 times what it was in 2002. Imports from Canada, mostly by rail, were 6.7% higher in quantity, and waterborne shipments from Mexico were 24.2% higher than those of 2002 (table 9). Imports from Venezuela were estimated to account for about 8.8% of all imported elemental sulfur. In addition to elemental sulfur, the United States also had significant trade in sulfuric acid. Sulfuric acid exports were 39.2% higher than those of 2002 (table 8). Acid imports were 4.42 times greater than exports (tables 8, 10). Canada and Mexico were the sources of about 61% of U.S. acid imports, most of which were probably byproduct acid from smelters. Canadian and some Mexican shipments to the United States came by rail, and the remainder of imports came primarily by ship from Europe. The tonnage of sulfuric acid imports was 14.2% lower than that of 2002, and the value of imported sulfuric acid decreased by 15.4%. World Industry Structure The global sulfur industry remained divided into two sectors―discretionary and nondiscretionary. In the discretionary sector, the mining of sulfur or pyrites is the sole objective; this voluntary production of native sulfur or pyrites is based on the orderly mining of discrete deposits with the objective of obtaining as nearly a complete recovery of the resource as economic conditions permit. In the nondiscretionary sector, sulfur or sulfuric acid is recovered as an involuntary byproduct; the quantity of output subject to demand for the primary product irrespective of sulfur demand. Nondiscretionary sources, once the primary sources of sulfur in all forms, represented 9.6% of the sulfur produced in all forms worldwide as listed in table 11. Poland was the only country that produced more than 500,000 t of native sulfur by using either the Frasch or conventional mining methods (table 11). Small quantities of native sulfur were produced in Asia, Europe, and South America. The importance of pyrites to the world sulfur supply has significantly decreased; China was the only country of the top producers with more than 500,000 t of sulfur produced whose primary sulfur source was from pyrites. About 74% of world pyrites production was in China. Of the 22 countries listed in table 11 that produced more than 500,000 t of sulfur, 14 obtained the majority of their production as recovered elemental sulfur. These 22 countries produced 91.5% of the total sulfur produced worldwide. The international sulfur trade was dominated, in descending order of quantity, by Canada, Russia, Saudi Arabia, the United Arab Emirates, and Japan; these countries exported more than 1 Mt of elemental sulfur each and accounted for 70.3% of total sulfur trade. Major sulfur importers, in descending order, were China, Morocco, the United States, Tunisia, Brazil, and India, all with imports of more than 1 Mt. World production of sulfur was slightly higher in 2003 than it was in 2002; consumption was believed to be slightly higher 74.4 U.S. GEOLOGICAL SURVEY MINERALS YEARBOOK—2003 also. Production exceeded consumption in 2003 for the 12th consecutive year, although surpluses were smaller than they had been in recent years (Kitto, 2004). Prices in most of the world were believed to have averaged higher throughout the year than in 2002, for the second consecutive year. Production of Frasch sulfur was slightly lower than that of 2002; production at the last mine in Poland remained steady. Recovered sulfur production was virtually the same, and byproduct sulfuric acid production increased by 3.5% compared with those of 2002. Supplies of sulfur in all forms continued to exceed demand; worldwide sulfur inventories increased, much of which was stockpiled in Canada, although Canadian stocks actually declined owing to the strong international demand for sulfur. Globally, production of sulfur from pyrites was about the same. Statistics compiled by the Oil & Gas Journal showed that the United States possessed 20% of the world’s total refining capacity and 42% of the world’s sulfur recovery capacity derived from oil refineries. The publication listed 717 oil refineries in 118 countries; only about one-half of these countries were reported to have sulfur recovery capacity (Stell, 2003§1). Although the sulfur recovery data appeared to be incomplete, analysis of the data showed that most of the countries that reported no sulfur recovery at refineries were small and had developing economies and limited refining industries. In general, as refining economies improve and the refining industries mature, additional efforts are made to reduce atmospheric emissions through installation of sulfur recovery units. Sulfur levels in motor fuels were being cut worldwide. In 2002, the European Council (EC) speeded up the deadline for mandatory sulfur-free fuels to 2009 from 2011. At that time, 10 ppm will be the maximum quantity of sulfur allowable in gasoline and diesel for all vehicles and equipment including off-road vehicles (Sulphur, 2002a). In 2003, environmental ministers of the EC encouraged member countries to use tax incentives to further speed the introduction of 10-ppm sulfur fuels. The Ministers would like to advance the deadline for compliance to 2005 (Sulphur, 2003j). Russia’s regulations limiting sulfur in fuels are not as strict as those in the European Union (EU). New legislation places the maximum sulfur content for diesel at 350 ppm and for gasoline at 150 ppm by 2004. Efforts were being made to make lower sulfur fuels available for vehicles that will be traveling in the EU to conform to regulations there (Sulphur, 2002b). Japan was working to limit the sulfur content of diesel and gasoline to 10 ppm from 50 ppm and 500 ppm, respectively, by 2008 (Sulphur, 2002a). The European Commission proposed new regulations limiting the sulfur content of ocean-going ship fuels to 1.5% sulfur for ships operating within three sulfur control areas. As proposed, these limits would reduce sulfur emissions from shipping by 10%. Shipping companies want alternative methods for reducing sulfur emissions, such as cleaning stack gases and allowance trading. Environmental groups argued that the reduction goals were too low, pushing for cuts of 80% through sulfur fuel limits of 0.5% for ships operating within 200 miles of the European Union’s coast and 0.2% within 12 miles of the coast (Sulphur, 2003h). Later in the year, the European Parliament proposed new sulfur limits of 0.5% in diesel fuels used in shipping and passenger vessels operating in European waters by 2008. Ships operating in European territory would be limited to 1.5% sulfur fuels by 2011 even if they do not enter European ports. A level of 0.5% would apply to these vessels after 2012. Marine diesel averaged about 3% sulfur in 2003 (Sulphur, 2003e). World Review Canada.—Canada was second only to the United States in production of byproduct sulfur and sulfur in all forms. It led the world in exports of elemental sulfur and stockpiled material. The majority of sulfur production came from natural gas plants in Alberta. Sulfur inventories were 14.2 Mt at the end of 2003, a slight decrease from those of 2002. Although some producers added to their stockpiles in some locations, others remelted inventories for shipment, resulting in a net decrease (North American Sulphur Review, 2004). Sulfur recovered from natural gas has declined in Canada for the past 3 years, and that trend is expected to continue. Recovery at refineries should increase, but the largest changes will be as a result of additional production from oil sands. Sulfur from oil sands may not be readily available to the market. Much of the production is at remote locations where market access is limited and the material has been poured to block, the term used for stockpiling sulfur (North American Sulphur Review, 2002a). Alberta has huge deposits of oil sands with estimated reserves of 300 million barrels (Mbbl) of recoverable crude oil that contain 4% to 5% sulfur (Stevens, 1998). The crude oil resource in oil sands in Alberta is larger than the proven reserves of crude oil in Saudi Arabia (Pok, 2002). As traditional petroleum production in Canada declined, oil sands became a more important source of petroleum for the North American market (Cunningham, 2001). The proportion of Canadian production from oil sands was expected to increase to 21% in 2005 and 30% in 2010 from 9% in 2001 (Pok, 2002). Expansions of oil sands operations were planned by several companies, several existing oil refineries were undergoing conversions to enable the processing of bitumen from oil sands, and pipelines were being built to deliver the bitumen to the refineries from the deposits. Canada’s ratification of the Kyoto Protocol, which limits carbon dioxide emissions, put the future of many oil sands operations in doubt. Large quantities of carbon dioxide are produced in the process of upgrading bitumen. The cost of reducing carbon dioxide emissions could increase the cost of producing oil sands too much for at least some of the projects to remain economically feasible. The Province of Alberta was concerned that ratifying the Kyoto Protocol could cost the industry many billions of dollars and many jobs (Cunningham, 2002). Rising costs and the Kyoto Protocol prompted some Canadian oil sands developers to reconsider additional investments. Petro-Canada considered delaying upgrading its Strathcona refinery near Edmonton, Alberta, to process bitumen from oil sands. Another company delayed oil-sands-related spending. Suncor Energy Inc. bought a U.S. refinery to process its high-sulfur synthetic crude, and Canadian Natural Resources Ltd. was considering a similar action. The United States did not intend to ratify the Kyoto Protocol (Sulphur, 2003g). 1A reference that includes a section mark (§) is found in the Internet Reference Cited section. SULFUR—2003 74.5 Kazakhstan.—The Tengiz oilfield and gasfield is the main source of current sulfur production in Kazakhstan. Located on the northeastern shore of the Caspian Sea in western Kazakhstan, Tengiz has been operated by Tengizchevroil (TCO) since 1993. The owners of TCO are ChevronTexaco (50%), ExxonMobil (25%), Kazakhoil National Oil and Gas Co. (Kazakhstan’s national oil and gas company) (20%), and LUKARCO (a joint venture between BP and Russian oil company LUKoil Oil Co.) (5%) (Chevron Corp., 2000). One of the world’s largest oilfields, Tengiz contains high-quality oil with 0.49% sulfur and associated natural gas that contains 12.5% hydrogen sulfide (Connell and others, 2000). Late in 2002, disagreements between the Government of Kazakhstan and TCO threatened further development of the Tengiz condensate-and-sour-gas field. Renegotiation of the original terms of the financial agreement between the Government and ChevronTexaco created doubts as to whether TCO would proceed with the second stage of development. In addition to the financial questions, local courts fined the company $73 million for environmental damage caused by the 6 Mt of elemental sulfur stockpiled at the site (Sulphur, 2003k). In 2003, the fine for exceeding the allowable sulfur stockpiling and causing ecological damage at Tengiz was reduced to $7 million from $73 million by the Supreme Court of Kazakhstan. The disagreement over financing of the expansion project prompted the consortium to suspend the expansion until agreement was reached. Following resolution of the conflicts with the Government of Kazakhstan, the expansion proceeded at Tengiz. The Tengiz expansion plan to nearly double oil production includes the reinjection of sour gas, limiting total recovery of sulfur at the site (Sulphur, 2003d). After some shipments of flaked sulfur were exported by rail to China in 2002, the first shipments of granulated sulfur following the installation of sulfur forming apparatus at Tengiz went to Israel, Spain, and Tunisia in 2003. Stockpiles of 6 Mt of blocked sulfur remain in place, and alternative disposal scenarios were being considered (Sulphur, 2003c). TCO proposed burying excess production in an old uranium mine. The Government rejected this proposal but countered with the possibility of using an old chromium mine (Sulphur, 2003j). Sulfur also is recovered from the Karachaganak gas-condensate field in Kazakhstan near the Russian border. Because it is close to the Russian gas processing operation in Orenberg, sour gas from Karachaganak is treated at Orenberg. No gas treatment facilities have been installed at the site (Sulfur, 2001a). Mexico.—Mexico was the second leading supplier of imported recovered sulfur to the United States. The majority of its sulfur is produced at petroleum refineries, and byproduct sulfuric acid is recovered at its smelters. Petrόleos Mexicanos S.A. de C.V. was pursuing a program to cut emissions from its refineries to improve the air quality in Mexico by increasing the efficiency of its sulfur recovery units to more than 99%. Nine sulfur recovery units have been completed with a total capacity of 3,440 metric tons per day (t/d) [1.26 million metric tons per year (Mt/yr)]. The improvement process was initiated in 1996 when the North American Free Trade Agreement was ratified and new Mexican environmental laws were enacted. After evaluating existing sulfur recovery units, plans were made to replace or upgrade facilities that did not meet new guidelines. Air quality improvements were to continue (Sulphur, 2003i). Philippines.—Crew Development Corp. went forward with its attempt to develop the Pamplona native sulfur deposit as a raw material source for a local fertilizer producer. Crew originally considered developing the sulfur deposit to supply its Mindinoro laterite nickel project but encountered difficulties getting permits for the pressure acid leach project. The company was considering commercial development of Pamplona prompted by increased sulfur prices. Pamplona contained 40 Mt of sulfur ore averaging 30% sulfur, as native sulfur and sulfides, that was amenable to open pit mining and another 80 Mt of inferred resources. Crew would produce between 2 and 4 Mt/yr (Sulphur, 2003a). Russia.—Russia is the second leading sulfur exporter in the world with more than 4 Mt of elemental sulfur exports in 2003 (International Fertilizer Industry Association, 2004). Gazprom’s gas processing plants in Astrakhan and Orenburg are the leading producers, totaling more than 5 Mt in 2002 (Sulphur, 2003i). MMC Norilsk Nickel started a cleanup project at its Siberian nickel smelter that will eventually result in the production of 1 Mt/yr of sulfur. The company also was working on another cleanup project on the Kola Peninsula in cooperation with the Government of Norway. Sulfur emissions by Norilsk’s Polar Division were to decrease by 70% by 2010. Sulfur emissions on the Kola Peninsula were to decrease by 90% by 2006 (Sulphur, 2003f). Outlook The sulfur industry continued on a path of increased production, slow growth in consumption, higher stocks, and expanded world trade. U.S. production from petroleum refineries is expected to increase substantially in the next few years as expansions, upgrades, and new facilities at existing refineries are completed, thus enabling refiners to increase throughput of crude oil and to process higher sulfur crudes. Production from natural gas operations was higher than it was in 2002, but that trend is not expected to continue. In fact, significant decreases are expected from gas operations in Wyoming, the State in which about 70% of all U.S. natural gas sulfur is recovered. Of four large gas operations in the State, three were expected to experience significant decreases in production beginning in 2003. Production at two operations was decreasing as a natural function of long-term extraction of natural gas. The operator of another gas plant was installing sour gas reinjection apparatus that would eliminate production at that site. The final company recently expanded its operation but was exploring the possibility of storing excess production underground. Theoretically, this material would be available to meet future needs. In reality, however, it represented an option for disposing of unwanted surplus material. Wyoming sulfur production is predicted to be 27% lower in 2005 than it was in 2002 even without disposal at the fourth operation (North American Sulphur Review, 2002c). If that company chooses to dispose of sulfur rather than market it, material recovered from natural gas processing could become a very small part of the domestic industry. Worldwide recovered sulfur output is expected to continue to increase. The largest increases in recovered sulfur production 74.6 U.S. GEOLOGICAL SURVEY MINERALS YEARBOOK—2003 through 2005 are expected to come from the Middle East’s and Russia’s growth in sulfur recovery from natural gas, Canada’s expanded oil sands operations, and Asia’s improved sulfur recovery at oil refineries. Refineries in developing countries should begin to improve environmental protection measures and eventually approach the environmental standards of plants in Japan, North America, and Western Europe. Experts from the natural gas industry estimated that the world demand for natural gas will grow by 2.5% per year during the next 20 years for a total 50% increase in demand. Producing 50% more gas means recovering at least an additional 50% in sulfur from that source. Future gas production, however, is likely to come from deeper, hotter, and more sour deposits that will result in even more excess sulfur production unless more efforts are made to develop new large-scale uses for sulfur. Other alternative technologies for reinjection and long-term storage to eliminate some of the excess sulfur supply will require further investigation to handle the quantity of surplus material anticipated (Hyne, 2000). Byproduct sulfuric acid production will remain depressed in the United States so long as the copper smelters remain idle. With the copper industry’s switch to lower cost production processes and producing regions, the four idle smelters may never reopen. BHP Billiton decided to permanently close its Magma, AZ, copper smelter that has been on a care-and-maintenance status since 1999 (North American Sulphur Review, 2003a). Other companies may make similar decisions. Worldwide, the outlook is different. Because copper production costs in many countries are lower than in the United States, acid production from those countries has not decreased as drastically, and increased production is likely. Environmental controls have been less of a concern in developing countries in the past. Many copper producers in developing and even in developed countries, however, are installing more efficient sulfuric acid plants to limit sulfur dioxide emissions at new and existing smelters. Planned and in-progress improvement projects could increase byproduct acid production significantly, although growth has been slower than previously expected. Frasch sulfur and pyrites production, however, have little chance of significant long-term increases, although higher sulfur prices have resulted in the temporary increases in pyrites consumption. Because of the continued growth of elemental sulfur recovery for environmental reasons rather than demand, discretionary sulfur has become increasingly less important as demonstrated by the decline of the Frasch sulfur industry. The Frasch process has become the high-cost process for sulfur production. Pyrites, with significant direct production costs, is an even higher cost raw material for sulfuric acid production when the environmental aspects are considered. Discretionary sulfur output should show a steady decline. The decreases will be more pronounced when large operations are closed outright for economic reasons, as was the case in 2000 and 2001. Sulfur and sulfuric acid will continue to be important in agricultural and industrial applications, although consumption will be less than production. World sulfur demand for fertilizer is forecast to increase by about 2.3% per year for the next 10 years; industrial demand is predicted to grow by 2.2% per year as a result of increased demand for copper and nickel leaching. The most important changes in sulfur consumption will be in location. Phosphate fertilizer production, where most sulfur is consumed, is projected to increase by about 2.0% per year through 2011. With new and expanding phosphate fertilizer capacity in Australia, China, and India, sulfur demand will grow in these areas at the expense of some phosphate operations elsewhere, thus transferring sulfur demand rather than creating new demand. The effects were already being felt by the U.S. phosphate industry as reflected in the permanent closure of some facilities and reduced production at others. U.S. phosphate products supply domestic requirements, but a large portion of U.S. production is exported. China and India are primary markets for U.S. phosphatic fertilizers. As the phosphate fertilizer industries develop in these countries, some of the markets for U.S. material could be lost. Sulfur will be required for phosphate production at new operations, and more producers will be competing for those markets. Use of sulfur directly or in compounds as fertilizer should increase, but this use will be dependent on agricultural economies and increased acceptance of the need for sulfur in plant nutrition. If widespread use of plant nutrient sulfur is adopted, then sulfur consumption in that application could be significant; thus far, however, growth has been slow. Industrial sulfur consumption has more prospects for growth than in recent years, but still not enough to consume all projected surplus production. Conversion to or increases in copper leaching by producers that require significantly more sulfuric acid for the leaching operations than was used in 2003 bode well for the sulfur industry. Nickel pressure acid leach operations were demanding increased quantities of sulfur. Changes in the preferred methods for producing oxygenated gasoline, especially in Canada and the United States, might result in additional alkylation capacity that would require additional sulfuric acid. Other industrial uses show less potential for expansion. Production is expected to surpass demand well into the future. Unless less traditional uses for elemental sulfur increase significantly, the oversupply situation will result in tremendous stockpiles accumulating around the world. In the 1970s and 1980s, research was conducted that showed the effectiveness of sulfur in several construction uses that held the promise of consuming huge quantities of sulfur in sulfur-extended asphalt and sulfur concretes. In many instances, these materials were found to be superior to the more conventional products, but their use thus far has been very limited. Interest in these materials seemed to be increasing but only in additional research. When sulfur prices are as high as they were in 2003, they are less attractive for unconventional applications where low-cost raw materials are the goal. Regardless of the prevailing price increases in 2003 that signaled tight supplies, the worldwide oversupply situation is likely to worsen. Unless measures are taken to use more sulfur, either voluntarily or through government mandate, large quantities of excess sulfur could be amassed in many more areas of the world, including the United States. SULFUR—2003 74.7 References Cited Chevron Corp., 2000, Chevron finalizes additional stake in Tengiz joint venture: San Francisco, CA, Chevron Corp. press release, August 29, 1 p. Connell, Dave, Ormiston, Bob, Amott, Nick, and Cullum, Irene, 2000, Gas-plant update moves Tengiz field toward 2004 producing target: Oil & Gas Journal, v. 98.24, June 12, p. 64-72. Cunningham, Chris, 2002, Kyoto debate makes Canada sour: Sulphur, no. 283, November-December, p. 23-24. Fertilizer Markets, 2000, Main Pass closed: Fertilizer Markets, v. 11, no. 4, September 4, p. 9. Fialka, J.J., 2003, Off-road vehicles fueled by diesel targeted by EPA: The Wall Street Journal, April 16, p. A4. Hyne, J.B., 2000, An invisible hill to climb: Sulphur, no. 269, July-August, p. 3. International Fertilizer Industry Association, 2004, Preliminary sulphur statistics―2003: Paris, France, International Fertilizer Industry Association A/04/79, 6 p. Kitto, Mike, 2004, Recent trends in the sulphur and sulphuric acid markets: Sulphur, no. 293, July-August, p. 24-28. McMoRan Exploration Co., 2002, McMoRan Exploration Co. completes sale of sulphur transportation & terminaling assets: New Orleans, LA, McMoRan Exploration Co. news release, June 14, 1 p. Nakamura, David, 2002, World refining capacity climbs to highest level ever: Oil & Gas Journal, v. 100.52, December 23, p. 62-66. Nakamura, David, 2003, Refining capacity creeps higher in 2003: Oil & Gas Journal, v. 101.52, December 22, p. 64-69. North American Sulphur Review, 2002a, Canadian production stabilizes, but an increasing proportion of it comes from oil sands: North American Sulphur Review, v. 13, no. 10, October, p. 1-2. North American Sulphur Review, 2002b, News & developments: North American Sulphur Review, v. 13, no. 10, October, p. 2. North American Sulphur Review, 2003a, In brief: North American Sulphur Review, v. 14, no. 11, November, p. 2-3. North American Sulphur Review, 2003b, News & developments: North American Sulphur Review, v. 14, no. 11, November, p. 1. North American Sulphur Review, 2004, North American sulphur balance is reviewed: North American Sulphur Review, v. 15, no. 7, July, p. 1. Oil & Gas Journal, 2000, EPA issues rule limiting US gasoline sulfur: Oil & Gas Journal, v. 98, no. 1, January 3, p. 26-27. Oil & Gas Journal, 2001, Government developments: Oil & Gas Journal, v. 99.1, January 1, p. 7. Oil & Gas Journal, 2003a, Industry trends: Oil & Gas Journal, v. 101.29, July 28, p. 7. Oil & Gas Journal, 2003b, US refiners on target to meet low-sulfur highway diesel rules: Oil & Gas Journal, v. 101.43, November 10, p. 34. Pok, Joe, 2002, Oil sands developments in Alberta and the implications on sulfur supply: International Fertilizer Industry Association Production and International Trade Conference, Quebec City, Quebec, Canada, October 16-18, 2002, Presentation, 15 p. Shelton, J.E., 1978, Sulfur and pyrites: U.S. Bureau of Mines Minerals Yearbook 1976, v. I, p. 1287-1307. Stevens, Jason, 1998, Oil sands projects gather pace: Sulphur, no. 254, January-February, p. 27-30. Sulphur, 2001a, Rising profile: Sulphur, no. 273, March-April, p. 17-19. Sulphur, 2001b, Three refiners ask for time out: Sulphur, no. 274, May-June, p. 30-31. Sulphur, 2002a, Government moves into line with EU on sulphur in auto fuels: Sulphur, no. 279, March-April, p. 10. Sulphur, 2002b, Russia can meet EU fuel spec—In part: Sulphur, no. 281, July-August, p. 9-10. Sulphur, 2003a, Crew wins go-ahead for native sulphur sampling: Sulphur, no. 288, September-October, p.10. Sulphur, 2003b, Kazahk granules: Sulphur, no. 285, March-April, p. 6. Sulphur, 2003c, Life less expensive for Tengiz S producers: Sulphur, no. 286, May-June, p. 11. Sulphur, 2003d, MEPs make waves over marine fuel S level: Sulphur, no. 287, July-August, p. 10. Sulphur, 2003e, Salaries and Kyoto make Petro-Canada rethink oil sands strategy: Sulphur, no. 287, July-August, p. 8. Sulphur, 2003f, Ship owners call for sulphur trading and stack clean-up: Sulphur, no. 285, March-April, p. 8-9. Sulphur, 2003g, Sulphur markets summit in Vienna: Sulphur, no. 284, January-February, p. 17-29. Sulphur, 2003h, TCO seeks out dump site for sulphur: Sulphur, no. 285, March-April, p. 10. Sulphur, 2003i, TCO settles differences with Kazakh government: Sulphur, no. 284, January-February, p. 10-12. Sulphur, 2003j, Transport ministers want sulphur-free fuels by 2005: Sulphur, no. 287, July-August, p. 10-11. U.S. Census Bureau, 2004, Inorganic fertilizer materials and related products— First quarter 2004: U.S. Census Bureau MQ325B(04)-1, July, 5 p. Internet Reference Cited Stell, Jeannie, 2003, 2002 worldwide refining survey, accessed February 23, 2004, at URL GENERAL SOURCES OF INFORMATION U.S. Geological Survey Publications Sulfur. Ch. in Mineral Commodity Summaries, annual. Sulfur. Ch. in United States Mineral Resources, Professional Paper 820, 1973. Sulfur. Mineral Industry Surveys, monthly. Other Chemical and Engineering News, weekly. Chemical Engineering, weekly. Chemical Market Reporter, weekly. Chemical Week, weekly. European Chemical News, weekly. Fertilizer International, monthly. Fertilizer Week America, weekly. Green Markets, weekly. Industrial Minerals, monthly. Oil & Gas Journal, weekly. PentaSul North America Sulphur Review, monthly. Sulfur. Ch. in Mineral Facts and Problems, U.S. Bureau of Mines Bulletin 675, 1985. Sulphur, bimonthly. 74.8 U.S. GEOLOGICAL SURVEY MINERALS YEARBOOK—2003 TABLE 2 RECOVERED SULFUR PRODUCED AND SHIPPED IN THE UNITED STATES, BY STATE1 (Thousand metric tons and thousand dollars) 2002 2003 Shipments Shipments State Production Quantity Valuee Production Quantity Valuee Alabama 269 271 3,880 234 231 7,710 California 965 962 3,590 1,070 1,060 20,600 Illinois 414 412 1,420 466 460 11,700 Louisiana 1,160 1,160 15,700 1,210 1,210 65,400 Michigan and Minnesota 35 34 119 39 39 195 Mississippi 545 547 24,900 534 548 19,700 New Mexico 43 43 (2) 42 42 (2) Ohio 115 116 1,260 104 105 4,070 Texas 2,750 2,730 41,600 2,900 2,910 81,600 Washington 105 106 (2) 122 122 (2) Wyoming 1,340 1,360 2,640 1,360 1,360 16,900 Other3 762 r 755 r 5,430 834 837 28,100 Total 8,500 8,490 100,000 8,920 8,920 256,000 See footnotes at end of table. TABLE 2--Continued RECOVERED SULFUR PRODUCED AND SHIPPED IN THE UNITED STATES, BY STATE1 eEstimated. rRevised. 1Data are rounded to no more than three significant digits; may not add to totals shown TABLE 1 SALIENT SULFUR STATISTICS1 (Thousand metric tons of sulfur content and thousand dollars unless otherwise specified) 1999 2000 2001 2002 2003 United States: Production: Frasch 1,780 e 900 e ------Recovered2 8,360 8,590 8,490 8,500 8,920 Other 1,320 1,030 982 772 683 Totale 11,500 10,500 9,470 9,270 9,600 Shipments: Frasch W W ------Recovered2 9,940 3 9,710 3 8,470 8,490 8,920 Other 1,320 1,030 982 772 683 Total 11,300 10,700 9,450 9,260 9,600 Exports: Elemental4 685 762 675 687 742 Sulfuric acid 51 62 69 48 67 Imports: Elemental 2,580 2,330 1,730 2,560 2,870 Sulfuric acid 447 463 462 346 297 Consumption, all forms5 13,600 r 12,700 10,900 11,400 12,000 Stocks, December 31, producer, Frasch and recovered 451 208 232 181 206 Value: Shipments, free on board (f.o.b.) mine or plant: Frasch W W ------Recovered2 $376,000 3 $240,000 3 $84,700 e $100,000 e $256,000 e Other $66,400 $55,100 $49,500 $35,500 $34,000 Total $442,000 $295,000 $134,000 $136,000 r $290,000 Exports, elemental6 $35,800 $53,700 $48,800 $40,000 $46,100 Imports, elemental $51,600 $39,400 $22,100 $26,800 $70,600 Price, elemental, f.o.b. mine or plant dollars per metric ton 37.81 24.73 10.01 e 11.84 e 28.71 e World, production, all forms (including pyrites) 58,500 r 59,700 r 60,400 r 60,500 r 61,800 e eEstimated. rRevised. W Withheld to avoid disclosing company proprietary data; included with "United States, value, recovered." -- Zero. 1Data are rounded to no more than three significant digits except prices; may not add to totals shown. 2Includes the U.S. Virgin Islands. 3Includes corresponding Frasch sulfur data. 4Includes exports from the U.S. Virgin Islands to foreign countries. 5Consumption is calculated as shipments minus exports plus imports. 6Includes value of exports from the U.S. Virgin Islands to foreign countries. SULFUR—2003 74.9 TABLE 3 RECOVERED SULFUR PRODUCED AND SHIPPED IN THE UNITED STATES, BY PETROLEUM ADMINISTRATION FOR DEFENSE (PAD) DISTRICT1 (Thousand metric tons) 2002 2003 District and source Production Shipments Production Shipments PAD 1: Petroleum and coke 233 233 229 232 Natural gas 27 27 26 26 Total 260 260 255 257 PAD 2: Petroleum and coke 852 850 904 896 Natural gas 48 47 44 44 Total 900 897 948 940 PAD 3:2 Petroleum and coke 4,440 4,420 4,430 4,470 Natural gas 428 429 617 613 Total 4,870 4,850 5,050 5,080 PAD 4 and 5: Petroleum and coke 1,220 1,220 1,410 1,380 Natural gas 1,250 1,260 1,260 1,260 Total 2,470 2,480 2,670 2,640 Grand total: 8,500 8,490 8,920 8,920 Of which: Petroleum and coke 6,750 6,720 6,970 6,970 Natural gas 1,760 1,770 1,950 1,940 1Data are rounded to no more than three significant digits; may not add to totals shown. 2Includes the U.S. Virgin Islands. Other 762 755 5,430 834 837 28,100 Total 8,500 8,490 100,000 8,920 8,920 256,000 See footnotes at end of table. TABLE 2--Continued RECOVERED SULFUR PRODUCED AND SHIPPED IN THE UNITED STATES, BY STATE1 eEstimated. rRevised. 1Data are rounded to no more than three significant digits; may not add to totals shown. 2Some sulfur producers in this State incur expenses to make their production available to consumers. 3Includes Arkansas, Colorado, Delaware, Florida, Indiana, Kansas, Kentucky, Montana, New Jersey, North Dakota, Pennsylvania, Utah, Virginia, Wisconsin, and the U.S. Virgin Islands. TABLE 4 BYPRODUCT SULFURIC ACID PRODUCED IN THE UNITED STATES1, 2 (Thousand metric tons of sulfur content and thousand dollars) Type of plant 2002 2003 Copper3 695 590 Zinc4 50 51 Lead and molybdenum4 28 42 Total: Quantity 772 683 Value 35,500 34,000 1May include acid produced from imported raw materials. 2Data are rounded to no more than three significant digits, may not add to totals shown. 3Excludes acid made from pyrites concentrates. 4Excludes acid made from native sulfur. 74.10 U.S. GEOLOGICAL SURVEY MINERALS YEARBOOK—2003 TABLE 5 CONSUMPTION OF SULFUR IN THE UNITED STATES1, 2, 3 (Thousand metric tons) 2002 2003 Elemental sulfur: Shipments4 8,490 8,920 Exports 687 742 Imports 2,560 2,870 Total 10,400 11,000 Byproduct sulfuric acid: Shipments4 772 683 Exports5 48 67 Imports5 346 297 Grand total 11,400 12,000 1Crude sulfur or sulfur content. 2Data are rounded to no more than three significant digits; may not add to totals shown. 3Consumption is calculated as shipments minus exports plus imports. 4Includes the U.S. Virgin Islands. 5May include sulfuric acid other than byproduct. SIC3 End use 2002 2003 2002 2003 2002 2003 102 Copper ores ----707 r 421 707 421 1094 Uranium and vanadium ores ----2 4 2 4 10 Other ores ----1 58 1 58 26, 261 Pulpmills and paper products W W 122 225 122 225 28, 285, Inorganic pigments paints and allied 286, 2816 products, industrial organic chemicals, other chemical products4 --5 27 71 27 76 281 Other inorganic chemicals W 188 50 97 50 285 282, 2822 Synthetic rubber and other plastic materials and synthetics ----66 82 66 82 2823 Cellulosic fibers including rayon ----6 1 6 1 283 Drugs ----2 2 2 2 284 Soaps and detergents W ----2 --2 286 Industrial organic chemicals ----4 22 4 22 2873 Nitrogenous fertilizers ----105 206 105 206 2874 Phosphatic fertilizers ----6,660 r 6,660 6,660 r 6,660 2879 Pesticides ----8 11 8 11 287 Other agricultural chemicals 1,650 1,590 29 46 1,680 1,630 2892 Explosives ----8 10 8 10 2899 Water-treating compounds ----59 98 59 98 28 Other chemical products ----21 45 21 45 29, 291 Petroleum refining and other petroleum and coal products 2,390 3,700 90 140 2,480 3,840 331 Steel pickling ----7 58 7 58 333 Nonferrous metals ----2 3 2 3 33 Other primary metals ----7 9 7 9 3691 Storage batteries (acid) ----3 13 3 13 Exported sulfuric acid ----334 1,420 334 1,420 Total identified 4,040 5,480 8,320 r 9,700 12,400 r 15,200 Unidentified 248 678 52 409 300 1,090 Grand total 4,290 6,160 8,380 r 10,100 12,700 r 16,300 1Data are rounded to no more than three significant digits; may not add to totals shown. Elemental sulfur2 (sulfur equivalent) Total rRevised. W Withheld to avoid disclosing company proprietary data; included with "Unidentified." -- Zero. See footnotes at end of table. TABLE 6--Continued SULFUR AND SULFURIC ACID SOLD OR USED IN THE UNITED STATES, BY END USE1 Sulfuric acid TABLE 6 SULFUR AND SULFURIC ACID SOLD OR USED IN THE UNITED STATES, BY END USE1 (Thousand metric tons of sulfur content) SULFUR—2003 74.11 Unidentified 248 678 52 409 300 1,090 Grand total 4,290 6,160 8,380 r 10,100 12,700 r 16,300 1Data are rounded to no more than three significant digits; may not add to totals shown. 2Does not include elemental sulfur used for production of sulfuric acid. 3Standard industrial classification. 4No elemental sulfur was used in inorganic pigments and paints and allied products. rRevised. W Withheld to avoid disclosing company proprietary data; included with "Unidentified." -- Zero. See footnotes at end of table. TABLE 6--Continued SULFUR AND SULFURIC ACID SOLD OR USED IN THE UNITED STATES, BY END USE1 TABLE 7 U.S. EXPORTS OF ELEMENTAL SULFUR, BY COUNTRY1, 2 (Thousand metric tons and thousand dollars) 2002 2003 Country Quantity Value Quantity Value Brazil 136 4,270 116 6,500 Canada 50 5,290 45 5,440 China 280 13,700 265 16,600 Mexico 41 2,800 31 2,220 Morocco 156 6,490 236 9,230 Other 24 r 7,500 r 49 6,070 Total 687 40,000 742 46,100 rRevised. 1Includes exports from the U.S. Virgin Islands. 2Data are rounded to no more than three significant digits; may not add to totals shown. Source: U.S. Census Bureau. TABLE 8 U.S. EXPORTS OF SULFURIC ACID (100% H2SO4), BY COUNTRY1 2002 2003 Quantity Value Quantity Value Country (metric tons) (thousands) (metric tons) (thousands) Canada 129,000 $6,670 164,000 $11,200 China 525 586 529 313 Dominican Republic 2,540 146 2,550 217 Israel 216 297 1,120 336 Japan 507 154 135 312 Korea, Republic of 472 154 337 78 Mexico 3,080 505 4,030 471 Netherlands Antilles 20 5 11,200 689 Saudi Arabia 1,020 1,170 861 1,340 Singapore 111 117 185 56 Taiwan 1,470 621 547 461 Trinidad and Tobago 1,990 277 6,450 326 United Kingdom 257 83 282 231 Venezuela ----2,700 211 Other 6,530 r 1,980 r 9,950 2,580 Total 147,000 12,800 205,000 18,800 rRevised. -- Zero. 1Data are rounded to no more than three significant digits; may not add to totals shown. Source: U.S. Census Bureau. 74.12 U.S. GEOLOGICAL SURVEY MINERALS YEARBOOK—2003 TABLE 9 U.S. IMPORTS OF ELEMENTAL SULFUR, BY COUNTRY1 (Thousand metric tons and thousand dollars) 2002 2003 Country Quantity Value2 Quantity Value2 Canada 1,950 e 9,450 2,080 e 32,000 Mexico 430 11,300 534 26,500 Other 180 6,050 253 12,000 Total 2,560 e 26,800 2,870 e 70,600 eEstimated. 1Data are rounded to no more than three significant digits; may not add to totals shown. 2Declared customs valuation. Source: U.S. Census Bureau as adjusted by the U.S. Geological Survey. TABLE 10 U.S. IMPORTS OF SULFURIC ACID (100% H2SO4), BY COUNTRY1 2002 2003 Quantity Value2 Quantity Value2 Country (metric tons) (thousands) (metric tons) (thousands) Canada 615,000 $20,700 386,000 $17,800 Germany 99,200 2,970 76,800 2,570 Mexico 97,400 7,900 167,000 2,450 Spain 10,300 493 62,400 3,140 Other 237,000 r 14,400 r 216,000 13,300 Total 1,060,000 46,400 908,000 39,200 rRevised. 1Data are rounded to no more than three significant digits; may not add to totals shown. 2Declared cost, insurance, and freight paid by shipper valuation. Source: U.S. Census Bureau. TABLE 11 SULFUR: WORLD PRODUCTION IN ALL FORMS, BY COUNTRY AND SOURCE1, 2 (Thousand metric tons) Country and source3 1999 2000 2001 2002 2003e Australia, byproduct:e Metallurgy 441 654 817 899 863 Petroleum 25 30 45 60 60 Total 466 684 862 959 923 Canada, byproduct: Metallurgy 1,159 r 1,167 1,124 r 1,109 r 969 4 Natural gas, petroleum, tar sands 8,656 r 8,621 r 8,620 r 7,816 r 8,061 4 Total 9,815 r 9,788 r 9,744 r 8,925 r 9,030 4 Chile, byproduct, metallurgye 1,040 1,100 1,160 1,275 4 1,300 China:e Elemental 280 290 290 290 290 Pyrites 3,860 3,370 3,090 3,240 3,400 Byproduct, metallurgy 1,630 1,900 2,000 2,200 2,400 Total 5,770 5,560 5,380 5,730 6,090 Finland:e Pyrites 380 260 r 270 r 359 r 341 Byproduct: Metallurgy 299 283 r 227 r 308 r 305 Petroleum 42 46 r 46 r 55 r 60 Total 721 589 r 543 r 722 r 706 France, byproduct:e Natural gas 600 600 600 500 500 Petroleum 250 250 250 250 250 See footnotes at end of table. SULFUR—2003 74.13 TABLE 11--Continued SULFUR: WORLD PRODUCTION IN ALL FORMS, BY COUNTRY AND SOURCE1, 2 (Thousand metric tons) Country and source3 1999 2000 2001 2002 2003e France, byproduct--Continued:e Unspecified 250 260 250 250 250 Total 1,100 1,110 1,100 1,000 1,000 Germany, byproduct: Pyrites 30 30 61 ---- 4 Byproduct: Metallurgy 504 r 618 r 684 r 754 r 697 4 Natural gas and petroleum 1,824 r 1,753 r 1,749 r 1,745 r 1,661 4 Unspecified -- r -- r -- r -- r -- 4 Total 2,358 r 2,401 r 2,494 r 2,499 r 2,358 4 India:e Pyrites 32 32 32 32 r 32 Byproduct: Metallurgy 261 359 458 458 r 539 Natural gas and petroleum 101 376 526 r 371 r 451 Total 394 767 1,020 r 861 r 1,020 Iran, byproduct:e Metallurgy 47 50 50 50 50 Natural gas and petroleum 963 963 880 r 1,200 r 1,310 Total 1,010 1,010 930 r 1,250 r 1,360 Italy, byproduct:e Metallurgy 193 203 203 142 119 Petroleum 485 490 540 560 565 Total 678 4 693 4 743 702 684 Japan: Pyritese 41 30 30 25 25 Byproduct: Metallurgy 1,361 1,384 1,319 1,326 r 1,281 4 Petroleum 2,060 2,072 2,424 1,865 2,000 Total 3,462 3,486 3,773 3,216 r 3,310 Kazakhstan, byproduct:e Metallurgy 245 300 310 r 260 r 325 Natural gas and petroleum 1,070 1,200 1,400 1,600 r 1,600 Total 1,320 1,500 1,710 r 1,860 r 1,930 Korea, Republic of, byproduct:e Metallurgy 528 572 665 680 r 690 Petroleum 600 600 600 610 610 Total 1,130 1,170 1,270 1,290 r 1,300 Kuwait, byproduct, natural gas and petroleume 639 512 524 634 714 Mexico, byproduct: Metallurgy 474 474 572 e 575 e 575 Natural gas and petroleum 860 851 878 877 r 1,034 4 Total 1,334 1,325 1,450 1,452 r 1,610 Netherlands, byproduct:e Metallurgy 129 123 126 124 119 Petroleum 445 428 4 384 373 408 Total 574 551 510 497 527 Poland:5 Frasch 1,172 1,482 942 760 750 Byproduct: Metallurgy 278 279 277 275 e 275 Petroleum 74 e 70 e 133 180 150 Total 1,524 1,831 1,352 1,220 e 1,180 Russia:e, 6 Native 50 50 50 50 50 Pyrites 300 350 400 400 450 Byproduct, natural gas 4,405 4 4,900 5,300 5,400 5,600 Other 510 600 500 500 500 Total 5,265 4 5,900 6,250 6,350 6,600 See footnotes at end of table. 74.14 U.S. GEOLOGICAL SURVEY MINERALS YEARBOOK—2003 TABLE 11--Continued SULFUR: WORLD PRODUCTION IN ALL FORMS, BY COUNTRY AND SOURCE1, 2 (Thousand metric tons) Country and source3 1999 2000 2001 2002 2003e Saudi Arabia, byproduct, all sources 1,940 2,101 2,350 e 2,360 r, e 2,400 Spain: Pyrites 388 138 71 e ----Byproduct:e Coal, lignite, gasification 2 1 1 1 1 Metallurgy 455 454 461 544 560 Petroleum 110 115 135 140 145 Total 955 708 668 685 706 United Arab Emirates, byproduct, natural gas and petroleume 1,089 4 1,120 1,490 1,900 1,900 United States: Frasch 1,780 e 900 e ------ 4 Byproduct: Metallurgy 1,320 1,030 982 772 683 4 Natural gas 2,150 2,230 2,000 1,760 1,940 4 Petroleum 6,210 6,360 6,480 6,750 6,970 4 Total 11,500 10,500 9,470 9,270 9,600 4 Other:e, 7 Frasch 23 24 24 23 r 23 Native 212 r 422 r 457 r 449 r 216 Pyrites 271 r 245 r 356 r 358 r 367 Byproduct: Metallurgy 914 949 1,120 r 1,390 r 1,320 Natural gas 160 r 196 r 226 r 226 r 226 Natural gas, petroleum, tar sands, undifferentiated 441 r 766 r 785 r 808 r 833 Petroleum 864 r 962 r 873 r 896 r 879 Unspecified 1,310 1,410 1,440 r 1,380 1,400 Total 4,190 r 4,970 r 5,280 r 5,530 r 5,260 Grand total: 58,500 r 59,700 r 60,400 r 60,500 r 61,800 Of which: Frasch 2,980 2,410 966 783 r 773 Native8 542 r 762 r 797 r 789 r 556 Pyrites 5,300 r 4,450 r 4,310 r 4,410 r 4,620 Byproduct: Coal, lignite, gasificatione 2 1 1 1 1 Metallurgy 11,400 r 12,000 r 12,700 r 13,200 r 13,200 Natural gas 7,310 r 7,920 r 8,130 r 7,880 r 8,270 Natural gas, petroleum, tar sands, undifferentiated 15,800 r 16,300 r 17,000 r 17,100 r 17,700 Petroleum 11,200 r 11,500 r 12,000 r 11,800 r 12,100 Unspecified 4,010 r 4,370 r 4,540 r 4,490 r 4,550 eEstimated. rRevised. -- Zero. 1World totals, U.S. data, and estimated data are rounded to no more than three significant digits; may not add to totals shown. 2Table includes data available through July 22, 2004. 3The term "Source" reflects the means of collecting sulfur and the type of raw material. Sources listed include the following: Frasch recovery; native comprising all production of elemental sulfur by traditional mining methods (thereby excluding Frasch); pyrites (whether or not the sulfur is recovered in the elemental form or as acid); byproduct recovery, either as elemental sulfur or as sulfur compounds from coal gasification, metallurgical operations including associated coal processing crude oil and natural gas extraction, petroleum refining, tar sand cleaning, and processing of spent oxide from stack-gas scrubbers; and recovery from processing mined gypsum. Recovery of sulfur in the form of sulfuric acid from artificial gypsum produced as a byproduct of phosphatic fertilizer production is excluded, because to include it would result in double counting. Production of Frasch sulfur, other native sulfur, pyrite-derived sulfur, mined gypsum derived sulfur, byproduct sulfur from extraction of crude oil and natural gas, and recovery from tar sands are all credited to the country of origin of the extracted raw materials. In contrast, byproduct recovery from metallurgical operations, petroleum refinieries, and spent oxides are credited to the nation where the recovery takes place, which is not the original source country of the crude product from which the sulfur is extracted. 4Reported figure. 5Official Polish sources report total Frasch and native mined elemental sulfur output annually, undifferentiated; this figure has been divided between Frasch and other native sulfur on the basis of information obtained from supplementary sources. 6Sulfur is believed to be produced from Frasch and as a petroleum byproduct; however, information is inadequate to formulate estimates. 7Except for the above mentioned countries, "Other" includes Albania, Algeria, Aruba, Austria, Bahrain, Belarus, Belgium, Bosnia and Herzegovina, Brazil, Bulgaria, Colombia, Croatia, Cuba, the Czech Republic, Denmark, Ecuador, Egypt, Greece, Hungary, Indonesia, Iraq, Israel, North Korea, Kuwait, Libya, Macedonia, Namibia, the Netherlands Antilles, Norway, Oman, Pakistan, Peru, the Philippines, Portugal, Qatar, Romania, Serbia and Montenegro, Singapore, Slovakia, South Africa, Sweden, Switzerland, Syria, Taiwan, Trinidad and Tobago, Turkey, Turkmenistan, Ukraine, the United Kingdom, Uruguay, Uzbekistan, Venezuela, Zambia, and Zimbabwe. 8Includes "China, elemental." SULFUR—2003 74.15 FIGURE 1 TRENDS IN SULFUR PRODUCTION IN THE UNITED STATES 0 2 4 6 8 10 12 14 1983 1985 1987 1989 1991 1993 1995 1997 1999 2001 2003 YEAR Frasch Recovered Byproduct acid Total MILLION METRIC TONS Includes 10 months of Frasch data for 1993; the other 2 months are included with the recovered sulfur data to conform with proprietary data requirements. Data are estimates for 1994 through 2000. FIGURE 2 ESTIMATED AVERAGE PRICE OF SULFUR IN ACTUAL AND CONSTANT DOLLARS1 0 20 40 60 80 100 120 140 160 180 1983 1985 1987 1989 1991 1993 1995 1997 1999 2001 2003 YEAR PRICE IN DOLLARS PER METRIC TON Actual prices Based on constant 2003 dollars 1Based on the reported average value for elemental sulfur (Frasch and recovered), free on board mine and/or plant. FIGURE 2 ESTIMATED AVERAGE PRICE OF SULFUR IN ACTUAL AND CONSTANT DOLLARS1 0 20 40 60 80 100 120 140 160 180 1983 1985 1987 1989 1991 1993 1995 1997 1999 2001 2003 YEAR PRICE IN DOLLARS PER METRIC TON Actual prices Based on constant 2003 dollars 1Based on the reported average value for elemental sulfur (Frasch and recovered), free on board mine and/or plant. 74.16 U.S. GEOLOGICAL SURVEY MINERALS YEARBOOK—2003 FIGURE 4 TRENDS IN SALIENT SULFUR STATISTICS 0 2 4 6 8 10 12 14 16 1983 1985 1987 1989 1991 1993 1995 1997 1999 2001 2003 YEAR MILLION METRIC TONS Production Stocks Exports Imports Consumption FIGURE 4 TRENDS IN SALIENT SULFUR STATISTICS 0 2 4 6 8 10 12 14 16 1983 1985 1987 1989 1991 1993 1995 1997 1999 2001 2003 YEAR MILLION METRIC TONS Production Stocks Exports Imports Consumption FIGURE 3 PERCENTAGE OF SULFUR PRODUCTION BY SOURCE 0 10 20 30 40 50 60 70 80 90 100 1983 1985 1987 1989 1991 1993 1995 1997 1999 2001 2003 YEAR PERCENTAGE Frasch Recovered Byproduct acid Includes 10 months of Frasch data for 1993; the other 2 months are included with the recovered sulfur data to conform with proprietary data requirements. Data are estimates for 1994 through 2000. FIGURE 3 PERCENTAGE OF SULFUR PRODUCTION BY SOURCE 0 10 20 30 40 50 60 70 80 90 100 1983 1985 1987 1989 1991 1993 1995 1997 1999 2001 2003 YEAR PERCENTAGE Frasch Recovered Byproduct acid Includes 10 months of Frasch data for 1993; the other 2 months are included with the recovered sulfur data to conform with proprietary data requirements. Data are estimates for 1994 through 2000.
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Исследовательское подразделение медиахолдинга "ЭКСПЕРТ" В июле нефтегазовые доходы российского бюджета сжались на 27 % Снижение цен на нефть и сильный рубль урезают нефтегазовые поступления в казну Нефтегазовая промышленность обеспечивает треть доходов федерального бюджета. Между тем этот источник находится под давлением. Согласно статистике,Минфина, в июле объем поступлений снизился в годовом выражении на 27% — до 787,3 млрд рублей против 1 079,3 млрд рублей аналогичного периода 2024 года. Таким образом, за семь месяцев нефтегазовые доходы уменьшились на 18,5%. Для этого источника критически важным является мировая конъюнктура и объемы экспорта. «В 2023 году нефтегазовые доходы поддержали два основных фактора: значительное ослабление рубля, тогда курс превысил 90 рублей за доллар, и рекордные объемы экспорта в Азию, достигшие 240 млн тонн нефти», — напоминает генеральный директор группы компаний 1Oil Ирек Хуснутдинов. «В этом году мы наблюдаем снижение цен на нефть, а рубль с начала года укрепился более, чем на 25%», — объясняет причины нынешнего спада аналитик ФГ «Финам» Николай Дудченко. По мнению управляющего эксперта Центра аналитики и экспертизы ПСБ Екатерины Крыловой, если ситуация не изменится, то к концу года бюджет не доберёт около 2 трлн. рублей от запланированного объема: «А это около 1% ВВП. По нашему прогнозу, дефицит государственного бюджета составит в 2025 году чуть более 2,5% ВВП. Впрочем, несмотря на ожидаемый рост дефицит бюджета, его уровень не несет угроз для макроэкономической стабильности в России». По мнению эксперта, определяющим условием поддержания стабильности бюджетной системы становится не внешняя конъюнктура, а сохранение устойчивого и сбалансированного экономического роста России. С такой оценкой соглашается Николай Дудченко: «Сейчас ситуация сложная, однако пока, на наш взгляд, не критичная. Напомним, что в РФ продолжает действовать бюджетное правило, согласно которому в случае, если цена опускается ниже цены отсечки, Минфин осуществляет расходование средств фонда национального благосостояния (ФНБ)». Николай Дудченко также обращает внимание на , что прогноз по цене на нефть на текущий год в плане бюджета был существенно снижен: «Сейчас прогнозная цена на нефть в текущем году составляет 56 долларов за баррель против 69,7 долларов в прежнем проекте бюджета. Соответственно, нефтегазовые доходы в настоящее время планируются на уровне 8,3 трлн рублей против 10,9 трлн. рублей ранее. План по бюджетному дефициту расширен с 1,2 трлн руб. до 3,8 трлн руб. Недобор нефтегазовых доходов бюджета может быть частично компенсирован ненефтегазовыми доходами». Фото: архив «Эксперт-Урал» Материалы по теме Средняя цена нефти в 2015 году составит 56,73 доллара за баррель Реализация нефтепродуктов в сети АЗС «Газпромнефть» выросла в 2014 году на 9% «Варьеганнефть» увеличила запасы нефти категории АВС1 на 7 925 тыс. тонн «Варьеганнефть» внедряет технологию зарезки боковых стволов на месторождениях Варьеганского нефтяного блока Standard & Poor’s прогнозирует восстановление экономики России в 2016 году Коррекция: укрепление рубля стало закономерным после резкого перегрева российской валюты Снижение ставки 12 сентября практически предрешено Сотни видеокамер установил «Ростелеком» на избирательных участках Среднего Урала Нефтегаз адаптируется к угрозам Чтоб не заржавело Возможности для продвижения благодаря «Платформе роста» получат уральские бренды Перевозчики Екатеринбурга получат 150 новых автобусов Уральский «цифровой юг»: «Ростелеком» в Екатеринбурге завершил создание телеком-инфраструктуры для крупнейшего делового комплекса В тройке уральских регионов-лидеров по темпам развития промышленности произошли изменения TNF 2025: Как сделать технологическое развитие устойчивым «Эксперт-Урал» №10 (906) в PDF Экзотическая страна станет страной-партнером крупнейшей на Урале международной туристической выставки «EXPOTRAVEL-2025» Около 60 тыс. пассажиров воспользовались туристскими поездами СвЖД за летний сезон Участок без торгов под строительство эко-деревни в Башкирии получит ММК Производство углеводородного волокна появится в «Титановой долине» Чистая прибыль группы ВТБ по МСФО остается на высоком уровне Опрос 2022 Экзотическая страна станет страной-партнером крупнейшей на Урале международной туристической выставки «EXPOTRAVEL-2025» Глава Свердловской области Денис Паслер уволил своего заместителя Половину зерновых культур обмолотили аграрии Курганской области Движение по новому мосту открыли в Свердловской области Жилой комплекс построят около главного корпуса УрФУ На 34% выросли продажи удмуртских компаний на Wildberries Кандидатуру первого замгубернатора одобрили депутаты Заксобрания Свердловской области Участок без торгов под строительство эко-деревни в Башкирии получит ММК Производство углеводородного волокна появится в «Титановой долине» Мы ждем ваших вопросов, пожеланий и отзывов. Сетевое издание «Эксперт-Урал» зарегистрировано Федеральной службой по надзору в сфере связи, информационных технологий и массовых коммуникаций 26.07.2022 года. Регистрационный номер: Эл № ФС77-83567. Учредитель: ООО «Аналитический центр «Эксперт». Главный редактор: Кобер П.А. Телефон редакции: +7(343)345-03-72. Адрес электронной почты редакции: expertural@expertural.com. Знак информационной продукции 12+.
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http://proceedings.mlr.press/v125/lin20a/lin20a.pdf
Proceedings of Machine Learning Research vol 125:1–42, 2020 33rd Annual Conference on Learning Theory Near-Optimal Algorithms for Minimax Optimization Tianyi Lin DARREN LIN@BERKELEY.EDU University of California, Berkeley Chi Jin CHIJ@PRINCETON.EDU Princeton University Michael. I. Jordan JORDAN@CS.BERKELEY.EDU University of California, Berkeley Editors: Jacob Abernethy and Shivani Agarwal Abstract This paper resolves a longstanding open question pertaining to the design of near-optimal first-order algorithms for smooth and strongly-convex-strongly-concave minimax problems. Current state-of-the-art first-order algorithms find an approximate Nash equilibrium using ˜ O(κx+κy) (Tseng, 1995) or ˜ O(min{κx√κy, √κxκy}) (Alkousa et al., 2019) gradient evaluations, where κx and κy are the condition numbers for the strong-convexity and strong-concavity assumptions. A gap still remains between these results and the best existing lower bound ˜ Ω(√κxκy) (Ibrahim et al., 2019; Zhang et al., 2019). This paper presents the first algorithm with ˜ O(√κxκy) gradient complexity, match-ing the lower bound up to logarithmic factors. Our algorithm is designed based on an accelerated proximal point method and an accelerated solver for minimax proximal steps. It can be easily ex-tended to the settings of strongly-convex-concave, convex-concave, nonconvex-strongly-concave, and nonconvex-concave functions. This paper also presents algorithms that match or outperform all existing methods in these settings in terms of gradient complexity, up to logarithmic factors. 1. Introduction Let Rm and Rn be finite-dimensional Euclidean spaces and let the function f : Rm × Rn →R be smooth. Let X and Y are two nonempty closed convex sets in Rm and Rn. Our problem of interest is the following minimax optimization problem: min x∈X max y∈Y f(x, y). (1.1) The theoretical study of solutions of problem (1.1) has been an focus of several decades of research in mathematics, statistics, economics and computer science (Basar and Olsder, 1999; Nisan et al., 2007; Von Neumann and Morgenstern, 2007; Facchinei and Pang, 2007; Berger, 2013). Recently, this line of research has become increasingly relevant to algorithmic machine learning, with ap-plications including robustness in adversarial learning (Goodfellow et al., 2014; Sinha et al., 2018), prediction and regression problems (Cesa-Bianchi and Lugosi, 2006; Xu et al., 2009) and distributed computing (Shamma, 2008; Mateos et al., 2010). Moreover, real-world machine-learning sys-tems are increasingly embedded in multi-agent systems or matching markets and subject to game-theoretic constraints (Jordan, 2018). Most existing work on minimax optimization focuses on the convex-concave setting, where the function f(·, y) is convex for each y ∈Rn and the function f(x, ·) is concave for each x ∈Rm. c ⃝2020 T. Lin, C. Jin & M.I. Jordan. NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION The best known convergence rate in a general convex-concave setting is O(1/ϵ) in terms of duality gap, which can be achieved by Nemirovski’s mirror-prox algorithm (Nemirovski, 2004) (a special case of which is the extragradient algorithm (Korpelevich, 1976)), Nesterov’s dual extrapolation algorithm (Nesterov, 2007) or Tseng’s accelerated proximal gradient algorithm (Tseng, 2008). This rate is known to be optimal for the class of smooth convex-concave problems (Ouyang and Xu, 2019). Furthermore, optimal algorithms are known for special instances of convex-concave setting; e.g., for the affinely constrained smooth convex problem (Ouyang et al., 2015) and problems with a composite bilinear objective function, f(x, y) = g(x) + x⊤Ay −h(y) (Chen et al., 2014). Very recently, the lower complexity bound of first-order algorithms have been established for solving general strongly-convex-strongly-concave and strongly-convex-concave minimax optimiza-tion problems (Ouyang and Xu, 2019; Ibrahim et al., 2019; Zhang et al., 2019). For the strongly-convex-strongly-concave setting, in which κx, κy > 0 are the condition numbers for f(·, y) and f(x, ·), respectively, the complexity bound is ˜ Ω(√κxκy) while the best known upper bounds are ˜ O(κx+κy) (Tseng, 1995; Gidel et al., 2019; Mokhtari et al., 2019b) and ˜ O(min{κx√κy, κy√κx}) (Alkousa et al., 2019). For the strongly-convex-concave setting in which κx > 0 and κy = 0, the lower complexity bound is ˜ Ω( p κx/ϵ) while the best known upper bound is O(κx/ϵ) (Thekumpara-mpil et al., 2019). The existing algorithms that obtain a rate of O( p κx/ϵ) in this context are only for special case of strongly-convex-linear, where x and y are connected only through a bilinear term x⊤Ay or f(x, ·) is linear for each x ∈Rm (see, e.g., Nesterov, 2005; Chambolle and Pock, 2016; Juditsky and Nemirovski, 2011; Hamedani and Aybat, 2018). Thus, a gap remains between the lower complexity bound and the upper complexity bound for existing algorithms in both the strongly-convex-strongly-concave setting and the strongly-convex-concave setting. Accordingly, we have the following open problem: Can we design first-order algorithms that achieve the lower bounds in these settings? This paper presents an affirmative answer by resolving the above open problem up to logarith-mic factors. More specifically, our contribution is as follows. We propose the first near-optimal algorithms for solving the strongly-convex-strongly-concave and strongly-convex-concave mini-max optimization problems. In the former setting, our algorithm achieves a gradient complexity of ˜ O(√κxκy) which matches the lower complexity bound (Ibrahim et al., 2019; Zhang et al., 2019) up to logarithmic factors. In the latter setting, our algorithm attains a gradient complexity of ˜ O( p κx/ϵ) which again matches the lower complexity bound (Ouyang and Xu, 2019) up to logarithmic factors. In addition, our algorithm extends to the general convex-concave setting, achieving a gradient com-plexity of ˜ O(ϵ−1), which matches the lower bound of Ouyang and Xu (2019) as well as the best existing upper bounds (Nemirovski, 2004; Nesterov, 2007; Tseng, 2008) up to logarithmic factors. Our second contribution is a class of accelerated algorithms for the smooth nonconvex-strongly-concave and nonconvex-concave minimax optimization problems. In the former setting, our algo-rithm achieves a gradient complexity bound of ˜ O(√κyϵ−2) which improves the best known bound ˜ O(κ2 yϵ−2) (Jin et al., 2019; Rafique et al., 2018; Lin et al., 2019; Lu et al., 2019). In the latter set-ting, our algorithms specialize to a range of different notions of optimality. In particular, expressing our results in terms of stationarity of f, our algorithm achieves a gradient complexity bound of ˜ O(ϵ−2.5), which improves the best known bound ˜ O(ϵ−3.5) (Nouiehed et al., 2019). In terms of stationarity of the function Φ(·) := maxy∈Y f(·, y), our algorithm achieves a gradient complexity bound of ˜ O(ϵ−3) which matches the current state-of-the-art results (Thekumparampil et al., 2019; Kong and Monteiro, 2019). 2 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Table 1: Comparison of gradient complexities to find an ϵ-saddle point (Definition 4) in the convex-concave setting. This table highlights only the dependency on error tolerance ϵ and the strong-convexity and strong-concavity condition numbers, κx, κy. Settings References Gradient Complexity Strongly-Convex-Strongly-Concave Tseng (1995) ˜ O(κx + κy) Nesterov and Scrimali (2006) Gidel et al. (2019) Mokhtari et al. (2019b) Alkousa et al. (2019) ˜ O(min{κx√κy, κy√κx}) This paper (Theorem 9) ˜ O(√κxκy) Lower bound (Ibrahim et al., 2019) ˜ Ω(√κxκy) Lower bound (Zhang et al., 2019) ˜ Ω(√κxκy) Strongly-Convex-Linear (special case of strongly-convex-concave) Juditsky and Nemirovski (2011) O( p κx/ϵ) Hamedani and Aybat (2018) Zhao (2019) Strongly-Convex-Concave Thekumparampil et al. (2019) ˜ O(κx/√ϵ) This paper (Corollary 10) ˜ O( p κx/ϵ) Lower bound (Ouyang and Xu, 2019) ˜ Ω( p κx/ϵ) Convex-Concave Nemirovski (2004) O(ϵ−1) Nesterov (2007) Tseng (2008) This paper (Corollary 11) ˜ O(ϵ−1) Lower bound (Ouyang and Xu, 2019) Ω(ϵ−1) We provide a head-to-head comparison between our results and existing results in the literature in Table 1 for convex-concave settings, and Table 2 for nonconvex-concave settings. 2. Preliminaries In this section, we clarify the notation used in this paper, review some background and provide formal definitions for the class of functions and optimality measure considered in this paper. Notation. We use bold lower-case letters to denote vectors, as in x, y, z and calligraphic upper case letters to denote sets, as in X and Y. For a differentiable function f(·) : Rn →R, we le4t ∇f(z) denote the gradient of f at z. For a function f(·, ·) : Rm × Rn →R of two variables, 3 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Table 2: Comparison of gradient complexities to find an ϵ-stationary point of f (Definition 5) or ϵ-stationary point of Φ(·) := maxy∈Y f(·, y) (Definition 14, 18) in the nonconvex-concave settings. This table only highlights the dependence on tolerance ϵ and the condition num-ber κy. Settings References Gradient Complexity Nonconvex-Strongly-Concave (stationarity of f or stationarity of Φ) Jin et al. (2019) ˜ O(κ2 yϵ−2) Rafique et al. (2018) Lin et al. (2019) Lu et al. (2019) This paper (Theorem 12 & 20) ˜ O(√κyϵ−2) Nonconvex-Concave (stationarity of f) Lu et al. (2019) ˜ O(ϵ−4) Nouiehed et al. (2019) ˜ O(ϵ−3.5) Ostrovskii et al. (2020) ˜ O(ϵ−2.5) This paper (Corollary 13) ˜ O(ϵ−2.5) Nonconvex-Concave (stationarity of Φ) Jin et al. (2019) ˜ O(ϵ−6) Rafique et al. (2018) Lin et al. (2019) Thekumparampil et al. (2019) ˜ O(ϵ−3) Zhao (2020) This paper (Corollary 21) ˜ O(ϵ−3) ∇xf(x, y) (or ∇yf(x, y)) to denote the partial gradient of f with respect to the first variable (or the second variable) at point (x, y). We also use ∇f(x, y) to denote the full gradient at (x, y) where ∇f(x, y) = (∇xf(x, y), ∇yf(x, y)). For a vector x, we denote ∥x∥as its ℓ2-norm. For constraint sets X and Y, we let Dx and Dy denote their diameters, where Dx = maxx,x′∈X ∥x −x′∥and Dy = maxy,y′∈Y ∥y −y′∥. We use the notation PX and PY to denote projections onto the sets X and Y. Finally, we use the notation O(·), Ω(·) to hide only absolute constants which do not depend on any problem parameter, and notation ˜ O(), ˜ Ω() to hide only absolute constants and log factors. 2.1. Minimax optimization We are interested in the ℓ-smooth minimax optimization problems in the form (1.1). The regularity conditions that we consider for the function f are as follows. Definition 1 A function f is L-Lipschitz if for ∀z, z′ ∈Rn, that |f(z) −f(z′)| ≤L∥z −z′∥. Definition 2 A function f is ℓ-smooth if for ∀z, z′ ∈Rn, that ∥∇f(z) −∇f(z′)∥≤ℓ∥z −z′∥. 4 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Definition 3 A differentiable function φ : Rd →R is µ-strongly-convex if for any x′, x ∈Rd: φ(x′) ≥φ(x) + (x′ −x)⊤∇φ(x) + (µ/2)∥x′ −x∥2 Furthermore, φ is µ-strongly-concave if −φ is µ-strongly-convex. If we set µ = 0, then we recover the definitions of convexity and concavity for a continuous differentiable function. Convex-concave setting: we assume that f(·, y) is convex for each y ∈Y and f(x, ·) is concave for each x ∈X. Here X and Y are both convex and bounded. Under these conditions, the Sion’s minimax theorem (Sion, 1958) guarantees that max y∈Y min x∈X f(x, y) = min x∈X max y∈Y f(x, y). (2.1) Furthermore, there exists at least one saddle point (or Nash equilibrium) (x⋆, y⋆) ∈X × Y such that the following equality holds true: min x∈X f(x, y⋆) = f(x⋆, y⋆) = max y∈Y f(x⋆, y). (2.2) Therefore, for any point (ˆ x, ˆ y) ∈X ×Y, the duality gap maxy∈Y f(ˆ x, y)−minx∈X f(x, ˆ y) forms the basis for a standard optimality criterion. Formally, we define Definition 4 A point (ˆ x, ˆ y) ∈X × Y is an ϵ-saddle point of a convex-concave function f(·, ·) if maxy∈Y f(ˆ x, y) −minx∈X f(x, ˆ y) ≤ϵ. If ϵ = 0, then (ˆ x, ˆ y) is a saddle point. In the case when f(·, y) is strongly convex for each y ∈Y and f(x, ·) is strongly concave for each x ∈X, we refer µx and µy to the strongly-convex or strongly-concave module. If f is further ℓ-smooth, we denote κx = ℓ/µx and κy = ℓ/µy as the condition numbers of f(·, y) and f(x, ·). Nonconvex-concave setting: we only assume that f(x, ·) is concave for each x ∈Rm. The function f(·, y) can be possibly nonconvex for some y ∈Y. Here X is convex but possibly un-bounded while Y is convex and bounded. In general, finding a global Nash equilibrium of f is intractable since in the special case where Y has only a single element, this problem reduces to a nonconvex optimization problem in which finding a global minimum is already NP-hard (Murty and Kabadi, 1987). Similar to the literature in nonconvex constrained optimization, we opt to find local surrogates—stationary points—whose gradient mappings are zero. Formally, we define our optimality criterion as follows. Definition 5 A point (ˆ x, ˆ y) ∈X × Y is an ϵ-stationary point of an ℓ-smooth function f(·, ·) if ℓ∥PX [ˆ x −(1/ℓ)∇xf(ˆ x, ˆ y)] −ˆ x∥≤ϵ, ℓ∥PY[ˆ y + (1/ℓ)∇yf(ˆ x, ˆ y)] −ˆ y∥≤ϵ. If ϵ = 0, then (ˆ x, ˆ y) is a stationary point. In the absence of constraints, Definition 5 reduces to the standard condition ∥∇xf(ˆ x, ˆ y)∥≤ϵ and ∥∇yf(ˆ x, ˆ y)∥≤ϵ for unconstrained problems. Intuitively, the norms of the vector PY[ˆ y + (1/ℓ)∇yf(ˆ x, ˆ y)] −ˆ y represent the distance to the point (ˆ x, ˆ y) when performing one step of pro-jected gradient ascent on y starting from that point. The vector also refers to gradient mapping at (ˆ x, ˆ y); see Nesterov (2013) for the details. We note that this notion of stationarity of f (Definition 5) is closely related to an optimality notion in terms of stationary points of the function Φ(·) := maxy∈Y f(·, y) for nonconvex-concave functions. We refer readers to Appendix B.1 for more discussion. 5 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Algorithm 1 AGD(g, X, x0, ℓ, µ, ϵ) 1: Input: initial point x0 ∈X, smoothness ℓ, strongly-convex module µ and tolerance ϵ > 0. 2: Initialize: set t ←0, ˜ x0 ←x0, η ←1/ℓ, κ ←ℓ/µ and θ ← √κ−1 √κ+1. 3: repeat 4: t ←t + 1 5: xt ←PX [˜ xt−1 −η∇g(˜ xt−1)]. 6: ˜ xt ←xt + θ(xt −xt−1). 7: until ∥xt −PX (xt −η∇g(xt))∥2 ≤ ϵ 2κ2(ℓ−µ) is satisfied. 8: Output: PX (xt −η∇g(xt)). 2.2. Nesterov’s accelerated gradient descent Nesterov’s Accelerated Gradient Descent (AGD) dates back to Nesterov’s seminal paper (Nesterov, 1983) where it is shown to be optimal among all the first-order algorithms for smooth and convex functions (Nesterov, 2018). We present a version of AGD in Algorithm 1 which is frequently used to minimize an ℓ-smooth and µ-strongly convex function g over a convex set X. The key steps of the AGD algorithm are Line 5-6, where Lines 5 performs a projected gradient descent step, while Line 6 performs a momentum step, which “overshoots” the iterate in the direction of momentum (xt−xt−1). Line 7 is the stopping condition to ensure that the output achieves the desired optimality. The following theorem provides an upper bound on the gradient complexity of AGD; i.e., the total number of gradient evaluations to find an ϵ-optimal point in terms of function value. Theorem 6 Assume that g is ℓ-smooth and µ-strongly convex, the output ˆ x = AGD(g, x0, ℓ, µ, ϵ) satisfies g(ˆ x) ≤minx∈X g(x) + ϵ and the total number of gradient evaluations is bounded by O √κ log κ3ℓ∥x0 −x⋆∥2 ϵ  , where κ = ℓ/µ is the condition number, and x⋆∈X is the unique global minimum of g. Compared with the classical result for Gradient Descent (GD), which requires ˜ O(κ) gradient evalu-ations in the same setting, AGD improves over GD by a factor of √κ. AGD will be used as a basic component for acceleration in this paper. 3. Algorithm Components In this section, we present two main algorithm components. Both of them are crucial for our final algorithms to achieve near-optimal convergence rates. 3.1. Inexact Accelerated Proximal Point Algorithm Our first component is the Accelerated Proximal Point Algorithm (APPA, Algorithm 2) for minimiz-ing a function g(·). Comparing APPA with classical AGD (Algorithm 1), we note that both of them have momentum steps which yield acceleration. The major difference is in Line 4 of Algorithm 2, where APPA solves a proximal subproblem xt ←argmin x∈X g(x) + ℓ∥x −˜ xt−1∥2. (3.1) 6 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Algorithm 2 INEXACT-APPA(g, x0, ℓ, µ, ϵ, T) 1: Input: initial point x0 ∈X, proximal parameter ℓ, strongly-convex module µ, tolerance ϵ > 0 and the maximum iteration number T > 0. 2: Initialize: set ˜ x0 ←x0, κ ←ℓ µ, δ ← ϵ (10κ)2 and θ ←2√κ−1 2√κ+1. 3: for t = 1, · · · , T do 4: find xt so that g(xt) + ℓ∥xt −˜ xt−1∥2 ≤minx∈X {g(x) + ℓ∥x −˜ xt−1∥2} + δ. 5: ˜ xt ←xt + θ(xt −xt−1). 6: end for 7: Output: xT . instead of performing a gradient-descent step as in AGD (Line 5 in Algorithm 1). We refer to the parameter ℓin (3.1) as the proximal parameter. We present an inexact version in Algorithm 2 where we tolerate a small error δ in terms of the function value in solving the proximal subproblem (3.1). That is, the solution xt satisfies g(xt) + ℓ∥xt −˜ xt−1∥2 ≤min x∈X{g(x) + ℓ∥x −˜ xt−1∥2} + δ. A theoretical guarantee for the inexact APPA algorithm is presented in the following theorem, which claims that as long as δ is sufficiently small, the algorithm finds an ϵ-optimal point of any µ-strongly-convex function g with proximal parameter ℓin ˜ O( p ℓ/µ) iterations. Theorem 7 Assume that g is µ-strongly convex, ϵ ∈(0, 1) and ℓ> µ. There exists T > 0 such that the output ˆ x = INEXACT-APPA(g, x0, ℓ, µ, ϵ, T) satisfies g(ˆ x) ≤minx∈X g(x) + ϵ and T > 0 satisfies the following inequality, T ≥c√κ log g(x0) −g(x⋆) + (µ/4)∥x0 −x⋆∥2 ϵ  , where κ = ℓ/µ is an effective condition number, x⋆∈X is the unique global minimum of g, and c > 0 is an absolute constant. Comparing with Theorem 6, the most important difference here is that Theorem 7 does not require the function g to have any smoothness property. In fact, ℓis only a proximal parameter in proximal subproblem (3.1), which does not necessarily relate to the smoothness of g. On the flip side, the proximal subproblem (3.1) can not be easily solved in general. Theorem 7 guarantees the iteration complexity of Algorithm 1 while the complexity for solving these proximal steps is not discussed. We conclude that APPA has a unique advantage over AGD in settings where g does not have a smoothness property but the proximal step (3.1) is easy to solve. These settings include LASSO (Beck and Teboulle, 2009), as well as minimax optimization problems (as we show in later sections). 3.2. Accelerated Solver for Minimax Proximal Steps In minimax optimization problems of the form (1.1), we are interested in solving the following proximal subproblem as follows, xt+1 ←argmin x∈X Φ(x) + ℓ∥x −˜ x∥2, where Φ(x) := max y∈Y f(x, y), (3.2) 7 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Algorithm 3 MAXIMIN-AG2(g, x0, y0, ℓ, µx, µy, ϵ) 1: Input: initial point x0, y0, smoothness ℓ, strongly convex module µx, µy and tolerance ϵ > 0. 2: Initialize: t ←0, ˜ x0 ←x0, η ← 1 2κxℓ, κx ← ℓ µx , κy ← ℓ µy , θ ←4√κxκy−1 4√κxκy+1, ˜ ϵ ← ϵ (10κxκy)7 . 3: repeat 4: t ←t + 1. 5: ˜ xt−1 ←AGD(g(·, ˜ yt−1), x0, ℓ, µx, ˜ ϵ). 6: yt ←PY[˜ yt−1 + η∇yg(˜ xt−1, ˜ yt−1)]. 7: ˜ yt ←yt + θ(yt −yt−1). 8: xt ←AGD(g(·, yt), x0, ℓ, µx, ˜ ϵ). 9: until ∥yt −PY(yt + η∇yg(xt, yt))∥2 ≤ ϵ (10κxκy)4ℓis satisfied. 10: Output: PX (xt −(1/2κyℓ)∇xg(xt, yt)). which is equivalent to solving the following minimax problem: min x∈X max y∈Y ˜ g(x, y) := f(x, y) + ℓ∥x −˜ x∥2. (3.3) For a generic strongly-convex-strongly-concave function g(·, ·), solving a minimax problem is equivalent to solving a maximin problem, due to Sion’s minimax theorem: min x∈X max y∈Y g(x, y) = max y∈Y min x∈X g(x, y). A straightforward way of solving the maximin problem is to use a double-loop algorithm which solves the maximization and minimization problems on two different time scales. Specifically, the inner loop performs AGD on function g(·, y) to solve the inner minimization; i.e., to compute Ψ(y) := minx∈X g(x, y) for each y, and the outer loop performs Accelerated Gradient Ascent (AGA) on the function Ψ(·) to solve the outer maximization. Since the algorithm aims to solve a maximin problem we use AGA-AGD, and we name the algorithm MAXIMIN-AG2. See Algorithm 3 for the formal version of this algorithm. We also incorporate Lines 8-9 to check termination conditions, which ensures that the output achieves the desired optimality. The theoretical guarantee for Algorithm 3 is given in the following theorem. Theorem 8 Assume that g(·, ·) is ℓ-smooth, g(·, y) is µx-strongly convex for each y ∈Y and g(x, ·) is µy-strongly concave for each x ∈X. Then ˆ x = MAXIMIN-AG2(g, x0, y0, ℓ, µx, µy, ϵ) satisfies that maxy∈Y g(ˆ x, y) ≤minx∈X maxy∈Y g(x, y) + ϵ and the total number of gradient evaluations is bounded by O κx√κy · log2 (κx + κy)ℓ( ˜ D2 x + D2 y) ϵ !! , where κx = ℓ/µx and κy = ℓ/µy are condition numbers, ˜ Dx = ∥x0 −x⋆ g(y0)∥is the initial distance where x⋆ g(y0) = argminx∈X g(x, y0) and Dy > 0 is the diameter of the constraint set Y. Theorem 8 claims that Algorithm 3 finds an ϵ-optimal point in ˜ O(κx√κy) iterations for strongly-convex-strongly-concave functions. This rate does not match the lower bound ˜ Ω(√κxκy) (Ibrahim 8 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION et al., 2019; Zhang et al., 2019). At a high level, it takes AGD ˜ O(√κx) steps to solve the inner minimization problem and compute Ψ(y) := minx∈X g(x, y). Despite the fact that the function g is ℓ-smooth, function Ψ is only guaranteed to be (κxℓ)-smooth in the worst case, which makes the condition number of Ψ be κxκy. Thus, AGA requires ˜ O(√κxκy) iterations in the outer loop to solve the maximization of Ψ, which gives a total gradient complexity ˜ O(κx√κy). The key observation here is that although Algorithm 3 is slow for general strongly-convex-strongly-concave functions, the functions ˜ g of the form (3.3) in the proximal steps have a crucial property that κx = O(1) if the proximal parameter ℓis chosen to be the smoothness parameter of function f. Therefore, when f(x, ·) is strongly concave, by Theorem 8, it only takes Algorithm 3 ˜ O(√κy) gradient evaluations to solve the proximal subproblem (3.3), which is very efficient. We will see the consequences of this fact in the following section. 4. Accelerating Convex-Concave Optimization In this section, we present our main results for accelerating convex-concave optimization. We first present our new near-optimal algorithm and its theoretical guarantee for optimizing strongly-convex-strongly-concave functions. Then, we use simple reduction arguments to obtain results for strongly-convex-concave and convex-concave functions. 4.1. Strongly-convex-strongly-concave setting With the algorithm components from Section 3 in hand, we are now ready to state our near-optimal algorithm. Algorithm 4 is a simple combination of Algorithm 2 and Algorithm 3. Its outer loop performs an inexact APPA to minimize the function Φ(·) := maxy∈Y f(·, y), while the inner loop uses Maximin-AG2 to solve the proximal subproblem (3.2), which is equivalent to solving (3.3). At the end, after finding a near-optimal xT , Algorithm 4 performs another AGD on the function −f(xT , ·) to find a near-optimal yT . The theoretical guarantee for the algorithm is given in the following theorem. Theorem 9 Assume that f is ℓ-smooth and µx-strongly-convex-µy-strongly-concave. Then there exists T > 0 such that the output (ˆ x, ˆ y) = MINIMAX-APPA(f, x0, y0, ℓ, µx, µy, ϵ, T) is an ϵ-saddle point, and the total number of gradient evaluations is bounded by O √κxκy log3 (κx + κy)ℓ(D2 x + D2 y) ϵ !! , where κx = ℓ/µx and κy = ℓ/µy are condition numbers. Theorem 9 asserts that Algorithm 4 finds ϵ-saddle points in ˜ O(√κxκy) gradient evaluations, match-ing the lower bound (Ibrahim et al., 2019; Zhang et al., 2019), up to logarithmic factors. At a high level, despite the function Φ having undesirable smoothness properties, APPA minimizes Φ in the outer loop using ˜ O(√κx) iterations according to Theorem 7, regardless of the smoothness of Φ. According to the discussion in Section 3.2, Maximin-AG2 solves the proximal step in the inner loop using ˜ O(√κy) gradient evaluations, since the condition number of gt(·, y) for any y ∈Y is O(1). This gives the total gradient complexity ˜ O(√κxκy). 9 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Algorithm 4 MINIMAX-APPA(f, x0, y0, ℓ, µx, µy, ϵ, T) 1: Input: initial point x0, y0, proximity ℓ, strongly-convex parameter µ, tolerance δ, iteration T. 2: Initialize: ˜ x0 ←x0, κx ← ℓ µx , θ ←2√κx−1 2√κx+1, δ ← ϵ (10κxκy)4 and ˜ ϵ ← ϵ 102κxκy . 3: for t = 1, · · · , T do 4: denote gt(·, ·) where gt(x, y) := f(x, y) + ℓ∥x −˜ xt−1∥2. 5: xt ←MAXIMIN-AG2(gt, x0, y0, 3ℓ, 2ℓ, µy, δ) 6: ˜ xt ←xt + θ(xt −xt−1). 7: end for 8: ˜ y ←AGD(−f(xT , ·), y0, ℓ, µy, ˜ ϵ). 9: yT ←PY (˜ y + (1/2κxℓ)∇yf(xT , ˜ y)). 10: Output: (xT , yT ). 4.2. Strongly-convex-concave setting Our result in the strongly-convex-strongly-concave setting readily implies a near-optimal result in the strongly-convex-concave setting. Consider the following auxiliary function for an arbitrary y0 ∈Y which is defined by fϵ,y(x, y) := f(x, y) −(ϵ/4D2 y)∥y −y0∥2. (4.1) By construction, it is clear that the difference between f and fϵ,y is small in terms of function value: max (x,y)∈X×Y |f(x, y) −fϵ,y(x, y)| ≤ϵ/4. This implies, according to Definition 4, that any (ϵ/2)-saddle point of function fϵ,y is also a ϵ-saddle point of function f, and thus it is sufficient to only solve the problem minx∈X maxx∈Y fϵ,y(x, y). Finally, when f is a µx-strongly-convex-concave function, fϵ,y becomes µx-strongly-convex-ϵ/(2D2 y)-strongly-concave, which can be fed into Algorithm 4 to obtain the following result. Corollary 10 Assume that f is ℓ-smooth and µx-strongly-convex-concave. Then there exists T > 0 such that the output (ˆ x, ˆ y) = MINIMAX-APPA(fϵ,y, x0, y0, ℓ, µx, ϵ/(4D2 y), ϵ/2, T) is an ϵ-saddle point, and the total number of gradient evaluations is bounded by O r κxℓ ϵ Dy log3 κxℓ(D2 x + D2 y) ϵ !! where κx = ℓ/µx is the condition number, and fϵ,y is defined as in (4.1). 4.3. Convex-concave setting Similar to the previous subsection, when f is only convex-concave, we can construct following strongly-convex-strongly-concave function fϵ: fϵ(x, y) = f(x, y) + (ϵ/8D2 x)∥x −x0∥2 −(ϵ/8D2 y)∥y −y0∥2, (4.2) which can be fed into Algorithm 4 to obtain the following result. 10 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Corollary 11 Assume function f is ℓ-smooth and convex-concave, then there exists T > 0, where the output (ˆ x, ˆ y) = MINIMAX-APPA(fϵ, x0, y0, ℓ, ϵ/(4D2 x), ϵ/(4D2 y), ϵ/2, T) will be an ϵ-saddle point, and the total number of gradient evaluations is bounded by O ℓDxDy ϵ log3 ℓ(D2 x + D2 y) ϵ !! , where fϵ is defined as in (4.2). 5. Accelerating Nonconvex-Concave Optimization In this section, we present methods for accelerating nonconvex-concave optimization. Similar to Section 4, we first present our algorithm and its theoretical guarantee for optimizing nonconvex-strongly-concave functions. We then use a simple reduction argument to obtain results for nonconvex-concave functions. This section present results using the stationarity of the function f (Definition 5) as an optimality measure. Please see Appendix B for additional results using the stationarity of the function Φ(·) := maxy∈Y f(·, y) as the optimality measure (Definition 14 and 18). 5.1. Nonconvex-strongly-concave setting Our algorithm for nonconvex-strongly-concave optimization is described in Algorithm 5. Similar to Algorithm 4, we still use our accelerated solver Maximin-AG2 for the same proximal subproblem in the inner loop. The only minor difference is that, in the outer loop, Algorithm 5 only uses the Proximal Point Algorithm (PPA) on function Φ(·) := maxy∈Y f(·, y) without acceleration (or momentum steps). This is due to fact that gradient descent is already optimal among all first-order algorithm for finding stationary points of smooth nonconvex functions (Carmon et al., 2019a). The standard acceleration technique will not help for smooth nonconvex functions. We presents the theoretical guarantees for Algorithm 5 in the following theorem. Theorem 12 Assume that f is ℓ-smooth and f(x, ·) is µy-strongly-concave for all x. Then there exists T > 0 such that the output (ˆ x, ˆ y) = MINIMAX-PPA(f, x0, y0, ℓ, µy, ϵ, T) is an ϵ-stationary point of f with probability at least 2/3, and the total number of gradient evaluations is bounded by O ℓ∆Φ ϵ2 · √κy log2 κyℓ( ˜ D2 x + D2 y) ϵ !! , where κy = ℓ/µy is the condition number, ∆Φ = Φ(x0) −minx∈Rm Φ(x) is the initial function value gap and ˜ Dx = ∥x0 −x⋆ g1(y0)∥is the initial distance where x⋆ g(y0) = argminx∈X g(x, y0). Theorem 12 claims that Algorithm 5 will find an ϵ-stationary point, with at least constant probability, in ˜ O(√κy/ϵ2) gradient evaluations. Similar to Theorem 9, the inner loop takes ˜ O(√κy) gradient evaluations to solve the proximal step since the condition number of gt(·, y) is O(1) for any y ∈Y. In the outer loop, regardless of the smoothness of Φ(·), PPA with proximal parameter ℓis capable of finding the stationary point in ˜ O(1/ϵ2) iterations. In total, the gradient complexity is ˜ O(√κy/ϵ2). 11 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Algorithm 5 MINIMAX-PPA(g, x0, y0, ℓ, µy, ϵ, T) 1: Input: initial point x0, y0, proximity ℓ, strongly-convex parameter µ, tolerance δ, iteration T. 2: Initialize: set δ ← ϵ2 (10κy)4ℓ· ( ϵ ℓDy )2. 3: for t = 1, · · · , T do 4: denote gt(·, ·) where gt(x, y) := f(x, y) + ℓ∥x −xt−1∥2. 5: xt ←MAXIMIN-AG2(gt, x0, y0, 3ℓ, ℓ, µ, δ). 6: end for 7: sample s uniformly from {1, 2, · · · , T}. 8: ys ←AGD(−f(xs, ·), y0, ℓ, µ, δ). 9: Output: (xs, ys). 5.2. Nonconvex-concave setting Our result in the nonconvex-strongly-concave setting readily implies a fast result in the nonconvex-concave setting. Consider the following auxiliary function for an arbitrary y0 ∈Y: ˜ fϵ(x, y) = f(x, y) −(ϵ/4Dy)∥y −y0∥2. (5.1) By construction, it is clear that the gradient of f and ˜ fϵ are close in the sense max (x,y)∈Rm×Y ∥∇f(x, y) −∇˜ fϵ(x, y)∥≤ϵ/4. This implies that any (ϵ/2)-stationary point of ˜ fϵ is also a ϵ-stationary point of f, and thus it is sufficient to solve the problem minx∈X maxx∈Y ˜ fϵ(x, y). Finally, the function ˜ fϵ(x, ·) is always ϵ/(2Dy)-strongly-concave, which can be fed into Algorithm 5 to obtain the following result. Corollary 13 Assume that f is ℓ-smooth and f(x, ·) is concave for all x. Then there exists T > 0 such that the output (ˆ x, ˆ y) = MINIMAX-PPA( ˜ fϵ, x0, y0, ℓ, ϵ/(2Dy), ϵ/2, T) is an ϵ-stationary point of f with probability at least 2/3, and the total number of gradient evaluations is bounded by O ℓ∆Φ ϵ2 · r ℓDy ϵ log2 ℓ( ˜ D2 x + D2 y) ϵ !! , where Dy > 0, ∆Φ = Φ(x0) −minx∈Rm Φ(x) is the initial function value gap and ˜ Dx = ∥x0 − x⋆ g1(y0)∥is the initial distance where x⋆ g(y0) = argminx∈X g(x, y0). 6. Conclusions This paper has provided the first set of near-optimal algorithms for strongly-convex-(strongly)-concave minimax optimization problems and the state-of-the-art algorithms for nonconvex-(strongly)-concave minimax optimization problems. For the former class of problems, our algorithms match the lower complexity bound for first-order algorithms (Ouyang and Xu, 2019; Ibrahim et al., 2019; Zhang et al., 2019) up to logarithmic factors. For the latter class of problems, our algorithms achieve the best known upper bound. In the future research, one important direction is to inves-tigate the lower complexity bound of first-order algorithms for nonconvex-(strongly)-concave min-imax problems. 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There are also several recent papers studying the convergence of EG and its variants; see Chambolle and Pock (2011); Malitsky (2015); Yadav et al. (2018) for reflected gradient descent ascent, Daskalakis et al. (2018); Mokhtari et al. (2019b,a) for optimistic gradient descent ascent (OGDA) and Rakhlin and Sridharan (2013a,b); Mertikopoulos et al. (2019); Chav-darova et al. (2019); Hsieh et al. (2019); Mishchenko et al. (2019) for other variants. In the bilinear setting, Daskalakis et al. (2018) established the convergence of the optimistic gradient descent as-cent (OGDA) method to a neighborhood of the solution; Liang and Stokes (2019) proved the linear convergence of the OGDA algorithm using a dynamical system approach. Very recently, Mokhtari et al. (2019b) have proposed a unified framework for achieving the sharpest convergence rates of both EG and OGDA algorithms. For the convex-concave minimax problem, Nemirovski (2004) proved that his mirror-prox al-gorithm returns an ϵ-saddle point within the gradient complexity of O(ϵ−1) when X and Y are bounded. This algorithm was subsequently generalized by Auslender and Teboulle (2005) to a class of distance-generating functions, and the complexity result was extended to unbounded sets and composite objectives (Monteiro and Svaiter, 2010, 2011) using the hybrid proximal extragradient algorithm with different error criteria. Nesterov (2007) developed a dual extrapolation algorithm which possesses the same complexity bound as in Nemirovski (2004). Tseng (2008) presented a unified treatment of these algorithms and a refined convergence analysis with same complexity result. Nedi´ c and Ozdaglar (2009) analyzed the (sub)gradient descent ascent algorithm for convex-concave saddle point problems when the (sub)gradients are bounded over the constraint sets. Aber-nethy et al. (2019) presented a Hamiltonian gradient descent algorithm with last-iterate convergence under a “sufficiently bilinear” condition. Several papers have studied special cases in the convex-concave setting. For the special case when the objective function is a composite bilinear form, f(x, y) = g(x) + x⊤Ay −h(y), Cham-bolle and Pock (2011) introduced a primal-dual algorithm that converges to a saddle point with the rate of O(1/ϵ) when the convex functions g and h are smooth. Nesterov (2005) proposed a smooth-ing technique and proved that the resulting algorithm achieves an improved rate with better depen-dence on Lipschitz constant of ∇g when h is the convex and smooth function and X, Y are both bounded. He and Monteiro (2016) and Kolossoski and Monteiro (2017) proved that such result also hold when X, Y are unbounded or the space is non-Euclidean. Chen et al. (2014, 2017) generalized Nesterov’s technique to develop optimal algorithms for solving a class of stochastic saddle point problems and stochastic monotone variational inequalities. For a class of certain purely bilinear games where g and h are zero functions, Azizian et al. (2020) demonstrated that linear convergence is possible for several algorithms and their new algorithm achieved the tight bound. The second case is the so-called affinely constrained smooth convex problem, i.e., minx∈X g(x), s.t. Ax = u. Esser et al. (2010) proposed a O(ϵ−1) primal-dual algorithm while Lan and Monteiro (2016) provided a first-order augmented Lagrangian method with the same O(ϵ−1) rate. By exploiting the struc-ture, Ouyang et al. (2015) proposed a near-optimal algorithm in this setting. 18 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION For the strongly convex-concave minimax problem, Tseng (1995) and Nesterov and Scrimali (2006) proved that their algorithms find an ϵ-saddle point with a gradient complexity of ˜ O(κx +κy) using a variational inequality. Using a different approach, Gidel et al. (2019) and Mokhtari et al. (2019b) derived the same complexity results for the OGDA algorithm. Very recently, Alkousa et al. (2019) proposed an accelerated gradient sliding algorithm with a gradient complexity of ˜ O(min{κx√κy, κy√κx}) while Ibrahim et al. (2019); Zhang et al. (2019) established a lower complexity bound of ˜ Ω(√κxκy) among all the first-order algorithms in this setting. For strongly-convex-concave minimax problems, the best known general lower bound for first-order algorithm is O( p κx/ϵ), as shown by Ouyang and Xu (2019). Several papers have studied strongly-convex-concave minimax problem with additional structures. This includex optimizing a strongly convex function with linear constraints (Goldstein et al., 2014; Xu and Zhang, 2018; Xu, 2019), the case when x and y are connected only through a bilinear term x⊤Ay (Nesterov, 2005; Chambolle and Pock, 2016; Xie and Shi, 2019) and the case when f(x, ·) is linear for each x ∈Rm (Juditsky and Nemirovski, 2011; Hamedani and Aybat, 2018; Zhao, 2019). The algorithms developed in these works were all guaranteed to return an ϵ-saddle point with a gradient complexity of ˜ O(1/√ϵ) and some of them even achieve a near-optimal gradient complexity of ˜ O( p κx/ϵ) (Nes-terov, 2005; Chambolle and Pock, 2016). However, the best known upper complexity bound for general strongly-convex-concave minimax problems is O(κx/√ϵ) which was shown using the dual implicit accelerated gradient algorithm (Thekumparampil et al., 2019). For nonconvex-concave minimax problems, a line of recent work (Jin et al., 2019; Rafique et al., 2018; Lin et al., 2019) has studied various algorithms and proved that they can find an approximate stationary point of Φ(·) := maxy∈Y f(·, y). In a deterministic setting, all of these algorithms guarantee a rate of ˜ O(κ2 yϵ−2) and ˜ O(ϵ−6) when f(x, ·) is strongly concave and con-cave respectively. Thekumparampil et al. (2019) consider the same setting as ours and proposed a proximal dual implicit accelerated gradient algorithm and proved that it finds an approximate stationary point of Φ(·) with the total gradient complexity of ˜ O(ϵ−3). Kong and Monteiro (2019) consider a general nonconvex minimax optimization model: minx h(x) + ρ(x), where h is a “sim-ple” proper, lower semi-continuous and convex function and ρ(x) = maxy∈Y f(x, y) with f sat-isfying that −f(x, ·) is proper, convex, and lower semi-continuous. They propose to smooth ρ to ρξ(x) = maxy∈Y f(x, y) −(1/2ξ)∥y −y0∥2 and apply an accelerated inexact proximal point method to solve the smoothed problem minx h(x) + ρξ(x). The resulting AIPP-S algorithm at-tains the iteration complexity of O(ϵ−3) using a slightly different but equivalent notion of sta-tionarity but requires the exact gradient of ρξ at each iteration. This amounts to assuming that maxy∈Y f(x, y) −(1/2ξ)∥y −y0∥2 can be solved exactly, which is restrictive due to the poten-tially complicated structure of f(x, ·) or Y. If f is further assumed to be smooth, Zhao (2020) developed a variant of AIPP-S algorithm which only requires an inexact gradient of ρξ at each it-eration and attains the total gradient complexity of ˜ O(ϵ−3). On the other hand, the stationarity of f(·, ·) is proposed for quantifying the efficiency in nonconvex-concave minimax optimization (Lu et al., 2019; Nouiehed et al., 2019; Kong and Monteiro, 2019; Ostrovskii et al., 2020). Using this notion of stationarity, Kong and Monteiro (2019) attains the rate of O(ϵ−2.5) but requires the exact gradient of ρξ at each iteration. Without this assumption, the current state-of-the-art rate is ˜ O(ϵ−2.5) achieved by our Algorithm 5 and the algorithm proposed by a concurrent work (Ostrovskii et al., 2020). Both algorithms are based on constructing an auxiliary function fϵ,y and applying an accel-erated solver for minimax proximal steps. Finally, several other algorithms have been developed 19 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION either for specific nonconvex-concave minimax problems or in stochastic setting; see Namkoong and Duchi (2016); Sinha et al. (2018); Sanjabi et al. (2018); Grnarova et al. (2018) for the details. Appendix B. Additional Results for Nonconvex-Concave Optimization In this section, we present our results for nonconvex-concave optimization using stationary of Φ(·) := maxy∈Y f(·, y) (Definition 14 and Definition 18) as the optimality measure. B.1. Optimality notion based on Moreau envelope We present another optimality notion based on Moreau envelope for nonconvex-concave setting in which f(·, y) is not necessarily convex for each y ∈Y but f(x, ·) is concave for each x ∈X. For simplicity, we let X = Rm and Y be convex and bounded. In general, finding a global saddle point of f is intractable since solving the special case with a singleton Y globally is already NP-hard (Murty and Kabadi, 1987) as mentioned in the main text. One approach, inspired by nonconvex optimization, is to equivalently reformulate problem (1.1) as the following nonconvex minimization problem: min x∈Rm  Φ(x) := max y∈Y f(x, y)  , (B.1) and define an optimality notion for the local surrogate of global optimum of Φ. In robust learning, x is the classifier while y is the adversarial noise. Practitioners are often only interested in finding a robust classifier x instead of an adversarial response y to each data point. Such a stationary point x precisely corresponds to a robust classifier that is stationary to the robust classification error. If f(x, ·) is further assumed to be strongly concave for each x ∈Rm, then Φ is smooth and a standard optimality notion is the stationary point. Definition 14 We call ˆ x an ϵ-stationary point of a smooth function Φ if ∥∇Φ(ˆ x)∥≤ϵ. If ϵ = 0, then ˆ x is called a stationary point. In contrast, when f(x, ·) is merely concave for each x ∈X, Φ is not necessarily smooth and even not differentiable. A weaker sufficient condition for the purpose of our paper is the weak convexity. Definition 15 A function Φ : Rd →R is L-weakly convex if Φ(·) + (L/2) ∥·∥2 is convex. First, a function Φ is ℓ-weakly convex if it is ℓ-smooth. Second, the subdifferential of a ℓ-weakly convex function Φ can be uniquely determined by the subdifferential of Φ(·) + (ℓ/2)∥· ∥2. This implies that the optimality notion can be defined by a point x ∈Rm with at least one small sub-gradient: minξ∈∂Φ(x) ∥ξ∥≤ϵ. Unfortunately, this notion can be restrictive if Φ is nonsmooth. Considering a one-dimensional function Φ(·) = | · |, a point x must be 0 if it satisfies the optimality notion with ϵ ∈[0, 1). This means that finding a sufficiently accurate solution under such optimality notion is as difficult as solving the minimization exactly. Another popular optimality notion is based on the Moreau envelope of Φ when Φ is weakly convex (Davis and Drusvyatskiy, 2019). Definition 16 A function Φλ is the Moreau envelope of Φ with λ > 0 if for ∀x ∈Rm, that Φλ(x) = min w∈Rm Φ(w) + (1/2λ)∥w −x∥2. 20 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Lemma 17 (Properties of Moreau envelopes) If the function Φ(·) is ℓ-weakly convex, its Moreau envelope Φ1/2ℓ(·) is 4ℓ-smooth with the gradient ∇Φ1/2ℓ(·) = 2ℓ(· −proxΦ/2ℓ(·)) in which a point proxΦ/2ℓ(·) = argminw∈Rm{Φ(w) + ℓ∥w −·∥2} is defined. Thus, an ϵ-stationary point of an ℓ-weakly convex function Φ can be alternatively defined as a point ˆ x satisfying that the gradient norm of Moreau envelope ∥∇Φ1/2ℓ(ˆ x)∥is small. Definition 18 We call ˆ x an ϵ-stationary point of a ℓ-weakly convex function Φ if ∇Φ1/2ℓ(ˆ x) ≤ ϵ. If ϵ = 0, then ˆ x is called a stationary point. Lemma 19 (Properties of ϵ-stationary point) If ˆ x is an ϵ-stationary point of a ℓ-weakly convex function Φ, then there exists ¯ x ∈Rm such that minξ∈∂Φ(¯ x) ∥ξ∥≤ϵ and ∥ˆ x −¯ x∥≤ϵ/2ℓ. Lemma 19 shows that an ϵ-stationary point defined by the Moreau envelope can be interpreted as the relaxation for a point with at least one small subgradient. In particular, if ˆ x is an ϵ-stationary point of a ℓ-weakly convex function Φ, then it is close to a point which has small subgradient. B.2. Nonconvex-strongly-concave setting In the setting of nonconvex-strongly-concave function, we still use Algorithm 5. Similar to Theorem 12, we can obtain a guarantee, which finds a point ˆ x satisfying ∥∇Φ(ˆ x)∥≤ϵ in the same number of iterations as in Theorem 12. Theorem 20 Assume that f is ℓ-smooth and f(x, ·) is µy-strongly-concave for all x. Then there ex-ists T > 0 such that the output (ˆ x, ˆ y) = MINIMAX-PPA(f, x0, y0, ℓ, µy, ϵ, T) satisfies ∥∇Φ(ˆ x)∥≤ ϵ with probability at least 2/3, and the total number of gradient evaluations is bounded by O ℓ∆Φ ϵ2 · √κy log2 κyℓ( ˜ D2 x + D2 y) ϵ !! where κy = ℓ/µy is the condition number, ∆Φ = Φ(x0) −minx∈Rm Φ(x) is the initial function value gap and ˜ Dx = ∥x0 −x⋆ g1(y0)∥is the initial distance where x⋆ g(y0) = argminx∈X g(x, y0). B.3. Nonconvex-concave setting We can similarly reduce the problem of optimizing a nonconvex-concave function to the problem of optimizing a nonconvex-strongly-concave function. The only caveat is that, in order to achieve the near-optimal point using Definition 18 as optimality measure, we can only add a O(ϵ2) term as follows: ¯ fϵ(x, y) = f(x, y) −(ϵ2/200ℓD2 y)∥y −y0∥2. (B.2) Now ¯ fϵ(x, ·) is only ϵ2/(100ℓD2 y)-concave, by feeding it to Algorithm 5 and through a slightly more complicated reduction argument, we can only obtain gradient complexity bound of ˜ O(ϵ−3) instead of ˜ O(ϵ−2.5) as in Corollary 13. Formally, we have 21 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Corollary 21 Assume that f is ℓ-smooth, and f(x, ·) is concave for all x. Then there exists T > 0 such that the output (ˆ x, ˆ y) = MINIMAX-PPA( ¯ fϵ, x0, y0, ℓ, ϵ2/(100ℓD2 y), ϵ/10, T) satis-fies ∥∇Φ1/2ℓ(ˆ x)∥≤ϵ with probability at least 2/3, and the total number of gradient evaluations is bounded by O ℓ2Dy∆Φ ϵ3 log2 ℓ( ˜ D2 x + D2 y) ϵ !! where Dy > 0, ∆Φ = Φ(x0) −minx∈Rm Φ(x) is the initial function value gap and ˜ Dx = ∥x0 − x⋆ g1(y0)∥is the initial distance where x⋆ g(y0) = argminx∈X g(x, y0). Appendix C. Proofs for Algorithm Components In this section, we present proofs for our algorithm components. C.1. Proof of Theorem 6 We divide the proof into three parts. In the first part, we show that the output ˆ x satisfies g(ˆ x) ≤ minx∈X g(x)+ϵ. In the second part, we derive the sufficient condition for guaranteeing the stopping criteria in Algorithm 1. In the third part, we derive the gradient complexity of the algorithm using the condition derived in the second part. Part I. Let ˜ xt = PX (xt −(1/ℓ)∇g(xt)) be defined as the point achieved by one-step projected gradient descent from xt. Since g is ℓ-smooth and µ-strongly convex, it is straightforward to derive from Nesterov (2018, Corollary 2.3.2) that g(x) ≥g(˜ xt) + ℓ(xt −˜ xt)⊤(x −xt) + ℓ 2∥xt −˜ xt∥2 + µ 2 ∥x −xt∥2, for all x ∈X. Using the Young’s inequality, we have (xt −˜ xt)⊤(x −xt) ≥−(1/2)(∥xt −˜ xt∥2 + ∥x −xt∥2). Putting these pieces together with x = x⋆yields that g(˜ xt) −min x∈X g(x) = g(˜ xt) −g(x⋆) ≤ ℓ−µ 2  ∥xt −x⋆∥2. Without loss of generality, we assume ℓ> µ. Indeed, if ℓ= µ, then one-step projected gradient de-scent from any points in X guarantees that g(˜ xt)−minx∈X g(x) = 0. Since ˆ x = ˜ xt in Algorithm 1, it suffices to show that the following statement holds true, ∥xt −PX (xt −(1/ℓ)∇g(xt))∥≤ r ϵ 2κ2(ℓ−µ) = ⇒∥xt −x⋆∥≤ r 2ϵ ℓ−µ. (C.1) Let ˜ xt = PX (xt −(1/ℓ)∇g(xt)) be defined as the point achieved by one-step projected gradient descent from xt, the ℓ-smoothness of g implies ∥˜ xt −x⋆∥≤∥xk −x⋆∥. (C.2) Using the definition of ˜ xt and x⋆, we have (x⋆−˜ xt)⊤(ℓ(˜ xt −xt) + ∇g(xt)) ≥0, (˜ xt −x⋆)⊤∇g(x⋆) ≥0. 22 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Summing up the above two inequalities and rearranging yields that (x⋆−xt)⊤(∇g(xt) −∇g(x⋆)) ≥ℓ(x⋆−˜ xt)⊤(xt −˜ xt) + (˜ xt −xt)⊤(∇g(xt) −∇g(x⋆)). Since g is ℓ-smooth and µ-strongly convex, we have −µ∥xt −x⋆∥2 ≥−ℓ∥xt −˜ xt∥(∥x⋆−˜ xt∥+ ∥x⋆−xt∥) (C.2) ≥ −2ℓ∥xt −˜ xt∥∥xt −x⋆∥. Therefore, we conclude that ∥xt −x⋆∥≤2κ∥xt −˜ xt∥= 2κ∥xt −PX (xt −(1/ℓ)∇g(xt))∥ (C.1) ≤ r 2ϵ ℓ−µ. Part II. We first show that ∥xt −x⋆∥≤ 1 3κ r ϵ 2(ℓ−µ) = ⇒∥xt −PX (xt −(1/ℓ)∇g(xt))∥≤ r ϵ 2κ2(ℓ−µ). By the definition of x⋆, we have x⋆= PX (x⋆−(1/ℓ)∇g(x⋆)). This equation together with the triangle inequality and the nonexpansiveness of PX yields that ∥xt −PX (xt −(1/ℓ)∇g(xt))∥≤ 3∥xt −x⋆∥which implies the desired result. Then we derive a sufficient condition for guaranteeing that ∥xt −x⋆∥≤(1/(3κ)) p ϵ/(2(ℓ−µ)). Since g is µ-strongly convex and xt ∈X, Nesterov (2018, Theorem 2.1.5) together with the fact that (xt −x⋆)⊤∇g(x⋆) ≥0 implies that ∥xt −x⋆∥2 ≤2 µ  g(xt) −min x∈X g(x)  . Putting these pieces together yields the desired sufficient condition as follows, g(xt) −min x∈X g(x) ≤ ϵ 36κ3 . (C.3) Part III. We proceed to derive the gradient complexity of the algorithm using the condition in Eq. (C.3). Since Algorithm 1 is exactly Nesterov’s accelerated gradient descent, standard arguments based on estimate sequence (Nesterov, 2018) implies g(xt) −min x∈X g(x) ≤  1 −1 √κ t  g(x0) −min x∈X g(x) + µ∥x⋆−x0∥2 2  . Therefore, the gradient complexity of Algorithm 1 to guarantee Eq. (C.3) is bounded by O  1 + √κ log κ3ℓ∥x0 −x⋆∥2 ϵ  . This completes the proof. 23 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION C.2. Proof of Theorem 7 Letting ˆ x = INEXACT-APPA(g, x0, ℓ, µ, ϵ, T). Since ˆ x = xT , it suffices for us to estimate an lower bound for the maximum number of iterations T such that g(xT ) ≤minx∈X g(x) + ϵ. The following technical lemma is crucial to the subsequent analysis. Lemma 22 For any x ∈X and {(xt, ˜ xt)}t≥0 generated by Algorithm 2, we have g(x) ≥g(xt) −2ℓ(x −˜ xt−1)⊤(xt −˜ xt−1) + 2ℓ∥xt −˜ xt−1∥2 + µ∥x −xt∥2 4 −7κδ. (C.4) Proof. Using the definition of xt in Algorithm 2, we have g(xt) + ℓ∥xt −˜ xt−1∥2 ≤min x∈X  g(x) + ℓ∥x −˜ xt−1∥2 + δ. Defining x⋆ t = argminx∈X {g(x) + ℓ∥x −˜ xt−1∥2} and using µ-strongly convexity of g, we have the following for any x ∈X: g(x) ≥g(x⋆ t ) + ℓ∥x⋆ t −˜ xt−1∥2 −ℓ∥x −˜ xt−1∥2 +  ℓ+ µ 2  ∥x −x⋆ t ∥2. Equivalently, we have g(x) ≥ g(xt) + ℓ∥xt −˜ xt−1∥2 −ℓ∥x −˜ xt−1∥2 +  ℓ+ µ 2  ∥x −x⋆ t ∥2 −δ ≥ g(xt) −2ℓ(x −xt)⊤(xt −˜ xt−1) −ℓ∥x −xt∥2 +  ℓ+ µ 2  ∥x −x⋆ t ∥2 −δ. On the other hand, we have  ℓ+ µ 2  ∥x−x⋆ t ∥2−ℓ∥x−xt∥2 = µ∥x −xt∥2 2 +(2ℓ+µ)(x−xt)⊤(xt−x⋆ t )+  ℓ+ µ 2  ∥xt−x⋆ t ∥2 Using Young’s inequality yields (x −xt)⊤(xt −x⋆ t ) ≥−µ∥x −xt∥2 4(2ℓ+ µ) −(1 + 2κ)∥xt −x⋆ t ∥2. Putting these pieces together yields that g(x) ≥g(xt) −2ℓ(x −xt)⊤(xt −˜ xt−1) + µ∥x −xt∥2 4 −(2ℓ+ µ)(1 + 2κ)∥xt −x⋆ t ∥2 −δ. Furthermore, we have (x −xt)⊤(xt −˜ xt−1) = (x −˜ xt−1)⊤(xt −˜ xt−1) −∥xt −˜ xt−1∥2, and ∥xt −x⋆ t ∥2 ≤ 2 µ + 2ℓ  g(xt) + ℓ∥xt −˜ xt−1∥2 −min x∈X  g(x) + ℓ∥x −˜ xt−1∥2  ≤ 2δ µ + 2ℓ. Putting these pieces together with κ ≥1 yields the desired inequality. □ 24 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION The remaining proof is based on Lemma 22. Indeed, we have  1 − 1 2√κ  g(xt−1) + 1 2√κ  g(x⋆) + 14κ3/2δ  Eq. (C.4) ≥  1 − 1 2√κ   g(xt) −2ℓ(xt−1 −˜ xt−1)⊤(xt −˜ xt−1) + 2ℓ∥xt −˜ xt−1∥2 + µ∥xt−1 −xt∥2 4 −7κδ  + 1 2√κ  g(xt) −2ℓ(x⋆−˜ xt−1)⊤(xt −˜ xt−1) + 2ℓ∥xt −˜ xt−1∥2 + µ∥x⋆−xt∥2 4 −7κδ  + 7κδ = g(xt) −2ℓ  1 − 1 2√κ  xt−1 + x⋆ 2√κ −˜ xt−1 ⊤ (xt −˜ xt−1) + 2ℓ∥xt −˜ xt−1∥2 + µ∥x⋆−xt∥2 8√κ . Equivalently, we have g(xt) −g(x⋆) ≤  1 − 1 2√κ  (g(xt−1) −g(x⋆)) + 2ℓ  1 − 1 2√κ  xt−1 + x⋆ 2√κ −˜ xt−1 ⊤ (xt −˜ xt−1) −2ℓ∥xt −˜ xt−1∥2 −µ∥x⋆−xt∥2 8√κ + 7κδ. (C.5) Consider ˜ xt = xt + 2√κ−1 2√κ+1(xt −xt−1), we let wt = ˜ xt + 2√κ(˜ xt −xt) and obtain that wt = (1 + 2√κ)˜ xt −2√κxt = 2√κxt −(2√κ −1)xt−1 =  1 − 1 2√κ  wt−1 + 2√κxt −4κ −1 2√κ ˜ xt−1 =  1 − 1 2√κ  wt−1 + 2√κ (xt −˜ xt−1) + ˜ xt−1 2√κ. This implies that ∥wt −x⋆∥2 =  1 − 1 2√κ  wt−1 + ˜ xt−1 2√κ −x⋆+ 2√κ (xt −˜ xt−1) 2 (C.6) =  1 − 1 2√κ  wt−1 + ˜ xt−1 2√κ −x⋆ 2 + 4√κ  1 − 1 2√κ  wt−1 + ˜ xt−1 2√κ −x⋆ ⊤ (xt −˜ xt−1) +4κ∥xt −˜ xt−1∥2. Since wt−1 = ˜ xt−1 + 2√κ(˜ xt−1 −xt−1), we have (1 − 1 2√κ)wt−1 + ˜ xt−1 2√κ = 2√κ˜ xt−1 −(2√κ −1)xt−1. (C.7) Using the Young’s inequality, we have  1 − 1 2√κ  wt−1 + ˜ xt−1 2√κ −x⋆ 2 (C.8) ≤  1 − 1 2√κ 2  1 + 5 8√κ −5  ∥wt−1 −x⋆∥2 + 1 4κ  1 + 8√κ −5 5  ∥˜ xt−1 −x⋆∥2 ≤  1 − 1 2√κ   1 + 1 8√κ −5  ∥wt−1 −x⋆∥2 + 2∥˜ xt−1 −x⋆∥2 5√κ ≤  1 − 1 6√κ  ∥wt−1 −x⋆∥2 + 2∥˜ xt−1 −x⋆∥2 5√κ . 25 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Using the Young’s inequality again, we have ∥˜ xt−1 −x⋆∥2 ≤5∥x⋆−xt∥2 4 + 5∥˜ xt−1 −xt∥2. (C.9) Putting Eq. (C.6)-Eq. (C.9) together with κ ≥1, we have ∥wt −x⋆∥2 ≤  1 − 1 6√κ  ∥wt−1 −x⋆∥2 + ∥x⋆−xt∥2 2√κ + 6κ∥xt −˜ xt−1∥2 (C.10) +8κ  ˜ xt−1 −  1 − 1 2√κ  xt−1 −x⋆ 2√κ ⊤ (xt −˜ xt−1) . Combining Eq. (C.5) and Eq. (C.10) yields that g(xt) −g(x⋆) + µ∥wt −x⋆∥2 4 ≤  1 − 1 2√κ  (g(xt−1) −g(x⋆)) +  1 − 1 6√κ  µ∥wt−1 −x⋆∥2 4 + 7κδ ≤  1 − 1 6√κ   g(xt−1) −g(x⋆) + µ∥wt−1 −x⋆∥2 4  + 7κδ. Repeating the above inequality yields that g(xT ) −g(x⋆) + µ∥wT −x⋆∥2 4 ≤  1 − 1 6√κ T  g(x0) −g(x⋆) + µ∥x0 −x⋆∥2 4  + 42κ3/2δ. Therefore, we conclude that g(xT ) −g(x⋆) ≤  1 − 1 6√κ T  g(x0) −g(x⋆) + µ∥x0 −x⋆∥2 4  + 42κ3/2δ. Since the tolerance δ ≤ϵκ−3/2/84, we conclude that the iteration complexity of Algorithm 2 to guarantee that g(xT ) −minx∈X g(x) ≤ϵ if there exists an absolute constant c > 0 such that T ≥c√κ log g(x0) −g(x⋆) + (µ/4)∥x0 −x⋆∥2 ϵ  . This completes the proof. C.3. Proof of Theorem 8 Before presenting the main proof, we define the following important functions: Φg(·) = maxy∈Y g(·, y), y⋆ g(·) = argmaxy∈Y g(·, y), Ψg(·) = minx∈X g(x, ·), x⋆ g(·) = argminx∈X g(x, ·). All the above functions are well defined since g(·, ·) is strongly convex-concave. We provide their complete characterization in the following structural lemma. Lemma 23 Under the assumptions imposed in Theorem 8, we have (a) A function y⋆ g(·) is κy-Lipschitz. 26 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION (b) A function Φg(·) is 2κyℓ-smooth and µx-strongly convex with ∇Φg(·) = ∇xg(·, y⋆ g(·)). (c) A function x⋆ g(·) is κx-Lipschitz. (d) A function Ψg(·) is 2κxℓ-smooth and µy-strongly concave with ∇Ψg(·) = ∇yg(x⋆ g(·), ·). where κx = ℓ/µx and κy = ℓ/µy are condition numbers. Now we are ready to prove Theorem 8. We divide the proof into three parts. In the first part, we show that the output ˆ x = MAXIMIN-AG2(g, x0, y0, ℓ, µx, µy, ϵ) satisfies max y∈Y g(ˆ x, y) ≤min x∈X max y∈Y g(x, y) + ϵ (C.11) In the second part, we get the sufficient condition for guaranteeing the stopping criteria in Algo-rithm 3. In the third part, we estimate an upper bound for the gradient complexity of the algorithm using the condition derived in the second part. For the ease of presentation, we denote (x⋆ g, y⋆ g) as the unique solution to the minimax optimization minx∈X maxy∈Y g(x, y). Part I. By the definition of Φg, the inequality in Eq. (C.11) can be rewritten as follows, Φg(ˆ x) ≤min x∈X Φg(x) + ϵ. Since ˆ x = PX (xT −(1/2κyℓ)∇xg(xT , yT )), we have 0 ≤ (x −ˆ x)⊤(2κyℓ(ˆ x −xT ) + ∇xg(xT , yT )) = (x −ˆ x)⊤(2κyℓ(ˆ x −xT ) + ∇Φg(xT )) + (x −ˆ x)⊤(∇xg(xT , yT ) −∇Φg(xT )). Since ∇Φg(xT ) = ∇xg(xT , y⋆ g(xT )), we have ∥∇xg(xT , yT ) −∇Φg(xT )∥≤ℓ∥yT −y⋆ g(xT )∥. Using the Young’s inequality, we have (x−ˆ x)⊤(∇xg(xT , yT )−∇Φg(xT )) ≤κyℓ∥ˆ x −xT ∥2 2 + κyℓ∥x −xT ∥2 2 +µy∥yT −y⋆ g(xT )∥2. Since Φg is 2κyℓ-smooth and µx-strongly convex, we have (x −ˆ x)⊤(2κyℓ(ˆ x −xT ) + ∇Φg(xT )) ≤ 2κyℓ(x −xT )⊤(ˆ x −xT ) + Φg(x) −Φg(ˆ x) −κyℓ∥ˆ x −xT ∥2 −µx∥x −xT ∥2 2 . Using the Young’s inequality, we have (x−xT )⊤(ˆ x−xT ) ≤∥x−xT ∥2+(1/4)∥ˆ x−xT ∥2. Putting these pieces together yields with x = x⋆ g yields that Φg(ˆ x) −min x∈X Φg(x) ≤3κyℓ∥xT −x⋆ g∥2 + µy∥yT −y⋆ g(xT )∥2. (C.12) In what follows, we prove that Φg(ˆ x) ≤minx∈X Φg(x) + ϵ if the following stopping conditions hold true, g(xT , yT ) −g(x⋆ g(yT ), yT ) ≤ ϵ 648κ3 xκ3 y , (C.13) ∥yT −PY(yT + (1/2κxℓ)∇yg(xT , yT ))∥≤ 1 24κ2 xκy r ϵ κyℓ. (C.14) 27 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Indeed, we observe that ∥xT −x⋆ g∥≤∥xT −x⋆ g(yT )∥+ ∥x⋆ g(yT ) −x⋆ g(y⋆ g)∥+ ∥x⋆ g(y⋆ g) −x⋆ g∥. By definition, we have x⋆ g(y⋆ g) = x⋆ g. Also, x⋆ g(·) is κx-Lipschitz. Therefore, we have ∥xT −x⋆ g∥≤∥xT −x⋆ g(yT )∥+ κx∥yT −y⋆ g∥. (C.15) By the similar argument, we have ∥yT −y⋆ g(xT )∥≤∥yT −y⋆ g∥+κy∥xT −x⋆ g∥≤κy∥xT −x⋆ g(yT )∥+κxκy∥yT −y⋆ g∥. (C.16) First, we bound the term ∥xT −x⋆ g(yT )∥. Since g(·, yT ) is µx-strongly convex, we have ∥xT −x⋆ g(yT )∥≤ s 2(g(xT , yT ) −g(x⋆(yT ), yT )) µx ≤ 1 18κxκy r ϵ κyℓ (C.17) It remains to bound the term ∥yT −y⋆ g∥. Indeed, we have ∇Ψg(yT ) = ∇yg(x⋆ g(yT ), yT ) and ∥yT −PY(yT + (1/2κxℓ)∇Ψg(yT ))∥ ≤ ∥yT −PY(yT + (1/2κxℓ)∇yg(xT , yT ))∥ +∥PY(yT + (1/2κxℓ)∇yg(xT , yT )) −PY(yT + (1/2κxℓ)∇Ψg(yT ))∥. Since PY is nonexpansive and ∇yg is ℓ-Lipschitz, we have ∥PY(yT + (1/2κxℓ)∇yg(xT , yT )) −PY(yT + (1/2κxℓ)∇Ψg(yT ))∥≤∥xT −x⋆ g(yT )∥ 2κx . Putting these pieces together with Eq. (C.14) and Eq. (C.17) yields that ∥yT −PY(yT + (1/2κxℓ)∇Ψg(yT ))∥≤ 1 18κ2 xκy r ϵ κyℓ. (C.18) Since y⋆ g = argmaxy∈Y Ψg(y) and ˜ yT = PY(yT + (1/2κxℓ)∇Ψg(yT )) is achieved by one-step projected gradient ascent from yT , we derive from the 2κxℓ-smoothness of Ψg, we have ∥˜ yT −y⋆ g∥≤∥yT −y⋆ g∥. (C.19) Using the definition of ˜ yT and y⋆ g, we have (y⋆ g −˜ yT )⊤(˜ yT −yT −(1/2κxℓ)∇Ψg(yT )) ≥0, (y⋆ g −˜ yT )⊤∇Ψg(y⋆ g) ≥0. Summing up the above two inequalities and rearranging yields that (y⋆ g−yT )⊤(∇Ψg(y⋆ g)−∇Ψg(yT )) ≥2κxℓ(y⋆ g−˜ yT )⊤(yT −˜ yT )+(˜ yT −yT )⊤(∇Ψg(y⋆ g)−∇Ψg(yT )). Since Ψg is 2κxℓ-smooth and µy-strongly concave, we have −µy∥y⋆ g−yT ∥2 ≥−2κxℓ∥˜ yT −yT ∥ ∥y⋆ g −˜ yT ∥+ ∥y⋆ g −yT ∥  (C.19) ≥ −4κxℓ∥˜ yT −yT ∥∥y⋆ g−yT ∥. This implies that ∥y⋆ g −yT ∥≤4κxκy∥yT −˜ yT ∥ (C.18) ≤ 1 4κx r ϵ κyℓ. (C.20) 28 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Plugging Eq. (C.17) and Eq. (C.20) into Eq. (C.15) yields that ∥xT −x⋆ g∥≤  1 18κxκy + 1 4  r ϵ κyℓ κx,κy≥1 ≤ 1 2 r ϵ κyℓ. Plugging Eq. (C.17) and Eq. (C.20) into Eq. (C.16) yields that ∥yT −y⋆ g(xT )∥≤  1 18κx + κy 4  r ϵ κyℓ κx,κy≥1 ≤ 1 2 rκyϵ ℓ. Putting these pieces together Eq. (C.12) yields the desired result. Part II. We first show that ∥yT −y⋆ g∥≤(1/216κ2 xκy) p ϵ/κyℓand Eq. (C.13) are sufficient to guarantee Eq. (C.14). Indeed, we have y⋆ g = PY(y⋆ g + (1/2κxℓ)∇Ψg(y⋆ g)). This together with the triangle inequality and the nonexpansiveness of PY yields ∥yT −PY(yT + (1/2κxℓ)∇yg(xT , yT ))∥≤2∥yT −y⋆ g∥+ ∥∇yg(xT , yT ) −∇Ψg(y⋆ g)∥ 2κxℓ . Furthermore, ∇Ψg(yT ) = ∇yg(x⋆(yT ), yT ) and ∥∇yg(xT , yT ) −∇Ψg(y⋆ g)∥≤∥∇yg(xT , yT ) −∇yg(x⋆ g(yT ), yT )∥+ ∥∇Ψg(yT ) −∇Ψg(y⋆ g)∥. Since g is ℓ-smooth and Ψg is 2κxℓ-smooth, we have ∥∇yg(xT , yT ) −∇Ψg(y⋆ g)∥≤ℓ∥xT −x⋆ g(yT )∥+ 2κxℓ∥yT −y⋆ g∥. Also, Eq. (C.13) guarantees that Eq. (C.17) holds true. Then we have ∥yT −PY(yT + (1/2κxℓ)∇yg(xT , yT ))∥≤3∥yT −y⋆ g∥+ 1 36κ2 xκy r ϵ κyℓ. The above inequality together with ∥yT −y⋆ g∥≤(1/216κ2 xκy) p ϵ/κyℓguarantees Eq. (C.14). Next we derive a sufficient condition for guaranteeing ∥yT −y⋆ g∥≤(1/216κ2 xκy) p ϵ/κyℓ. Since Ψg is µy-strongly concave, Nesterov (2018, Theorem 2.1.5) implies that ∥yT −y⋆ g∥2 ≤ 2 µy  max y∈Y Ψg(y) −Ψg(yT )  . Putting these pieces together yields the desired condition as follows, max y∈Y Ψg(y) −Ψg(yT ) ≤ ϵ 93312κ4 xκ4 y . (C.21) Part III. We proceed to estimate an upper bound for the gradient complexity of Algorithm 3 using Eq. (C.21). Note that ˜ ϵ ≤ϵ/(4477676(κxκy)11/2) and we provide a key technical lemma which is crucial to the subsequent analysis. Lemma 24 For any y ∈Y and {(yt, ˜ yt)}t≥0 generated by Algorithm 3, we have Ψg(y) ≤2κxℓ(y−˜ yt−1)⊤(yt−˜ yt−1)+Ψg(yt)−κxℓ∥yt −˜ yt−1∥2 2 −µy∥y −˜ yt−1∥2 4 +3κxκy˜ ϵ. 29 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Proof. For any y ∈Y, the update formula yt ←PY(˜ yt−1 + (1/2κxℓ)∇yg(˜ xt−1, ˜ yt−1)) implies that 0 ≤ (y −yt)⊤(2κxℓ(yt −˜ yt−1) −∇yg(˜ xt−1, ˜ yt−1)) = (y −yt)⊤(2κxℓ(yt −˜ yt−1) −∇Ψg(˜ yt−1)) + (y −yt)⊤(∇Ψg(˜ yt−1) −∇yg(˜ xt−1, ˜ yt−1)). Since ∇Ψg(˜ yt−1) = ∇yg(x⋆ g(˜ yt−1), ˜ yt−1), we have ∥∇Ψg(˜ yt−1) −∇yg(˜ xt−1, ˜ yt−1)∥≤ℓ∥x⋆ g(˜ yt−1) −˜ xt−1∥. Since g(·, ˜ yt−1) is µx-strongly convex, we have ∥x⋆ g(˜ yt−1) −˜ xt−1∥≤ s 2(g(˜ xt−1, ˜ yt−1) −g(x⋆ g(˜ yt−1), ˜ yt−1)) µx ≤ s 2˜ ϵ µx . Using Young’s inequality, we have (y −yt)⊤(∇Ψg(˜ yt−1) −∇yg(˜ xt−1, ˜ yt−1)) ≤κxℓ∥yt −˜ yt−1∥2 2 + µy∥y −˜ yt−1∥2 4 + 3κxκy˜ ϵ. Since Ψg is 2κxℓ-smooth and µy-strongly concave, we have (y −yt)⊤(2κxℓ(yt −˜ yt−1) −∇Ψg(˜ yt−1)) ≤ 2κxℓ(y −˜ yt−1)⊤(yt −˜ yt−1) + Ψg(yt) −Ψg(y) −κxℓ∥yt −˜ yt−1∥2 −µy∥y −˜ yt−1∥2 2 . Putting these pieces together yields the desired inequality. □ The remaining proof is based on the modification of Nesterov’s techniques (Nesterov, 2018, Sec-tion 2.2.5). Indeed, we define the estimate sequence as follows, Γ0(y) = Ψg(y0) −µy∥y −y0∥2 2 , Γt+1(y) = 1 4√κxκy  Ψg(yt+1) + 2κxℓ(y −˜ yt)⊤(yt+1 −˜ yt) −κxℓ∥yt+1 −˜ yt∥2 2 −µy∥y −˜ yt∥2 4 −12(κxκy)3/2˜ ϵ  +  1 − 1 4√κxκy  Γt(y) for all t ≥0. We apply the inductive argument to prove, max y∈Rn Γt(y) ≤Ψg(yt) for all t ≥0. (C.22) Eq. (C.22) holds trivially when t = 0. In what follows, we show that Eq. (C.22) holds true when t = T if Eq. (C.22) holds true for all t ≤T −1. Let vt = argmaxy∈Rn Γt(y) and Γ⋆ t = maxy∈Rn Γt(y), we have the canonical form Γt(y) = Γ⋆ t −(µy/4)∥y −vt∥2. The following recursive rules hold for vt and Γ⋆ t : vt+1 =  1 − 1 4√κxκy  vt + ˜ yt 4√κxκy + √κxκy(yt+1 −˜ yt), Γ⋆ t+1 =  1 − 1 4√κxκy  Γ⋆ t + 1 4√κxκy  Ψg(yt+1) −12(κxκy)3/2˜ ϵ  −  ℓ 8 rκx κy −κxℓ 4  ∥yt+1 −˜ yt∥2 − 1 4√κxκy  1 − 1 4√κxκy  µy∥˜ yt −vt∥2 4 −2κxℓ(vt −˜ yt)⊤(yt+1 −˜ yt)  . 30 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION It follows from the recursive rule for Γt and its canonical form that ∇Γt+1(y) = −  1 − 1 4√κxκy  µy(y −vt) 2 + 1 4√κxκy  2κxℓ(yt+1 −˜ yt) −µy(y −˜ yt) 2  . The recursive rule for vt can be achieved by solving ∇Γt+1(vt+1) = 0. Then we have Γ⋆ t+1 = Γt+1(vt+1) =  1 − 1 4√κxκy  Γ⋆ t −  1 − 1 4√κxκy  µy∥vt+1 −vt∥2 4 + 1 4√κxκy  Ψg(yt+1) −12(κxκy)3/2˜ ϵ −κxℓ∥yt+1 −˜ yt∥2 2  + 1 4√κxκy  2κxℓ(vt+1 −˜ yt)⊤(yt+1 −˜ yt) −µy∥vt+1 −˜ yt∥2 4  . Then we conclude the recursive rule for Γ⋆ t by plugging the recursive rule for vk into the above equality. By the induction, Eq. (C.22) holds true when t = T −1 which implies Γ⋆ T ≤  1 − 1 4√κxκy  Ψg(yT−1) + 1 4√κxκy  Ψg(yT ) −12(κxκy)3/2˜ ϵ  −  ℓ 8 rκx κy −κxℓ 4  ∥yT −˜ yT−1∥2 − 1 4√κxκy  1 − 1 4√κxκy  µy∥˜ yT−1 −vT−1∥2 2 −2κxℓ(vT−1 −˜ yT−1)⊤(yT −˜ yT−1)  . Applying Lemma 24 with t = T and y = yT−1 further implies that Ψg(yT−1) ≤ 2κxℓ(yT−1 −˜ yT−1)⊤(yT −˜ yT−1) + Ψ(yT ) −κxℓ∥yT −˜ yT−1∥2 2 −µy∥yT−1 −˜ yT−1∥2 2 + 3κxκy˜ ϵ. Putting these pieces together yields that Γ⋆ T ≤ Ψg(yT ) +  1 − 1 4√κxκy  2κxℓ(yT −˜ yT−1)⊤  (yT−1 −˜ yT−1) + 1 4√κxκy (vT−1 −˜ yT−1)  . Using the update formula ˜ yt = yt + 4√κxκy−1 4√κxκy+1(yt −yt−1) and the recursive rule for vt with the inductive argument, it is straightforward that (yt −˜ yt) + 1 4√κxκy (vt −˜ yt) = 0 for all t ≥0. This implies that Γ⋆ T ≤Ψg(yT ). Therefore, we conclude that Eq. (C.22) holds true for all t ≥0. On the other hand, Lemma 24 and the update formula for Γt implies that Γt+1(y) ≥ 1 4√κxκy  Ψg(y) −12(κxκy)3/2˜ ϵ −3κxκy˜ ϵ  +  1 − 1 4√κxκy  Γt(y). Since κx, κy ≥1, we have Ψg(y) −Γt+1(y) ≤  1 − 1 4√κxκy  (Ψg(y) −Γt(y)) + 6κxκy˜ ϵ. 31 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Repeating the above inequality yields that Ψg(y) −ΓT (y) ≤  1 − 1 4√κxκy T (Ψg(y) −Γ0(y)) + 24(κxκy)3/2˜ ϵ. Therefore, we conclude that max y∈Y Ψg(y) −Ψg(yT ) ≤  1 − 1 4√κxκy T (2κxℓ+ ¯ µ)D2 y 2 + 24(κxκy)3/2˜ ϵ. Since the tolerance ˜ ϵ ≤ ϵ 4477676(κxκy)11/2 , we conclude that the iteration complexity Algorithm 3 to guarantee Eq. (C.21) is bounded by O(√κxκy log(ℓD2 y/ϵ)). Now it suffices to establish the gradient complexity of the two AGD subroutines at each iteration. In particular, we use the gradient complexity of the AGD subroutine to guarantee that g(ˆ x) ≤ minX g(x) + ϵ is bounded by O  1 + √κ log κ3ℓ∥x0 −x⋆∥2 ϵ  , where κ is the condition number of g and x⋆is the global optimum of g over X. Since Y is a convex and bounded set, {yt}t≥0 is a bounded sequence. Hence {˜ yt}t≥0 is also a bounded sequence. Since x⋆ g(·) is κx-Lipschitz (cf. Lemma 23), the sequences {x⋆ g(˜ yt)}t≥0 and {x⋆ g(yt)}t≥0 are bounded. Thus, we have ∥x0 −x⋆ g(yt)∥2 = ∥x0 −x⋆ g(˜ yt)∥2 = O(∥x0 −x⋆ g(y0)∥2 + κ2 xD2 y). Putting these pieces together yields that the gradient complexity of every AGD subroutines at each iteration is bounded by O(√κx log((κ3 xℓ(∥x0 −x⋆ g(y0)∥2 + κ2 xD2 y)/˜ ϵ)). Therefore, the gradient complexity of Algorithm 3 to guarantee Eq. (C.21) is bounded by O κx√κy · log2 (κx + κy)ℓ( ˜ D2 x + D2 y) ϵ !! , where κx = ℓ/µx and κy = ℓ/µy are condition numbers, ˜ Dx = ∥x0 −x⋆ g(y0)∥is the initial distance where x⋆ g(y0) = argminx∈X g(x, y0) and Dy > 0 is the diameter of the constraint set Y. Appendix D. Proofs for Convex-Concave Settings In this section, we present proofs for all results in Section 4. D.1. Proof of Theorem 9 We first show that there exists T > 0 such that (ˆ x, ˆ y) = MINIMAX-APPA(f, x0, y0, ℓ, µx, µy, ϵ, T) is an ϵ-saddle point. Then we estimate the total number of gradient evaluations required to output an ϵ-approximate saddle point. 32 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION First, we note that MINIMAX-APPA in Algorithm 4 can be interpreted as an inexact accelerated proximal point algorithm INEXACT-APPA with the inner loop solver MAXIMIN-AG2 and AGD. Using Theorem 6 and Theorem 7, the point (ˆ x, ˆ y) satisfies max y∈Y f(ˆ x, y)−min x∈X max y∈Y f(x, y) ≤  1 − 1 6√κx T  Φ(x0) −Φ(x⋆) + µx∥x⋆−x0∥2 4  +42κ3/2 x δ. and ˆ y ←PY (˜ y + (1/2κxℓ)∇yf(ˆ x, ˜ y)) where ˜ y ∈Y satisfies that max y∈Y f(ˆ x, y) −f(ˆ x, ˜ y) ≤˜ ϵ. We let Φ(·) = maxy∈Y f(·, y) and note that Φ is µx-strongly convex function. Since f is µx-strongly-convex-µy-strongly-concave, the Nash equilibrium (x⋆, y⋆) is unique and x⋆= argminx∈X Φ(x). Therefore, we have ∥ˆ x −x⋆∥2 ≤ 2 µx  max y∈Y f(ˆ x, y) −min x∈X max y∈Y f(x, y)  . Since f(ˆ x, ·) is µy-strongly concave, Nesterov (2018, Theorem 2.1.5) implies that ∥˜ y −y⋆(ˆ x)∥2 ≤ 2 µy  max y∈Y f(ˆ x, y) −f(ˆ x, ˜ y)  ≤2˜ ϵ µy . Since y⋆(·) = argmaxy∈Y f(·, y) is κy-Lipschitz (cf. Lemma 23), ∥y⋆−y⋆(ˆ x)∥2 = ∥y⋆(x⋆) − y⋆(ˆ x)∥2 ≤κ2 y∥ˆ x −x⋆∥2. Thus, we have ∥˜ y −y⋆∥2 ≤2κ2 y∥ˆ x −x⋆∥2 + 4˜ ϵ µy . Let Ψ(·) = minx∈X f(x, ·). By the definition of ˆ y, the following inequality holds true for any y ∈Y, 0 ≤ (y −ˆ y)⊤(2κxℓ(ˆ y −˜ y) −∇yf(ˆ x, ˜ y)) = (y −ˆ y)⊤(2κxℓ(ˆ y −˜ y) −∇Ψ(˜ y)) + (y −ˆ y)⊤(∇Ψ(˜ y) −∇yf(ˆ x, ˜ y)). Since ∇Ψ(˜ y) = ∇yf(x⋆(˜ y), ˜ y), we have ∥∇Ψ(˜ y) −∇yf(ˆ x, ˜ y)∥≤ℓ∥x⋆(˜ y) −ˆ x∥. Using the Young’s inequality, we have (y −ˆ y)⊤(∇Ψ(˜ y) −∇yf(ˆ x, ˜ y)) ≤κxℓ∥ˆ y −˜ y∥2 2 + κxℓ∥y −˜ y∥2 2 + µx∥x⋆(˜ y) −ˆ x∥2. Since Ψ is µy-strongly concave and 2κxℓ-smooth, we have (y −ˆ y)⊤(2κxℓ(ˆ y −˜ y) −∇Ψ(˜ y)) ≤ 2κxℓ(y −˜ y)⊤(ˆ y −˜ y) + Ψ(ˆ y) −Ψ(y) −κxℓ∥ˆ y −˜ y∥2 −µx∥y −˜ y∥2 2 . 33 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Using the Young’s inequality, we have (y −˜ y)⊤(ˆ y −˜ y) ≤∥y −˜ y∥2 + (1/4)∥ˆ y −˜ y∥2. Putting these pieces together with y = y⋆yields that min x∈X max y∈Y f(x, y) −min x∈X f(x, ˆ y) = Ψ(y⋆) −Ψ(ˆ y) ≤3κxℓ∥˜ y −y⋆∥2 + µx∥x⋆(˜ y) −ˆ x∥2 ≤3κxℓ∥˜ y −y⋆∥2 + 2µx∥x⋆(˜ y) −x⋆(y⋆)∥2 + 2µx∥x⋆−ˆ x∥2 ≤5κxℓ∥˜ y −y⋆∥2 + 2µx∥x⋆−ˆ x∥2. Therefore, we conclude that max y∈Y f(ˆ x, y) −min x∈X f(x, ˆ y) ≤20κxκy˜ ϵ + (20κ2 xκ2 y + 5)  max y∈Y f(ˆ x, y) −min x∈X max y∈Y f(x, y)  . Note that ˜ ϵ ≤ϵ/80κxκy and δ ≤ϵ/4200κ7/2 x κ2 y. This together with the above inequality implies that max y∈Y f(ˆ x, y)−min x∈X f(x, ˆ y) ≤3ϵ 4 +(20κ2 xκ2 y+5)  1 − 1 6√κx T  Φ(x0) −Φ(x⋆) + µx∥x⋆−x0∥2 4  . To this end, there exists an absolute constant c > 0 such that maxy∈Y f(ˆ x, y)−minx∈X f(x, ˆ y) ≤ϵ if the maximum number of iterations T ≥c√κx log(κ2 xκ2 yℓ∥x⋆−x0∥2/ϵ). This implies that the total number of iterations is bounded by O √κx log κ2 xκ2 yℓ∥x⋆−x0∥2 ϵ !! . Furthermore, we call the solver MAXIMIN-AG2 at each iteration. Using Theorem 8 and δ = ϵ/(10κxκy)4, the number of gradient evaluations at each iteration is bounded by O √κy log κ7/2 x κ3 yℓ( ˜ D2 x + D2 y) ϵ ! log κ4 xκ4 yℓD2 y ϵ !! . Recalling D = max{Dx, Dy} < +∞, we conclude that the total number of gradient evaluations is bounded by O √κxκy log3 κxκyℓD2 ϵ  . This completes the proof. D.2. Proof of Corollary 10 We first show that (ˆ x, ˆ y) = MINIMAX-APPA(fϵ,y, x0, y0, ℓ, µx, ϵ/(4D2 y), ϵ/2, T) is an ϵ-saddle point. Then we estimate the number of gradient evaluations to output an ϵ-saddle point using Theo-rem 9. By the definition of fϵ, the output (ˆ x, ˆ y) satisfies max y∈Y  f(ˆ x, y) −ϵ∥y −y0∥2 4D2 y  −min x∈X  f(x, ˆ y) −ϵ∥ˆ y −y0∥2 4D2 y  ≤ϵ 2. 34 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Since the function f(x, ·) is concave for each x ∈X, we have max y∈Y  f(ˆ x, y) −ϵ∥y −y0∥2 4D2 y  ≥max y∈Y f(xT+1, y) −ϵ 4. On the other hand, we have min x∈X  f(x, ˆ y) −ϵ∥ˆ y −y0∥2 4D2 y  ≤min x∈X f(x, ˆ y) + ϵ 4. Putting these pieces together yields that maxy∈Y f(ˆ x, y) −minx∈X f(x, ˆ y) ≤ϵ. Furthermore, letting κy = 2ℓD2 y/ϵ in the gradient complexity bound presented in Theorem 9, we conclude that the total number of gradient evaluations is bounded by O r κxℓ ϵ Dy log3 κxℓD2 ϵ ! . This completes the proof. D.3. Proof of Corollary 11 We first show that (ˆ x, ˆ y) = MINIMAX-APPA(fϵ, x0, y0, ℓ, ϵ/(4D2 x), ϵ/(4D2 y), ϵ/2, T) is an ϵ-saddle point. Then we estimate the number of gradient evaluations to output an ϵ-saddle point using Theorem 9. By the definition of fϵ, the output (ˆ x, ˆ y) satisfies max y∈Y  f(ˆ x, y) + ϵ∥ˆ x −x0∥2 8D2 x −ϵ∥y −y0∥2 8D2 y  −min x∈X  f(x, ˆ y) + ϵ∥x −x0∥2 8D2 x −ϵ∥ˆ y −y0∥2 8D2 y  ≤ϵ 2. Since the function f(x, ·) is concave for each x ∈X, we have max y∈Y  f(ˆ x, y) + ϵ∥ˆ x −x0∥2 8D2 x −ϵ∥y −y0∥2 8D2 y  ≥max y∈Y f(ˆ x, y) −ϵ 4. On the other hand, min x∈X  f(x, ˆ y) + ϵ∥x −x0∥2 8D2 x −ϵ∥ˆ y −y0∥2 8D2 y  ≤min x∈X f(x, ˆ y) + ϵ 4. Putting these pieces together yields that maxy∈Y f(ˆ x, y) −minx∈X f(x, ˆ y) ≤ϵ. Furthermore, letting κx = 4ℓD2 x/ϵ and κy = 2ℓD2 y/ϵ in the gradient complexity bound pre-sented in Theorem 9, we conclude that the total number of gradient evaluations is bounded by O ℓDxDy ϵ log3 ℓD2 ϵ  . This completes the proof. Appendix E. Proofs for Nonconvex-Concave Settings In this section, we present proofs for all results in Section 5 and Section B 35 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION E.1. Proof of Theorem 12 Using the definition of gt, we have max y∈Y f(xt+1, y) + ℓ∥xt+1 −xt∥2 ≤min x∈X  max y∈Y f(x, y) + ℓ∥x −xt∥2  + δ. This implies that Φ(xt+1) + ℓ∥xt+1 −xt∥2 ≤min x∈X  Φ(x) + ℓ∥x −xt∥2 + δ ≤Φ(xt) + δ. Equivalently, we have ∥xt+1 −xt∥2 ≤Φ(xt) −Φ(xt+1) + δ ℓ . (E.1) Note that the function Φ(·) + ℓ∥· −xt∥2 is ℓ-strongly convex and its minimizer x∗ t is well defined and unique (Davis and Drusvyatskiy, 2019). Since the function Φ(·) + ℓ∥· −xt∥2 is ℓ-strongly convex, we derive from Nesterov (2018, Theorem 2.1.5) that ∥xt+1 −x∗ t ∥2 ≤2 ℓ  Φ(xt+1) + ℓ∥xt+1 −xt∥2 −min x∈Rm  Φ(x) + ℓ∥x −xt∥2  ≤2δ ℓ. (E.2) Since Φ is differentiable, we have x∗ t −PX  x∗ t −∇Φ(x∗ t ) + 2ℓ(x∗ t −xt) ℓ  = 0. Therefore, we have xt+1 −PX  xt+1 −∇Φ(xt+1) ℓ  ≤2∥xt+1 −x∗ t ∥+ 2∥xt −x∗ t ∥+ ∥∇Φ(x∗ t ) −∇Φ(xt+1)∥ ℓ . Since Φ(·) is 2κyℓ-smooth, we have ∥∇Φ(xt+1) −∇Φ(x∗ t )∥≤2κyℓ∥xt+1 −x∗ t ∥. Putting these pieces together yields that xt+1 −PX  xt+1 −∇Φ(xt+1) ℓ  ≤ (2κy + 2)∥xt+1 −x∗ t ∥+ 2∥xt −x∗ t ∥ (E.3) κy≥1 ≤ 6κy∥xt+1 −x∗ t ∥+ 2∥xt+1 −xt∥. Putting Eq. (E.1), Eq. (E.2) and Eq. (E.3) together with the Cauchy-Schwarz inequality yields (ℓ∥xt+1 −PX (xt+1 −(1/ℓ)∇Φ(xt+1))∥)2 ≤ 72κ2 yℓ2∥xt+1 −x∗ t ∥2 + 8ℓ2∥xt+1 −xt∥2 ≤ 8ℓ(Φ(xt) −Φ(xt+1) + δ) + 144κ2 yℓδ. Summing up the above inequality over t = 0, 1, . . . , T −1 and dividing it by T yields that 1 T T−1 X t=0 (ℓ∥xt+1 −PX (xt+1 −(1/ℓ)∇Φ(xt+1))∥)2 ! ≤ 8ℓ(Φ(x0) −Φ(xT )) T + 8ℓδ + 144κ2 yℓδ κy≥1 ≤ 8ℓ(Φ(x0) −Φ(xT )) T + 152κ2 yℓδ. 36 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Since ˆ x = xs is uniformly chosen from {xs}1≤s≤T and δ ≤ϵ2/(10κy)4ℓ, we have E (ℓ∥ˆ x −PX (ˆ x −(1/ℓ)∇Φ(ˆ x))∥)2 = 1 T T−1 X t=0 (ℓ∥xt+1 −PX (xt+1 −(1/ℓ)∇Φ(xt+1))∥)2 ! ≤ 8ℓ(Φ(x0) −Φ(xT )) T + 152κ2 yℓδ ≤8ℓ∆Φ T + ϵ2 8 . Using the Markov inequality, we conclude that there exists T > cℓ∆Φϵ−2, where the output ˆ x will satisfy ℓ∥ˆ x −PX (ˆ x −(1/ℓ)∇Φ(ˆ x))∥≤ϵ/2 with probability at least 2/3. Since ˆ y is obtained by running AGD on −f(ˆ x, ·) to optimal with tolerance δ ≤ϵ2/(10κy)4ℓ, and f(ˆ x, ·) is µy-concave function, we know that δ-optimality guarantees: ℓ∥PY[ˆ y + (1/ℓ)∇yf(ˆ x, ˆ y)] −ˆ y∥ ≤ ϵ, ∥ˆ y −y⋆(ˆ x)∥ ≤ ϵ 2ℓ. Putting these pieces together yields that ℓ∥ˆ x −PX (ˆ x −(1/ℓ)∇xf(ˆ x, ˆ y))∥ ≤ ℓ∥ˆ x −PX (ˆ x −(1/ℓ)∇Φ(ˆ x))∥+ ∥∇Φ(ˆ x) −∇xf(ˆ x, ˆ y)∥ ≤ ℓ∥ˆ x −PX (ˆ x −(1/ℓ)∇Φ(ˆ x))∥+ ℓ∥ˆ y −y⋆(ˆ x)∥ ≤ ϵ. This implies that (ˆ x, ˆ y) is an ϵ-stationary point. Furthermore, we call the solver MAXIMIN-AG2 at each iteration. Using Theorem 8 and δ ≤ϵ2/(10κy)4ℓ, the number of gradient evaluations at each iteration is bounded by O √κy log κ5 yℓ2( ˜ D2 x + D2 y) ϵ2 ! log κ4 yℓ2D2 y ϵ2 !! . Therefore, we conclude that the total number of gradient evaluations is bounded by O ℓ∆Φ ϵ2 · √κy log2 κyℓ( ˜ D2 x + D2 y) ϵ !! . This completes the proof. E.2. Proof of Corollary 13 Recall that the function ˜ fϵ is defined by ˜ fϵ(x, y) = f(x, y) −ϵ∥y −y0∥2 4Dy . This implies that the following statement holds for all (x, y) ∈X × Y that ∇xf(x, y) −∇x ˜ fϵ(x, y) = 0, ∥∇yf(x, y) −∇y ˜ fϵ(x, y)∥ ≤ ϵ 2. 37 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Since (ˆ x, ˆ y) = MINIMAX-PPA( ˜ fϵ, x0, y0, ℓ, ϵ/(2Dy), ϵ/2, T), we have ℓ∥PX [ˆ x −(1/ℓ)∇x ˜ fϵ(ˆ x, ˆ y)] −ˆ x∥ ≤ ϵ 2, ℓ∥PY[ˆ y + (1/ℓ)∇y ˜ fϵ(ˆ x, ˆ y)] −ˆ y∥ ≤ ϵ 2. Putting these pieces together yields that ℓ∥PX [ˆ x −(1/ℓ)∇xf(ˆ x, ˆ y)] −ˆ x∥ ≤ ϵ 2 ≤ϵ, ℓ∥PY[ˆ y + (1/ℓ)∇yf(ˆ x, ˆ y)] −ˆ y∥ ≤ ℓ∥PY[ˆ y + (1/ℓ)∇y ˜ fϵ(ˆ x, ˆ y)] −ˆ y∥+ ∥∇yf(x, y) −∇y ˜ fϵ(x, y)∥ ≤ ϵ. Therefore, we conclude that (ˆ x, ˆ y) is an ϵ-stationary point of f. Furthermore, letting κy = 2ℓDy/ϵ in the gradient complexity bound presented in Theorem 12, we conclude that the total number of gradient evaluations is bounded by O ℓ∆Φ ϵ2 · r ℓDy ϵ log2 ℓ( ˜ D2 x + D2 y) ϵ !! . This completes the proof. E.3. Proof of Theorem 20 Using the same argument as in Theorem 12, we have ∥xt+1 −xt∥2 ≤Φ(xt) −Φ(xt+1) + δ ℓ . (E.4) and ∥xt+1 −x∗ t ∥2 ≤2 ℓ  Φ(xt+1) + ℓ∥xt+1 −xt∥2 −min x∈Rm  Φ(x) + ℓ∥x −xt∥2  ≤2δ ℓ. (E.5) Since Φ is differentiable, we have ∇Φ(x∗ t )+2ℓ(x∗ t −xt) = 0 which implies ∥∇Φ(x∗ t )∥= 2ℓ∥x∗ t − xt∥. Since Φ(·) is 2κyℓ-smooth, we have ∥∇Φ(xt+1) −∇Φ(x∗ t )∥≤2κyℓ∥xt+1 −x∗ t ∥. Putting these pieces together yields that ∥∇Φ(xt+1)∥ ≤ 2κyℓ∥xt+1 −x∗ t ∥+ 2ℓ∥x∗ t −xt∥≤(2κyℓ+ 2ℓ)∥xt+1 −x∗ t ∥+ 2ℓ∥xt+1 −xt∥ κy≥1 ≤ 4κyℓ∥xt+1 −x∗ t ∥+ 2ℓ∥xt+1 −xt∥. (E.6) Putting Eq. (E.4), Eq. (E.5) and Eq. (E.6) together with the Cauchy-Schwarz inequality yields ∥∇Φ(xt+1)∥2 ≤32κ2 yℓ2∥xt+1−x∗ t ∥2+8ℓ2∥xt+1−xt∥2 ≤8ℓ(Φ(xt) −Φ(xt+1) + δ)+64κ2 yℓδ. Summing up the above inequality over t = 0, 1, . . . , T −1 and dividing it by T yields that 1 T T−1 X t=0 ∥∇Φ(xt+1)∥2 ! ≤8ℓ(Φ(x0) −Φ(xT )) T +8ℓδ+64κ2 yℓδ κy≥1 ≤ 8ℓ(Φ(x0) −Φ(xT )) T +72κ2 yℓδ. 38 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Since ˆ x = xs is uniformly chosen from {xs}1≤s≤T and δ ≤ϵ2/144κ2 yℓ, we have E ∥∇Φ(ˆ x)∥2 = 1 T T−1 X t=0 ∥∇Φ(xt+1)∥2 ! ≤8ℓ(Φ(x0) −Φ(xT )) T + 72κ2 yℓδ ≤8ℓ∆Φ T + ϵ2 2 . Using the Markov inequality, we conclude that there exists T > cℓ∆Φϵ−2, where the output ˆ x will satisfy ∥∇Φ(ˆ x)∥≤ϵ with probability at least 2/3. Furthermore, we call the solver MAXIMIN-AG2 at each iteration. Using Theorem 8 and δ ≤ϵ2/144κ2 yℓ, the number of gradient evaluations at each iteration is bounded by O √κy log κ3 yℓ2( ˜ D2 x + D2 y) ϵ2 ! log κ2 yℓ2D2 y ϵ2 !! . Therefore, we conclude that the total number of gradient evaluations is bounded by O ℓ∆Φ ϵ2 · √κy log2 κyℓ( ˜ D2 x + D2 y) ϵ !! . This completes the proof. E.4. Proof of Corollary 21 Recall that the function ¯ fϵ is defined by ¯ fϵ(x, y) = f(x, y) −ϵ2∥y −y0∥2 200ℓD2 y . This implies that the following statement holds for all (x, y) ∈X × Y that ∇xf(x, y) −∇x ¯ fϵ(x, y) = 0, ∥∇yf(x, y) −∇y ¯ fϵ(x, y)∥ ≤ ϵ2 100ℓDy . Using Theorem 20 and letting y⋆ ϵ(·) = argminy∈Y ¯ fϵ(·, y), we have ∥∇x ¯ fϵ(ˆ x, y⋆ ϵ(ˆ x))]∥ ≤ ϵ 10, ℓ∥PY[y⋆ ϵ(ˆ x) + (1/ℓ)∇y ¯ fϵ(ˆ x, y⋆ ϵ(ˆ x))] −y⋆ ϵ(ˆ x)∥ = 0. Putting these pieces together yields that ∥∇xf(ˆ x, y⋆ ϵ(ˆ x))∥ ≤ ϵ 10, ℓ∥PY[y⋆ ϵ(ˆ x) + (1/ℓ)∇yf(ˆ x, y⋆ ϵ(ˆ x))] −y⋆ ϵ(ˆ x)∥ ≤ ϵ2 50ℓDy . Now let x⋆(ˆ x) = argminx∈Rm Φ1/2ℓ(x) := Φ(x) + ℓ∥x −ˆ x∥2, we have ∥∇Φ1/2ℓ(ˆ x)∥2 = 4ℓ2∥ˆ x −x⋆∥2 39 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Since Φ(·) + ℓ∥· −ˆ x∥2 is ℓ/2-strongly-convex, we have max y∈Y f(ˆ x, y)−max y∈Y f(x∗(ˆ x), y)−ℓ∥x∗(ˆ x)−ˆ x∥2 = Φ(ˆ x)−Φ(x∗(ˆ x))−ℓ∥x∗(ˆ x)−ˆ x∥2 ≥ℓ∥ˆ x −x∗(ˆ x)∥2 4 . Since (ˆ x, y⋆ ϵ(ˆ x)) satisfies ∥∇xf(ˆ x, y⋆ ϵ(ˆ x))∥≤ϵ/10 and ∥PY(y⋆ ϵ(ˆ x) + (1/ℓ)∇yf(ˆ x, y⋆ ϵ(ˆ x))) − y⋆ ϵ(ˆ x)∥≤ϵ2/(50ℓDy), we have max y∈Y f(ˆ x, y) −max y∈Y f(x∗(ˆ x), y) −ℓ∥x∗(ˆ x) −ˆ x∥2 = max y∈Y f(ˆ x, y) −f(ˆ x, y⋆ ϵ(ˆ x)) + f(ˆ x, y⋆ ϵ(ˆ x)) −max y∈Y f(x∗(ˆ x), y) −ℓ∥x∗(ˆ x) −ˆ x∥2 ≤ max y∈Y f(ˆ x, y) −f(ˆ x, y⋆ ϵ(ˆ x)) + f(ˆ x, y⋆ ϵ(ˆ x)) −f(x∗(ˆ x), y⋆ ϵ(ˆ x)) −ℓ∥x∗(ˆ x) −ˆ x∥2 ≤ ℓDy∥PY(y⋆ ϵ(ˆ x) + (1/ℓ)∇yf(ˆ x, y⋆ ϵ(ˆ x))) −y⋆ ϵ(ˆ x)∥+ ∥ˆ x −x∗(ˆ x)∥∥∇xf(ˆ x, y⋆ ϵ(ˆ x))∥−ℓ∥ˆ x −x∗(ˆ x)∥2 4 ≤ ϵ2 50ℓ+ ∥∇xf(ˆ x, y⋆ ϵ(ˆ x))∥2 ℓ ≤ϵ2 16ℓ. Putting these pieces together yields that ∥∇Φ1/2ℓ(ˆ x)∥≤ϵ. Furthermore, letting κy = 100ℓ2D2 y/ϵ2 in the gradient complexity bound presented in Theorem 20, we conclude that the total number of gradient evaluations is bounded by O ℓ2Dy∆Φ ϵ3 log2 ℓ( ˜ D2 x + D2 y) ϵ !! . This completes the proof. Appendix F. Proof of Technical Lemmas In this section, we provide complete proofs for the lemmas in the paper. F.1. Proof of Lemma 17 We provide a proof for an expanded version of Lemma 17. Lemma 25 If Φ is ℓ-weakly convex, we have (a) Φ1/2ℓ(x) and proxΦ/2ℓ(x) = argmin Φ(w) + ℓ∥w −x∥2 are well defined for any x ∈Rm. (b) Φ(proxΦ/2ℓ(x)) ≤Φ(x) for any x ∈Rm. (c) Φ1/2ℓis 4ℓ-smooth with ∇Φ1/2ℓ(x) = 2ℓ(x −proxΦ/2ℓ(x)). Proof. Since Φ is ℓ-weakly convex, Φ(·) + (ℓ/2) ∥· −x∥2 is convex for any x ∈Rm. This implies that Φ(·) + ℓ∥· −x∥2 is (ℓ/2)-strongly convex and Φ1/2ℓ(x) and proxΦ/2ℓ(x) are well defined. For any x ∈Rm, the definition of proxΦ/2ℓ(x) implies that Φ(proxΦ/2ℓ(x)) ≤Φ1/2ℓ(proxΦ/2ℓ(x)) ≤Φ(x). Using Davis and Drusvyatskiy (2019, Lemma 2.2), Φ1/2ℓis differentiable with ∇Φ1/2ℓ(x) = 2ℓ(x−proxΦ/2ℓ(x)). Since proxΦ/2ℓis 1-Lipschitz, we ∥∇Φ1/2ℓ(x)−∇Φ1/2ℓ(x′)∥≤4ℓ∥x−x′∥. Therefore, the function Φ1/2ℓis 4ℓ-smooth. □ 40 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION F.2. Proof of Lemma 19 Denote ˆ x := proxΦ/2ℓ(x), part (c) in Lemma 17 implies ∥ˆ x −x∥= ∥∇Φ1/2ℓ(x)∥ 2ℓ . Furthermore, we have 2ℓ(x −ˆ x) ∈∂Φ(ˆ x). Putting these pieces together yields the desired result. F.3. Proof of Lemma 23 Part (a): Let x, x′ ∈Rm, the points y∗ g(x) and y∗ g(x′) satisfy that (y −y∗ g(x))⊤∇yg(x, y∗ g(x)) ≤0, ∀y ∈Y, (F.1) (y −y∗ g(x′))⊤∇yg(x′, y∗ g(x′)) ≤0, ∀y ∈Y. (F.2) Summing up Eq. (F.1) with y = y∗ g(x′) and Eq. (F.2) with y = y∗ g(x) yields (y∗ g(x′) −y∗ g(x))⊤(∇yg(x, y∗ g(x)) −∇yg(x′, y∗ g(x′))) ≤0. Since g(x, ·) is µy-strongly concave, we have (y∗ g(x′) −y∗ g(x))⊤(∇yg(x, y∗ g(x′)) −∇yg(x, y∗ g(x))) + µy∥y∗ g(x′) −y∗ g(x)∥2 ≤0. Summing up the above two inequalities yields that (y∗ g(x′) −y∗ g(x))⊤(∇yg(x, y∗ g(x′)) −∇yg(x′, y∗ g(x′))) + µy∥y∗ g(x′) −y∗ g(x)∥2 ≤0. Since ∇yg is ℓ-Lipschitz, we have µy∥y∗ g(x′) −y∗ g(x)∥2 ≤ℓ∥y∗ g(x′) −y∗ g(x)∥∥x′ −x∥. Therefore, we conclude that the function y∗ g(·) is κy-Lipschitz. Part (b): Since the function y∗ g(·) is unique, Danskin’s theorem (Rockafellar, 1970) implies that Φg is differentiable and ∇Φg(·) = ∇xg(·, y∗ g(·)). Let x, x′ ∈Rm, we have ∥∇Φg(x) −∇Φg(x′)∥ = ∥∇xg(x, y∗ g(x)) −∇xg(x′, y∗ g(x′))∥≤ℓ∥x −x′∥+ ℓ∥y∗ g(x) −y∗ g(x′)∥ ¯ κ≥1 ≤ κyℓ∥x −x′∥+ ℓ∥y∗ g(x) −y∗ g(x′)∥. Since y∗ g(·) is κy-Lipschitz, the function Φg is 2κyℓ-smooth. Furthermore, let x, x′ ∈Rm, we have Φg(x′) −Φg(x) −(x′ −x)⊤∇Φg(x) = g(x′, y∗ g(x′)) −g(x, y∗ g(x)) −(x′ −x)⊤∇xg(x, y∗ g(x)) ≥ g(x′, y∗ g(x)) −g(x, y∗ g(x)) −(x′ −x)⊤∇xg(x, y∗ g(x)). Since g(·, y) is µx-strongly convex for each y ∈Y, we have Φg(x′) −Φg(x) −(x′ −x)⊤∇Φg(x) ≥µx∥x′ −x∥2 2 . Therefore, the function Φg is µx-strongly convex. 41 NEAR-OPTIMAL ALGORITHMS FOR MINIMAX OPTIMIZATION Part (c): Let y, y′ ∈Rn, the points x∗ g(y) and x∗ g(y′) satisfy that (x −x∗ g(y))⊤∇xg(x∗ g(y), y) ≥0, ∀x ∈X, (F.3) (x −x∗ g(y′))⊤∇xg(x∗ g(y′), y′) ≥0, ∀x ∈X. (F.4) Summing up Eq. (F.3) with x = x∗ g(y′) and Eq. (F.4) with x = x∗ g(y) yields (x∗ g(y′) −x∗ g(y))⊤(∇xg(x∗ g(y), y) −∇xg(x∗ g(y′), y′)) ≥0. Since g(·, y) is µx-strongly convex, we have (x∗ g(y′) −x∗ g(y))⊤(∇xg(x∗ g(y′), y′) −∇xg(x∗ g(y), y′)) −µx∥x∗ g(y′) −x∗ g(y)∥2 ≥0. Summing up the above two inequalities yields that (x∗ g(y′) −x∗ g(y))⊤(∇xg(x∗ g(y), y) −∇xg(x∗ g(y), y′)) −µx∥x∗ g(y′) −x∗ g(y)∥2 ≥0. Since ∇xg is ℓ-smooth, we have µx∥x∗ g(y′) −x∗ g(y)∥2 ≤ℓ∥x∗ g(y′) −x∗ g(y)∥∥y′ −y∥. Therefore, we conclude that the function x∗ g is κx-Lipschitz. Part (d): Since the function x∗ g(·) is unique, Danskin’s theorem (Rockafellar, 1970) implies that Ψg is differentiable and ∇Ψg(·) = ∇yg(x∗ g(·), ·). Let y, y′ ∈Rn, we have ∥∇Ψg(y)−∇Ψg(y′)∥= ∥∇yg(x∗ g(y), y)−∇yg(x∗ g(y′), y)∥≤ℓ∥x∗ g(y)−x∗ g(y′)∥+ℓ∥y−y′∥. Since x∗ g(·) is κx-Lipschitz, the function Ψg is 2κxℓ-smooth. Furthermore, let y, y′ ∈Rn, we have Ψg(y) −Ψg(y′) −(y −y′)⊤∇Ψg(y) = g(x∗ g(y), y) −g(x∗ g(y′), y′) −(y −y′)⊤∇yg(x∗ g(y), y) ≥ g(x∗ g(y), y) −g(x∗ g(y), y′) −(y −y′)⊤∇yg(x∗ g(y), y). Since g(x, ·) is µy-strongly concave for each x ∈X, we have Ψg(y) −Ψg(y′) −(y −y′)⊤∇Ψg(y) ≥µy∥y′ −y∥2 2 . Therefore, the function Ψg is µy-strongly concave. 42
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Pincer Grasp: Significant Milestone in Child Development Health Conditions Health Conditions All Breast Cancer Cancer Care Caregiving for Alzheimer's Disease Chronic Kidney Disease Chronic Obstructive Pulmonary Disease (COPD) Digestive Health Eye Health Heart Health Menopause Mental Health Migraine Multiple Sclerosis (MS) Parkinson’s Disease Psoriasis Rheumatoid Arthritis (RA) Sleep Health Type 2 Diabetes Weight Management Condition Spotlight All Controlling Ulcerative Colitis Navigating Life with Bipolar Disorder Mastering Geographic Atrophy Managing Type 2 Diabetes Wellness Wellness Topics All CBD Fitness Healthy Aging Hearing Mental Well-Being Nutrition Parenthood Recipes Sexual Health Skin Care Sleep Health Vitamins and Supplements Women's Wellness Product Reviews All At-Home Testing Men's Health Mental Health Nutrition Sleep Vitamins and Supplements Women's Health Featured Programs All Your Guide to Glucose Health Inflammation and Aging Cold & Flu Season Survival Guide She’s Good for Real Tools Featured Video Series Pill Identifier FindCare Drugs A-Z Lessons All Crohn’s and Ulcerative Colitis Essentials Diabetes Nutrition High Cholesterol Taming Inflammation in Psoriasis Taming Inflammation in Psoriatic Arthritis Newsletters All Anxiety and Depression Digestive Health Heart Health Migraine Nutrition Edition Type 2 Diabetes Wellness Wire Lifestyle Quizzes Find a Diet Find Healthy Snacks Weight Management How Well Do You Sleep? 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Join Bezzy on the web or mobile app. All Breast Cancer Multiple Sclerosis Depression Migraine Type 2 Diabetes Psoriasis Follow us on social media Can't get enough? Connect with us for all things health. Subscribe Why a Pincer Grasp Is Crucial for a Baby’s Development Written by Rachel Nall, MSN, CRNA on November 13, 2018 Development Activities Developmental delay Summary Pincer grasp definition The pincer grasp is the coordination of the index finger and thumb to hold an item. Each time you hold a pen or button your shirt, you’re using the pincer grasp. While it may seem like second nature to an adult, to a baby this is an important milestone in fine motor development. The pincer grasp represents the coordination of brain and muscles that’s necessary to help them gain increasing independence. A baby will typically develop this skill between the ages of9 and 10 months, although this can vary. Children develop at different rates. If a child doesn’t develop this milestone over time, doctors may interpret this as a delayed development sign. Doctors can recommend activities and therapies that can help a child improve their use of the pincer grasp. Pincer grasp development A pincer grasp represents further development of fine motor skills. These are movements that require precise control of small muscles in the hands. They require multiple skills, including strength and hand-eye coordination. Fine motor skills are the foundation that will later allow your child to write and use a computer mouse. A child will usually start to develop a pincer grasp around 9 months of age, according to the Children’s Hospital of Orange County. However, you may observe this earlier or later depending on your child’s unique development. Other milestones that occur around this time include learning how to bang two objects together and clapping their hands. Stages of pincer grasp development Pincer grasp development is usually the result of building on several grasping and coordination milestones. Some of the early developmental milestones that later allow a child to perform the pincer grasp include: palmar grasp:bringing the fingers in toward the palm, allowing babies to curl their fingers around an object raking grasp: using the fingers other than the thumb like a rake, curling the top of the fingers over the object to bring items toward them inferior pincer grasp:using the pads of the thumb and index finger to pick up and hold objects; this precursor to the pincer grasp usually takes place between 7 and 8 months of age A true pincer grasp is when a child uses the tips of their fingers to pick up objects. This is also called a superior or “neat” pincer grasp. Children are able to pick up smaller, thinner objects when they can accomplish a pincer grasp. Allowing a child to grasp items, make contact with their hands, and engage with items is a step toward the pincer grasp. Pincer grasp toys and activities Parents and caregivers can foster a child’s pincer grasp development through these activities. Put differently sized small items in front of your baby and watch how they try to pick up various things. Examples could include play coins, marbles, or buttons. Babies at this age put everything in their mouths, so supervise this activity carefully to ensure your child doesn’t choke or try to swallow them. Place soft finger foods like pieces of banana or cooked carrots in front of your baby and have them reach to pick them up and eat them. Using spoons, forks, markers, crayons, and anything else that is held in the fingers can help your child develop a pincer grasp. Eating with the hands and playing with balls and toys of varying sizes can also help. ADVERTISEMENT Find nannies you can trust Care․com requires all caregivers to complete background checks before interacting with families. Create a free account to start browsing. LEARN MORE Trusted by 3M+ people 4.6 out of 5 ★ 474K+ caregivers What if a child shows no interest in picking up toys? Motor development milestones such as the pincer grasp represent the development of motor tracts in the nervous system. If your 8- to 12-month-old child shows no interest picking up objects, talk to your child’s doctor. Sometimes this is an indicator of a known condition that can affect motor development, such as a developmental coordination disorder. A doctor may recommend interventions such as occupational therapy. An occupational therapist can work with your child to encourage developmental milestones. They can also teach you how to foster these efforts. Takeaway If your child is older than 12 months and hasn’t shown signs of a pincer grasp yet, talk to their pediatrician. Your child’s pediatrician can evaluate their fine motor skills as well as discuss a timeline for such milestones given your child’s overall development. ADVERTISEMENT Explore these resources for weight loss Great for sustainable weight loss plans Prioritizes a balanced relationship with food Uses a quiz to create a custom plan and estimated timeline Now pairs GLP-1 treatments with psychological support $209 per year with other plans available 21-Day Free Trial TAKE QUIZ Great for a modern approach to healthcare $9/month with your Prime membership Accessible and affordable care designed around your needs Offers 24/7 care, in-app health records, messaging, and more Accepts insurance for in-office and virtual visits START 14-DAY FREE TRIAL End the weight loss roller coaster with Wellos Offers personalized nutrition support based on smart tracking Provides a quiz to determine what plan fits you best Includes 100+ lessons from dieticians and behavior change experts Only $25 per month Free 14-Day Trial TAKE QUIZ Share this article Written by Rachel Nall, MSN, CRNA on November 13, 2018 HEALTHLINE NEWSLETTER The best of health and wellness We do the research so you don't have to. 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