Math 304, Differential Forms, Lecture 1 May 26, 2011

Similar documents
SUPPLEMENTARY READING FOR MATH 53: DIFFERENTIAL FORMS AND THE GENERAL STOKES FORMULA

(1) Recap of Differential Calculus and Integral Calculus (2) Preview of Calculus in three dimensional space (3) Tools for Calculus 3

Math 234. What you should know on day one. August 28, You should be able to use general principles like. x = cos t, y = sin t, 0 t π.

Overview of vector calculus. Coordinate systems in space. Distance formula. (Sec. 12.1)

Sec. 1.1: Basics of Vectors

Chapter 1. Vector Algebra and Vector Space

Math 241, Exam 1 Information.

1.1 Single Variable Calculus versus Multivariable Calculus Rectangular Coordinate Systems... 4

MATH 280 Multivariate Calculus Fall Integrating a vector field over a curve

Math 212-Lecture 8. The chain rule with one independent variable

Past Exam Problems in Integrals, Solutions

Figure 25:Differentials of surface.

MATH 280 Multivariate Calculus Fall Integrating a vector field over a surface

Vectors and Fields. Vectors versus scalars

Practice Problems for Exam 3 (Solutions) 1. Let F(x, y) = xyi+(y 3x)j, and let C be the curve r(t) = ti+(3t t 2 )j for 0 t 2. Compute F dr.

Vector Calculus handout

Topic 2-2: Derivatives of Vector Functions. Textbook: Section 13.2, 13.4

Dr. Allen Back. Dec. 3, 2014

LB 220 Homework 4 Solutions

Name: SOLUTIONS Date: 11/9/2017. M20550 Calculus III Tutorial Worksheet 8

12.1. Cartesian Space

M273Q Multivariable Calculus Spring 2017 Review Problems for Exam 3

Math 67. Rumbos Fall Solutions to Review Problems for Final Exam. (a) Use the triangle inequality to derive the inequality

Course Notes Math 275 Boise State University. Shari Ultman

e x2 dxdy, e x2 da, e x2 x 3 dx = e

Topic 5.1: Line Element and Scalar Line Integrals

2 dimensions Area Under a Curve Added up rectangles length (f lies in 2-dimensional space) 3-dimensional Volume

Dr. Allen Back. Sep. 10, 2014

Vector calculus background

MA 441 Advanced Engineering Mathematics I Assignments - Spring 2014

Figure 21:The polar and Cartesian coordinate systems.

Line and Surface Integrals. Stokes and Divergence Theorems

Vector Calculus, Maths II

25. Chain Rule. Now, f is a function of t only. Expand by multiplication:

Calculus: Several Variables Lecture 27

Jim Lambers MAT 280 Spring Semester Lecture 26 and 27 Notes

EE2007: Engineering Mathematics II Vector Calculus

MAS 4156 Lecture Notes Differential Forms

Lecture 2: Vector-Vector Operations

MATH Topics in Applied Mathematics Lecture 12: Evaluation of determinants. Cross product.

Math 20C Homework 2 Partial Solutions

Math 265H: Calculus III Practice Midterm II: Fall 2014

Chapter 4. The First Fundamental Form (Induced Metric)

Math Final Exam

12. Lines in R 3. through P 0 and is parallel to v is written parametrically as a function of t: Using vector notation, the same line is written

Topic 5.6: Surfaces and Surface Elements

x + ye z2 + ze y2, y + xe z2 + ze x2, z and where T is the

Exam 2 Solutions. (a) Is W closed under addition? Why or why not? W is not closed under addition. For example,

Archive of Calculus IV Questions Noel Brady Department of Mathematics University of Oklahoma

Summary of various integrals

Lecture II: Vector and Multivariate Calculus

f dr. (6.1) f(x i, y i, z i ) r i. (6.2) N i=1

Chain Rule. MATH 311, Calculus III. J. Robert Buchanan. Spring Department of Mathematics

CURRENT MATERIAL: Vector Calculus.

How to Use Calculus Like a Physicist

EE2007: Engineering Mathematics II Vector Calculus

(b) Find the range of h(x, y) (5) Use the definition of continuity to explain whether or not the function f(x, y) is continuous at (0, 0)

MATH The Chain Rule Fall 2016 A vector function of a vector variable is a function F: R n R m. In practice, if x 1, x n is the input,

Jim Lambers MAT 280 Summer Semester Practice Final Exam Solution. dy + xz dz = x(t)y(t) dt. t 3 (4t 3 ) + e t2 (2t) + t 7 (3t 2 ) dt

Mo, 12/03: Review Tu, 12/04: 9:40-10:30, AEB 340, study session

Maxima and Minima. (a, b) of R if

MAT 211 Final Exam. Spring Jennings. Show your work!

Contents. MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables. 1 Multiple Integrals 3. 2 Vector Fields 9

********************************************************** 1. Evaluate the double or iterated integrals:

Linear Algebra 1 Exam 2 Solutions 7/14/3

Math 1272 Solutions for Fall 2005 Final Exam

MAC2313 Final A. (5 pts) 1. How many of the following are necessarily true? i. The vector field F = 2x + 3y, 3x 5y is conservative.

Practice problems. 1. Given a = 3i 2j and b = 2i + j. Write c = i + j in terms of a and b.

MATH 2433 Homework 1

Ideas from Vector Calculus Kurt Bryan

In general, the formula is S f ds = D f(φ(u, v)) Φ u Φ v da. To compute surface area, we choose f = 1. We compute

Final Exam Review Sheet : Comments and Selected Solutions

LECTURE 1: LINES IN R 3

Introduction to Vector Spaces Linear Algebra, Fall 2008

Laplace equation in polar coordinates

SOME PROBLEMS YOU SHOULD BE ABLE TO DO

Chapter 3 - Vector Calculus

Chapter 2. Matrix Arithmetic. Chapter 2

Mathematics (Course B) Lent Term 2005 Examples Sheet 2

Math 53: Worksheet 9 Solutions

3. The Theorems of Green and Stokes

MATH 304 Linear Algebra Lecture 19: Least squares problems (continued). Norms and inner products.

What is a Linear Space/Vector Space?

The Calculus of Vec- tors

MULTIVARIABLE INTEGRATION

MAY THE FORCE BE WITH YOU, YOUNG JEDIS!!!

Linear Algebra. Hoffman & Kunze. 2nd edition. Answers and Solutions to Problems and Exercises Typos, comments and etc...

Math 425 Lecture 1: Vectors in R 3, R n

Divergence Theorem December 2013

Vectors. The standard geometric definition of vector is as something which has direction and magnitude but not position.

Math 309 Notes and Homework for Days 4-6

n=0 ( 1)n /(n + 1) converges, but not

MULTIVARIABLE CALCULUS

Higher Dimensions. Introduction. Differential Forms

Arc Length and Surface Area in Parametric Equations

Lecture 7 - Separable Equations

The exam will have 5 questions covering multiple

Sections minutes. 5 to 10 problems, similar to homework problems. No calculators, no notes, no books, no phones. No green book needed.

Problem Points S C O R E

PRACTICE PROBLEMS. Please let me know if you find any mistakes in the text so that i can fix them. 1. Mixed partial derivatives.

Transcription:

Math 304, Differential Forms, Lecture 1 May 26, 2011 Differential forms were developed and applied to vector fields in a decisive way by the French Mathematician Élie artan (1869 1951) in the first quarter of the 20th century. They provide an elegant framework for the theorems of vector calculus in 3 dimensions and can be used to formulate useful generalizations to higher dimensions. The development below is based upon the treatments in olley, Bressoud, and Bachman (in decreasing order of importance). What is a Differential Form? According to Harley Flanders, the author of a popular Dover book on applications of differential forms 1, the answer to this question is a differential form is what appears inside an integral. That is to say, f(x) dx, f(x, y) dx dy, f(x, y, z) dx dy dz, and P (x, y, z) dx+q(x, y, z) dy+r(x, y, z) dz, are all examples of differential forms. While this is true, and it does tell us what a differential form looks like, it does not tell us what a differential form actually is. artan s answer to the question is the definition of a differential form as a function that is built up from basic linear forms that project a vector to its components on each of the coordinate axes. Basic One-Forms on R 3 There are three basic one-forms in 3 dimensional space, dx, dy, and dz. ach one is a function from R 3 to R. Given a vector a = (a 1, a 2, a 3 ) in R 3, dx( a) = a 1, dy( a) = a 2, and dz( a) = a 3. It is easy to see that each of these functions is linear. That is, dx( a + b) = dx( a) + dx( b) and dx(λ a) = λdx( a), the same being true for dy and dz. In the mathematical literature they are referred to as linear forms. Geometry Geometrically, the basic 1-forms dx, dy, and dz map a vector a to its signed projection in the direction of each of the coordinate axes. If a is regarded as a displacement vector, then dx( a) is the displacement in the x-direction, dy( a) is the displacement in the y-direction, and dz( a) is the displacement in the z-direction. xample 1. If a = (2, 0, 1), then dx( a) = 2, dy( a) = 0, and dz( a) = 1. onstant oefficient One-Forms on R 3 A constant coefficient 1-form on R 3 is an expression of the form ω =b 1 dx + b 2 dy + b 3 dz, where b = (b 1, b 2, b 3 ) is a vector in R 3. That is, ω is a linear combination of 1-forms with constant coefficients. By defining ω( a) as ω( a) = b 1 dx( a) + b 2 dy( a) + b 3 dz( a), for each a = (a 1, a 2, a 3 ) in R 3, ω is also a linear form 2 on R 3. Observe that ω( a) = b 1 a 1 + b 2 a 2 + b 3 a 3 = b a, the work done by the force vector b in moving an object from the tail of a to its head. The number ω( a) can also be interpreted as the circulation along a induced by a constant flow with velocity vector b. 1 Differential Forms with Applications to the Physical Sciences, ISBN 10-0486661695 2 It is not difficult to show that every linear form on R 3 is a constant coefficient linear combination of basic 1-forms. 1

Variable oefficient One-Forms on R 3 A variable coefficient 1-form on R 3 is an expression of the form ω =P (x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz, (1) where F (x, y, z) = (P (x, y, z), Q(x, y, z), R(x, y, z)) is a vector field on R 3. That is, ω is a linear combination of 1-forms with variable coefficients. By defining ω (x,y,z) ( a) as ω (x,y,z) ( a) = P (x, y, z) dx( a) + Q(x, y, z) dy( a) + R(x, y, z) dz( a), for each (x, y, z) and a = (a 1, a 2, a 3 ) in R 3, ω determines is a vector field of linear forms on R 3. Observe that ω (x,y,z) ( a) = F (x, y, z) a (2) which approximates the work done by the vector field F in moving an object from the point (x, y, z), the tail of a, to its head at the point (x + a 1, y + a 2, z + a 3 ). ω (x,y,z) ( a) can also be interpreted as an approximation to the circulation along the line from (x, y, z) to (x + a 1, y + a 2, z + a 3 ) induced by the flow associated with the vector field F. xample 2. If F = (x 2, x z, xyz), then ω = x 2 dx + (x z) dy + xyz dz and ω (1,2,3) (3, 2, 1) = (1 2 ) 3 + (1 3) 2 + (1 2 3) 1 = 5. The Integral of a One-Form If is a curve in R 3 parametrized by r(t) = (x(t), y(t), z(t)), a t b, then quation (2) implies that onsequently, b This motivates the following definition. a ω r(t) ( r (t)) = F ( r(t)) r (t). ω r(t) ( r (t)) dt = b a F ( r(t)) r (t) dt. Definition. Given an oriented curve in R 3 parametrized by r(t), a t b, and a differential 1-form ω = P dx + Q dy + R dz, the integral of ω over is defined as ω = b a ω r(t) ( r (t)) dt. This does not define anything new, it simply introduces new notation for the line integral of a vector field in R 3. As immediate consequences note that when ω = P dx + Q dy + R dz is the differential 1-form associated with the vector field F = (P, Q, R) then ω = F d r = F T ds and, of course, ω = P dx + Q dy + R dz. 2

The Algebra of One-Forms Given two 1-forms in R 3, ω = P dx + Q dy + R dz and η = A dx + B dy + dz their sum is defined to be the 1-form ω + η = (P + A) dx + (Q + B) dy + (R + ) dz. If f(x, y, z) is a scalar field, officially referred to as a 0-form, (yes, that s right, a zero form) then fω is defined to be the 1-form fω = fp dx + fq dy + fr dz. It is easy to see that addition and scalar multiplication of 1-forms have the following algebraic properties. ω + η = η + ω ω + (η + β) = (ω + η) + β f(ω + η) = fω + fη (f + g)ω = fω + fω Multiplication of One-Forms: Basic Two-Forms It is also possible to multiply two 1-forms. The product of the two basic 1-forms dx and dy is defined to operate on an pair of vectors a = (a 1, a 2, a 3 ) and b = (b 1, b 2, b 3 ) in R 3 to produce a real number as follows. dx dy( a, dx( a) dx( b) b) = dy( a) dy( b) = a 1 b 1 a 2 b 2 = a 1b 2 a 2 b 1 dx dy is called a basic 2-form. 3 Geometry Geometrically, the basic 2-form dx dy maps the pair of vectors ( a, b) to the third component of their cross-product a b. This is the signed area of the projection of the parallelogram formed by a and b to the xy-plane. Similarly, the basic 2-form dz dx maps ( a, b) to a 3 b 1 a 1 b 3, which is the second component of their cross-product, and dy dz maps ( a, b) to a 2 b 3 a 3 b 2, which is the first component of their cross-product. xample 3. If a = (1, 2, 3) and b = (3, 2, 1), then dx dy( a, b ) = 1 3 2 2 = 1 2 2 3 = 4. Properties of 2 2 determinants imply that, as bilinear forms, the 9 basic 2-forms satisfy the following relations. dx dx = 0, dx dy = dy dx, dx dz = dz dx dy dx = dx dy, dy dy = 0, dy dz = dz dy dz dx = dx dz, dz dy = dy dz, dz dz = 0 These relations imply that any linear combination of basic 2-forms can be expressed uniquely as a linear combination of the three fundamental 2-forms dydz, dzdx, and dxdy. The fact that reversing the order of multiplication of two 1-forms changes the sign of the product is referred to as anti-commutativity of multiplication for two basic 1-forms. onstant oefficient Two-Forms on R 3 A constant coefficient 2-form on R 3 is an expression of the form ω =c 1 dydz + c 2 dzdy + c 3 dxdy, 3 2-forms are maps from R 3 R 3 to R that are linear in each variable making them bilinear. 3

where c = (c 1, c 2, c 3 ) is a vector in R 3. By defining ω( a, b) as ω( a, b) = c 1 dydz( a, b) + c 2 dzdz( a, b) + c 3 dxdy( a, b), for each a = (a 1, a 2, a 3 ) and b = (b 1, b 2, b 3 ) in R 3, ω is also a bilinear form 4 on R 3. The geometry of the fundamental 2-forms implies that ω( a, b) = c ( a b), which is the signed volume of the parallelopiped formed by the vectors a, b, and c. It is worth noting that this scalar triple product is also equal to the rate that fluid that will flow across the parallelogram formed by a and b in the direction of a b when the flow is determined by the constant velocity field F (x, y, z) = c. Variable oefficient Two-Forms on R 3 A variable coefficient 2-form on R 3 is an expression of the form ω =P (x, y, z) dydz + Q(x, y, z) dzdx + R(x, y, z) dxdy, (3) where F (x, y, z) = (P (x, y, z), Q(x, y, z), R(x, y, z)) is a vector field on R 3. That is, ω is a linear combination of 2-forms with variable coefficients. By defining ω (x,y,z) ( a, b ) as ω (x,y,z) ( a, b ) = P (x, y, z) dydz( a, b ) + Q(x, y, z) dzdx( a, b ) + R(x, y, z) dxdy( a, b ), for each (x, y, z), a = (a 1, a 2, a 3 ), and b = (b 1, b 2, b 3 ) in R 3, ω generates a vector field of bilinear forms on R 3. Observe that ω (x,y,z) ( a, b ) = F (x, y, z) ( a b ) (4) which approximates the rate that fluid that will flow across the parallelogram formed by a and b in the direction of a b when the flow is determined by the velocity field F. xample 4. If F = (x 2, x z, xyz), then ω = x 2 dydz + (x z) dzdx + xyz dxdy and ω (1,2,3) ((3, 2, 1), (1, 0, 1)) = (1 2, 1 3, 1 2 3) ((3, 2, 1) (1, 0, 1)) = 22. The Integral of a Two-Form If is an orientable surface in R 3 parametrized by Φ(u, v) = (x(u, v), y(u, v), z(u, v)), (u, v) in the parameter domain D, then quation (4) implies that onsequently, This motivates the following definition. D ω Φ(u,v) (Φ u, Φ v ) = F (Φ(u, v)) (Φ u Φ v ). ω Φ(u,v) (Φ u, Φ v ) du dv = D F (Φ(u, v)) (Φ u Φ v ) du dv. Definition. Given an oriented surface in R 3 parametrized by Φ(u, v), (u, v) in the parameter domain D, and a differential 2-form ω = P dydz + Q dzdx + R dxdy, the integral of ω over is defined as ω = ω Φ(u,v) (Φ u, Φ v ) du dv. D 4 It is not difficult to show that every bilinear form on R 3 that satisfies the nine relations listed above is a constant coefficient linear combination of the three fundamental 2-forms. 4

Once more, this does not define anything new. It simply introduces new notation for the surface integral of a vector field in R 3. As immediate consequences note that when ω = P dydz + Q dzdx + R dxdy is the differential 2-form associated with the vector field F = (P, Q, R) then ω = F ds = F n ds and, of course, ω = P dydz + Q dzdx + R dxdy. The Algebra of Two-Forms Given two 2-forms in R 3, ω = P dydz + Q dzdx + R dxdy and η = A dydz + B dzdx + dxdy their sum is defined to be the 2-form ω + η = (P + A) dydz + (Q + B) dzdx + (R + ) dxdy. If f(x, y, z) is a 0-form (i.e. a scalar field), then fω is defined to be the 2-form fω = fp dydz + fq dzdx + fr dxdy. Addition and scalar multiplication of 2-forms have the same algebraic properties as 1-forms. ω + η = η + ω ω + (η + β) = (ω + η) + β f(ω + η) = fω + fη (f + g)ω = fω + fω Addition is not defined for a 1-form and a 2-form, but the product of two differential forms can be defined in a natural way. Multiplication of Forms We already know that the product of a 0-form and a 1-form is a 1-form and the product of a 0-form and a 2-form is a 2-form. Products of 1-forms with 1-forms and 1-forms with 2-forms are obtained by expanding parentheses just as one would with polynomials, and collecting terms. Interesting things happen. xample 5. Let ω = P dx + Q dy + R dz be the 1-form associated with the vector field F = (P, Q, R) and η = A dx + B dy + dz the 1-form associated with G = (A, B, ). The following calculation shows that the product ω η is the 2-form associated with the cross product F G of F and G. In the computation the product forms dxdx, dydy, and dzdz were deleted and the anti-commutativity property of the product of two 1-forms was used to collect terms. ωη = (P dx + Q dy + R dz)(a dx + B dy + dz) = (Q RB) dydz + (RA P ) dzdx + (P B AQ) dxdy Similarly, the product η ω is associated with the cross product G F implying that general 1-forms have the same anti-commutativity property as basic 1-forms, namely η ω = ω η. The product of the basic 2-form dxdy and the basic 1-form dz produces what is referred to as the fundamental basic 3-form dxdydz. This is a trilinear form defined on ordered triples of vectors in R 3 in the way that generalizes the definition of a 2-form. dx( a ) dx( b ) dx( c ) dxdydz( a, b, c ) = dy( a ) dy( a 1 b 1 c 1 b ) dy( c ) = a 2 b 2 c 2 dz( a ) dz( b ) dz( c ) a 3 b 3 c 3 5

Geometrically, this is the volume of the parallelopiped spanned by the vectors a, b, and c. If the product of dxdy and dz is formed in the opposite order to yield dzdxdy, then the value of the determinant is unchanged. 5 That is, as trilinear forms, dxdydz = dzdxdy. On the other hand, it can readily be shown by direct calculation that dxdzdy = dxdydz and, more generally, any three form (that does not simply evaluate to 0) is equal to dxdydz or to dxdydz. onsequently, the general 3-form in R 3 with a variable coefficient looks like this: ω = f(x, y, z) dxdydz. Here is one more example involving multiplication. This time we multiply a general 1-form and a general 2-form. xample 6. Let ω = P dx + Q dy + R dz be the 1-form associated with the vector field F = (P, Q, R) and η = A dydz + B dzdx + dxdy the 2-form associated with G = (A, B, ). The following calculation shows that the product ω η is the 3-form associated with the dot product F G of F and G. In the computation the trivial product forms dxdxdy, dxdxdx, and so on, were deleted, and the commutativity property of the product of a basic 1-form and a basic 2-form was used to write all terms as scalar multiples of dxdydz. ωη = (P dx + Q dy + R dz)(a dydz + B dzdx + dxdy) = (P A + QB + R) dxdydz Similarly, the product η ω is associated with the dot product G F implying that general 1-forms always commute with general 2-forms: η ω = ω η. Properties of Multiplication of Differential Forms It can be shown that the following properties of multiplication hold for differential forms of any dimension. Distributivity If ω 1 and ω 1 are k-forms and η is an l-form, then (ω 1 + ω 2 ) η = ω 1 η + ω 2 η. Anti-commutativity If ω is a k-form and η is an l-form, then ω η = ( 1) kl η ω. Associativity If ω is a k-form, η an l-form, and β a p-form, then (ω η) β = ω (η β). Homogeneity If ω is a k-form, η an l-form, and f a 0-form, then f(ω η) = (fω) η = ω (f η). xercises 1. valuate the differential form ω = 2 dx 3 dy 3 dz at the vector a = (2, 1, 0). 2. If η = x dx + x 2 y dy (x + z) dz, then what is η (2,0,1) (0, 2, 1)? 3. Find ω (1, 1,3) ( a, b ) for ω = xy dxdy + xz dzdx and a = (1, 2, 3), b = (3, 2, 1). 4. Let ω and η be the 1-forms defined in xercises 1 and 2. Find (a) (2ω η) (1,2,3) (α, β, γ) (b) ω η (c) ω ω (How can we tell immediately that this is the zero 2-form?) 5. If ω is the 2-form on R 4 given by ω = x 2 x 3 dx 2 dx 4 x 4 dx 1 dx 3, then how would you evaluate the expression ω (0,1,2,3) ( a, b ) in terms of the components of a and b? 6. If ω = x dxdy and η = y dydz, then what is (y ω + x η) (a,b,c) ((1, 0, 0), (0, 1, 0))? 7. Referring to ω as defined in xercise 5, what is ω ω? (You will have to invent the definition for a 4-form to answer this.) 5 Moving the third row from the bottom of a 3 3 matrix to the top does not change the sign of its determinant because this matrix transformation is equivalent to two row interchanges. 6