* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download MA352_Differential_Geometry_CIIT_VU
Survey
Document related concepts
Duality (projective geometry) wikipedia , lookup
Surface (topology) wikipedia , lookup
Analytic geometry wikipedia , lookup
Euclidean geometry wikipedia , lookup
History of geometry wikipedia , lookup
Affine connection wikipedia , lookup
CR manifold wikipedia , lookup
Tensors in curvilinear coordinates wikipedia , lookup
Covariance and contravariance of vectors wikipedia , lookup
Cartesian tensor wikipedia , lookup
Cartan connection wikipedia , lookup
Differential geometry of surfaces wikipedia , lookup
Lie derivative wikipedia , lookup
Line (geometry) wikipedia , lookup
Transcript
Lecture notes on MTH352: Differential Geometry MTH352: Differential Geometry Module Handbook Credit Hours: 3(3,0) For Master of Mathematics By Dr. SOHAIL IQBAL Assistant Professor Department of Mathematics, CIIT Islamabad, Pakistan. 1 of 162 Lecture notes on MTH352: Differential Geometry The following lecture notes are designed for the course of MTH-352 Differential Geometry The notes are not intended to be an independent script and are designed to go together with the videos and the recommended book. 2 of 162 Lecture notes on MTH352: Differential Geometry Chapter 1 Calculus On Euclidean Space 3 of 162 Lecture notes on MTH352: Differential Geometry Lecture # 1 History From the very beginning of the evolution human beings started observing geometry of objects around. The first written work is by Euclid. He compiled of his and others work into volume form known as Euclid's Elements. His work is one of the most influential works in the history of mathematics. The βElementsβ have been serving as the main textbook for teaching geometry from the 300 B.C. until the 20th century. For his contributions he is often referred as βFather of geometryβ. The next name we came across in the history of geometry is Archimedes. He understood the geometry of different objects about him, for example, he calculated the area and volumes of different objects, and the techniques were later formulated in calculus. He understood geometry to such an extent that once he said give me a place to stand on and I will move the Earth. The next major development in geometry was done in Muslim Era. The need to predict the phases of the Moon for Ramadan and other religious festivals led to great steps forward in geometry of celestial objects (astronomy). Introduction of βCartesian coordinatesβ marked a new stage for geometry, since geometric figures, could now be represented analytically, that is, with functions and equations. It is said that Cartesian coordinates were invented by René Descartes and Pierre de Fermat independently. In Cartesian coordinates we can associate a point in plane with an ordered pair. In a coordinate plane we can associate a curve starting from an equation. Hence 4 of 162 Lecture notes on MTH352: Differential Geometry the Cartesian coordinates built a bridge between algebra and geometry. The following example reminds us the procedure. Example: Graph of equation π¦ = π₯ 2 in Euclidean plane. We can sketch the graph by first finding the images of different values of π₯ under the equation π¦ = π₯ 2 . Then we can sketch the ordered pair in Cartesian plane to get the graph of the equation π¦ = π₯ 2 . X 0 1 -1 2 -2 Y 0 1 1 4 4 On the same lines we can have Cartesian coordinates in three dimensional space. On the same lines we can associate a point in space with an ordered triple. Using these coordinates we can associate curves and surfaces in space with an equation. There is a particular emphasis on surfaces in geometry. Mainly because there are many examples of surfaces around us, for example, surface of earth and the geometry of space due to some heavy object. 5 of 162 Lecture notes on MTH352: Differential Geometry Aim of the course: In this course we are mainly interested in doing calculus on surfaces. For this we aim at doing the followings. ο Review of differential calculus. ο Develop tools to study curves and surfaces in space. ο Proper definition of surface. How to do calculus on surface. ο A detailed study of geometry of surface. Let us start the course with some review of basics. Some definitions Set A set S is a collection of objects that are called the elements of S. Example1: S = {1,2,3,4,β¦} is a set of natural numbers. Example2: S = { x : x is an integer ^ x is divisible by 2} = Set of even integers. Subset A set A is a subset of S provided each element of A is also an element of S. Example1: A = {1,2,3} is a subset of S = {1,2,3,4,β¦}. Example2: A = { x : x is an integer ^ x is divisible by 2 } is a subset of integers. Function A function from set π· to set π , written as π : π· β π , is a rule that assigns each element element of π· to a unique element of π 6 of 162 Lecture notes on MTH352: Differential Geometry Example1: Let π· = {π, π, π}, π = {1,2,3,4,5,6} Domain π· is called domain of π. Range π is called range of π. Image For x β π·, π(π₯) is called image of π₯ under π. In above example π(π) = 1. Image of the function π is the set {π(π₯): π₯ β π·}. In above example image of π is {1,2,4}. Example: π: πΉ β πΉ, π(π₯) = 1 + π₯, Image of π = πΉ. Composite Function Let π: π· β π and π: πΈ β π such that g(πΈ) β π· then π β π(π₯) = π(π(π₯)). Example: π(π₯) = 1 + π₯, π(π₯) = π₯ 2 , then π β π(π₯) = π(π(π₯)) = π(π₯ 2 ) = 1 + π₯ 2 . One-to-one π: π· β π , π(π₯) = π(π¦) )π₯ = π¦. Example: π: π΄ β π΅, for π΄ = {1,2,3,4}, π΅ = {π, π, π, π}. Example: π(π₯) = 1 + π₯. 7 of 162 Lecture notes on MTH352: Differential Geometry Onto functions A function π: π΄ β π΅, is onto if π(π΄) = π΅. Example: π: π΄ β π΅, for π΄ = {1,2,3,4}, π΅ = {π, π, π, π} Inverse π: π· β π is one-to-one and onto, then π has inverse π β1 : π β π·, such that π¦ β π₯ where π¦ = π(π₯). Or π is inverse of π if π β π = π β π = πΌ. Example: π(π₯) = 1 + π₯, has inverse π(π₯) = π₯ β 1. Euclidean 3-Space Euclidean 3-space πΉ is the set of all ordered triples of real numbers. Such a triple π = (π1 , π2 , π3 ) is called a point of πΉπ . For π = (π1 , π2 , π3 ), π = (π1 , π2 , π3 ) then π + π = (π1 + π1 , π2 + π2 , π3 + π3 ) For π = (π1 , π2 , π3 ), and πΌ a scalar then πΌπ = (πΌπ1 , πΌπ2 , πΌπ3 ) πΉπ is a vector space . The point πΆ = (0,0,0) is called the origin of πΉπ . Review of Fields and vector spaces Field A Field is a set πΉ such that for any πΌ, π½, πΎ in πΉ the following holds: 8 of 162 Lecture notes on MTH352: Differential Geometry ο Closure of πΉ under addition and multiplication: πΌ + π½ β πΉ, and πΌ. π½ β πΉ ο Associativity of addition and multiplication: πΌ + (π½ + πΎ) = (πΌ + π½) + πΎ and πΌ. (π½. πΎ) = (πΌ. π½). πΎ ο Commutativity of addition and multiplication: πΌ + π½ = π½ + πΌ and πΌ. π½ = π½. πΌ ο Existence of additive and multiplicative identity elements: πΌ + 0 = 0 + πΌ = πΌ, πΌ. 1 = πΌ ο Existence of additive inverses and multiplicative inverses: βπΌ β πΉ , πΌ + (βπΌ) = (βπΌ) + πΌ = 0 for πΌ β 0, there exists an element πΌ β1 β πΉ , such that πΌ. πΌ β1 = 1 ο Distributivity of multiplication over addition: πΌ. (π½ + πΎ) = πΌ. π½ + πΌπΎ Example: Set of real numbers and set of rational numbers are examples of field. Vector Space A vector space over a field is a set π such that: for any πΌ, π½ β πΉ and π£, π€ β π, the following hold. ο Closed under addition and scalar multiplication ο Associativity of addition ο Commutativity of addition u + v β V , Ξ±v β V π’ + (π£ + π€) = (π’ + π£) + π€ π’+π£ =π£+π’ ο Identity element of addition 0 β π such that π£ + 0 = 0 + π£ = π£ ο Inverse elements of addition For every π£ β π there exist βπ£ β π such that π£ + (βπ£) = (βπ£) + π£ = 0 9 of 162 Lecture notes on MTH352: Differential Geometry ο Distributivity of scalar multiplication with respect to vector addition πΌ(π’ + π£) = πΌπ’ + πΌπ£ ο Distributivity of scalar multiplication with respect to field addition (πΌ + π½)π£ = πΌπ£ + π½π£ ο Compatibility of scalar multiplication with field multiplication πΌ(π½π£) = (πΌπ½)π£ ο Identity element of scalar multiplication 1π£ = π£, where 1 is the multiplicative identity in πΉ. Example: The Euclidean 3-space π 3 is a vector space. Coordinate functions Let π₯, π¦, and π§ be real-valued functions on π 3 such the for each point π = (π1 , π2 , π3 ) we have π₯(π) = π1 , π¦(π) = π2 , π§(π) = π3 We can use π₯1 = π₯, π₯2 = π¦, π₯3 = π§ π = (π1, π2 , π3 ) = (π₯1 (π), π₯2 (π), π₯3 (π)) Differentiable OR πͺβ functions A real-valued function π on πΉπ is differentiable (or infinitely differentiable, or smooth, or of class πΆ β ) provided all partial derivatives of π, of all orders, exist and are continuous. Example: Consider a function π: π 2 β π , defined as π(π₯, π¦) = π₯ 2 π¦. Is π a smooth function? Solution: Since π is a polynomial function, hence π is differentiable. Now the first partial derivatives of π are : ππ = 2π₯π¦, ππ₯ ππ = π₯2. ππ¦ Which are continuous. Now the second partial derivatives are: π 2π = 2π¦, ππ₯ 2 π 2π = 0, ππ¦ 2 π 2π = 2π₯. ππ₯ππ¦ 10 of 162 Lecture notes on MTH352: Differential Geometry Which are continuous again. Similarly we can see that all the partial derivatives of π(π₯, π¦) are continuous so π(π₯, π¦) is a smooth function. Arithmetic of differentiable functions Differentiable real-valued functions π and π may be added and multiplied in the following way ο (π + π)(π) = π(π) + π(π), ο (ππ)(π) = π(π)π(π) Q1(a) (Exercise 1.1): If π = π₯ 2 π¦ and π = π¦ π ππ π§ then find ππ2 . Solution: Here ππ2 represents the point wise multiplication. So ππ2 = (π₯ 2 π¦)(π¦ sin π§) = π₯ 2 π¦π§ sin π§ . Chain rule If g is differentiable at x and f is differentiable at g(x) then the composition π β π is differentiable at x. Moreover, if y ο½ f ( g ( x)) and u ο½ g ( x) Then π¦ = π(π’) and dy dy du ο½ ο΄ dx du dx Alternatively d dx ο f ο¨ g ο¨ x ο©ο©ο ο½ ο¨ f ο’ ο― g ο© ο¨ x ο© ο½ f ο’ο¨ g ο¨ x ο©ο©g ο’ο¨ x ο© Derivative of outside function Derivative of inside function Example: ππ¦ Find ππ₯ if π¦ = πππ (π₯ 3 ). 11 of 162 Lecture notes on MTH352: Differential Geometry Solution: Let π’ = π₯3 Then π¦ = cos π’ . Now ππ¦ = β sin π’. ππ’ And ππ’ = 3π₯ 2 ππ₯ So by using chain rule ππ¦ ππ¦ ππ’ = × ππ₯ ππ’ ππ₯ Rates of change multiply ππ¦ = (β sin π’)(3π₯ 2 ) ππ₯ ππ¦ = β3π₯ 2 sin π₯ 3 ππ₯ Q1(d) (Exercise 1.1): Find π ππ¦ (sin π) where π = π₯ 2 π¦. Solution: Using chain rule π π ππ sin π = ( sin π ) ( ). ππ¦ ππ ππ¦ π sin π = (cos π )(π₯ 2 ). ππ¦ 12 of 162 Lecture notes on MTH352: Differential Geometry Lecture # 2 The lecture contents ο Vectors in π 3 ο Tangent vectors ο Vector field ο Natural frame field on π 3 ο Differentiable vector fields ο Some questions from Exercise 1.2 Vectors in πΉπ Intuitively a vector in π 3 is an oriented line segment or βarrowβ. Vectors are used to describe vector quantities such as force, velocities, angular momenta etc. 13 of 162 Lecture notes on MTH352: Differential Geometry For practical purposes there is a problem with this definition and needs to be improved. According to above definition two translated vectors are same, but in real life two vectors can have different effect if there initial points are different (see picture below). So we need to incorporate the initial point in the definition of vector. To define a vector v precisely we mention the starting point p of the vector as well. Definition: A vector in R3 is p + v, where p is initial point of vector v. Strictly speaking v is a point in R3 . Tangent vectors Definition: A tangent vector π£π to π 3 consists of two points of π 3; its vector part π£ and its point of application π. A tangent vector π£π is drawn as an arrow from the point π to the point π + π£. Example: If π£ = (2,3,2), and π = (1,1,3) then the tangent vector π£π = (2,3,2)(1,1,3) 14 of 162 Lecture notes on MTH352: Differential Geometry Starts in (1,1,3) and ends in (3,4,5). Equality of tangent vectors ο Two tangent vectors π£π and π€π are parallel if and only in π£ = π€. ο Two tangent vectors π£π and π€π are equal if and only in π£ = π€ and π = π. ο π£π and π£π are different tangent vectors if π β π. Definition: Let π be a point of πΉπ. The set ππ (πΉπ) consisting of all tangent vectors that have π as point of application is called the tangent space of πΉπ at π. Note: Each point of πΉπ has its own tangent space. They are all different from each other. Here we give a recall of the method of parallelogram law, scalar multiple of a vector, linear transformation, and isomorphism of vector spaces. ο Given two vectors π and π in π 3 we add them by parallelogram law. ο Let π be scalar, then for vector π΄, ππ΄ stretches or shrinks π΄ by the factor π, and reverse the direction if π < 0. For π£π , π€π β ππ (π 3 ) then we can define: π£π + π€π = (π£ + π€)π And for any scalar c β π we define ππ£π = (ππ£)π Fact : ππ (π 3) is vector space under the above addition and scalar multiplication. 15 of 162 Lecture notes on MTH352: Differential Geometry Linear Transformation: A linear transformation L: π β π from a vector space π to vector space π is a function that satisfy the following conditions: for any vectors π£, π€ β π and any scalar π πΏ(π£ + π€) = πΏ(π£) + πΏ(π€) L(ππ£) = ππΏ(π£). Isomorphism: Two vector spaces π and π are isomorphic if there exist a bijective linear transformation between them L: π β π. A bijective linear transformation πΏ is also called an isomorphism. Fact: Tp (R3) is isomorphic to R3. The isomorphism L: Tp (R3 ) β R3 between them such that for vp β Tp (R3 ) Using above definition we can conclude that the map L: Tp (R3 ) β R3 defined as L(vp ) = v. is an isomorphism. Vector field There are many examples of vector field around us, for example, the gravitational field etc. In such vector fields each point of the domain space represents a vector, for example, in case of gravitation vector field each point of the space represents a vector, directed towards the center of the earth and the magnitude representing the amount of force with which earth attracts the object towards itself. As the force of attraction of earth reduced as the object moves away from the center of earth so there will be vectors of different length in the gravitational vector field. Hence we define a vector field as. Definition: A vector field π on π 3 is a function that assigns to each point π of π 3 a tangent vector π(π) to π 3 at π. 16 of 162 Lecture notes on MTH352: Differential Geometry Since a vector field is a function so we can 3 add two vector fields just like we add functions. So if π and π are two vector fields on π then (π + π)(π) = π(π) + π(π). For any real-valued function π: π 3 β π we define (ππ)(π) = π(π)π(π). Natural frame field on ππ To investigate vector fields on π 3 we introduce a set of basic vector fields on π 3 . Definition: Let π1 , π2 , π3 be the vector fields on π 3 such that π1 (π) = (1,0,0)π π2 (π) = (0,1,0)π π3 (π) = (0,0,1)π For each point π of π 3 . We call π1 , π2 , π3 -collectively-the natural frame field on π 3 . The following result shows how to factorize given vector field into combination of natural frame: Lemma: If π is a vector field on π 3, there are three uniquely determined real-valued functions, π£1 , π£2 , π£3 on π 3 such that π = π£1 π1 + π£2 π2 + π£3 π3 The functions π£1 , π£2 , π£3 are called 17 of 162 Lecture notes on MTH352: Differential Geometry the Euclidean coordinate functions of π. Differentiable vector field Definition: A vector field π is differentiable if each of its three Euclidean coordinate function π£1 , π£2 , π£3 , is differentiable function (smooth, of class πΆ β ). We always assume vector fields are differentiable. So from now on vector field means differentiable vector field End of the lecture 18 of 162 Lecture notes on MTH352: Differential Geometry Lecture # 3 Contents: ο ο ο ο ο ο Directional derivatives How to differentiate composite functions (Chain rule) How to compute directional derivatives more efficiently The main properties of directional derivatives Operation of a vector field Basic properties of operations of vector fields The main of the lecture is to discuss directional derivative of real-valued functions on π 3 . Recall that a real-valued function on π 3 , written as π: π 3 β π , has domain π 3 and range π . Graph of such a function is a surface, for example, the graph in the following is a graph of a real-valued function on π 3 . To define directional derivative we first need the following. Associated with each tangent vector π£π to πΉπ is the straight line π β π + ππ 19 of 162 Lecture notes on MTH352: Differential Geometry The line is parallel to vector π£ and passes through point π. Definition: Let π be a differentiable real-valued function on πΉπ, and let π£π be a tangent vector to πΉπ . Then the number π£π [π] = π (π(π + π‘π£))|π‘=0 ππ‘ Is called the derivative of π with respect to π£π . From the picture shows the geometrically interpretation of the directional derivative. The surface is the graph of the function π: π 3 β π . The directional derivative of π corresponding to the tangent vector is calculated in the following mannar: ο First calculate the line corresponding to the tangent vector π£π . ο The line corresponds to a curve on the surface. ο The directional derivative π£π [π] is then the derivative of the curve at point π. Here the derivative of the curve means the same thing as we did in calculus of one variable (In fact the green curve in above graph is contained in a plane which is not shown in the graph). The following are worth noting: ο The number π£π [π] is called the derivative of π with respect to π£π ο We also say that π£π [π] is a directional derivative. ο The directional derivative π£π [π] informs us about the change in the value of π as we move away from π in the direction π£π . Example: Compute π£π [π] for the function π = π₯ 2 π¦π§, with π = (1,1,0) and π£ = (1,0, β3). Solution: First we compute the line defined by π and π£ π + π‘π£ = (1,1,0) + π‘(1,0, β3) = (1 + π‘, 1, β3π‘) Now evaluating π along the line. π(π + π‘π£) = π(1 + π‘, 1, β3π‘) = (1 + π‘)2 (1)(β3π‘) = β3π‘ β 6π‘ 2 β 3π‘ 3 20 of 162 Lecture notes on MTH352: Differential Geometry Now π π (π(π + π‘π£)) = (β3π‘ β 6π‘ 2 β 3π‘ 3 ) = β3 β 12π‘ β 9π‘ 2 ππ‘ ππ‘ Now applying definition ofπ£π [π] we get π π£π [π] = ππ‘ (π(π + π‘π£))|π‘=0 = (β3 β 12π‘ β 9π‘ 2 )|π‘=0 = β3. For next discussion we need to recall the following. The following results can be used to calculate directional derivatives more efficiently. Lemma: If π£π = (π£1, π£2, π£3 ) is a tangent vector to πΉπ, then π£π [π] = βπ£π ππ (π) ππ₯π Proof: Let π = (π1, π2, π3 ); then π + π‘π£ = (π1 + π‘π£1 , π2 + π‘π£2 , π3 + π‘π£3 ) Now π (π + π‘π£) = π((π1 + π‘π£1 , π2 + π‘π£2 , π3 + π‘π£3 )) Since π (π + π‘π£π ) = π£π ππ‘ π By using chain rule, we get π£π [π] = π ππ π ππ (ππ + π‘π£π )|π‘=0 = βπ£π (π) π(π + π‘π£ )|π‘=0 = β ππ‘ ππ₯π ππ‘ ππ₯π 21 of 162 Lecture notes on MTH352: Differential Geometry The same example as before: Example: Compute π£π [π] for the function π = π₯ 2π¦π§, with π = (1,1,0) and π£ = (1,0, β3). Solution: We know that If π£π = (π£1 , π£2 , π£3 ) is a tangent vector to πΉπ , then π£π [π] = βπ£π ππ (π). ππ₯π So we calculate ππ ππ ππ = 2π₯π¦π§, = π₯ 2 π§, = π₯2π¦ ππ₯ ππ¦ ππ§ Thus at point π = (1,1,0) ππ ππ ππ = 0, = 0, =1 ππ₯ ππ¦ ππ§ So π£π [π] = βπ£π ππ (π) = 1(0) + 0(0) β 3(1) = β3 ππ₯π Theorem: Let π and π be functions on πΉπ, π£π and π€π tangent vectors, π and π numbers. Then 1)- (ππ£π + ππ€π )[π] = ππ£π [π] + ππ€π [π]. 2)- π£π [ππ + ππ] = ππ£π [π] + ππ£π [π]. 3)- π£π [ππ] = π£π [π]. π(π) + π(π). π£π [π] Proof (3): 22 of 162 Lecture notes on MTH352: Differential Geometry Operation of a vector field Operation of a vector field π on a function π is defined as follows: Definition: Let π be a vector field, and let π: πΉπ β πΉ be a function. Then π[π] be the function defined by π[π]: πΉπ β πΉ, π[π](π) = π(π)[π] So π[π] is a function which to each point π in πΉπ gives us the derivative of π with respect to the tangent vector π(π). Example: For the natural fame field π1, π2, π3, we get π1 [π](π) = π1 (π)[π] = (1,0,0)π [π] Lemma: If π£π = (π£1, π£2, π£3 ) is a tangent vector to πΉπ, then π£π [π] = βπ£π ππ (π) ππ₯π So we get π1 [π](π) = π1 (π)[π] = (1,0,0)π [π] = 1 ππ ππ ππ ππ (π) + 0. (π) + 0 (π) = (π). ππ₯ ππ¦ ππ₯ ππ₯ Basic properties of operations of vector fields Corollary: Let π and π be vector fields on πΉπ and π, π and β are real-valued function, then (1)- (ππ + ππ)[β] = ππ[β] + ππ[β] (2)- π[ππ + ππ] = ππ[π] + ππ[π] (3)- π[ππ] = π[π]. π + ππ[π] Proof (2): π[ππ + ππ] = ππ[π] + ππ[π] For the proof we need to recall the following: 23 of 162 Lecture notes on MTH352: Differential Geometry Proof (3): π[ππ] = π[π]. π + ππ[π] Example: Let π be a vector field π = π₯ 2 π1 + π¦π3 and let πbe the function π = π₯ 2 π¦ β π¦ 2π§. Calculate π[π]. Solution: π[π] = (π₯ 2 π1 + π¦π3 )(π₯ 2 π¦ β π¦ 2 π§) Using the following two properties: 1)- (ππ + ππ)[β] = ππ[β] + ππ[β] 2)- π[ππ + ππ] = ππ[π] + ππ[π] We get π[π] = (π₯ 2 π1 + π¦π3 )(π₯ 2 π¦ β π¦ 2 π§) = (π₯ 2 π1 + π¦π3 )π₯ 2 π¦ β (π₯ 2 π1 + π¦π3 )π¦ 2 π§ = π₯ 2 π1 [π₯ 2 π¦] + π¦π3 [π₯ 2 π¦]βπ₯ 2 π1 [π¦ 2 π§] β π¦π3 [π¦ 2 π§] = π₯ 2 (2π₯π¦) + π¦. 0 β π₯ 2 . 0 β π¦(π¦ 2 ) = 2π₯ 3 π¦ β π¦ 3 . 1)- (ππ£π + ππ€π )[π] = ππ£π [π] + ππ€π [π]. 2)- π£π [ππ + ππ] = ππ£π [π] + ππ£π [π]. 3)- π£π [ππ] = π£π [π]. π(π) + π(π). π£π [π] 24 of 162 Lecture notes on MTH352: Differential Geometry End of the lecture Lecture # 4 Contents: ο Curves in π 3 ο Velocity Vectors ο Reparametrization of Curves in π 3 ο Derivative with respect to Velocity ο Properties of Curves ο Conclusion Curves in πΉπ Open intervals in π An open interval in π is a set πΌ of real numbers of one of the four forms β’ {π‘| π < π‘ < π} β’ {π‘| π < π‘} β’ {π‘|π‘ < π} β’ π Where π and π are real numbers. 25 of 162 Lecture notes on MTH352: Differential Geometry What is a curve in space intuitively? The path followed by the bird in π 3 gives a trajectory in space. Let for time π‘ the bird is located at πΌ(π‘) = (πΌ1 (π‘), πΌ2 (π‘), πΌ3 (π‘)), in πΉπ for π‘ β (0,8). This path together with some extra conditions gives us a curve. In rigorous terms: πΌ is a function from πΌ to πΉπ , where πΌ is an open interval. Thus we write πΌ(π‘) = (πΌ1 (π‘), πΌ2 (π‘), πΌ2 (π‘)) πππ πππ π‘ ππ πΌ. The real-valued functions πΌ1 (π‘), πΌ2 (π‘), and πΌ3 (π‘) are called Euclidean coordinate functions. Differentiable functions: We define the function πΌ to be differentiable provided its coordinate functions πΌ1 (π‘), πΌ2 (π‘), and πΌ3 (π‘) are differentiable. Definition: A curve in π 3 is a differentiable function πΆ: π° β πΉπ from an open interval πΌ to πΉπ . Example 1: Straight line A line through point π and in the direction π is a curve πΌ: πΉ β πΉπ defined as 26 of 162 Lecture notes on MTH352: Differential Geometry πΌ(π‘) = π + π‘π = (π1 + π‘π1 , π2 + π‘π2 , π3 + π‘π3 ). Circle: The curve π‘ β (acos π‘, ππ πππ‘, 0) travels around a circle of radius a > 0 in the π₯π¦-plane of π 3 . Helix: If we allow this curve to rise (or fall) at a constant rate, we obtain a helix πΌ: πΉ β πΉπ defined as πΌ(π‘) = (π cos π‘ , asin π‘ , ππ‘) where π > 0, π β 0. Example: The curve πΌ: πΉ β πΉπ defined by π‘ πΌ(π‘) = (1 + cos π‘ , sin π‘ , 2 sin ) 2 is a curve. The green line in the figure shows the curve. 27 of 162 Lecture notes on MTH352: Differential Geometry Example: The curve πΌ: πΉ β πΉπ defined as πΆ(π) = (ππ , πβπ , βπ π ) is shown in the picture. Example: The curve πΌ: πΉ β πΉπ defined by πΌ(π‘) = (3π‘ β π‘ 3 , 3π‘ 2 , 3π‘ + π‘ 3 ) is shown in the figure. This particular curve is called 3-curve. 28 of 162 Lecture notes on MTH352: Differential Geometry Velocity vector Definition: Let πΌ: πΌ β πΉπ be a curve in πΉπ with πΌ = (πΌ1, πΌ2, πΌ3). For each number π‘ in πΌ, the velocity vector of πΌ at π‘ is the tangent vector πΌ β² (π‘) = ( ππΌ1 ππΌ2 ππΌ3 (π‘), (π‘), (π‘)) ππ‘ ππ‘ ππ‘ πΌ(π‘) at the point πΌ(π‘) in πΉπ . Example: Let πΌ: πΉ β πΉπ is defined by πΌ(π‘) = (3π‘ β π‘ 3 , 3π‘ 2 , 3π‘ + π‘ 3 ). Geometric interpretation: 29 of 162 Lecture notes on MTH352: Differential Geometry Q1 (Exercise 1.4): Compute the velocity vector of the curve in Example 4.2(3) for arbitrary π π 4 2 π‘ and for π‘ = 0, π‘ = , π‘ = visualizing these on figure 1.8. Solution: Here πΌ: πΉ β πΉπ ππ πππ£ππ ππ π‘ πΌ(π‘) = (1 + cos π‘ , sin π‘ , 2 sin 2). Now by definition 30 of 162 Lecture notes on MTH352: Differential Geometry ππΌ1 πΌ β² (π‘) = ( ππ‘ (π‘), ππΌ2 ππ‘ (π‘), ππΌ3 ππ‘ (π‘)) πΌ(π‘) which in this case is π‘ πΌ β² (π‘) = (β sin π‘, cos π‘ , cos 2) πΌ(π‘) Now at π = π πΌ(0) = (2,0 , 0 ). and hence πΌ β² (0) = (0,1,1 )(2,0 ,0 ) . Now at π = π π π πΌ (4 ) = ( 1+β2 2 , 1 β2 , 0.76537). and hence π πΌ β² (4 ) = (β Now at π = 1 , 1 β2 β2 , 0.92388 ) 1+β2 1 , 2 β2 ( ,0.76537) . π π π πΌ (2 ) = (1 , 1 , β2 ) and hence π πΌ β² (2 ) = (β1,0, 1 β2 ) (1 ,1 ,β2 ) . Reparametrization of Curves in πΉπ Definition: Let πΌ: πΌ β πΉπ be a curve. If β: π½ β πΌ is a differentiable function on an open interval π½, then the composite function π½ = πΌ(β): π½ β πΉπ Is a curve called a reparametrization of πΌ by β. 31 of 162 Lecture notes on MTH352: Differential Geometry Example: Let the curve πΌ is defined as πΌ(π‘) = (βπ‘, π‘βπ‘, 1 β π‘) ππ πΌ = (0,4) If β: π½ β πΌ be the function given as β(π ) = π 2 , where π½ = (0,2). Then the reparametrization is π½(π ) = πΌ(β(π )) = πΌ(π 2 ) = (π , π 3 , 1 β π 2 ) Lemma: If π½ is the reparametrization of πΌ by β, then π½ β² (π ) = πβ (π )πΌ β² (β(π )). ππ Proof: If πΌ = (πΌ1, πΌ2, πΌ3), then π½(π ) = πΌ(β(π )) = (πΌ1 (β(π )), πΌ2 (β(π )), πΌ3 (β(π ))). Now π½β²(π ) = πΌβ²(β(π )) = (πΌ1 (β(π ))β², πΌ2 (β(π ))β², πΌ3 (β(π ))β²). By applying chain rule on πΌπ β²(β(π )), for π = 1,2,3 we get β² πΌπ (β(π )) = πΌπβ² (β(π ))ββ² (π ) πππ π = 1,2,3. Which yields π½β²(π‘) = πΌβ² ( β(π )) = (πΌ1β² (β(π ))ββ² (π ), πΌ2β² (β(π ))ββ² (π ), πΌ3β² (β(π ))ββ² (π )) = ββ² (π ) (πΌ1β² (β(π )), πΌ2β² (β(π )), πΌ3β² (β(π ))) = ββ² (π )πΌ β² (π ) . Q3 (Exercise 1.4): Find the coordinate function of the curve π½ = πΌ(β), where πΌ is the curve in example 4.2(3) and β(π ) = cosβ1 (π ) on π½: 0 < π < 1. 32 of 162 Lecture notes on MTH352: Differential Geometry Solution: The curve πΌ from 4.2(3) is given by: π‘ πΌ(π‘) = (1 + cos π‘ , sin π‘ , 2 sin ). 2 Lemma: Let πΌ be a curve in πΉπ and let π be a differentiable function on πΉπ. Then πΌ β² (π‘)[π] = π(π(πΌ)) (π‘) ππ‘ Proof: In this case ππΌ1 ππΌ2 ππΌ3 πΌβ² = ( , , ) ππ‘ ππ‘ ππ‘ So πΌ β² (π‘)[π] = β ππ ππΌπ (πΌ(π‘)) (π‘) ππ₯π ππ‘ But by chain rule. πΌ β² (π‘)[π] = π(π(πΌ)) (π‘). ππ‘ Properties of Curves One-to-one curves: The function πΌ: πΌ β πΉπ is one-to one if πΌ(π‘1 ) β πΌ(π‘2 )πππ π‘1 β π‘2 If πΌ is one-to-one then the curve does not intersect itself, or stay in the same point without moving on, or repeat itself over after some time. Periodic curves: A curve πΌ: πΉ β πΉπ is called periodic if there is a positive number π such that πΌ(π‘ + π) = πΌ(π‘) πππ πππ π‘ β πΉ 33 of 162 Lecture notes on MTH352: Differential Geometry The smallest such number π is the period of πΌ. Regular curves: A curve πΌ: πΌ β πΉπ is called regular if πΌ β² (π‘) β (0,0,0)πΌ(π‘) πππ πππ π‘ β πΌ Example: The curve πΌ: πΉ β πΉπ given as πΌ(π‘) = (π‘ 2 , π‘ 3 , 0) is not regular. End of the lecture Lecture # 5 Contents: ο 1-forms ο Differentials ο Properties of differentials Classical differentials: If π is a real-valued function on πΉπ , then in elementary calculus the differential of π is usually defined as ππ ππ ππ ππ = ππ₯ + ππ¦ + ππ§. ππ₯ ππ¦ ππ§ ππ here calculates the small change in the value of π when there is small change (dx, ππ¦, ππ§) in π₯, π¦ and π§. 34 of 162 Lecture notes on MTH352: Differential Geometry What does π π = ππ ππ π π + ππ ππ π π + ππ ππ π π means? We will see the meaning of this expression using the notion of 1-forms. Definition: A 1 βform π on πΉπ is a real-valued function on the set of all tangent vectors to πΉπ such that π is linear at each point, that is, π(ππ + ππ) = π π(π) + ππ(π) For any numbers π, π and tangent vectors π π at the same point of πΉπ . Sometimes we write π£π instead of π£ for tangent vector to πΉπ at π. Fact: For fix point p in πΉπ the resulting fucntion ππ : ππ (πΉπ ) β πΉ is linear. Sum of two π-forms: Let π and π be two 1 βforms then their sum is defined as: (π + π)(π) = π(π) + π (π) for all tangent vectors π. Product of a π-form with a function π: πΉπ β πΉ Let π: πΉπ β πΉ be a real valued function and π be a one form. Then for any tangent vector π£π to πΉπ we define: (ππ)(π£π ) = π(π)π(π£π ) Evaluating a π-form on a vector form Let π be a vector field, then for any point β πΉπ , π(π) is a tangent vector to πΉπ at point π. So we can evaluate a 1-form π on a vector field π in the following way At each point π β πΉπ the value of π(π) is the number π(π(π)). Differentiable π-form 35 of 162 Lecture notes on MTH352: Differential Geometry A 1-form π is differentiable if π(π) is differentiable whenever π is differentiable then π is differentiable . Properties of π(π½) The linearity of π and vector field properties imply that: π(ππ + ππ) = ππ(π) + ππ(π) and (ππ + ππ)(π) = ππ(π) + ππ(π). Where π and π are functions from πΉπ to πΉ. Definition: If π is a differentiable real-valued function on πΉπ , the differential ππ of π is the 1-form such that ππ(π£π ) = π£π [π] For all tangent vectors π£π . The 1-form ππ satisfies all the conditions of 1-form: ο ππ maps every tangent vector to a real number. ο ππ is linear because of the properties of directional derivative. Fact: For any real-valued function π on πΉπ the differential of π knows directional derivative of π in every direction π£π . Let us consider some examples of 1-forms on π 3 Example 1 ( Differential of Natural coordinate functions ) Natural coordinate functions: Natural coordinate functions π₯1 , π₯2 and π₯3 are functions from πΉπ to πΉ are defined as: π₯1 (π1 , π2 , π3 ) = π1 36 of 162 Lecture notes on MTH352: Differential Geometry π₯2 (π1 , π2 , π3 ) = π2 π₯3 (π1 , π2 , π3 ) = π3 . Now we calculate differentials of these natural coordinate functions. Using definition of ππ we get, for any tangent vector π£π : ππ₯π (π£π ) = π£π [π₯π ] = β π ππ₯π (π)π£π = β πΏππ π£π = π£π ππ₯π Where πΏππ is Kronecker delta defined as: 1 πΏππ = { 0, π€βππ π = π π€βππ π β π Thus ππ₯π only depends on π£ and not on the point of application π. Example 2 (Linear combination of 1-forms) Let π1 , π2 , π3 : πΉπ β πΉ be three functions. Let π = π1 ππ₯1 + π2 ππ₯2 + π3 ππ₯3 Then π is a 1-form, and for any vector π£π we have π(π£π ) = π1 (π) ππ₯1 (π£) + π2 (π) ππ₯2 (π£) + π3 (π) ππ₯3 (π£) = π1 (π) π£1 + π2 (π) π£2 + π3 (π) π£3 = β ππ (π)π£π π Next: we show that in fact each 1-form can be written in the form of π above. Euclidean coordinate functions of a π-form Lemma: If π is a 1-form on πΉπ , then π = β ππ ππ₯π , where ππ = π(ππ ). These functions π1 , π2 , and π3 are called the Euclidean coordinate functions of π. Proof: 37 of 162 Lecture notes on MTH352: Differential Geometry The classical differentials: Corollary: Let π: π 3 β π be a differentiable function. Then ππ = ππ ππ ππ ππ₯ + ππ¦ + ππ§. ππ₯ ππ¦ ππ§ Proof: Product rule: Lemma: Let ππ be the product of differentiable functions π and π on πΉπ. Then π(ππ) = πππ + πππ Proof: Chain rule: 38 of 162 Lecture notes on MTH352: Differential Geometry Lemma: Let π: πΉπ β πΉ and β: πΉ β πΉ be differentiable functions, so the composite function β(π): πΉπ β πΉ is also differentiable. Then π(β(π)) = ββ² (π)ππ. Proof: We know that chain rule for β(π) is Now Example: Calculate ππ, where π = (π₯ 2 β 1)π¦ + (π¦ 2 + 2)π§. Solution: ππ = (2π₯ππ₯)π¦ + (π₯ 2 β 1)ππ¦ + (2π¦ππ¦)π§ + (π¦ 2 + 2)ππ§ = 2π₯π¦ ππ₯ + (π₯ 2 β 1 + 2π¦π§)ππ¦ + (π¦ 2 + 2)ππ§ Now we evaluate ππ on the tangent vector π£π where π = (π1 , π2 , π3 ) and π£ = (π£1 , π£2 , π£3 ). ππ(π£π ) = π£π [π] = 2π1 π2 π£1 + (π12 β 1 + 2π2 π3 )π£2 + (π22 + 2 )π£3 39 of 162 Lecture notes on MTH352: Differential Geometry End of the lecture Lecture # 6 Contents ο Differential Forms ο π-forms ο Exterior derivatives ο Conclusion Differential Forms ο 1 βforms (we discussed in last lecture) are part of a larger system called differential forms. ο Will discuss some properties of differential forms rather than full description. To construct a differential form we follow the following procedure: 40 of 162 Lecture notes on MTH352: Differential Geometry ο§ Take real-valued functions π1 , π2 , π3 , β¦ , ππ : π 3 β π ο§ Take differential 1-forms of the coordinate functions ππ₯, ππ¦, ππ§. ο§ Take sum and product of above. Examples: 1)- π₯ 2 ππ₯ππ¦ + π₯π¦ππ¦ππ§ + (π₯π§ + π¦ 2 )ππ§ππ¦ 2)- π₯π¦π§ ππ₯ + (π₯ 2 π§ 2 + π¦)ππ¦ + π₯π¦π§ππ§ 3)- π₯ 2 π¦ 2 ππ₯ππ¦ππ§ . ο The multiplication of forms ππ₯ ππ¦ is not the usual multiplication and is not commutative. In fact ππ₯ ππ¦ = βππ¦ ππ₯. called the alternation rule. This is a special kind of multiplication called βWedge productβ, we denote it by " β§ " (called wedge). Hence the above expression becomes: ππ₯ β§ ππ¦ = βππ¦ β§ ππ₯. Other properties of wedge product are as follows. For differential forms πΌ, π½ and πΎ we have: 1) πΌ β§ (π½ β§ πΎ) = (πΌ β§ π½) β§ πΎ 2) πΌ β§ (π½ + πΎ) = πΌ β§ π½ + πΌ β§ πΎ We will ignore the " β§ " sign where ever it is clear that what kind of multiplication we are considering. An important consequence of the wedge product is that βrepeats are zeroβ, due to βalternation ruleβ. Which means ππ₯ β§ ππ₯ = 0. Similarly ππ¦ β§ ππ¦ = 0 and ππ§ β§ ππ§ = 0. In fact ππ₯ β§ ππ₯ = βππ₯ β§ ππ₯ β 2 ππ₯ β§ ππ₯ = 0 β ππ₯ β§ ππ₯ = 0 Similarly we can show 41 of 162 Lecture notes on MTH352: Differential Geometry ππ¦ β§ ππ¦ = 0 and ππ§ β§ ππ§ = 0. Exercise: Calculate (π₯ 2 ππ₯ππ¦ + π₯π¦ ππ¦ππ§) β§ (π₯π§ ππ₯ + (π§ 2 + π¦)ππ₯ππ¦). π-forms: A π-form, for π one of 0,1,2,3, is defined as: 1) A 0-form is just a differentiable function π. 2) A 1-form is an expression π₯ + πππ¦ + βππ§ . 3) A 2-form is an expression πππ₯ β§ ππ¦ + πππ₯ β§ ππ§ + βππ¦ β§ ππ§ (or simply πππ₯ππ¦ + πππ₯ππ§ + βππ¦ππ§) . 4) A 3-form is an expression πππ₯ β§ ππ¦ β§ ππ§ ( or simply πππ₯ β§ ππ¦ β§ ππ§). Notice that on π 3 , where we have three coordinates namely π₯, π¦, π§ the 4-forms and higher π-forms are zero due to the βalternation ruleβ. For instance ππ₯ β§ ππ¦ β§ ππ§ β§ ππ¦ = 0. βProduct of a π-form and a π-form is a (π + π)-formβ. Lemma: If π and π are 1-forms, then π β§ π = βπ β§ π. Proof: Write π = β ππ ππ₯π and π = β ππ ππ₯π . Then 42 of 162 Lecture notes on MTH352: Differential Geometry Exterior derivative Recall: Definition: If π is a differentiable real-valued function on π 3 , the differential ππ of π is the 1-form such that ππ(π£π ) = π£π [π] For all tangent vectors π£π . We can generalize this concept to define an apply it on a π-form and get a (π + 1)-form. For example for π = 1 we define Definition: If π = β ππ ππ₯π is a 1-form on π 3 , the exterior derivative of π is the 2-form ππ = β πππ β§ ππ₯π . We know that: Corollary: Let π: πΉπ β πΉ be a differentiable function. Then ππ = ππ ππ ππ ππ₯ + ππ¦ + ππ§. ππ₯ ππ¦ ππ§ So if we expand ππ = β πππ β§ ππ₯π using above then using: ππ1 = ππ1 ππ1 ππ1 ππ₯1 + ππ₯2 + ππ₯3 . ππ₯1 ππ₯2 ππ₯3 43 of 162 Lecture notes on MTH352: Differential Geometry ππ2 = ππ2 ππ2 ππ2 ππ₯1 + ππ₯2 + ππ₯3 . ππ₯1 ππ₯2 ππ₯3 ππ3 = ππ3 ππ3 ππ3 ππ₯1 + ππ₯2 + ππ₯3 . ππ₯1 ππ₯2 ππ₯3 ππ = β πππ β§ ππ₯π using above then using: ππ1 = ππ1 ππ1 ππ1 ππ₯1 + ππ₯2 + ππ₯3 . ππ₯1 ππ₯2 ππ₯3 ππ2 = ππ2 ππ2 ππ2 ππ₯1 + ππ₯2 + ππ₯3 . ππ₯1 ππ₯2 ππ₯3 ππ3 = ππ3 ππ3 ππ3 ππ₯1 + ππ₯2 + ππ₯3 . ππ₯1 ππ₯2 ππ₯3 Definition: Let π = πππ₯ππ¦ + πππ₯ππ§ + βππ¦ππ§ be a 2-form. Then the exterior derivative of π is the 3-form ππ = ππ β§ ππ₯ππ¦ + ππ β§ ππ₯ππ§ + πβ β§ ππ¦ππ§. It is easy to check the following π (ππ + ππ) = πππ + πππ. Where π and π are arbitrary forms and a and b are numbers. Theorem: Let π and π be functions, π and π be 1-forms. Then 1) π(ππ) = ππ π + πππ 44 of 162 Lecture notes on MTH352: Differential Geometry 2) π(ππ) = ππ β§ π + π ππ 3) π(π β§ π) = ππ β§ π β π β§ ππ Proof (1): π (ππ) = π π π + ππ π Proof (3): π(π β§ π) = ππ β§ π β π β§ ππ Lets first consider the case where π = π ππ₯ and π = π ππ¦ then: and and Lets first consider now π = β ππ ππ₯π and π = β ππ ππ₯π . 45 of 162 Lecture notes on MTH352: Differential Geometry End of the lecture Lecture # 7 Contents: ο Introduction to Mappings ο Tangent Maps Introduction to Mappings Our aim: To understand the mappings π: π π β π π for different values of π and π. 46 of 162 Lecture notes on MTH352: Differential Geometry In particular we are interested in the following cases π: π 2 β π 2 , π: π 2 β π 3, π: π 3 β π 3 Definition: Given a function πΉ: π π β π π , let π1 , π2 , β¦ , ππ denote the real-valued functions on π π such that πΉ(π) = (π1 (π), π2 (π), β¦ , ππ (π)) for all points π in π π . These function are called the Euclidean coordinate functions of πΉ, and we write πΉ = (π1 , π2 , β¦ , ππ ). Mappings: A function πΉ: π π β π π is differentiable if all its Euclidean coordinate functions are differentiable. A differentiable function from π π to π π is called a mapping. Example: Consider πΉ: π 3 β π 3 defined by πΉ(π₯, π¦, π§) = (π₯ 2 , π¦π§, π₯π¦) is differentiable and hence a mapping. Definition: If πΌ: πΌ β π π is a curve in π π and πΉ: π π β π π is a mapping then the composite function π½ = πΉ(πΌ): πΌ β π π is a curve in π π called the image of πΌ under πΉ. Example: Consider the mapping πΉ: π 2 β π 2 define by: πΉ(π’, π£) = (π’2 β π£ 2 , 2π’π£) To see the effect of this mapping we consider images of a special curve πΌ in π 2 under this mapping, namely circle of radius r, given as 47 of 162 Lecture notes on MTH352: Differential Geometry πΌ(π‘) = (π cos π‘ , π sin π‘) where 0 β€ π‘ β€ 2π Example: Consider the mapping πΉ: π 3 β π 3 such that F(π₯, π¦, π§) = (π₯ β π¦, π₯ + π¦, 2π§) Observe that πΉ is a linear transformation. Tangent Maps Our next goal is to find analogous linear approximation for the mapping point π β π π . πΉ: π π β π π near a For this we once again used special curves in π π , namely lines associated to tangent vectors π£π . We approximate πΉ near π by the map πΉβ . Definition: Let πΉ: π π β π π be a mapping. If π£ is a tangent vector to π π at π, let πΉβ (π£) be the initial velocity vector of the curve π‘ β πΉ(π + π‘π£). The resulting function πΉβ sends tangent vectors to π π to tangent vectors to π π , and is called the tangent map of πΉ. How to calculate tangent maps: Proposition: Let πΉ = (π1, π2, β¦ , ππ ) be a mapping from π π to π π . If π£ is a tangent vector to π π at π, then 48 of 162 Lecture notes on MTH352: Differential Geometry πΉβ (π£) = (π£[π1 ], π£[π2 ], β¦ , π£[ππ ]) at πΉ(π) See book for proof. Tangent map of a mapping at a point π The tangent map πΉβ sends tangent vectors at π to tangent vectors at πΉ(π). Thus for each π in πΉπ , the map πΉβ give rise to a function πΉβ : ππ (πΉπ ) β ππΉ(π) (πΉπ ) Called the tangent map of πΉ at π. Tangent are linear maps at a point An important maps which is a proposition is: property of tangent consequence of Corollary: If πΉ: π π β π π is a mapping, then at each point π of π 3 the tangent map πΉβ: ππ (π π ) β ππΉ(π) (π π ) is a linear transformation. See book for proof. The velocity of an image curve Another important consequence of the proposition is that mappings preserved velocities. Corollary: Let πΉ: π π β π π be a mapping. If π½ = πΉ(πΌ) is the image of a curve πΌ in π π , then π½β² = πΉβ (πΌβ²). Proof: We take π = 3 By proposition above we have: we have 49 of 162 Lecture notes on MTH352: Differential Geometry Tangent maps on vector fields Μ π } (1 β€ π β€ π) be the natural frame fields of π π and π π respectively. Let {ππ } (1 β€ π β€ π) and {π Then: Corollary: If πΉ = (π1, π2 , β¦ , ππ ) is a mapping from π π to π π , then π πΉβ (ππ (π)) = β π=1 πππ Μ π (π) (π) π ππ₯π for 1 β€ π β€ π . See book for proof. The matrix appearing in the preceding formula, ( πππ (π)) πππ 1β€πβ€π,1β€πβ€π Is called the Jacobian matrix of πΉ at point π. In expanded form it is given as: We study mappings πΉ using the tangent map. Definition: A mapping πΉ: π π β π π is regular provided that at every point π of π π the tangent map πΉβπ is one-to-one. For π β π π the following are equivalent: ο πΉβ π is one-to-one 50 of 162 Lecture notes on MTH352: Differential Geometry ο πΉβπ (vp ) = 0 implies π£π = 0 ο The rank of the Jacobian matrix of πΉ at π is π. Diffeomorphism Definition: if πΉ: π π β π π has an inverse mapping, then πΉ is called diffeomorphism. Inverse function theorem. Theorem: Let πΉ: π π β π π be a mapping. If πΉβπ is one-to-one at π, then there is a restriction of πΉ to an open set containing π which is a diffeomorphism. End of the Lecture Chapter 2 Lecture #8 Contents: ο The Dot Product ο Frames The Dot Product 51 of 162 Lecture notes on MTH352: Differential Geometry We study distances and angles in geometry. We will see that geometry of Euclidean spaces can be derived from dot product. Definition: Let π = (π1 , π2 , π3 ) and π = (π1 , π2 , π3 ) be two points in π 3 . The dot product of π and π is the real number π β π = π1 π1 + π2 π2 + π3 π3 Dot product has the following properties If π, q and π are points in πΉπ and π, π are real numbers, then 1. Bilinearity: (ππ + ππ). π = π(π. π) + π(π. π) π. (ππ + ππ) = π(π. π) + π(π. π) 2. Symmetry: π. π = π. π 3. Positive definiteness π. π β₯ 0 and π. π = 0 β π = 0 The norm of a point The norm of a point π = (π1 , π2 , π3 ) is the number 1 βπβ = (π β π)2 = βπ12 + π22 + π32 The norm is a real-valued function on πΉπ . It has the following properties ο βπ + πβ β€ βπβ + βπβ ο βππβ = |π|βπβ for vectors π and π of π 3 and real number π. 52 of 162 Lecture notes on MTH352: Differential Geometry Distances in πΉπ If π and π are two points of π 3 , the Euclidean distance from π to π is the number π(π, π) = βπ β πβ So we calculate π(π, π) = β(π1 β π1 )2 + (π2 β π2 )2 + (π3 β π3 )2 Open sets The distance formula can be used to give a precise definition of open sets. Definition: If π is a point in π 3 and π is a positive real number, then the π βneighborhood N π (π) of π in π 3 is the set of all points π in π 3 such that π(π, π) < π: N π (π) = {π β π 3 : π(π, π) < π} Definition: A subset O of π 3 is open if every point of O has some π βneighborhood that is completely contained in O : βπ β O β π > 0: N π (π) β O This definition of open set can be used on any π π . We can also replace π with some other distance function. Dot product on tangent vectors We know that tangent space at any point of π 3 , that is, ππ (π 3 ) is isomorphic to π 3 . We can use this relation to define dot product on any tangent space ππ (π 3 ). 53 of 162 Lecture notes on MTH352: Differential Geometry Definition: The dot product of tangent vectors π£π and π€π at the same point of π 3 is the number π£π . π€π = π£. π€. Similarly we can define norm of any tangent vector in the following way. Definition: The norm, or length, of a tangent vector π£π is the norm of its vector part, that is, βπ£π β = βπ£β Schwarz inequality: A fundamental result of linear algebra is Schwarz inequality given by |π£. π€| β€ βπ£ββπ€β This permits us to define the cosine of the angel between two vectors as: Definition: To any non-zero vectors π£ and π€ define the cosine of the angle between π£ and π€ by π£. π€ = βπ£ββπ€β cos π Definition: Two vectors are orthogonal if π£. π€ = 0. Definition: A vector of length 1 is called a unit vector. Frames Definition: A set π1 , π2 , π3 of three mutually orthogonal unit vectors tangent to π 3 at point π is called a frame at the point π. We can say that a set of three tangent vectors π1 , π2 , π3 in ππ (π 3 ) is a frame at π if the vectors satisfy π1 . π2 = π1 . π3 = π2 . π3 = 0 54 of 162 Lecture notes on MTH352: Differential Geometry π1 . π1 = π2 . π2 = π3 . π3 = 1 We can write down the same definition using notion of Kronecker delta function πΏππ defined as 0, ππ π β π πΏππ = { 1, ππ π = π So the alternate definition becomes: Definition: The set of tangent vectors π1, π2, π3 to π 3 at π is a frame at π if and only if ππ . ππ = πΏππ for all π, π β {1,2,3} Theorem: Let π1, π2, π3 be a frame at a point π of π 3 . If π£ is any tangent vector to π 3 at π, then π£ = (π£. π1 )π1 + (π£. π2 )π2 + (π£. π3 )π3 See book for proof. Now we know that the tangent space ππ (πΉπ ) is isomorphic to πΉπ . So three linearly independent vectors form a basis of ππ (πΉπ ). End of the lecture 55 of 162 Lecture notes on MTH352: Differential Geometry Lecture # 9 Contents: ο The Attitude Matrix ο Cross Product The Attitude Matrix The Attitude Matrix Of A Frame Let {π1 , π2 , π3 } be a frame at point π in πΉπ If π1 = (π11 , π12 , π13 ) 56 of 162 Lecture notes on MTH352: Differential Geometry π2 = (π21 , π22 , π23 ) π3 = (π31 , π32 , π33 ) Then π11 π π΄ = ( 21 π31 π12 π22 π32 π13 π23 ) π33 is called the attitude matrix of the frame {π1 , π2 , π3 }. Definition: If the rows of a square matrix are orthonormal, then the matrix is called orthogonal. Hence the attitude matrix of any frame is an orthogonal matrix. Definition: The transpose of 3 × 3 matrix is defined by π11 π΄ = (π21 π31 π12 π22 π32 π13 π11 π23 ) β π‘π΄ = (π12 π33 π13 π21 π22 π23 π31 π32 ) π33 Property of orthogonal matrices If π΄ is a 3 × 3 orthogonal matrix, then the matrix product of π΄ with π΄π‘ is equal to 1 0 π΄π΄π‘ = (0 1 0 0 So π΄π‘ 0 0) 1 is inverse of π΄. Cross Product Cross Product: 57 of 162 Lecture notes on MTH352: Differential Geometry Let π£ and π€ be tangent vectors to π 3 at a point π. The cross product of π£ and π€ is the determinant π1 (π) π2 (π) π£2 π£ × π€ = | π£1 π€1 π€2 π3 (π) π£3 | π€3 Which is also element of ππ (π 3 ). Properties Of Cross Product Linearity: Let π’, v and π€ be tangent vectors to π 3 at the same point π, and let π, π be real numbers. Then (ππ’ + ππ£) × π€ = π(π’ × π€ ) + π(π£ × π€) and π’ × (ππ£ + ππ€) = π(π’ × π£) + π(π’ × π€) Properties Of Cross Product Alternation Rule: If π£, π€ are tangent vectors to π 3 at a point π, then π£ × π€ = βπ€ × π£ It is also easy to see that π£ × π£ = (0,0,0). Direction of the Cross Product Lemma: The cross product of π£ and π€ is orthogonal to both π£ and π€. See book for the proof. The length of the cross product: 58 of 162 Lecture notes on MTH352: Differential Geometry The cross product of π£ and π€ has length such that βπ£ × π€β2 = (π£. π£)(π€. π€) β (π£. π€)2 See book for the proof. A more intuitive description of the length of a cross product is βπ£ × π€β = βπ£ββπ€β sin π Triple Scalar Product Let π’, π£, π€ be three vectors. The triple scalar product of π’, π£, π€ is the number π’. π£ × π€ We can also write the above as π’1 π’. π£ × π€ = | π£1 π€1 π’2 π£2 π€2 π’3 π£3 | π€3 End of the lecture 59 of 162 Lecture notes on MTH352: Differential Geometry Lecture # 10 Contents: ο Speed Of A Curve ο Vector Fields On Curves ο Differentiation of Vector Fields Speed Of A Curve Speed Let a car moves between Islamabad and Quetta and follow the path πΌ = (πΌ1 , πΌ2 ): πΌ β π 2 60 of 162 Lecture notes on MTH352: Differential Geometry given by πΌ(π‘) = (π‘ 2 , π‘ 3 ) What is the speed of the car after 3 hours. To answer such questions we need the following. Definition: Let πΌ = (πΌ1 , πΌ2 , πΌ3 ): πΌ β π 3 be a curve The speed of πΌ is defined as the length of the velocity of : π£= βπΌ β² β ππΌ1 2 ππΌ2 2 ππΌ3 2 β = ( ) +( ) +( ) ππ‘ ππ‘ ππ‘ Arc length Let a car moves between Islamabad and Quetta and follow the path πΌ = (πΌ1 , πΌ2 , πΌ3 ): πΌ β π 2 given by πΌ(π‘) = (2π‘, π‘ 2 ) Question: What is the distance travelled in 3 hours? To answer such questions we need the following. Definition: The arc length of πΌ from π‘ = π to π‘ = π is equal to π π β« βπΌβ²βππ‘ = β« π£(π‘)ππ‘ π π 61 of 162 Lecture notes on MTH352: Differential Geometry Definition: The restriction of a curve πΌ: πΌ β π 3 to a closed interval [π, π] is called a curve segment. If π is a curve segment of πΌ, π: [π, π] β π 3 then the length of π is written as πΏ(π). Definition: A curve π½ has unit speed if its speed is 1 at every point, that is, βπ½ β² β = 1. Example: Let π½: πΌ β π 3 be defined by π½(π‘) = (cos π‘, sin π‘ , π‘) Arc-Length parameterization Theorem: Let πΌ: πΌ β π 3 be a regular curve. Then there exist a reparameterization π½ of πΌ such that βπ½ β² β = 1. See book for proof. Orientation of a reparameterization A reparameterization of a curve πΌ given by π½(π‘) = πΌ(β(π )) Is orientation preserving if ββ² (π ) β₯ 0 It is orientation reversing if ββ² (π ) β€ 0 FACT: Unit speed reparameterization is always orientation preserving. Let πΌ: πΌ β π 3 be a curve. We define: Definition: A vector field π on curve πΌ is a function that assigns to each number π‘ in πΌ a tangent vector π(π‘) to π 3 at the point πΌ(π‘). 62 of 162 Lecture notes on MTH352: Differential Geometry Vector Fields On Curves Example: For curve πΌ the velocity πΌβ² satisfies the definition of vector field on curve. a Properties of vector fields on curves Definition: If π is a vector field on πΌ: πΌ β πΉπ , then for each π‘ in πΌ we can write π(π‘) = (π¦1 (π‘), π¦2 (π‘), π¦3 (π‘))πΌ(π‘) = β π¦π (π‘)ππ (πΌ(π‘)). The real-valued function π¦1 , π¦2 π¦3 : πΉ β πΉ are called Euclidean coordinate functions. Addition of vector fields Let π(π‘) = π¦1 π1 + π¦2 π2 + π¦3 π3 and π(π‘) = π§1 π1 + π§2 π2 + π§3 π3 be two vector valued functions on πΌ: πΌ β πΉπ . Then the addition of π(π‘) and π(π‘) is defined as: Scalar Multiplication Let π(π‘) = π¦1 π1 + π¦2 π2 + π¦3 π3 be a vector field on πΌ: πΌ β πΉπ and π(π‘) be a real valued functions. Then we define ππ Dot product of two vector fields Let π(π‘) = π¦1 π1 + π¦2 π2 + π¦3 π3 and π(π‘) = π§1 π1 + π§2 π2 + π§3 π3 be two vector valued functions on πΌ: πΌ β πΉπ . Then the dot product of of π(π‘) and π(π‘) is defined as: Cross product of two vector fields Let π(π‘) = π¦1 π1 + π¦2 π2 + π¦3 π3 and π(π‘) = π§1 π1 + π§2 π2 + π§3 π3 be two vector valued functions on πΌ: πΌ β πΉπ . Then the cross product of of π(π‘) and π(π‘) is defined as: Differentiation of a vector field on a curve 63 of 162 Lecture notes on MTH352: Differential Geometry Definition: Let π(π‘) be a vector field on a curve , with π(π‘) = β π¦π ππ . Then the new vector filed on πΌ π β² (π‘) = β ππ¦π π ππ‘ π is the derivative of π(π‘). Definition: Let πΌ: πΌ β πΉπ be a curve with coordinates πΌ(π‘) = (πΌ1(π‘), πΌ2(π‘), πΌ3(π‘)) then the acceleration of πΆ is defined as π2 πΌ1 π 2 πΌ2 π 2 πΌ3 πΌ =( 2 , , ) ππ‘ ππ‘ 2 ππ‘ 2 β²β² Properties of differentiation Let π(π‘) = π¦1 π1 + π¦2 π2 + π¦3 π3 and π(π‘) = π§1 π1 + π§2 π2 + π§3 π3 be two vector valued functions on πΌ: πΌ β πΉπ . Let π: π β π be a real-valued function. Let π and π be two real numbers then Linearity: (ππ + ππ)β² = ππ β² + ππ β² Leibnizian properties (π. π)β² = π β² . π + π. π β² (ππ) = π β² π + ππ β² Parallel vector fields A vector field on a curve is parallel if all its values are parallel tangent vectors. That is π(π‘) = (π1 , π2 , π3 )πΌ(π‘) = β ππ ππ (πΌ(π‘)) Vanishing derivatives: 64 of 162 Lecture notes on MTH352: Differential Geometry Lemma: Let πΌ: πΌ β π 3 be a curve then: 1. πΌ is constant if and only if πΌ β² = 0. 2. If πΌ is not constant, then πΌ is a straight line if and only if πΌ β²β² = 0. 3. A vector field π on a curve πΌ is parallel if and only if π β² = 0. See book for proof. End of the lecture Lecture #11 Contents: ο Curvature ο Frenet Frame Field ο Frenet Formulas ο Unit-Speed Helix 65 of 162 Lecture notes on MTH352: Differential Geometry Aim of todayβs lecture is to understand the geometry of the curves. In daily life we can understand the geometry of the curve by looking at them, for example, in the following figure there are three curves given. We can easily see that which curves have more curvature and which curve is turning very fast. Today we will define these properties mathematically. Curvature Definition: The rate of change of tangent, that is, π β² = π½ β²β² is called the curvature vector filed of π½. Since we have that βπβ = 1 So π. π = 1 And this gives us π β². π = 0 Definition: A vector field π on π½ is called normal to π½ if π. π½ β² = 0 . We can see that curvature vector field is normal to π½. Definition (Curvature): The function π : πΌ β πΉ defined by π (π‘) = βπ β² (π‘)β Is called the curvature function of π½. The curvature function satisfies π (π‘) β₯ 0 for all π β πΌ . 66 of 162 Lecture notes on MTH352: Differential Geometry Note: ο When π is small then the curve is close to straight. ο When π is large then the turn in the curve is sharp. Principal normal Let π½ be a curve. Let > 0 , that is, curve is never straight. We define principal normal vector π as: π= Tβ² π We can see that βπβ = 1, hence a unit vector. The principal normal vector tells us the direction in which the curve turns. Binormal vector field The vector field on curve π½ π΅ = π×π is called the binormal vector field of π½. Lemma: If π½ is a unit-speed curve in πΉπ with π > 0. Then the three vector fields π, π, π΅ on π½ are unit vectors that are mutually orthogonal at each point. Proof: See book. 67 of 162 Lecture notes on MTH352: Differential Geometry Definition (Frenet Frame ) The triplet (π, π΅, π) is called the Frenet frame field on π½. The derivatives of a Frenet field We can express derivatives of π, π, π΅ in terms of π, π, π΅. The derivative of π» We defined π= πβ² π So π β² = π π The derivatives of a Frenet field We can express derivatives of π, π, π΅ in terms of π, π, π΅. The derivative of B We can write π΅β² in the following form π΅ β² = (π΅ β² . π)π + (π΅β² . π)π + (π΅ β² . π΅)π΅ The torsion of a curve The function π: πΌ β πΉ such that π΅ = βππ is called the torsion of the curve. 68 of 162 Lecture notes on MTH352: Differential Geometry The Frenet Formulas Theorem: If π½ is a unit speed curve with curvature πβ² = π and torsion π then π π π β² = βπ π π΅β² = + ππ΅ β ππ Proof: see book. End of the lecture Lecture # 12 Contents: ο Frenet Approximation ο Plane Curves ο Planes In this lecture we are mainly interested in approximating the curve using Frenet Frame and Frenet formulas. So we want to approximate the curve using the geometric information, for example, using curvature, torsion, normal, etc. First we recall definition of a plane in π 3 . A plane through π and orthogonal to π β 0 consists of all points π such that (π β π). π = π 69 of 162 Lecture notes on MTH352: Differential Geometry forms a plane. Next we are going to recall Taylor series. Taylor Series Let π(π₯) be a function from π to π . If π(π₯) is infinitely differentiable near 0. Then the functional values of π(π₯) near π₯ = 0 are given by the Taylor series. ππ π2π π₯ 2 π3π π₯3 (0) π₯ + 2 (0) (0) π(π₯) = π(0) + + +β― ππ₯ ππ₯ 2 ππ₯ 3 3! To approximate a curve near an arbitrary point on the curve. We are going to achieve this by using Taylor series and Frenet Frame at a point. Frenet Approximation Approximation of Euclidean Coordinate Functions The Taylor series around origin is given. Taylor series around point π± = π 70 of 162 Lecture notes on MTH352: Differential Geometry ππ π2π π 2 π3π π 3 (0) π + 2 (0) + 3 (0) + β― π(π₯) = π(0) + ππ ππ 2 ππ 3! Where π₯ is any point near π₯ = 0. Now let π½ = (π½1 , π½2 , π½3 ) be a unit speed curve. We can approximate each of the functions π½π , for π = 1,2,3 in the following way: π½π ~ π½π (0) + ππ½π π2 π½ π 2 π3 π½ π 3 (0) π + 2 (0) + 3 (0) ππ ππ 2 ππ 3! In coordinate form we can write. π 2 β²β² π 3 β²β²β² π½(π ) = (π½1 (π ), π½2 (π ), π½3 (π )) ~ π½(0) + π π½β²(0) + π½ (0) + π½ (0) 2 3! Now the above is an approximation to the curve. Now we want to see what kind of geometrical properties can be obtained from this geometrical approximation of the curve. For this we apply Frenet formulas. Applying Frenet Formulas We know that π(π ) = π½β²(π ) so this gives us π½ β² (0) = π(0) = π0 We also know that π β² (π ) = π (π )π(π ) β π½ β²β² (π ) = π (π )π(π ) π½0β²β² = π 0 π0 β π½ β²β² (0) = π (0) π(0) β To calculate π½ β²β²β² (0) we use Leibniz rule π½ β²β²β² = π(π π) ππ ππ = π+π ππ ππ ππ Frenet formulas imply that π β² (π ) = βπ π + π π΅ Using this in above expression we get π½ β²β²β² = ππ π β π 2 π + ππ π΅ ππ So we get π½0β²β²β² = π½ β²β²β² (0) = ππ (0) π0 β π 02 π0 + π0 π 0 π΅0 ππ 71 of 162 Lecture notes on MTH352: Differential Geometry Definition: Let π½ be a unit speed curve then π½Μ (π ) = π½(0) + π π0 + π 0 π 2 π 3 π + π0 π 0 π΅ 2 0 6 0 is called Frenet approximation of π½ near π = 0. Next question: Can we do the same process for a point π 0 other than π = 0. The Frenet approximation at point ππ For this we use Taylor series around point π = π 0 . Taylor series around point π± = ππ π(π) = π(ππ ) + (π β ππ )π π π π π π (π β ππ )π π π π (ππ ) (π β ππ ) + π (ππ ) (π ) + π π +β― π π π π π π π π! Where π₯ is any point near π₯ = π 0. Using above and Frenet Formulas we can show that for a general point π 0 on the curve the approximation is given by π½Μπ 0 (π ) = π½(π 0 ) + (π β π 0 )ππ 0 + π π 0 (π β π 0 )2 (π β π 0 )3 ππ 0 + π π 0 ππ 0 π΅π 0 2 3! Important: We have triplet (π, π΅, π) at each and every point of the curve. So we have at each point of curve π½Μ (π ) = π½(0) + π π0 + π 0 π 2 π 3 π + π0 π 0 π΅ 2 0 6 0 Now we discuss the terms of approximation one by one. 72 of 162 Lecture notes on MTH352: Differential Geometry The curve defined by π β π½(0) + π π0 is the tangent line to π½ at π = 0. It is the best linear approximation to π½ near π½(0). The approximation π½(0) + π π0 + π 0 π 2 2 π0 is in fact a quadric function in π and in particular defines a parabola. The parabola lies in the plane with axis π0 and π0 . This plane is called Osculating plane. The approximation π 2 π 3 π½(0) + π π0 + π 0 π + π0 π 0 π΅ 2 0 6 0 is curve which is moving the parabola in the third dimension of π΅0. Lets recall what is a planar curve. Planar curve A plane curve in πΉπ is the set of all point which lie in the same plane in πΉπ . See figure below for a geometrical idea. Torsion Free Curves Let π½ be a curve in πΉπ with positive curvature π > 0. Then π½ is a plane curve if and only if π = 0. 73 of 162 Lecture notes on MTH352: Differential Geometry Theorem: Let π½ be a curve in πΉπ with positive curvature π > 0. Then π½ is a plane curve if and only if π = 0. See book for proof. End of the lecture Lecture #13 Contents: ο Frenet Approximation ο Conclusion ο Frenet Frame For Arbitrary Speed Curves ο Velocity And Acceleration Curvature Of Circle Consider the following circle π π π½(π ) = (acos , π sin ) π π 74 of 162 Lecture notes on MTH352: Differential Geometry 1 Then it is a unit speed curve. And its curvature is . π Now the question is: every curve with torsion zero (contained in plane) and constant curvature is a circle? Lemma: If π½ is a unit speed curve with constant curvature π and torsion zero, then π½ is part of a 1 circle of radius π . See book for proof. Arbitrary Speed Curves We use the following result. Theorem: Let πΌ: πΌ β πΉπ be a regular curve. Then there exist a reparameterization π½ of πΌ such that βπ½ β² β = 1. Arc length parameter Let πΌ: I β π be a regular curve. We can find arc-length parameter π in the following way π‘ π = β« βπΌβ²(π’)βππ’ π Let πΌΜ be the arc-length reparametrization of πΌ such that πΌ(π‘) = πΌ(π (π‘)) π‘βπΌ Example: Consider a helix πΌ: πΉ β πΉ defined by πΌ(π‘) = (π cos π‘ , π sin π‘, ππ‘) 75 of 162 Lecture notes on MTH352: Differential Geometry Now πΌΜ is a unit speed curve. If π Μ > 0 then we can define the following Μ , π΅Μ π Μ , πΜ , πΜ , π The Frenet Frame of πΆ Definition: Let πΌ be a regular curve and let πΆΜ be its arc-length parametrization. Then we define The curvature function of : π = π Μ (π ) The torsion function of : π = πΜ (π ) Unit tangent vector field of πΌ: π = πΜ (π ) Μ (π ) Principal normal vector field of : π = π Binormal vector field of : π΅ = π΅Μ (π ) For arbitrary speed curve πΌ the speed π£(π‘) = βπΌβ²(π‘)β is an important factor in the Frenet formulas for arbitrary speed curves. Lemma: If πΌ is a regular curve in πΉπ with π > 0, then πβ² = π π£π π β² = βπ π£π π΅β² = + ππ£π΅ β ππ If a object (e.g. bird, airplane, etc ) is moving on a constant speed then velocity and acceleration are perpendicular. If the object follow the path πΌ: πΌ β πΉπ . Since πΌ has constant speed so βπΌβ²β = π Definition: Let πΌ: πΌ β πΉπ be a curve with coordinates πΌ(π‘) = (πΌ1(π‘), πΌ2(π‘), πΌ3(π‘)) then the acceleration of πΆ is defined as 76 of 162 Lecture notes on MTH352: Differential Geometry π2 πΌ1 π 2 πΌ2 π 2 πΌ3 πΌ =( 2 , , ) ππ‘ ππ‘ 2 ππ‘ 2 β²β² So πΌ.β² πΌ β² = π 2 πΌ β² . πΌ β²β² = 0 So velocity and acceleration are perpendicular. We can analyze the situation by expressing velocity and acceleration in terms of Frenet Frame field. Lemma: If πΌ is a regular curve with speed function π£, then its velocity and acceleration are expressed by πΌ β² = π£π, πΌ β²β² = ππ£ ππ‘ π + π π£ 2 π Definition: The term ππ£ ππ‘ π is called the tangential component of acceleration. Definition: The term π π£ 2 π is called normal component of πΌβ²β². End of the lecture 77 of 162 Lecture notes on MTH352: Differential Geometry Lecture # 14 Contents: ο Frenet Apparatus For A Regular Curve ο Computing Frenet Frame ο The Spherical Image ο Cylindrical Helix ο Conclusion 78 of 162 Lecture notes on MTH352: Differential Geometry In this lecture we will learn to calculate the Frenet Frame for arbitrary speed curves. For this purpose the following result is useful. Theorem: Let πΌ be a curve in πΉπ. Then π= πΌ , βπΌ β² β π =π΅×π π΅= πΌβ² × πΌ β²β² , βπΌ β² × πΌ β²β² β βπΌ β² × πΌ β²β² β π = βπΌ β² β3 (πΌ β² × πΌ β²β² ). πΌ β²β²β² π= βπΌ β² × πΌ β²β² β2 See lecture video of the book for the proof of this. Exercise: Consider πΌ a curve in πΉπ given by πΌ(π‘) = (3π‘ β π‘ 3 , 3π‘ 2 . 3π‘ + π‘ 3 ) Find the Frenet apparatus for πΌ. Applications Of Frenet Formulas Given a curve π½ we construct a new curve π½Μ . Such that π½Μ has some properties of π½ and helps us to understand those properties in a better way. Spherical Image: Let π½: πΌ β πΉπ be a unit speed curve. The spherical image π½ is the curve π β π. That is π: πΌ β πΉπ is given by π(π‘) = (π£1 (π‘), π£2 (π‘), π£3 (π‘)) where (π£1 (π‘), π£2 (π‘), π£3 (π‘))π½(π‘) 79 of 162 Lecture notes on MTH352: Differential Geometry is tangent at point π½(π‘) of the curve. Example: The unit helix is given by π π ππ π½(π‘) = (π πππ , π π ππ , ) π‘ π‘ π where π = βπ2 + π 2 . Then the spherical image π of π½ is given by: Curvature Of The Spherical Image Let π be the spherical image of the unit speed curve π½. Let us calculate the curvature of π. 80 of 162 Lecture notes on MTH352: Differential Geometry End of the lecture Lecture # 15 Contents: ο Cylindrical Helix ο Covariant Derivatives ο Euclidean Coordinate Representation ο Properties Of The Covariant Derivative ο The Vector Field Of Covariant Derivatives 81 of 162 Lecture notes on MTH352: Differential Geometry Cylindrical Helix Sometimes the geometrical information like torsion and curvature are enough to tell what the curve is exactly going to be. Cylindrical is one of those examples where, by just looking at the π torsion and curvature, you can tell with exactness. In fact the ration π is some times very decisive in the case of some curves. We start with the definition of the cylindrical helix. Definition: A regular curve πΌ in πΉπ is a cylindrical helix provided the unit tangent π of πΌ has constant angle π with some fixed unit vector π; that is, π(π‘). π = πππ π. Movement of a cylindrical helix Assume that π½ is a unit-speed curve which is a cylindrical helix, that is, some fixed unit vector π. Lets consider the following function: π. π’ = cos π, for β(π ) = (π½(π ) β π½(0)). π π Theorem: A regular curve πΌ with π > 0 is a cylindrical helix if and only if the ration π is constant. 82 of 162 Lecture notes on MTH352: Differential Geometry See book for proof. In summary we have the following information directly about the curve by just looking at the Frenet apparatus. Covariant Derivative We have seen the definition of vector field. For an example of a vector field see the diagram. As we can see that the arrows are different at each and every point. Our next aim is understand the rate of change of vector field at any point. For this we define covariant derivative. Definition: Let π be a vector field on πΉπ . Let π£π be a tangent vector to πΉπ at point π. The covriant derivative of π with respect to ππ is the tangent vector π»π£ π = π(π + π‘π)β² (0) at the point π. The symbol π» is called nabla. 83 of 162 Lecture notes on MTH352: Differential Geometry In fact π»π π measures the initial rate of change of π(π) as π moves in the direction of π. Example: Wind Vector field One of the practical examples of a vector fields are wind vector field, which tells us the direction and magnitude of the wind at any point. Calculating the rate of change of this vector field is a crucial part of weather prediction. We use covariant derivative to calculate such quantities. Now lets see how to calculate the covariant derivative. Exercise: Consider a vector field π = π₯ 2 π1 + π¦π§π2 and π£ = (β1,0,2), Lemma: If ππ‘ π = (2,1,0). π = β π€π ππ is a vector field on πΉπ , and π is a tangent vector at π, then βv W = β π£[π€π ]ππ (π) See book for proof. From the properties of directional derivative we can drive Theorem: Let π and π be tangent vectors to πΉπ at π, and let π and π be vector fields on πΉπ . Then for numbers π, π and functions π. (1) βππ£+ππ€ π = πβπ£ π + π βπ€ π (2) βπ£ (ππ + ππ) = π βπ£ π + π βπ£ π (3) βπ£ (πβ) = π£[π]π(π) + π(π)βπ£ π 84 of 162 Lecture notes on MTH352: Differential Geometry (4) π£[π. π] = βπ£ π. π (π) + π(π). βπ£ (π) The vector field of covariant derivatives Definition: Let π and π be vector fields on πΉπ. The covariant derivative βπ π of π with respect to π is a vector field defined by (βπ π)(π) = βπ(π) π Corollary: If = β π€π ππ , then βπ π = β π[π€π ]ππ ππ Example: We always use βπ π = β π[π€π ]ππ and ππ [π] = ππ₯ : π If π = π₯π¦π1 β π π§ π3 and π = π§π1 + (π₯ β π¦)π2 , then π[π₯π¦] = π§π1 [π₯π¦] + (π₯ β π¦)π2 [π₯π¦] = π¦π§ + π₯(π₯ β π¦) π[π π§ ] = π§π1 [π π§ ] + (π₯ β π¦)π2 [π π§ ] = 0 and therefore βπ π = (π¦π§ + π₯(π₯ β π¦))π1 Corollary: Let π, π, π and π be vector fields on πΉπ . Then (1) βππ+ππ π = πβπ£ π + π βπ π for all functions π and π (2) βπ (ππ + ππ) = πβπ π + π βπ π for all numbers π and π (3) βπ (ππ) = π[π]π + π βV Y for all functions π (4) π[π. π] = βπ π. π + π. βπ π See book for proof. 85 of 162 Lecture notes on MTH352: Differential Geometry End of the lecture Lecture #16 Contents: ο From Curves to Space ο Frame Fields ο Coordinate Functions From Curves To Space 86 of 162 Lecture notes on MTH352: Differential Geometry In 1847 Frenet formulas were discovered by Jean Frédéric Frenet (7 February 1816 β 12 June 1900) in 1847. In 1851 Joseph Alfred Serret (August 30, 1819 - March 2, 1885) discovered the same formulas independently. So we can call them FrenetβSerret Formulas. From Curves to Surfaces 0 πβ² (πβ²) = (βπ π£ 0 π΅β² π π£ 0 βππ£ 0 π ππ£ ) (π) 0 π΅ So far we have discussed the geometry of curves in π 3 . We mainly use the Frenet frame π, π , π΅. Then we calculated the rate of change π β² , π β² , π΅ β² using Frenet formulas. Now to go from curves to surfaces we use the same idea, that is, Assign a frame to each point of the surface And then we calculate the changes in the frame in terms of the frame itself. The following theorem helps Theorem: Let π1 , π2 , π3 be a frame at a point π of πΉπ . If π£ is any tangent vector to πΉπ at π, then 87 of 162 Lecture notes on MTH352: Differential Geometry π£ = (π£. π1 )π1 + (π£. π2 )π2 + (π£. π3 )π3 . Instead of defining frame field on the geometrical object, like sphere, we carry on this idea for the Euclidean space πΉπ . Then the restriction of the idea to geometries in πΉπ is simple. Frame Fields We can extend the idea of point wise operation to define the following terms for vector fields. Definition: Let π and π be vector fields in πΉπ , then we can define their dot product π. π as follows: π. π (π) = π(π). π(π) Similarly we can define the following: Definition: Let π be a vector field then we define βπβ to be a real valued function on π 3 βπβ: π 3 β π given by βπ(π)β = βπ(π). π(π) The norm βπβ is not defined around 0. Definition: Three vector fields πΈ1, πΈ2 , πΈ3 on πΉπ forms a frame field on πΉπ if they satisfy πΈπ . πΈπ = πΏππ for 1 β€ π, π β€ 3 Thus at each point π the vectors πΈ1 (π), πΈ2 (π), πΈ3 (π) do in fact form a frame since they have unit length and are mutually orthogonal. 88 of 162 Lecture notes on MTH352: Differential Geometry Example ( Cylindrical Frame Field) Before defining the cylindrical frame we recall what are cylindrical coordinates. Cylindrical Coordinates: Let π , π, π§ be the cylindrical coordinate functions of πΉπ . We pick unit vector field in the direction in which each coordinate increases πΈ1 = cos π π1 + sin π π2 π π πΈ2 = cos (π + ) π1 + sin (π + ) 2 2 = β sin π π1 + cos π π2 πΈ3 = π3 It is easy to check that πΈπ . πΈπ = πΏππ So (πΈ1 , πΈ2 , πΈ3 ) = (cos π π1 + sin π π2 , β sin π π1 + cos π π2 , π3 ) forms a frame field on πΉπ called cylindrical frame field. Now the question is are there points where the cylindrical frame is not defined? 89 of 162 Lecture notes on MTH352: Differential Geometry Yes, the points on π§ βaxis does not have a cylindrical frame defined on them. Cylindrical coordinates are useful in many practical situations, like, designing of a robotic arm etc. Spherical Coordinates: Why spherical coordinates? Just like cylindrical coordinates, there are situations in real like that Could be dealt in a good way if we use the spherical coordiantes. From Cylindrical Frame to Spherical Frame πΉ2 = πΈ2 πΉ1 = cos π πΈ1 + sin π πΈ3 π π πΉ3 = cos (π + 2 ) πΈ1 + sin (π + 2 ) πΈ3 = β sin π πΈ1 + cos π πΈ3 90 of 162 Lecture notes on MTH352: Differential Geometry So we have πΉ2 = πΈ2 πΉ1 = cos π πΈ1 + sin π πΈ3 πΉ3 = β sin π πΈ1 + cos π πΈ3 Lemma: Let πΈ1, πΈ2, πΈ3 be a frame field on πΉπ. 1) Let π be a vector field, then π = β ππ πΈπ . Where the functions ππ are called the coordinate functions of π with respect to πΈ1 , πΈ2 , πΈ3 . 2) If π = β ππ πΈπ and π = β ππ πΈπ then π. π = β ππ ππ And βπβ = 1 (π. π)2 = 1 2 2 (β ππ ) See book for proof. So far we have seen many frame fields. But for a given problem we would like to choose a frame field that is especially easy to work with. The calculations could be very easy with the right selection of frame field. 91 of 162 Lecture notes on MTH352: Differential Geometry End of the lecture Lecture # 17 Contents: ο Connection Form ο Connection Equations 92 of 162 Lecture notes on MTH352: Differential Geometry ο How To Calculate Connection Forms Connection Form Covariant derivative of a frame field Let {πΈ1 , πΈ2 , πΈ3 } be a frame field on πΉπ . Let π£ be an arbitrary tangent vector at point π of πΉπ . Then we can write βv E1 = c11 E1 + c12 E2 + c13 E3 βv E2 = c21 E1 + c22 E2 + c23 E3 βv E3 = c31 E1 + c32 E2 + c33 E3 We can use orthonormal expansion: Theorem: Let π1, π2, π3 be a frame at a point π of πΉπ . If π£ is any tangent vector to πΉπ at π, then π£ = (π£. π1 )π1 + (π£. π2 )π2 + (π£. π3 )π3 . To express each coefficient πππ as πππ = βπ£ πΈπ . πΈπ (π) for 1 β€ π, π β€ 3 Note that the coefficients πππ depend only the vector π£, so we can write them πππ (π£) = πππ (π£) = βπ£ πΈπ . πΈπ (π) for 1 β€ π, π β€ 3 The function πππ (π£) given by πππ (π£) = βπ£ πΈπ . πΈπ (π) for 1 β€ π, π β€ 3 is a function of tangents vectors on πΉπ , and the output is a real number. Have we met any function that satisfy this property? Yes, we have. Definition: A 1 βform π on πΉπ is a real-valued function on the set of all tangent vectors to πΉπ such that π is linear at each point, that is, 93 of 162 Lecture notes on MTH352: Differential Geometry π(ππ + ππ) = π π(π) + ππ(π) For any numbers π, π and tangent vectors π , π at the same point of πΉπ . Lemma: Let πΈ1, πΈ2, πΈ3 be a frame field on πΉπ. For each tangent vector π£ to πΉπ at point π, let πππ (π£) = βπ£ πΈπ . πΈπ (π) for 1 β€ π, π β€ 3 Then each πππ is a one form, and πππ = βπππ . See book for proof. Theorem: Let πππ be the connection forms of the frame field πΈ1, πΈ2 , πΈ3 on πΉπ. Then for any vector field π on πΉπ , βπ πΈπ = β πππ (π)πΈπ We call the equations π»π πΈπ = β πππ (π)πΈπ Connection equations. The matrix of connection forms We can write the connection forms πππ in matrix form π11 π = (π21 π31 π12 π22 π32 π13 π13 ) π33 But since πππ = βπππ , so for π = π we have πππ = βπππ . Hence we have πππ = 0 for 1 β€ π β€ 3. Hence the matrix of connection forms becomes 0 π = (π21 π31 π12 0 π32 π13 π13 ) 0 94 of 162 Lecture notes on MTH352: Differential Geometry Hence we can write connection equations as + π12 (V)E2 + π13 (π)E3 π»V E1 = π»V E2 = π21 (π )E1 + π23 (π)E3 π»π E3 = π31 (π )E1 + π32 (π ) E2 How To Calculate Connection Forms Let {πΈ1 , πΈ2 , πΈ3 } be an arbitrary frame field and let {π1 , π2 , π3 } be natural frame field. We can write πΈ1 = π11 π1 + π12 π2 + π13 π3 πΈ2 = π21 π1 + π22 π2 + π23 π3 πΈ3 = π31 π1 + π32 π2 + π33 π3 here the coefficients can be calculated using orthonormal expansion πππ = πΈπ . ππ Here πππ are real valued functions on πΉπ . From here we can define the attitude matrix of the frame. π11 π π΄ = ( 21 π31 π12 π22 π32 π13 π23 ) π33 Now we define differential of π΄ = (πππ ) to be ππ΄ = (ππππ ). So ππ΄ is a matrix whose entries are 1forms. We can calculate the connection forms in terms the attitude matrix as follows: Theorem: If π΄ = (πππ ) is an attitude matrix of a frame field {πΈ1, πΈ2, πΈ3}. Let πππ be the connection forms of this frame field, then πππ = ππ΄ π‘π΄ Or equivalently πππ = β ππππ πππ See book for proof. 95 of 162 Lecture notes on MTH352: Differential Geometry Connection forms for the cylindrical frame We know that So the attitude matrix of the cylindrical frame is cos π π΄ = (β sin π 0 sin π cos π 0 0 0) 1 Hence 0 π = (βπ π 0 ππ 0 0 0 0) 0 ππ 0 0 0 0) 0 Now the connection forms for cylindrical frame are 0 π = (βπ π 0 Hence the connection equations become βπ πΈ1 = ππ π2 βπ πΈ2 = βπ π π1 βπ πΈ3 = 0 Since π is a real valued function on πΉπ . So ππ(π) = π[π] Hence the equations βπ πΈ1 = ππ π2 βπ πΈ2 = βπ π π1 βπ πΈ3 = 0 Becomes βπ πΈ1 = π[π] π2 βπ πΈ2 = βπ[π] π1 βπ πΈ3 = 0 96 of 162 Lecture notes on MTH352: Differential Geometry End of the lecture Lecture # 18 Contents: 97 of 162 Lecture notes on MTH352: Differential Geometry ο Dual Forms ο Cartan Structural Equations ο Structural Equations For Spherical Frame Dual Forms The dual forms of a frame field Any frame field can be described in terms of 1-forms. Definition: If πΈ1, πΈ2, πΈ3 is a frame field on π 3, then the dual 1-forms π1, π2, π3 of the frame field are the 1-forms such that ππ (π) = π£. πΈπ (π) For each tangent vector π to πΉπ at point π. Here ππ are 1-forms since 1)- ππ takes vectors to real numbers 2)- ππ are linear Another property of π½π ππ (πΈπ ) = πΏππ for 1 β€ π, π β€ 3 here πΏππ is Kronecker delta function. Orthonormal expansion formula for vector field Let π be a vector field then we can write in terms of frame field {πΈ1 , πΈ2 , πΈ3 } π = β ππ πΈπ In terms of dual 1-forms the orthonormal expansion can be written as π = β ππ (π)πΈπ 98 of 162 Lecture notes on MTH352: Differential Geometry Expressing 1-forms in terms of dual forms In case of natural frame field π1 , π2 , π3 the dual 1-forms are ππ₯1 , ππ₯2 , ππ₯3 . We also know that any 1-form π can be written as π = β π(ππ ) ππ₯π Question: Is it true in general? Answer: Yes, by the following result. Theorem: Let π1 , π2 , π3 be the dual 1-forms of a frame field πΈ1, πΈ2, πΈ3 . Then any 1-form π on π 3 has a unique expression π = β π(πΈπ )ππ Formula for π½π We know that any frame field {πΈ1 , πΈ2 , πΈ3 } can be written in terms of natural frame field, πΈπ = β πππ ππ for 1 β€ π, π β€ 3 We have the following result for dual 1-forms π1 , π2 , π3 of πΈ1 , πΈ2 , πΈ3 as: Fact: ππ = β πππ ππ₯π πππ 1 β€ π, π β€ 3 Cartan Structural Equations Exterior derivatives Exterior derivatives of 1-forms and connection forms were given by Cartan. Theorem (Cartan Structural Equations) Let πΈ1 , πΈ2 , πΈ3 be a frame field on πΉπ with dual forms π1 , π2 , π3 and connection forms πππ (1 β€ π, π β€ 3). 99 of 162 Lecture notes on MTH352: Differential Geometry The exterior derivatives of these forms satisfy: (1) The first structural equations: πππ = β πππ β§ ππ . (2) The second structural equation: π πππ = β πππ β§ πππ . Proof is not included. Matrix notation We define some notations and remarks to make these structural equations more easy to understand. We define π11 π = (π21 π31 π12 π22 π32 π13 π23 ) π33 π11 π΄ = (π21 π31 π12 π22 π32 π13 π23 ) π33 and we also know the attitude matrix And we can write the dual forms π1 , π2 , π3 in matrix form as π1 π = (π2 ) π3 And the dual of natural frame fields as: π π₯1 ππ = (π π₯2 ) π π₯3 Using these notations we can write the following fact Fact: ππ = β πππ ππ₯π πππ 1 β€ π, π β€ 3 In the following form: 100 of 162 Lecture notes on MTH352: Differential Geometry π11 π1 (π2 ) = (π21 π31 π3 π12 π22 π32 π π₯1 π13 π23 ) (π π₯2 ) π33 π π₯3 In short form it becomes: π =π΄ππ So we can write the Cartan structural equations as First structural equation πππ = β πππ β§ ππ can be written as π π1 π11 (π π2 ) = π π = β πππ β§ ππ = (π21 π31 π π3 π12 π22 π32 π13 π1 π23 ) (π2 ) π33 π3 Similarly The second structural equation π πππ = β πππ β§ πππ . can be written as ππ =ππ Example: The Spherical Frame Field The spherical Frame field is given by πΉ1 = cos π cos π π1 + sin π sin π π2 + sin π π3 πΉ2 = β sin π π1 + cos π π2 πΉ3 = β cos π sin π π1 β sin π sin π π2 + cos π π3 So the attitude matrix is 101 of 162 Lecture notes on MTH352: Differential Geometry cos π cos π π΄ = ( β sin π β cos π sin π sin π sin π cos π β sin π sin π sin π 0 ) cos π Now if π1 , π2 , π3 are dual 1-forms of the spherical frame field then: π = π΄ ππ Calculation of π½ Now to calculate ππ₯1 , ππ₯2 , ππ₯3 we consider x1 = π cos π cos π π₯2 = π cos π sin π π₯3 = sin π We calculate ππ₯1 ππ₯2 , ππ₯3 use the following to get π. ππ = ππ ππ ππ ππ₯1 + ππ₯2 + ππ₯3 ππ₯1 ππ₯2 ππ₯3 Calculation of π Now we will cos π cos π π΄ = ( β sin π β cos π sin π sin π sin π cos π β sin π sin π sin π 0 ) cos π Now we after calculating ππ΄, the matrix of π is given by 0 π=( cos π ππ 0 ππ sin π ππ ) 0 First Structural Equation for Spherical Frame So we have (1) The first structural equations: πππ = β πππ β§ ππ . (2) The second structural equation: π πππ = β πππ β§ πππ . 102 of 162 Lecture notes on MTH352: Differential Geometry Where ππ π = (π cos π) π ππ 0 πππ π=( cos π ππ 0 ππ sin π ππ) 0 Recipe for dealing with geometrical problems in πΉπ Procedure 1) Select a frame field 2) Find its dual 1-forms 3) Calculate connection forms 4) Apply structural equations 5) Interpret the results End of the lecture Lecture # 19 103 of 162 Lecture notes on MTH352: Differential Geometry Contents: ο Surfaces in π 3 ο The Sphere ο Construction Of A Surface Surfaces in πΉπ Coordinate patches A coordinate patch π₯ : π· β π 3 is a ο Mapping ο one-to-one ο regular of an open set π· of π 2 into π 3 . Proper Patches A coordinate patch π: π· β πΉπ is called proper if its inverse function πβ1 : π(π·) β πΉπ is continuous. 104 of 162 Lecture notes on MTH352: Differential Geometry What is a surface roughly? Every small region of surface looks like a region in π 2 , or image of patch. Definition: Let π: π· β πΉπ be a patch with π(π·) β π is a patch in π. Definition: A neighborhood in π β πΉπ of a point π is the subset of all points in π at distance less that π, for some real number π > 0. Surface A surface in πΉπ is a subset π of πΉπ such that for each point π of π there exist a proper patch in π whose image contains a neighborhood of π in π. Sphere: A sphere Ξ£ is the set of all point in πΉπ whose distance from origin is 1, that is all points π such that π βπβ = (πππ + πππ + πππ )π = π Patch containing north pole Consider the point (0,0,1) on sphere, known as North pole. Let π· = {(π’, π£): π’2 + π£ 2 < 1} Define 105 of 162 Lecture notes on MTH352: Differential Geometry π₯: π· β Ξ£ such that x(u, v) = (u, v, β1 β π’2 β π£ 2 ) Lets see if π₯ is a patch or not. For this check the following: ο Mapping: since the coordinate functions are differentiable ο One to one To check if the patch π₯ is regular we need to calculate the Jacobian of this map: Similarly we can define the patch for the south pole, that is, the point (0,0, β1). From the sphere example we can guess the following The map π₯: π· β π 3 given by π₯(π’, π£) = (π’, π£, π(π’, π£)) for any differentiable function π(π’, π£) is a patch. These kind of patches are called the Monge patches. Simple Surface Let π₯: π· β πΉπ be patch. Then the image π₯(π·) satisfies definition of a surface. This type of surface is called simple surface. Example: Let π(π’, π£) be a real valued differentiable function from π 2 to π 3 . The graph of z = π(π’, π£) is the set of all points in π 3 satisfying π§ β π(π’, π£) = 0. In fact π is the image of the Monge patch. 106 of 162 Lecture notes on MTH352: Differential Geometry So π is a simple surface. End of the lecture Lecture 20 Contents: 107 of 162 Lecture notes on MTH352: Differential Geometry ο Implicitly Defined Surfaces ο Surfaces of Revolution ο Properties Of Patches Implicitly Defined Surfaces Let π be a real valued function from π 3 to π . For any real number π, the equation π(π₯, π¦, π§) = π defines the set of all points π of π 3 such that π(π) = π Example: Let g(π₯, π¦, π§) = π₯ 2 + π¦ 2 + π§ 2 and let π = 1 then the set of all points π such that π(π) = 1 forms a sphere of radius 1. Example: Let π(π₯, π¦, π§) = π§ and π = 1 then the set of all points π in πΉπ such that π(π) = 1 is π§=1 forms a plane parallel to π₯π¦ βplane. An Implicit Construction Theorem: Let π: πΉπ β πΉ be a differentiable real valued function on πΉπ , let π be a number. The subset π: π(π₯, π¦, π§) = π of πΉπ is a surface if the differential ππ is not zero at any point of π. See book for proof. 108 of 162 Lecture notes on MTH352: Differential Geometry Monge Patches: The map π₯: π· β π 3 given by π₯(π’, π£) = (π’, π£, π(π’, π£)) for any differentiable function π(π’, π£) is a patch. These kind of patches are called the Monge patches. Sphere is a surface A sphere Ξ£ with center π = (ππ , ππ , ππ ) and radius r is the set of all point in πΉπ whose distance from π is π, that is all points π = (π, π, π) such that π βπ β πβ = ((π β ππ )π + (π β ππ )π + (π β ππ )π )π = ππ Let π(π₯, π¦, π§) = (π β ππ )π + (π β ππ )π + (π β ππ )π Then π(π₯, π¦, π§) = π 2 defines sphere with center π and radius π. We apply the previous theorem to shwo if this sphere is a surface or not. Surfaces of Revolution If πΆ is defined by π(π₯, π¦) = π, where π: π 2 β π is a differentiable function. Also πΆ is regular curve. Then the surface of revolution is defined by π(π₯, π¦, π§) = π (π₯, βπ¦ 2 + π§ 2 ) = π More surfaces of revolution 109 of 162 Lecture notes on MTH352: Differential Geometry We can generate a torus using the idea of surface of revolution. Lets see why we need properness in defining surface patches. Example: Let π· = {(π’, π£): βπ < π’ < βπ, < 0π£ < 1} and π₯: π· β π 3 πππππππ ππ¦ π₯(π’, π£) = (sin π’, sin 2π’ , π£) Here π₯ is a patch. But the image is not a surface. End of the lecture Lecture 21 110 of 162 Lecture notes on MTH352: Differential Geometry Contents: ο Parameter Curves on Surfaces ο Parametrizations ο Torus ο Ruled Surface Parameter Curves on Surfaces Parameter Curves Let π₯: π· β π 3 be a coordinate patch. For each point (π’0 , π£0 ) in π· the curve π’ β π₯(π’, π£0 ) is called the π’ βparameter curve, π£ = π£0 , of π₯. Similarly the curve π£ β π(ππ , π) Is called π£ βparameter curve, π’ = π’0 , of π. ο The image π₯(π·) of π· is the union of all such curves. ο Each point of π₯(π·) lies on exactly one curve of each type. Partial Velocities If π₯: π· β π 3 is a patch, for each point (π’0 , π£0 ) in π·. ο The velocity vector at π’0 of the π’ βparameter curve, π£ = π£0 , is denoted by π₯π’ (π’π , π£0 ). 111 of 162 Lecture notes on MTH352: Differential Geometry ο The velocity vector at π£0 of the v βparameter curve, u = π’0 , is denoted by ππ (π’π , π£0 ). The vectors ππ (π’0 , π£0 ) and ππ (π’0 , π£0 ) are called the partial velocities of π at (π’0 , π£0 ). Calculation of partial velocities Given a coordinate patch π₯: π· β π 3 π₯(π’, π£) = (π₯1 (π’, π£), π₯2 (π’, π£), π₯3 (π’, π£)) Then the partial velocities functions are ππ₯1 ππ₯2 ππ₯3 π₯π’ = ( , , ) ππ’ ππ’ ππ’ (π’0 ,π£0) π₯π£ = ( ππ₯1 ππ₯2 ππ₯3 , , ) ππ£ ππ£ ππ£ (π’0 ,π£0 ) Where (π’0 , π£0 ) is point of application and can be omitted. Geographical patch in a sphere Let Ξ£ be a sphere of radius π and center (0,0,0) in π 3 . Let π₯ be a mapping defined on π· = {(π’, π£): βπ < π’ < π, β π π <π£< } 2 2 Defined by π₯(π’, π£) = (π cos π£ cos π’, π cos π£ sin π’ , π sin π£) 112 of 162 Lecture notes on MTH352: Differential Geometry Parametric Curves on the sphere ο The π’ βparameter curve, π£ = π£0 , is a circle with constant value of π§. It is the parallel at latitude π£0 . ο The π£ βparameter curve, π’ = π’0 , is a half circle. It is the meridian of longitude π’0 . Parametrizations Definition: A regular mapping π₯: π· β π whose image lies in a surface π is called a parametrization of the region π₯(π·) in π. Properties of parametrization ο A 1-1 parametrization is a patch ο In parametrization of a curve mean a surjective map onto the whole curve. But in case of surface parametrization it might be only the region π(π·) of π. When a mapping is a parametrization? Given a mapping π: π· β πΉπ . To check whether it is a parametrization of a surface π β πΉπ we check the following: ο π(π«) β π΄ ? First determine that π₯(π·) β π or not. For example if surface is defined by π: π = π. Then 113 of 162 Lecture notes on MTH352: Differential Geometry "π₯(π·) is contained in π if and only if π(π₯(π·)) = π.β ο π is regular? The map π₯: π· β π is regular if π1 ππ₯1 | ππ × ππ = ππ’ | ππ₯1 ππ£ π2 ππ₯2 ππ’ ππ₯2 ππ£ π3 ππ₯3 | ππ’ | ππ₯3 ππ£ is non zero. Parametrization for sphere Given mapping π₯(π’, π£) = (π cos π£ cos π’, π cos π£ sin π’ , π sin π£) Question: Is it parametrization of sphere Ξ£ of radius π and center (0,0,0)? Lets check the two conditions to see. ο π(π«) β πΊ ? The equation of sphere is π(π₯, π¦, π§) = π₯ 2 + π¦ 2 + π§ 2 = π 2 Then π₯(π·) β π if and only if (π(π₯)) = π 2 . ο π is regular? π π Here π· = {(π’, π£): βπ < π’ < π, β 2 < π£ < 2 } To check if π is a patch we need to only check if π is 1-1 or not. Surface of revolution Let the curve πΆ has parametrization πΌ(π’) = (π(π’), β(π’), 0) here β > 0. As the point on the curve is rotated then it reaches the point π₯(π’, π£) = (π(π’), β(π’) cos π£ , β(π’) sin π£) 114 of 162 Lecture notes on MTH352: Differential Geometry Question: Is it a parametrization of the surface of revolution? Answer: check the following two. ο π(π«) β π΄? ο π is regular? Torus Definition: A torus of revolution π is the surface of revolution of a circle πΆ. Example: Suppose that πΆ is the circle in the π₯π§ βplane with radius π > 0 and center (π , 0,0). We rotate the circle around π§ βaxis here π > π. Here parametrization of πΆ is πΌ(π’) = (π + π cos π’ , π sin π’) This yields the parametrization π₯(π’, π£) = ((π + π cos π’ ) cos π£ , (π + π cos π’ ) sin π£ , π sin π’) Definition: A ruled swept out by a straight line πΏ moving along a curve π½. surface is a surface The various positions of the ruling line πΏ are called the rulings of the surface. Such a surface always has a ruled parametrization π₯(π’, π£) = π½(π’) + π£πΏ(π’) here π½ is base curve and πΏ is the director curve. 115 of 162 Lecture notes on MTH352: Differential Geometry We put conditions on π½ and πΏ to make π₯ a parametrization. End of the lecture Lecture # 22 116 of 162 Lecture notes on MTH352: Differential Geometry Contents: ο Coordinate Expressions ο Curves on a Surface ο Differentiable Functions Definition: A regular mapping π₯: π· β π whose image lies in a surface π is called a parametrization of the region π₯(π·) in π. In parametrization of a curve mean a surjective map onto the whole curve. But in case of surface parametrization it might be only the region π(π·) of π. Coordinate Expressions Suppose that π is a real valued function on a surface π. If π₯: π· β π is a coordinate patch of π then the composite function π(π₯) is called the coordinate expression for π. The coordinate expression π is just a real-valued function π(π₯): π· β π The function π is differentiable provided all its coordinate expressions are differentiable. 117 of 162 Lecture notes on MTH352: Differential Geometry For a function πΉ: π π β π each patch of π π₯: π· β π gives a coordinate expression of πΉ, given by π₯ β1 (πΉ): π β π· here π is an open set containing points from the set {π β π π : πΉ(π) β π₯(π·)} The function πΉ is differentiable if every coordinate expression for πΉ is differentiable. Every set π is open set. Definition: A curve on a surface π is a differentiable function πΌ: πΌ β π from open interval πΌ to π. Coordinate functions of a curve Lemma: If πΌ is a curve πΌ: πΌ β π whose route lies in the image π₯(π·) of a single patch π₯, then there exist unique differentiable functions π1 , π2 on πΌ such that πΌ(π‘) = π₯(π1 (π‘), π2 (π‘)) πππ πππ π‘ Or πΌ = π₯(π1 , π2 ). See book for proof 118 of 162 Lecture notes on MTH352: Differential Geometry Map of the region Let π₯: π· β π be a patch in π. We take the domain π· as the map of the region π₯(π·). The functions π₯ and π₯ β1 establish a one-to-one correspondence between objects in D and the objects in π(π·). Too many patches To check the differentiability of functions πΉ: π π β π we need to check differentiability at each and every patch. The following theorem simplify the situation. Theorem: Let π be a surface in π 3. If πΉ: π π β π 3 is a (differentiable) mapping whose image lies in π, then considered as a function πΉ: π π β π into π, πΉ is differentiable. Overlapping patches Corollary: Let π₯: π· β π and π¦: πΈ β π be patches in π, where π β π 3. Assume that images of π₯ and π¦ overlap, that is, π₯(π·) β π¦(πΈ) β π. Then each of π₯ β1 π¦ and π¦ β1 π₯ is a differentiable mapping defined on an open set in π 2 . Here the mapping π₯ β1 π¦ is defined on the set π₯ β1 (π₯(π·) β π¦(πΈ)). Similarly the mapping π¦ β1 π₯ is defined on the set π¦ β1 (π₯(π·) β π¦(πΈ)). 119 of 162 Lecture notes on MTH352: Differential Geometry How to show that π: π β πΉ is differentiable By definition π: π β πΉ is differentiable if π(π₯) is differentiable for all patches π₯: π· β π. Assume that we show that π(π₯) is differentiable for a patch π₯. Let π¦ be another patch. Then π(π₯) is differentiable and π₯ β1 π¦ is also differentiable. Therefore the composition ππ₯π₯ β1 π¦ is differentiable. The function ππ₯π₯ β1 π¦ is the restriction of π(π¦) to only a small region of π, where π₯ and π¦ overlap. But π is covered by such regions, therefore π(π¦) is differentiable in an open region around every point of π End of the lecture Lecture #23 120 of 162 Lecture notes on MTH352: Differential Geometry Contents: ο Tangents ο Tangent Vector Fields ο Gradient Vector Field Tangents Tangents to a surface Definition: Given an surface π. Let π is a point of π. Then a tangent vector to π 3 at π is tangent vector to π if π£ is velocity of some curve in π. The set of all tangent vectors to π at π is called the tangent plane of π at point π and is denoted by ππ (π). Fact: At each point the tangent plane ππ (π) is actually a 2-dimensional vector subspace of the tangent space ππ (π 3 ). Lemma: Let π be a surface in π 3 and π be a point on π. Let π₯ be a patch of π such that π₯(π’0 , π£0 ) = π. A tangent vector to π 3 is a tangent vector to π at point π if and only if π£ can be written as a linear combination of π₯π’ (π’0 , π£0 ) and π₯π£ (π’0 , π£0 ). See book for proof. Tangent plane as approximation to π΄ at point π We can think ππ (π) is linear approximation of surface near point π. 121 of 162 Lecture notes on MTH352: Differential Geometry Euclidean and tangent vector fields Definition: A Euclidean vector field on π is a function π that maps every point π of π to a tangent vector π(π) to πΉπ at point π. Definition: A tangent vector field on π is a Euclidean vector field π on π that maps every point π of π to a tangent vector V(π) at π. Most of the time a Euclidean vector field are defined only on a small region of π. Normal vector field A tangent vector π§ to π 3 at point π is normal to π if π§ is orthogonal to the tangent plane ππ (π 3 ). A Euclidean vector field π is normal vector field on π if π(π) is normal to π for every point π of π. It is easy to deal with tangent and normal vector fields for surfaces given in implicit form. To see this we define: Gradient Vector Field Gradient vector field Let π be a surface given in implicit form π: π = π. The gradient vector field is the Euclidean vector field βπ that maps every point π of π to 122 of 162 Lecture notes on MTH352: Differential Geometry βπ (π) = β ππ (π) ππ (π) ππ₯π Question: What does the gradient vector field look like? Answer is given by the following Lemma. Lemma: Let π: π = π be a surface in implicit form. Then the gradient vector field has the following properties: ο A normal vector field on π. ο Nonzero at every point of π. Directional derivatives ο Let π β π be a point on surface and π£ β ππ (π) be a tangent vector. Let π: π β π be a real valued function on π. ο The directional derivative of π with respect to π£, written as π£[π], is defined as ο π£[π] = ππ(πΌ) ππ‘ (0) ο for any curve πΌ in π with πΌ(0) = π and πΌ β² (0) = π£. ο Directional derivative has the same properties as we studied for π 3 . End of the lecture Lecture # 24 123 of 162 Lecture notes on MTH352: Differential Geometry Contents: ο Differential Forms ο Exterior Derivatives ο Differential Forms On The Euclidean Plane ο Closed And Exact Forms Differential Forms Let π be a surface in π 3 ο A 0 βform on π is a function π: π β π ο A 1 βform on π is a real-valued function π on the tangent vectors on π, such that π is linear on ππ (π) for every π β π. We defined 2-form and 3-form on π 3 in the following way. But we didβt define them properly. Intuitively what is a 2-form. Definition A 2 βform π on a surface π is a real-valued function on all ordered pairs of tangent vectors π£, π€ to π such that 1. π(π£, π€) is linear in π£ and in π€ 2. π(π£, π€) = βπ(π€, π£) Definition: A π βform is a form of degree π. Fact: Since surface is two dimensional hence all π βforms with π > 2 are zero. Properties of forms ο Two π βforms, for π = 0,1,2, can be added. 124 of 162 Lecture notes on MTH352: Differential Geometry ο A 1 βform can be evaluated on a vector field π. ο Similarly a 2 βform can be added evaluated on pair of vector fields π, π. ο Second axiom of the definition of the 2 βform implies that π(π£, π£) = 0. π βforms and determinants Let π be a 2 βform on surface π. Let π£, π€ be two linearly independent tangent vectors at some point of π. Then π(ππ£ + ππ€, ππ£ + ππ€) = | π π π |. π See book for proof. Wedge product ο Product of a π βform and a π βform is a (π + π) βform. ο The product of a 0 βform and a 1 βform and 2 βform is usual multiplication. ο On surface the wedge product is zero if π + π > 2. So we need only the following definition: Definition: If π and π are 1 βforms on a surface π, the wedge product π β§ π is a 2 βform on π such that (π β§ π)(π£, π€) = π(π£)π(π€) β π(π€)π(π£) for all pairs π£, π€ of tangent vectors to π. Question : Why π β§ π is a 2 βform? Rule for interchanging the factors in a wedge product If π is a π1 βform and π is a π2 βform, then π β§ π = (β1)(π+π) π β§ π In case of surfaces sign changes only occur in case of multiplication of two 1 βforms. In particular for 1 βform π we have π β§ π = 0. Exterior Derivatives The differential calculus of forms is based on the exterior derivatives π. ο For 0 βform, that is a real-valued function π , the exterior ππ is defined as 125 of 162 Lecture notes on MTH352: Differential Geometry ππ = π£[π]. ο The exterior derivative of a π βform is a (π + 1) βform. ο For surface we need the following: Definition: Let π be a 1 βform on a surface π. Then the exterior derivative π π of π is a 2 βform such that for each patch π₯: π· β π, π π(π₯π’ , π₯π£ ) = π π (π(π₯π£ )) β (π(π₯π’ )). ππ’ ππ£ Definition: Let π be a 1 βform on a surface π. Then the exterior derivative π π of π is a 2 βform such that for each patch π₯: π· β π, π π(π₯π’ , π₯π£ ) = π π (π(π₯π£ )) β (π(π₯π’ )). ππ’ ππ£ Lemma: Let π be a 1 βform on π. If π₯ and π¦ are patches in π. If π₯ and π¦ are patches in π, then dx π = ππ¦ π on the overlap of π₯(π·) and π¦(πΈ). Differentiation Rule Let π be a surface in π 3 . Let π: π β π and π: π β π be a function. Then π(π(π)) = πβ² (π) ππ Differentiation of wedge product Let π, π, β be three functions on π and π is a 1 βform. Then ο π(ππβ) = πβ ππ + πβ ππ + ππ πβ ο π(π π) = π ππ β π β§ ππ 126 of 162 Lecture notes on MTH352: Differential Geometry ο (ππ β§ ππ)(π£, π€) = π£[π] π€[π] β π£[π] π€[π] Theoerem: If π is a real valued function on π, then π (π (π)) = 0. How to show that the forms are equal Two criteria Let π₯: π· β π be a patch in π. Then ο Two 1 βforms, π, π are equal on π₯(π·) if and only if π(π₯π’ ) = π(π₯π’ ) πππ π(π₯π£ ) = π(π₯π£ ) ο Two 2 βforms π, π are equal on π₯(π·) if and only if π(π₯π’ , π₯π£ ) = π(π₯π’ , π₯π£ ). Closed and exact forms Let π be a differential form. If the exterior derivative of π is zero, ππ = 0, then π is closed. If π is the exterior derivative of some form π, so that π can be written as π = π π, then π is exact. End of the lecture Lecture # 25 127 of 162 Lecture notes on MTH352: Differential Geometry Contents: ο Mappings of Surfaces ο Tangent Maps of Mappings ο Diffeomorphism Mappings of Surfaces Differential functions Let π and π be two surfaces in π 3 . Let πΉ: π β π be a function. Then πΉ is differentiable if the following condition is satisfied For each patch π₯: π· β π in π and π¦: πΈ β π in π the composite function π¦ β1 πΉπ₯ is Euclidean differentiable (defined on an open subset of π 2 ). πΉ is then called a mapping. The function π¦ β1 πΉπ₯ is defined at all points (π’, π£) of π· such that πΉ(π’, π£) lies in the image of π¦. FACT: It suffices to check enough patches to cover both π and π. Cylindrical Projection Let Ξ£ be a unit sphere in πΉπ with center at origin with north and south poles removed, that is, πΊ = {π© β πΉπ : π (π, π) = π\{(π, π, π), (π, π, βπ)} }. 128 of 162 Lecture notes on MTH352: Differential Geometry Let πΆ be the cylinder bases on the unit circle in the π₯π¦ plane. So πΆ is in contact with the sphere along the equator, that is πΆ is given by πΆ = {π β π 3 : π12 + π22 = 1} Cylindrical projection Define πΉ: Ξ£ β πΆ as follows. Start from a point π = (π1, π2 , π3 ) β Ξ£. Draw the line πΏ through (0,0, π3 ) and = (π1 , π2 , π3 ) . Let πΉ(π) be the intersection of πΏ with πΆ as seen in the figure. Question: Is πΉ a mapping? Consider the following patch π₯: π· β Ξ£ of the sphere which we discussed earlier π₯(π’, π£) = (π cos π£ cos π’, π cos π£ sin π’ , π sin π£) where π· = {(π’, π£): βπ < π’ < π, β π π < π£ < }. 2 2 Now consider a patch π¦: πΈ β πΆ of the cylinder given by π¦(π’, π£) = (cos π’ , sin π’ , π£) where πΈ = {(π’, π£): π < π’ < βπ, β 1 < π£ < 1}. Now from the definition of πΉ we get πΉ(π₯(π’, π£)) = (cos π’, sin π’ , sin π£). But π¦(π’, sin π£) = (cos π’, sin π’ , sin π£). Hence πΉ(π₯(π’, π£)) = π¦(π’, sin π£). Applying πΉ β1 we get π¦ β1 πΉπ₯(π’, π£) = (π’, sin π£). 129 of 162 Lecture notes on MTH352: Differential Geometry Hence πΉ: Ξ£ β πΆ is a mapping. Stereographic projection Let Ξ£ be a unit sphere with center (0,0,1). Delete the north pole from Ξ£, that is, consider Ξ£ = {π = (π1 , π2 , π3 ) β π 3 : π(π, (0,0,1)) = 1} \{(0,0,2)}. Let us identify π₯π¦-plane with π 2 , that is, π 2 = {(π1 , π2 , 0): π1 , π2 β π }. The function πΉ is given by π π1 π π2 πΉ(π1 , π2 , π3 ) = ( , ) π π where π and π are shown in the figure below. Using the similar triangles in the figure below (left), we get, π π = 2 2 β π3 So 2π1 2π2 πΉ(π1 , π2 , π3 ) = ( , ). 2 β π3 2 β π3 The function 2π1 2π2 πΉ(π1 , π2 , π3 ) = ( , ). 2 β π3 2 β π3 is differentiable. For any patch π₯: π· β Ξ£ in Ξ£ the function πΉ(π₯) is differentiable. Let π¦: π 2 β π 3 given by π¦(π1 , π2 ) = (π1 , π2 , π3 ) be the patch for the π₯π¦ β plane, then π¦ β1 is differentiable. Hence π¦ β1 πΉπ₯ is differentiable, making πΉ a mapping. 130 of 162 Lecture notes on MTH352: Differential Geometry Definition Let πΉ: π β π be a mapping of surfaces. Let π£ be a tangent vector to π such that there exist a curve πΌ in π and π£ = πΌ β² (0). Then F(πΌ) is a curve in π. We define the tangent map πΉβ of πΉ as πΉβ (π£) = πΉ(πΌ)β² (0) that takes tangent vectors of π to the tangent vectors of π. Fact1: The tangent map πΉβ is a linear map from ππ (π) to ππ(π) (π). Fact 2: Let πΉ: π β π and πΊ: π β π be mappings of surfaces. Then πΊ(πΉ): π β π is a mapping of surfaces and (πΊ(πΉ))β = πΊβ (πΉβ ). Computation of tangent map Let πΉ: π β π be a mapping. Let π₯: π· β π be a parameterization in π. Let π¦ = πΉ(π₯): π· β π. Diffeomorphisms Definition: Let πΉ: π β π be a mapping of surfaces. If πΉ has an inverse πΉ β1 : π β π which is also a mapping of surfaces then πΉ is a diffeomorphism. Inverse function theorem Let πΉ: π β π be a mapping Let π β π be a point. of the surface. Assume that πΉβ : ππ (π) β ππ(π) (π) 131 of 162 Lecture notes on MTH352: Differential Geometry is an isomorphism. Then there are neighborhoods π β π of π and π β π of πΉ(π) such that πΉ|π : π β π is a diffeomorphism. Corollary: If πΉ is regular and bijective, then πΉ is a diffeomorphism. Definition: Two surfaces are diffeomorphic if there is a diffeomorphism between them. End of the lecture Lecture # 26 132 of 162 Lecture notes on MTH352: Differential Geometry Contents: ο Diffeomorphic Surfaces ο Mapping of Differential Forms Diffeomorphic Surfaces Examples of diffeomorphic surfaces Example1: An open rectangle in πΉπ is isomorphic to the entire plane πΉπ Let π be an open set given as π = {(π’, π£): π’ < β π π π π < π’ < , β < π£ < }. 2 2 2 2 Then the function πΉ: π β π 2 defines as πΉ(π’, π£) = (tan π’, tan π£) is a mapping of π into π 2 . The inverse πΉ β1 is given by πΉ β1 (π’, π£) = (tanβ1 π’, tanβ1 π£). So πΉ is a diffeomorphism. Similarly every open rectangle in π 2 is diffeomorphic to π 2 . Example2: The sphere minus one point to πΉπ 133 of 162 Lecture notes on MTH352: Differential Geometry Let Ξ£0 be a sphere minus one point, namely the north pole. A parametrization of Ξ£0 is given by: π₯(π’, π£) = (cos π£ cos π’ , cos π£ sin π’ , 1 + sin π£). where (π’, π£) β π × (β 3π π , ) 2 2 Let πΉ from Ξ£0 to πΉπ be the projection map, so a point (π1 , π2 , π3 ) of Ξ£0 has image: πΉ(π1 , π2 , π3 ) = ( 2π1 2π2 , , 0). 2 β π3 2 β π3 So π¦(π’, π£) = πΉ(π₯(π’, π£)) = 2 cos π£ (cos π’ , sin π’ , 0). 1 β sin π£ Example 3: A cylinder and πΉπ minus a point Let πΆ: π₯ 2 + π¦ 2 = 1 be a cylinder in π 3 . Define πΉ: πΆ β π 2 by πΉ(π₯, π¦, π§) = π π§ (π₯, π¦) Then πΉ maps objectively πΆ to π 2 \(0,0). The inverse of πΉ is gives as: πΉ β1 (π’, π£) = ( π’ βπ’2 + π£2 , π£ βπ’2 + π£2 , ln βπ’2 + π£ 2 ). This is again a mapping, therefore πΉ is a diffeomorphism. Mappings of Differential Forms Mapping of a π βform Let πΉ: π β π be a mapping of surface. 134 of 162 Lecture notes on MTH352: Differential Geometry Question: If g: M β π is a function then how to move π to a function defined on π? This is not possible in general, but we take functions π: π β π to functions on π. For functions π: π β π we define: π(πΉ): π β π , π β π(πΉ(π)). We call π(πΉ) the pull back of π. Pull-back of π βforms and π βforms Definition: Let πΉ: π β π be a mapping of surfaces. ο Let π be a 1 βform on π, the pull-back of π, written as πΉ β (π) is defined as: For every tangent vector π£ to π define (πΉ β π)(π£) = π(πΉβ π£). ο Let π be a 2 βform on π. The pull-back of π, denoted by πΉ β (π), is defined as: For every pair of tangent vectors π’, π£ to π πΉ β π(π’, π£) = π(πΉβ π’, πΉβ π£). The essential operations on forms are sum, wedge product, and exterior derivative, are all preserved by mapping. Theorem: Let πΉ: π β π be a mapping of surfaces, and let π and π be forms on π. Then 1. πΉ β (π + π) = πΉ β π + πΉ β π. 2. πΉ β (π β§ π) = πΉ β π β§ πΉ β π. 3. πΉ β (ππ) = π (πΉ β π). End of the lecture Lecture # 27 135 of 162 Lecture notes on MTH352: Differential Geometry Contents: ο Line Integrals ο Integration of 2 βforms ο Integral Over a Boundary Line Integrals Pull-back of π βform on a curve Let πΌ: [π, π] β π be a closed curve segment on surface π. Let π be a 1 βform on π. The pull-back of π is a 1 βform on [π, π] is of the form π(π‘) ππ‘ where π(π‘) = (πΌ β π)(π1 (π‘)) = π (πΌβ (π1 (π‘))) = π(πΌ β² (π‘)). Definition: Let π be a 1 βform on π. Let πΌ: [π, π] β π be a curve segment. Then the integral of π over πΌ is equal to π β« π=β« πΌ Another integral of π [π,π] πΌ β π = β« π(πΌ β² (π‘)) ππ‘ . π name is the over πΌ. 136 of 162 Lecture notes on MTH352: Differential Geometry Force field: Particle moving in a force field Example: Let π be a vector field on a surface π. We assume that π is a force field. Assume that a particle moves on a curve πΌ: [π, π] β π. We want to know how much work is done by the force field on the particle as it moves between points πΌ(π) and πΌ(π). Work is done on the particle only by the component of force tangent to πΌ, that is, πΌβ² π(πΌ). βπΌβ² β = βπ(πΌ)β cos π The work done by the force during time Ξπ‘ is approximately given as (π(πΌ). πΌβ² ) βπΌ β² (π‘)βΞπ‘ . βπΌ β² β Adding these on the whole time interval [π, π] we get π π = β« π(πΌ(π‘)). πΌ β² (π‘) ππ‘ . π To express the following more easily π π = β« π(πΌ(π‘)). πΌ β² (π‘) ππ‘ π we introduce the following Definition: The dual to the vector field π on π is the 1 β form π defined by π(π€π ) = π€. π(π) for every π β π and every π€π β ππ (π). The total work done by the particle 137 of 162 Lecture notes on MTH352: Differential Geometry becomes π=β« π πΌ where π is the dual 1 βform of π. The integral of a differential Theorem: Let π: π β π be a function on π. Let πΌ: [π, π] β π be a curve segment on π, where πΌ(π) = π and πΌ(π) = π. Then β« ππ = π(π) β π(π). πΌ See the book for proof. Theorem: Let π: π β π be a function on π. Let πΌ: [π, π] β π be a curve segment on π, where πΌ(π) = π and πΌ(π) = π. Then β« ππ = π(π) β π(π). πΌ The above can also be restated as: Let the βboundaryβ of the curve be π β π, where π is the starting point and π is the end point. Then the integral of ππ over πΌ equals the βintegralβ of π over the boundary of πΌ, namely π(π) β π(π). Extending to surface maps Definition: Let π = [π, π] × [π, π] = {(π’, π£): π β€ π’ β€ π πππ π β€ π£ β€ π} . be a closed rectangle in π 2 . A 2 βsegment is a function π₯: π β π which is differentiable, that is, there exist an open set π· β π 2 such that π β π·, and π₯: π· β π is differentiable. 138 of 162 Lecture notes on MTH352: Differential Geometry Note: We do not assume that a 2 βsegment π₯ is regular or injective. The partial velocities π₯π’ and π₯π£ are defined as usual. But they may behave in some strange way if π₯ is not regular. Pull-back of a π βform Let π₯: π β π be a 2 βsegment on a surface π. Let π be a 2 βform on π. Then the pull-back by π₯ of π has the form β ππ’ ππ£ Where β = (π₯ β π)(π1 , π2 ) = π(π₯β π1 , π₯β π2 ) = π(π₯π’ , π₯π£ ). Definition: The integral of π over π₯ is π π β β¬ π = β¬ π₯ π = β« β« π(π₯π’ , π₯π£ ) ππ’ ππ£ . π₯ π π π Definition: Let = [π, π] × [π, π] . Let π₯: π β π be a 2 βsegment in π. Let πΌ, π½, πΎ, πΏ be the curve segments such that πΌ(π’) = π₯(π’, π), π½(π£) = π₯(π, π£), πΎ(π’) = (π’, π ), πΏ(π£) = (π, π£ ) . 139 of 162 Lecture notes on MTH352: Differential Geometry The boundary of π₯ is a formal expression ππ₯ = πΌ + π½ β πΎ β πΏ Let π be a 1 βform on π then the integral of π on ππ₯ is defined as β« π =β« π+β« πββ« πββ« π . ππ₯ πΌ π½ πΎ πΏ End of the lecture Lecture # 28 140 of 162 Lecture notes on MTH352: Differential Geometry Contents: ο Stokes Theorem ο Reparametrization Stokes Theorem Theorem: Let π be a 1 βform on π. Let π be a closed rectangle π = [π, π] × [π, π]. Let π₯: π β π be a 2 βsegment in π. Then β« β«π π = β« π . π₯ ππ₯ Proof: Idea of the proof: We will start from the double integral and show that it turns into the integral over the boundary of π₯. We have π π β« β«π π = β« β« ππ(π₯π’ , π₯π£ )ππ’ ππ£ π₯ π π By definition of exterior derivative: π π π π β« β« ππ(π₯π’ , π₯π£ )ππ’ ππ£ = β« β« ( π π π π π π π(π₯π£ ) β π(π₯π’ )) ππ’ ππ£ ππ’ ππ’ So 141 of 162 Lecture notes on MTH352: Differential Geometry π π β« β«π π = β« β« ( π₯ π π π π π(π₯π£ ) β π(π₯π’ )) ππ’ ππ£ . ππ’ ππ’ Let π = π(π₯π’ ): π β πΉ and π = π(π₯π£ ): π β πΉ. . So π π β« β« ( π π π π π π ππ ππ π(π₯π£ ) β π(π₯π’ )) ππ’ ππ£ = β« β« ( β ) ππ’ ππ£ ππ’ ππ’ ππ’ ππ’ π π π π =β« β« ( π π π π ππ ππ ) ππ’ ππ£ β β« β« ( ) ππ’ ππ£ ππ’ ππ’ π π Now we will calculate: π π π ππ β« β« ( ) ππ’ ππ£ ππ’ π π πππ π β« β« ( π π ππ ) ππ’ ππ£ ππ’ Now we will calculate: π π π ππ β« β« ( ) ππ’ ππ£ π π ππ’ πππ π β« β« ( π π ππ ) ππ’ ππ£ ππ’ Consider π π π π π ππ ππ β« β« ( ) ππ’ ππ£ = β« (β« ( ) ππ’) ππ£ = β« (πΌ) ππ£ π π ππ’ π π ππ’ π Where π ππ πΌ = β« ( ) ππ’ π ππ’ By fundamental theorem of calculus: π πΌ=β« π ππ(π’, π£) ππ’ = π(π, π£) β π(π, π£). ππ’ If π½(π£) = π₯(π, π£) and πΏ(π£) = π₯(π, π£) then π π π β« π(π, π£) ππ£ = β« π(π₯π£ (π, π£)) ππ£ = β« π ( π π π π π π₯(π, π£)) ππ£ = β« π(π½ β² (π£)) ππ£ = β« π ππ£ π π½ Now π β« β«π π = (β« π(π½ π π π β² (π£)) ππ£ β β« π(πΏ π π β² (π£)) ππ£ ) β (β« π(πΎ π π β² (π£)) ππ£ β β« π(πΌ β² (π£)) ππ£ ) π 142 of 162 Lecture notes on MTH352: Differential Geometry β« β«π π = (β« π β β«π ) β (β«π β β« π ) π₯ π½ πΏ πΎ πΌ =β« π ππ₯ Reparametrization Line integrals of reparametrized curves Let πΌ(β): [π, π] β π be a reparametrization of a curve segment πΌ: [π, π] β π by β: [π, π] β [π, π]. Then for any 1 βform on π 1) If β is orientation preserving that is β(π) = π and β(π) = π then β« π = β«π π(β) πΌ 2) β is orientation reversing that is β(π) = π and β(π) = π then β« π = β β«π πΌ(β) πΌ See the book for proof. Integral of the boundary without the minus sign Now if π: [π‘1 , π‘2 ] β π be any curve segment, define βπ: [π‘1 , π‘2 ] β π by βπ(π‘) = π(π‘1 + π‘2 β π‘) Then π is orientation reversing reparametrization of π So by previous result β« π = β β«π βπ π We have seen that β« β«π π = β« π = (β« π β β«π ) β (β«π β β« π ) π₯ ππ₯ π½ πΏ πΎ πΌ 143 of 162 Lecture notes on MTH352: Differential Geometry So we can write β« β«π π = β« π = β« π + β« π + β« π + β« π π₯ ππ₯ πΌ π½ βπΏ βπΎ End of the lecture Lecture # 29 144 of 162 Lecture notes on MTH352: Differential Geometry Contents: ο Connectedness ο Compactness ο Orientability Connectedness Definition: Let π be a surface. The surface π is connected if for any two points π and π of π there exists a curve segment in π from π to π. Example: Let π be implicitly defined surface π: π§ 2 = π₯ 2 + π¦ 2 + 1 Then π is disconnected. There are no curves in π between points with π§ > 0 and points with π < 0. Compactness Definition: Let π΄ be a set in a topological space. If for every covering of π΄ by open sets π΄ = β ππΌ , πΌβπ½ 145 of 162 Lecture notes on MTH352: Differential Geometry there exists π β π΅ and there exist πΌ1 , πΌ2 , β¦ , πΌπ β π½ such that π π΄ββ π=1 ππΌπ , then π΄ is compact. Examples: ο Closed and bounded intervals [π, π] β πΉ. ο Rectangles [π, π] × [π, π] β π 2 . ο Closed and bounded subsets of π π . Compactness of surfaces Lemma: A surface π is compact if and only if there exist a π β π΅ and 2 βsegments π₯π : π π β π πππ π = 1,2, β¦ , π such that π π=β π=1 π₯π (π π ) Proof: 146 of 162 Lecture notes on MTH352: Differential Geometry Compact regions on surfaces If follows from the previous result that βA region π in π in compact if it is composed of the images of finitely many 2 βsegments in πβ. Examples: 147 of 162 Lecture notes on MTH352: Differential Geometry ο A sphere is compact. ο A torus is compact. ο Surface of revolution whose profile curve is closed. How to check compactness A continuous real valued function on a closed rectangle π in the plane takes a maximum at some point of π . Lemma: A continuous function π on a compact region in a surface π takes on a maximum at some point of π. See the book for proof. Examples of non-compact surfaces ο A cylinder is not compact. ο The disk in πΉπ is not compact. Orientability Definition: A surface π is orientable if there exists a continuous 2 βform on π that is non-zero at every point of π. Example: ο The plane πΉπ is a surface. The product ππ’ β§ ππ£ is a 2 βform on π 2 that is non-zero at every point of πΉπ . Therefore πΉπ is an orientable surface. Unit normal A unit normal π on π is a differentiable Euclidean vector field on π that has unit length and is everywhere normal to π. 148 of 162 Lecture notes on MTH352: Differential Geometry Proposition: A surface π β π 3 is orientable if and only if there exists a unit normal vector field on π. Examples of orientable surfaces ο Spheres ο Cylinder ο Surface of revolution ο Implicitly defined surfaces. End of the lecture Lecture # 30 149 of 162 Lecture notes on MTH352: Differential Geometry Contents: ο Homotopy ο Simply Connectd Surfaces ο Poincare Lemma ο Conditions of Orientability Homotopy Definition: 150 of 162 Lecture notes on MTH352: Differential Geometry Simply Connected Surfaces Definition: A surface π is simply connected provided it is connected and every loop in π in homotopic to a constant. Example: The plane π 2 is simply connected. In general every Euclidean space is simply connected. Example: A Sphere is simply connected 151 of 162 Lecture notes on MTH352: Differential Geometry Not simply connected surfaces Definition: If π is any differential form, then π is called closed if π π = 0. Lemma: Let π be a surface and π a closed 1 βform on π. Let πΌ be a loop in π. If πΌ is homotopic to a constant then β« π = 0. πΌ Proof: Example: πΉπ minus a point is not simply connected Definition: A differential form is exact if it is exterior derivative of another form. Lemma: Every closed 1 βform on a simply connected space is exact. 152 of 162 Lecture notes on MTH352: Differential Geometry Compactness and orientability Theorem: Every compact surface in π 3 is orientable. The theorem is a consequence of the following result. Jordan-Brouwe-Separation theorem A compact surface in π 3 separates π 3 into two non-empty sets, an exterior and an interior. 153 of 162 Lecture notes on MTH352: Differential Geometry Connectedness and orientability Theorem: Every simply connected surface in π 3 is orientable. End of the lecture 154 of 162 Lecture notes on MTH352: Differential Geometry Lecture # 31 Contents: ο Abstract Surfaces ο Manifolds Abstract Surfaces Abstract patch Let π be any set. An abstract patch in π is a function π₯: π· β π that satisfies the following ο π₯ is injective. ο π· is an open set in π 2 . Definition An abstract surface is a pair (π, π) such that π is a set of abstract patches in π, and such that the following three axioms are satisfied: ο The covering axiom: The union of the images of the patches in π is equal to π. ο The smooth overlap axiom: If π₯, π¦ are elements of π, then π¦ β1 (π₯) and π₯ β1 (π¦) are differentiable functions from open sets in π 2 into π 2 . ο The and π Hausdorff axiom: If π are different points 155 of 162 Lecture notes on MTH352: Differential Geometry in π then there are disjoint patches π₯, π¦ β π such that π is in the image of π₯ and π is in the image of π¦. Open sets of a surface Let (π, π) be an abstract surface. An open set of π is any union of sets π₯(π) such that π₯ β π and π is open subset of the domain of π₯. The projective plane Let π₯: π· β Ξ£ be a patch in Ξ£. Call π₯ a small patch if π₯(π·) is contained in a hemisphere of Ξ£. So {π, βπ} is not subset of π₯(π·) for all π β Ξ£. If π₯ is small, then πΉ(π₯): π· β π is injective. Therefore πΉ(π₯) is an abstract patch in π. (πΊ, π·)is an abstract surface It is clear that (π, π) satisfies the covering axiom (1). Is also satisfies the Hausdorff axiom (3). This is easy to show, because the domain π· of every patch has the Hausdorff property: for any two given distinct points there are disjoint open sets around them. What about the smooth overlap axiom? (π, π) satisfies the smooth overlap axiom 156 of 162 Lecture notes on MTH352: Differential Geometry Assume that πΉ(π₯), πΉ(π¦) β π and there is a point {π, βπ} which is in the image of both πΉ(π₯) and πΉ(π¦). Then either the images of π₯ and π¦ in Ξ£ both contain π or βπ or one contains π and the other contains βπ. If π₯ and π¦ overlap in Ξ£, then πΉ(π¦)β1 (πΉ(π₯)) = π¦ β1 π₯ and we know that π¦ β1 π₯ is differentiable. If π₯ and π¦ do not overlap, then π₯ and π΄(π¦) overlap instead. β1 (πΉ(π¦)) (πΉ(π₯)) = (πΉ(π΄(π¦))) β1 (πΉ(π₯)) = π¦ β1 π₯ which proves that πΉ(π₯) and πΉ(π¦) satisfies the smooth overlap condition. Surfaces in πΉπ A surface in π 3 is an example of an abstract surface, if we explain π as the set of patches that is used to construct π. By definition (π, π1 ) and (π, π2 ) are different surfaces if π1 β π2 . We will assume that the set π of patches in an abstract surface (π, π) is as large as possible. So all patches in π that satisfies the overlap condition with all elements of π also have to be in π. Sometimes we only speak of the surface π when we mean the abstract surface (π, π). Curves in abstract surface Let π be an abstract surface. A curve in π is a function πΌ: πΌ β π Such that πΌ has a velocity vector πΌ β² (π‘) for every π‘ β πΌ. Definition: A velocity vector of a function πΌ: πΌ β π is a function πΌ β² (π‘) such that for every differentiable function π: π β πΉ πΌ β² (π‘)(π) = πΌ β² (π‘)[π] = ππ(πΌ) (π‘) πππ πππ π‘ β πΌ . ππ‘ We notice that differential forms and integration on surfaces can all be explained in terms of velocity of curves. 157 of 162 Lecture notes on MTH352: Differential Geometry Manifold An π βdimensional manifold is a pair (π, π), where π is a set of points, and π is a set of abstract patches π₯: π· β π injective, with π· β π π an open set, satisfying: ο The covering axiom: The union of the images of the patches in π is equal to π. ο The smooth overlap axiom: If π₯, π¦ are elements of π, then π¦ β1 (π₯) and π₯ β1 (π¦) are differentiable functions from open sets in π 2 into π 2 . ο The Hausdorff axiom: If π and π are different points in π then there are disjoint patches π₯, π¦ β π such that π is in the image of π₯ and π is in the image of π¦. Example: ο An abstract surface is precisely the same as a 2 βdimensional manifold. ο π π is an example of an π βdimensional manifold. Tangent bundle of a surface Let π be a surface in π 3 (or abstract surface, or 2 βdimensional manifold). Let π(π) be the set of all tangent vectors to π at all points. Then π(π) is the tangent bundle of π. Example: An element of π(π) corresponds to a pair (π, π£) β π × ππ (π). Since π and ππ (π) each has dimension 2. We expect that π(π) has dimension 4. Let π₯: π· β π be any patch in π. Μ = {(π1 , π2 , π3 , π4 ): (π1 , π2 ) β π·} = π· × π 2 Let π· Μ β π(π) be defined as Let π₯Μ: π· (π1 , π2 , π3 , π4 ) β π3 π₯π’ (π1 , π2 ) + π4 π₯π£ (π1 , π2 ). Then π₯Μ is injective, and if π is the set of all derived patches π₯Μ of patches in π, then (π(π), π) satisfies the definition of a manifold. If follows that π(π) is a 4 βdimensional manifold. End of the lecture 158 of 162 Lecture notes on MTH352: Differential Geometry Lecture # 32 Contents: ο Geodesic Curves ο Examples Geodesic Curves Definition: Let π be a surface in π. The πΌ is a geodesic of π if πΌβ²β² is normal to π. Remark: Geodesics is a curve that does not turn right or left on π. It only moves up and down to stay on π itself. Properties of geodesics ο A geodesic has constant speed βπΌ β² β ο Every straight line is a geodesic Geodesics on spheres ο Let Ξ£ be a sphere with center (0,0,0) in π 3 . Let π be a plane in π 3 through (0,0,0). Then π β© Ξ£ is a great circle. 159 of 162 Lecture notes on MTH352: Differential Geometry Great circles are geodesics Assume that πΌ is a parametrization of a great circle on Ξ£. We know that πΌ β²β² at every point of πΌ points towards the (0,0,0) of the circle. This implies that πΌ β²β² is normal to Ξ£. It follows that πΌ is a geodesic. Geodesics on cylinders Geodesics are helices on cylinders Assume that π is a circular cylinder π: π₯ 2 + π¦ 2 = π 2 in π 3 . Let πΌ(π‘) = (π cos π(π‘) , π sin π(π‘) , β(π‘)) be any curve in π. For every π β π we have (0,0,1) β ππ (π). So if πΌ β²β² is normal to π, then ββ²β² (π‘) = 0. This implies β(π‘) = ππ‘ + π For constants π, π β π . The speed of πΌ is βπ 2 π β² + π 2 . The speed is constant if and only if π β² is constant π. Thus if and only in π(π‘) = ππ‘ + π. It follows that a geodesic is a curve of the form πΌ(π‘) = (π cos (ππ‘ + π) , π sin (ππ‘ + π) , ππ‘ + π) If π and π are non-zero then this is a helix. 160 of 162 Lecture notes on MTH352: Differential Geometry If π = 0 and π β 0 then πΌ is a circle around the π§ βaxis. Closed geodesics Assume that πΌ: [π, π] β π is a segment of a geodesic. If πΌ(π) = πΌ(π) and πΌ β² (π) = πΌ β² (π) then πΌ is a closed geodesic. Example: Every geodesic on a sphere is closed. Example: A geodesic on a cylinder is closed only if it is a circle around the π§-axis. Geodesics in orthogonal planes Theorem: Assume that πΌ is a unit speed curve in π. Let there exist a plane π such that πΌ βπβ©π. If π is orthogonal to π(π) at every point of πΌ, then πΌ is a geodesic. Proof: Let πΌ has a constant speed. This implies that πΌ β² . πΌ β²β² = 0 . We know that πΌ β² β π β© π(π) And πΌ β²β² β π. Since π is orthogonal to π(π), this implies that πΌ β²β² is orthogonal to π(π). So πΌ is a geodesic. Surface of revolution If π is a surface of revolution, then every meridian in π is a geodesic. Proof: Every meridian is the intersection of π with a plane π that contains the rotation axis. 161 of 162 Lecture notes on MTH352: Differential Geometry Such a plane is orthogonal to π at every point. It follows from the previous result that a meridian is a geodesic. End of the lecture 162 of 162