Lecture 8: The First Fundamental Form Table of contents

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1 Math 38 Fall 2016 Lecture 8: The First Fundamental Form Disclaimer. As we have a textbook, this lecture note is for guidance and sulement only. It should not be relied on when rearing for exams. In this lecture we study how to measure distance on a surface atch. The required textbook sections are 6.1. The otional sections are I try my best to make the examles in this note dierent from examles in the textbook. Please read the textbook carefully and try your hands on the exercises. During this lease don't hesitate to contact me if you have any questions. Table of contents Lecture 8: The First Fundamental Form Measurements on a surface atch Motivation The rst fundamental form First fundamental form under re-arametrization Examles

2 Dierential Geometry of Curves & Surfaces Review of midterm. Fun alication of curve theory. Which way does the bicycle go. Very nicely exlained in this youtube video: htts://youtu.be/etnbfuw8zy. Also see htt:// for related questions. Exercise 1. Given the track of the rear wheel, calculate the curvature of the front wheel. Question 1. If all we have is a icture of the bicycle tracks (so that there is deformation), can we still determine which way the bicycle goes? Question 2. What if the bicycle is on a curved surface? Overview of local theory of surfaces. local: We are focusing on surface atches. First fundamental form: Measuring distance, angle, area. kx 0 (t)k for surfaces. Second fundamental form: Measuring curving. and for surfaces. Turns out that the two fundamental forms totally determine the surface (u to a rigid motion). Exercise 2. Consider a curve in R 3. Is it determined (u to a rigid motion) by kx 0 (t)k, (t); (t)? Can we dro kx 0 (t)k from the list? 1. Measurements on a surface atch 1.1. Motivation Our goal is to measure length, angle, and area on a surface atch without having to re-write everything in the ambient sace R 3. Recall that in R 3, we do measurements through inner roduct of vectors: Let w; w~ be vectors in R 3, then we can comute their inner roduct as w w~ = w 1 w~ 1 + w 2 w~ 2 + w 3 w~ 3 : (1) Using this we can comute length: kwk = w w; (2) angle: cos\(w; w~) = w w~ kwk kw~k ; (3) area of the arallelogram formed by w; w~: A = kw w~k = kwk 2 kw~k 2 (w w~) 2 : () Exercise 3. Prove that kw w~k = kwk 2 kw~k 2 (w w~) 2. 2

3 Math 38 Fall 2016 Note. Kee in mind that when we are talking about vectors in R 3, we are talking about the R 3 that is the velocity overlay sace of the location sace R 3. In other words, the R 3 here is in fact T R 3. When we need to measure in the location sace, we integrate: b L = kx 0 (t)k dt; A = k u v k du dv: (5) a U Now we ask the following natural question: How to measure on surfaces? Consider a oint 2 S, a surface atch given by : U 7! R 3. How should measurements be done within the tangent lane T S? Let w; w~ 2 T S. Let = (u 0 ; v 0 ). Then we have w = w 1 u (u 0 ; v 0 ) + w 2 v (u 0 ; v 0 ); w~ = w~ 1 u (u 0 ; v 0 ) + w~ 2 v (u 0 ; v 0 ): (6) It now follows w w~ = E(u 0 ; v 0 ) w 1 w~ 1 + F(u 0 ; v 0 ) (w 2 w~ 1 + w 1 w~ 2 ) + G(u 0 ; v 0 ) w 2 w~ 2 (7) where E(u; v) := k u (u; v)k 2 ; (8) F(u; v) := u (u; v) v (u; v); (9) G(u; v) := k v (u; v)k 2 : (10) The oint here is that, once we know E; F; G, we can do all the measurements without leaving the surface The rst fundamental form Definition 3. (Definition of the textbook) Let be a oint of a surface S. The rst fundamental form of S at is the bilinear form hw; w~i ;S := w w~: (11) Theorem. Let h; i ;S be the rst fundamental form of S at. Let be a surface atch of S covering. Let w = w 1 u + w 2 v, w~ = w~ 1 u + w~ 2 v be two vectors in T S. Then hw; w~i ;S = E(u 0 ; v 0 ) w 1 w~ 1 + F(u 0 ; v 0 ) (w 2 w~ 1 + w 1 w~ 2 ) + G(u 0 ; v 0 ) w 2 w~ 2 (12) where E; F; G are given by ( 810), and = (u 0 ; v 0 ). Remark 5. There is now the natural question: What haens if we take a dierent rst fundamental form? It can be shown that h; i ;S remains the same if we choose a dierent surface atch for S. Exercise. Try to make sense of the above statement. 3

4 Dierential Geometry of Curves & Surfaces Theorem 6. (Measurements on S) Let S be a surface atch with rst fundamental form E; F; G. Then, i. if (u(t); v(t)) is a curve on S, its arc length from t = a to t = b is given by R b h 0 (t); 0 (t)i (t);s dt which equals a a E((t)) u 0 (t) 2 + F((t)) u 0 (t) v 0 (t) + G((t)) v 0 (t) 2 : (13) b ii. if (u(t); v(t)) and (u~(t); v~(t)) are two curves on S intersecting at (u 0 ; v 0 ), then the angle between the two tangent vectors w := cos = iii. the area of the surface atch is given by 2 A = hw; wi :S hw~; w~i ;S hw; w~i ;S U d(u(t); v(t)) dt and w~ := d(u~(t); v~(t)) dt is given by hw; w~i ;S hw; wi 1/2 :S hw~; w~i : (1) 1/2 ;S = U 1.3. First fundamental form under re-arametrization E G F 2 du dv: (15) A common situation is the following. Say we have the rst fundament form of a surface atch (u; v) as E(u; v); F(u; v); G(u; v). Now the surface is re-arametrized through u = U(u~:v~) and v = V (u~; v~). How to obtain the new rst fundamental form? The simlest way is to do the following: Substitute u = U(u~:v~), v = V (u~; v~), and into the formula to obtain 2. Examles Examle 7. Consider the lane We easily calculate which gives Thus the rst fundament form is du = U u~ du~ + U v~ dv~; dv = V u~ du~ + V v~ dv~ (16) E du F du dv + G dv 2 (17) E~ du~ F~ du~ dv~+ G~ dv~ 2 : (18) (u; v) = (u; v; u + 3 v): (19) u = (1; 0; 2); v = (0; 1; 3) (20) E = 5; F = 6; G = 10: (21) 5 du du dv + 10 dv 2 : (22)

5 Math 38 Fall 2016 Examle 8. Consider the cylinder This gives Thus the rst fundamental form is (u; v) = (cos u; sin u; v): (23) u = ( sin u; cos u; 0); v = (0; 0; 1): (2) E = 1; F = 0; G = 1: (25) du 2 + dv 2 : (26) Examle 9. Consider the unit shere (u; v) = u; v; 1 u 2 v 2 : (27) This gives E = u = v = u 1; 0; ; (28) 1 u 2 v 2 v 0; 1; : (29) 1 u 2 v 2 1 v2 1 u 2 v 2; F = u v 1 u2 1 u 2 v2; G = (30) 1 u 2 v 2 which gives the rst fundamental form as 1 v 2 u v 1 1 u 2 v 2 du2 u2 + 2 du dv + 1 u 2 v2 1 u 2 v 2 dv2 : (31) Examle 10. Consider the torus Therefore (u; v) = ((a + b cos u) cos v; (a + b cos u) sin v; b sin u): (32) u = ((a b sin u) cos v; (a b sin u) sin v; b cos u) (33) v = ( (a + b cos u) sin v; (a + b cos u) cos v; 0): (3) E = a 2 + b 2 2 a b cos u; F = 0; G = a 2 + b 2 cos 2 u + 2 a b cos u: (35) The rst fundamental form is then given by (a 2 + b 2 2 a b cos u) du 2 + (a 2 + b 2 cos 2 u + 2 a b cos u) dv 2 (36) We notice that E; F; G are all indeendent of v. Examle 11. Consider the surface atch 2 u (u; v) = u 2 + v ; 2 v u 2 + v ; u2 + v 2 1 u 2 + v : (37) 5

6 Dierential Geometry of Curves & Surfaces u = v = Consequently we have 2 (v 2 u 2 + 1) (u 2 + v 2 + 1) ; u v 2 (u 2 + v 2 + 1) 2; u ; (38) (u 2 + v 2 + 1) 2 u v (u 2 + v 2 + 1) 2; 2 (u2 v 2 + 1) (u 2 + v 2 + 1) ; v : (39) 2 (u 2 + v 2 + 1) 2 Thus we see that the rst fundamental form is E = (u 2 + v 2 + 1) 2; (0) F = 0; (1) G = (u 2 + v 2 + 1) 2: (2) (u 2 + v 2 + 1) 2[du2 + dv 2 ]: (3) In articular, we notice that the curves (u; v 0 ) and (u 0 ; v) are always orthogonal at their intersection. 6

Lectures 12: The Second Fundamental Form. Table of contents

Lectures 12: The Second Fundamental Form. Table of contents ectures 1: The Second Fundamental Form Disclaimer. As we have a textbook, this lecture note is for guidance sulement only. It should not be relied on when rearing for exams. In this lecture we introduce