Lectures 18: Gauss's Remarkable Theorem II. Table of contents

Transcription

1 Math 348 Fall 27 Lectures 8: Gauss's Remarkable Theorem II Disclaimer. As we have a textbook, this lecture note is for guidance and supplement only. It should not be relied on when preparing for exams. Much of the material in this lecture is optional. On the other hand, it is benecial to work through the notes as the calculations etc. here can serve as good review of the concepts/formulas we studied in the past two months. The required textbook sections are..2. The optional sections are.3.4. I try my best to make the examples in this note dierent from examples in the textbook. Please read the textbook carefully and try your hands on the exercises. During this please don't hesitate to contact me if you have any questions. Table of contents Lectures 8: Gauss's Remarkable Theorem II More discussions on Gauss and Codazzi-Mainradi equations Surfaces of constant Gaussian curvature Surfaces with K = Surfaces with K > constant Surfaces with K < constant Compact surfaces

2 Dierential Geometry of Curves & Surfaces. More discussions on Gauss and Codazzi-Mainradi equations Remark. As k ij can be calculated using E; F; G, the Codazzi-Mainradi and Gauss equations can be seen as equations for L; M; N given E; F; G. Remark 2. It can be shown that as long as E; F; G; L; M; N satisfy the Codazzi-Mainradi and Gauss equations, with k ij calculated from E; F; G as we have seen before, and with L N M2 K :=, then there is a unique surface patch with E du F du dv + G dv 2, E G F 2 L du M du dv + N dv 2 as its rst and second fundamental forms, k ij as its Christoel symbols, and K as its Gaussian curvature. Remark 3. (Relation to Riemann Curvature Tensors) We have the following formulas for the Riemann curvature tensor: R k ijl j k l k ij + k j il + k j2 2 il k l ij k l2 2 ij : () In our context, we have i; j; k; l = ; 2. Thus there are 6 possible R k ijl 's. However it is easy to notice R k ijl = R k ilj. Thus we only need to calculate R k i2. Setting u = u ; v = u 2, we see that the Gauss equations (2) become F K = R 2 = R ; R 22 = G K; R 2 = E K: (2) Now if we dene R ijkl = g im R m jkl, that is R2 R 22 E F = R 22 R 222 F G We see that R2 R 22 R 22 R 222 = (E G F 2 ) R 2 R R 2 R 22! ; (3) : (4) We see that, among the 6 possible R ijkl, there is exactly one non-trivial value: R 22 = (L N M 2 ): (5) One now also see that the sectional curvature K( u ; v )=K is exactly the Gaussian curvature of the surface. In general, sectional curvatures are generalizations of Gaussian curvatures. Example 4. 2 Is there a surface with the rst and second fundamental forms du 2 + cos 2 u dv 2 and cos 2 u du 2 + dv 2? We see that E = ; F = ; G= cos 2 u; L = cos 2 u; M= ; N =. Now we calculate k ij. As it is too hard to remember the formulas, we start from the rst principle (2): We have uu = u + v + L N ; uv = 2 u + 2 v + M N ; vv = 22 u + 22 v + N N: = E + 2 F = uu u = 2 E u = : (7) cos 2 u 2 = F + 2 G = uu v = ( u v ) u 2 ( u u ) v = (8). Theorem..3 of the textbook. 2. Exercise..2 of the textbook. (6) 2

3 Math 348 Fall 27 Thus = 2 =. Similarly we can calculate Now we see that while 2 = ; 2 2 = tan u; 22 = cos u sin u; 2 22 = : (9) M v N u = () L 22 + M ( ) N 2 2 = cos 3 u sin u + tan u =/ : () Thus the second Codazzi-Mainradi equation is not satised and consequently there is no surface with the rst and second fundamental forms du 2 + cos 2 u dv 2 and cos 2 u du 2 + dv 2. Remark 5. One can calculate an explicit formula for K using E; F; G only 3 : K = det 2 E vv + F uv 2 G uu 2 E u F u 2 E v B v 2 G u E F C 2 G v F G A det 2. Surfaces of constant Gaussian curvature 2 E v 2 G u 2 E v E F 2 G u F G (E G F 2 ) 2 : (2) By Gauss's Theorem we know that if two surfaces S ; S 2 are isometric, that is there is a mapping f: S 7! S 2 such that if (u; v) parametrizes S, and 2 (u; v) := f( (u; v)), then E (u; v) = E 2 (u; v), F (u; v) = F 2 (u; v), G (u; v) = G 2 (u; v), then K (u; v) = K 2 (u; v) at every (u; v). The natural question to ask now is Does the converse hold? More specically, we ask, if (u; v) parametrizes S and 2 (u; v) parametrizes S 2, so that K (u; v) = K 2 (u; v) at every (u; v). Are S ; S 2 isometric? The answer is no. Example 6. 4 Let (u; v) = a u; b v; 2 (a u2 + b v 2 ) and 2 (u; v) = a u; b v; 2 (au2 + b v 2. Let jaj > jbj; jaj > jb j. Then a) K (u; v) = K 2 (u; v) if a b = ab. b) The two surfaces are not locally isometric unless jaj = jaj and jbj = jb j. Exercise. Prove that the two surfaces are locally isometry when jaj = jaj and jbj = jb j. Proof. a) For we calculate and Thus 3. Corollary.2.2 of the textbook. 4. Taken from Chen99. E = a 2 ( + u 2 ); F = a b u v; G = b 2 ( + v 2 ) (3) a b L = p ; M = ; N = p : (4) + u 2 + v 2 + u 2 + v 2 K = a b ( + u 2 + v 2 ) 2: (5) C A 3

4 Dierential Geometry of Curves & Surfaces Similarly we have K 2 = The conclusion now follows. ab ( + u 2 + v 2 ) 2: (6) b) Assume the contrary, that is there is an isometry f between the two surfaces. For simplicity of presentation assume a > b > and a> b >. Other cases are left as exercises. Now as K (; ) = < K a b 2(u; v) whenever (u; v) =/ (; ), we must have f( (; )) = 2 (; ). We write f a u; b v; 2 (a u2 + b v 2 ) = au ; b V ; 2 (au 2 + b V 2 ) (7) where U ; V are functions of u; v with U(; ) = V (; ) =. As K (u; v) = K 2 (U ; V ) we must have u 2 + v 2 = U 2 + V 2. Taking derivatives we reach At (; ), we have E = a 2 ; F = ; G = b 2, and U u 2 + V u 2 = ; (8) U u U v + V u V v = ; (9) U v 2 + V v 2 = : (2) E 2 = a 2 U u 2 + b 2 V u 2 ; F 2 = a 2 U u U v + b 2 V u V v ; G 2 = a 2 U v 2 + b 2 V v 2 : (2) There now must hold a 2 U u 2 + b 2 V u 2 = a 2 ; (22) a 2 U u U v + b 2 V u V v = ; (23) a 2 U v 2 + b 2 V v 2 = b 2 : (24) As a> b >, (9) and (23) now lead to U u U v = and V u V v =. Thus there are two cases i. U u = V v = ; ii. U v = V u =. The conclusion now easily follows in both cases. Remark 7. It is easy to convince ourselves that a b = a b are not sucient for the isometry. Consider " (u; v) := " u; 4 v; " 2 " u v2. Now as " &, the surface " approaches a folded plane. Now consider all the geodesics of length emanating from the the origin on 2 (u; v), their other end points form a circle with radius <. If there is an isometry between 2 (u; v) and this folded plane (u; v), then this circle would be mapped to the union of two semicircles (one on the front side, one on the back) with radius on the folded plane. But these two circles must have the same parameter. Contradiction. 4

5 Math 348 Fall 27 On the other hand, if K (u; v) = K 2 (u; v) = constant, then the answer is yes. More specically, a) Any surface with K = everywhere is locally isometric to the plane; b) Any surface with K = constant > everywhere is locally isometric to a sphere; c) Any surface with K = constant < everywhere is locally isometric to a pseudosphere. 2.. Surfaces with K = Theorem 8. Let S be a surface covered by one single surface patch. Assume that its Gaussian curvature K = everywhere. Then S is isometric to an open subset of a plane. Proof. We have seen in Lecture 4 (Proposition ) that such S must be a ruled surface. Thus we can assume the surface patch to be (u; v) = (u) + v l(u) with klk = k_ k =. Then we have Consequently (2) now gives u = _(u) + v l _ (u); v = l(u): (25) E = + 2 v _ l _ + v 2 l _ 2 ; F = _ l; G = : (26) B = det@ We discuss two cases. l _ 2 E 2 u F u E 2 v E F F 2 2 E v = l _ 2 (E F 2 ) + C A 2 E v 2 E v E F F = l _ v _ l _ + v 2 l _ 2 (_ l) 2 + _ l _ + v l _ 2 2 = l _ 2 ( (_ l) 2 ) + _ l _2 : (27) i. l _ =. In this case is an open subset of a generalized cylinder and is isometric to an open subset of a plane. ii. l _ =/. Let l _^:= l _ l _, from (27) we have (_ l) 2 C A _ l _^2 = : (28) As _ is a unit vector and so are l; l _^ and furthermore l?l _^, we see that _ l l _^= and therefore _ l l _ =. From 3 of Lecture we see that this implies S is developable and consequently is isometric to an open subset of a plane Surfaces with K > constant Theorem 9. Let S be a surface covered by one single surface patch. Assume that its Gaussian curvature K is a positive constant. Then S is isometric to an open subset of a sphere. 5

6 Dierential Geometry of Curves & Surfaces Proof. Clearly it suces to consider the case K =. Exercise 2. Rigorously justify this. We rst re-parametrize the surface so that the fundamental forms are simple. Geodesic coordinates Proposition. (Proposition 9.5. of the textbook) Let p 2 S. Then there is a neighborhood of p that can be parametrized so that the rst fundamental form is du 2 + G(u; v) dv 2 where G(u; v) is a smooth function satisfying (p = (; )) G(; v) = ; G u (; v) = : (29) Proof. Let (v) be a geodesic passing p with v the arc length parameter. At each point on, let ~ v (u) be the geodesic passing that point and is perpendicular to. Further assume that u is also the arc length parameter. Exercise 3. Explain why there is a neighborhood of p that is fully covered by these ~ v (u)'s. Exercise 4. Explain why each point in this neighborhood belongs to exactly one ~ v (u). Thus we obtain a surface patch (u; v) = ~ v (u). As for each v, ~ v (u) is arc length parametrized, there holds k u k = =) E=. To show that F=, we notice that by construction u (; v ) v (; v ) = : (3) On the other hand, we parametrize ~ v (u) by u(t) = t; v(t) =v. Thus we have u_ = ; v_ =. As E =, we see that this is arc length parametrization. The geodesic equations now become d (E u_ + F v_) = dt d (F u_ + G v_) = dt 2 E F (u_ ; v_) F G (u_ ; v_) 2 E F F G u_ u v_ u_ v_ v (3) (32) = and F_ = : (33) Thus F = along ~ v, and consequently for all u; v as v is arbitrary. Exercise 5. Explain, using the intuition geodesics are shortest paths, why (u ; v) has to be perpendicular to (u; v ). Finally, v (; v) = d so G(; v) =. Then (3) applied to the geodesic dv (v) = (; v) gives G u (; v) =. (2) now gives 2 G uu G + 4 G u 2 = G 2 : (34) 6

7 Math 348 Fall 27 Setting G = g 2 we reach g uu + g = =) g(u; v) = A(v) cos u + B(v) sin u: (35) Thanks to (29) there holds g(;v)=; g u (;v)= which give A(v)=;B(v)=. Consequently we have the rst fundamental form to be du 2 + cos 2 u dv 2 (36) which is exactly the rst fundamental form of S 2 in spherical coordinates Surfaces with K < constant Theorem. Let S be a surface covered by one single surface patch. Assume that its Gaussian curvature K is a negative constant. Then S is isometric to an open subset of a pseudosphere. Consider a surface of revolution (u; v) = (f(u) cos v; f(u) sin v; g(u)) where f > and (f ) 2 + (g ) 2 =. Exercise 6. Show that K = f f. The pseudosphere When K =, we have f f ==) f(u)=a e u + b e u. Such a surface is called a pseudosphere. For example when a = ; b = we have Z p p g(u) = e 2u du = e 2u ln p e u + e 2u : (37) Exercise 7. Draw a sketch of the pseudosphere in this case. Proof. (of Theorem ) Same idea as the proof of Theorem 9. See p.258 of the textbook Compact surfaces Theorem 2. 5 Every connected compact surface whose Gaussian curvature is constant is a sphere. Exercise 8. Every compact surface has a point where K >. (Hint: Consider p 2 S that is furthest away from the origin) Proof. Consider the function J = ( 2 ) 2. Let p be where J reaches maximumthis is possible as J is continuous. If this maximum is then we have = 2 and S must be part of a sphere. If not, thanks to 2 = K being constant, there must hold that is at local maximum and 2 at local minimum at p. We reach contradiction using the following lemma. 5. Theorem.3.4 of the textbook. 7

8 Dierential Geometry of Curves & Surfaces Lemma 3. 6 Let : U 7! R 3 be a surface patch containing a point p that is not an umbilic 7. Let > 2 be the principal curvatures of and suppose that has a local maximum at p and 2 has a local minimum at p. Then K(p) 6. Proof. (of Lemma 3) Taking a small neighborhood of p such that > 2 in it. Take the coordinate system along the principal vectors so that the two fundamental forms are Thus we have = L E and 2 = N G. Exercise 9. What about > 2? E du 2 + G dv 2 ; L du 2 + N dv 2 : (38) The Codazzi-Mainradi equations now become L v = L 2 E v E + N G and consequently E v = 2 E 2 ; N u = 2 G u ( ) v ; G u = L E + N G 2 G 2 At p we have ( ) v = ( 2 ) u = =) E v = G u = which leads to K(p) Gu p p Ev p 2 E E E G = G uu + E vv 2 E G = G ( 2) uu E ( ) vv E G ( 2 ) (39) ( 2 ) u : (4) 6 ; (4) where the last inequality comes from taking a local maximum and 2 taking a local minimum at p. Remark 4. Similar to the proof of Theorem 2 we can show that a compact surface with K > everywhere and constant H is a sphere. 6. Lemma.3.5 of the textbook. 7. that is =/ 2. 8