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1 Differential Geometry MT451 Problems/Homework Recommended Reading: Morgan F. Riemannian geometry, a beginner s guide Klingenberg W. A Course in Differential Geometry do Carmo M.P. Differential geometry of curves and surfaces Oprea. Differential geometry and its applications O Neill B. Elementary differential geometry 1. Recall that a differentiable map f : U R n, U R k open, is an immersion if for all x U the differential d x f is injective. Let f : R 3 R 4 be the map with f(x, y, z) = (xyz, x + y + z, x 3 + y 3 + z 3, x). Compute the differential of f. Is f an immersion? Solution: d (x,y,z) f = yz 1 3x 1 xz 1 3y xy 1 3z For x = y = z = this has only rank, hence d (,,) f is not injective, f is not an immersion.. Consider the (ordered) basis (v 1,..., v 5 ) for R 5, v 1 = (,,,, ), v = (, 3,,, ), v 3 = (,, 3,, 4), v 4 = (,,,, 1), v 5 = (, 1, 1, 1, ). Find an orthonormal basis (b 1,..., b 5 ) for R 5 such that for j = 1,..., 5 we have. span{b 1,..., b j } = span{v 1,..., v j } Hint: you need compute almost nothing for this problem! 3. Sketch the following subsets of R : { (x, y) x, y R, x 99 + y 99 = 1 }, {(x, y) x, y R, xy = 4} { (x, y) x, y R, y = x 3} {, (x, y) x, y R, y = x } { (x, y) x, y R, y = x } {, (x, y) x, y R, ((y 1) + x 1)((y + 1) + x 1) = } 4. Let c: [, ] R 3 be the (parametrized) curve given by c(t) = (cos(t), sin(t), t) for all t [, ]. Compute the length of c. Solution: L(c) = c (t) dt = ( sin(t), cos(t), 1) dt = 5. The standard scalar product on R n is the bilinear function dt =. R n R n R (x, y) x y := n x i y i i=1 The (standard) norm of x R n is x := x x = n i=1 x i

2 where x = (x 1,..., x n ), y = (y 1,..., y n ). Let c: R R n be differentiable and so that t R : c (t) = 3. Prove that t R : c(t) c (t) =. please hand up 1-5 Monday, 9/9 in class 6. Find a diffeomorphism Φ: R R such that Φ( { (x, x ) x R } ) = {(x, ) x R}. Solution: Φ: R R, (x, y) (x, x y) is differentiable and Φ = id R, hence Φ is a diffeomorphism. Also Φ(x, x ) = (x, x x ) = (x, ) for all x R. 7. Let c: R R be smooth and C = {(t, c(t)) t R} R R = R 3 be the graph of c. Prove that C is a (unparametrized) curve. Solution: The map Φ: R 3 R 3 (x, y) (x, c(x) y), x R, y R is smooth and Φ = id R 3, hence Φ is a diffeomorphism. We also have Φ(C) = {(t, ) please hand up 6-7 Tuesday, 1/1 in class t R}. 8. Let C be the curve (a) Sketch C. C = } {(x, y) R x + y 4 = 1 (b) Find a parametrization for all of C, i.e. a regular curve c: R C with c(r) = C. Solution: c: R R t (cos(t), sin(t)) (c) Compute the curvature of C at an arbitrary point. Hint: The parametrization you give in 8b most likely is not by arc length. Solution: We first compute the derivatives of c from 8b. Hence κ C (cos(t), sin(t)) = c (t) Jc (t) = c (t) 3 Thus for (x, y) C we have c (t) = ( sin(t), cos(t)) c (t) = ( cos(t), sin(t)) Jc (t) = ( cos(t), sin(t)) c (t) = sin(t) + 4 cos(t) = cos(t) + sin(t) (sin(t) + 4 cos(t) ) 3 ( cos(t), sin(t)) ( cos(t), sin(t)) ( sin(t), cos(t)) 3 = (1 + 3 cos(t) ) 3 κ C (x, y) = ±. (1 + 3x ) 3

3 9. Let c: R R 3 be a unit speed curve and r R + be such that for each t R we have c(t) = r. Thus the curve lies in the sphere of radius r. Find and prove a positive lower bound for the curvature of c. Thus you need to specify b R + such that Hint: Differentiate the equation c(t) = r. Solution: Differentiating c(t) = c(t) κ(c(t)) = c (t) b for all t R. c(t) = r for t we get c (t) c(t) = c (t) c(t) + c (t) c (t) = Since c is unit speed, we have c (t) c (t) = 1 and therefore c (t) c(t) = 1. (1) By the Cauchy Schwarz inequality c (t) c(t) c (t) c(t) = r c (t) Together with (1) we get hence r c (t) 1 κ(c(t)) = c (t) 1 r. 1. Compute max { x 3 + y 3 + 3z 3 x, y, z R, x + y + z = 1 } Solution: By convexity of the function t t 3/ we get max { x 3 + y 3 + 3z 3 x, y, z R, x + y + z = 1 } = max { x 3 + y 3 + 3z 3 x, y, z R, x + y + z = 1, x, y, z } = max { x 3/ + y 3/ + 3z 3/ x, y, z R, x + y + z = 1, x, y, z } max { 3(x + y + z) 3/ x, y, z R, x + y + z = 1, x, y, z } = 3 and taking (x, y, z) = (,, 1) we see that 1 {x 3 + y 3 + 3z 3 x, y, z R, x + y + z = 1}, hence max { x 3 + y 3 + 3z 3 x, y, z R, x + y + z = 1 } = 3 please hand up 8-1 Tuesday, 19/1 in class 11. Compute max { x 3 x R, y R : x 4 + y 4 x y = 1 } Solution: This is the same as max { x x R, y R : x 4 + y 4 x y = 1 } and y R : x 4 + y 4 x y 1 = is equivalent to Thus max { x 3 x 4 4 x4 + 1 and x + x 4 4 x x R, y R : x 4 + y 4 x y = 1 } = max {x x4 = max } {x 1 3x4 = } 4 x4 + 1

4 1. Compute max { x 3 + y 3 + 3z 3 x, y, z R, x 4 + y 4 + z 4 = 1 } Solution: The maximum of f(x, y, z) = x 3 +y 3 +3z 3 given h(x, y, z) = x 4 +y 4 +z 4 = 1 obviously(?) is assumed for x y z. Now grad (x,y,z) f = (3x, 6y, 9z ) grad (x,y,z) h = (4x 3, 4y 3, 4z 3 ). Since grad (x,y,z) h = (4x 3, 4y 3, 4z 3 ) if h(x, y, z) = 1, we have grad (x,y,z) f = λgrad (x,y,z) h for some λ R if (x, y, z) maximizes f(x, y, z) on h 1 (1). Assuming this from now on, we thus have Since x y z we have the cases (a) z = : Then x = y =, h(x, y, z) = 1 (b) z > y = : Then x =, z = 1, f(x, y, z) = 3 3x = 4x 3 λ 6y = 4y 3 λ 9z = 4z 3 λ. (c) z > y > x = : Then λ = 9/4z = 6/4y, y = z/3, (z/3) 4 + z 4 = 1, z = ( 4 / ) 1/4 hence ) ( ) f(x, y, z) = ( 3 4 3/ (d) z > y > x > : Then λ = 9/4z = 6/4y = 3/4x, y = z/3, x = z/3, (z/3) 4 + (z/3) 4 + z 4 = 1, z = (1/ / ) 1/4 hence ( ) ( ) 3/4 1 1 f(x, y, z) = The last one is the maximum. 13. Let r R +, λ R and c: R R 3 be the curve with c(t) = (r cos(t), r sin(t), λt) for all t R. Compute the Frenet frame along this curve, i.e. vector fields T, N, B : R R 3 such that for all t R, (a) {T (t), N(t), B(t)} is an orthonormal basis for R 3, (b) span {T (t)} = span {c (t)} (c) span {T (t), N(t)} = span {c (t), c (t)} (d) B(t) = T (t) N(t). Solution: The first derivatives of c are c(t) = (r cos(t), r sin(t), λt) c (t) = ( r sin(t), r cos(t), λ) c (t) = ( r cos(t), r sin(t), ) Hence T (t) = c (t) c (t) N(t) = c (t) c (t) T (t) T (t) c (t) c (t) T (t) T (t) B(t) = T (t) N(t) = = ( r sin(t), r cos(t), λ) r + λ = ( cos(t), sin(t), ) (λ sin(t), λ cos(t), r) r + λ

5 Compute a matrix valued function K : R R 3 so that T (t) = K(t)T (t), N (t) = K(t)N(t), B (t) = K(t)B(t) for all t R. Solution: We compute the derivatives of the Frenet frame above. T (t) = ( r cos(t), r sin(t), ) r + λ = r r + λ N N (t) = (sin(t), cos(t), ) = r r + λ T + λ r + λ B B (t) = (λ cos(t), λ sin(t), ) r + λ = λ r + λ N Hence K(t) := (T (t), N(t), B(t)) r r +λ r r λ +λ r +λ λ r +λ (T (t), N(t), B(t)) t where (T (t), N(t), B(t)) is the 3 3 matrix with columns T (t), N(t), B(t). 14. Let {b 1,..., b n } and {c 1,..., c n } be orthonormal basis for R n and K Hom(R n, R n ) be a linear map with Kb i = c i for all i = 1,..., n. Prove that K is orthogonal, i.e. Kx Ky = x y for all x, y R n. Solution: Let x, y R n, x = n i=1 x ib i and y = n j=1 y jb j with x i, y j R. Then n n n n Kx Ky = K x i b i K y j b j = x i Kb i y j Kb j = i=1 j=1 i=1 j=1 i=1...n,j=1...n x i y j Kb i Kb j = x i y j c i c j = x i y j δ i,j = i=1...n,j=1...n i=1...n,j=1...n n x i y i = x y. i=1 15. Let K : R Hom(R n, R n ) be smooth so that for all t R, K(t) is orthogonal. Prove that K(t) 1 K (t) is skew symmetric for all t, i.e. K(t) 1 K (t)x y = x K(t) 1 K (t)y for all t R and x, y R n. Solution: Differentiate x y = K(t)x K(t)y. Thus = d dt x y = d dt K(t)x K(t)y = K (t)x K(t)y + K(t)x K (t)y = K(t) t K (t)x y + x K(t) t K (t)y = K(t) 1 K (t)x y + x K(t) 1 K (t)y. please hand up Tuesday, /11 in class 16. Consider the cone C = { (x, y, z) R 3 x + y = z } (a) Prove that S := C \ {(,, )} R 3 is a surface. Hint: Recall that S R 3 is a surface if for each point p S there are open subsets U, V R 3, p U and a diffeomorphism φ: U V and a plane E R 3 such that φ(u S) = V E. The set S given in the problem here has two components S ±. Find open sets U ±, V ± and diffeomorphisms φ ± : U ± V ± so that φ ± (S U ± ) = V ± {(x, y, ) x, y R}.

6 Solution: U ± := { (x, y, z) R 3 ± z > }, V ± := {(x, y, z) R 3 ± z > } x + y and φ ± : U ± V ± (x, y, z) (x, y, z x + y ) (b) Prove that C is not a surface. Hint: The problematic point is p = (,, ). Find three curves c 1, c, c 3 : ( 1, 1) C so that c i () = p and c 1(), c (), c 3() are linearly independent. Can a diffeomorphism map such curves in a plane? You might also prove that if S is a path connected surface and p S then S \ {p} is path connected. Solution: For the curves c 1, c, c 3 : ( ɛ, ɛ) C, we have c 1 (t) = (t,, t), c (t) = (, t, t), c 3 (t) = ( t,, t) c 1() = (1,, 1), c () = (, 1, 1), c 3() = ( 1,, 1) In particular, c 1(), c (), c 3() are linearly independent. Assume φ: U V, p = (,, ) U, U, V R 3 open, were a diffeomorphism so that φ(c U) = V E for some plane E R 3. We may assume that φ(,, ) = (,, ) V and E R 3 is a linear subspace. Since φ is a diffeomorphism, d p φ: R 3 R 3 is an isomorphism of vector spaces. Hence {d p φc 1(), d p φc (), d p φc 3()} is a basis for R 3. But for ɛ sufficiently small we have c i (t) U C for all i = 1,, 3 and t ( ɛ, ɛ). Hence φ(c i (t)) E and d p φc i() = d dt φ(c i (t)) E t= Since dim E = the three vectors d p φc 1(), d p φc (), d p φc 3() can not be linearly independent. 17. For the following subsets X i R 3 find minimal subsets S i X i such that X i \ S i is a surface. (a) X 1 = {(x, y, z) R 3 x 4 + y 4 = e z } Solution: X 1 is a surface, the graph of the function R \ {(, )} R, (x, y) ln(x 4 + y 4 ). So S 1 =. (b) X = {(x, y, z) R 3 x + y = sin(z) } Solution: S = {(,, kπ) k Z} (c) X 3 = {(x, y, ) R 3 x + y 1} Solution: S 3 = {(x, y, ) R 3 x + y = 1} (d) X 4 = {(x, y, z) R 3 (x 1) + z = 1} Solution: S 4 = {(, y, ) R 3 y R} 18. Find an immersion φ: R R 3 with image Solution: has rank for all (u, z) R. please hand up Monday, 15/11 in class φ(r ) = { (x, y, z) R 3 x + y = cosh(z) }. φ(u, z) = ( cosh(z) cos(u), cosh(z) sin(u), z) cosh(z) sin(u) d (u,z) φ = cosh(z) cos(u) 1

7 19. Let X R 3 be the surface X = {(x, y, z) x = y + yz + 1} and p = (1,, ) X. (a) Find a unit normal vector field N on X. Solution: With h: R 3 R, h(x, y, z) = x + y + yz + 1, we have X = h 1 (). A unit normal field N : X S is given by N(x, y, z) = grad (x,y,z)h grad (x,y,z) h = ( 1, 1 + z, y) 1 + (1 + z) + y in particular N(p) = ( 1, 1, ). (b) Find an orthonormal basis B = (b 1, b ) for T p X. ( ) Solution: B = (b 1, b ) = (1,1,), (,, 1) (c) Compute the matrix of the shape operator of X in p wrt to N and B. Solution: We compute d p N = hence d p Nb 1 = d p Nb = (,, 1 ) = 1 b, ( ) 1, 1, = 1 b 1, and the matrix of the shape operator is d p N TpX (b 1, b ) = ( 1 1 ) (d) Find the Gauss curvature and the mean curvature wrt to N. Solution: The Gauss curvature is K X (p) = det d p N TpX = 1 and the mean curvature is 4 H X (p) = 1 trace d pn TpX =. (e) Determine the principal curvatures and the principal curvature directions of X in p. Solution: The principal curvatures k 1, k are the eigenvalues of the shape operator d p N TpX. The characteristic polynomial of this is p dpn(t) = det d p N TpX t = t 1/4 which has zeros k 1, = ± 1. The principal curvature directions are corresponding eigenspaces ( ) 1 1 ker(d p N TpX k 1, ) \ {} = ker = span {(1, ±1)} B. Thus the vectors v 1 = b 1 + b = ( ) 1 1,, 1 span the principal curvature directions. 1 1 B and v = b 1 b = ( ) 1 1,, 1

8 (f) Give a neighbourhood U T p X of in T p X, a neighbourhood V R 3 of p in R 3, and a function f : U R such that V X = {p + u + f(u)n(p) u U}. () Solution: T p X = {(x, x, y) x, y R}. Now solve ( ) ( 1, 1, ) ( h (1,, ) + (x, x, y) + f(x, x, y) = h (1 + x f/, x + f/ ), y) =, i.e. for f = f(x, x, y). We get 1 x + f/ + (x + f/ )(1 + y) + 1 = f : U = {(x, x, y) x, y R, y } R With we have (). xy f(x, x, y) = + y. V := { (x, y, z) R 3 z } please hand up 19 by Thursday, 5/11. Assume that a surface contains a straight line. Prove that then the Gauss curvature of the surface is non positive at all points of this line. Solution: Let X be a surface and c: R X, c(t) = a+vt with a, v R 3, v, be a parametrization of a straight line. Let p = c(t) be a point on the line. We have c =, hence = c (t) N(c(t)) = d c(t) Nc (t) c (t) = d c(t) Nv v. (3) If w T p X \ {}, w v then the matrix of d p N wrt the orthogonal basis (v, w) of T p X is symmetric and because of (3) we have a zero on the diagonal, hence ( ) d p N a (v, w) = a b where a = 1 w d pnv w and b = 1 w d pnw w. Hence the Gauss curvature at p is ( ) a det d p N = det = a. a b 1. Let C R be a curve and X = C R = { (x, y, z) R 3 (x, y) C }. (a) Prove that X is a surface. Solution: Let p = (x, y, z) X, hence (x, y) C. Let U R be an open neighbourhood of (x, y) in R, (, ) V R open and φ : U V a diffeomorphism so that φ (C U) = {(x, ) x R} V. Then define U := {(x, y, z) z R, (x, y) U}, V := {(x, y, z) z R, (x, y) V } and φ := φ id R : U V, (x, y, z) (φ (x, y), z). This is a diffeomorphism, φ 1 = φ 1 id R, and φ(x U) = φ((c U ) R) = ({(x, ) x R} V ) R = {(x,, z) x, z R} V

9 (b) Assume that c: R C = c(r) is a parametrization of all of C by arc length, c (t) = 1 for all t R. What are the geodesics of X? Solution: Let (p, z ) C R = X, p = c(t ), be a given point and v = (αc (t ), β) T p X, α + β = 1, be a given unit vector. We claim that the curve γ : R X, γ(t) = (c(αt + t ), βt) is the geodesic with initial conditions γ() = p, γ () = v. That γ satisfies the initial conditions is immediate. Also γ is unit speed Also γ (t) = (αc (αt + t ), β) hence γ = α c (αt + t ) + β = α + β = 1. is perpendicular to because c (s) c (s) for all s. please hand up 1 by Thursday, 9/1 γ (t) = α c (αt + t ), ) T γ(t) X = span {(c (αt + t ), ), (,, 1)}. Let M be the Riemann surface shown in the picture below. Compute the integral K over its M Gauss curvature K. Hint: Near the boundary circles C 1 and C, the metric of M is that of a cylinder. Closing M by two caps/hemishperes gives a surface of genus 3. C 1 M C please hand up by Tuesday, 14/1

10 Some Problems for the Study Week 3. For each of the following subsets S i R 3 either prove with a short argument that S i is a surface or find a subset C i S i, minimal so that S i \ C i is not a surface. (a) S a = {(x, y, z) R 3 x + y + z = 1 or x + 16y + 16z = 4} {( ) } Solution: C a = ± 5, y, z R 3 y + z = 3 15 (b) S b = {(x, y, z) R 3 x + y = z 1} Solution: This the level set h 1 () of the function For (x, y, z) S b we have h: R 3 R, h(x, y, z) = x + y z + 1. d (x,y,z) h = (x, y, z). Hence S b is a surface. (c) S c = { } (x, y, z) R 3 k Z : (x k) + y + z = 1 4 Solution: S c is a union of spheres, S c = { (x, y, z) R 3 (x k) + y + z = 1 } 4 k Z = S 1 (k,, ) k Z So away from the intersection points of these spheres, S c is a surface. Removing any intersection point (k + 1,, ), k Z, however always makes S c disconnected. Hence C c = { (k + 1,, ) k Z}. (d) S d = {(( 1 + cos(v)) sin(u), ( 1 + cos(v)) cos(u), sin(v) ) R 3 u, v R } {( Solution: C d =,, ± 3 )} (e) S e = {(x, y, z) R 3 x + xy + xz 3 + y 4 + y z 3 + z 6 = } Solution: S e = {(x, y, z) R 3 x = y z 3 } is a graph. 4. (Surface of revolution) Let f : R R + be a smooth function and X = { (x, y, z) R 3 f(x) = y + z } Compute the Gauss curvature and the mean curvature of X. Solution: A unit normal field N : X R 3 is given by N(x, y, z) = ( f(x)f (x), y, z) f(x) f (x) + y + z It suffices to compute at points in the (x, y)-plane. At a point p = (x, f(x), ), the tangent space is N(x, f(x), ) = ( f (x), 1, ) f (x) + 1, T (x,f(x),) X = span {(1, f (x), ), (,, 1)}. We now compute the shape operator in this basis. With L := f (x) + 1, d (x,f(x),) N = 1 f L + f f f f L = 1 f f f L L 3 L L 3 L Hence d (x,f(x),) N(1, f (x), ) = f L 3 (1, f (x), ) d (x,f(x),) N(,, 1) = 1 (,, 1) L

11 which gives the matrix for the shape operator, and the Gauss curvature is d p N (1, f (x), ), (,, 1) = K = f L 4 = ( f L 3 1 L f (1 + f ) ) 5. Let X R 3 be a surface, p X, and assume that the Gauss curvature of X at p is not. Prove that X can not contain more than two different straight lines through p. Hint: A straight line through p R 3 is a set of the form {p + tv t R} for some v R 3, v. Solution: Let S be the shape operator of X at p and assume that X contains three lines spanned by u, v, w T p X no two of which lie on the same line. We then have Su u = = Sv v = Sw w. Since {u, v} is a basis for T p X there are a, b R so that w = au + bv, and since w does not lie on the lines spanned by u resp. v, we have a, b. From bilinearity of the scalar product and the symmetry of the shape operator we compute = Sw w = asu + bsv au + bv = asu au + asu bv + bsv au + bsv bv = a Su u + ab ( Su v + Sv u ) + b Sv v = a Su u + ab Su v + b Sv v = ab Su v. Since ab we get Su v = u Sv =. It follows that Su = Sv =, hence S =. 6. Can a surface have Gauss curvature and mean curvature 1 at one of its points?

12 Some Problems with solutions. Some of these might be outside the context of this course. 7. Consider the surface S = { (u + v, u, uv) u, v R } R 3 and the point p = (1,, ) S. (a) Compute a unit normal vector field N on S. Solution: Let f(u, v) := (u + v, u, uv), so S = f(r ). We normalize f u f v = (u, 1, v) (1,, u) = (u, v u, 1), N(f(u, v)) = (u, v u, 1) u + (v u ) + 1 (b) Find an orthonormal basis for T p S. Solution: We use Gram-Schmidt for (f u, f v )(, 1) = ((, 1, 1), (1,, )) to get the orthonormal basis ((, 1, 1)/, (1,, )). (c) Compute the matrix of the Shape operator (corresponding to N) with respect to a basis of your choice. Say which basis you use! Solution: We use the basis (f u, f v )(, 1) = ((, 1, 1), (1,, )). dnf u (, 1) = (1,, ) dnf v (, 1) = (, 1, 1) Hence S {(,1,1),(1,,)} = ( 1/ 1/ ). (d) Compute Gauss curvature and mean curvature (corresponding to N) of S at p. Solution: The Gauss curvature is det S = 1 and the mean curvature is 1 trace S = Prove that S = { (s t, s + 3t 4, t ) s, t R +} is a regular surface in R 3. Solution: We can rewrite S as a level set h 1 () of the function h: { (x, y, z) R 3 z >, y > 3z } R, h(x, y, z) = x (y 3z )z = x yz + 3z 3. The differential of h is dh (x,y,z) = (1, z, y + 9z ) (,, ) for all (x, y, z) R For k = 1,, 1 give a two variable polynomial p(x, y) R[x, y] such that the graph S = {(x, y, p(x, y)) x, y R} of p has Gauss-curvature k at the point (,, p(, )). Solution: p(x, y) = 1 x + k y. 3. Determine the unit speed curve c: R R 3 with c() = (,, 3), ċ() = (1,, ), c() = (,, ) and curvature κ(c(t)) =, torsion τ(c(t)) = for all t R.

13 Solution: Since the torsion vanishes, the curve lies in the (affine) plane at c() spanned by ċ() and c(). The Frenet formulas simplify to T = κn, Ṅ = κt, Ḃ =, hence T = κ T and therefore Integration gives c(t) = c() + T (t) = T () cos(κt) + T () sin(κt) κ t ( sin(t) T = (,, 3) Compute the curvature and the torsion of the curve = (cos(t), sin(t), )., 1 cos(t) ) ( sin(t), = c: R R 3 with c(t) = (t, t, t ) for all t R. Solution: The torsion is since the curve lies in a plane. The curvature at c(t) is ċ c ċ 3 = 3. Consider the surface (1, t, 4t) (,, 4) (1 + 4t + 16t ) 3/ = (1,, ) (,, 4) (1 + 4t + 16t ) = (, 4, ) 3/ (1 + 4t + 16t ) = 3/, 1 cos(t) ), 3 S = {(x, y, z) R 3 x + y + z 3 = 1} and the point p = (,, 1) S. (a) Compute a unit normal vector field N on S. Solution: We normalize the gradient of h, hence N(x, y, z) = (1, y, 3z ) 1 + 4y + 9z 4 (1 + 4t + 16t ) 3/.. (b) Find an orthonormal basis for T p S. Solution: S is the zero set of h(x, y, z) = x + y + z 3 1 whose differential (at p) is dh = (1, y, 3z ), d p h = (1,, 3). An orthonormal basis for T p S = ker d p h is therefore {(, 1, ), (3/ 1,, 1/ 1)}. (c) Compute the matrix of the Shape operator (corresponding to N) with respect to this basis. Solution: Abbreviating L := 1 + 4y 4 + 9z 4 we get dn = 1 4y 18z 3 L 8y 36yz 3 L 3 1yz 6zL 54z 5 d (,,1) N = The matrix elements of the shape operator S are b i dnb j where b i, i = 1, are the two basis vectors of 3b. Hence S {(,1,),(3/ 1,, 1/ 1)} = 1 ( ) (d) Compute Gauss curvature and mean curvature (corresponding to N) of S at p. Solution: The Gauss curvature is det S = 3 and the mean curvature is 1 13 trace S =

14 33. For each of the following two subsets S R 3 decide, whether S is a surface. If S is a surface, write it as a level set, i.e. find a function h: R 3 R such that S = h 1 (). If S is not a surface, find a subset C S as small as possible, such that S \ C is a surface. Hint: You need not prove that S resp. S \ C actually is a surface. (a) S = {(v cos(u), v sin(u), v ) u, v R}, Solution: h(x, y, z) = x + y z. (b) S = {(u, (u 1) cos(v), (u 1) sin(v)) u, v R}, Solution: C = {(±1,, )} 34. Let S R 3 be a surface, p S, and assume that the Gauss curvature of S at p is 1 and the mean curvature vanishes. Prove that there are two perpendicular asymptotic directions in the tangent space T p S. Solution: The shape operator has two orthonormal eigenvectors e ± with eigenvalues ±λ. The asymptotic directions are e + ± e.