CURVATURE. 1. Problems (1) Let S be a surface with a chart (φ, U) so that. + g11 g 22 x 1 and g 22 = φ

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1 CURVTURE NDRÉ NEVES 1. Problems (1) Let S be a surface with a chart (φ, U) so that. = 0 x 1 x 2 for all (x 1, x 2 ) U. Show that [ 1 K φ = 2 g 11 g 22 x 2 where g 11 = x 1. x 1 ( ) x2 g 11 + g11 g 22 x 1 and g 22 = x 2. x 2 ( )] x1 g 22, g11 g 22 (2) Let S be a compact surface with no boundary and let P = {x S : K(x) 0}, where K is the gaussian curvature. a) Show that the the Gauss map of S restricted to P is surjective. In other words, for every p S 2 = {x 2 + y 2 + z 2 = 1}, there is x P so that N(x) = p. b) Show that S K + d 4π, where K + (x) = max{k(x), 0} for all x S. HINT: If N, M are two surfaces and F is a surjective smooth map F : N M, then area(m) detdf d. (3) (Do Carmo, 4.2, Ex 8) Let G : R 3 R 3 be a map such that N G(p) G(q) = p q for all p, q R 3. Show the existence of p 0 R 3 and B a matrix in O(3) such that (4) (Do Carmo, 4.2, Ex 11) G(p) = B(p) + p 0, for all p R 3. 1

2 2 NDRÉ NEVES (a) Let S R 3 be a regular surface and let F : R 3 R 3 be a distance preserving map (like the one in the previous exercise) such that F (S) S. Prove that the restriction of F to S is an isometry of S. (c) Use (a) to show that the group of isometries of {x 2 +y 2 +z 2 = 1} is contained in the group of orthogonal linear maps of R 3, which is O(3). (c) Give an example to show that there are isometries φ : S 1 S 2 which cannot be extended into distance preserving maps F : R 3 R 3. (5) (Do Carmo, 4.2, Ex 12) Let C = {x 2 + y 2 = 1}. Construct an isometry φ : C C such that the set of fixed points {p C : φ(p) = p} contains exactly two points. (6) (Do Carmo, 4.3, Ex 2) Show that if we have a parametrization of a surface such that g 11 = g 22 = λ(x 1, x 2 ) and g 12=0 then K = 1 (log λ). 2λ Conclude that if g 11 = g 22 = (x x2 2 + c) 2, then K = 4c, where c is a constant. (7) (Do Carmo, 4.3, Ex 3) Verify that the surfaces parametrized by x(u, v) = (u cos v, u sin v, log u) x(u, v) = (u cos v, u sin v, volume) have equal Gaussian curvature at points x(u, v) and x(u, v) but x x is not an isometry. This shows that the converse to Gauss Egregium s Theorem is not true. (8) Show that there exist no surface with g 11 = g 22 = 1 and 11 = 1, 22 = 1, and 12 = 0. (9) Let S be a compact surface in R 3 with no boundary which has positive Gaussian curvature and denote the Gauss map by N : S {x 2 + y 2 + z 2 = 1}. Show that if γ is a geodesic in S which divides S into a set and another set B (i.e., S = B and = B = γ), then area(n()) = area(n(b)). Hint: If you have a diffeomorphism F : S {x 2 + y 2 + z 2 = 1} R 3, what is the formula for area(f ()) in terms of DF and the set?

3 CURVTURE 3 (10) Let S be the cylinder {x 2 + y 2 = 1} in R 3, with the chart φ : [0, 2π] R S, φ(θ, z) = (cos θ, sin θ, z). and let γ be a simple closed curve in S. (a) Let γ be a simple closed curve in S for which there is a path α : [0, 1] [0, 2π] R so that α(0) = (0, z 0 ), α(1) = (2π, z 0 ), and γ = φ α. In other words, γ loops once around the z-axis. If ν denotes a unit normal vector to γ show that k. νdσ = 0. γ (b) Let γ be a simple closed curve in S for which there is a path α : [0, 1] [0, 2π] R so that α(0) = (θ 0, z 0 ) = α(1). In other words, γ is the boundary of a disc D in S. Show that γ is not a geodesic. (11) (Do Carmo, 4.4, Ex 5 a) and 6) Consider the torus of revolution generated by rotating the circle (x a) 2 + z 2 = r 2 0, y = 0 about the z-axis (a > r 0 > 0). The parallels generated by the points (a + r 0, 0), (a r 0, 0), (a, r 0 ) are called the maximum parallel, the minimum parallel, and the upper parallel, respectively. (a) Check which of these parallels is a geodesic (i.e., k g = 0). (b) Compute the geodesic curvature of the upper parallels. (12) (Do Carmo, 4.4, Ex 7) Intersect the cylinder {x 2 + y 2 = 1} with a plane passing through the x-axis and making an angle θ with the xy-plane. (a) Show that the intersecting curve is an ellipse C. (b) Compute the absolute value of the geodesic curvature of C in the cylinder at the point where C meets the axes. (13) Let S be a surface and suppose there is F : U R 2 S so that for every (t 0, s 0 ) U i) the curves α(t) = F (t, s 0 ) and α(s) = F (t 0, s) are geodesics and ii) t F (t 0, s 0 ). s F (t 0, s 0 ) = 0 for every s 0, t 0 fixed. Show that the Gaussian curvature is zero. (14) Consider the surface S parametrized by φ(u, v) = (u sin(v), u cos(v), v), (u, v) R 2.

4 4 NDRÉ NEVES (a) Show that φ is a chart, i.e., injective and u φ, v φ are linearly independent. (b) Compute the Gaussian curvature and mean curvature of S. (15) Show that if S is a connected surface with nonnegative Gaussian curvature and zero mean curvature, then it must be a plane. (16) Given a positive function r : [a, b] (0, + ) consider the surface of revolution S = {(r(s) cos θ, r(s) sin θ, s) : s [a, b], 0 < θ < 2π}. (a) Make a picture of S to make sure you understand what it is. (b) If the function r has a local minimum at s 0 with r (s 0 ) > 0, show that the Gaussian curvature of S at p = (r(s 0 ) cos θ, r(s 0 ) sin θ, θ) is negative. (c) Suppose the Gaussian curvature of S is nonnegative and r(s) r(a) = r(b) for all s. Show that S must be a cylinder. 2. Solutions (1) During the proof of Gauss s Egregium Theorem we saw that K = x 1 (φ T x 2 x 2 ).φ x1 x2 (φ T x 1 x 2 ).φ x1 g 11 g 22, where φ x1 = x1 φ, φ x1 x 2 = 2 x 1 x 2 φ and so on. The thing we want to show is [ 1 K = 2 g 11 g 22 x 2 ( ) x2 g 11 + ( )] x1 g 22. g11 g 22 x 1 g11 g 22 Let N denote a unit normal to the surface. We have and so φ T x 1 x 2 = φ x1 x 2 (φ x1 x 2.N)N x2 (φ T x 1 x 2 ) = φ x1 x 2 x 1 x2 (φ x1 x 2.N)N (φ x1 x 2.N) x2 N. Thus, using the fact that x1 φ, x2 φ are orthogonal x2 (φ T x 1 x 2 ).φ x1 = x2 (φ T x 1 x 2. x1 ) φ T x 1 x 2.φ x1 x 1 = x2 (φ x1 x 2. x1 ) φ T x 1 x 2.φ x1 x 1 = x2 (φ x1 x 2. x1 ) (φ x 1 x 2.φ x1 )(φ x1.φ x1 x 1 ) (φ x 1 x 2.φ x2 )(φ x2.φ x1 x 1 ) g 11 g 22 = 2 x 2 g 11 ( x 2 g 11 ) 2 ( x 2 g 22 ) g 11 4g 22 Now φ T x 2 x 2 = φ x1 x 2 (φ x1 x 2.N)N

5 CURVTURE 5 and so, using again the fact that x1 φ. x2 φ = 0, we have x1 (φ T x 2 x 2 ).φ x1 = x1 (φ x2 x 2.φ x1 ) φ T x 2 x 2.φ x1 x 1 = x1 x2 (φ x2.φ x1 ) x1 (φ x2.φ x2 x 1 ) φ T x 2 x 2.φ x1 x 1 = x1 (φ x2.φ x2 x 1 ) φ T x 2 x 2.φ x1 x 1 = x1 (φ x2.φ x2 x 1 ) (φ x 2 x 2.φ x1 )(φ x1.φ x1 x 1 ) (φ x 2 x 2.φ x2 )(φ x2.φ x1 x 1 ) g 11 g 22 = x1 (φ x2.φ x2 x 1 ) + (φ x 1 x 2.φ x2 )(φ x1.φ x1 x 1 ) + (φ x 2 x 2.φ x2 )(φ x1.φ x2 x 1 ) g 11 g 22 = 2 x 1 g x 2 g 22 x2 g 11 4g 22 + x 1 g 22 x1 g 11 4g 11. From this last two identities direct computation shows that [ ( ) 1 = 2 x2 g 11 + g 11 g 22 x 2 g11 g 22 x 1 x1 (φ T x 2 x 2 ).φ x1 x2 (φ T x 1 x 2 ).φ x1 g 11 g 22 ( )] x1 g 22. g11 g 22 (2) Let s solve (a) first. Choose N to be the exterior unit normal and this part is important because if N were the interior unit normal the argument would need to have some (trivial) changes. Choose p S 2 we want to find x S with N(x) = p. Let P t = { x R 3 : p.x = t}. For t very large and positive we have P t S = and clearly there is some s so that P s S is not empty. So we can consider t 0 = inf{t : P s S = for all s t}. We have P t0 S, otherwise we could find ε small so that P t0 ε S = and this contradicts the minimality property of t 0. Let x P t0 S. We want to argue that N(x) = p. From the graphical decomposition lemma we know that in a neighborhood of x we can write S as being the graph of a function defined on a set of T x S. More rigorously, there is an open set U T x S, an open neighborhood V of x in R 3, and a function f : U R so that S V = {f(y)n(x) + y + x : y U}. The function f has f(0, 0) = 0 and f(0, 0) = 0 necessarily (why?). Now we must have z.p t 0 for all z S and x.p = t 0 (why?). Thus using the graphical parametrization t 0 (f(y)n(x) + y + x).p = f(y)n(x).p + y.p + t 0. In other words g(y) = f(y)n(x).p + y.p + t 0 has a global maximum at y = 0 and, because f(0, 0) = 0, this implies p is orthogonal to T x S (because x g(0, 0) = y g(0, 0) = 0). The fact that N was the exterior unit normal implies N(x) = p instead of N(x) = p.

6 6 NDRÉ NEVES We just need to conclude that K(x) 0. Well, by what we have done we have T x S + s = P t0 and S lies all in one side of P t0. Thus if K(x) < 0, the surface would lie on both sides of P t0, which is impossible. For part (b) consider L ε = {x S : K(x) > ε} which is a surface as well (the set {x : K(x) 0}) is not surface strictus sensus). Restrict the Gauss map to N : L ε S 2 which is surjective. Then by the hint 4π det N d = K d = Kd + K d L ε L ε L 0 L ε\l 0 Kd + εarea(s) = K + d + εarea(s). L 0 S Making ε tend to zero the result follows. (3) The first step is to show that G is a linear map. Fix a point p and consider α 1 (t) = p + tx, where X is any vector in R 3. Set F (t) = G α 1 (t) G(q). By differentiation in the t variable when t = 0 we have F (t) 2 = tx + p q 2 = F (0).F (0) = X.(p q) = DG p (X).(G(p) G(q)) = X.(p q) = X.(DG p ) (G(p) G(q)) = X.(p q) for all X and all q. In the last line we use the fact that if is a 3 3 matrix with transpose T, then (X).Y = X. T (Y ) for every vectors X, Y. Because the identity we derived is valid for all X it is simple to conclude that (DG p ) (G(p) G(q)) = p q for every q R 3. In particular the linear map (DG p ) T is surjective and thus injective. Denoting by B its inverse we have from the above formula that G(p) = G(q) + B(p q)for every p, q R 3. This show that G is a linear map (just take p 0 = G(0) B(0) for some q fixed.) Using again the hypothesis we obtain that B(p q) = G(p) G(q) = p q. Thus B preserves distances and so it is an isometry of R 3 as well. It is easy to conclude that B O(3) by differentiating the identity below and set t = 0 B(X + ty ).B(X + ty ) = X + ty 2 = X 2 + 2tX.Y + t 2 Y 2. (4) First a). From the previous exercise we know that F is a linear isometry of R 3, which means that DF p = DF 0 for all p in R 3 (i.e. F is linear) and DF 0 (X).DF 0 (Y ) = X.Y for all vectors X, Y in

7 CURVTURE 7 R 3. In particular for every p S we have that DF p = DF 0 and DF p (X).DF p (Y ) = X.Y for all X, Y in T p S, which means that F is an isometry of S. Technically speaking one should show that if F is an ambient map of R 3, then the restriction of F to S has the property that for every X T p S, then DF p (X) as it was defined in class is nothing but DF p (X) as it was defined in multivariable calculus. This is just the chain rule and so I will not say anything else to avoid the risk of just ending up making it more confusing. Now b). The orthogonal linear transformations are just those 3 3 matrices which have T = Id, i.e., 1 = T. In this case we have (X) 2 = (X).(x) = X. T (X) = X.X = X 2 and so send the unit sphere into the unit sphere. Moreover (X) (Y ) = (X Y ) = X Y, and so its distance preserving in the sense of Exercise 8. Thus a) can be applied to conclude is an isometry of the sphere. Finally c). Take F (x, y, 0) = (cos x, sin x, y) which is an isometry from part of the plane {z = 0} into part of the cylinder {x 2 +y 2 = 1}. F is not distance preserving because F (0, 0, 0) F (θ, 0, 0) = 2(1 cos θ) and (0, 0, 0) (θ, 0, 0) = θ. (5) Consider F (x, y, z) = (x, y, z). This is an isometry of the cylinder S = {x 2 + y 2 = 1} because F send the S into S and F is an orthogonal linear transformation. Finally F (x, y, z) = (x, y, z) then z = 0 and y = 0. But x 2 + y 2 = 1, which means x = ±1, i.e., the only fixed points are (1, 0, 0) and ( 1, 0, 0). (6) Let s use exercise 1. Set λ = e 2u for convenience. Then we have g11 g 22 = e 2u, x2 g 11 (g 11 g 22 ) 1/2 = 2 x 2 u, x1 g 22 (g 11 g 22 ) 1/2 = 2 x 1 u. Thus, recalling that h = x1 x 1 h + x2 x 2 h, we obtain K = e 2u u = 1 ln λ. 2λ (Recall that ln λ = 1 2 ln λ). To be consistent with the rest of the notation I should use λ = (x x2 2 + c) 2 for the last part. Then and xi ln λ = 2 xi ln(x x 2 4x i 2 + c) = x x2 2 + c x 2 4 i x i ln λ = x x2 2 + c + 8x 2 i (x x c)2

8 8 NDRÉ NEVES Thus 8 ln λ = x x2 2 + c + 8(x2 1 + x2 2 ) (x x2 2 + = 8c c)2 (x = 8cλ x2 2 + c)2 and so K = 1 ln λ = 4c. 2λ (7) We have x u = (cos v, sin v, u 1 ), x = ( u sin v, u cos v, 0) v and so g 11 = x u. x u = 1 + u 2, g 12 = x u. x v = 0, g 22 = x v. x v = u2. Because v g 11 = 0 we obtain from Exercise 1 ( ) 1 K = 2 2u 1 = 1 + u 2 u 1 + u 2 (1 + u 2 ) 2 We have x x = (cos v, sin v, 0), = ( u sin v, u cos v, 0) u v and so ḡ 11 = x. x u u = 1, ḡ 12 = x. x u v = 0, ḡ 22 = x. x v v = 1 + u2. Using the formula in Exercise 1 we have ( ) 1 K = 2 2u 1 = 1 + u 2 u 1 + u 2 (1 + u 2 ) 2. Therefore the surfaces have the same curvature. If the map x x 1 were an isometry then by Proposition 1, page 220, we would have g 11 = ḡ 11, g 12 = ḡ 12, and g 22 = ḡ 22. Because this is false the surfaces are not isometric. (8) If there was such a surface then from the definition of Gaussian curvature we would have K = ( )(g 11g 22 g 2 12 ) 1 = 1. But from Gauss s Theorem we know that K is intrinsic, i.e., depends only on g 11, g 12 and g 12. Thus K must be zero because we can surely put coordinates on a plane which have g 11 = g 22 = 1, g 12 = 0 and a plane has Gaussian curvature zero. (9) Because K > 0, we have from Gauss-Bonnet that S is diffeomorphic to a sphere. Thus andb are both diffeomorphic to a disc and

9 CURVTURE 9 = B = γ is a geodesic. Hence, if ν denotes the interior unit normal to we have Kd + k.νdσ = 2π = Kd = 2π. Likewise B Kd = 2π. I will show now that area(n()) = Kd = 2π and this completes the exercise. The first remark is that the Gaussian map N must be a diffeomorphism. Why? Well, because K = det = detdn, we have from the inverse function theorem that N is locally a diffeomorphism. nd it happens that locally diffeomorphisms from a sphere into a sphere must be global diffeomorphism. The second remark is to show the hint, namely that area(f ()) = detdf d. Suppose that φ : D R 2 is a chart for. Then ψ = F φ is a chart for N = {x 2 + y 2 + z 2 = 1} because F is a diffeomorphism. From the lectures we know that area(ψ(d)) = ψ x ψ y dxdy. D Now let {e 1, e 2 } be a basis for T φ(p) which has df p (e 1 ) = λ 1 e 1 and df p (e 2 ) = λ 2 e 2, i.e., {e 1, e 2 } is an eigenbasis for df. Then and thus x (p) = a 1e 1 + b 1 e 2 y (p) = a 2e 1 + b 2 e 2 x y = a 1b 2 a 2 b 1 e 1 e 2. Furthermore ( ) ψ x (p) = df φ(p) x (p) = λ 1 a 1 e 1 + λ 2 b 1 e 2 and ( ) ψ y (p) = df φ(p) y (p) = λ 1 a 2 e 1 + λ 2 b 2 e 2. Thus ψ x ψ y = λ 1λ 2 a 1 b 2 a 2 b 1 e 1 e 2 = detdf φ(p) x y.

10 10 NDRÉ NEVES (This formula was long to derive but it should be highly expected). Therefore we have area(f (φ(d))) = area(ψ(d)) = ψ D x ψ y dxdy. = detdf φ(p) x y dxdy = detdf d. D φ(d) Covering with a finite number of charts we arrive at area(f ()) = detdf d. Using this formula with F = N the Gauss map we obtain area(n()) = detdn d = K d = Kd = 2π. (10) If the path α(t) = (θ(t), z(t)), choose z 1 so that z(t) > z 1 for all t [0, 1] and consider the path β(t) = (2πt, z 1 ). Note that φ β is a geodesic and there is a region R in S so that S = γ φ β. To recognize this last fact just let R = φ(), where is the region in [0, 2π] R below α and above β. Note that R is diffemorphic to a cylinder and thus has zero Euler characteristic. Using Gauss-Bonnet we obtain that Kd + k. νdσ + k. νdσ = 0, R γ where ν is the interior unit normal. Because K = 0 and φ β is a geodesic we obtain that γ φ β k. νdσ = 0, which is what we wanted to show. For the second one note that the Euler characteristic of D is one and so by the Gauss-Bonnet Theorem we have Kd + k. νdσ = 2π = k. νdσ = 2π. D Thus, γ cannot be a geodesic. (11) Consider a parametrization of S γ φ(s, θ) = (r(s) cos θ, r(s) sin θ, z(s)). Then s = (r cos θ, r sin θ, z ), θ γ = ( r sin θ, r cos θ, 0),

11 CURVTURE 11 2 φ ( s) 2 = (r cos θ, r sin θ, z ), 2 φ = ( r cos θ, r sin θ, 0), ( θ) 2 and so 2 φ θ s = ( r sin θ, r cos θ, 0). g 11 = (r ) 2 + (z ) 2 = 1, g 12 = 0, g 22 = r 2, where I am assuming, without loss of generality, that (r(s), z(s)) is parametrized by arc-length. If α(t) = φ(s(t), θ(t)) is a curve parametrized by arc-length then, denoting differentiation with respect to t with a dot, we have α = d dt (ṡ sφ + θ θ φ) = s s φ + θ θ φ + (ṡ) 2 2 ssφ + 2ṡ θ 2 sθ φ + ( θ) 2 2 θθ φ. Consider the orthonormal basis X 1, X 2 for the tangent plane of S where We have X 1 = s, X 1 2 = r = ( sin θ, cos θ, 0). θ α.x 1 = s + (ṡ) 2 (r r + z z ) ( θ) 2 rr = s ( θ) 2 rr, where we use the fact that and (r ) 2 + (z ) 2 = 1 = r r + z z = 0, α.x 2 = θ + 2ṡ θrr. The normal vector N is given by N = X 1 X 2 and E = N α is given by E = N (ṡ s φ + θ θ φ) = N (ṡx 1 + θrx 2 ) Therefore = (X 1 X 2 ) (ṡx 1 + θrx 2 ) = ṡx 2 θrx 1. k g = α.e = ṡ( θ + 2ṡ θrr ) θr( s ( θ) 2 rr ) For the case in point we have r(s) = a + r 0 cos(s/r 0 ), z(s) = r 0 sin(s/r 0 ), where (r ) 2 + (z ) 2 = 1. The curves are maximum parallel: α 1 (t) = φ(0, t/r 0 ), minimum parallel: α 2 (t) = φ(r 0 π, t/r 0 ), upper parallel: α 3 (t) = φ(r 0 π/2, t/r 0 )

12 12 NDRÉ NEVES I use the term t/r 0 instead of t because I want the curves to be parametrized by arc-length. In any case we have θ = r0 1, ṡ = θ = 0, and r = sin(s/r 0 ). Using the formula above for k we get k g (α 1 ) = r 3 0 r(0)r (0) = 0 and k g (α 1 ) = r 3 0 r(r 0π)r (r 0 π) = 0 k g (α 3 ) = r0 3 r(r 0π/2)r (r 0 π/2) = a r0 2. Note that k g (α 3 ) compute the geodesic curvature of the upper parallels. (12) We use the coordinate chart k g = d du φ(std, z) = (cos s, sin s, z) which makes it into a local isometry with Euclidean plane, i.e., g 11 = 1 = g 22, F = 0. The plane which intersects the cylinder is P = {z = y tan θ} and so the curve C is parametrized by C(t) = (cos t, sin t, sin t tan θ) = φ(t, sin t tan θ). If e 1 = (1, 0, 0), e 2 = (0, cos θ, sin θ) is an orthonormal basis for P, then C = {xe 1 + ye 2 x 2 + cos 2 θy 2 = 1} and so we see that C is indeed an ellipse in the plane P. To compute its curvature vector we have the little annoying thing that C(t) is not a parametrization by arc-length. Suppose then that t = t(u) is a change of variable which makes C(u) parametrized by arc length. Then, if E denotes a unit vector lying in the tangent plane to to the cylinder which is orthogonal to C (t), we know that k g = k.e. Now and thus ( C (t) dt du dc du = C (t) dt du = ).E = = C (t) 2 C (t).e. dt du = C (t) 1 ( ) dt 2 C (t).e + d2 t du (du) 2 C (t).e = ( ) dt 2 C (t).e du Let s use this formula. C intersects the axis when t = 0, t = π. In this case C (t) = C(t) and so C (0) = (1, 0, 0), C (π) = (1, 0, 0). In both cases C has the direction of the normal vector to the cylinder and so the term C (t).e is zero because E lies in the tangent plane of the cylinder. Thus k g = 0 at those points.

13 CURVTURE 13 (13) One obvious consequence of i) is that s F and t F are never zero. Hence from ii) we get that s F (t, s), t F (t, s) are linearly independent which means that for every (t 0, s 0 ) U, there is a small neighborhood V so that F restricted to V is a chart. Therefore we can use the formula from the first exercise in this problem sheet. From problem 1) it suffices to see that s g 11 = t g 22 = 0. Condition ii) is saying that g 12 = 0. The fact that t F (t, s) is a geodesic in S implies that 2 ttf.e = 0, where E is a unit vector lying in the tangent plane of S and perpendicular to t F (the tangent vector to the curve). Thus, because g 12 = 0 we have that E is a multiple of s F and so 2 ttf. s F = 0. Therefore s g 11 = 2 2 F st. t F = 0. Same reasoning shows that t g 22 = 0. (14) (a) φ is clearly smooth. Suppose φ(u, v) = φ(u, v ). Then we must have v = v by looking at the third component of φ. Thus sin(v) = sin(v ) and cos(v) = cos(v ). If sin(v) 0 we conclude that u = u by looking at the first component of φ. If sin(v) = 0 then cos(v) 0 and so we conclude that u = u by looking at the second component of φ. Note that = (sin v, cos v, 0), u v = (u cos v, u sin v, 1) and thus u φ. v φ = 0. Because u φ, v φ are never zero, this implies they must be linearly independent if they are orthogonal. (b) The matrix that represents the metric g is given by [ ] g11 g 12 = g 21 g 22 The unit normal vector is We have N = uφ v φ u φ v φ [ ] u 2. ( cos v, sin v, u) =. 1 + u 2 2 uuφ = (0, 0, 0), 2 uvφ = (cos v, sin v, 0), 2 vvφ = ( u sin v, u cos v, 0) and so the matrix that represents the second fundamental form is [ ] [ ] = u 2 1 0

14 14 NDRÉ NEVES Therefore, the matrix that represents dn in the basis u φ, v φ is given by [ ] [ ] σ = g = 1 + u 2 0 (1 + u 2 ) [ ] = 1 + u 2 (1 + u 2 ) 1. 0 Hence K = detσ = (1 + u 2 ) 3/2, H = 1 trσ = 0. 2 (15) If H = 0, then λ 1 + λ 2 = 0, which means λ 1 = λ 2. Thus the Gaussian curvature becomes K = λ 1 λ 2 = λ The condition K 0 implies then that K = λ 2 1 = 0, which means λ 1 = λ 2 = 0. Therefore, the surface is umbilic and by a Theorem in the notes we know that it must be contained in a plane (the case of being contained in a sphere it is impossible because K = 0.) (16) (b) We have r φ = (r (s) cos θ, r (s) sin θ, 1), θ φ = ( r(s) sin θ, r(s) cos θ, 0) and so the matrix that represents the metric g is given by [ ] [ g11 g 12 (r = ) 2 ] g 21 g 22 0 r 2. The normal vector is given by Finally, N = rφ θ φ r φ θ φ = ( cos θ, sin θ, r ) 1 + (r ) 2. rr φ = (r cos θ, r sin θ, 0), rθ φ = ( r sin θ, r cos θ, 0) θθ φ = ( r cos θ, r sin θ, 0) and so the matrix that represents the second fundamental form is [ ] [ ] r 0 = (r ) 2 0 r The matrix that represents dn is given by [ σ = g 1 1 ((r = ) 2 + 1) 1 ] [ ] 0 r (r ) 2 0 r 2 0 r [ 1 ((r = ) 2 + 1) 1 r ] (r ) 2 0 r 1.

15 CURVTURE 15 Therefore K = (r ) 2 + 1) 3/2 r 1 r. t a local minimum with r (s 0 ) > 0 we have K = r (s 0 ) r(s 0 ) < 0. If the Gaussian curvature is nonnegative we have K = (r ) 2 + 1) 3/2 r 1 r 0 = r 0. Thus r is a concave function with r(a) = r(b) and this implies that r(s) = r(a) for all a s b. quick way to see this is to note that r 0 implies r is nonincreasing, which means r (a) r (s) r (b). But r(s) r(a) and r(s) r(b) implies r (a) 0 and r (b) 0. Therefore r (s) = 0 for all s and this means r is constant.