HOMEWORK 2 - SOLUTIONS

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1 HOMEWORK 2 - SOLUTIONS ANDRÉ NEVES Exercise 15 of Chapter 2.3 of Do Carmo s book: Okay, I have no idea why I set this one because it is similar to another one from the previous homework. I might have meant something else but has been to long to know what it was... Anyho, assuming that h is indeed a diffeomorphism then β is just a reparametrization of α so the length of the curves must be the same, as we saw in the other homework sheet. The fact that h is indeed a diffeo could require some details if we are into... details. The map h is one to one because when Do Carmo says the curve is regular he is assuming the parametrizations α β are already one to one onto its image thus h is bijective. The issue is to show that h is differentiable its derivative never zero. This is not totally formal because β 1 is not defined on an open set of R 3 but only in β(j). The trick is to pick p = β(t 0 ) choose two vectors e 1, e 2 so that {β (t 0 ), e 1, e 2 } is a basis of R 3. We then set F (t, u, v) = β(t) + ue 1 + ve 2. The important property is that DF (t 0, 0, 0) is a 3 3 matrix where the columns (or the rows, I always get confused) are β (t 0 ), e 1 e 2. But these vectors are linearly independent, which means DF (t 0, 0, 0) is bijective. By the inverse function theorem it must have an inverse F 1 defined in a neighbourhood of V of p = F (t 0, 0, 0). Finally, consider the projection π(t, u, v) = t. Then π F 1 (β(t)) = π F 1 (F (t, 0, 0)) = π(t, 0, 0) = t so, near p, β 1 = π F 1 which means h = π F 1 α. Thus h is a composition of three differentiable maps which means it must also be differentiable. I leave you to check that h (s) 0 for all s. Exercise: Let φ : U R 2 R 3 be a chart, h : W R 2 U R 2 a diffeomorphishm, set ψ = φ h. Show that v (u, v) = φ v (h(u, v)) det Dh (u, v) From the cain rule, using the notation (u, v ) = h(u, v) we have = φ + φ v v v = φ v + φ v v v. I am keeping the notation simple but where is φ you should really see φ (h(u, v)), where is you should really see (u, v), so on. 1

2 2 ANDRÉ NEVES Hence v = ( v v v ) φ v v = (detdh) φ v. Therefore, taking the norm on both sides, we obtain v (u, v) = φ v (h(u, v)) det Dh (u, v). Exercise: Let S, M be two surfaces in R 3, F : S M a smooth map, φ : U R 2 R 3 a chart with φ(0) = p. Recall that the definition of was DF p (α (0)) = DF p : T p S T F (p) M d(f α) (0) where α : ( ε, ε) S is such that α(0) = p. a) Using this definition argue that DF p (0) = (F φ) (0). b) Use the previous exercise to compute DF 0 when S = {(x, y, z) z = 2x y 3 } F (x, y, z) = (z, y, 0). a) Set e 1 = (1, 0), e 2 = (0, 1). Consider the path α i (t) = te i in U the path γ i (t) = φ α i (t) in S. From the chain rule we have Thus DF p (0) ( ) dγi = DF p (0) dγ i φ (0) = (0). = d(f γ i) (0) where the last identity follows from the chain rule. = d((f φ) α i) (F φ) (0) = (0), b) Use the previous exercise to compute DF 0 when S = {(x, y, z) z = 2x y 3 } F (x, y, z) = (z, y, 0). In this case M = {z = 0} F : S M R 3. Because F (0) = 0 we need to compute T 0 S T 0 M. The latter one is easy because M is a vector space so T 0 M = {z = 0}. The first one is not hard as well because we have the chart φ : R 2 S, φ(x, y) = (x, y, 2x y 3 )

3 so T 0 S = span HOMEWORK 2 - SOLUTIONS { } φ φ (0), x y (0) = span{(1, 0, 2), (0, 1, 0)} = {z = 2x}. To determine DF 0 it suffices to determine DF 0 (1, 0, 2) DF 0 (0, 1, 0). Note that F φ = (2x y 3, y, 0) so, using the previous exercise, we have DF 0 (1, 0, 2) = DF 0 x (0) (F φ) = (0) = (2, 0, 0) x DF 0 (0, 1, 0) = DF 0 y (0) (F φ) = (0) = (0, 1, 0). y Exercise: Let α : I S be a curve in S parametrized by arc length so that λ 1 λ 2 > 0 at the point α(0). Show that α (0) min{ λ 1, λ 2 }, where λ 1, λ 2 are the principal curvatures of the second fundamental form. Let p = α(0) denote by e 1, e 2 in T p S the principal directions of A, the second fundamental form, meaning that A(e 1 ) = λ 1 e 1 A(e 2 ) = λ 1 e 2 e 1, e 2 form an orthonormal basis for T p S. Because α (0) T p S α (0).α (0) = 1, one has the existence of some θ for which Therefore α (0) = cos θe 1 + sin θe 2. A(α (0)).α (0) = cos θa(e 1 ).α (0) + sin θa(e 2 ).α (0) = λ 1 cos θe 1.α (0) + λ 2 sin θe 2.α (0) = λ 1 cos 2 θ + λ 2 sin 2 θ + (λ 1 + λ 2 )e 1.e 2 cos θ sin θ = λ 1 cos 2 θ + λ 2 sin 2 θ. Either λ 1, λ 2 are both positive or both negative so in either case we have A(α (0)).α (0) = λ 1 cos 2 θ + λ 2 sin 2 θ. If N denotes the unit normal to S we saw in class that α (0).N = A(α (0)).α (0) thus, denoting by σ = min{ λ 1, λ 2 }, we have α (0) α (0).N = A(α (0)).α (0) = λ 1 cos 2 θ + λ 2 sin 2 θ σ cos 2 θ + σ sin 2 θ = σ = min{ λ 1, λ 2 }. Exercise 4 of Chapter 3.2 of Do Carmo s book: No. The plane P = {z = 0} has principal curvatures zero but the curve α(t) = (cos t, sin t, 0) doe not have curvature zero. Exercise 8 of Chapter 3.2 of Do Carmo s book:

4 4 ANDRÉ NEVES Before I solve it let me make some remarks. If R θ : R 3 R 3 denotes a rotation by angle θ which fixes the z-axis, then all surfaces we consider have that R θ (S) = S, i.e., the surfaces are invariant under rotations which fix the z-axis. The consequence of this remark is that if N(p) denotes the unit normal vector at the point p S, then N(R θ (p)) = R θ (N(p)). In other words, the image of the Gauss map is invariant under rotations which fix the z-axis. Each of the surfaces is of the form S = {f(x, y, z) = 0} thus N = f f 1. Our previous remark means it suffices to look at N(x, 0, z). a) f(x, y, z) = z x 2 y 2 thus N(x, 0, z) = ( 2x, 0, 1)(4x 2 + 1) 1/2, where z = x 2 x R. It is simple to see that x ( 2x, 0, 1)(4x 2 + 1) 1/2 maps the real line into S 2 {y = 0, z > 0}, where S 2 denotes the unit sphere. Because the image of the Gauss map must be invariant under rotations which fix the z-axis we obtain that the image must be S 2 {z > 0}, i..e, the northern hemisphere. b) f(x, y, z) = z x 2 y 2 thus N(x, 0, z) = ( x, 0, z)(x 2 + z 2 ) 1/2, where 1 + z 2 = x 2. The curve 1 + z 2 = x 2, y = 0 has two connected components, without loss of generality, we can consider just the one given by x = 1 + z 2, y = 0. Why is that? Well, if C 1, C 2 are the two connected components then C 2 = R π (C 1 ), i.e, one connected component can be obtained by the other if we rotate it 180 degrees around the z-axis so, for the purpose of computing the image of the Gauss map, it is enough to just consider C 1 because we already know that the image will be invariant by rotations which keep the z-axis fixed. In sum, it suffices to see what is the image of (recall x = 1 + z 2 ) We have N(x, 0, z) = ( 1 + z 2, 0, z)(1 + 2z 2 ) 1/2, z R. lim N(x, 0, z) = ( 1, 0, z + 1)2 1/2, lim N(x, 0, z) = ( 1, 0, 1)2 1/2 z 1 2 < z < 1 for all z R z 2 2 Therefore the image of N(x, 0, z) is the part of the great circle S 2 {y = 0} which lies in the sector {(cos θ, 0, sin θ) 3π/4 < θ < 5π/4}.

5 HOMEWORK 2 - SOLUTIONS The image of the Gauss map is then {x 2 + y 2 + z 2 = 1} { z < 2 1/2 }. c) f(x, y, z) = cosh 2 z x 2 y 2 thus N(x, 0, z) = ( x, 0, sinh z cosh z)(x 2 + sinh 2 z cosh 2 z) 1/2, where cosh 2 z = x 2. The curve cosh 2 z = x 2, y = 0 has two connected components so it is enough to consider the one given by x = cosh z, y = 0. In this case the expression simplifies to N(x, 0, z) = ( 1, 0, sinh z)(1 + sinh 2 z) 1/2 = ( 1, 0, sinh z) cosh 1 z. We have lim N(x, 0, z) = (0, 0, 1), lim N(x, 0, z) = (0, 0, 1) z + z 1 < sinh z < 1 for all z R. cosh z Thus the image of N(x, 0, z) is S 2 {y = 0} x < 0 the image of the Gauss map is the sphere minus the north pole minus the south pole.