HOMEWORK 3 - GEOMETRY OF CURVES AND SURFACES. where ν is the unit normal consistent with the orientation of α (right hand rule).
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1 HOMEWORK 3 - GEOMETRY OF CURVES AND SURFACES ANDRÉ NEVES ) If α : I R 2 is a curve on the plane parametrized by arc length and θ is the angle that α makes with the x-axis, show that α t) = dθ dt ν, where ν is the unit normal consistent with the orientation of α right hand rule). The bridge with the framework of what I have done in class is that we can see R 2 as the plane P = {z = 0} in R 3 and so α is a curve on P where Dα k = = α ). dt But N = 0, 0, ) is the unit normal vector to P and α = x, y, 0), which implies α ) = α. In other words k is just α for a curve on the plane. If E =, 0, 0) and E 2 = 0,, 0), then it is simple to see that DE = 0 dt E.α = cos θ, ν.e = α.e 2 = sin θ. Thus sin θ dθ dt = d cos θ = d dt dt α.e = α.e +α. DE = α.ν)ν.e = sin θα.ν) dt and so dθ dt = α.ν. Chapter 4.2, Exercise 8: The first step is to show that G is a linear map. Fix a point p and consider α t) = p + tx, where X is any vector in R 3. Set F t) = G α t) Gq). 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).Gp) Gq)) = X.p q) = X.DG p ) Gp) Gq)) = X.p q)
2 2 ANDRÉ NEVES for all X and all q. In the last line we use the fact that if A is a 3 3 matrix with transpose A T, then AX).Y = X.A 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 ) Gp) Gq)) = 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 Gp) = Gq) + Bp q)for every p, q R 3. Using again the hypothesis we obtain that Bp q) = Gp) Gq) = p q. Using Exercise 7 of Chapter 4.2 more precisely, that b) is equivalent to a) which is simple to show) it follows at once that G is a linear isometry of R 3. Chapter 4.2, Exercise : 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 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 A T A = Id, i.e., A = A T. In this case we have AX) 2 = AX).Ax) = X.A T AX) = X.X = X 2 and so A send the unit sphere into the unit sphere. Moreover AX) AY ) = AX Y ) = X Y, and so its distance preserving in the sense of Exercise 8. Thus a) can be applied to conclude A 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 = }. F is not distance preserving because F 0, 0, 0) F θ, 0, 0) = 2 cos θ) and 0, 0, 0) θ, 0, 0) = θ. Chapter 4.2, Exercise 2: Consider F x, y, z) = x, y, z). This is an isometry of the cylinder S = {x 2 + y 2 = } 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.
3 HOMEWORK 3 - GEOMETRY OF CURVES AND SURFACES 3 But x 2 + y 2 =, which means x = ±, i.e., the only fixed points are, 0, 0) and, 0, 0). Chapter 4.2, Exercise 5: To be in accordance with the stuff we have been learning in class, the slight modified set up should be P = {z = 0} and F is a map from P into P which is given by To see that is conformal we have F x, y, 0) = ux, y), vx, y), 0). F x. F y = xu, x v, 0). y u, y v, 0) = x u y u+ x v y v = x v y v+ x v y v = 0 F x. F x = xu) 2 + x v) 2 F y. F y = yu) 2 + y v) 2 = x v) 2 + x u) 2 = F x. F x. Thus E = G and F = 0, which means the map is conformal. Chapter 4.3, Exercise : It becomes slightly simpler if we use E = e 2u and G = e 2v. Note that my u, v are different from the u, v in do Carmo. The do Carmo u, v correspond to my x, x 2. In this case, using the formula 2) in page 232 of Do Carmo we have Γ = x u, Γ 2 = e 2u v) x2 u, Γ 2 = x2 u, Γ 2 2 = x v, Γ 22 = e 2v u) x u, Γ 2 22 = x2 v. Using formula 5) in page 234 of Do Carmo we have K = E x Γ 2 2 x2 Γ 2 + Γ 2Γ 2 + Γ 2 2Γ 2 2 Γ 2 Γ 2 22 Γ Γ 2 ) 2 = e 2u x 2 x v + x2 e 2u v) x2 u) e 2u v) x2 u) 2 + x v) 2 ) +e 2u v) x2 u x2 v x u x v ) = e 2u x 2 x v + e 2u v) x2 u v) x2 u + e 2u v) x 2 2 x 2 u + x v) 2 x u x v ) = e 2u x 2 x v + e 2u v) x2 u v) x2 u + e 2u v) x 2 2 x 2 u + x v u) x v = e 2u x 2 x v e 2v x2 u v) x2 u e 2v x 2 2 x 2 u e 2u x v u) x v.
4 4 ANDRÉ NEVES On the other hand 2EG) /2 x2 ) x2 E EG) /2 + = e u+v) x2 x x2 u e v u) )) x G EG) ) /2 + x x v e u v) ) = e u+v) e u v) x 2 2 x 2 u + e u v) x2 u v) x2 u + e v u) x 2 x v + e v u) x v u) x v = e 2v 2 x 2 x 2 u e 2v x2 u v) x2 u e 2u 2 x x v e 2u x v u) x v. Thus, from the previous exercise, K = 2EG) /2 Chapter 4.3, Exercise 2: x2 x2 E EG) /2 )) ) + )) x G x EG) /2. We follow the previous exercise and write λ = e 2u, i.e.,define u = ln λ. Note that this u is not the u of do Carmo.). Then we have EG = e 2u, Thus x2 E EG) /2 = 2 x 2 u, x G EG) /2 = 2 x u. K = e 2u u = ln λ. 2λ Recall that ln λ = 2 ln λ). To be consistent with the rest of the notation I should use λ = x 2 + x2 2 + c) 2 for the last part. Then and Thus and so xi ln λ = 2 xi lnx 2 + x 2 4x i 2 + c) = x 2 + x2 2 + c x 2 4 i x i ln λ = x 2 + x2 2 + c + 8x 2 i x 2 + x c)2 8 ln λ = x 2 + x2 2 + c + 8x2 + x2 2 ) x 2 + x2 2 + = 8c c)2 x 2 + = 8cλ x2 2 + c)2 K = ln λ = 4c. 2λ Chapter 4.3, Exercise 2: We have x u = cos v, sin v, u ), x = u sin v, u cos v, 0) v
5 and so HOMEWORK 3 - GEOMETRY OF CURVES AND SURFACES 5 E = x u. x u = + u 2, F = x u. x v Because v E = 0 we obtain from Exercise 2u + u 2 We have and so K = 2 + u 2 u = 0, G = x v. x v = u2. ) = + u 2 ) 2 x x = cos v, sin v, 0), = u sin v, u cos v, 0) u v Ē = x. x u u =, F = x. x x = 0, G =. x u v v v = + u2. Using the formula in Exercise we have ) K = 2 2u = + u 2 u + u 2 + u 2 ) 2. Therefore the surfaces have the same curvature. If the map x x were an isometry then by Proposition, page 220, we would have E = Ē, F = F, and G = Ḡ. Because this is false the surfaces are not isometric. Chapter 4.3, Exercise 6: Let s ignore the part about using Bonnet s Theorem because I did not talk about that. If there were a surface with E = G =, F = 0 and e = g =, f = 0, then from the definition of Gaussian curvature we would have K = eg f)eg F ) =. But from Gauss s Theorem we know that K is intrinsic, i.e., depends only on E, F and G. Thus K must be zero because we can surely put coordinates on a plane which have E = G =, F = 0 and a plane has Gaussian curvature zero. Chapter 4.4, Exercise 5 a) and 6: Let s do thing s a bit more generally here because it will be useful. Consider a parametrization of S Then φs, θ) = rs) cos θ, rs) sin θ, zs)). φ s = r cos θ, r sin θ, z ), 2 φ s) 2 = r cos θ, r sin θ, z ), 2 φ θ s = r sin θ, r cos θ, 0). φ = r sin θ, r cos θ, 0), θ 2 φ = r cos θ, r sin θ, 0), θ) 2
6 6 ANDRÉ NEVES and so E = r ) 2 + z ) 2 =, F = 0, G = r 2, where I am assuming, without loss of generality, that rs), zs)) is parametrized by arc-length. If αt) = φst), θt)) 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, X 2 for T S where Then X = φ s, X φ 2 = r = r sin θ, r cos θ, 0). θ α.x = s + ṡ) 2 r r + z z ) θ) 2 rr = s θ) 2 rr, where we use the fact that and Therefore r ) 2 + z ) 2 = = r r + z z = 0, α.x 2 = θ + 2ṡ θrr. D α k = dt = α)t = α.x )X + α.x 2 )X 2 = s θ) 2 rr ) φ s + θ + 2ṡ θrr φ r θ. For the case in point we have rs) = a + r 0 coss/r 0 ), zs) = r 0 sins/r 0 ) so that r ) 2 +z ) 2 = ). Note that my r 0 is the parameter r in Do Carmo s notation for this exercise. The curves are maximum parallel: α t) = φ0, t/r 0 ), minimum parallel: α 2 t) = φr 0 π, t/r 0 ), upper parallel: α 3 t) = φr 0 π/2, t/r 0 ) 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, ṡ = θ = 0, and r = sins/r 0 ). Using the formula above for k we get kα ) = r 2 0 r0)r 0) φ s = 0 kα ) = r0 2 rr 0π)r r 0 π) φ s = 0 and kα3 ) = r0 2 rr 0π/2)r r 0 π/2) φ s = a φ r0 2 s. Note that kα 3 ) compute the geodesic curvature of the upper parallel in Exercise 6. Chapter 4.4, Exercise 7:
7 HOMEWORK 3 - GEOMETRY OF CURVES AND SURFACES 7 We use the coordinate chart φstd, z) = cos s, sin s, z) which makes it into a local isometry with Euclidean plane, i.e., E = = G, F = 0. The plane which intersects the cylinder is P = {z = y tan θ} and so the curve C is parametrized by Ct) = cos t, sin t, sin t tan θ) = φt, sin t tan θ). If e =, 0, 0), e 2 = 0, cos θ, sin θ) is an orthonormal basis for P, then C = {xe + ye 2 x 2 + cos 2 θy 2 = } 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 Ct) is not a parametrization by arc-length. Suppose then that t = tu) is a change of variable which makes Cu) parametrized by arc length. Then, if ν denotes a unit vector lying in the tangent plane to to the cylinder which is orthogonal to C t), we know that k has the direction of ν and so k = k.ν)ν. Now and thus k.ν = d du C t) dt ).ν = du = C t) 2 C t).ν. dc du = C t) dt du = dt du = C t) ) dt 2 C t).ν + d2 t du du) 2 C t).ν = ) dt 2 C t).ν du Let s use this formula. C intersects the axis when t = 0, t = π. In this case C t) = Ct) and so C 0) =, 0, 0), C π) =, 0, 0). In both cases C has the direction of the normal vector to the cylinder and so the term C t).ν is zero because ν lies in the tangent plane of the cylinder. Thus k = 0 at those points. Chapter 4.4, Exercise 7: There is some terminology that needs further explanation. Essentially what the hypothesis of the exercise is saying is that there is F : U R 2 : S so that i) the curves αt) = F t, s 0 ) or αs) = 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. 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. We will show that for this chart we have E and G constant and F = 0. This implies K must be zero for the following reason: If E = a 2 and G = b 2, then we consider the map φx, y) = ax, by, 0) from R 2 into P = {z = 0}. For this map we have E = a 2, G = b 2 and F = 0 as well. Thus S is locally isometric to P and so the intrinsic invariance of K implies that the Gaussian
8 8 ANDRÉ NEVES curvature of S must be the same as the Gaussian curvature of P which is zero. That F = s F. t F is zero is essentially condition ii). We show that E is constant. t E = t t F. t )F = 2 2 ttf. t F = 2 D tf t. t F = 0 because t F t, s) is a geodesic for every s fixed. Likewise s E = s t F. t F ) = 2 2 stf. t F = 2 t s F. t F ) 2 s F. 2 ttf = 2 D tf t. s F = 0. Thus E is constant. The same reasoning shows that G is constant which is what we wanted to show.
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