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Enneper's Surface GEOMETRY HW 4 CLAY SHONKWILER 3.3.5 Consider the parametrized surface (Enneper’s surface) u3 v3 φ(u, v) = x − + uv2, v − + vu2, u2 − v2 3 3 and show that (a) The coefficients of the first fundamental form are E = G = (1 + u2 + v2)2,F = 0. Answer: We know that E = hΦ1, Φ1i, F = hΦ1, Φ2i and G = hΦ2, Φ2i, so we need to calculate Φ1 and Φ2. Now, 2 2 Φ1 = (1 − u + v , 2uv, 2u) and 2 2 Φ2 = (2uv, 1 − v + u , −2v). Hence, 2 2 2 2 2 2 E = hΦ1, Φ1i = (1 − u + v ) + 4u v + 4u = 1 + 2u2 + 2v2 + u4 + v42u2v2 = (1 + u2 + v2)2 2 2 2 2 F = hΦ1, Φ2i = 2uv(1 − u + v ) + 2uv(1 − v + u ) − 4uv = 2uv − 2u3v + 2uv3 + 2uv − 2uv3 + 2u3v − 4uv = 0 and 2 2 2 2 2 2 G = hΦ2, Φ2i = 4u v + (1 − v + u ) + 4v = 1 + 2u2 + 2v2 + u4 + v42u2v2 . = (1 + u2 + v2)2 ♣ (b) The coefficients of the second fundamental form are e = 2, g = −2, f = 0. Answer: Recall that Φ × Φ Φ × Φ N = 1 2 = √ 1 2 ||Φ1 × Φ2|| EG − F 2 (−2uv2 − 2u − 2u3, 2u2v + 2v + 2v3, 1 − 2u2v2 − u4 − v4) = p(1 + u2 + v2)4 1 2 CLAY SHONKWILER Hence, D ∂2φ E e = N, ∂u2 1 2 3 2 3 2 2 4 4 = (1+u2+v2)2 h(−2uv − 2u − 2u , 2u v + 2v + 2v , 1 − 2u v − u − v ), (−2u, 2v, 2)i 1 2 2 4 4 2 2 = (1+u2+v2)2 2(1 + 2u + 2v + u + v + 2u v ) (1+u2+v2)2 = 2 (1+u2+v2)2 = 2. Now, ∂2φ f = hN, ∂u∂v i 1 2 3 2 3 = (1+u2+v2)2 h(−2uv − 2u − 2u , 2u v + 2v + 2v , 1 − 2u2v2 − u4 − v4), (2v, 2u, 0)i 1 3 3 3 3 = (1+u2+v2)2 2(−2uv − 2uv + 2u v − 2u v + 2uv + 2uv ) 1 = (1+u2+v2)2 (0) = 0. and ∂2φ g = hN, ∂v2 i 1 2 3 2 3 = (1+u2+v2)2 h(−2uv − 2u − 2u , 2u v + 2v + 2v , 1 − 2u2v2 − u4 − v4), (2u, −2v, −2)i 1 2 3 2 3 = − (1+u2+v2)2 h(−2uv − 2u − 2u , 2u v + 2v + 2v , 1 − 2u2v2 − u4 − v4), (2u, −2v, −2)i = −e = −2. That is to say, e = 2, g = −2, f = 0. ♣ (c) The principal curvatures are 2 −2 k = , k = . 1 (1 + u2 + v2)2 2 (1 + u2 + v2)2 Answer: Now, −fF + eG −fF + gE k = −a = , k = −a = . 1 11 EG − F 2 2 22 EG − F 2 Since f = F = 0, eG e gE g k = = k = = . 1 EG E 2 EG G Hence, 2 −2 k = k = . 1 (1 + u2 + v2)2 2 (1 + u2 + v2)2 ♣ (d) The lines of curvature are the coordinate curves. GEOMETRY HW 4 3 Proof. As given on pg. 161, the differential equation giving the lines of curvature is (α0 )2 −α0 α0 (α0 )2 1 1 2 2 EFG = 0 e f g where α(t) = φ(α1(t), α2(t)) is a regular curve. Now, recalling that F = f = 0, this reduces to the equation 0 0 0 0 2 2 2 2 2 0 = −α1α2(Eg − eG) = −α1α2(−2(1 + u + v ) − 2(1 + u + v )). 2 2 2 0 0 Since −4(1+u +v ) 6= 0, it follows that either α1 = 0 or α2 = 0. However, this occurs precisely when α is a coordinate curve, so we see that the lines of curvature are the coordinate curves. (e) The asymptotic curves are u + v = const., u − v = const. Proof. From pg. 160, we see that a connected regular curve C is an asymp- totic curve if and only if, for any parametrization α(t) = φ(α1(t), α2(t)), 0 2 0 0 0 2 e(α1) + fα1α2 + g(α2) = 0. Now, plugging in our values for e, f and g, we see that this reduces to the condition that 0 2 0 2 2(α1) − 2(α2) = 0. Reducing, we see that (u0)2 = (v0)2, which is to say that u0 = ±v0. Integrating both sides of the above, we see that, u = ±v + C. In other words, u ∓ v = C. Hence, curves of this form are precisely the asymptotic curves. 3.3.7 φ(u, v) = (γ(v) cos u, γ(v) sin u, ψ(v)) is given as a surface of revolution with constant Gaussian curvature K. To determine the functions γ and ψ, choose the parameter v in such a way that (γ0)2 + (ψ0)2 = 1. Show that (a) γ satisfies γ00 + Kγ = 0 and ψ is given by ψ = R p1 − (γ0)2dv; thus, 0 < u < 2π, and the domain of v is such that the last integral makes sense. Proof. We start by computing Φ1,Φ2: 0 0 0 Φ1 = (−γ(v) sin u, γ(v) cos u, 0), Φ2 = (γ (v) cos u, γ (v) sin u, ψ (v)). From this, then, we can compute 2 2 2 2 E = hΦ1, Φ1i = (γ(v)) (sin u + cos u) = (γ(v)) , 0 0 F = hΦ1, Φ2i = −γ(v)γ (v) sin u cos u + γ(v)γ (v) sin u cos u = 0 4 CLAY SHONKWILER and 0 2 2 2 0 2 0 2 0 2 G = hΦ2, Φ2i = (γ (v)) (sin u + cos u) + (ψ (v)) = (γ (v)) + (ψ (v)) = 1. Recall that ∂2φ e = 1/M(Φ , Φ , ) 1 2 ∂u2 ∂2φ f = 1/M(Φ , Φ , ) 1 2 ∂u∂v and ∂2φ g = 1/M(Φ , Φ , ) 1 2 ∂v2 √ where M = EG − F 2 and (a, b, c) denotes the determinant of the matrix with columns given by a, b and c. Now, p p EG − F 2 = (γ(v)2 = φ(v). Hence, −γ sin u γ0 cos u −γ cos u 1 e = γ cos u γ0 sin u −γ sin u γ(v) 0 ψ0 0 1 0 2 2 2 2 = γ(v) − ψ (γ sin u + γ cos u) −γ2ψ0 = γ = −γψ0. Furthermore, −γ sin u γ0 cos u −γ0 sin u 1 f = γ cos u γ0 sin u γ0 cos u γ(v) 0 ψ0 0 1 0 0 0 = γ − ψ (−γγ sin u cos u + γγ sin u cos u) = 0. and −γ sin u γ0 cos u γ00 cos u 1 g = γ cos u γ0 sin u γ00 sin u γ(v) 0 ψ0 ψ00 1 0 00 2 00 2 00 0 2 0 2 = γ [−ψ (−γγ sin u + γγ cos u) + ψ (−γγ sin u − γγ cos u)] 1 00 00 0 = γ (ψγγ − ψ γγ ) = ψγ00 − ψ00γ0. GEOMETRY HW 4 5 Now, if we differentiate both sides of the equation (γ0)2 + (ψ0)2 = 1, we see that γ0γ00 = −ψ0ψ00. Hence, eg−f 2 K = EG−F 2 −γψ0(ψ0γ00−ψ00γ0) = γ2 −ψ0(ψ0γ00−ψ00γ0) = γ −(ψ0)2γ00+ψ0ψ00γ0 = γ (ψ0)2γ00+(γ0)2γ00 = − γ γ00 = − γ . Re-arranging, we see that γ satisfies γ00 + kγ = 0. Now, on the other hand, since (γ0)2 + (ψ0)2 = 1, we can re-arrange to see that p ψ0 = 1 − (γ0)2. Hence, Z p ψ = 1 − (γ0)2dv, where v is such that this integral makes sense. (b) All surfaces of revolution with constant curvature K = 1 which inter- sect perpindicularly the plane xOy are given by Z v p γ(v) = C cos v, ψ(v) = 1 − C2 sin2 vdv, 0 where C is a constant. Determine the domain of v and draw a rough sketch of the profile o f the surface in the xz plane for the cases C = 1, C > 1, C < 1. Observe that C = 1 gives the sphere. Proof. Recall that in part (a) we showed that γ satisfies the equation γ00 + Kγ = 0. Plugging in K = 1, we see that γ00 + γ = 0. 2 This has√ characteristic equation λ + 1 = 0, meaning the solutions are λ = ± −1. Hence, 0v 0v γ(v) = C1e cos v + C2e sin v = C1 cos v + C2 sin v. After translating the curve, we may assume that at v = 0 the curve (γ(v), ψ(v)) passes through (C, 0) and (γ0(0), ψ0(0)) = (0, z). Then 0 0 = γ (0) = −C1 sin(0) + C2 cos(0) = C2. Hence, γ(v) = C1 cos v. Now, using our formula for ψ given in part (a) and plugging in the above for γ, we see that Z v p Z v p ψ(v) = 1 − (γ0)2dv = 1 − C2 sin2 vdv. 0 0 6 CLAY SHONKWILER −1 −1 where v ∈ [− sin (1/C), sin (1/C)]. (c) All surfaces of revolution with constant curvature K = −1 may be given by one of the following types: p 1. γ(v) = C cosh v, ψ(v) = R v 1 − C2 sinh2 vdv. 0 p 2. γ(v) = C sinh v, ψ(v) = R v 1 − C2 cosh2 vdv. √ 0 v R v 2v 3. γ(v) = e , ψ(v) = 0 1 − e dv. Determine the domain of v and draw a rough sketch of the profile of the surface in the xz plane. Proof. Plugging the value K = −1 into the expression derived in (a), we see that γ00 − γ = 0. The characteristic equation, then, is λ2 − 1 = 0, so the solution of the differential equation is v −v γ(v) = C1e + C2e since λ = ±1 is the solution of the characteristic equation. Since there are no initial conditions given, we cannot necessarily reduce γ to one of the above three forms.
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