Exercise 1.1. Verify that if we apply GS to the coordinate basis Gauss form ds 2 = E(u, v)du 2 + 2F (u, v)dudv + G(u, v)dv 2

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1 Math 209 Riemannian Geometry Jeongmin Shon Problem. Let M 2 R 3 be embedded surface. Then the induced metric on M 2 is obtained by taking the standard inner product on R 3 and restricting it to the tangent planes T m M R 3 to the surface. In this way we obtain a smoothly varying inner product on the tangent bundle of M: a Riemannian metric. xercise 0.. Standard spherical coordinates on M 2 = S 2, the unit two-sphere, are x θ, φ = cosθ sinφ, sinθ sinφ, cosφ. Compute that d x d x = dφ 2 +sin 2 φ dθ 2. d x = sinθ sinφdθ+cosθ cosφdφ, cosθ sinφdθ+sinθ cosφdφ, sinφdφ. Thus, d x d x = sin 2 θ sin 2 φdθ 2 + cos 2 θ cos 2 φdφ 2 2 sinθ sinφ cosθ cosφdθdφ + cos 2 θ sin 2 φdθ 2 + sin 2 θ cos 2 φdφ sinθ sinφ cosθ cosφdθdφ + sin 2 φdφ. = dφ 2 + sin 2 φ dθ 2. xercise.. Verify that if we apply GS to the coordinate basis Gauss form ds 2 = u, vdu 2 + 2F u, vdudv + Gu, vdv 2 u, v associated to then we get a smooth frame e, e 2. xpress e, e 2 in terms of, F, G and u, v. Argue the resulting vector fields are smooth. Note that ds 2 is a positive-definite quadratic form on the tangent bundle. G F 2 > 0 everywhere. Let Hence, v = u v 2 = v v u u u u = v F u.

2 2 Then v and v 2 are orthogonal. We can see that Let < v, v >= < v 2, v 2 >= F 2 2 2F + G = F 2 + G. e = u e 2 = G F 2 v F u. Then e and e 2 are orthonormal. Since ds 2 u, u = 0, e is a smooth vector field. Since G F 2 > 0 everywhere, is a smooth function. Hence, e 2 is also a smooth vector field. G F 2 xercise.2. Let, 2 be a frame field and θ, θ 2 the dual coframe field. Show that, 2 is orthonormal if and only if ds 2 = θ 2 + θ 2 2. Note that ds 2 = eθ 2 + 2fθ θ 2 + gθ 2 2. On the tangent space T p M for each p M, the metric tensor ds 2 : T p M T p M R is a quadratic symmetric bilinear form. For v, w T p M, let v = v, v 2 and w = w, w 2 with respect to the basis p, 2 p. [ ] e f Then there is a symmetric matrix GL2, R such that f g [ ] [ ] e f ds 2 w pv, w = [v v 2 ]. f g w 2 Then ds 2 pv, w = ev w + fv w 2 + v 2 w + gv 2 w 2 = [eθp 2 + 2fθpθ p 2 + gθp 2 2 ]v, w. Thus, = ds 2 p p, p = eθp pθp p = e 0 = ds 2 p p, 2 p = 2f 2 [θ pθ 2 2 p + θ 2 pθ 2 p] = f = ds 2 p 2 p, 2 p = gθp 2 pθp 2 p = g.

3 3 Thus, ds 2 p = θp 2 + θp 2 2 on T p M.q At each point p M ds 2 p i p, j p = θp i pθp j p + θp 2 i pθp 2 j p = δ ij. Thus,, 2 is orthonormal. xercise.3. Let M R 3 be an oriented surface with unit normal vector N. Show that the two-form Ω = i N dx dy dz restricted to M is the area form da on M. Hint: show that if v, w T m M then Ω v, w = N v w. Let N = N x, N y, N z. Ω = i N dx dy dz = i N dxdy dz dx i N dy dz = dxndy dz dx dyndz dzndy = N x dy dz N y dx dz + N z dx dy. Let v = v x, v y, v z and w = w x, w y, w z in T m M. Then Ω v, w = N x dy dz v, w N y dx dz v, w + N z dx dy v, w = N x v y w z w y v z N y v x w z w x v z + N z v x w y v y w x = N x, N y, N z v y w z w y v z, v x w z w x v z, v x w y v y w x = N v w. For any distinct vectors v, w T m M, the vector v w = cn for a nonzero constant c R. Thus, Ω is a nowhere-vanishing 2-form. Hence, Ω is a volume form on M. xercise.4. Suppose that θ, θ 2 is an oriented orthonormal coframe for M 2, ds 2 defined on a neighborhood V M. Show that the pair of one-forms θ, θ 2, also defined on V, is also an oriented orthonormal coframe if and only if there is a circle-valued function ψ : V S such that on V we have that: θ = cosψθ + sinψθ 2 θ 2 = sinψθ + cosψθ 2

4 4 Let θ = f θ + f 2 θ 2 θ 2 = g θ + g 2 θ 2 Since ds 2 = θ 2 + θ 2 2 and ds 2 = θ 2 + θ 2 2, f 2 + g 2 =, f g 2 2 = and f f 2 + g g 2 = 0. Also, f g 2 g f 2 = because θ θ 2 = θ θ 2. Thus, for each p M and on a neighborhood V of p, the matrix [ ] f p g p SO2, R. f 2 p g 2 p Then there is a circle-valued function ψ : V S such that [ ] [ f p g p cosψp sinψp = f 2 p g 2 p sinψp cosψp ]. Thus, on V θ = cosψθ + sinψθ 2 θ 2 = sinψθ + cosψθ 2. We can easily see that θ 2 + θ 2 2 = θ 2 + θ 2 2 = ds 2 and θ θ 2 = θ θ 2. xercise 2.. Given the oriented orthonormal coframe θ, θ 2, show that the Cartan s structure formula uniquely determines ω. Let ω = αθ + βθ 2 for scalar function α, β. Then dθ = ω θ 2 = αθ + βθ 2 θ 2 = α θ θ 2 dθ 2 = ω θ 2 = αθ + βθ 2 θ = β θ θ 2. Let ω = α θ +β θ 2 be another one-form satisfying the Cartan s structure equation. Then α = α and β = β. xercise 2.2. Show that if θ, θ 2 is another oriented orthonormal coframe then its connection one-form ω is given by ω = ω + dψ with ψ as in exercise.4.

5 5 By xercise.4, there is a circle-valued function ψ such that θ = cosψθ + sinψθ 2 θ 2 = sinψθ + cosψθ 2. Then dθ = sinψdψ θ + cosψdθ + cosψdψ θ 2 + sinψdθ 2 = sinψdψ + ω θ + cosψdψ + ω θ 2 = dψ + ω sinψθ + cosψθ 2 = dψ + ω θ 2 dθ 2 = cosψdψ θ sinψdθ sinψdψ θ 2 + cosψdθ 2 = cosψdψ + ω θ sinψdψ + ω θ 2 = dψ + ω cosψθ + sinψθ 2 = dψ + ω θ. Thus, the connection one-form ω is given by ω = ω + dψ. xercise 2.3. Show that K does not depend on the orientation. If we reverse the orientation of M then K remains unchanged. In particular M does not need to be oriented for the Gaussian curvature to be defined. Assume that M is oriented. Then M has an oriented orthonormal coframe θ, θ 2 on a connected neighborhood U of p, for p M.The 2-form θ θ 2 determines the orientation of M. Also, θ θ 2 determines the reversing orientation of M. Let θ = θ and θ 2 = θ 2. Then ds 2 = θ 2 + θ 2 2. Also, dθ = ω θ 2 dθ 2 = ω θ 2 = ω θ. Let ω = ω. Then dω = Kθ θ 2. Thus, K remains unchanged.

6 6 xercise 2.4. Compute the Cartan structure equations and find the Gaussian curvature K in the following two cases : ds 2 = dr 2 + fr 2 dθ 2 ds 2 = λu, v 2 du 2 + dv 2. 2 Assume that fr is a nowhere vanishing function. Let θ = dr and θ 2 = frdθ. dθ = 0 dθ 2 = f r dθ dr. Let ω = f r dθ. Then dω = 2 f r 2 dr dθ = 2 f 2 f r 2 fr. r 2 fr θ θ 2. Thus, the curvature is 2 Assume that λu, v is a nowhere vanishing function. Let θ = λu, vdu and θ 2 = λu, vdv. dθ = λ v du dv = dθ 2 = λ du dv = u λ v λ u λ dv λ du λdv = λ v λ λdu = Let ω = λ v λ λ du + u λ dv = log λ u dv log λ v du. Then dω = [log λ uu + log λ vv ] du dv = log λ du + λ u λ dv λ v λ du + λ u λ dv λ 2 θ θ 2. θ 2 θ. Thus, K = log λ λ 2. xercise 2.6. Find f as in of exer 2.4 such that K = and f has first order Taylor expansion fr = r + Or 2. From in exercise 2.4, 2 f r 2 fr =. Then 2 f = fr. Hence, r2 fr = c e r + c 2 e r

7 7 for any constants c, c 2. By the assumption, f0 = 0 and f 0 =. 0 = f0 = c + c 2 = f 0 = c c 2. Hence, c = 2 and c 2 = 2. Thus, fr = sinhr.