DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

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60 Chapter 2. Surfaces: Local Theoryand so E vv ≥ 0 and G uu ≥ 0atP . Using the equation (∗) for the Gaussian curvature on p. 57, wesee that−2KEG = E vv + G uu + a(u, v)E v + b(u, v)G ufor some functions a(u, v) and b(u, v). So we conclude that K(P ) ≤ 0, as desired.Proof of Theorem 3.5. By Proposition 3.4, there is a point where M is positively curved,and since the Gaussian curvature is constant, we must have K>0. If every point is umbilic, thenby Exercise 2.2.12, we know that M is a sphere. If there is some non-umbilic point, the largerprincipal curvature, k 1 ,achieves its maximum value at some point P because M is compact. Then,since K = k 1 k 2 is constant, the function k 2 = K/k 1 must achieve its minimum at P . Since Pis necessarily a non-umbilic point (why?), it follows from Lemma 3.6 that K(P ) ≤ 0, which is acontradiction. □Remark. Hopf proved a stronger result, which requires techniques from complex analysis: IfM is a compact surface topologically equivalent to a sphere and having constant mean curvature,then M must be a sphere.We conclude this section with the analogue of Theorem 3.1 of Chapter 1.Theorem 3.7 (Fundamental Theorem of Surface Theory). Uniqueness: Two parametrized surfacesx, x ∗ : U → R 3 are congruent (i.e., differ by a rigid motion) if and only if I = I ∗ andII = ±II ∗ . Existence: Moreover, given differentiable functions E, F , G, l, m, and n with E>0 andEG− F 2 > 0 and satisfying the Codazzi and Gauss equations, there exists (locally) a parametrizedsurface x(u, v) with the respective I and II.Proof. The existence statement requires some theorems from partial differential equationsbeyond our reach at this stage. The uniqueness statement, however, is much like the proof ofTheorem 3.1 of Chapter 1. (The main technical difference is that we no longer are lucky enoughto be working with an orthonormal basis at each point, as we were with the Frenet frame.)First, suppose x ∗ =Ψ◦x for some rigid motion Ψ: R 3 → R 3 (i.e., Ψ(x) =Ax + b for someb ∈ R 3 and some 3×3 orthogonal matrix A). Since a translation doesn’t change partial derivatives,we may assume that b = 0. Now, since orthogonal matrices preserve length and dot product, wehave E ∗ = ‖x ∗ u‖ 2 = ‖Ax u ‖ 2 = ‖x u ‖ 2 = E, etc., so I = I ∗ . If det A>0, then n ∗ = An, whereas ifdet A0 and the negative when det A0 and II ∗ = −II ifdet A
§3. The Codazzi and Gauss Equations and the Fundamental Theorem of Surface Theory 61v ′ j(t) =3∑q ij v i (t) and vj ∗′ (t) =i=13∑q ij vi ∗ (t), j =1, 2, 3.(Note that the coefficient functions p i and q ij are the same for both the starred and unstarredequations.) If α(0) = α ∗ (0) and v i (0) = vi ∗(0), i =1, 2, 3, then α(t) =α∗ (t) and v i (t) =vi ∗ (t) forall t ∈ [0,b], i =1, 2, 3.Fix a point u 0 ∈ U. By composing x ∗ with a rigid motion, we may assume that at u 0 we havex = x ∗ , x u = x ∗ u, x v = x ∗ v, and n = n ∗ (why?). Choose an arbitrary u 1 ∈ U, and join u 0 to u 1by a path u(t), t ∈ [0,b], and apply the lemma with α = x◦u, v 1 = x u ◦u, v 2 = x v ◦u, v 3 = n◦u,p i = u ′ i , and the q ij prescribed by the equations (†) and (††). Since I = I ∗ and II = II ∗ , the sameequations hold for α ∗ = x ∗ ◦u, and so x(u 1 )=x ∗ (u 1 )asdesired. That is, the two parametrizedsurfaces are identical. □Proof of Lemma 3.8. Introduce the matrix function of t⎡⎤| | |⎢⎥M(t) = ⎣ v 1 (t) v 2 (t) v 3 (t) ⎦ ,| | |and analogously for M ∗ (t). Then the displayed equations in the statement of the Lemma can bewritten asi=1M ′ (t) =M(t)Q(t) and M ∗′ (t) =M ∗ (t)Q(t).On the other hand, we have M(t) T M(t) =G(t). Since the v i (t) form a basis for R 3 for eacht, weknow the matrix G is invertible. Now, differentiating the equation G(t)G −1 (t) =I yields(G −1 ) ′ (t) =−G −1 (t)G ′ (t)G −1 (t), and differentiating the equation G(t) =M(t) T M(t) yields G ′ (t) =M ′ (t) T M(t)+M(t) T M ′ (t) =Q(t) T G(t)+G(t)Q(t). Now consider(M ∗ G −1 M T ) ′ (t) =M ∗′ (t)G(t) −1 M(t) T + M ∗ (t)(G −1 ) ′ (t)M(t) T + M ∗ (t)G(t) −1 M ′ (t) T= M ∗ (t)Q(t)G(t) −1 M(t) T + M ∗ (t) ( −G(t) −1 G ′ (t)G(t) −1) M(t) T+ M ∗ (t)G(t) −1 Q(t) T M(t) T= M ∗ (t)Q(t)G(t) −1 M(t) T − M ∗ (t)G(t) −1 Q(t) T M(t) T − M ∗ (t)Q(t)G(t) −1 M(t) T+ M ∗ (t)G(t) −1 Q(t) T M(t) T = O.Since M(0) = M ∗ (0), we have M ∗ (0)G(0) −1 M(0) T = M(0)M(0) −1 M(0) T−1 M(0) T = I, and soM ∗ (t)G(t) −1 M(t) T = I for all t ∈ [0,b]. It follows that M ∗ (t) =M(t) for all t ∈ [0,b], and soα ∗′ (t) − α ′ (t) =0 for all t as well. Since α ∗ (0) = α(0), it follows that α ∗ (t) =α(t) for all t ∈ [0,b],as we wished to establish. □EXERCISES 2.31. Calculate the Christoffel symbols for a cone, x(u, v) =(u cos v, u sin v, u), both directly and byusing the formulas (‡).
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DIFFERENTIALGEOMETRY:A First Course
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4 Chapter 1. Curves2πb2πaFigure 1
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6 Chapter 1. CurvesAlternatively, s
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8 Chapter 1. CurvesAn important obs
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Page 14 and 15: 12 Chapter 1. CurvesSummarizing,κ(
Page 16 and 17: 14 Chapter 1. Curvescurvature κ. I
Page 18 and 19: 16 Chapter 1. CurvesFigure 2.2Conve
Page 20 and 21: 18 Chapter 1. Curvesalso Exercise 2
Page 22 and 23: 20 Chapter 1. Curvesto the curve β
Page 24 and 25: 22 Chapter 1. Curvesthe osculating
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Page 28 and 29: 26 Chapter 1. Curvesintersect C eit
Page 30 and 31: 28 Chapter 1. Curvesh(s,t)α(t)α(s
Page 32 and 33: 30 Chapter 1. Curvesy( x(s) ,y(s))
Page 34 and 35: 32 Chapter 1. Curvesɛ2ɛ 1ɛ 3CFig
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Page 38 and 39: 36 Chapter 2. Surfaces: Local Theor
Page 78 and 79: 76 Chapter 3. Surfaces: Further Top
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110 Appendix. Review of Linear Alge
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112 Appendix. Review of Linear Alge
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ANSWERS TO SELECTED EXERCISES1.1.1
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116 SELECTED ANSWERS2.4.8 Only circ
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118 INDEXisometry, 107K, 48k-point
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DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

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