DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

98 Chapter 3. Surfaces: Further TopicsGauss equation: dω 12 = −ω 13 ∧ ω 23Codazzi equations: dω 13 = ω 12 ∧ ω 23dω 23 = −ω 12 ∧ ω 13Example 1. To illustrate the power of the moving frame approach, we reprove Proposition 3.3of Chapter 2: Suppose K =0and M has no planar points. Then we claim that M is ruled and thetangent plane of M is constant along the rulings. We work in a principal moving frame with k 1 =0,so ω 13 =0. Therefore, by the first Codazzi equation, dω 13 =0=ω 12 ∧ ω 23 = ω 12 ∧ k 2 ω 2 . Sincek 2 ≠0,wemust have ω 12 ∧ ω 2 =0,and so ω 12 = fω 2 for some function f. Therefore, ω 12 (e 1 )=0,and so de 1 (e 1 )=ω 12 (e 1 )e 2 + ω 13 (e 1 )e 3 = 0. It follows that e 1 stays constant as we move in thee 1 direction, so following the e 1 direction gives us a line. Moreover, de 3 (e 1 )=0 (since k 1 = 0), sothe tangent plane to M is constant along that line. ▽The Gauss equation is particularly interesting. First, note thatω 13 ∧ ω 23 =(h 11 ω 1 + h 12 ω 2 ) ∧ (h 12 ω 1 + h 22 ω 2 )=(h 11 h 22 − h 2 12)ω 1 ∧ ω 2 = KdA,where K = det [ h αβ]= det SP is the Gaussian curvature. So, the Gauss equation really reads:(⋆)dω 12 = −KdA.(How elegant!) Note, moreover, that, by Lemma 3.3, for any two moving frames e 1 , e 2 , e 3 ande 1 , e 2 , e 3 ,wehave dω 12 = dω 12 (which is good, since the right-hand side of (⋆) doesn’t depend onthe frame field). Next, we observe that, because of the first equations in Theorem 3.1, ω 12 canbe computed just from knowing ω 1 and ω 2 , hence depends just on the first fundamental form ofthe surface. (If we write ω 12 = Pω 1 + Qω 2 , then the first equation determines P and the seconddetermines Q.) We therefore arrive at a new proof of Gauss’s Theorema Egregium, Theorem 3.1of Chapter 2.Example 2. Let’s go back to our usual parametrization of the unit sphere,Then we havex(u, v) =(sin u cos v, sin u sin v, cos u), 0