DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

76 Chapter 3. Surfaces: Further TopicsSince (much as in the case of curves) e 1 and e 2 give an orthonormal basis for the tangent space ofour surface at each point, all the intrinsic curvature information (such as given by the Christoffelsymbols) is encapsulated in knowing how e 1 twists towards e 2 as we move around the surface. Inparticular, if α(t) =x(u(t),v(t)), a ≤ t ≤ b, isaparametrized curve, we can setφ 12 (t) = d dt(e1 (u(t),v(t)) ) · e 2 (u(t),v(t)),which we may write more casually as e ′ 1 (t) · e 2(t), with the understanding that everything must bedone in terms of the parametrization. We emphasize that φ 12 depends in an essential way on theparametrized curve α. Perhaps it’s better, then, to writeφ 12 = ∇ α ′e 1 · e 2 .Note, moreover, that the proof of Proposition 4.2 of Chapter 2 shows that ∇ α ′e 2 · e 1 = −φ 12 and∇ α ′e 1 · e 1 = ∇ α ′e 2 · e 2 =0. (Why?)Remark. Although the notation seems cumbersome, it reminds us that φ 12 is measuring howe 1 twists towards e 2 as we move along the curve α. This notation will fit in a more general contextin Section 3.Suppose now that α is a closed curve and we are interested in the holonomy around α. Ife 1 happens to be parallel along α, then the holonomy will, of course, be 0. If not, let’s considerX(t) tobethe parallel translation of e 1 along α(t) and write X(t) =cos ψ(t)e 1 + sin ψ(t)e 2 , takingψ(0) = 0. Then X is parallel along α if and only if0 = ∇ α ′X = ∇ α ′(cos ψe 1 + sin ψe 2 )= cos ψ∇ α ′e 1 + sin ψ∇ α ′e 2 +(− sin ψe 1 + cos ψe 2 )ψ ′= cos ψφ 12 e 2 − sin ψφ 12 e 1 +(− sin ψe 1 + cos ψe 2 )ψ ′=(φ 12 + ψ ′ )(− sin ψe 1 + cos ψe 2 ).Thus, X is parallel along α if and only if ψ ′ (t) =−φ 12 (t). We therefore conclude:Proposition 1.1. The holonomy around the closed curve C equals ∆ψ = −∫ baφ 12 (t)dt.Example 1. Back to our example of the latitude circle u = u 0 on the unit sphere. Thene 1 = x u and e 2 =(1/ sin u)x v . Ifweparametrize the curve by taking t = v, 0≤ t ≤ 2π, then wehave (see Example 1 of Chapter 2, Section 3)∇ α ′e 1 = ∇ α ′x u = x uv = cot u 0 x v = cos u 0 e 2 ,and so φ 12 = cos u 0 . Therefore, the holonomy around the latitude circle (oriented counterclockwise)is ∆ψ = −∫ 2π0cos u 0 dt = −2π cos u 0 , confirming our previous results.Note that if we wish to parametrize the curve by arclength (as will be important shortly),we take s =(sin u 0 )v, 0≤ s ≤ 2π sin u 0 . Then, with respect to this parametrization, we haveφ 12 (s) =cot u 0 . (Why?) ▽