54 Chapter 2. <strong>Surfaces</strong>: Local Theoryc. Suppose M is represented locally near P as <strong>in</strong> Exercise 20. Show that for small positivevalues of c, the <strong>in</strong>tersection of M with the plane z = c “looks like” the Dup<strong>in</strong> <strong>in</strong>dicatrix.How can you make this statement more precise?22. Suppose the surface M is given near P as a level surface of a smooth function f : R 3 → R with∇f(P ) ≠0.Al<strong>in</strong>e L ⊂ R 3 is said to have (at least) k-po<strong>in</strong>t contact with M at P if, given anyl<strong>in</strong>ear parametrization α of L with α(0) = P , the function f = f ◦α vanishes to order k − 1,i.e., f(0) = f ′ (0) = ···= f (k−1) (0) = 0. (Such a l<strong>in</strong>e is to be visualized as the limit of l<strong>in</strong>es that<strong>in</strong>tersect M at P <strong>and</strong> at k − 1 other po<strong>in</strong>ts that approach P .)a. Show that L has 2-po<strong>in</strong>t contact with M at P if <strong>and</strong> only if L is tangent to M at P , i.e.,L ⊂ T P M.b. Show that L has 3-po<strong>in</strong>t contact with M at P if <strong>and</strong> only if L is an asymptotic directionat P . (H<strong>in</strong>t: It may be helpful to follow the setup of Exercise 20.)c. (Challenge) What does it mean for L to have 4-po<strong>in</strong>t contact with M at P ?3. The Codazzi <strong>and</strong> Gauss Equations <strong>and</strong> the Fundamental Theorem of SurfaceTheoryWe now wish to proceed towards a deeper underst<strong>and</strong><strong>in</strong>g of Gaussian curvature. We have tothis po<strong>in</strong>t considered only the normal components of the second derivatives x uu , x uv , <strong>and</strong> x vv .Nowlet’s consider them <strong>in</strong> toto. S<strong>in</strong>ce {x u , x v , n} gives a basis for R 3 , there are functions Γuu, u Γuu,vΓuv u =Γvu, u Γuv v =Γvu, v Γvv, u <strong>and</strong> Γvv v so that(†)x uu =Γ u uux u +Γ v uux v + lnx uv =Γ u uvx u +Γ v uvx v + mnx vv =Γ uvvx u +Γ vvvx v + nn.(Note that x uv = x vu dictates the symmetries Γ uv • =Γ vu.) • The functions Γ •• • are called Christoffelsymbols.Example 1. Let’s compute the Christoffel symbols for the usual parametrization of the sphere.By straightforward calculation we obta<strong>in</strong>x u = (cos u cos v, cos u s<strong>in</strong> v, − s<strong>in</strong> u)x v =(− s<strong>in</strong> u s<strong>in</strong> v, s<strong>in</strong> u cos v, 0)x uu =(− s<strong>in</strong> u cos v, − s<strong>in</strong> u s<strong>in</strong> v, − cos u) =−x(u, v)x uv =(− cos u s<strong>in</strong> v, cos u cos v, 0)x vv =(− s<strong>in</strong> u cos v, − s<strong>in</strong> u s<strong>in</strong> v, 0) = − s<strong>in</strong> u(cos v, s<strong>in</strong> v, 0).(Note that the u-curves are great circles, parametrized by arclength, so it is no surprise that theacceleration vector x uu is <strong>in</strong>ward-po<strong>in</strong>t<strong>in</strong>g of length 1. The v-curves are latitude circles of radiuss<strong>in</strong> u, so, similarly, the acceleration vector x vv po<strong>in</strong>ts <strong>in</strong>wards towards the center of the respectivecircle.)
§3. The Codazzi <strong>and</strong> Gauss Equations <strong>and</strong> the Fundamental Theorem of Surface Theory 55S<strong>in</strong>ce x uu lies entirely <strong>in</strong> the direction of n, wehave Γuu u =Γuu v =0. Now, by<strong>in</strong>spection,x uv = cot ux v ,soΓuv u =0<strong>and</strong> Γuv v = cot u. Last, as we can see <strong>in</strong> Figure 3.1, we have x vv =nx vvs<strong>in</strong> u cos uus<strong>in</strong> uux uFigure 3.1− s<strong>in</strong> u cos ux u − s<strong>in</strong> 2 un, soΓ uvv = − s<strong>in</strong> u cos u <strong>and</strong> Γ vvv =0.▽Now, dott<strong>in</strong>g the equations <strong>in</strong> (†) with x u <strong>and</strong> x v givesx uu · x u =ΓuuE u +ΓuuFvx uu · x v =ΓuuF u +ΓuuGvx uv · x u =ΓuvE u +ΓuvFvx uv · x v =ΓuvF u +ΓuvGvNow observe thatx vv · x u =Γ uvvE +Γ vvvFx vv · x v =Γ uvvF +Γ vvvG.x uu · x u = 1 2 (x u · x u ) u = 1 2 E ux uv · x u = 1 2 (x u · x u ) v = 1 2 E vx uv · x v = 1 2 (x v · x v ) u = 1 2 G ux uu · x v =(x u · x v ) u − x u · x uv = F u − 1 2 E vx vv · x u =(x u · x v ) v − x uv · x v = F v − 1 2 G ux vv · x v = 1 2 (x v · x v ) v = 1 2 G vThus, we can rewrite our equations as follows:[ ][ ] [ ]E F Γuuu 1F G Γuuv =2 E uF u − 1 2 E =⇒v[ ][ ] [ ]E F Γuvu 1F G Γuvv =2 E v(‡)12 G =⇒u[ ][ ] [ ]E F Γvvu F v − 1F G Γvvv =2 G u12 G =⇒v[Γ u uuΓ v uu[Γ u uvΓ v uv[Γ uvvΓ vvv]=[E=F] [E=F] [EF] −1 [ ]1F2 E uG F u − 1 2 E v] −1 [ ]F 12 E v1G2 G u] −1 [ ]F F v − 1 2 G u1G2 G .v